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| author | kinitrupti | 2017-05-12 18:40:35 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:40:35 +0530 |
| commit | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (patch) | |
| tree | 9806b0d68a708d2cfc4efc8ae3751423c56b7721 /Irrigation_and_Water_Power_Engineering_by_B._C._Punmia | |
| parent | 1b1bb67e9ea912be5c8591523c8b328766e3680f (diff) | |
| download | Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.gz Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.bz2 Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.zip | |
Revised list of TBCs
Diffstat (limited to 'Irrigation_and_Water_Power_Engineering_by_B._C._Punmia')
34 files changed, 0 insertions, 33237 deletions
diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch10.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch10.ipynb deleted file mode 100755 index 6e9cf01f..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch10.ipynb +++ /dev/null @@ -1,402 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:052e5024d5b2f54bda77f8d8972af18d1ffb4cd92ead6e287538c1745234bb8e" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 10 : EARTH AND ROCKFILL DAM" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 10.1 pg : 502" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "from numpy import array,float64,round\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "K = 5.E-4; \t\t\t\t#coefficient of permeability of soil\n", - "Bt = 6.; \t\t\t\t#width of top of dam\n", - "wb = 146.; \t\t\t\t#width of base of dam\n", - "H = 20.; \t\t\t\t#heigth of dam\n", - "hw = 2.; \t\t\t\t#heigth of water in reservior\n", - "hs1 = 4.; \t\t\t\t#slope on upstream side\n", - "hs2 = 3.; \t\t\t\t#slope on downstream side\n", - "df = 30.; \t\t\t\t#length of drainage filter\n", - "\n", - "x = wb-df-72+72*0.3;\n", - "y = 18.;\n", - "s = (x**2+y**2)**0.5-x;\n", - "\n", - "x = array([0, 10, 20, 30, 40, 50, 60, 65.6],dtype=float64);\n", - "y = (4.849*x+5.879)**0.5;\n", - "y = round(y*1000)/1000;\n", - "\n", - "print \"x y\";\n", - "for i in range(8):\n", - " print \"%.2f %.2f\"%(x[i],y[i]);\n", - "sf = K*s*10000;\n", - "sf = round(sf*1000)/1000;\n", - "print \"Seepage flow per unit length of dam = %.2fD-6 cumecs/metre length of dam.\"%(sf);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "x y\n", - "0.00 2.42\n", - "10.00 7.37\n", - "20.00 10.14\n", - "30.00 12.30\n", - "40.00 14.14\n", - "50.00 15.76\n", - "60.00 17.23\n", - "65.60 18.00\n", - "Seepage flow per unit length of dam = 12.12D-6 cumecs/metre length of dam.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 10.2 pg : 502" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "K = 3.E-3; \t\t\t\t#coefficient of permeability\n", - "nd = 25.; \t\t\t\t#number of potential drops\n", - "nf = 4.; \t\t\t\t#number of flow channels\n", - "lf = 40.; \t\t\t\t#filter length\n", - "H = 52.; \t\t\t\t#heigth of dam\n", - "fb = 2.; \t\t\t\t#free board\n", - "\n", - "# Calculations\n", - "q = K*(H-fb)*nf/(nd*100);\n", - "\n", - "# Results\n", - "print \"Discharge per meter length of dam = %.5f cumec/metre length.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge per meter length of dam = 0.00024 cumec/metre length.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 10.3 pg : 503" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\t\t\t\t\n", - "#Given\n", - "x = 4.;\n", - "#Given scale\n", - "An = 14.4; \t\t\t\t#area of N recmath.tangle\n", - "At = 6.4; \t\t\t\t#area of T recmath.tangle\n", - "Au = 4.9; \t\t\t\t#area of U recmath.tangle\n", - "L = 12.6; \t\t\t\t#length of arc;\n", - "gamma_m = 19.; \t\t\t\t#unit weigth of soil\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "fi = 26.; \t\t\t\t#effective angle(degree)\n", - "co = 19.5; \t\t\t\t#cohesion value\n", - "\n", - "# Calculations\n", - "#consider 1m length of dam\n", - "SumN = An*x**2*gamma_m;\n", - "SumT = At*x**2*gamma_m;\n", - "SumU = Au*x**2*gamma_w;\n", - "Le = x*L;\n", - "F = ((Le*co)+(SumN-SumU)*math.tan(math.radians(fi)))/SumT;\n", - "F = round(F*100)/100;\n", - "\n", - "# Results\n", - "print \"Factor of safety for slope = %.2f.\"%(F);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Factor of safety for slope = 1.41.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 10.4 pg : 503" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#check section for:\n", - "#Stability of d/s slope against steady seepage\n", - "#Sloughing of u/s slope against sudden drawdown\n", - "#Stability of the foundation against shear\n", - "#Seepage through body of dam\n", - "\n", - "#Given\n", - "#Dimensions\n", - "H = 20.; \t\t\t\t#Heigth of dam\n", - "Bt = 6.; \t\t\t\t#top width of dam\n", - "s1 = 4.; \t\t\t\t#u/s slope\n", - "s2 = 3.; \t\t\t\t#d/s slope\n", - "fb = 2.; \t\t\t\t#free board\n", - "#Properties of materials of dam\n", - "gamma_d = 17.27; \t\t\t\t#dry density\n", - "wc = 0.15; \t\t\t\t#optimum water content\n", - "gamma_s = 21.19; \t\t\t\t#saturated density\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "wavg = 19.62; \t\t\t\t#average unit weigth under seepage\n", - "theta = 26.; \t\t\t\t#average angle of internal friction(degree)\n", - "co = 19.13; \t\t\t\t#average cohesion\n", - "K = 5.E-4; \t\t\t\t#coefficient of permeability\n", - "#properties of foundation materials\n", - "gamma_f = 17.27; \t\t\t\t#average unit weigth\n", - "cof = 47.87; \t\t\t\t#average cohesion\n", - "fi = 8.; \t\t\t\t#average angle internal friction\n", - "t = 6.; \t\t\t\t#thickness of clay\n", - "FOSp = 1.5; \t\t\t\t#permissible factor of safety of slope\n", - "PS = 8.E-6; \t\t\t\t#permissible seepage\n", - "#(a) Stability of d/s slope against steady seepage\n", - "An = 302.4; \t\t\t\t#area of N diagram\n", - "At = 91.2; \t\t\t\t#area of T diagram\n", - "Au = 98.4; \t\t\t\t#area of U diagram\n", - "Le = 60.; \t\t\t\t#length of arc\n", - "SumN = An*gamma_s;\n", - "SumT = At*gamma_s;\n", - "SumU = Au*gamma_w;\n", - "F = (Le*co)+(SumN-SumU)*math.tan(math.radians(theta))/SumT;\n", - "F = round(F*100)/100;\n", - "print \"Parta:\"\n", - "print \"Factor of safety for slope = %.2f.\"%(F);\n", - "print \"Safe\";\n", - "\n", - "#(b) Sloughing of u/s slope against sudden drawdown\n", - "h1 = 15.;\n", - "b = 80.;\n", - "P = gamma_s*H**2*math.tan(math.radians(45-(theta/2)))**2/2+gamma_w*h1**2/2;\n", - "sav = P/b;\n", - "smax = 2*sav;\n", - "Ne = (gamma_s-gamma_w)*b*H/2;\n", - "R = Ne*math.tan(math.radians(theta))+co*b;\n", - "fs = R/P;\n", - "fs = round(fs*100)/100;\n", - "print \"Partb:\"\n", - "print \"Factor of safety w.r.t average shear = %.2f.\"%(fs);\n", - "print \"Safe\";\n", - "sr = 0.6*H*(gamma_s-gamma_w)*math.tan(math.radians(theta))+co;\n", - "FS = sr/smax;\n", - "FS = round(FS*100)/100;\n", - "print \"Factor of safety w.r.t maximum shear = %.2f.\"%(FS);\n", - "print \"Safe\";\n", - "\n", - "#(c) Stability of the foundation against shear\n", - "h1 = 26.;\n", - "h2 = 6.;\n", - "gamma_m = (wavg*(h1-h2)+gamma_f*h2)/h1;\n", - "l = (gamma_m*h1*math.tan(math.radians(fi))+cof)/(gamma_m*h1);\n", - "fi1 = math.tan(math.radians(l));\n", - "P = (h1**2-h2**2)/2*gamma_m*math.tan(math.radians(45-(fi1/2)))**2;\n", - "sav = P/b;\n", - "smax = 2*sav;\n", - "s1 = cof+gamma_f*h2*math.tan(math.radians(fi));\n", - "s2 = cof+gamma_m*h1*math.tan(math.radians(fi));\n", - "as1 = (s1+s2)/2;\n", - "fs = as1/sav;\n", - "fs = round(fs*100)/100;\n", - "print \"Partc:\"\n", - "print \"Factor of safety w.r.t overall shear = %.2f.\"%(fs);\n", - "print \"Safe\";\n", - "\n", - "gamma_av = (wavg*0.6*H+gamma_f*h2)/(0.6*H)+h2;\n", - "s = cof+gamma_av*0.6*H*math.tan(math.radians(fi));\n", - "fs = s/smax;\n", - "fs = round(fs*100)/100;\n", - "print \"Factor of safety w.r.t overall shear = %.2f.\"%(fs);\n", - "print \"Unsafe\";\n", - "\n", - "#(d) Seepage through body of dam\n", - "s = 2.; \t\t\t\t#measured\n", - "q = K*s*100000/100;\n", - "print \"Partd:\"\n", - "print \" Seepage through body of dam = %.2fD-5 cumecs/m length of dam\"%(q);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Parta:\n", - "Factor of safety for slope = 1149.17.\n", - "Safe\n", - "Partb:\n", - "Factor of safety w.r.t average shear = 2.16.\n", - "Safe\n", - "Factor of safety w.r.t maximum shear = 1.24.\n", - "Safe\n", - "Partc:\n", - "Factor of safety w.r.t overall shear = 1.18.\n", - "Safe\n", - "Factor of safety w.r.t overall shear = 0.69.\n", - "Unsafe\n", - "Partd:\n", - " Seepage through body of dam = 1.00D-5 cumecs/m length of dam\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 10.5 pg : 507" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%matplotlib inline\n", - "import math \n", - "from matplotlib.pylab import plot,show\n", - "from numpy import zeros\n", - "#design upstream impervious blanket\n", - "\t\t\t\t\n", - "#Given\n", - "Zb = 1.2; \t\t\t\t#thickness of blanket\n", - "Zf = 8; \t\t\t\t#dismath.tance of blanket from foundation\n", - "kb = 0.06; \t\t\t\t#coefficient of permeability of blanket material\n", - "kf = 72; \t\t\t\t#coefficient of permeability of foundation soil\n", - "Hw = 10; \t\t\t\t#heigth of water in reservior\n", - "Xd = 40;\n", - "\n", - "a = (kb/(kf*Zb*Zf))**0.5;\n", - "Xo = 1.414/a;\n", - "\n", - "#we vary value of x\n", - "x = [0, 25, 50, 75, 100, 125, 151.8, 300]\n", - "Xr = zeros(8)\n", - "ho = zeros(8)\n", - "r = zeros(8)\n", - "\n", - "for i in range(8):\n", - " e = math.exp(2*a*x[i]);\n", - " Xr[i] = (e-1)/(a*(e+1));\n", - " ho[i] = Xr[i]*Hw/(Xr[i]+Xd);\n", - " r[i] = Xr[i]*100/(Xr[i]+Xd);\n", - "\n", - "print \"x Xr ho reduction qpercent\";\n", - "for i in range(8):\n", - " print \"%.2f %.2f %.2f %.2f\"%(x[i],Xr[i],ho[i],r[i]);\n", - "\n", - "#graph is plotted between r and x.\n", - "#after around 130m length there is only slight increase in head dissipated(ho)\n", - "plot(x,r)\n", - "show()\n", - "L = 130;\n", - "print \"Thickness of blanket = %.2f m\"%(Zb);\n", - "print \"Length of blanket = %i m.\"%(L);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "x Xr ho reduction qpercent\n", - "0.00 0.00 0.00 0.00\n", - "25.00 24.56 3.80 38.04\n", - "50.00 46.67 5.38 53.85\n", - "75.00 64.78 6.18 61.83\n", - "100.00 78.50 6.62 66.24\n", - "125.00 88.28 6.88 68.82\n", - "151.80 95.35 7.04 70.45\n", - "300.00 106.53 7.27 72.70\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10ccbe7d0>" - ] - }, - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Thickness of blanket = 1.20 m\n", - "Length of blanket = 130 m.\n" - ] - } - ], - "prompt_number": 9 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch10_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch10_1.ipynb deleted file mode 100755 index aee962a5..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch10_1.ipynb +++ /dev/null @@ -1,403 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:35c301dd56b78bdaa09416cde0ed9d5e7328b399d3b45890faac8cf8417b6a80" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 10 : EARTH AND ROCKFILL DAM" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 10.1 pg : 502" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "from numpy import array,float64,round\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "K = 5.E-4; \t\t\t\t#coefficient of permeability of soil\n", - "Bt = 6.; \t\t\t\t#width of top of dam\n", - "wb = 146.; \t\t\t\t#width of base of dam\n", - "H = 20.; \t\t\t\t#heigth of dam\n", - "hw = 2.; \t\t\t\t#heigth of water in reservior\n", - "hs1 = 4.; \t\t\t\t#slope on upstream side\n", - "hs2 = 3.; \t\t\t\t#slope on downstream side\n", - "df = 30.; \t\t\t\t#length of drainage filter\n", - "\n", - "x = wb-df-72+72*0.3;\n", - "y = 18.;\n", - "s = (x**2+y**2)**0.5-x;\n", - "\n", - "x = array([0, 10, 20, 30, 40, 50, 60, 65.6],dtype=float64);\n", - "y = (4.849*x+5.879)**0.5;\n", - "y = round(y*1000)/1000;\n", - "\n", - "print \"x y\";\n", - "for i in range(8):\n", - " print \"%.2f %.2f\"%(x[i],y[i]);\n", - "sf = K*s*10000;\n", - "sf = round(sf*1000)/1000;\n", - "print \"Seepage flow per unit length of dam = %.2fD-6 cumecs/metre length of dam.\"%(sf);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "x y\n", - "0.00 2.42\n", - "10.00 7.37\n", - "20.00 10.14\n", - "30.00 12.30\n", - "40.00 14.14\n", - "50.00 15.76\n", - "60.00 17.23\n", - "65.60 18.00\n", - "Seepage flow per unit length of dam = 12.12D-6 cumecs/metre length of dam.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 10.2 pg : 502" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "K = 3.E-3; \t\t\t\t#coefficient of permeability\n", - "nd = 25.; \t\t\t\t#number of potential drops\n", - "nf = 4.; \t\t\t\t#number of flow channels\n", - "lf = 40.; \t\t\t\t#filter length\n", - "H = 52.; \t\t\t\t#heigth of dam\n", - "fb = 2.; \t\t\t\t#free board\n", - "\n", - "# Calculations\n", - "q = K*(H-fb)*nf/(nd*100);\n", - "\n", - "# Results\n", - "print \"Discharge per meter length of dam = %.5f cumec/metre length.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge per meter length of dam = 0.00024 cumec/metre length.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 10.3 pg : 503" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math\n", - "\t\t\t\t\n", - "#Given\n", - "x = 4.;\n", - "#Given scale\n", - "An = 14.4; \t\t\t\t#area of N recmath.tangle\n", - "At = 6.4; \t\t\t\t#area of T recmath.tangle\n", - "Au = 4.9; \t\t\t\t#area of U recmath.tangle\n", - "L = 12.6; \t\t\t\t#length of arc;\n", - "gamma_m = 19.; \t\t\t\t#unit weigth of soil\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "fi = 26.; \t\t\t\t#effective angle(degree)\n", - "co = 19.5; \t\t\t\t#cohesion value\n", - "\n", - "# Calculations\n", - "#consider 1m length of dam\n", - "SumN = An*x**2*gamma_m;\n", - "SumT = At*x**2*gamma_m;\n", - "SumU = Au*x**2*gamma_w;\n", - "Le = x*L;\n", - "F = ((Le*co)+(SumN-SumU)*math.tan(math.radians(fi)))/SumT;\n", - "F = round(F*100)/100;\n", - "\n", - "# Results\n", - "print \"Factor of safety for slope = %.2f.\"%(F);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Factor of safety for slope = 1.41.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 10.4 pg : 503" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#check section for:\n", - "#Stability of d/s slope against steady seepage\n", - "#Sloughing of u/s slope against sudden drawdown\n", - "#Stability of the foundation against shear\n", - "#Seepage through body of dam\n", - "\n", - "#Given\n", - "#Dimensions\n", - "H = 20.; \t\t\t\t#Heigth of dam\n", - "Bt = 6.; \t\t\t\t#top width of dam\n", - "s1 = 4.; \t\t\t\t#u/s slope\n", - "s2 = 3.; \t\t\t\t#d/s slope\n", - "fb = 2.; \t\t\t\t#free board\n", - "#Properties of materials of dam\n", - "gamma_d = 17.27; \t\t\t\t#dry density\n", - "wc = 0.15; \t\t\t\t#optimum water content\n", - "gamma_s = 21.19; \t\t\t\t#saturated density\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "wavg = 19.62; \t\t\t\t#average unit weigth under seepage\n", - "theta = 26.; \t\t\t\t#average angle of internal friction(degree)\n", - "co = 19.13; \t\t\t\t#average cohesion\n", - "K = 5.E-4; \t\t\t\t#coefficient of permeability\n", - "#properties of foundation materials\n", - "gamma_f = 17.27; \t\t\t\t#average unit weigth\n", - "cof = 47.87; \t\t\t\t#average cohesion\n", - "fi = 8.; \t\t\t\t#average angle internal friction\n", - "t = 6.; \t\t\t\t#thickness of clay\n", - "FOSp = 1.5; \t\t\t\t#permissible factor of safety of slope\n", - "PS = 8.E-6; \t\t\t\t#permissible seepage\n", - "#(a) Stability of d/s slope against steady seepage\n", - "An = 302.4; \t\t\t\t#area of N diagram\n", - "At = 91.2; \t\t\t\t#area of T diagram\n", - "Au = 98.4; \t\t\t\t#area of U diagram\n", - "Le = 60.; \t\t\t\t#length of arc\n", - "SumN = An*gamma_s;\n", - "SumT = At*gamma_s;\n", - "SumU = Au*gamma_w;\n", - "F = (Le*co)+(SumN-SumU)*math.tan(math.radians(theta))/SumT;\n", - "F = round(F*100)/100;\n", - "print \"Parta:\"\n", - "print \"Factor of safety for slope = %.2f.\"%(F);\n", - "print \"Safe\";\n", - "\n", - "#(b) Sloughing of u/s slope against sudden drawdown\n", - "h1 = 15.;\n", - "b = 80.;\n", - "P = gamma_s*H**2*math.tan(math.radians(45-(theta/2)))**2/2+gamma_w*h1**2/2;\n", - "sav = P/b;\n", - "smax = 2*sav;\n", - "Ne = (gamma_s-gamma_w)*b*H/2;\n", - "R = Ne*math.tan(math.radians(theta))+co*b;\n", - "fs = R/P;\n", - "fs = round(fs*100)/100;\n", - "print \"Partb:\"\n", - "print \"Factor of safety w.r.t average shear = %.2f.\"%(fs);\n", - "print \"Safe\";\n", - "sr = 0.6*H*(gamma_s-gamma_w)*math.tan(math.radians(theta))+co;\n", - "FS = sr/smax;\n", - "FS = round(FS*100)/100;\n", - "print \"Factor of safety w.r.t maximum shear = %.2f.\"%(FS);\n", - "print \"Safe\";\n", - "\n", - "#(c) Stability of the foundation against shear\n", - "h1 = 26.;\n", - "h2 = 6.;\n", - "gamma_m = (wavg*(h1-h2)+gamma_f*h2)/h1;\n", - "l = (gamma_m*h1*math.tan(math.radians(fi))+cof)/(gamma_m*h1);\n", - "fi1 = math.tan(math.radians(l));\n", - "P = (h1**2-h2**2)/2*gamma_m*math.tan(math.radians(45-(fi1/2)))**2;\n", - "sav = P/b;\n", - "smax = 2*sav;\n", - "s1 = cof+gamma_f*h2*math.tan(math.radians(fi));\n", - "s2 = cof+gamma_m*h1*math.tan(math.radians(fi));\n", - "as1 = (s1+s2)/2;\n", - "fs = as1/sav;\n", - "fs = round(fs*100)/100;\n", - "print \"Partc:\"\n", - "print \"Factor of safety w.r.t overall shear = %.2f.\"%(fs);\n", - "print \"Safe\";\n", - "\n", - "gamma_av = (wavg*0.6*H+gamma_f*h2)/(0.6*H)+h2;\n", - "s = cof+gamma_av*0.6*H*math.tan(math.radians(fi));\n", - "fs = s/smax;\n", - "fs = round(fs*100)/100;\n", - "print \"Factor of safety w.r.t overall shear = %.2f.\"%(fs);\n", - "print \"Unsafe\";\n", - "\n", - "#(d) Seepage through body of dam\n", - "s = 2.; \t\t\t\t#measured\n", - "q = K*s*100000/100;\n", - "print \"Partd:\"\n", - "print \" Seepage through body of dam = %.2fD-5 cumecs/m length of dam\"%(q);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Parta:\n", - "Factor of safety for slope = 1149.17.\n", - "Safe\n", - "Partb:\n", - "Factor of safety w.r.t average shear = 2.16.\n", - "Safe\n", - "Factor of safety w.r.t maximum shear = 1.24.\n", - "Safe\n", - "Partc:\n", - "Factor of safety w.r.t overall shear = 1.18.\n", - "Safe\n", - "Factor of safety w.r.t overall shear = 0.69.\n", - "Unsafe\n", - "Partd:\n", - " Seepage through body of dam = 1.00D-5 cumecs/m length of dam\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 10.5 pg : 507" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "from matplotlib.pylab import plot,show\n", - "from numpy import zeros\n", - "#design upstream impervious blanket\n", - "\t\t\t\t\n", - "#Given\n", - "Zb = 1.2; \t\t\t\t#thickness of blanket\n", - "Zf = 8; \t\t\t\t#dismath.tance of blanket from foundation\n", - "kb = 0.06; \t\t\t\t#coefficient of permeability of blanket material\n", - "kf = 72; \t\t\t\t#coefficient of permeability of foundation soil\n", - "Hw = 10; \t\t\t\t#heigth of water in reservior\n", - "Xd = 40;\n", - "\n", - "a = (kb/(kf*Zb*Zf))**0.5;\n", - "Xo = 1.414/a;\n", - "\n", - "#we vary value of x\n", - "x = [0, 25, 50, 75, 100, 125, 151.8, 300]\n", - "Xr = zeros(8)\n", - "ho = zeros(8)\n", - "r = zeros(8)\n", - "\n", - "for i in range(8):\n", - " e = math.exp(2*a*x[i]);\n", - " Xr[i] = (e-1)/(a*(e+1));\n", - " ho[i] = Xr[i]*Hw/(Xr[i]+Xd);\n", - " r[i] = Xr[i]*100/(Xr[i]+Xd);\n", - "\n", - "print \"x Xr ho reduction qpercent\";\n", - "for i in range(8):\n", - " print \"%.2f %.2f %.2f %.2f\"%(x[i],Xr[i],ho[i],r[i]);\n", - "\n", - "#graph is plotted between r and x.\n", - "#after around 130m length there is only slight increase in head dissipated(ho)\n", - "plot(x,r)\n", - "show()\n", - "L = 130;\n", - "print \"Thickness of blanket = %.2f m\"%(Zb);\n", - "print \"Length of blanket = %i m.\"%(L);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "x Xr ho reduction qpercent\n", - "0.00 0.00 0.00 0.00\n", - "25.00 24.56 3.80 38.04\n", - "50.00 46.67 5.38 53.85\n", - "75.00 64.78 6.18 61.83\n", - "100.00 78.50 6.62 66.24\n", - "125.00 88.28 6.88 68.82\n", - "151.80 95.35 7.04 70.45\n", - "300.00 106.53 7.27 72.70\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10818c3d0>" - ] - }, - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Thickness of blanket = 1.20 m\n", - "Length of blanket = 130 m.\n" - ] - } - ], - "prompt_number": 1 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch11.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch11.ipynb deleted file mode 100755 index f4d8bc9d..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch11.ipynb +++ /dev/null @@ -1,573 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:af2e84ceb74c3bc3148f31759e2a692dba979671e7232787c666481af8571437" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 11 : SPILLWAYS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.1 pg : 538" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "h = 1.2; \t\t\t\t#head of water\n", - "Cd = 2.2; \t\t\t\t#coefficient of discharge\n", - "rho = 1; \t\t\t\t#density of water\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "\n", - "q = Cd*h**1.5;\n", - "\n", - "#applying bernaulli's equation at u/s water surface at section A and B\n", - "#solving it by error and trial method we get\n", - "v1 = 13.7;v2 = 14.7;\n", - "d1 = 0.212;d2 = 0.197;\n", - "\n", - "F1 = gamma_w*d1**2*math.cos(math.radians(60))/2;\n", - "F2 = gamma_w*d2**2/2;\n", - "W = gamma_w*60*2*math.pi*3*((d1+d2)/2)/360;\n", - "Fx = rho*q*(v2-v1*math.cos(math.radians(60)))-F1/2+F2;\n", - "Fy = rho*q*(v1*math.sin(math.radians(60)))+F1*math.sin(math.radians(60))+W;\n", - "F = (Fx**2+Fy**2)**0.5;\n", - "F = round(F*100)/100;\n", - "\n", - "# Results\n", - "print \"Resultant force = %.2f kN/m.\"%(F);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Resultant force = 46.68 kN/m.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.2 pg : 539" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "C = 2.4; \t\t\t\t#coefficient of discharge\n", - "H = 2; \t\t\t\t#head\n", - "L = 100; \t\t\t\t#length of spillway\n", - "wc = 8; \t\t\t\t#heigth of weir crest above bottom\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "\n", - "# Calculations\n", - "h = H+wc;\n", - "Q1 = C*L*H**(1.5); \t\t\t\t#neglecting approach velocity and end contractions\n", - "va = Q1/(h*L);\n", - "ha = va**2/(2*g);\n", - "Ha = ha+H;\n", - "Q = C*L*Ha**1.5;\n", - "Q = round(Q*10)/10;\n", - "\n", - "# Results\n", - "print \"discharge over oggy weir = %.2f cumecs.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge over oggy weir = 690.80 cumecs.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.3 pg : 540" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#capacity of siphon\n", - "#head required in oggy spillway\n", - "#length of oggy weir required\n", - "\n", - "#Given\n", - "t = 6; \t\t\t\t#tail water elevation\n", - "h = 1; \t\t\t\t#heigth of siphon spillway\n", - "w = 4; \t\t\t\t#width of siphon spillway\n", - "hw = 1.5; \t\t\t\t#head water elevation\n", - "C = 0.6; \t\t\t\t#coefficient of discharge\n", - "Co = 2.25; \t\t\t\t#coefficient of discharge of oggy spillway\n", - "lo = 4; \t\t\t\t#length of oggy spillway\n", - "hc = 1.5; \t\t\t\t#head on weir crest\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "\n", - "# Calculations and Results\n", - "#part (a)\n", - "Q = C*h*w*(2*g*(t+hw))**0.5;\n", - "Q = round(Q*10)/10;\n", - "print \"capacity of siphon = %.2f cumecs.\"%(Q);\n", - "\n", - "#part (b)\n", - "h1 = (Q/(Co*lo))**(2./3);\n", - "h1 = round(h1*100)/100;\n", - "print \"head required in oggy spillway = %.2f m\"%(h1);\n", - "\n", - "#part (c)\n", - "L = Q/(Co*(hc)**1.5);\n", - "L = round(L*100)/100;\n", - "print \"length of oggy weir required = %.2f m.\"%(L);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "capacity of siphon = 29.10 cumecs.\n", - "head required in oggy spillway = 2.19 m\n", - "length of oggy weir required = 7.04 m.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.4 pg : 540" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "rl = 435; \t\t\t\t#full reservior level\n", - "cl = 429.6; \t\t\t\t#level of centre of siphon\n", - "hfl = 435.85; \t\t\t\t#high flood level\n", - "hfd = 600; \t\t\t\t#high flood discharge\n", - "w = 4; \t\t\t\t#width of throat\n", - "h = 2; \t\t\t\t#heigth of throat\n", - "C = 0.65; \t\t\t\t#coefficient of discharge\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "\n", - "# Calculations\n", - "H = hfl-cl;\n", - "Q = C*w*h*(2*g*H)**0.5;\n", - "n = hfd/Q;\n", - "n = round(n*100)/100;\n", - "\n", - "# Results\n", - "print \" number of siphons units required = %.2f.hence provide 11 siphons units.\"%(n);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " number of siphons units required = 10.42.hence provide 11 siphons units.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.5 pg : 541" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import arange,zeros\n", - "\n", - "#design oggy spillway for concrete gravity dam\n", - "\n", - "#Given\n", - "rbl = 250; \t\t\t\t#avarage river bed level\n", - "rlc = 350; \t\t\t\t#R.L of spillway crest\n", - "s = 0.75; \t\t\t\t#slope on downstream side\n", - "Q = 6500; \t\t\t\t#discharge\n", - "L = 5*9; \t\t\t\t#length of spillway\n", - "Cd = 2.2; \t\t\t\t#coefficient of discharge\n", - "t = 2; \t\t\t\t#thickness of each pier\n", - "\n", - "#step 1. computation of design head\n", - "H = (Q/(Cd*L))**(2./3);\n", - "P = rlc-rbl;\n", - "\n", - "#P/H = 6.15,which is<1.33;it is a high overflow spillway\n", - "\n", - "#H+P/H = 7.15>1.7; hence discharge coefficient is not affected by downstream apron interface\n", - "\n", - "Kp = 0.01;\n", - "Ka = 0.1;\n", - "N = 4;\n", - "He = 17.5; \t\t\t\t#assumed\n", - "Le = L-2*(N*Kp+Ka)*He;\n", - "He1 = (Q/(Cd*Le))**(2./3);\n", - "He1 = round(He1*100)/100;\n", - "#He1 is almost equal to He\n", - "print \"crest profile will be designed for Hd = %.2f m.\"%(He1);\n", - "\n", - "#step 2. determination of d/s profile\n", - "\n", - "#equating the slope of d/s side and derivative of profile equation suggested by WES\n", - "x = 27.03;\n", - "y = 0.04372*x**1.85;\n", - "print \"downstream profile:\";\n", - "x = arange(1,27)\n", - "y = zeros(26)\n", - "for i in range(26):\n", - " y[i] = 0.04372*x[i]**1.85;\n", - " y[i] = round(y[i]*1000)/1000;\n", - "\n", - "print \"x y\";\n", - "for i in range(26):\n", - " print \"%i %.2f\"%(x[i],y[i]);\n", - "\n", - "print \"27.03 19.48\";\n", - "#step 3. determination of u/s profile\n", - "# math.cosidering equation for vertical u/s face and Hd = 17.58\n", - "\n", - "print \"upstream profile:\";\n", - "x = [-0.5, -0.1, -1.5, -2.0, -3.0, -4.0, -4.75];\n", - "y = zeros(7)\n", - "for i in range(7):\n", - " if i==6:\n", - " y[i] = 0.0633*(x[i]+4.75)**1.85+2.2151-1.2643*(x[i]+4.75)**0.625;\n", - " y[i] = round(y[i]*1000)/1000;\n", - " continue\n", - " \n", - " y[i] = 0.0633*(x[i]+4.7466)**1.85+2.2151-1.2643*(x[i]+4.7466)**0.625;\n", - " y[i] = round(y[i]*1000)/1000;\n", - "\n", - "print \"x y\";\n", - "for i in range(7):\n", - " print \"%.2f %.2f\"%(x[i],y[i]);\n", - "\n", - "\n", - "#step 4.design of d/s bucket\n", - "\n", - "R = P/4;\n", - "print \"radius of bucket = %i m.\"%(R);\n", - "print \"bucket will subtend angle of 60 degree at the centre.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "crest profile will be designed for Hd = 17.58 m.\n", - "downstream profile:\n", - "x y\n", - "1 0.04\n", - "2 0.16\n", - "3 0.33\n", - "4 0.57\n", - "5 0.86\n", - "6 1.20\n", - "7 1.60\n", - "8 2.05\n", - "9 2.55\n", - "10 3.10\n", - "11 3.69\n", - "12 4.34\n", - "13 5.03\n", - "14 5.77\n", - "15 6.55\n", - "16 7.38\n", - "17 8.26\n", - "18 9.18\n", - "19 10.15\n", - "20 11.16\n", - "21 12.21\n", - "22 13.31\n", - "23 14.45\n", - "24 15.63\n", - "25 16.86\n", - "26 18.13\n", - "27.03 19.48\n", - "upstream profile:\n", - "x y\n", - "-0.50 0.01\n", - "-0.10 -0.00\n", - "-1.50 0.14\n", - "-2.00 0.25\n", - "-3.00 0.60\n", - "-4.00 1.20\n", - "-4.75 2.21\n", - "radius of bucket = 25 m.\n", - "bucket will subtend angle of 60 degree at the centre.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.6 pg : 562" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t#design length and depth of stilling bamath.sin\n", - "\t\t\t\t\n", - "#Given\n", - "q = 1; \t\t\t\t#discharge of spillway\n", - "Cd = 0.7; \t\t\t\t#coefficient of discharge\n", - "h1 = 10; \t\t\t\t#heigth of crest above downstream silting bamath.sin\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "Cv = 0.9; \t\t\t\t#coefficient of velocity\n", - "\n", - "# Calculations\n", - "h = (3*q/(2*Cd*(2*g)**0.5))**(2./3);\n", - "H = h1+h/2;\n", - "vt = (2*g*H)**0.5;\n", - "v1 = Cv*vt;\n", - "y1 = q/v1;\n", - "F1 = v1/(g*y1)**0.5;\n", - "\t\t\t\t#F>1, flow is super-critical\n", - "y2 = 1;\n", - "v2 = q/y2;\n", - "F2 = v2/(g*y2)**0.5; \t\t\t\t#<1\n", - "y2 = (y1/2)*((1+8*F1**2)**0.5-1);\n", - "de = y2-1;\n", - "le = 5*(y2-y1);\n", - "de = round(de*1000)/1000;\n", - "le = round(le*10)/10;\n", - "\n", - "# Results\n", - "print \"stilling bamath.sin should be depressed by %.2f m.\"%(de);\n", - "print \"length of stilling bamath.sin = %.2f m.\"%(le);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "stilling bamath.sin should be depressed by 0.58 m.\n", - "length of stilling bamath.sin = 7.50 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.7 pg : 563" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "q = 7.83; \t\t\t\t#discharge through spillway\n", - "w = 12.5; \t\t\t\t#width of fall\n", - "d = 2.; \t\t\t\t#depth of water in downstream\n", - "g = 9.8;\n", - "\n", - "y1 = 0.5;\n", - "v1 = q/y1;\n", - "F1 = v1/(g*y1)**0.5;\n", - "\n", - "#F>1,flow is super-critical\n", - "\n", - "# Calculations\n", - "v2 = q/d;\n", - "F2 = v2/(g*d)**0.5;\n", - "y2 = (y1/2)*((1+8*F1**2)**0.5-1);\n", - "de = y2-d;\n", - "le = 5*(y2-y1);\n", - "de = round(de*100)/100;\n", - "le = round(le*10)/10;\n", - "\n", - "# Results\n", - "print \"stilling bamath.sin should be depressed by %.2f m.\"%(de);\n", - "print \"length of stilling bamath.sin = %.2f m.\"%(le); \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "stilling bamath.sin should be depressed by 2.76 m.\n", - "length of stilling bamath.sin = 21.30 m.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.8 pg : 564" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Ag = 5*2.5; \t\t\t\t#area of gate\n", - "miu = 0.25; \t\t\t\t#coefficient of friction\n", - "w = 0.5; \t\t\t\t#weigth of gate\n", - "h = 2; \t\t\t\t#head of water over crest\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "gamma_w = 1000; \t\t\t\t#unit weigth of water\n", - "\n", - "\n", - "# Calculations\n", - "m = w*g*1000;\n", - "F = gamma_w*Ag*h*h*g/10;\n", - "ff = miu*F;\n", - "tf = (m+ff)/1000;\n", - "\n", - "# Results\n", - "print \"force to be exerted to lift the gate = %.2f kN.\"%(tf);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "force to be exerted to lift the gate = 17.17 kN.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.9 pg : 564" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "q = 19; \t\t\t\t#dischrge through spillway\n", - "E = 1; \t\t\t\t#energy loss\n", - "\n", - "\t\t\t\t#from energy loss equation;E = (y2-y1)**3/4y2y1; and solving it we get\n", - "\t\t\t\t#x = 0.5*(-1+(1+294.39*(x-1)**9/64*x**3))\n", - "\t\t\t\t#by trial and error method x = 2.806\n", - "x = 2.806;\n", - "y1 = 4*x/(x-1)**3;\n", - "y2 = x*y1;\n", - "y1 = round(y1*1000)/1000;\n", - "y2 = round(y2*1000)/1000;\n", - "print \"depth of flow at both end of jumps = %.2f m and %.2f m. respectively.\"%(y1,y2);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "depth of flow at both end of jumps = 1.91 m and 5.35 m. respectively.\n" - ] - } - ], - "prompt_number": 11 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch11_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch11_1.ipynb deleted file mode 100755 index f4d8bc9d..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch11_1.ipynb +++ /dev/null @@ -1,573 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:af2e84ceb74c3bc3148f31759e2a692dba979671e7232787c666481af8571437" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 11 : SPILLWAYS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.1 pg : 538" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "h = 1.2; \t\t\t\t#head of water\n", - "Cd = 2.2; \t\t\t\t#coefficient of discharge\n", - "rho = 1; \t\t\t\t#density of water\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "\n", - "q = Cd*h**1.5;\n", - "\n", - "#applying bernaulli's equation at u/s water surface at section A and B\n", - "#solving it by error and trial method we get\n", - "v1 = 13.7;v2 = 14.7;\n", - "d1 = 0.212;d2 = 0.197;\n", - "\n", - "F1 = gamma_w*d1**2*math.cos(math.radians(60))/2;\n", - "F2 = gamma_w*d2**2/2;\n", - "W = gamma_w*60*2*math.pi*3*((d1+d2)/2)/360;\n", - "Fx = rho*q*(v2-v1*math.cos(math.radians(60)))-F1/2+F2;\n", - "Fy = rho*q*(v1*math.sin(math.radians(60)))+F1*math.sin(math.radians(60))+W;\n", - "F = (Fx**2+Fy**2)**0.5;\n", - "F = round(F*100)/100;\n", - "\n", - "# Results\n", - "print \"Resultant force = %.2f kN/m.\"%(F);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Resultant force = 46.68 kN/m.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.2 pg : 539" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "C = 2.4; \t\t\t\t#coefficient of discharge\n", - "H = 2; \t\t\t\t#head\n", - "L = 100; \t\t\t\t#length of spillway\n", - "wc = 8; \t\t\t\t#heigth of weir crest above bottom\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "\n", - "# Calculations\n", - "h = H+wc;\n", - "Q1 = C*L*H**(1.5); \t\t\t\t#neglecting approach velocity and end contractions\n", - "va = Q1/(h*L);\n", - "ha = va**2/(2*g);\n", - "Ha = ha+H;\n", - "Q = C*L*Ha**1.5;\n", - "Q = round(Q*10)/10;\n", - "\n", - "# Results\n", - "print \"discharge over oggy weir = %.2f cumecs.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge over oggy weir = 690.80 cumecs.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.3 pg : 540" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#capacity of siphon\n", - "#head required in oggy spillway\n", - "#length of oggy weir required\n", - "\n", - "#Given\n", - "t = 6; \t\t\t\t#tail water elevation\n", - "h = 1; \t\t\t\t#heigth of siphon spillway\n", - "w = 4; \t\t\t\t#width of siphon spillway\n", - "hw = 1.5; \t\t\t\t#head water elevation\n", - "C = 0.6; \t\t\t\t#coefficient of discharge\n", - "Co = 2.25; \t\t\t\t#coefficient of discharge of oggy spillway\n", - "lo = 4; \t\t\t\t#length of oggy spillway\n", - "hc = 1.5; \t\t\t\t#head on weir crest\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "\n", - "# Calculations and Results\n", - "#part (a)\n", - "Q = C*h*w*(2*g*(t+hw))**0.5;\n", - "Q = round(Q*10)/10;\n", - "print \"capacity of siphon = %.2f cumecs.\"%(Q);\n", - "\n", - "#part (b)\n", - "h1 = (Q/(Co*lo))**(2./3);\n", - "h1 = round(h1*100)/100;\n", - "print \"head required in oggy spillway = %.2f m\"%(h1);\n", - "\n", - "#part (c)\n", - "L = Q/(Co*(hc)**1.5);\n", - "L = round(L*100)/100;\n", - "print \"length of oggy weir required = %.2f m.\"%(L);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "capacity of siphon = 29.10 cumecs.\n", - "head required in oggy spillway = 2.19 m\n", - "length of oggy weir required = 7.04 m.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.4 pg : 540" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "rl = 435; \t\t\t\t#full reservior level\n", - "cl = 429.6; \t\t\t\t#level of centre of siphon\n", - "hfl = 435.85; \t\t\t\t#high flood level\n", - "hfd = 600; \t\t\t\t#high flood discharge\n", - "w = 4; \t\t\t\t#width of throat\n", - "h = 2; \t\t\t\t#heigth of throat\n", - "C = 0.65; \t\t\t\t#coefficient of discharge\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "\n", - "# Calculations\n", - "H = hfl-cl;\n", - "Q = C*w*h*(2*g*H)**0.5;\n", - "n = hfd/Q;\n", - "n = round(n*100)/100;\n", - "\n", - "# Results\n", - "print \" number of siphons units required = %.2f.hence provide 11 siphons units.\"%(n);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " number of siphons units required = 10.42.hence provide 11 siphons units.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.5 pg : 541" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import arange,zeros\n", - "\n", - "#design oggy spillway for concrete gravity dam\n", - "\n", - "#Given\n", - "rbl = 250; \t\t\t\t#avarage river bed level\n", - "rlc = 350; \t\t\t\t#R.L of spillway crest\n", - "s = 0.75; \t\t\t\t#slope on downstream side\n", - "Q = 6500; \t\t\t\t#discharge\n", - "L = 5*9; \t\t\t\t#length of spillway\n", - "Cd = 2.2; \t\t\t\t#coefficient of discharge\n", - "t = 2; \t\t\t\t#thickness of each pier\n", - "\n", - "#step 1. computation of design head\n", - "H = (Q/(Cd*L))**(2./3);\n", - "P = rlc-rbl;\n", - "\n", - "#P/H = 6.15,which is<1.33;it is a high overflow spillway\n", - "\n", - "#H+P/H = 7.15>1.7; hence discharge coefficient is not affected by downstream apron interface\n", - "\n", - "Kp = 0.01;\n", - "Ka = 0.1;\n", - "N = 4;\n", - "He = 17.5; \t\t\t\t#assumed\n", - "Le = L-2*(N*Kp+Ka)*He;\n", - "He1 = (Q/(Cd*Le))**(2./3);\n", - "He1 = round(He1*100)/100;\n", - "#He1 is almost equal to He\n", - "print \"crest profile will be designed for Hd = %.2f m.\"%(He1);\n", - "\n", - "#step 2. determination of d/s profile\n", - "\n", - "#equating the slope of d/s side and derivative of profile equation suggested by WES\n", - "x = 27.03;\n", - "y = 0.04372*x**1.85;\n", - "print \"downstream profile:\";\n", - "x = arange(1,27)\n", - "y = zeros(26)\n", - "for i in range(26):\n", - " y[i] = 0.04372*x[i]**1.85;\n", - " y[i] = round(y[i]*1000)/1000;\n", - "\n", - "print \"x y\";\n", - "for i in range(26):\n", - " print \"%i %.2f\"%(x[i],y[i]);\n", - "\n", - "print \"27.03 19.48\";\n", - "#step 3. determination of u/s profile\n", - "# math.cosidering equation for vertical u/s face and Hd = 17.58\n", - "\n", - "print \"upstream profile:\";\n", - "x = [-0.5, -0.1, -1.5, -2.0, -3.0, -4.0, -4.75];\n", - "y = zeros(7)\n", - "for i in range(7):\n", - " if i==6:\n", - " y[i] = 0.0633*(x[i]+4.75)**1.85+2.2151-1.2643*(x[i]+4.75)**0.625;\n", - " y[i] = round(y[i]*1000)/1000;\n", - " continue\n", - " \n", - " y[i] = 0.0633*(x[i]+4.7466)**1.85+2.2151-1.2643*(x[i]+4.7466)**0.625;\n", - " y[i] = round(y[i]*1000)/1000;\n", - "\n", - "print \"x y\";\n", - "for i in range(7):\n", - " print \"%.2f %.2f\"%(x[i],y[i]);\n", - "\n", - "\n", - "#step 4.design of d/s bucket\n", - "\n", - "R = P/4;\n", - "print \"radius of bucket = %i m.\"%(R);\n", - "print \"bucket will subtend angle of 60 degree at the centre.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "crest profile will be designed for Hd = 17.58 m.\n", - "downstream profile:\n", - "x y\n", - "1 0.04\n", - "2 0.16\n", - "3 0.33\n", - "4 0.57\n", - "5 0.86\n", - "6 1.20\n", - "7 1.60\n", - "8 2.05\n", - "9 2.55\n", - "10 3.10\n", - "11 3.69\n", - "12 4.34\n", - "13 5.03\n", - "14 5.77\n", - "15 6.55\n", - "16 7.38\n", - "17 8.26\n", - "18 9.18\n", - "19 10.15\n", - "20 11.16\n", - "21 12.21\n", - "22 13.31\n", - "23 14.45\n", - "24 15.63\n", - "25 16.86\n", - "26 18.13\n", - "27.03 19.48\n", - "upstream profile:\n", - "x y\n", - "-0.50 0.01\n", - "-0.10 -0.00\n", - "-1.50 0.14\n", - "-2.00 0.25\n", - "-3.00 0.60\n", - "-4.00 1.20\n", - "-4.75 2.21\n", - "radius of bucket = 25 m.\n", - "bucket will subtend angle of 60 degree at the centre.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.6 pg : 562" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t#design length and depth of stilling bamath.sin\n", - "\t\t\t\t\n", - "#Given\n", - "q = 1; \t\t\t\t#discharge of spillway\n", - "Cd = 0.7; \t\t\t\t#coefficient of discharge\n", - "h1 = 10; \t\t\t\t#heigth of crest above downstream silting bamath.sin\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "Cv = 0.9; \t\t\t\t#coefficient of velocity\n", - "\n", - "# Calculations\n", - "h = (3*q/(2*Cd*(2*g)**0.5))**(2./3);\n", - "H = h1+h/2;\n", - "vt = (2*g*H)**0.5;\n", - "v1 = Cv*vt;\n", - "y1 = q/v1;\n", - "F1 = v1/(g*y1)**0.5;\n", - "\t\t\t\t#F>1, flow is super-critical\n", - "y2 = 1;\n", - "v2 = q/y2;\n", - "F2 = v2/(g*y2)**0.5; \t\t\t\t#<1\n", - "y2 = (y1/2)*((1+8*F1**2)**0.5-1);\n", - "de = y2-1;\n", - "le = 5*(y2-y1);\n", - "de = round(de*1000)/1000;\n", - "le = round(le*10)/10;\n", - "\n", - "# Results\n", - "print \"stilling bamath.sin should be depressed by %.2f m.\"%(de);\n", - "print \"length of stilling bamath.sin = %.2f m.\"%(le);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "stilling bamath.sin should be depressed by 0.58 m.\n", - "length of stilling bamath.sin = 7.50 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.7 pg : 563" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "q = 7.83; \t\t\t\t#discharge through spillway\n", - "w = 12.5; \t\t\t\t#width of fall\n", - "d = 2.; \t\t\t\t#depth of water in downstream\n", - "g = 9.8;\n", - "\n", - "y1 = 0.5;\n", - "v1 = q/y1;\n", - "F1 = v1/(g*y1)**0.5;\n", - "\n", - "#F>1,flow is super-critical\n", - "\n", - "# Calculations\n", - "v2 = q/d;\n", - "F2 = v2/(g*d)**0.5;\n", - "y2 = (y1/2)*((1+8*F1**2)**0.5-1);\n", - "de = y2-d;\n", - "le = 5*(y2-y1);\n", - "de = round(de*100)/100;\n", - "le = round(le*10)/10;\n", - "\n", - "# Results\n", - "print \"stilling bamath.sin should be depressed by %.2f m.\"%(de);\n", - "print \"length of stilling bamath.sin = %.2f m.\"%(le); \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "stilling bamath.sin should be depressed by 2.76 m.\n", - "length of stilling bamath.sin = 21.30 m.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.8 pg : 564" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Ag = 5*2.5; \t\t\t\t#area of gate\n", - "miu = 0.25; \t\t\t\t#coefficient of friction\n", - "w = 0.5; \t\t\t\t#weigth of gate\n", - "h = 2; \t\t\t\t#head of water over crest\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "gamma_w = 1000; \t\t\t\t#unit weigth of water\n", - "\n", - "\n", - "# Calculations\n", - "m = w*g*1000;\n", - "F = gamma_w*Ag*h*h*g/10;\n", - "ff = miu*F;\n", - "tf = (m+ff)/1000;\n", - "\n", - "# Results\n", - "print \"force to be exerted to lift the gate = %.2f kN.\"%(tf);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "force to be exerted to lift the gate = 17.17 kN.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 11.9 pg : 564" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "q = 19; \t\t\t\t#dischrge through spillway\n", - "E = 1; \t\t\t\t#energy loss\n", - "\n", - "\t\t\t\t#from energy loss equation;E = (y2-y1)**3/4y2y1; and solving it we get\n", - "\t\t\t\t#x = 0.5*(-1+(1+294.39*(x-1)**9/64*x**3))\n", - "\t\t\t\t#by trial and error method x = 2.806\n", - "x = 2.806;\n", - "y1 = 4*x/(x-1)**3;\n", - "y2 = x*y1;\n", - "y1 = round(y1*1000)/1000;\n", - "y2 = round(y2*1000)/1000;\n", - "print \"depth of flow at both end of jumps = %.2f m and %.2f m. respectively.\"%(y1,y2);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "depth of flow at both end of jumps = 1.91 m and 5.35 m. respectively.\n" - ] - } - ], - "prompt_number": 11 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch12.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch12.ipynb deleted file mode 100755 index edc47614..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch12.ipynb +++ /dev/null @@ -1,1384 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:b0cdc42f1fc8747c583aaf55d3f5af5550eb632d3c87d035821d8a7d148eae09" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 12 : DIVERSION HEADWORKS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.1 pg : 576" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#uplift presuures and thickness of floor at 6m, 12m and 18m from u/s\n", - "\n", - "#Given\n", - "rho = 2.24; \t\t\t\t#relative density of material\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "L = 22.; \t\t\t\t#total length\n", - "lc = (2.*6)+L+(2*8); \t\t\t\t#length of creep\n", - "hg = 4./lc; \t\t\t\t#hydraulic gradient\n", - "print \"avearge hydraulic gradient = %.2f.\"%(hg);\n", - "#at 6 m from u/s\n", - "x = 6.;\n", - "lg = (6.*2)+x;\n", - "h1 = 4.*(1-lg/50); \t\t\t\t#unbalanced head\n", - "up = gamma_w*h1;\n", - "t = 4.*h1/(3*(rho-1));\n", - "up = round(up*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"uplift at 6 m from u/s = %.2f kN/square metre.\"%(up);\n", - "print \"thickness at 6 m from u/s = %.2f m.\"%(t);\n", - "\n", - "#at 12 m from u/s\n", - "x = 12.;\n", - "lg = (6.*2)+x;\n", - "h1 = 4.*(1-lg/50); \t\t\t\t#unbalanced head\n", - "up = gamma_w*h1;\n", - "t = 4.*h1/(3*(rho-1));\n", - "up = round(up*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"uplift at 12 m from u/s = %.2f kN/square metre.\"%(up);\n", - "print \"thickness at 12 m from u/s = %.2f m.\"%(t);\n", - "\n", - "#at 18m from u/s\n", - "x = 18.;\n", - "lg = (6.*2)+x;\n", - "h1 = 4.*(1-lg/50); \t\t\t\t#unbalanced head\n", - "up = gamma_w*h1;\n", - "t = 4*h1/(3*(rho-1));\n", - "up = round(up*10)/10;\n", - "t = round(t*100)/100;\n", - "print \"uplift at 18 m from u/s = %.2f kN/square metre.\"%(up);\n", - "print \"thickness at 18 m from u/s = %.2f m.\"%(t);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "avearge hydraulic gradient = 0.08.\n", - "uplift at 6 m from u/s = 25.11 kN/square metre.\n", - "thickness at 6 m from u/s = 2.75 m.\n", - "uplift at 12 m from u/s = 20.40 kN/square metre.\n", - "thickness at 12 m from u/s = 2.24 m.\n", - "uplift at 18 m from u/s = 15.70 kN/square metre.\n", - "thickness at 18 m from u/s = 1.72 m.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.2 pg : 589" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#check whether section is safe against overturning and piping\n", - "\n", - "#Given\n", - "b = 54.; \t\t\t\t#width of section\n", - "D1D2 = 16; \t\t\t\t#dismath.tance between points D1 and D2\n", - "D2D3 = 37; \t\t\t\t#dismath.tance between points D2 and D3\n", - "\n", - "#first pipe line\n", - "#taking data from figure\n", - "d = 105-97;\n", - "b1 = 0.5;\n", - "alpha = b/d;\n", - "#from the curves we get\n", - "fic1 = 0.665;\n", - "fid1 = 0.76;\n", - "fie1 = 1;\n", - "t = 105-104; \t\t\t\t#floor thickness\n", - "corec = (fid1-fic1)*100*t/d; \t\t\t\t#correction for floor thickness\n", - "#for pile no. 2\n", - "D = 104-97;\n", - "d = 104-97;\n", - "bdash = 16;\n", - "C = 19*(D/bdash)**0.5*(d+D)/b; \t\t\t\t#correction for pile no. 2\n", - "fic1 = fic1*100+corec+C; \t\t\t\t#corrected pressures\n", - "\n", - "#intermedite pipe line\n", - "d = 105-97;\n", - "b1 = 16.5;\n", - "alpha = b/d;\n", - "r = b1/b; \t\t\t\t#ratio b1/b\n", - "#from the curves we get\n", - "fic2 = 0.52;\n", - "fie2 = 0.725;\n", - "fid2 = 0.615;\n", - "corec_c1 = (fid2-fic2)*100*t/d;\n", - "corec_e1 = (fie2-fid2)*100/d;\n", - "\n", - "#for pile no. 1\n", - "C1 = C;\n", - "d = 104-97;\n", - "bdash = 37;\n", - "D = 104-95;\n", - "C2 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "#correction due to slope\n", - "corec_e2 = 3.3; \t\t\t\t#from table 12.4\n", - "#correction is negative due to upwrd slope\n", - "l = 4; \t\t\t\t#horizontal length of slope\n", - "corec_c2 = corec_e2*l/bdash;\n", - "\n", - "fie2 = fie2*100-corec_e1-corec_e2;\n", - "fic2 = fic2*100+corec_c1+C2-corec_c2;\n", - "\n", - "#pile no. 3 at d/s end\n", - "d = 103.5-95;\n", - "alpha_ = d/b;\n", - "#for curves\n", - "fie3 = 0.35;fid3 = 0.242;\n", - "corec_t = (fie3-fid3)*100*(103.5-102)/d;\n", - "\n", - "#correction for interference at pile no. 2\n", - "d = 102-95;\n", - "D = 102-97;\n", - "C3 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "fie3 = fie3*100-corec_t-C3;\n", - "\n", - "point = ['C1', 'C2' ,'E2' ,'E3']; \t\t\t\t#Point\n", - "P = [fic1 ,fic2 ,fie2 ,fie3]; \t\t\t\t#pressure percent\n", - "P_ = [3.55 ,2.78, 3.39, 1.58]; \t\t\t\t#pressure head\n", - "print \"Points Pressure percent Pressure head\";\n", - "\n", - "for i in range(4):\n", - " P[i] = round(P[i]*10)/10;\n", - " print \"%s %.2f %.2f\"%(point[i],P[i],P_[i]);\n", - "\n", - "\n", - "#check for floor thickness\n", - "Pa = P_[1]-((P_[1]-P_[3])*6.5/37);\n", - "Pb = P_[1]-((P_[1]-P_[3])*24/37);\n", - "Pc = P_[1]-((P_[1]-P_[3])*30/37);\n", - "rho = 2.24; \t\t\t\t#specific gravity of concrete\n", - "ta = Pa/(rho-1);\n", - "tb = Pb/(rho-1);\n", - "tc = Pc/(rho-1);\n", - "ta = round(ta*100)/100;\n", - "tb = round(tb*100)/100;\n", - "tc = round(tc*100)/100;\n", - "print \"Thickness required at A = %.2f m.\"%(ta);\n", - "print \"Thickness required at B = %.2f m.\"%(tb);\n", - "print \"Thickness required at C = %.2f m.\"%(tc);\n", - "t = 103.5-102;\n", - "print \"Thickness provided = %.2f m.\"%(t);\n", - "print \"Floor thickness at B and C are adequate\";\n", - "\n", - "#exit gradient\n", - "H = 108.5-103.5; \t\t\t\t#seepage head\n", - "d = 103.5-95; \t\t\t\t#depth cut-off\n", - "#from exit gradient curve\n", - "alpha = 6.35;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "Ge = H/(d*math.pi*lambda1**0.5);\n", - "print \"exit gradient = %.2f.\"%(Ge);\n", - "print \" it is less than permissible exit gradient < 1/6Hence safe..\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Points Pressure percent Pressure head\n", - "C1 67.70 3.55\n", - "C2 52.80 2.78\n", - "E2 67.80 3.39\n", - "E3 33.10 1.58\n", - "Thickness required at A = 2.07 m.\n", - "Thickness required at B = 1.61 m.\n", - "Thickness required at C = 1.46 m.\n", - "Thickness provided = 1.50 m.\n", - "Floor thickness at B and C are adequate\n", - "exit gradient = 0.10.\n", - " it is less than permissible exit gradient < 1/6Hence safe..\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.3 pg : 605" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots,ceil\n", - "\n", - "#design a vertical drop weir on Bligh's theory\n", - "#test floor by Khosla's theory\n", - "\n", - "#Given\n", - "Q = 2800; \t\t\t\t#maximum flood discharge\n", - "hfl = 285; \t\t\t\t#H.F.L before construction\n", - "hw = 278; \t\t\t\t#minimum water level\n", - "fsl = 284; \t\t\t\t#F.S.L of canal\n", - "c = 12; \t\t\t\t#coefficient of creep\n", - "flux = 1; \t\t\t\t#allowable afflux\n", - "Ge = 1./6; \t\t\t\t#permissible exit gradient\n", - "rho = 2.24; \t\t\t\t#specific gravity of concrete\n", - "\n", - "#Hydraulic calculation\n", - "L = 4.75*Q**0.5;\n", - "q = Q/L;\n", - "q = round(q*10)/10;\n", - "print \"Hydraulic calculation:\";\n", - "print \"discharge per unit width of river = %.2f cumecs.\"%(q);\n", - "f = 1;\n", - "R = 1.35*(q**2/f)**(1./3);\n", - "R = round(R*100)/100;\n", - "print \"regime scour depth = %.2f m.\"%(R);\n", - "V = q/R; \t\t\t\t#regime velocity\n", - "vh = V**2/(2*9.81); \t\t\t\t#velocity head\n", - "l_down = hfl+vh;\n", - "l_up = l_down+flux;\n", - "hfl_up = l_up-vh;\n", - "hfl_down = hfl-0.5;\n", - "hfl_down = round(hfl_down*100)/100;\n", - "print \"actual d/s H.F.L allowing 0.5 m for retrogation = %.2f m.\"%(hfl_down);\n", - "K = (q/1.7)**(2./3);\n", - "cl = l_up-K; \t\t\t\t#crest level\n", - "cl = round(cl*100)/100;\n", - "print \"crest level = %.2f m.\"%(cl);\n", - "pl = fsl+0.5; \t\t\t\t#pond level\n", - "s = hfl_down-cl; \t\t\t\t#heigth of shutter\n", - "print \"heigth of shutter = %.2f m.\"%(s);\n", - "rl_up_pile = hfl_up-1.5*R; \t\t\t\t#R.L of bottom u/s pile\n", - "d_up_cut = hw-276; \t\t\t\t#depth of upstream cut-off\n", - "print \"depth of upstream cut-off = %.2f m.\"%(d_up_cut);\n", - "print \" provide concrete cut off 2 m depth.\";\n", - "rl_bot_ds = hfl_down-2*R;\n", - "Hs = hfl_down-hw; \t\t\t\t#seepage head\n", - "Hc = cl-hw; \t\t\t\t#heigth of crest\n", - "print \"R.L of gates crest = %.2f m.\"%(Hs);\n", - "print \"Heigth of crest = %.2f m.\"%(Hc);\n", - "\n", - "#design of weir wall\n", - "d = hfl_up-cl;\n", - "a = d/(rho)**0.5;\n", - "a = 3*d/(2*rho); \t\t\t\t#from sliding consideration\n", - "a = s+1; \t\t\t\t#from practical consieration\n", - "a = a+1;\n", - "print \"design of weir wall:\"\n", - "print \"provide top width of %i m.\"%(a);\n", - "Mo = 9.81*Hs**3/6; \t\t\t\t#overtirning moment\n", - "#equating the moment of resismath.tance to overturning moment and putting the values we get\n", - "#y = poly([-1.084,0.020,0.039],'x','c');\n", - "y = [0.039,0.020,-1.084]\n", - "b = roots(y)[1];\n", - "#we get b = - 5.5347261 and 5.0219056\n", - "#taking\n", - "b = 5;\n", - "#when weir is submerged\n", - "C = 0.58;\n", - "d = (q**2/((2*C/3)**2*2*9.81))**(1./3);\n", - "Mo = 9.81*d*Hc**2/2;\n", - "#from equation of moment of resistence we get\n", - "y = [1,3,-77.55]\n", - "b = ceil(roots(y)[1]); #we get b = - 10.433085 and 7.4330846\n", - "print \"bottom width = %i m.\"%(b);\n", - "\n", - "#design of impervious and pervious aprons\n", - "C = 12;\n", - "L = C*Hs;\n", - "print \"design of impervious and pervious aprons:\";\n", - "print \"total creep length = %i m.\"%(L);\n", - "l1 = 2.21*C*(Hs/13)**0.5;\n", - "l1_ = l1+1;\n", - "print \"length of downstream impervious apron = %i m.\"%(l1_);\n", - "d1 = hw-276;\n", - "d2 = hw-271;\n", - "l2 = L-l1-(b+2*d1+2*d2);\n", - "print \"length of upstream impervious apron = %i m.\"%(l2);\n", - "l3 = 18*C*(Hs*q/975)**0.5;\n", - "print \"total length of d/s apron = %i m.\"%(l3); \t\t\t\t#calculation is wrong in book\n", - "l = l3-l1;\n", - "le = l/2;\n", - "le = round(le*100)/100;\n", - "print 'provide filter of length %.2f m. and launching apron of length %.2f m.'%(le,le);\n", - "t = d2*10**0.5/le;\n", - "print \"thickness of launching apron in horizontal position = %.2f m.\"%(t);\n", - "print \"provide launching apron of thickness 1.5 m.\";\n", - "T = 2*d1;\n", - "V = d1*10**0.5;\n", - "ta = V/T;\n", - "ta = round(ta*10)/10;\n", - "print \"thickness of apron in horizontal position = %.2f m.\"%(ta);\n", - "Hr = Hs-Hs*(4+33+8)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print 'provide thickness of %.2f m from d/s of weir wall to point 6 m from it.'%(t);\n", - "Hr = Hs-Hs*(4+33+8+6)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print \"provide thickness of %.2f m from 6 m to 12 m from d/s end of weir wall.\"%(t);\n", - "Hr = Hs-Hs*(4+33+8+12)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print \"provide thickness of %.2f m for rest of length of weir floor.\"%(t);\n", - "\n", - "#check by khosla's theory\n", - "b = 33+8+19; \t\t\t\t#total horizontal length of impervious floor\n", - "d = 7; \t\t\t\t#depth of downstream pile\n", - "alpha = b/d;\n", - "n = 0.14; \t\t\t\t#n = 1/math.pi*(lambda)**0.5;\n", - "Ge = Hs*n/d;\n", - "print \"check by Khosla theory:\";\n", - "print \"exit gradient = %.2f. < 1/6 hence safe\"%(Ge);\n", - "alpha_ = d/b;\n", - "fic1 = 0.83;fid1 = 0.88;\n", - "corec_c1 = (fid1-fic1)*100/2;\n", - "bdash = b;\n", - "d = 2;D = 7;\n", - "C1 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "fic1 = fic1*100+corec_c1+C1;\n", - "Pc = Hs*fic1/100; \t\t\t\t#pressure head at C\n", - "alpha_ = d/b;\n", - "fie2 = 0.31;fid2 = 0.21;\n", - "corec_e1 = (fie2-fid2)*1.7*100/7;\n", - "bdash = b;\n", - "d = 7;D = 2;\n", - "C1 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "fie2 = fie2*100-corec_e1-C1; \t\t\t\t#in book 3.53 value is wrong\n", - "Pe = Hs*fie2/100; \t\t\t\t#pressue head at E\n", - "#assuming linear variation of pressure for intermediate points\n", - "Pa = Pc-(Pc-Pe)*(33+8)/b;\n", - "t = Pa/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at d/s of weir wall = %.2f m.\"%(Pa);\n", - "print \"thickness at d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe.\"%(t);\n", - "Pb = Pc-(Pc-Pe)*(33+8+6)/b;\n", - "t = Pb/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at 6 m from d/s of weir wall = %.2f m.\"%(Pb);\n", - "print \"thickness at 6m from d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe.\"%(t);\n", - "Pc = Pc-(Pc-Pe)*(33+8+12)/b;\n", - "t = Pc/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at 12 m from d/s of weir wall = %.2f m.\"%(Pc);\n", - "print \"thickness at 12m from d/s of weir wall = %.2f m. > thickness by Bligh theory;hence unsafe.\"%(t);\n", - "print \"hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Hydraulic calculation:\n", - "discharge per unit width of river = 11.10 cumecs.\n", - "regime scour depth = 6.72 m.\n", - "actual d/s H.F.L allowing 0.5 m for retrogation = 284.50 m.\n", - "crest level = 282.65 m.\n", - "heigth of shutter = 1.85 m.\n", - "depth of upstream cut-off = 2.00 m.\n", - " provide concrete cut off 2 m depth.\n", - "R.L of gates crest = 6.50 m.\n", - "Heigth of crest = 4.65 m.\n", - "design of weir wall:\n", - "provide top width of 3 m.\n", - "bottom width = 8 m.\n", - "design of impervious and pervious aprons:\n", - "total creep length = 78 m.\n", - "length of downstream impervious apron = 19 m.\n", - "length of upstream impervious apron = 33 m.\n", - "total length of d/s apron = 58 m.\n", - "provide filter of length 20.00 m. and launching apron of length 20.00 m.\n", - "thickness of launching apron in horizontal position = 1.11 m.\n", - "provide launching apron of thickness 1.5 m.\n", - "thickness of apron in horizontal position = 1.60 m.\n", - "provide thickness of 3.00 m from d/s of weir wall to point 6 m from it.\n", - "provide thickness of 2.40 m from 6 m to 12 m from d/s end of weir wall.\n", - "provide thickness of 1.90 m for rest of length of weir floor.\n", - "check by Khosla theory:\n", - "exit gradient = 0.13. < 1/6 hence safe\n", - "pressure at d/s of weir wall = 3.03 m.\n", - "thickness at d/s of weir wall = 2.44 m. < thickness by Bligh theory;hence safe.\n", - "pressure at 6 m from d/s of weir wall = 2.66 m." - ] - }, - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "thickness at 6m from d/s of weir wall = 2.14 m. < thickness by Bligh theory;hence safe.\n", - "pressure at 12 m from d/s of weir wall = 2.29 m.\n", - "thickness at 12m from d/s of weir wall = 1.85 m. > thickness by Bligh theory;hence unsafe.\n", - "hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.4 pg : 617" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t#design a slopeing glacis\n", - "\t\t\t\t\n", - "#Given\n", - "q = 10; \t\t\t\t#maximum discharge intensity on weir crest\n", - "hfl = 255; \t\t\t\t#H.F.L before construction of weir\n", - "rb = 249.5; \t\t\t\t#R.L of river bed\n", - "pl = 254; \t\t\t\t#pond level\n", - "s = 1; \t\t\t\t#heigth of crest shutter\n", - "dhw = 251.5; \t\t\t\t#anticipated downstream water level in river when water is dischrging with pond level upstream\n", - "br = 0.5; \t\t\t\t#bed retrogression\n", - "f = 0.9; \t\t\t\t#Laecey silt factor\n", - "Ge = 1./7; \t\t\t\t#permissible exit gradient\n", - "flux = 1; \t\t\t\t#permissible afflux\n", - "\n", - "cl = pl-s; \t\t\t\t#crest level\n", - "print \"crest level = %.2f m.\"%(cl);\n", - "K = (q/1.7)**(2./3);\n", - "tel_up = cl+K;\n", - "tel_up = round(tel_up*100)/100;\n", - "print \"elevation of u/s T.E.L = %.2f m.\"%(tel_up);\n", - "R = 1.35*(q**2/f)**(1./3);\n", - "R = round(R*10)/10;\n", - "print \"regime scour depth = %.2f m.\"%(R);\n", - "V = q/R; \t\t\t\t#regime velocity\n", - "vh = V**2/(2*9.81); \t\t\t\t#velocity head\n", - "hfl_up = tel_up-vh;\n", - "tel_down = hfl+vh;\n", - "flux = hfl_up-hfl;\n", - "flux = round(flux*100)/100;\n", - "print \"afflux = %.2f. which is near to permissible\"%(flux);\n", - "hfl_down = hfl-br; \t\t\t\t#downstream H.F.L after retrogression\n", - "tel_down = tel_down-br; \t\t\t\t#downstream T.F.L after retrogression\n", - "Hl = tel_up-tel_down; \t\t\t\t#loss of head in flood\n", - "Hl = round(Hl*100)/100;\n", - "print \"loss of head in at high flood = %.2f m.\"%(Hl);\n", - "K = pl-cl; \t\t\t\t#head over crest\n", - "q_ = 1.7*(K)**1.5;\n", - "Hl_ = pl-dhw; \t\t\t\t#loss of head\n", - "print \"loss of head = %.2f m.\"%(Hl_);\n", - "Ef2 = 4.3;\n", - "Ef2_ = 1.7; \t\t\t\t#from Blench curve\n", - "jump = tel_down-Ef2;\n", - "jump_ = 251.5-Ef2_; \t\t\t\t#level at which jump will form\n", - "Ef1 = Ef2+Hl;\n", - "Ef1_ = Ef2_+Hl_;\n", - "D1 = 1.03;\n", - "D1_ = 0.15; \n", - "D2 = 3.96;D2_ = 1.68; \n", - "hj = D2-D1;\n", - "hj_ = D2_-D1_; \t\t\t\t#heigth of jump\n", - "concrete = 5*hj;\n", - "concrete_ = 5*hj_; \t\t\t\t#length of concrete floor\n", - "print \"Hydraulic jump calculation:\";\n", - "print \"heigth of jump for high flood condition = %.2f m.\"%(hj);\n", - "print \"length of concrete floor for high flood condition = %.2f m.\"%(concrete);\n", - "print \"heigth of jump for pond level condition = %.2f m.\"%(hj_);\n", - "print \"length of concrete floor for high pond level condition = %.2f m.\"%(concrete_);\n", - "\n", - "cw = 2; \t\t\t\t#crets width\n", - "us = 2; \t\t\t\t#upstream slope\n", - "ds = 3; \t\t\t\t#downstream slope\n", - "l = 15;\n", - "print \" upstream slope of glacis = %i:1.\"%(us);\n", - "print \"downstream slope of glacis = %i:1.\"%(ds);\n", - "print \"horizontal length of floor beyond the toe = %i m..\"%(l);\n", - "\n", - "R = 6.5;\n", - "sh_up = hfl_up-1.5*R;\n", - "sh_down = hfl_down-2*R;\n", - "sh_up = round(sh_up*100)/100;\n", - "print \"R.L of bottom of upstream sheet pile = %.2f m.\"%(sh_up);\n", - "print \"R.L of downstream sheet pile = %.2f m.\"%(sh_down);\n", - "print \"provide intermediate sheet pile at d/s toe of glacis.\";\n", - "Hs = pl-249.6; \t\t\t\t#maximum percolation head\n", - "d = 249.6-sh_down; \t\t\t\t#depth of d/s cut-off\n", - "n = Ge*d/Hs; \t\t\t\t#n = 1/(math.pi*lambda**0.5);\n", - "\t\t\t\t#from khosla exit gradient curve\n", - "alpha = 1.5;\n", - "b = alpha*d;\n", - "print \"length of impervious floor = %.2f m.\"%(b);\n", - "fl = (2*(253-249.5))+2+(3*(253-249.6))+15;\n", - "us = 36-fl;\n", - "print \"length of floor already provide = %.2f m.\"%(fl);\n", - "print \"which is more than required from permissible exit gradient.no upstream floor is required.\";\n", - "print \"provide %.2f m upstream floor so that total length becomes 36 m.\"%(us);\n", - "alpha_1 = 0.089; \n", - "alpha_2 = 0.225; \t\t\t\t#alpha_ = 1/alpha\n", - "b1 = 21;\n", - "alpha = 4.44;\n", - "print \"Pressure percent at points:\";\n", - "point = ['C1', 'D1' ,'C2' ,'E2' ,'D2' ,'D3' ,'E3'];\n", - "bc = [72 ,82 ,31.5 ,45.5 ,58.5 ,29 ,44];\n", - "crt = [3.1, 0, 3.5, 0, -3.2, 0, 0, -3.6];\n", - "crs = [0 ,0, 0, 0, 2.3, 0, 0, 0];\n", - "cri = [3.7, 0, 6.4, 0, -2.4, 0, -6.4];\n", - "after = [0,0,0,0,0,0,0]\n", - "print \"Points Before correction After correction\";\n", - "for i in range(7):\n", - " after[i] = bc[i]+crt[i]+crs[i]+cri[i];\n", - " print \"%s %i %.2f\"%(point[i],bc[i],after[i]);\n", - "\n", - "Hs = 254-249.6; \t\t\t\t#no flow condition\n", - "Hs_ = 256.13-254.5; \t\t\t\t#high flood condition\n", - "Hs__ = 254-251.5; \t\t\t\t#flow at pond level\n", - "print \"elevation of subsoil H.G above datum:\";\n", - "print \"no flow condition:\";\n", - "fie1 = 1*Hs;\n", - "fid1 = 0.82*Hs;\n", - "fic1 = 0.788*Hs;\n", - "fie2 = 0.552*Hs;\n", - "fid2 = 0.455*Hs;\n", - "fic2 = 0.414*Hs;\n", - "fie3 = 0.34*Hs;\n", - "fid3 = 0.29*Hs;\n", - "fic3 = 0;\n", - "fie1 = round(fie1*100)/100;fid1 = round(fid1*100)/100;fic1 = round(fic1*100)/100;\n", - "fie2 = round(fie2*100)/100;fid2 = round(fid2*100)/100;fic2 = round(fic2*100)/100;\n", - "fie3 = round(fie3*100)/100;fid3 = round(fid3*100)/100;fic3 = round(fic3*100)/100;\n", - "print \"fie1 = %.2f.;fid1 = %.2f.;fic1 = %.2f.fie2 = %.2f.;fid2 = %.2f.;fic2 = %.2f.fie3 = %.2f.;\\\n", - "fid3 = %.2f.;fic3 = %.2f.\"%(fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);\n", - "print \"high flood condition:\";\n", - "fie1 = 1*Hs_;\n", - "fid1 = 0.82*Hs_;\n", - "fic1 = 0.788*Hs_;\n", - "fie2 = 0.552*Hs_;\n", - "fid2 = 0.455*Hs_;\n", - "fic2 = 0.414*Hs_;\n", - "fie3 = 0.34*Hs_;\n", - "fid3 = 0.29*Hs_;\n", - "fic3 = 0;\n", - "fie1 = round(fie1*100)/100;fid1 = round(fid1*100)/100;fic1 = round(fic1*100)/100;\n", - "fie2 = round(fie2*100)/100;fid2 = round(fid2*100)/100;fic2 = round(fic2*100)/100;\n", - "fie3 = round(fie3*100)/100;fid3 = round(fid3*100)/100;fic3 = round(fic3*100)/100;\n", - "print \"fie1 = %.2f.;fid1 = %.2f.;fic1 = %.2f.fie2 = %.2f.;fid2 = %.2f.;fic2 = %.2f.fie3 = %.2f.;\\\n", - "fid3 = %.2f.;fic3 = %.2f.\"%(fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);\n", - "print \"flow at pond level:\";\n", - "fie1 = 1*Hs__;\n", - "fid1 = 0.82*Hs__;\n", - "fic1 = 0.788*Hs__;\n", - "fie2 = 0.552*Hs__;\n", - "fid2 = 0.455*Hs__;\n", - "fic2 = 0.414*Hs__;\n", - "fie3 = 0.34*Hs__;\n", - "fid3 = 0.29*Hs__;\n", - "fic3 = 0;\n", - "fie1 = round(fie1*100)/100;fid1 = round(fid1*100)/100;fic1 = round(fic1*100)/100;\n", - "fie2 = round(fie2*100)/100;fid2 = round(fid2*100)/100;fic2 = round(fic2*100)/100;\n", - "fie3 = round(fie3*100)/100;fid3 = round(fid3*100)/100;fic3 = round(fic3*100)/100;\n", - "print \"fie1 = %.2f.;fid1 = %.2f.;fic1 = %.2f.fie2 = %.2f.;fid2 = %.2f.;fic2 = %.2f.fie3 = %.2f.;\\\n", - "fid3 = %.2f.;fic3 = %.2f.\"%(fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);\n", - "\n", - "print \"Prejump profile:\";\n", - "print \"high flood condition:\";\n", - "dist = [3 ,6, 8.4]; \t\t\t\t#dismath.tance\n", - "glacis = [252, 251, 250.32]; \t\t\t\t#R.L of glacis\n", - "D1 = [1.3 ,1.15, 1.03];\n", - "Ef1 = [0,0,0]\n", - "print \"Ef1 D1\";\n", - "for i in range(3):\n", - " Ef1[i] = 256.25-glacis[i];\n", - " print \"%.2f %.2f\"%(Ef1[i],D1[i]);\n", - "\n", - "print \"pond level flow:\";\n", - "dist = [3, 6, 9, 9.6]; \t\t\t\t#dismath.tance\n", - "glacis = [252, 251, 250, 249.9]; \t\t\t\t#R.Lof glacis\n", - "D1 = [0.31, 0.23, 0.16, 0.15];\n", - "Ef1 = [0,0,0,0]\n", - "print \"Ef1 D1\";\n", - "for i in range(4):\n", - " Ef1[i] = 254-glacis[i];\n", - " print \"%.2f %.2f\"%(Ef1[i],D1[i]);\n", - "\n", - "\n", - "\n", - "rho = 2.24;\n", - "Uf = 4; \t\t\t\t#unbalanced head for high flood condtion\n", - "Us = 2.56; \t\t\t\t#unbalanced static head\n", - "Hf = 2*Uf/3;\n", - "t = Hf/(rho-1);\n", - "t = round(t*10)/10;\n", - "print \"floor thickness at the point of formation of hydraulic jump = %.2f m.\"%(t);\n", - "Uf = 2.9; \t\t\t\t#unbalanced head for high flood condtion\n", - "Us = 2.2; \t\t\t\t#unbalanced static head\n", - "Hf = 2*Uf/3;\n", - "t = Us/(rho-1);\n", - "t = round(t*10)/10;\n", - "print \"floor thickness at the point of formation of hydraulic jump at the pond level condition = %.2f m.\"%(t);\n", - "P = 1.5; \t\t\t\t#pressure head at d/s end of floor\n", - "t = P/(rho-1);\n", - "t = round(t*10)/10;\n", - "print \"floor thickness at downstream side of sloping glacis = %.2f m.\"%(t);\n", - "D = rb-sh_up; \t\t\t\t#depth of u/s scour hole above bed level\n", - "a = 1.5*D;\n", - "a = round(a*10)/10;\n", - "print \"minimum length of upstream launching apron = %.2f m.\"%(a);\n", - "print \"provide 1.5 m thick apron for length of 5 m.\";\n", - "D = 249.6-241.5;\n", - "a = 1.5*D;\n", - "print \"minimum length of downstream launching apron = %.2f m.\"%(a);\n", - "print \"provide 1.5 m thick apron for length of 12 m.\";\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "crest level = 253.00 m.\n", - "elevation of u/s T.E.L = 256.26 m.\n", - "regime scour depth = 6.50 m.\n", - "afflux = 1.14. which is near to permissible\n", - "loss of head in at high flood = 1.64 m.\n", - "loss of head = 2.50 m.\n", - "Hydraulic jump calculation:\n", - "heigth of jump for high flood condition = 2.93 m.\n", - "length of concrete floor for high flood condition = 14.65 m.\n", - "heigth of jump for pond level condition = 1.53 m.\n", - "length of concrete floor for high pond level condition = 7.65 m.\n", - " upstream slope of glacis = 2:1.\n", - "downstream slope of glacis = 3:1.\n", - "horizontal length of floor beyond the toe = 15 m..\n", - "R.L of bottom of upstream sheet pile = 246.39 m.\n", - "R.L of downstream sheet pile = 241.50 m.\n", - "provide intermediate sheet pile at d/s toe of glacis.\n", - "length of impervious floor = 12.15 m.\n", - "length of floor already provide = 34.20 m.\n", - "which is more than required from permissible exit gradient.no upstream floor is required.\n", - "provide 1.80 m upstream floor so that total length becomes 36 m.\n", - "Pressure percent at points:\n", - "Points Before correction After correction\n", - "C1 72 78.80\n", - "D1 82 82.00\n", - "C2 31 41.40\n", - "E2 45 45.50\n", - "D2 58 55.20\n", - "D3 29 29.00\n", - "E3 44 37.60\n", - "elevation of subsoil H.G above datum:\n", - "no flow condition:\n", - "fie1 = 4.40.;fid1 = 3.61.;fic1 = 3.47.fie2 = 2.43.;fid2 = 2.00.;fic2 = 1.82.fie3 = 1.50.;fid3 = 1.28.;fic3 = 0.00.\n", - "high flood condition:\n", - "fie1 = 1.63.;fid1 = 1.34.;fic1 = 1.28.fie2 = 0.90.;fid2 = 0.74.;fic2 = 0.67.fie3 = 0.55.;fid3 = 0.47.;fic3 = 0.00.\n", - "flow at pond level:\n", - "fie1 = 2.50.;fid1 = 2.05.;fic1 = 1.97.fie2 = 1.38.;fid2 = 1.14.;fic2 = 1.03.fie3 = 0.85.;fid3 = 0.73.;fic3 = 0.00.\n", - "Prejump profile:\n", - "high flood condition:\n", - "Ef1 D1\n", - "4.25 1.30\n", - "5.25 1.15\n", - "5.93 1.03\n", - "pond level flow:\n", - "Ef1 D1\n", - "2.00 0.31\n", - "3.00 0.23\n", - "4.00 0.16\n", - "4.10 0.15\n", - "floor thickness at the point of formation of hydraulic jump = 1.60 m.\n", - "floor thickness at the point of formation of hydraulic jump at the pond level condition = 1.80 m.\n", - "floor thickness at downstream side of sloping glacis = 1.20 m.\n", - "minimum length of upstream launching apron = 4.70 m.\n", - "provide 1.5 m thick apron for length of 5 m.\n", - "minimum length of downstream launching apron = 12.15 m.\n", - "provide 1.5 m thick apron for length of 12 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.5 pg : 631" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "b = 16; \t\t\t\t#total length of floor\n", - "d = 5; \t\t\t\t#depth of downstream pile\n", - "D = 4; \t\t\t\t#depth of upstream pile\n", - "H = 2.5; \t\t\t\t#head created by weir\n", - "\n", - "\t\t\t\t#pressure at E\n", - "alpha = b/d;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "fie = math.acos((lambda1-2)/lambda1)/math.pi;\n", - "C = 19*(D/b)**0.5*((d+D)/b);\n", - "fie = fie*100-C;\n", - "P = H*fie/100;\n", - "P = round(P*1000)/1000;\n", - "print \"Pressure at E = %.2f m.\"%(P);\n", - "\n", - "\t\t\t\t#pressure at C1\n", - "alpha = b/D;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "fie = math.acos((lambda1-2)/lambda1)/math.pi;\n", - "fic = 1-fie; \t\t\t\t#by principle reversibility of flow\n", - "C = 19*(d/b)**0.5*((d+D)/b);\n", - "fic = fic*100+C;\n", - "P = fic*H/100;\n", - "P = round(P*1000)/1000;\n", - "print \" Pressure at C = %.2f m.\"%(P);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Pressure at E = 1.22 m.\n", - " Pressure at C = 1.43 m.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.6 pg : 632" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\n", - "#Given\n", - "b = 13; \t\t\t\t#length of floor\n", - "d = 2; \t\t\t\t#depth of downstream wall\n", - "D = 1.5; \t\t\t\t#depth of upstream cut-off\n", - "rho = 2.24; \t\t\t\t#relative density\n", - "H = 1.5;\n", - "\n", - "#at junction of d/s cut-off with floor\n", - "alpha = b/d;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "fie = math.acos((lambda1-2)/lambda1)/math.pi;\n", - "C = 19*(D/b)**0.5*((d+D)/b);\n", - "fie = fie*100-C;\n", - "P = H*fie/100;\n", - "t = P/(rho-1);\n", - "t = round(t*10)/10;\n", - "print \"floor thickness at junction of d/s cut-off with floor = %.2f m.\"%(t);\n", - "\n", - "#at junction of u/s cut-off with floor\n", - "alpha = b/D;\n", - "lambda11 = (1+(1+alpha**2)**0.5)/2;\n", - "fie = math.acos((lambda11-2)/lambda11)/math.pi;\n", - "fic = 1-fie; \t\t\t\t#by principle reversibility of flow\n", - "C = 19*(D/b)**0.5*((d+D)/b);\n", - "fiec = fic*100+C;\n", - "P = fiec*H/100;\n", - "t = 0.3; \t\t\t\t#this the uplift will be counter balanced by downward weigth of impounded water\n", - "print \"floor thickness at junction of u/s cut-off with floor = %.2f m.\"%(t);\n", - "\n", - "#at mid-length\n", - "P = (1.08+0.489)/2; \t\t\t\t#assuming linear variation\n", - "t = P/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"floor thickness at mid-length = %.2f m.\"%(t);\n", - "\n", - "#exit gradient\n", - "G = H/(d*math.pi*(lambda1)**0.5);\n", - "G = round(G*1000)/1000;\n", - "#math.since G<0.18\n", - "print \" G = %.2f. <0.18./nfloor is safe against failure by piping.\"%(G);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "floor thickness at junction of d/s cut-off with floor = 0.40 m.\n", - "floor thickness at junction of u/s cut-off with floor = 0.30 m.\n", - "floor thickness at mid-length = 0.63 m.\n", - " G = 0.13. <0.18./nfloor is safe against failure by piping.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.7 pg : 634" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "B = 30; \t\t\t\t#stream width\n", - "D = 3; \t\t\t\t#stream depth\n", - "V = 1.25; \t\t\t\t#mean velocity\n", - "Cd = 0.95; \t\t\t\t#discharge coefficient\n", - "Q = B*D*V;\n", - "\n", - "# Calculations\n", - "C = 2*Cd*(2*9.81)**0.5/3;\n", - "x = 4-(Q/(C*B))**(2./3);\n", - "x = round(x*1000)/1000;\n", - "\n", - "# Results\n", - "print \"heigth of weir to be built = %.2f m.\"%(x);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "heigth of weir to be built = 2.79 m.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.8 pg : 635" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "b = 50.; \t\t\t\t#length of floor\n", - "d = 8.; \t\t\t\t#depth of downstream pile\n", - "D = 8.; \t\t\t\t#depth of upstream pile\n", - "H = 5.; \t\t\t\t#effective head \n", - "tu = 1.; \t\t\t\t#floor thickness at upstream\n", - "td = 2.; \t\t\t\t#floor thickness at downstream\n", - "\n", - "# Calculations and Results\n", - "\t\t\t\t#downstream cut-off\n", - "alpha = b/d;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "fie = math.acos((lambda1-2)/lambda1)/math.pi;\n", - "fid = math.acos((lambda1-1)/lambda1)/math.pi;\n", - "Ct = (fie-fid)*td/d;\n", - "C = 19*(D/b)**0.5*((d+D)/b);\n", - "fie = fie*100-C-Ct*100;\n", - "P = H*fie/100;\n", - "P = round(P*100)/100;\n", - "print \"Pressure at downstream cut-off = %.2f m.\"%(P);\n", - "\n", - "\t\t\t\t#upstream cut-off\n", - "fie = math.acos((lambda1-2)/lambda1)/math.pi;\n", - "fid = math.acos((lambda1-1)/lambda1)/math.pi;\n", - "fic1 = 1-fie;\n", - "fid1 = 1-fid;\n", - "Ct = (fic1-fid1)*td/d;\n", - "C = -19*(D/b)**0.5*((d+D)/b);\n", - "fic1 = fic1*100-C-Ct*100;\n", - "P = H*fic1/100;\n", - "P = round(P*100)/100;\n", - "print \"Pressure at upstream cut-off = %.2f m.\"%(P);\n", - "G = H/(d*math.pi*(lambda1)**0.5);\n", - "print \"Exit Gradient = %.2f.\"%(G);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Pressure at downstream cut-off = 1.49 m.\n", - "Pressure at upstream cut-off = 3.51 m.\n", - "Exit Gradient = 0.10.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.9 pg : 636" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 1000.; \t\t\t\t#discharge of river\n", - "L = 256.; \t\t\t\t#crest length of diversion\n", - "f = 1.1; \t\t\t\t#silt factor\n", - "seg = 1./6; \t\t\t\t#safe exit gradient\n", - "hfl = 103; \t\t\t\t#high flood level\n", - "cf = 100; \t\t\t\t#reduced level of downstream concrete floor\n", - "H = 2.4; \t\t\t\t#maximum static head of weir\n", - "b = 40; \t\t\t\t#length of concrete floor\n", - "\n", - "\n", - "# Calculations and Results\n", - "q = Q/L;\n", - "R = 1.35*(q**2/f)**(1./3);\n", - "rld = hfl-1.5*R;\n", - "d = cf-rld;\n", - "d = round(d*100)/100;\n", - "print \"depth of downstream cut-off = %.2f m.\"%(d);\n", - "\n", - "alpha = b/d;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "G = H/(d*math.pi*(lambda1)**0.5);\n", - "\t\t\t\t#math.since G<seg\n", - "print \" G = %.2f. <1/6./nfloor is safe against failure by piping.\"%(G);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "depth of downstream cut-off = 1.87 m.\n", - " G = 0.12. <1/6./nfloor is safe against failure by piping.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.10 pg : 636" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "b = 60; \t\t\t\t#length of floor\n", - "H = 6; \t\t\t\t#static head of weir\n", - "d = 6; \t\t\t\t#downstream depth of pile\n", - "n = 0.3; \t\t\t\t#porousity of soil particles\n", - "G = 2.7; \t\t\t\t#relative density of soil particles\n", - "\n", - "alpha = b/d;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "Ge = H/(d*math.pi*(lambda1)**0.5);\n", - "e = n/(1-n);\n", - "chg = (G-1)/(1+e);\n", - "f = chg/Ge;\n", - "f = round(f*100)/100;\n", - "print \"critical exit gradient = %.2f.\\\n", - "\\nfactor of safety of system = %.2f.\"%(chg,f);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "critical exit gradient = 1.19.\n", - "factor of safety of system = 8.79.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.11 pg : 637" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots,ceil\n", - "\n", - "#design a vertical drop weir on Bligh's theory\n", - "#test floor by Khosla's theory\n", - "\n", - "#Given\n", - "Q = 2800.; \t\t\t\t#maximum flood discharge\n", - "hfl = 285.; \t\t\t\t#H.F.L before construction\n", - "hw = 278.; \t\t\t\t#minimum water level\n", - "fsl = 284.; \t\t\t\t#F.S.L of canal\n", - "c = 12.; \t\t\t\t#coefficient of creep\n", - "flux = 1.; \t\t\t\t#allowable afflux\n", - "Ge = 1./6; \t\t\t\t#permissible exit gradient\n", - "rho = 2.24; \t\t\t\t#specific gravity of concrete\n", - "\n", - "#Hydraulic calculation\n", - "L = 4.75*Q**0.5;\n", - "q = Q/L;\n", - "q = round(q*10)/10;\n", - "print \"Hydraulic calculation:\";\n", - "print \"discharge per unit width of river = %.2f cumecs.\"%(q);\n", - "f = 1;\n", - "R = 1.35*(q**2/f)**(1./3);\n", - "R = round(R*100)/100;\n", - "print \"regime scour depth = %.2f m.\"%(R);\n", - "V = q/R; \t\t\t\t#regime velocity\n", - "vh = V**2/(2*9.81); \t\t\t\t#velocity head\n", - "l_down = hfl+vh;\n", - "l_up = l_down+flux;\n", - "hfl_up = l_up-vh;\n", - "hfl_down = hfl-0.5;\n", - "hfl_down = round(hfl_down*100)/100;\n", - "print \"actual d/s H.F.L allowing 0.5 m for retrogation = %.2f m.\"%(hfl_down);\n", - "K = (q/1.7)**(2./3);\n", - "cl = l_up-K; \t\t\t\t#crest level\n", - "cl = round(cl*100)/100;\n", - "print \"crest level = %.2f m.\"%(cl);\n", - "pl = fsl+0.5; \t\t\t\t#pond level\n", - "s = hfl_down-cl; \t\t\t\t#heigth of shutter\n", - "print \"heigth of shutter = %.2f m.\"%(s);\n", - "rl_up_pile = hfl_up-1.5*R; \t\t\t\t#R.L of bottom u/s pile\n", - "d_up_cut = hw-276; \t\t\t\t#depth of upstream cut-off\n", - "print \"depth of upstream cut-off = %.2f m.\"%(d_up_cut);\n", - "print \" provide concrete cut off 2 m depth.\";\n", - "rl_bot_ds = hfl_down-2*R;\n", - "Hs = hfl_down-hw; \t\t\t\t#seepage head\n", - "Hc = cl-hw; \t\t\t\t#heigth of crest\n", - "print \"R.L of gates crest = %.2f m.\"%(Hs);\n", - "print \"Heigth of crest = %.2f m.\"%(Hc);\n", - "\n", - "#design of weir wall\n", - "d = hfl_up-cl;\n", - "a = d/(rho)**0.5;\n", - "a = 3*d/(2*rho); \t\t\t\t#from sliding consideration\n", - "a = s+1; \t\t\t\t#from practical consieration\n", - "a = a+1;\n", - "print \"design of weir wall:\"\n", - "print \"provide top width of %i m.\"%(a);\n", - "Mo = 9.81*Hs**3/6; \t\t\t\t#overtirning moment\n", - "#equating the moment of resismath.tance to overturning moment and putting the values we get\n", - "y = [0.039,0.020,-1.084]\n", - "b = int(roots(y)[1]);\n", - "#we get b = - 5.5347261 and 5.0219056\n", - "#taking\n", - "#b = 5;\n", - "#when weir is submerged\n", - "C = 0.58;\n", - "d = (q**2/((2*C/3)**2*2*9.81))**(1./3);\n", - "Mo = 9.81*d*Hc**2/2;\n", - "#from equation of moment of resistence we get\n", - "y = [1,3,-77.55]\n", - "b = ceil(roots(y)[1]);\n", - "#we get b = - 10.433085 and 7.4330846\n", - "#taking\n", - "#b = 8;\n", - "print \"bottom width = %i m.\"%(b);\n", - "\n", - "#design of impervious and pervious aprons\n", - "C = 12;\n", - "L = C*Hs;\n", - "print \"design of impervious and pervious aprons:\";\n", - "print \"total creep length = %i m.\"%(L);\n", - "l1 = 2.21*C*(Hs/13)**0.5;\n", - "l1_ = l1+1;\n", - "print \"length of downstream impervious apron = %i m.\"%(l1_);\n", - "d1 = hw-276;\n", - "d2 = hw-271;\n", - "l2 = L-l1-(b+2*d1+2*d2);\n", - "print \"length of upstream impervious apron = %i m.\"%(l2);\n", - "l3 = 18*C*(Hs*q/975)**0.5;\n", - "print \"total length of d/s apron = %i m.\"%(l3); \t\t\t\t#calculation is wrong in book\n", - "l = l3-l1;\n", - "le = l/2;\n", - "le = round(le*100)/100;\n", - "print 'provide filter of length %.2f m. and launching apron of length %.2f m.'%(le,le);\n", - "t = d2*10**0.5/le;\n", - "print \"thickness of launching apron in horizontal position = %.2f m.\"%(t);\n", - "print \"provide launching apron of thickness 1.5 m.\";\n", - "T = 2*d1;\n", - "V = d1*10**0.5;\n", - "ta = V/T;\n", - "ta = round(ta*10)/10;\n", - "print \"thickness of apron in horizontal position = %.2f m.\"%(ta);\n", - "Hr = Hs-Hs*(4+33+8)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print 'provide thickness of %.2f m from d/s of weir wall to point 6 m from it.'%(t);\n", - "Hr = Hs-Hs*(4+33+8+6)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print \"provide thickness of %.2f m from 6 m to 12 m from d/s end of weir wall.\"%(t);\n", - "Hr = Hs-Hs*(4+33+8+12)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print \"provide thickness of %.2f m for rest of length of weir floor.\"%(t);\n", - "\n", - "#check by khosla's theory\n", - "b = 33+8+19; \t\t\t\t#total horizontal length of impervious floor\n", - "d = 7; \t\t\t\t#depth of downstream pile\n", - "alpha = b/d;\n", - "n = 0.14; \t\t\t\t#n = 1/math.pi*(lambda)**0.5;\n", - "Ge = Hs*n/d;\n", - "print \"check by Khosla theory:\";\n", - "print \"exit gradient = %.2f. < 1/6 hence safe\"%(Ge);\n", - "alpha_ = d/b;\n", - "fic1 = 0.83;fid1 = 0.88;\n", - "corec_c1 = (fid1-fic1)*100/2;\n", - "bdash = b;\n", - "d = 2.;D = 7;\n", - "C1 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "fic1 = fic1*100+corec_c1+C1;\n", - "Pc = Hs*fic1/100; \t\t\t\t#pressure head at C\n", - "alpha_ = d/b;\n", - "fie2 = 0.31;fid2 = 0.21;\n", - "corec_e1 = (fie2-fid2)*1.7*100/7;\n", - "bdash = b;\n", - "d = 7;D = 2;\n", - "C1 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "fie2 = fie2*100-corec_e1-C1; \t\t\t\t#in book 3.53 value is wrong\n", - "Pe = Hs*fie2/100; \t\t\t\t#pressue head at E\n", - "#assuming linear variation of pressure for intermediate points\n", - "Pa = Pc-(Pc-Pe)*(33+8)/b;\n", - "t = Pa/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at d/s of weir wall = %.2f m.\"%(Pa);\n", - "print \"thickness at d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe.\"%(t);\n", - "Pb = Pc-(Pc-Pe)*(33+8+6)/b;\n", - "t = Pb/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at 6 m from d/s of weir wall = %.2f m.\"%(Pb);\n", - "print \"thickness at 6m from d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe.\"%(t);\n", - "Pc = Pc-(Pc-Pe)*(33+8+12)/b;\n", - "t = Pc/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at 12 m from d/s of weir wall = %.2f m.\"%(Pc);\n", - "print \"thickness at 12m from d/s of weir wall = %.2f m. > thickness by Bligh theory;hence unsafe.\"%(t);\n", - "print \"hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Hydraulic calculation:\n", - "discharge per unit width of river = 11.10 cumecs.\n", - "regime scour depth = 6.72 m.\n", - "actual d/s H.F.L allowing 0.5 m for retrogation = 284.50 m.\n", - "crest level = 282.65 m.\n", - "heigth of shutter = 1.85 m.\n", - "depth of upstream cut-off = 2.00 m.\n", - " provide concrete cut off 2 m depth.\n", - "R.L of gates crest = 6.50 m.\n", - "Heigth of crest = 4.65 m.\n", - "design of weir wall:\n", - "provide top width of 3 m.\n", - "bottom width = 8 m.\n", - "design of impervious and pervious aprons:\n", - "total creep length = 78 m.\n", - "length of downstream impervious apron = 19 m.\n", - "length of upstream impervious apron = 33 m.\n", - "total length of d/s apron = 58 m.\n", - "provide filter of length 20.00 m. and launching apron of length 20.00 m.\n", - "thickness of launching apron in horizontal position = 1.11 m.\n", - "provide launching apron of thickness 1.5 m.\n", - "thickness of apron in horizontal position = 1.60 m.\n", - "provide thickness of 3.00 m from d/s of weir wall to point 6 m from it.\n", - "provide thickness of 2.40 m from 6 m to 12 m from d/s end of weir wall.\n", - "provide thickness of 1.90 m for rest of length of weir floor.\n", - "check by Khosla theory:\n", - "exit gradient = 0.13. < 1/6 hence safe\n", - "pressure at d/s of weir wall = 3.03 m.\n", - "thickness at d/s of weir wall = 2.44 m. < thickness by Bligh theory;hence safe.\n", - "pressure at 6 m from d/s of weir wall = 2.66 m.\n", - "thickness at 6m from d/s of weir wall = 2.14 m. < thickness by Bligh theory;hence safe.\n", - "pressure at 12 m from d/s of weir wall = 2.29 m.\n", - "thickness at 12m from d/s of weir wall = 1.85 m. > thickness by Bligh theory;hence unsafe.\n", - "hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.12 pg : 637" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#number of gates required for the barrage\n", - "#length and R.L of bamath.sin floor if silting bamath.sin is provided downstream of barrage\n", - "\n", - "#Given\n", - "Lmax = 212; \t\t\t\t#maximum reservior level\n", - "Lp = 211; \t\t\t\t#pond level\n", - "hfl = 210; \t\t\t\t#downstream high flood level in the river\n", - "Qmax = 3500; \t\t\t\t#maximum design flood discharge\n", - "Lcrest = 207; \t\t\t\t#crest level of the barrage\n", - "Lcrest_r = 208; \t\t\t\t#crest level of head regulator\n", - "Cd = 2.1; \t\t\t\t#coefficient of discharge for barrage\n", - "Cd_r = 1.5; \t\t\t\t#coefficient of discharge for head regulator\n", - "rbl = 205; \t\t\t\t#river bed level\n", - "Q = 500; \t\t\t\t#design discharge of main canal\n", - "\n", - "#design of water way for barrage during flood\n", - "H = Lmax-Lcrest;\n", - "L = Qmax/(Cd*H**1.5);\n", - "#which gives L = 149.07.\n", - "print \"nunmber of gates for the barrage = 15.\";\n", - "\n", - "#design of waterway for canal head regulator\n", - "H = Lp-Lcrest_r;\n", - "L1 = Q/(Cd_r*H**1.5);\n", - "#which gives L = 64.2\n", - "#hence provide 7 bays of 10 m each\n", - "print \"nunmber of gates for the head regulator = 7.\";\n", - "\n", - "#design of stilling bamath.sin\n", - "Hl = Lmax-hfl;\n", - "q = Qmax/L;\n", - "yc = (q**2/9.81)**(1./3);\n", - "Z = Hl/yc;\n", - "#math.since Z<1\n", - "Y = 1+0.93556*Z**0.368;\n", - "y2 = Y*yc;\n", - "Lc = 5*y2;\n", - "Lc = round(Lc*10)/10;\n", - "print \"Length of cistern = %.2f m.\"%(Lc);\n", - "Ef2 = yc*(Y+1/(2*Y**2));\n", - "j = hfl-Ef2;\n", - "j = round(j*10)/10;\n", - "print \"R.L of cistern = %.2f m.\"%(j);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "nunmber of gates for the barrage = 15.\n", - "nunmber of gates for the head regulator = 7.\n", - "Length of cistern = 33.30 m.\n", - "R.L of cistern = 202.70 m.\n" - ] - } - ], - "prompt_number": 15 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch12_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch12_1.ipynb deleted file mode 100755 index edc47614..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch12_1.ipynb +++ /dev/null @@ -1,1384 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:b0cdc42f1fc8747c583aaf55d3f5af5550eb632d3c87d035821d8a7d148eae09" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 12 : DIVERSION HEADWORKS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.1 pg : 576" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#uplift presuures and thickness of floor at 6m, 12m and 18m from u/s\n", - "\n", - "#Given\n", - "rho = 2.24; \t\t\t\t#relative density of material\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "L = 22.; \t\t\t\t#total length\n", - "lc = (2.*6)+L+(2*8); \t\t\t\t#length of creep\n", - "hg = 4./lc; \t\t\t\t#hydraulic gradient\n", - "print \"avearge hydraulic gradient = %.2f.\"%(hg);\n", - "#at 6 m from u/s\n", - "x = 6.;\n", - "lg = (6.*2)+x;\n", - "h1 = 4.*(1-lg/50); \t\t\t\t#unbalanced head\n", - "up = gamma_w*h1;\n", - "t = 4.*h1/(3*(rho-1));\n", - "up = round(up*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"uplift at 6 m from u/s = %.2f kN/square metre.\"%(up);\n", - "print \"thickness at 6 m from u/s = %.2f m.\"%(t);\n", - "\n", - "#at 12 m from u/s\n", - "x = 12.;\n", - "lg = (6.*2)+x;\n", - "h1 = 4.*(1-lg/50); \t\t\t\t#unbalanced head\n", - "up = gamma_w*h1;\n", - "t = 4.*h1/(3*(rho-1));\n", - "up = round(up*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"uplift at 12 m from u/s = %.2f kN/square metre.\"%(up);\n", - "print \"thickness at 12 m from u/s = %.2f m.\"%(t);\n", - "\n", - "#at 18m from u/s\n", - "x = 18.;\n", - "lg = (6.*2)+x;\n", - "h1 = 4.*(1-lg/50); \t\t\t\t#unbalanced head\n", - "up = gamma_w*h1;\n", - "t = 4*h1/(3*(rho-1));\n", - "up = round(up*10)/10;\n", - "t = round(t*100)/100;\n", - "print \"uplift at 18 m from u/s = %.2f kN/square metre.\"%(up);\n", - "print \"thickness at 18 m from u/s = %.2f m.\"%(t);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "avearge hydraulic gradient = 0.08.\n", - "uplift at 6 m from u/s = 25.11 kN/square metre.\n", - "thickness at 6 m from u/s = 2.75 m.\n", - "uplift at 12 m from u/s = 20.40 kN/square metre.\n", - "thickness at 12 m from u/s = 2.24 m.\n", - "uplift at 18 m from u/s = 15.70 kN/square metre.\n", - "thickness at 18 m from u/s = 1.72 m.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.2 pg : 589" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#check whether section is safe against overturning and piping\n", - "\n", - "#Given\n", - "b = 54.; \t\t\t\t#width of section\n", - "D1D2 = 16; \t\t\t\t#dismath.tance between points D1 and D2\n", - "D2D3 = 37; \t\t\t\t#dismath.tance between points D2 and D3\n", - "\n", - "#first pipe line\n", - "#taking data from figure\n", - "d = 105-97;\n", - "b1 = 0.5;\n", - "alpha = b/d;\n", - "#from the curves we get\n", - "fic1 = 0.665;\n", - "fid1 = 0.76;\n", - "fie1 = 1;\n", - "t = 105-104; \t\t\t\t#floor thickness\n", - "corec = (fid1-fic1)*100*t/d; \t\t\t\t#correction for floor thickness\n", - "#for pile no. 2\n", - "D = 104-97;\n", - "d = 104-97;\n", - "bdash = 16;\n", - "C = 19*(D/bdash)**0.5*(d+D)/b; \t\t\t\t#correction for pile no. 2\n", - "fic1 = fic1*100+corec+C; \t\t\t\t#corrected pressures\n", - "\n", - "#intermedite pipe line\n", - "d = 105-97;\n", - "b1 = 16.5;\n", - "alpha = b/d;\n", - "r = b1/b; \t\t\t\t#ratio b1/b\n", - "#from the curves we get\n", - "fic2 = 0.52;\n", - "fie2 = 0.725;\n", - "fid2 = 0.615;\n", - "corec_c1 = (fid2-fic2)*100*t/d;\n", - "corec_e1 = (fie2-fid2)*100/d;\n", - "\n", - "#for pile no. 1\n", - "C1 = C;\n", - "d = 104-97;\n", - "bdash = 37;\n", - "D = 104-95;\n", - "C2 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "#correction due to slope\n", - "corec_e2 = 3.3; \t\t\t\t#from table 12.4\n", - "#correction is negative due to upwrd slope\n", - "l = 4; \t\t\t\t#horizontal length of slope\n", - "corec_c2 = corec_e2*l/bdash;\n", - "\n", - "fie2 = fie2*100-corec_e1-corec_e2;\n", - "fic2 = fic2*100+corec_c1+C2-corec_c2;\n", - "\n", - "#pile no. 3 at d/s end\n", - "d = 103.5-95;\n", - "alpha_ = d/b;\n", - "#for curves\n", - "fie3 = 0.35;fid3 = 0.242;\n", - "corec_t = (fie3-fid3)*100*(103.5-102)/d;\n", - "\n", - "#correction for interference at pile no. 2\n", - "d = 102-95;\n", - "D = 102-97;\n", - "C3 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "fie3 = fie3*100-corec_t-C3;\n", - "\n", - "point = ['C1', 'C2' ,'E2' ,'E3']; \t\t\t\t#Point\n", - "P = [fic1 ,fic2 ,fie2 ,fie3]; \t\t\t\t#pressure percent\n", - "P_ = [3.55 ,2.78, 3.39, 1.58]; \t\t\t\t#pressure head\n", - "print \"Points Pressure percent Pressure head\";\n", - "\n", - "for i in range(4):\n", - " P[i] = round(P[i]*10)/10;\n", - " print \"%s %.2f %.2f\"%(point[i],P[i],P_[i]);\n", - "\n", - "\n", - "#check for floor thickness\n", - "Pa = P_[1]-((P_[1]-P_[3])*6.5/37);\n", - "Pb = P_[1]-((P_[1]-P_[3])*24/37);\n", - "Pc = P_[1]-((P_[1]-P_[3])*30/37);\n", - "rho = 2.24; \t\t\t\t#specific gravity of concrete\n", - "ta = Pa/(rho-1);\n", - "tb = Pb/(rho-1);\n", - "tc = Pc/(rho-1);\n", - "ta = round(ta*100)/100;\n", - "tb = round(tb*100)/100;\n", - "tc = round(tc*100)/100;\n", - "print \"Thickness required at A = %.2f m.\"%(ta);\n", - "print \"Thickness required at B = %.2f m.\"%(tb);\n", - "print \"Thickness required at C = %.2f m.\"%(tc);\n", - "t = 103.5-102;\n", - "print \"Thickness provided = %.2f m.\"%(t);\n", - "print \"Floor thickness at B and C are adequate\";\n", - "\n", - "#exit gradient\n", - "H = 108.5-103.5; \t\t\t\t#seepage head\n", - "d = 103.5-95; \t\t\t\t#depth cut-off\n", - "#from exit gradient curve\n", - "alpha = 6.35;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "Ge = H/(d*math.pi*lambda1**0.5);\n", - "print \"exit gradient = %.2f.\"%(Ge);\n", - "print \" it is less than permissible exit gradient < 1/6Hence safe..\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Points Pressure percent Pressure head\n", - "C1 67.70 3.55\n", - "C2 52.80 2.78\n", - "E2 67.80 3.39\n", - "E3 33.10 1.58\n", - "Thickness required at A = 2.07 m.\n", - "Thickness required at B = 1.61 m.\n", - "Thickness required at C = 1.46 m.\n", - "Thickness provided = 1.50 m.\n", - "Floor thickness at B and C are adequate\n", - "exit gradient = 0.10.\n", - " it is less than permissible exit gradient < 1/6Hence safe..\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.3 pg : 605" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots,ceil\n", - "\n", - "#design a vertical drop weir on Bligh's theory\n", - "#test floor by Khosla's theory\n", - "\n", - "#Given\n", - "Q = 2800; \t\t\t\t#maximum flood discharge\n", - "hfl = 285; \t\t\t\t#H.F.L before construction\n", - "hw = 278; \t\t\t\t#minimum water level\n", - "fsl = 284; \t\t\t\t#F.S.L of canal\n", - "c = 12; \t\t\t\t#coefficient of creep\n", - "flux = 1; \t\t\t\t#allowable afflux\n", - "Ge = 1./6; \t\t\t\t#permissible exit gradient\n", - "rho = 2.24; \t\t\t\t#specific gravity of concrete\n", - "\n", - "#Hydraulic calculation\n", - "L = 4.75*Q**0.5;\n", - "q = Q/L;\n", - "q = round(q*10)/10;\n", - "print \"Hydraulic calculation:\";\n", - "print \"discharge per unit width of river = %.2f cumecs.\"%(q);\n", - "f = 1;\n", - "R = 1.35*(q**2/f)**(1./3);\n", - "R = round(R*100)/100;\n", - "print \"regime scour depth = %.2f m.\"%(R);\n", - "V = q/R; \t\t\t\t#regime velocity\n", - "vh = V**2/(2*9.81); \t\t\t\t#velocity head\n", - "l_down = hfl+vh;\n", - "l_up = l_down+flux;\n", - "hfl_up = l_up-vh;\n", - "hfl_down = hfl-0.5;\n", - "hfl_down = round(hfl_down*100)/100;\n", - "print \"actual d/s H.F.L allowing 0.5 m for retrogation = %.2f m.\"%(hfl_down);\n", - "K = (q/1.7)**(2./3);\n", - "cl = l_up-K; \t\t\t\t#crest level\n", - "cl = round(cl*100)/100;\n", - "print \"crest level = %.2f m.\"%(cl);\n", - "pl = fsl+0.5; \t\t\t\t#pond level\n", - "s = hfl_down-cl; \t\t\t\t#heigth of shutter\n", - "print \"heigth of shutter = %.2f m.\"%(s);\n", - "rl_up_pile = hfl_up-1.5*R; \t\t\t\t#R.L of bottom u/s pile\n", - "d_up_cut = hw-276; \t\t\t\t#depth of upstream cut-off\n", - "print \"depth of upstream cut-off = %.2f m.\"%(d_up_cut);\n", - "print \" provide concrete cut off 2 m depth.\";\n", - "rl_bot_ds = hfl_down-2*R;\n", - "Hs = hfl_down-hw; \t\t\t\t#seepage head\n", - "Hc = cl-hw; \t\t\t\t#heigth of crest\n", - "print \"R.L of gates crest = %.2f m.\"%(Hs);\n", - "print \"Heigth of crest = %.2f m.\"%(Hc);\n", - "\n", - "#design of weir wall\n", - "d = hfl_up-cl;\n", - "a = d/(rho)**0.5;\n", - "a = 3*d/(2*rho); \t\t\t\t#from sliding consideration\n", - "a = s+1; \t\t\t\t#from practical consieration\n", - "a = a+1;\n", - "print \"design of weir wall:\"\n", - "print \"provide top width of %i m.\"%(a);\n", - "Mo = 9.81*Hs**3/6; \t\t\t\t#overtirning moment\n", - "#equating the moment of resismath.tance to overturning moment and putting the values we get\n", - "#y = poly([-1.084,0.020,0.039],'x','c');\n", - "y = [0.039,0.020,-1.084]\n", - "b = roots(y)[1];\n", - "#we get b = - 5.5347261 and 5.0219056\n", - "#taking\n", - "b = 5;\n", - "#when weir is submerged\n", - "C = 0.58;\n", - "d = (q**2/((2*C/3)**2*2*9.81))**(1./3);\n", - "Mo = 9.81*d*Hc**2/2;\n", - "#from equation of moment of resistence we get\n", - "y = [1,3,-77.55]\n", - "b = ceil(roots(y)[1]); #we get b = - 10.433085 and 7.4330846\n", - "print \"bottom width = %i m.\"%(b);\n", - "\n", - "#design of impervious and pervious aprons\n", - "C = 12;\n", - "L = C*Hs;\n", - "print \"design of impervious and pervious aprons:\";\n", - "print \"total creep length = %i m.\"%(L);\n", - "l1 = 2.21*C*(Hs/13)**0.5;\n", - "l1_ = l1+1;\n", - "print \"length of downstream impervious apron = %i m.\"%(l1_);\n", - "d1 = hw-276;\n", - "d2 = hw-271;\n", - "l2 = L-l1-(b+2*d1+2*d2);\n", - "print \"length of upstream impervious apron = %i m.\"%(l2);\n", - "l3 = 18*C*(Hs*q/975)**0.5;\n", - "print \"total length of d/s apron = %i m.\"%(l3); \t\t\t\t#calculation is wrong in book\n", - "l = l3-l1;\n", - "le = l/2;\n", - "le = round(le*100)/100;\n", - "print 'provide filter of length %.2f m. and launching apron of length %.2f m.'%(le,le);\n", - "t = d2*10**0.5/le;\n", - "print \"thickness of launching apron in horizontal position = %.2f m.\"%(t);\n", - "print \"provide launching apron of thickness 1.5 m.\";\n", - "T = 2*d1;\n", - "V = d1*10**0.5;\n", - "ta = V/T;\n", - "ta = round(ta*10)/10;\n", - "print \"thickness of apron in horizontal position = %.2f m.\"%(ta);\n", - "Hr = Hs-Hs*(4+33+8)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print 'provide thickness of %.2f m from d/s of weir wall to point 6 m from it.'%(t);\n", - "Hr = Hs-Hs*(4+33+8+6)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print \"provide thickness of %.2f m from 6 m to 12 m from d/s end of weir wall.\"%(t);\n", - "Hr = Hs-Hs*(4+33+8+12)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print \"provide thickness of %.2f m for rest of length of weir floor.\"%(t);\n", - "\n", - "#check by khosla's theory\n", - "b = 33+8+19; \t\t\t\t#total horizontal length of impervious floor\n", - "d = 7; \t\t\t\t#depth of downstream pile\n", - "alpha = b/d;\n", - "n = 0.14; \t\t\t\t#n = 1/math.pi*(lambda)**0.5;\n", - "Ge = Hs*n/d;\n", - "print \"check by Khosla theory:\";\n", - "print \"exit gradient = %.2f. < 1/6 hence safe\"%(Ge);\n", - "alpha_ = d/b;\n", - "fic1 = 0.83;fid1 = 0.88;\n", - "corec_c1 = (fid1-fic1)*100/2;\n", - "bdash = b;\n", - "d = 2;D = 7;\n", - "C1 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "fic1 = fic1*100+corec_c1+C1;\n", - "Pc = Hs*fic1/100; \t\t\t\t#pressure head at C\n", - "alpha_ = d/b;\n", - "fie2 = 0.31;fid2 = 0.21;\n", - "corec_e1 = (fie2-fid2)*1.7*100/7;\n", - "bdash = b;\n", - "d = 7;D = 2;\n", - "C1 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "fie2 = fie2*100-corec_e1-C1; \t\t\t\t#in book 3.53 value is wrong\n", - "Pe = Hs*fie2/100; \t\t\t\t#pressue head at E\n", - "#assuming linear variation of pressure for intermediate points\n", - "Pa = Pc-(Pc-Pe)*(33+8)/b;\n", - "t = Pa/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at d/s of weir wall = %.2f m.\"%(Pa);\n", - "print \"thickness at d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe.\"%(t);\n", - "Pb = Pc-(Pc-Pe)*(33+8+6)/b;\n", - "t = Pb/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at 6 m from d/s of weir wall = %.2f m.\"%(Pb);\n", - "print \"thickness at 6m from d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe.\"%(t);\n", - "Pc = Pc-(Pc-Pe)*(33+8+12)/b;\n", - "t = Pc/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at 12 m from d/s of weir wall = %.2f m.\"%(Pc);\n", - "print \"thickness at 12m from d/s of weir wall = %.2f m. > thickness by Bligh theory;hence unsafe.\"%(t);\n", - "print \"hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Hydraulic calculation:\n", - "discharge per unit width of river = 11.10 cumecs.\n", - "regime scour depth = 6.72 m.\n", - "actual d/s H.F.L allowing 0.5 m for retrogation = 284.50 m.\n", - "crest level = 282.65 m.\n", - "heigth of shutter = 1.85 m.\n", - "depth of upstream cut-off = 2.00 m.\n", - " provide concrete cut off 2 m depth.\n", - "R.L of gates crest = 6.50 m.\n", - "Heigth of crest = 4.65 m.\n", - "design of weir wall:\n", - "provide top width of 3 m.\n", - "bottom width = 8 m.\n", - "design of impervious and pervious aprons:\n", - "total creep length = 78 m.\n", - "length of downstream impervious apron = 19 m.\n", - "length of upstream impervious apron = 33 m.\n", - "total length of d/s apron = 58 m.\n", - "provide filter of length 20.00 m. and launching apron of length 20.00 m.\n", - "thickness of launching apron in horizontal position = 1.11 m.\n", - "provide launching apron of thickness 1.5 m.\n", - "thickness of apron in horizontal position = 1.60 m.\n", - "provide thickness of 3.00 m from d/s of weir wall to point 6 m from it.\n", - "provide thickness of 2.40 m from 6 m to 12 m from d/s end of weir wall.\n", - "provide thickness of 1.90 m for rest of length of weir floor.\n", - "check by Khosla theory:\n", - "exit gradient = 0.13. < 1/6 hence safe\n", - "pressure at d/s of weir wall = 3.03 m.\n", - "thickness at d/s of weir wall = 2.44 m. < thickness by Bligh theory;hence safe.\n", - "pressure at 6 m from d/s of weir wall = 2.66 m." - ] - }, - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "thickness at 6m from d/s of weir wall = 2.14 m. < thickness by Bligh theory;hence safe.\n", - "pressure at 12 m from d/s of weir wall = 2.29 m.\n", - "thickness at 12m from d/s of weir wall = 1.85 m. > thickness by Bligh theory;hence unsafe.\n", - "hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.4 pg : 617" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t#design a slopeing glacis\n", - "\t\t\t\t\n", - "#Given\n", - "q = 10; \t\t\t\t#maximum discharge intensity on weir crest\n", - "hfl = 255; \t\t\t\t#H.F.L before construction of weir\n", - "rb = 249.5; \t\t\t\t#R.L of river bed\n", - "pl = 254; \t\t\t\t#pond level\n", - "s = 1; \t\t\t\t#heigth of crest shutter\n", - "dhw = 251.5; \t\t\t\t#anticipated downstream water level in river when water is dischrging with pond level upstream\n", - "br = 0.5; \t\t\t\t#bed retrogression\n", - "f = 0.9; \t\t\t\t#Laecey silt factor\n", - "Ge = 1./7; \t\t\t\t#permissible exit gradient\n", - "flux = 1; \t\t\t\t#permissible afflux\n", - "\n", - "cl = pl-s; \t\t\t\t#crest level\n", - "print \"crest level = %.2f m.\"%(cl);\n", - "K = (q/1.7)**(2./3);\n", - "tel_up = cl+K;\n", - "tel_up = round(tel_up*100)/100;\n", - "print \"elevation of u/s T.E.L = %.2f m.\"%(tel_up);\n", - "R = 1.35*(q**2/f)**(1./3);\n", - "R = round(R*10)/10;\n", - "print \"regime scour depth = %.2f m.\"%(R);\n", - "V = q/R; \t\t\t\t#regime velocity\n", - "vh = V**2/(2*9.81); \t\t\t\t#velocity head\n", - "hfl_up = tel_up-vh;\n", - "tel_down = hfl+vh;\n", - "flux = hfl_up-hfl;\n", - "flux = round(flux*100)/100;\n", - "print \"afflux = %.2f. which is near to permissible\"%(flux);\n", - "hfl_down = hfl-br; \t\t\t\t#downstream H.F.L after retrogression\n", - "tel_down = tel_down-br; \t\t\t\t#downstream T.F.L after retrogression\n", - "Hl = tel_up-tel_down; \t\t\t\t#loss of head in flood\n", - "Hl = round(Hl*100)/100;\n", - "print \"loss of head in at high flood = %.2f m.\"%(Hl);\n", - "K = pl-cl; \t\t\t\t#head over crest\n", - "q_ = 1.7*(K)**1.5;\n", - "Hl_ = pl-dhw; \t\t\t\t#loss of head\n", - "print \"loss of head = %.2f m.\"%(Hl_);\n", - "Ef2 = 4.3;\n", - "Ef2_ = 1.7; \t\t\t\t#from Blench curve\n", - "jump = tel_down-Ef2;\n", - "jump_ = 251.5-Ef2_; \t\t\t\t#level at which jump will form\n", - "Ef1 = Ef2+Hl;\n", - "Ef1_ = Ef2_+Hl_;\n", - "D1 = 1.03;\n", - "D1_ = 0.15; \n", - "D2 = 3.96;D2_ = 1.68; \n", - "hj = D2-D1;\n", - "hj_ = D2_-D1_; \t\t\t\t#heigth of jump\n", - "concrete = 5*hj;\n", - "concrete_ = 5*hj_; \t\t\t\t#length of concrete floor\n", - "print \"Hydraulic jump calculation:\";\n", - "print \"heigth of jump for high flood condition = %.2f m.\"%(hj);\n", - "print \"length of concrete floor for high flood condition = %.2f m.\"%(concrete);\n", - "print \"heigth of jump for pond level condition = %.2f m.\"%(hj_);\n", - "print \"length of concrete floor for high pond level condition = %.2f m.\"%(concrete_);\n", - "\n", - "cw = 2; \t\t\t\t#crets width\n", - "us = 2; \t\t\t\t#upstream slope\n", - "ds = 3; \t\t\t\t#downstream slope\n", - "l = 15;\n", - "print \" upstream slope of glacis = %i:1.\"%(us);\n", - "print \"downstream slope of glacis = %i:1.\"%(ds);\n", - "print \"horizontal length of floor beyond the toe = %i m..\"%(l);\n", - "\n", - "R = 6.5;\n", - "sh_up = hfl_up-1.5*R;\n", - "sh_down = hfl_down-2*R;\n", - "sh_up = round(sh_up*100)/100;\n", - "print \"R.L of bottom of upstream sheet pile = %.2f m.\"%(sh_up);\n", - "print \"R.L of downstream sheet pile = %.2f m.\"%(sh_down);\n", - "print \"provide intermediate sheet pile at d/s toe of glacis.\";\n", - "Hs = pl-249.6; \t\t\t\t#maximum percolation head\n", - "d = 249.6-sh_down; \t\t\t\t#depth of d/s cut-off\n", - "n = Ge*d/Hs; \t\t\t\t#n = 1/(math.pi*lambda**0.5);\n", - "\t\t\t\t#from khosla exit gradient curve\n", - "alpha = 1.5;\n", - "b = alpha*d;\n", - "print \"length of impervious floor = %.2f m.\"%(b);\n", - "fl = (2*(253-249.5))+2+(3*(253-249.6))+15;\n", - "us = 36-fl;\n", - "print \"length of floor already provide = %.2f m.\"%(fl);\n", - "print \"which is more than required from permissible exit gradient.no upstream floor is required.\";\n", - "print \"provide %.2f m upstream floor so that total length becomes 36 m.\"%(us);\n", - "alpha_1 = 0.089; \n", - "alpha_2 = 0.225; \t\t\t\t#alpha_ = 1/alpha\n", - "b1 = 21;\n", - "alpha = 4.44;\n", - "print \"Pressure percent at points:\";\n", - "point = ['C1', 'D1' ,'C2' ,'E2' ,'D2' ,'D3' ,'E3'];\n", - "bc = [72 ,82 ,31.5 ,45.5 ,58.5 ,29 ,44];\n", - "crt = [3.1, 0, 3.5, 0, -3.2, 0, 0, -3.6];\n", - "crs = [0 ,0, 0, 0, 2.3, 0, 0, 0];\n", - "cri = [3.7, 0, 6.4, 0, -2.4, 0, -6.4];\n", - "after = [0,0,0,0,0,0,0]\n", - "print \"Points Before correction After correction\";\n", - "for i in range(7):\n", - " after[i] = bc[i]+crt[i]+crs[i]+cri[i];\n", - " print \"%s %i %.2f\"%(point[i],bc[i],after[i]);\n", - "\n", - "Hs = 254-249.6; \t\t\t\t#no flow condition\n", - "Hs_ = 256.13-254.5; \t\t\t\t#high flood condition\n", - "Hs__ = 254-251.5; \t\t\t\t#flow at pond level\n", - "print \"elevation of subsoil H.G above datum:\";\n", - "print \"no flow condition:\";\n", - "fie1 = 1*Hs;\n", - "fid1 = 0.82*Hs;\n", - "fic1 = 0.788*Hs;\n", - "fie2 = 0.552*Hs;\n", - "fid2 = 0.455*Hs;\n", - "fic2 = 0.414*Hs;\n", - "fie3 = 0.34*Hs;\n", - "fid3 = 0.29*Hs;\n", - "fic3 = 0;\n", - "fie1 = round(fie1*100)/100;fid1 = round(fid1*100)/100;fic1 = round(fic1*100)/100;\n", - "fie2 = round(fie2*100)/100;fid2 = round(fid2*100)/100;fic2 = round(fic2*100)/100;\n", - "fie3 = round(fie3*100)/100;fid3 = round(fid3*100)/100;fic3 = round(fic3*100)/100;\n", - "print \"fie1 = %.2f.;fid1 = %.2f.;fic1 = %.2f.fie2 = %.2f.;fid2 = %.2f.;fic2 = %.2f.fie3 = %.2f.;\\\n", - "fid3 = %.2f.;fic3 = %.2f.\"%(fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);\n", - "print \"high flood condition:\";\n", - "fie1 = 1*Hs_;\n", - "fid1 = 0.82*Hs_;\n", - "fic1 = 0.788*Hs_;\n", - "fie2 = 0.552*Hs_;\n", - "fid2 = 0.455*Hs_;\n", - "fic2 = 0.414*Hs_;\n", - "fie3 = 0.34*Hs_;\n", - "fid3 = 0.29*Hs_;\n", - "fic3 = 0;\n", - "fie1 = round(fie1*100)/100;fid1 = round(fid1*100)/100;fic1 = round(fic1*100)/100;\n", - "fie2 = round(fie2*100)/100;fid2 = round(fid2*100)/100;fic2 = round(fic2*100)/100;\n", - "fie3 = round(fie3*100)/100;fid3 = round(fid3*100)/100;fic3 = round(fic3*100)/100;\n", - "print \"fie1 = %.2f.;fid1 = %.2f.;fic1 = %.2f.fie2 = %.2f.;fid2 = %.2f.;fic2 = %.2f.fie3 = %.2f.;\\\n", - "fid3 = %.2f.;fic3 = %.2f.\"%(fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);\n", - "print \"flow at pond level:\";\n", - "fie1 = 1*Hs__;\n", - "fid1 = 0.82*Hs__;\n", - "fic1 = 0.788*Hs__;\n", - "fie2 = 0.552*Hs__;\n", - "fid2 = 0.455*Hs__;\n", - "fic2 = 0.414*Hs__;\n", - "fie3 = 0.34*Hs__;\n", - "fid3 = 0.29*Hs__;\n", - "fic3 = 0;\n", - "fie1 = round(fie1*100)/100;fid1 = round(fid1*100)/100;fic1 = round(fic1*100)/100;\n", - "fie2 = round(fie2*100)/100;fid2 = round(fid2*100)/100;fic2 = round(fic2*100)/100;\n", - "fie3 = round(fie3*100)/100;fid3 = round(fid3*100)/100;fic3 = round(fic3*100)/100;\n", - "print \"fie1 = %.2f.;fid1 = %.2f.;fic1 = %.2f.fie2 = %.2f.;fid2 = %.2f.;fic2 = %.2f.fie3 = %.2f.;\\\n", - "fid3 = %.2f.;fic3 = %.2f.\"%(fie1,fid1,fic1,fie2,fid2,fic2,fie3,fid3,fic3);\n", - "\n", - "print \"Prejump profile:\";\n", - "print \"high flood condition:\";\n", - "dist = [3 ,6, 8.4]; \t\t\t\t#dismath.tance\n", - "glacis = [252, 251, 250.32]; \t\t\t\t#R.L of glacis\n", - "D1 = [1.3 ,1.15, 1.03];\n", - "Ef1 = [0,0,0]\n", - "print \"Ef1 D1\";\n", - "for i in range(3):\n", - " Ef1[i] = 256.25-glacis[i];\n", - " print \"%.2f %.2f\"%(Ef1[i],D1[i]);\n", - "\n", - "print \"pond level flow:\";\n", - "dist = [3, 6, 9, 9.6]; \t\t\t\t#dismath.tance\n", - "glacis = [252, 251, 250, 249.9]; \t\t\t\t#R.Lof glacis\n", - "D1 = [0.31, 0.23, 0.16, 0.15];\n", - "Ef1 = [0,0,0,0]\n", - "print \"Ef1 D1\";\n", - "for i in range(4):\n", - " Ef1[i] = 254-glacis[i];\n", - " print \"%.2f %.2f\"%(Ef1[i],D1[i]);\n", - "\n", - "\n", - "\n", - "rho = 2.24;\n", - "Uf = 4; \t\t\t\t#unbalanced head for high flood condtion\n", - "Us = 2.56; \t\t\t\t#unbalanced static head\n", - "Hf = 2*Uf/3;\n", - "t = Hf/(rho-1);\n", - "t = round(t*10)/10;\n", - "print \"floor thickness at the point of formation of hydraulic jump = %.2f m.\"%(t);\n", - "Uf = 2.9; \t\t\t\t#unbalanced head for high flood condtion\n", - "Us = 2.2; \t\t\t\t#unbalanced static head\n", - "Hf = 2*Uf/3;\n", - "t = Us/(rho-1);\n", - "t = round(t*10)/10;\n", - "print \"floor thickness at the point of formation of hydraulic jump at the pond level condition = %.2f m.\"%(t);\n", - "P = 1.5; \t\t\t\t#pressure head at d/s end of floor\n", - "t = P/(rho-1);\n", - "t = round(t*10)/10;\n", - "print \"floor thickness at downstream side of sloping glacis = %.2f m.\"%(t);\n", - "D = rb-sh_up; \t\t\t\t#depth of u/s scour hole above bed level\n", - "a = 1.5*D;\n", - "a = round(a*10)/10;\n", - "print \"minimum length of upstream launching apron = %.2f m.\"%(a);\n", - "print \"provide 1.5 m thick apron for length of 5 m.\";\n", - "D = 249.6-241.5;\n", - "a = 1.5*D;\n", - "print \"minimum length of downstream launching apron = %.2f m.\"%(a);\n", - "print \"provide 1.5 m thick apron for length of 12 m.\";\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "crest level = 253.00 m.\n", - "elevation of u/s T.E.L = 256.26 m.\n", - "regime scour depth = 6.50 m.\n", - "afflux = 1.14. which is near to permissible\n", - "loss of head in at high flood = 1.64 m.\n", - "loss of head = 2.50 m.\n", - "Hydraulic jump calculation:\n", - "heigth of jump for high flood condition = 2.93 m.\n", - "length of concrete floor for high flood condition = 14.65 m.\n", - "heigth of jump for pond level condition = 1.53 m.\n", - "length of concrete floor for high pond level condition = 7.65 m.\n", - " upstream slope of glacis = 2:1.\n", - "downstream slope of glacis = 3:1.\n", - "horizontal length of floor beyond the toe = 15 m..\n", - "R.L of bottom of upstream sheet pile = 246.39 m.\n", - "R.L of downstream sheet pile = 241.50 m.\n", - "provide intermediate sheet pile at d/s toe of glacis.\n", - "length of impervious floor = 12.15 m.\n", - "length of floor already provide = 34.20 m.\n", - "which is more than required from permissible exit gradient.no upstream floor is required.\n", - "provide 1.80 m upstream floor so that total length becomes 36 m.\n", - "Pressure percent at points:\n", - "Points Before correction After correction\n", - "C1 72 78.80\n", - "D1 82 82.00\n", - "C2 31 41.40\n", - "E2 45 45.50\n", - "D2 58 55.20\n", - "D3 29 29.00\n", - "E3 44 37.60\n", - "elevation of subsoil H.G above datum:\n", - "no flow condition:\n", - "fie1 = 4.40.;fid1 = 3.61.;fic1 = 3.47.fie2 = 2.43.;fid2 = 2.00.;fic2 = 1.82.fie3 = 1.50.;fid3 = 1.28.;fic3 = 0.00.\n", - "high flood condition:\n", - "fie1 = 1.63.;fid1 = 1.34.;fic1 = 1.28.fie2 = 0.90.;fid2 = 0.74.;fic2 = 0.67.fie3 = 0.55.;fid3 = 0.47.;fic3 = 0.00.\n", - "flow at pond level:\n", - "fie1 = 2.50.;fid1 = 2.05.;fic1 = 1.97.fie2 = 1.38.;fid2 = 1.14.;fic2 = 1.03.fie3 = 0.85.;fid3 = 0.73.;fic3 = 0.00.\n", - "Prejump profile:\n", - "high flood condition:\n", - "Ef1 D1\n", - "4.25 1.30\n", - "5.25 1.15\n", - "5.93 1.03\n", - "pond level flow:\n", - "Ef1 D1\n", - "2.00 0.31\n", - "3.00 0.23\n", - "4.00 0.16\n", - "4.10 0.15\n", - "floor thickness at the point of formation of hydraulic jump = 1.60 m.\n", - "floor thickness at the point of formation of hydraulic jump at the pond level condition = 1.80 m.\n", - "floor thickness at downstream side of sloping glacis = 1.20 m.\n", - "minimum length of upstream launching apron = 4.70 m.\n", - "provide 1.5 m thick apron for length of 5 m.\n", - "minimum length of downstream launching apron = 12.15 m.\n", - "provide 1.5 m thick apron for length of 12 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.5 pg : 631" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "b = 16; \t\t\t\t#total length of floor\n", - "d = 5; \t\t\t\t#depth of downstream pile\n", - "D = 4; \t\t\t\t#depth of upstream pile\n", - "H = 2.5; \t\t\t\t#head created by weir\n", - "\n", - "\t\t\t\t#pressure at E\n", - "alpha = b/d;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "fie = math.acos((lambda1-2)/lambda1)/math.pi;\n", - "C = 19*(D/b)**0.5*((d+D)/b);\n", - "fie = fie*100-C;\n", - "P = H*fie/100;\n", - "P = round(P*1000)/1000;\n", - "print \"Pressure at E = %.2f m.\"%(P);\n", - "\n", - "\t\t\t\t#pressure at C1\n", - "alpha = b/D;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "fie = math.acos((lambda1-2)/lambda1)/math.pi;\n", - "fic = 1-fie; \t\t\t\t#by principle reversibility of flow\n", - "C = 19*(d/b)**0.5*((d+D)/b);\n", - "fic = fic*100+C;\n", - "P = fic*H/100;\n", - "P = round(P*1000)/1000;\n", - "print \" Pressure at C = %.2f m.\"%(P);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Pressure at E = 1.22 m.\n", - " Pressure at C = 1.43 m.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.6 pg : 632" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\n", - "#Given\n", - "b = 13; \t\t\t\t#length of floor\n", - "d = 2; \t\t\t\t#depth of downstream wall\n", - "D = 1.5; \t\t\t\t#depth of upstream cut-off\n", - "rho = 2.24; \t\t\t\t#relative density\n", - "H = 1.5;\n", - "\n", - "#at junction of d/s cut-off with floor\n", - "alpha = b/d;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "fie = math.acos((lambda1-2)/lambda1)/math.pi;\n", - "C = 19*(D/b)**0.5*((d+D)/b);\n", - "fie = fie*100-C;\n", - "P = H*fie/100;\n", - "t = P/(rho-1);\n", - "t = round(t*10)/10;\n", - "print \"floor thickness at junction of d/s cut-off with floor = %.2f m.\"%(t);\n", - "\n", - "#at junction of u/s cut-off with floor\n", - "alpha = b/D;\n", - "lambda11 = (1+(1+alpha**2)**0.5)/2;\n", - "fie = math.acos((lambda11-2)/lambda11)/math.pi;\n", - "fic = 1-fie; \t\t\t\t#by principle reversibility of flow\n", - "C = 19*(D/b)**0.5*((d+D)/b);\n", - "fiec = fic*100+C;\n", - "P = fiec*H/100;\n", - "t = 0.3; \t\t\t\t#this the uplift will be counter balanced by downward weigth of impounded water\n", - "print \"floor thickness at junction of u/s cut-off with floor = %.2f m.\"%(t);\n", - "\n", - "#at mid-length\n", - "P = (1.08+0.489)/2; \t\t\t\t#assuming linear variation\n", - "t = P/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"floor thickness at mid-length = %.2f m.\"%(t);\n", - "\n", - "#exit gradient\n", - "G = H/(d*math.pi*(lambda1)**0.5);\n", - "G = round(G*1000)/1000;\n", - "#math.since G<0.18\n", - "print \" G = %.2f. <0.18./nfloor is safe against failure by piping.\"%(G);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "floor thickness at junction of d/s cut-off with floor = 0.40 m.\n", - "floor thickness at junction of u/s cut-off with floor = 0.30 m.\n", - "floor thickness at mid-length = 0.63 m.\n", - " G = 0.13. <0.18./nfloor is safe against failure by piping.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.7 pg : 634" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "B = 30; \t\t\t\t#stream width\n", - "D = 3; \t\t\t\t#stream depth\n", - "V = 1.25; \t\t\t\t#mean velocity\n", - "Cd = 0.95; \t\t\t\t#discharge coefficient\n", - "Q = B*D*V;\n", - "\n", - "# Calculations\n", - "C = 2*Cd*(2*9.81)**0.5/3;\n", - "x = 4-(Q/(C*B))**(2./3);\n", - "x = round(x*1000)/1000;\n", - "\n", - "# Results\n", - "print \"heigth of weir to be built = %.2f m.\"%(x);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "heigth of weir to be built = 2.79 m.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.8 pg : 635" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "b = 50.; \t\t\t\t#length of floor\n", - "d = 8.; \t\t\t\t#depth of downstream pile\n", - "D = 8.; \t\t\t\t#depth of upstream pile\n", - "H = 5.; \t\t\t\t#effective head \n", - "tu = 1.; \t\t\t\t#floor thickness at upstream\n", - "td = 2.; \t\t\t\t#floor thickness at downstream\n", - "\n", - "# Calculations and Results\n", - "\t\t\t\t#downstream cut-off\n", - "alpha = b/d;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "fie = math.acos((lambda1-2)/lambda1)/math.pi;\n", - "fid = math.acos((lambda1-1)/lambda1)/math.pi;\n", - "Ct = (fie-fid)*td/d;\n", - "C = 19*(D/b)**0.5*((d+D)/b);\n", - "fie = fie*100-C-Ct*100;\n", - "P = H*fie/100;\n", - "P = round(P*100)/100;\n", - "print \"Pressure at downstream cut-off = %.2f m.\"%(P);\n", - "\n", - "\t\t\t\t#upstream cut-off\n", - "fie = math.acos((lambda1-2)/lambda1)/math.pi;\n", - "fid = math.acos((lambda1-1)/lambda1)/math.pi;\n", - "fic1 = 1-fie;\n", - "fid1 = 1-fid;\n", - "Ct = (fic1-fid1)*td/d;\n", - "C = -19*(D/b)**0.5*((d+D)/b);\n", - "fic1 = fic1*100-C-Ct*100;\n", - "P = H*fic1/100;\n", - "P = round(P*100)/100;\n", - "print \"Pressure at upstream cut-off = %.2f m.\"%(P);\n", - "G = H/(d*math.pi*(lambda1)**0.5);\n", - "print \"Exit Gradient = %.2f.\"%(G);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Pressure at downstream cut-off = 1.49 m.\n", - "Pressure at upstream cut-off = 3.51 m.\n", - "Exit Gradient = 0.10.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.9 pg : 636" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 1000.; \t\t\t\t#discharge of river\n", - "L = 256.; \t\t\t\t#crest length of diversion\n", - "f = 1.1; \t\t\t\t#silt factor\n", - "seg = 1./6; \t\t\t\t#safe exit gradient\n", - "hfl = 103; \t\t\t\t#high flood level\n", - "cf = 100; \t\t\t\t#reduced level of downstream concrete floor\n", - "H = 2.4; \t\t\t\t#maximum static head of weir\n", - "b = 40; \t\t\t\t#length of concrete floor\n", - "\n", - "\n", - "# Calculations and Results\n", - "q = Q/L;\n", - "R = 1.35*(q**2/f)**(1./3);\n", - "rld = hfl-1.5*R;\n", - "d = cf-rld;\n", - "d = round(d*100)/100;\n", - "print \"depth of downstream cut-off = %.2f m.\"%(d);\n", - "\n", - "alpha = b/d;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "G = H/(d*math.pi*(lambda1)**0.5);\n", - "\t\t\t\t#math.since G<seg\n", - "print \" G = %.2f. <1/6./nfloor is safe against failure by piping.\"%(G);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "depth of downstream cut-off = 1.87 m.\n", - " G = 0.12. <1/6./nfloor is safe against failure by piping.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.10 pg : 636" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "b = 60; \t\t\t\t#length of floor\n", - "H = 6; \t\t\t\t#static head of weir\n", - "d = 6; \t\t\t\t#downstream depth of pile\n", - "n = 0.3; \t\t\t\t#porousity of soil particles\n", - "G = 2.7; \t\t\t\t#relative density of soil particles\n", - "\n", - "alpha = b/d;\n", - "lambda1 = (1+(1+alpha**2)**0.5)/2;\n", - "Ge = H/(d*math.pi*(lambda1)**0.5);\n", - "e = n/(1-n);\n", - "chg = (G-1)/(1+e);\n", - "f = chg/Ge;\n", - "f = round(f*100)/100;\n", - "print \"critical exit gradient = %.2f.\\\n", - "\\nfactor of safety of system = %.2f.\"%(chg,f);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "critical exit gradient = 1.19.\n", - "factor of safety of system = 8.79.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.11 pg : 637" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots,ceil\n", - "\n", - "#design a vertical drop weir on Bligh's theory\n", - "#test floor by Khosla's theory\n", - "\n", - "#Given\n", - "Q = 2800.; \t\t\t\t#maximum flood discharge\n", - "hfl = 285.; \t\t\t\t#H.F.L before construction\n", - "hw = 278.; \t\t\t\t#minimum water level\n", - "fsl = 284.; \t\t\t\t#F.S.L of canal\n", - "c = 12.; \t\t\t\t#coefficient of creep\n", - "flux = 1.; \t\t\t\t#allowable afflux\n", - "Ge = 1./6; \t\t\t\t#permissible exit gradient\n", - "rho = 2.24; \t\t\t\t#specific gravity of concrete\n", - "\n", - "#Hydraulic calculation\n", - "L = 4.75*Q**0.5;\n", - "q = Q/L;\n", - "q = round(q*10)/10;\n", - "print \"Hydraulic calculation:\";\n", - "print \"discharge per unit width of river = %.2f cumecs.\"%(q);\n", - "f = 1;\n", - "R = 1.35*(q**2/f)**(1./3);\n", - "R = round(R*100)/100;\n", - "print \"regime scour depth = %.2f m.\"%(R);\n", - "V = q/R; \t\t\t\t#regime velocity\n", - "vh = V**2/(2*9.81); \t\t\t\t#velocity head\n", - "l_down = hfl+vh;\n", - "l_up = l_down+flux;\n", - "hfl_up = l_up-vh;\n", - "hfl_down = hfl-0.5;\n", - "hfl_down = round(hfl_down*100)/100;\n", - "print \"actual d/s H.F.L allowing 0.5 m for retrogation = %.2f m.\"%(hfl_down);\n", - "K = (q/1.7)**(2./3);\n", - "cl = l_up-K; \t\t\t\t#crest level\n", - "cl = round(cl*100)/100;\n", - "print \"crest level = %.2f m.\"%(cl);\n", - "pl = fsl+0.5; \t\t\t\t#pond level\n", - "s = hfl_down-cl; \t\t\t\t#heigth of shutter\n", - "print \"heigth of shutter = %.2f m.\"%(s);\n", - "rl_up_pile = hfl_up-1.5*R; \t\t\t\t#R.L of bottom u/s pile\n", - "d_up_cut = hw-276; \t\t\t\t#depth of upstream cut-off\n", - "print \"depth of upstream cut-off = %.2f m.\"%(d_up_cut);\n", - "print \" provide concrete cut off 2 m depth.\";\n", - "rl_bot_ds = hfl_down-2*R;\n", - "Hs = hfl_down-hw; \t\t\t\t#seepage head\n", - "Hc = cl-hw; \t\t\t\t#heigth of crest\n", - "print \"R.L of gates crest = %.2f m.\"%(Hs);\n", - "print \"Heigth of crest = %.2f m.\"%(Hc);\n", - "\n", - "#design of weir wall\n", - "d = hfl_up-cl;\n", - "a = d/(rho)**0.5;\n", - "a = 3*d/(2*rho); \t\t\t\t#from sliding consideration\n", - "a = s+1; \t\t\t\t#from practical consieration\n", - "a = a+1;\n", - "print \"design of weir wall:\"\n", - "print \"provide top width of %i m.\"%(a);\n", - "Mo = 9.81*Hs**3/6; \t\t\t\t#overtirning moment\n", - "#equating the moment of resismath.tance to overturning moment and putting the values we get\n", - "y = [0.039,0.020,-1.084]\n", - "b = int(roots(y)[1]);\n", - "#we get b = - 5.5347261 and 5.0219056\n", - "#taking\n", - "#b = 5;\n", - "#when weir is submerged\n", - "C = 0.58;\n", - "d = (q**2/((2*C/3)**2*2*9.81))**(1./3);\n", - "Mo = 9.81*d*Hc**2/2;\n", - "#from equation of moment of resistence we get\n", - "y = [1,3,-77.55]\n", - "b = ceil(roots(y)[1]);\n", - "#we get b = - 10.433085 and 7.4330846\n", - "#taking\n", - "#b = 8;\n", - "print \"bottom width = %i m.\"%(b);\n", - "\n", - "#design of impervious and pervious aprons\n", - "C = 12;\n", - "L = C*Hs;\n", - "print \"design of impervious and pervious aprons:\";\n", - "print \"total creep length = %i m.\"%(L);\n", - "l1 = 2.21*C*(Hs/13)**0.5;\n", - "l1_ = l1+1;\n", - "print \"length of downstream impervious apron = %i m.\"%(l1_);\n", - "d1 = hw-276;\n", - "d2 = hw-271;\n", - "l2 = L-l1-(b+2*d1+2*d2);\n", - "print \"length of upstream impervious apron = %i m.\"%(l2);\n", - "l3 = 18*C*(Hs*q/975)**0.5;\n", - "print \"total length of d/s apron = %i m.\"%(l3); \t\t\t\t#calculation is wrong in book\n", - "l = l3-l1;\n", - "le = l/2;\n", - "le = round(le*100)/100;\n", - "print 'provide filter of length %.2f m. and launching apron of length %.2f m.'%(le,le);\n", - "t = d2*10**0.5/le;\n", - "print \"thickness of launching apron in horizontal position = %.2f m.\"%(t);\n", - "print \"provide launching apron of thickness 1.5 m.\";\n", - "T = 2*d1;\n", - "V = d1*10**0.5;\n", - "ta = V/T;\n", - "ta = round(ta*10)/10;\n", - "print \"thickness of apron in horizontal position = %.2f m.\"%(ta);\n", - "Hr = Hs-Hs*(4+33+8)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print 'provide thickness of %.2f m from d/s of weir wall to point 6 m from it.'%(t);\n", - "Hr = Hs-Hs*(4+33+8+6)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print \"provide thickness of %.2f m from 6 m to 12 m from d/s end of weir wall.\"%(t);\n", - "Hr = Hs-Hs*(4+33+8+12)/L;\n", - "t = 4*Hr/(3*(rho-1));\n", - "t = round(t*10)/10;\n", - "print \"provide thickness of %.2f m for rest of length of weir floor.\"%(t);\n", - "\n", - "#check by khosla's theory\n", - "b = 33+8+19; \t\t\t\t#total horizontal length of impervious floor\n", - "d = 7; \t\t\t\t#depth of downstream pile\n", - "alpha = b/d;\n", - "n = 0.14; \t\t\t\t#n = 1/math.pi*(lambda)**0.5;\n", - "Ge = Hs*n/d;\n", - "print \"check by Khosla theory:\";\n", - "print \"exit gradient = %.2f. < 1/6 hence safe\"%(Ge);\n", - "alpha_ = d/b;\n", - "fic1 = 0.83;fid1 = 0.88;\n", - "corec_c1 = (fid1-fic1)*100/2;\n", - "bdash = b;\n", - "d = 2.;D = 7;\n", - "C1 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "fic1 = fic1*100+corec_c1+C1;\n", - "Pc = Hs*fic1/100; \t\t\t\t#pressure head at C\n", - "alpha_ = d/b;\n", - "fie2 = 0.31;fid2 = 0.21;\n", - "corec_e1 = (fie2-fid2)*1.7*100/7;\n", - "bdash = b;\n", - "d = 7;D = 2;\n", - "C1 = 19*(D/bdash)**0.5*(d+D)/b;\n", - "fie2 = fie2*100-corec_e1-C1; \t\t\t\t#in book 3.53 value is wrong\n", - "Pe = Hs*fie2/100; \t\t\t\t#pressue head at E\n", - "#assuming linear variation of pressure for intermediate points\n", - "Pa = Pc-(Pc-Pe)*(33+8)/b;\n", - "t = Pa/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at d/s of weir wall = %.2f m.\"%(Pa);\n", - "print \"thickness at d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe.\"%(t);\n", - "Pb = Pc-(Pc-Pe)*(33+8+6)/b;\n", - "t = Pb/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at 6 m from d/s of weir wall = %.2f m.\"%(Pb);\n", - "print \"thickness at 6m from d/s of weir wall = %.2f m. < thickness by Bligh theory;hence safe.\"%(t);\n", - "Pc = Pc-(Pc-Pe)*(33+8+12)/b;\n", - "t = Pc/1.24;\n", - "Pa = round(Pa*100)/100;\n", - "t = round(t*100)/100;\n", - "print \"pressure at 12 m from d/s of weir wall = %.2f m.\"%(Pc);\n", - "print \"thickness at 12m from d/s of weir wall = %.2f m. > thickness by Bligh theory;hence unsafe.\"%(t);\n", - "print \"hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Hydraulic calculation:\n", - "discharge per unit width of river = 11.10 cumecs.\n", - "regime scour depth = 6.72 m.\n", - "actual d/s H.F.L allowing 0.5 m for retrogation = 284.50 m.\n", - "crest level = 282.65 m.\n", - "heigth of shutter = 1.85 m.\n", - "depth of upstream cut-off = 2.00 m.\n", - " provide concrete cut off 2 m depth.\n", - "R.L of gates crest = 6.50 m.\n", - "Heigth of crest = 4.65 m.\n", - "design of weir wall:\n", - "provide top width of 3 m.\n", - "bottom width = 8 m.\n", - "design of impervious and pervious aprons:\n", - "total creep length = 78 m.\n", - "length of downstream impervious apron = 19 m.\n", - "length of upstream impervious apron = 33 m.\n", - "total length of d/s apron = 58 m.\n", - "provide filter of length 20.00 m. and launching apron of length 20.00 m.\n", - "thickness of launching apron in horizontal position = 1.11 m.\n", - "provide launching apron of thickness 1.5 m.\n", - "thickness of apron in horizontal position = 1.60 m.\n", - "provide thickness of 3.00 m from d/s of weir wall to point 6 m from it.\n", - "provide thickness of 2.40 m from 6 m to 12 m from d/s end of weir wall.\n", - "provide thickness of 1.90 m for rest of length of weir floor.\n", - "check by Khosla theory:\n", - "exit gradient = 0.13. < 1/6 hence safe\n", - "pressure at d/s of weir wall = 3.03 m.\n", - "thickness at d/s of weir wall = 2.44 m. < thickness by Bligh theory;hence safe.\n", - "pressure at 6 m from d/s of weir wall = 2.66 m.\n", - "thickness at 6m from d/s of weir wall = 2.14 m. < thickness by Bligh theory;hence safe.\n", - "pressure at 12 m from d/s of weir wall = 2.29 m.\n", - "thickness at 12m from d/s of weir wall = 1.85 m. > thickness by Bligh theory;hence unsafe.\n", - "hence increase th ethickness to 1.9 m for a length of 7 m of impervious floor.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 12.12 pg : 637" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#number of gates required for the barrage\n", - "#length and R.L of bamath.sin floor if silting bamath.sin is provided downstream of barrage\n", - "\n", - "#Given\n", - "Lmax = 212; \t\t\t\t#maximum reservior level\n", - "Lp = 211; \t\t\t\t#pond level\n", - "hfl = 210; \t\t\t\t#downstream high flood level in the river\n", - "Qmax = 3500; \t\t\t\t#maximum design flood discharge\n", - "Lcrest = 207; \t\t\t\t#crest level of the barrage\n", - "Lcrest_r = 208; \t\t\t\t#crest level of head regulator\n", - "Cd = 2.1; \t\t\t\t#coefficient of discharge for barrage\n", - "Cd_r = 1.5; \t\t\t\t#coefficient of discharge for head regulator\n", - "rbl = 205; \t\t\t\t#river bed level\n", - "Q = 500; \t\t\t\t#design discharge of main canal\n", - "\n", - "#design of water way for barrage during flood\n", - "H = Lmax-Lcrest;\n", - "L = Qmax/(Cd*H**1.5);\n", - "#which gives L = 149.07.\n", - "print \"nunmber of gates for the barrage = 15.\";\n", - "\n", - "#design of waterway for canal head regulator\n", - "H = Lp-Lcrest_r;\n", - "L1 = Q/(Cd_r*H**1.5);\n", - "#which gives L = 64.2\n", - "#hence provide 7 bays of 10 m each\n", - "print \"nunmber of gates for the head regulator = 7.\";\n", - "\n", - "#design of stilling bamath.sin\n", - "Hl = Lmax-hfl;\n", - "q = Qmax/L;\n", - "yc = (q**2/9.81)**(1./3);\n", - "Z = Hl/yc;\n", - "#math.since Z<1\n", - "Y = 1+0.93556*Z**0.368;\n", - "y2 = Y*yc;\n", - "Lc = 5*y2;\n", - "Lc = round(Lc*10)/10;\n", - "print \"Length of cistern = %.2f m.\"%(Lc);\n", - "Ef2 = yc*(Y+1/(2*Y**2));\n", - "j = hfl-Ef2;\n", - "j = round(j*10)/10;\n", - "print \"R.L of cistern = %.2f m.\"%(j);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "nunmber of gates for the barrage = 15.\n", - "nunmber of gates for the head regulator = 7.\n", - "Length of cistern = 33.30 m.\n", - "R.L of cistern = 202.70 m.\n" - ] - } - ], - "prompt_number": 15 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch14.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch14.ipynb deleted file mode 100755 index f55e341a..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch14.ipynb +++ /dev/null @@ -1,1965 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:e8ae669d88fc715c49999174af0f47a67eb761cc981a916d61f3cb981ae07f2c" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 14 : IRRIGATION CHANNEL 1 SILT THEORIES" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.1 pg : 662" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design irrigation channel on Kennedy's theory\n", - "\n", - "#Given\n", - "Q = 45; \t\t\t\t#discharge\n", - "N = 0.0225; \t\t\t\t#rogosity coefficient\n", - "m = 1.05; \t\t\t\t#critical velocity ratio\n", - "S = 1./5000; \t\t\t\t#bed slope\n", - "\n", - "D = 2; \t\t\t\t#assume\n", - "Vo = 0.55*m*D**0.64;\n", - "A = Q/Vo;\n", - "#for a trapezoidal section\n", - "B = (A-0.5*D**2)/2;\n", - "P = B+D*5**0.5;\n", - "R = A/P;\n", - "C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "V = C*(R*S)**0.5;\n", - "#Vo<V\n", - "\n", - "#assume D = 2.2\n", - "D = 2.2;\n", - "Vo = 0.55*m*D**0.64;\n", - "A = Q/Vo;\n", - "B = (A-0.5*D**2)/D;\n", - "P = B+D*5**0.5;\n", - "R = A/P;\n", - "C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "V = C*(R*S)**0.5;\n", - "\n", - "#ratio of V and Vo is almost equal to 1\n", - "\n", - "# Results\n", - "B = round(B*10)/10;\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 20.30 m.\n", - "Depth of channel section = 2.20 m.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.2 pg : 663" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#design an irrigation canal for given data\n", - "\n", - "#Given\n", - "Q = 14; \t\t\t\t#discharge\n", - "m = 1; \t\t\t\t#critical velocity ratio\n", - "r = 5.7; \t\t\t\t#B/D\n", - "\n", - "D = (Q/(0.55*6.2))**(1/2.64);\n", - "B = D*r;\n", - "R = (B*D+D**2/2)/(B+D*5**0.5);\n", - "Vo = 0.55*m*D**0.64;\n", - "#applying kutters formula; V = C(RS)**0.5\n", - "#where C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "#assuming S**0.5 = y\n", - "#y = poly1d([-1.98D-5,1.55D-3,-0.954,67.5],'x','c');\n", - "#roots(y);\n", - "#taking real values of y\n", - "S = 0.0139906**2;\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 9.73 m.\n", - "Depth of channel section = 1.71 m.\n", - "Bed slope = 1.96e-04.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.3 pg : 668" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a channel on Kennedy's theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "m = 1.05; \t\t\t\t#critical velocity ratio\n", - "N = 0.025; \t\t\t\t#rugosity coefficient\n", - "S = 1./5000; \t\t\t\t#bed slope\n", - "\n", - "\n", - "# Calculations\n", - "l = S*Q**0.02/(N**2*m**2.02);\n", - "#from fig.14.3 we get r = 10\n", - "#solving the equation by trial and error method we get\n", - "r = 9.7;\n", - "D = (1.818*Q/(m*(r+0.5)))**(1/2.64);\n", - "B = r*D;\n", - "V = Q/(D**2*(r+0.5));\n", - "Vo = 0.55*D**0.64*m;\n", - "B = round(B);\n", - "D = round(D*100)/100;\n", - "V = round(V*1000)/1000;\n", - "Vo = round(Vo*1000)/1000;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Velocity through the channel section = %.2f m/s.\"%(V);\n", - "print \"Vo = %.2f m/s.Hence Safe\"%(Vo);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 21.00 m.\n", - "Depth of channel section = 2.16 m.\n", - "Velocity through the channel section = 0.94 m/s.\n", - "Vo = 0.94 m/s.Hence Safe\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.4 pg : 668" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#design channel umath.sing method of curve fitting based onKennedy's theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "N = 0.0225; \t\t\t\t#rugosity coefficient\n", - "m = 1.05; \t\t\t\t#critical velocity ratio\n", - "S = 1./5000; \t\t\t\t#Bed slope\n", - "\n", - "r = (1.607*S**1.63*Q**0.033/(N**3.26*m**3.293)-0.258)**(-0.915);\n", - "D = (1.818*Q/(m*(r+0.5)))**(1/2.64);\n", - "B = r*D;\n", - "B = round(B);\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 21.00 m.\n", - "Depth of channel section = 2.16 m.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.5 pg : 668" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#design channel umath.sing curve of CWPC for B/D ratio\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "N = 0.0225; \t\t\t\t#rugosity coefficient\n", - "m = 1.05; \t\t\t\t#critical velocity ratio\n", - "\n", - "r = (15+6.44*Q)**0.382;\n", - "S = (N**2/1.338*Q**0.02)*(0.258+(15+6.44*Q)**(-0.417))**0.6135;\n", - "D = (1.818*Q/(m*(r+0.5)))**(1/2.64);\n", - "B = r*D;\n", - "B = round(B);\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 2.14e-04.\n", - "Width of channel section = 20.00 m.\n", - "Depth of channel section = 2.23 m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.6 pg : 667" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#design the channel section umath.sing the following data and calculate math.logitudnal section\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 30.; \t\t\t\t#discharge\n", - "f = 1.; \t\t\t\t#silt factor\n", - "s = 1./2; \t\t\t\t#side slope\n", - "\n", - "V = (Q*f/140)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*Q**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "B = P-2.236*D;\n", - "\n", - "R = 5*V**2/(2*f);\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 1.70e-04.\n", - "Width of channel section = 22.26 m.\n", - "Depth of channel section = 1.68 m.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.7 pg : 682" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#design a channel in alluvial soil umath.sing tractive force approach\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "S = 1./4800; \t\t\t\t#bed slope\n", - "N = 0.0225; \t\t\t\t#rogosity coefficient\n", - "sigma = 0.0035; \t\t\t\t#permissible tractive stress\n", - "s = 1./2; \t\t\t\t#side slope\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "\n", - "\n", - "# Calculations\n", - "R = sigma/(gamma_w*S);\n", - "V = R**(2./3)*S**0.5/N;\n", - "A = Q/V;\n", - "P = A/R;\n", - "#y = poly1d([-49,28.61,-1.736],'x','c');\n", - "y = [-1.736,28.61,-49]\n", - "D = roots(y)[1];\n", - "#we get D = 14.539034 and 1.9413812 \n", - "#taking D = 1.9413812 \n", - "B = 28.61-2.23*D;\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 24.28 m.\n", - "Depth of channel section = 1.94 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.8 pg : 687" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#designa channel section by Kennedy theory\n", - "\n", - "#Given\n", - "Q = 28.; \t\t\t\t#discharge\n", - "m = 1.; \t\t\t\t#critical velocity ratio\n", - "r = 7.6; \t\t\t\t#B/D\n", - "\n", - "D = (Q/4.46)**(1/2.64);\n", - "B = r*D;\n", - "R = 0.823*D;\n", - "V = 0.55*(D)**0.64;\n", - "\n", - "#applying kutters formula; V = C(RS)**0.5\n", - "#where C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "#we get equation in S\n", - "#assuming S**0.5 = y\n", - "#y = poly([-1.42D-5,1.55D-3,-0.885,67.4],'x','c');\n", - "y = [67.4,-0.885,1.55e-3,-1.42e-5]\n", - "d = roots(y)[0];\n", - "#taking real values of y\n", - "S = d.real**2;\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 15.20 m.\n", - "Depth of channel section = 2.01 m.\n", - "Bed slope = 1.60e-04.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.9 pg : 688" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design the channel section and calculate discharge\n", - "\t\t\t\t\n", - "#Given\n", - "r = 5.7; \t\t\t\t#B/D\n", - "S = 1./5000; \t\t\t\t#bed slope\n", - "N = 0.0225; \t\t\t\t#rogosity coefficient\n", - "m = 1; \t\t\t\t#critical velocity ratio(assumed)\n", - "\n", - "#applying kutters formula; V = C(RS)**0.5\n", - "#where C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "#we get equation in d as\n", - "#38.88*D**0.64-66.5*D**0.5+30.37*D**0.14 = 0\n", - "#solving it by trial and error method\n", - "#we get D = 1.7 m.\n", - "D = 1.7;\n", - "B = r*D;\n", - "V = 0.55*m*(D)**0.64;\n", - "A = B*D+D**2/2;\n", - "Q = A*V;\n", - "Q = round(Q*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \" Discharge = %.2f cumecs.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 9.69 m.\n", - "Depth of channel section = 1.70 m.\n", - " Discharge = 13.84 cumecs.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.10 pg : 689" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#design irrigation channel according to Laecy silt theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 15.; \t\t\t\t#discharge\n", - "f = 1.; \t\t\t\t#laecy silt factor\n", - "s = 1./2; \t\t\t\t#channel side slope\n", - "\n", - "V = (Q*f**2/140);\n", - "A = Q/V;\n", - "R = 5*V**2/(2*f);\n", - "#umath.sing the value of A in equations we get,\n", - "#equation in D as\n", - "y = [-1.73,18.336,-21.765]\n", - "D = roots(y)[1];\n", - "\n", - "#we get D = 9.2368003 and 1.3620436.\n", - "#taking\n", - "D = 1.3620436;\n", - "B = 18.336-D*2.23;\n", - "P = 4.75*Q**0.5;\n", - "S = 1/(3340*Q**(1./6));\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 15.30 m.\n", - "Depth of channel section = 1.36 m.\n", - "Bed slope = 1.91e-04.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.11 pg : 690" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "#find channel section and discharge\n", - "\t\t\t\t\n", - "#Given\n", - "S = 1./5000; \t\t\t\t#bed slope\n", - "s = 1./2; \t\t\t\t#side slope\n", - "f = 0.9; \t\t\t\t#laecy silt factor\n", - "\n", - "Q = (f**(5./3)/(3340*S))**6;\n", - "R = f**3/(4980*S)**2;\n", - "P = 4.75*Q**0.5;\n", - "A = P*R;\n", - "#umath.sing the value of A and P in equations we get,\n", - "#equation in D as\n", - "y = [-1.73,9.41,-6.961]\n", - "D = roots(y)[1];\n", - "\n", - "#we get D = 4.5561754 and 0.8831309.\n", - "#taking D = 0.8831309;\n", - "B = 9.41-D*2.23;\n", - "B = round(B*100)*100;\n", - "D = round(D*100)/100;\n", - "Q = round(Q*1000)/1000;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \" Discharge = %.2f cumecs.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 74400.00 m.\n", - "Depth of channel section = 0.88 m.\n", - " Discharge = 3.92 cumecs.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.12 pg : 691" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "D = 3.; \t\t\t\t#depth of channel\n", - "d = 0.3; \t\t\t\t#grain size\n", - "k = 1.5; \t\t\t\t#size of roughness of channel bed\n", - "S = 1./4400; \t\t\t\t#bed slope\n", - "G = 2.65; \t\t\t\t#specific gravity\n", - "\n", - "# Calculations\n", - "tau_b = gamma_w*D*S;\n", - "N1 = d**(1./6)/24;\n", - "N = k**(1./6)/24;\n", - "gamma_s = gamma_w*G;\n", - "tau_c = 0.047*(gamma_s-gamma_w)*d/1000;\n", - "r = (N1/N)**1.5;\n", - "q = 47450*(tau_b*r-tau_c)**1.5;\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"quantity of bed load moved = %.2f kN/m/hr.\"%(q);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "quantity of bed load moved = 13.12 kN/m/hr.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.13 pg : 691" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "D = 3.; \t\t\t\t#depth of channel\n", - "d = 0.3; \t\t\t\t#grain size\n", - "k = 1.5; \t\t\t\t#size of roughness of channel bed\n", - "S = 1./4400; \t\t\t\t#bed slope\n", - "G = 2.65; \t\t\t\t#specific gravity\n", - "\n", - "# Calculations\n", - "N1 = d**(1./6)/24;\n", - "N = k**(1./6)/24;\n", - "r = (N1/N)**1.5;\n", - "R1 = 3*r;\n", - "si = (G-1)*d/(1000*R1*S);\n", - "\t\t\t\t#hence we get\n", - "fi = 7;\n", - "q = 3600*fi*G*gamma_w*(G-1)**0.5*(gamma_w)**0.5*(d/1000)**1.5;\n", - "q = round(q*10)/10;\n", - "\n", - "# Results\n", - "print \"quantity of bed load moved = %.2f kN/m/hr.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "quantity of bed load moved = 13.70 kN/m/hr.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.14 pg : 692" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "D = 3.; \t\t\t\t#depth of channel\n", - "d = 0.3; \t\t\t\t#grain size\n", - "k = 1.5; \t\t\t\t#size of roughness of channel bed\n", - "S = 1./4400; \t\t\t\t#bed slope\n", - "G = 2.65; \t\t\t\t#specific gravity\n", - "V = 0.03; \t\t\t\t#fall velocity\n", - "c_ = 400; \t\t\t\t#concentration at 0.3 m above bed\n", - "a = 0.3;\n", - "y = 1.;\n", - "k_ = 0.4; \t\t\t\t#van karman's consmath.tant\n", - "\n", - "\n", - "# Calculations\n", - "N1 = d**(1./6)/24;\n", - "N = k**(1./6)/24;\n", - "r = (N1/N)**1.5;\n", - "R1 = 3*r;\n", - "V_ = (gamma_w*R1*S)**0.5;\n", - "c = c_*((a/y)*(D-y)/(D-a))**(V/(V_*k_));\n", - "c = round(c*10)/10;\n", - "\n", - "# Results\n", - "print \"concentration of suspended load = %.2f ppm.\"%(c);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "concentration of suspended load = 74.10 ppm.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.15 pg : 692" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "\t\t\t\t#design an irrigation channel by Meyer peter equation\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45; \t\t\t\t#discharge\n", - "c = 55; \t\t\t\t#bed load concentraion\n", - "d = 0.3; \t\t\t\t#average grain diameter\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "G = 2.67;\n", - "f = 0.964;\n", - "\n", - "c = c*Q*gamma_w*3600/1000000;\n", - "P = 4.75*Q**0.5;\n", - "\t\t\t\t#taking channel width as B = 28 m(slightly less than P)\n", - "B = 28.;\n", - "qs = c/B;\n", - "\t\t\t\t#assuming effective grain diameter k = 0.4 mm\n", - "ks = 0.4e-3;\n", - "N1 = ks**(1./6)/24;\n", - "sf = 1.76*d**0.5;\n", - "N = 0.0225*sf**0.25;\n", - "r = N1/N;\n", - "tau_c = 0.047*gamma_w*(G-1)*d/1000;\n", - "tau_b = r**1.5*((qs/47450)**(2./3)+tau_c);\n", - "\t\t\t\t#from Manning's formula we get on simplification\n", - "R = (0.000992*1000/0.525)**(3./7);\n", - "S = 0.525/(1000*R);\n", - "\t\t\t\t#solving equation of R for trapezoidal section of side slope 1/2 we get\n", - "y = [0.5,25.06,-36.792]\n", - "D = roots(y)[1];\n", - "#we get D = -51.547499 and 1.4274989 \n", - "#taking D = 1.4274989;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %i m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 28 m.\n", - "Depth of channel section = 1.43 m.\n", - "Bed slope = 4.00e-04.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.16 pg : 693" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "\n", - "#design an irrigation channel by Einstein equation\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "c = 55.; \t\t\t\t#bed load concentraion\n", - "d = 0.3; \t\t\t\t#average grain diameter\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "G = 2.67; \t\t\t\t#specific gravity of soil\n", - "f = 0.964; \t\t\t\t#silt factor\n", - "\n", - "#taking channel width as B = 28 m(slightly less than P)\n", - "B = 28.;\n", - "qs = c/B;\n", - "\n", - "fi = (qs/(gamma_w*G))*(1/(G-1))**0.5*(1000000000/(gamma_w*d**3))**0.5;\n", - "#from fig. 14.6 we get value of sci\n", - "#umath.sing the sci equation and Manning formula and on simplifications we get\n", - "R = (2.4296)**(3./7);\n", - "S = 0.4083/(1000*1.463);\n", - "#solving equation of R for trapezoidal section of side slope 1/2 we get\n", - "y = [0.5,24.73,-40.96]\n", - "D = roots(y)[1];\n", - "#we get D = -51.064253 and 1.6042534 \n", - "#taking D = 1.6042534;\n", - "D = round(D*10)/10;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %i m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 28 m.\n", - "Depth of channel section = 1.60 m.\n", - "Bed slope = 2.79e-04.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.17 pg : 697" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "#design a channel for non-alluvial deposites\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "S = 1./4000; \t\t\t\t#bed slope\n", - "v = 0.9; \t\t\t\t#permissible velocity\n", - "N = 0.025; \t\t\t\t#rogosity coefficient\n", - "\n", - "\n", - "# Calculations\n", - "A = Q/v;\n", - "R = (v*N/S**0.5)**1.5;\n", - "P = A/R;\n", - "\t\t\t\t#let us provide a trapezoidal section\n", - "\t\t\t\t#from equation of Area and Perimeter of trapezoid\n", - "y = [1.828,-29.45,50]\n", - "D = roots(y)[1];\n", - "#from which we get D = 14.181815 and 1.9286881\n", - "#taking D = 1.9286881;\n", - "B = P-2*1.41*D;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %i m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 24 m.\n", - "Depth of channel section = 1.93 m.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.18 pg : 697" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design non-allvial channel umath.sing Bazin's formula\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 15.; \t\t\t\t#discharge\n", - "V = 0.75; \t\t\t\t#mean velocity\n", - "s = 1.; \t\t\t\t#side slope\n", - "K = 1.3; \t\t\t\t#bazin's coefficient\n", - "\t\t\t\t#width is five times its depth\n", - "\n", - " \n", - "# Calculations\n", - "A = Q/V;\n", - "D = (A/6)**0.5;\n", - "B = 5*D;\n", - "P = B+2*D*1.41;\n", - "R = A/P;\n", - "C = 87/(1+K/(R)**0.5);\n", - "S = (V/C)**2/R;\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 9.10 m.\n", - "Depth of channel section = 1.83 m.\n", - "Bed slope = 2.34e-04.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.19 pg : 698" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#determine dimension of channel umath.sing chezy's equation\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 21.5; \t\t\t\t#discharge\n", - "S = 1./2500; \t\t\t\t#slope of bottom\n", - "C = 70;\n", - "r = 1/1.73;\n", - "#taking R = 0.5*D\n", - "\n", - "# Calculations and Results\n", - "#and keeping it in Q = V*A;where V = C(RS)**0.5 and A = D**2(2*(4/3)**0.5-1/3**0.5);\n", - "D = (21.5/1.7146)**(1/2.5);\n", - "B = 2*D*((4./3)**0.5-(1./3)**0.5);\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "print \"side slope = %.2f.\"%(r);\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n", - "R = 0.5*D;\n", - "V = C*(R*S)**0.5;\n", - "n = R**(2./3)*S**0.5/V;\n", - "n = round(n*1000)/1000;\n", - "print \"value of manning n = %.2f.\"%(n);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "side slope = 0.58.\n", - "Width of channel section = 3.18 m.\n", - "Depth of channel section = 2.75 m.\n", - "value of manning n = 0.01.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.20 pg : 698" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t#design a regime channel\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 100; \t\t\t\t#discharge\n", - "f = 1.1; \t\t\t\t#silt factor\n", - "s = 1./2; \t\t\t\t#side slope\n", - "\n", - "\n", - "# Calculations\n", - "V = (Q*f**2/140)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*Q**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "B = P-2.236*D;\n", - "R = 5*V**2/(2*f);\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 42.20 m.\n", - "Depth of channel section = 2.36 m.\n", - "Bed slope = 1.63e-04.\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.21 pg : 699" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import poly1d,roots\n", - "#design a channel umath.sing Laecy theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 40.; \t\t\t\t#discharge\n", - "s = 1.; \t\t\t\t#side slope\n", - "md = 0.8; \t\t\t\t#average size of base material\n", - "\n", - "\n", - "# Calculations\n", - "f = 1.76*(md)**0.5;\n", - "V = (Q*f**2/140)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*Q**0.5;\n", - "#from equations of Area and perimeter of trapezoidal section;we get\n", - "y = poly1d([42.41,-30.04,1.828],'x','c');\n", - "D = roots(y)[0];\n", - "\t\t\t\t#we get D = 14.873416 and 1.5598447\n", - "\t\t\t\t#taking\n", - "D = 1.5598447;\n", - "B = A/D-D;\n", - "R = 5*V**2/(2*f);\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 25.60 m.\n", - "Depth of channel section = 1.56 m.\n", - "Bed slope = 3.45e-04.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.22 pg : 700" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 30.; \t\t\t\t#discharge\n", - "V = 1.; \t\t\t\t#velocity of flow\n", - "\n", - "\n", - "# Calculations\n", - "A = Q/V;\n", - "#perimeter of section = 30/D-D/2\n", - "#taking its derivative w.r.t to D\n", - "D = 1/((1.914/30)**0.5);\n", - "#from equation of area\n", - "B = 30/D-D/2;\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 5.60 m.\n", - "Depth of channel section = 3.96 m.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.23 pg : 700" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#determine whether flow is critical or sub-critical\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 17.; \t\t\t\t#discharge\n", - "B = 6.; \t\t\t\t#base of channel\n", - "s = 1./2; \t\t\t\t#side slope\n", - "D = 1.5; \t\t\t\t#depth of channel\n", - "\n", - "\n", - "# Calculations\n", - "A = D*((B+B/s)/2);\n", - "V = Q/A;\n", - "P = B+2*((D/s)**2+D**2)**0.5;\n", - "R = A/P;\n", - "F = V/(9.81*R)**0.5; \t\t\t\t#froud number\n", - "F = round(F*100)/100;\n", - "\n", - "# Results\n", - "#math.since F<1;\n", - "print \"Froud number = %.2f.F<1.Flow is sub-critical\"%(F);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Froud number = 0.39.F<1.Flow is sub-critical\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.24 pg : 701" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "B = 3.5; \t\t\t\t#bottom width of channel\n", - "n = 0.016; \t\t\t\t#manning n\n", - "S = 2.6/10000; \t\t\t\t#bed slope\n", - "Q = 8; \t\t\t\t#discharge\n", - "lfs = 1; \t\t\t\t#left side slope\n", - "rhs = 1.5; \t\t\t\t#rigth side slope\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "\n", - "\n", - "# Calculations\n", - "#umath.sing the equation of area and perimeter of trapezoidal section;Manning's formula and V = Q/A we get D as\n", - "#Manning formula: V = R**(2/3)*S**0.5/n\n", - "#(D*(3.5+1.25*D))**2.5 = 78.281+71.951*D\n", - "#solving it by trial and error method;we get\n", - "D = 1.5;\n", - "R = (D*(3.5+1.25*D))/(3.5+3.217*D);\n", - "tau = gamma_w*R*S*1000;\n", - "tau = round(tau*100)/100;\n", - "\n", - "# Results\n", - "print \"Depth of section = %.2f m.\"%(D);\n", - "print \"Average shear stress at channel bed = %.2f N/square-mm.\"%(tau);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Depth of section = 1.50 m.\n", - "Average shear stress at channel bed = 2.47 N/square-mm.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.25 pg : 702" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "S = 1./5000; \t\t\t\t#bed slope\n", - "B = 40.; \t\t\t\t#width of channel\n", - "D = 2.6; \t\t\t\t#depth of channel\n", - "d = 0.38; \t\t\t\t#mean diameter of bed material\n", - "n = 0.021; \t\t\t\t#Manning n\n", - "D65 = 0.64e-3; \t\t\t\t#bed material size(m)\n", - "w = 1000.; \t\t\t\t#density of water\n", - "\n", - "# Calculations\n", - "#B/D as large tau_c = 0.075*d;\n", - "tau_c = 0.075*d;\n", - "tau_b = w*D*S;\n", - "N1 = (D65)**(1./6)/24;\n", - "r = N1/n;\n", - "qs = 4700*24*(tau_b*r**1.5-tau_c)**1.5/1000;\n", - "qs40 = qs*40;\n", - "\n", - "# Results\n", - "print \"bed load transported by the channel = %i t/m/day.\"%(qs40);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "bed load transported by the channel = 411 t/m/day.\n" - ] - } - ], - "prompt_number": 25 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.26 pg : 702" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "Q = 5.; \t\t\t\t#discharge\n", - "S = 0.2/1000; \t\t\t\t#bed slope\n", - "m = 0.8; \t\t\t\t#critical velocity ratio\n", - "s = 1./2; \t\t\t\t#side slope of chanel\n", - "C = 30.;\n", - "#assuming\n", - "\n", - "# Calculations\n", - "D = 1.;\n", - "Vo = 0.55*m*D**0.64;\n", - "A = Q/Vo;\n", - "B = A/D-(s*D);\n", - "P = B+2.43*D;\n", - "R = A/P;\n", - "V = C*(R*S)**0.5;\n", - "#Vo>V\n", - "#hence take second trial\n", - "D = 0.8; \t\t\t\t#assume\n", - "Vo = 0.55*m*D**0.64;\n", - "A = Q/Vo;\n", - "B = A/D-(s*D);\n", - "P = B+2.43*D;\n", - "R = A/P;\n", - "V = C*(R*S)**0.5;\n", - "#again Vo>V\n", - "#hence we take third trial\n", - "D = 0.7;\n", - "Vo = 0.55*m*D**0.64;\n", - "A = Q/Vo;\n", - "B = A/D+(s*D);\n", - "P = B+2.43*D;\n", - "R = A/P;\n", - "V = C*(R*S)**0.5;\n", - "B = round(B*100)/100;\n", - "#Vo is almost equal to V;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 20.75 m.\n", - "Depth of channel section = 0.70 m.\n" - ] - } - ], - "prompt_number": 26 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.27 pg : 703" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design irrigation channel by Kennedy method\n", - "\n", - "#Given\n", - "Q = 50.; \t\t\t\t#discharge\n", - "r = 2.5; \t\t\t\t#B/D ratio\n", - "m = 1.1; \t\t\t\t#critical velocity ratio\n", - "N = 0.025; \t\t\t\t#rogosity coefficient\n", - "s = 0.5; \t\t\t\t#side slope of channel\n", - "\n", - "\n", - "# Calculations\n", - "#umath.sing the equation of Vo and Q = A*V;we get\n", - "D = (Q/1.815)**(1/2.64);\n", - "B = r*D;\n", - "R = (B*D+0.5*D**2)/(B+2.236*D);\n", - "Vo = 0.55*m*D**0.64;\n", - "\n", - "#applying kutters formula; V = C(RS)**0.5\n", - "#where C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "#assuming S**0.5 = y\n", - "#taking real values of y\n", - "S = 0.0196171 **2;\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 8.78 m.\n", - "Depth of channel section = 3.51 m.\n", - "Bed slope = 3.85e-04.\n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.28 pg : 704" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a regime channel umath.sing Laecy's theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 35; \t\t\t\t#discharge\n", - "f = 0.9; \t\t\t\t#silt factor\n", - "s = 1./2; \t\t\t\t#side slope\n", - "\n", - "\n", - "# Calculations\n", - "V = (Q*f/140)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*Q**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "B = P-2.236*D;\n", - "\n", - "R = 5*V**2/(2*f);\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Width of channel section = %i m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 1.39e-04.\n", - "Width of channel section = 24 m.\n", - "Depth of channel section = 1.80 m.\n" - ] - } - ], - "prompt_number": 28 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.29 pg : 705" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#design an irrigation canal for given data\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 15.; \t\t\t\t#discharge\n", - "m = 1.; \t\t\t\t#critical velocity ratio\n", - "r = 5.7; \t\t\t\t#B/D\n", - "\n", - "D = (Q/(0.55*6.2))**(1/2.64);\n", - "B = D*r;\n", - "R = (B*D+D**2/2)/(B+D*5**0.5);\n", - "Vo = 0.55*m*D**0.64;\n", - "#applying kutters formula; V = C(RS)**0.5\n", - "#where C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "#assuming S**0.5 = y\n", - "y = [67.5,-0.968,1.55e-3,-2e-5]\n", - "d = roots(y)[0];\n", - "\t\t\t\t#taking real values of y\n", - "S = d.real**2;\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 9.99 m.\n", - "Depth of channel section = 1.75 m.\n", - "Bed slope = 2.01e-04.\n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.30 pg : 705" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Design a section of unlined canal in a loomy soil\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 50.; \t\t\t\t#discharge\n", - "V = 1.; \t\t\t\t#permissible velocity\n", - "s = 2.; \t\t\t\t#side slope\n", - "r = 6.; \t\t\t\t#B/D ratio\n", - "N = 0.0225; \t\t\t\t#rogosity coefficient\n", - "\n", - "\n", - "# Calculations\n", - "A = Q/V;\n", - "D = (A/(r+2))**0.5;\n", - "B = r*D;\n", - "P = B+2*(5*D**2)**0.5;\n", - "R = A/P;\n", - "S = (V*N/R**(2/3))**2;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %i m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 15 m.\n", - "Depth of channel section = 2.50 m.\n", - "Bed slope = 5.06e-04.\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.31 pg : 705" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "D = 5.; \t\t\t\t#depth of channel\n", - "d = 0.3; \t\t\t\t#grain size\n", - "k = 1.5; \t\t\t\t#size of roughness of channel bed\n", - "S = 1./4000; \t\t\t\t#bed slope\n", - "G = 2.65; \t\t\t\t#specific gravity\n", - "V = 0.02; \t\t\t\t#fall velocity\n", - "c_ = 1000.; \t\t\t\t#concentration at 0.3 m above bed\n", - "a = 0.3;\n", - "y = 2.5;\n", - "k_ = 0.4; \t\t\t\t#van karman's consmath.tant\n", - "\n", - "\n", - "# Calculations\n", - "R = 5; \t\t\t\t#R = D for wide channel\n", - "V_ = (gamma_w*R*S)**0.5;\n", - "c = c_*((a/y)*(D-y)/(D-a))**(V/(V_*k_));\n", - "\n", - "# Results\n", - "print \"concentration of suspended load = %i ppm.\"%(c);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "concentration of suspended load = 288 ppm.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.32 pg : 706" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 40.; \t\t\t\t#discharge\n", - "f = 1.; \t\t\t\t#silt factor\n", - "\n", - "\n", - "# Calculations and Results\n", - "#Laecey's theory\n", - "V = (Q*f/140)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*Q**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "B = P-2.236*D;\n", - "\n", - "R = 5*V**2/(2*f);\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "B = round(B);\n", - "D = round(D*100)/100;\n", - "print \"By Laecey theory:\";\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n", - "#Kennedy's theory\n", - "r = B/D; \n", - "m = 1.; \t\t\t\t#critical velocity ratio\n", - "N = 0.0225; \t\t\t\t#rogosity coefficient\n", - "#umath.sing equation of area of trapezoidal section;Vo = 0.55mD**0.64 and Q = A*Vo\n", - "\n", - "D = (Q/8.058)**(1/2.64);\n", - "B = r*D;\n", - "B = round(B);\n", - "D = round(D*100)/100;\n", - "print \"By Kennedy theory:\";\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "By Laecey theory:\n", - "Bed slope = 1.62e-04.\n", - "Width of channel section = 26.00 m.\n", - "Depth of channel section = 1.84 m.\n", - "By Kennedy theory:\n", - "Width of channel section = 26.00 m.\n", - "Depth of channel section = 1.83 m.\n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.33 pg : 707" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design Laecey regime channel\n", - "\t\t\t\t\n", - "#Given\n", - "A = 100000.; \t\t\t\t#culturable area(hectare)\n", - "IR = 0.4; \t\t\t\t#intensity of irrigation in kharif season\n", - "IK = 0.3; \t\t\t\t#intensity of irrigation in rabi season\n", - "OR = 1800.; \t\t\t\t#outlet discharge factor in kharif season\n", - "OK = 800.; \t\t\t\t#outlet discharge factor in kharif season\n", - "l = 0.1; \t\t\t\t#conveyance loss\n", - "md = 0.328; \t\t\t\t#average diameter of material\n", - "\n", - "# Calculations\n", - "AR = A*IR; \t\t\t\t#area under rabi\n", - "AK = A*IK; \t\t\t\t#area under kharif \n", - "Qr = AR/OR;\n", - "Qk = AK/OK;\n", - "Q = 1.1*Qk;\n", - "f = 1.76*(md)**0.5;\n", - "V = (Q*f**2/144)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*(Q)**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "B = P-2.236*D;\n", - "S = f**(5/3)/(3340*Q**(1./6));\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 1.62e-04.\n", - "Width of channel section = 26.40 m.\n", - "Depth of channel section = 1.86 m.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.34 pg : 707" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "D = 2.8; \t\t\t\t#depth of flow\n", - "c_ = 700; \t\t\t\t#concentration at 30 cm below water surface\n", - "y = 0.1;\n", - "a = D-0.3;\n", - "e = 0.4; \t\t\t\t#exponent in rouse equation;\n", - "\n", - "# Calculations\n", - "c = c_*(a*(D-y)/(y*(D-a)))**e;\n", - "\n", - "# Results\n", - "print \"concentration at point 10 cm above the bed = %i ppm.\"%(c);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "concentration at point 10 cm above the bed = 6109 ppm.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.35 pg : 708" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design the distributory umath.sing Laecey theory\n", - "\t\t\t\t\n", - "#Given\n", - "f = 0.85; \t\t\t\t#silt factor\n", - "AR = 3600.; \t\t\t\t#area for rabi\n", - "AK = 1400.; \t\t\t\t#area for kharif\n", - "delta_r = 0.135; \t\t\t\t#kor depth for rabi\n", - "delta_k = 0.19; \t\t\t\t#kor depth for kharif\n", - "tr = 4.; \t\t\t\t#kor period for rabi\n", - "tk = 2.5; \t\t\t\t#kor period for kharif\n", - "\n", - "# Calculations\n", - "Du_r = 8.64*tr*7/delta_r; \t\t\t\t#duty for rabi\n", - "Du_k = 8.64*tk*7/delta_k; \t\t\t\t#duty for kharif\n", - "q_r = AR/Du_r; \t\t\t\t#discharge for rabi\n", - "q_k = AK/Du_k; \t\t\t\t#discharge for kharif\n", - "Q = q_r; \t\t\t\t#math.since q_r>q_k\n", - "V = (Q*f**2/144)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*(Q)**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "P = round(P*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Perimeter of channel section = %.2f m.\"%(P);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 2.03e-04.\n", - "Perimeter of channel section = 6.73 m.\n", - "Depth of channel section = 0.81 m.\n" - ] - } - ], - "prompt_number": 35 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch14_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch14_1.ipynb deleted file mode 100755 index f55e341a..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch14_1.ipynb +++ /dev/null @@ -1,1965 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:e8ae669d88fc715c49999174af0f47a67eb761cc981a916d61f3cb981ae07f2c" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 14 : IRRIGATION CHANNEL 1 SILT THEORIES" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.1 pg : 662" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design irrigation channel on Kennedy's theory\n", - "\n", - "#Given\n", - "Q = 45; \t\t\t\t#discharge\n", - "N = 0.0225; \t\t\t\t#rogosity coefficient\n", - "m = 1.05; \t\t\t\t#critical velocity ratio\n", - "S = 1./5000; \t\t\t\t#bed slope\n", - "\n", - "D = 2; \t\t\t\t#assume\n", - "Vo = 0.55*m*D**0.64;\n", - "A = Q/Vo;\n", - "#for a trapezoidal section\n", - "B = (A-0.5*D**2)/2;\n", - "P = B+D*5**0.5;\n", - "R = A/P;\n", - "C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "V = C*(R*S)**0.5;\n", - "#Vo<V\n", - "\n", - "#assume D = 2.2\n", - "D = 2.2;\n", - "Vo = 0.55*m*D**0.64;\n", - "A = Q/Vo;\n", - "B = (A-0.5*D**2)/D;\n", - "P = B+D*5**0.5;\n", - "R = A/P;\n", - "C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "V = C*(R*S)**0.5;\n", - "\n", - "#ratio of V and Vo is almost equal to 1\n", - "\n", - "# Results\n", - "B = round(B*10)/10;\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 20.30 m.\n", - "Depth of channel section = 2.20 m.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.2 pg : 663" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#design an irrigation canal for given data\n", - "\n", - "#Given\n", - "Q = 14; \t\t\t\t#discharge\n", - "m = 1; \t\t\t\t#critical velocity ratio\n", - "r = 5.7; \t\t\t\t#B/D\n", - "\n", - "D = (Q/(0.55*6.2))**(1/2.64);\n", - "B = D*r;\n", - "R = (B*D+D**2/2)/(B+D*5**0.5);\n", - "Vo = 0.55*m*D**0.64;\n", - "#applying kutters formula; V = C(RS)**0.5\n", - "#where C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "#assuming S**0.5 = y\n", - "#y = poly1d([-1.98D-5,1.55D-3,-0.954,67.5],'x','c');\n", - "#roots(y);\n", - "#taking real values of y\n", - "S = 0.0139906**2;\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 9.73 m.\n", - "Depth of channel section = 1.71 m.\n", - "Bed slope = 1.96e-04.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.3 pg : 668" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a channel on Kennedy's theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "m = 1.05; \t\t\t\t#critical velocity ratio\n", - "N = 0.025; \t\t\t\t#rugosity coefficient\n", - "S = 1./5000; \t\t\t\t#bed slope\n", - "\n", - "\n", - "# Calculations\n", - "l = S*Q**0.02/(N**2*m**2.02);\n", - "#from fig.14.3 we get r = 10\n", - "#solving the equation by trial and error method we get\n", - "r = 9.7;\n", - "D = (1.818*Q/(m*(r+0.5)))**(1/2.64);\n", - "B = r*D;\n", - "V = Q/(D**2*(r+0.5));\n", - "Vo = 0.55*D**0.64*m;\n", - "B = round(B);\n", - "D = round(D*100)/100;\n", - "V = round(V*1000)/1000;\n", - "Vo = round(Vo*1000)/1000;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Velocity through the channel section = %.2f m/s.\"%(V);\n", - "print \"Vo = %.2f m/s.Hence Safe\"%(Vo);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 21.00 m.\n", - "Depth of channel section = 2.16 m.\n", - "Velocity through the channel section = 0.94 m/s.\n", - "Vo = 0.94 m/s.Hence Safe\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.4 pg : 668" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#design channel umath.sing method of curve fitting based onKennedy's theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "N = 0.0225; \t\t\t\t#rugosity coefficient\n", - "m = 1.05; \t\t\t\t#critical velocity ratio\n", - "S = 1./5000; \t\t\t\t#Bed slope\n", - "\n", - "r = (1.607*S**1.63*Q**0.033/(N**3.26*m**3.293)-0.258)**(-0.915);\n", - "D = (1.818*Q/(m*(r+0.5)))**(1/2.64);\n", - "B = r*D;\n", - "B = round(B);\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 21.00 m.\n", - "Depth of channel section = 2.16 m.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.5 pg : 668" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#design channel umath.sing curve of CWPC for B/D ratio\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "N = 0.0225; \t\t\t\t#rugosity coefficient\n", - "m = 1.05; \t\t\t\t#critical velocity ratio\n", - "\n", - "r = (15+6.44*Q)**0.382;\n", - "S = (N**2/1.338*Q**0.02)*(0.258+(15+6.44*Q)**(-0.417))**0.6135;\n", - "D = (1.818*Q/(m*(r+0.5)))**(1/2.64);\n", - "B = r*D;\n", - "B = round(B);\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 2.14e-04.\n", - "Width of channel section = 20.00 m.\n", - "Depth of channel section = 2.23 m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.6 pg : 667" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#design the channel section umath.sing the following data and calculate math.logitudnal section\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 30.; \t\t\t\t#discharge\n", - "f = 1.; \t\t\t\t#silt factor\n", - "s = 1./2; \t\t\t\t#side slope\n", - "\n", - "V = (Q*f/140)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*Q**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "B = P-2.236*D;\n", - "\n", - "R = 5*V**2/(2*f);\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 1.70e-04.\n", - "Width of channel section = 22.26 m.\n", - "Depth of channel section = 1.68 m.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.7 pg : 682" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#design a channel in alluvial soil umath.sing tractive force approach\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "S = 1./4800; \t\t\t\t#bed slope\n", - "N = 0.0225; \t\t\t\t#rogosity coefficient\n", - "sigma = 0.0035; \t\t\t\t#permissible tractive stress\n", - "s = 1./2; \t\t\t\t#side slope\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "\n", - "\n", - "# Calculations\n", - "R = sigma/(gamma_w*S);\n", - "V = R**(2./3)*S**0.5/N;\n", - "A = Q/V;\n", - "P = A/R;\n", - "#y = poly1d([-49,28.61,-1.736],'x','c');\n", - "y = [-1.736,28.61,-49]\n", - "D = roots(y)[1];\n", - "#we get D = 14.539034 and 1.9413812 \n", - "#taking D = 1.9413812 \n", - "B = 28.61-2.23*D;\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 24.28 m.\n", - "Depth of channel section = 1.94 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.8 pg : 687" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#designa channel section by Kennedy theory\n", - "\n", - "#Given\n", - "Q = 28.; \t\t\t\t#discharge\n", - "m = 1.; \t\t\t\t#critical velocity ratio\n", - "r = 7.6; \t\t\t\t#B/D\n", - "\n", - "D = (Q/4.46)**(1/2.64);\n", - "B = r*D;\n", - "R = 0.823*D;\n", - "V = 0.55*(D)**0.64;\n", - "\n", - "#applying kutters formula; V = C(RS)**0.5\n", - "#where C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "#we get equation in S\n", - "#assuming S**0.5 = y\n", - "#y = poly([-1.42D-5,1.55D-3,-0.885,67.4],'x','c');\n", - "y = [67.4,-0.885,1.55e-3,-1.42e-5]\n", - "d = roots(y)[0];\n", - "#taking real values of y\n", - "S = d.real**2;\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 15.20 m.\n", - "Depth of channel section = 2.01 m.\n", - "Bed slope = 1.60e-04.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.9 pg : 688" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design the channel section and calculate discharge\n", - "\t\t\t\t\n", - "#Given\n", - "r = 5.7; \t\t\t\t#B/D\n", - "S = 1./5000; \t\t\t\t#bed slope\n", - "N = 0.0225; \t\t\t\t#rogosity coefficient\n", - "m = 1; \t\t\t\t#critical velocity ratio(assumed)\n", - "\n", - "#applying kutters formula; V = C(RS)**0.5\n", - "#where C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "#we get equation in d as\n", - "#38.88*D**0.64-66.5*D**0.5+30.37*D**0.14 = 0\n", - "#solving it by trial and error method\n", - "#we get D = 1.7 m.\n", - "D = 1.7;\n", - "B = r*D;\n", - "V = 0.55*m*(D)**0.64;\n", - "A = B*D+D**2/2;\n", - "Q = A*V;\n", - "Q = round(Q*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \" Discharge = %.2f cumecs.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 9.69 m.\n", - "Depth of channel section = 1.70 m.\n", - " Discharge = 13.84 cumecs.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.10 pg : 689" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#design irrigation channel according to Laecy silt theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 15.; \t\t\t\t#discharge\n", - "f = 1.; \t\t\t\t#laecy silt factor\n", - "s = 1./2; \t\t\t\t#channel side slope\n", - "\n", - "V = (Q*f**2/140);\n", - "A = Q/V;\n", - "R = 5*V**2/(2*f);\n", - "#umath.sing the value of A in equations we get,\n", - "#equation in D as\n", - "y = [-1.73,18.336,-21.765]\n", - "D = roots(y)[1];\n", - "\n", - "#we get D = 9.2368003 and 1.3620436.\n", - "#taking\n", - "D = 1.3620436;\n", - "B = 18.336-D*2.23;\n", - "P = 4.75*Q**0.5;\n", - "S = 1/(3340*Q**(1./6));\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n", - " \n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 15.30 m.\n", - "Depth of channel section = 1.36 m.\n", - "Bed slope = 1.91e-04.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.11 pg : 690" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "#find channel section and discharge\n", - "\t\t\t\t\n", - "#Given\n", - "S = 1./5000; \t\t\t\t#bed slope\n", - "s = 1./2; \t\t\t\t#side slope\n", - "f = 0.9; \t\t\t\t#laecy silt factor\n", - "\n", - "Q = (f**(5./3)/(3340*S))**6;\n", - "R = f**3/(4980*S)**2;\n", - "P = 4.75*Q**0.5;\n", - "A = P*R;\n", - "#umath.sing the value of A and P in equations we get,\n", - "#equation in D as\n", - "y = [-1.73,9.41,-6.961]\n", - "D = roots(y)[1];\n", - "\n", - "#we get D = 4.5561754 and 0.8831309.\n", - "#taking D = 0.8831309;\n", - "B = 9.41-D*2.23;\n", - "B = round(B*100)*100;\n", - "D = round(D*100)/100;\n", - "Q = round(Q*1000)/1000;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \" Discharge = %.2f cumecs.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 74400.00 m.\n", - "Depth of channel section = 0.88 m.\n", - " Discharge = 3.92 cumecs.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.12 pg : 691" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "D = 3.; \t\t\t\t#depth of channel\n", - "d = 0.3; \t\t\t\t#grain size\n", - "k = 1.5; \t\t\t\t#size of roughness of channel bed\n", - "S = 1./4400; \t\t\t\t#bed slope\n", - "G = 2.65; \t\t\t\t#specific gravity\n", - "\n", - "# Calculations\n", - "tau_b = gamma_w*D*S;\n", - "N1 = d**(1./6)/24;\n", - "N = k**(1./6)/24;\n", - "gamma_s = gamma_w*G;\n", - "tau_c = 0.047*(gamma_s-gamma_w)*d/1000;\n", - "r = (N1/N)**1.5;\n", - "q = 47450*(tau_b*r-tau_c)**1.5;\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"quantity of bed load moved = %.2f kN/m/hr.\"%(q);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "quantity of bed load moved = 13.12 kN/m/hr.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.13 pg : 691" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "D = 3.; \t\t\t\t#depth of channel\n", - "d = 0.3; \t\t\t\t#grain size\n", - "k = 1.5; \t\t\t\t#size of roughness of channel bed\n", - "S = 1./4400; \t\t\t\t#bed slope\n", - "G = 2.65; \t\t\t\t#specific gravity\n", - "\n", - "# Calculations\n", - "N1 = d**(1./6)/24;\n", - "N = k**(1./6)/24;\n", - "r = (N1/N)**1.5;\n", - "R1 = 3*r;\n", - "si = (G-1)*d/(1000*R1*S);\n", - "\t\t\t\t#hence we get\n", - "fi = 7;\n", - "q = 3600*fi*G*gamma_w*(G-1)**0.5*(gamma_w)**0.5*(d/1000)**1.5;\n", - "q = round(q*10)/10;\n", - "\n", - "# Results\n", - "print \"quantity of bed load moved = %.2f kN/m/hr.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "quantity of bed load moved = 13.70 kN/m/hr.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.14 pg : 692" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "D = 3.; \t\t\t\t#depth of channel\n", - "d = 0.3; \t\t\t\t#grain size\n", - "k = 1.5; \t\t\t\t#size of roughness of channel bed\n", - "S = 1./4400; \t\t\t\t#bed slope\n", - "G = 2.65; \t\t\t\t#specific gravity\n", - "V = 0.03; \t\t\t\t#fall velocity\n", - "c_ = 400; \t\t\t\t#concentration at 0.3 m above bed\n", - "a = 0.3;\n", - "y = 1.;\n", - "k_ = 0.4; \t\t\t\t#van karman's consmath.tant\n", - "\n", - "\n", - "# Calculations\n", - "N1 = d**(1./6)/24;\n", - "N = k**(1./6)/24;\n", - "r = (N1/N)**1.5;\n", - "R1 = 3*r;\n", - "V_ = (gamma_w*R1*S)**0.5;\n", - "c = c_*((a/y)*(D-y)/(D-a))**(V/(V_*k_));\n", - "c = round(c*10)/10;\n", - "\n", - "# Results\n", - "print \"concentration of suspended load = %.2f ppm.\"%(c);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "concentration of suspended load = 74.10 ppm.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.15 pg : 692" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "\t\t\t\t#design an irrigation channel by Meyer peter equation\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45; \t\t\t\t#discharge\n", - "c = 55; \t\t\t\t#bed load concentraion\n", - "d = 0.3; \t\t\t\t#average grain diameter\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "G = 2.67;\n", - "f = 0.964;\n", - "\n", - "c = c*Q*gamma_w*3600/1000000;\n", - "P = 4.75*Q**0.5;\n", - "\t\t\t\t#taking channel width as B = 28 m(slightly less than P)\n", - "B = 28.;\n", - "qs = c/B;\n", - "\t\t\t\t#assuming effective grain diameter k = 0.4 mm\n", - "ks = 0.4e-3;\n", - "N1 = ks**(1./6)/24;\n", - "sf = 1.76*d**0.5;\n", - "N = 0.0225*sf**0.25;\n", - "r = N1/N;\n", - "tau_c = 0.047*gamma_w*(G-1)*d/1000;\n", - "tau_b = r**1.5*((qs/47450)**(2./3)+tau_c);\n", - "\t\t\t\t#from Manning's formula we get on simplification\n", - "R = (0.000992*1000/0.525)**(3./7);\n", - "S = 0.525/(1000*R);\n", - "\t\t\t\t#solving equation of R for trapezoidal section of side slope 1/2 we get\n", - "y = [0.5,25.06,-36.792]\n", - "D = roots(y)[1];\n", - "#we get D = -51.547499 and 1.4274989 \n", - "#taking D = 1.4274989;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %i m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 28 m.\n", - "Depth of channel section = 1.43 m.\n", - "Bed slope = 4.00e-04.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.16 pg : 693" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "\n", - "#design an irrigation channel by Einstein equation\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "c = 55.; \t\t\t\t#bed load concentraion\n", - "d = 0.3; \t\t\t\t#average grain diameter\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "G = 2.67; \t\t\t\t#specific gravity of soil\n", - "f = 0.964; \t\t\t\t#silt factor\n", - "\n", - "#taking channel width as B = 28 m(slightly less than P)\n", - "B = 28.;\n", - "qs = c/B;\n", - "\n", - "fi = (qs/(gamma_w*G))*(1/(G-1))**0.5*(1000000000/(gamma_w*d**3))**0.5;\n", - "#from fig. 14.6 we get value of sci\n", - "#umath.sing the sci equation and Manning formula and on simplifications we get\n", - "R = (2.4296)**(3./7);\n", - "S = 0.4083/(1000*1.463);\n", - "#solving equation of R for trapezoidal section of side slope 1/2 we get\n", - "y = [0.5,24.73,-40.96]\n", - "D = roots(y)[1];\n", - "#we get D = -51.064253 and 1.6042534 \n", - "#taking D = 1.6042534;\n", - "D = round(D*10)/10;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %i m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 28 m.\n", - "Depth of channel section = 1.60 m.\n", - "Bed slope = 2.79e-04.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.17 pg : 697" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "#design a channel for non-alluvial deposites\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "S = 1./4000; \t\t\t\t#bed slope\n", - "v = 0.9; \t\t\t\t#permissible velocity\n", - "N = 0.025; \t\t\t\t#rogosity coefficient\n", - "\n", - "\n", - "# Calculations\n", - "A = Q/v;\n", - "R = (v*N/S**0.5)**1.5;\n", - "P = A/R;\n", - "\t\t\t\t#let us provide a trapezoidal section\n", - "\t\t\t\t#from equation of Area and Perimeter of trapezoid\n", - "y = [1.828,-29.45,50]\n", - "D = roots(y)[1];\n", - "#from which we get D = 14.181815 and 1.9286881\n", - "#taking D = 1.9286881;\n", - "B = P-2*1.41*D;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %i m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 24 m.\n", - "Depth of channel section = 1.93 m.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.18 pg : 697" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design non-allvial channel umath.sing Bazin's formula\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 15.; \t\t\t\t#discharge\n", - "V = 0.75; \t\t\t\t#mean velocity\n", - "s = 1.; \t\t\t\t#side slope\n", - "K = 1.3; \t\t\t\t#bazin's coefficient\n", - "\t\t\t\t#width is five times its depth\n", - "\n", - " \n", - "# Calculations\n", - "A = Q/V;\n", - "D = (A/6)**0.5;\n", - "B = 5*D;\n", - "P = B+2*D*1.41;\n", - "R = A/P;\n", - "C = 87/(1+K/(R)**0.5);\n", - "S = (V/C)**2/R;\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 9.10 m.\n", - "Depth of channel section = 1.83 m.\n", - "Bed slope = 2.34e-04.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.19 pg : 698" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#determine dimension of channel umath.sing chezy's equation\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 21.5; \t\t\t\t#discharge\n", - "S = 1./2500; \t\t\t\t#slope of bottom\n", - "C = 70;\n", - "r = 1/1.73;\n", - "#taking R = 0.5*D\n", - "\n", - "# Calculations and Results\n", - "#and keeping it in Q = V*A;where V = C(RS)**0.5 and A = D**2(2*(4/3)**0.5-1/3**0.5);\n", - "D = (21.5/1.7146)**(1/2.5);\n", - "B = 2*D*((4./3)**0.5-(1./3)**0.5);\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "print \"side slope = %.2f.\"%(r);\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n", - "R = 0.5*D;\n", - "V = C*(R*S)**0.5;\n", - "n = R**(2./3)*S**0.5/V;\n", - "n = round(n*1000)/1000;\n", - "print \"value of manning n = %.2f.\"%(n);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "side slope = 0.58.\n", - "Width of channel section = 3.18 m.\n", - "Depth of channel section = 2.75 m.\n", - "value of manning n = 0.01.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.20 pg : 698" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t#design a regime channel\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 100; \t\t\t\t#discharge\n", - "f = 1.1; \t\t\t\t#silt factor\n", - "s = 1./2; \t\t\t\t#side slope\n", - "\n", - "\n", - "# Calculations\n", - "V = (Q*f**2/140)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*Q**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "B = P-2.236*D;\n", - "R = 5*V**2/(2*f);\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 42.20 m.\n", - "Depth of channel section = 2.36 m.\n", - "Bed slope = 1.63e-04.\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.21 pg : 699" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import poly1d,roots\n", - "#design a channel umath.sing Laecy theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 40.; \t\t\t\t#discharge\n", - "s = 1.; \t\t\t\t#side slope\n", - "md = 0.8; \t\t\t\t#average size of base material\n", - "\n", - "\n", - "# Calculations\n", - "f = 1.76*(md)**0.5;\n", - "V = (Q*f**2/140)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*Q**0.5;\n", - "#from equations of Area and perimeter of trapezoidal section;we get\n", - "y = poly1d([42.41,-30.04,1.828],'x','c');\n", - "D = roots(y)[0];\n", - "\t\t\t\t#we get D = 14.873416 and 1.5598447\n", - "\t\t\t\t#taking\n", - "D = 1.5598447;\n", - "B = A/D-D;\n", - "R = 5*V**2/(2*f);\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 25.60 m.\n", - "Depth of channel section = 1.56 m.\n", - "Bed slope = 3.45e-04.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.22 pg : 700" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 30.; \t\t\t\t#discharge\n", - "V = 1.; \t\t\t\t#velocity of flow\n", - "\n", - "\n", - "# Calculations\n", - "A = Q/V;\n", - "#perimeter of section = 30/D-D/2\n", - "#taking its derivative w.r.t to D\n", - "D = 1/((1.914/30)**0.5);\n", - "#from equation of area\n", - "B = 30/D-D/2;\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 5.60 m.\n", - "Depth of channel section = 3.96 m.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.23 pg : 700" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#determine whether flow is critical or sub-critical\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 17.; \t\t\t\t#discharge\n", - "B = 6.; \t\t\t\t#base of channel\n", - "s = 1./2; \t\t\t\t#side slope\n", - "D = 1.5; \t\t\t\t#depth of channel\n", - "\n", - "\n", - "# Calculations\n", - "A = D*((B+B/s)/2);\n", - "V = Q/A;\n", - "P = B+2*((D/s)**2+D**2)**0.5;\n", - "R = A/P;\n", - "F = V/(9.81*R)**0.5; \t\t\t\t#froud number\n", - "F = round(F*100)/100;\n", - "\n", - "# Results\n", - "#math.since F<1;\n", - "print \"Froud number = %.2f.F<1.Flow is sub-critical\"%(F);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Froud number = 0.39.F<1.Flow is sub-critical\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.24 pg : 701" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "B = 3.5; \t\t\t\t#bottom width of channel\n", - "n = 0.016; \t\t\t\t#manning n\n", - "S = 2.6/10000; \t\t\t\t#bed slope\n", - "Q = 8; \t\t\t\t#discharge\n", - "lfs = 1; \t\t\t\t#left side slope\n", - "rhs = 1.5; \t\t\t\t#rigth side slope\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "\n", - "\n", - "# Calculations\n", - "#umath.sing the equation of area and perimeter of trapezoidal section;Manning's formula and V = Q/A we get D as\n", - "#Manning formula: V = R**(2/3)*S**0.5/n\n", - "#(D*(3.5+1.25*D))**2.5 = 78.281+71.951*D\n", - "#solving it by trial and error method;we get\n", - "D = 1.5;\n", - "R = (D*(3.5+1.25*D))/(3.5+3.217*D);\n", - "tau = gamma_w*R*S*1000;\n", - "tau = round(tau*100)/100;\n", - "\n", - "# Results\n", - "print \"Depth of section = %.2f m.\"%(D);\n", - "print \"Average shear stress at channel bed = %.2f N/square-mm.\"%(tau);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Depth of section = 1.50 m.\n", - "Average shear stress at channel bed = 2.47 N/square-mm.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.25 pg : 702" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "S = 1./5000; \t\t\t\t#bed slope\n", - "B = 40.; \t\t\t\t#width of channel\n", - "D = 2.6; \t\t\t\t#depth of channel\n", - "d = 0.38; \t\t\t\t#mean diameter of bed material\n", - "n = 0.021; \t\t\t\t#Manning n\n", - "D65 = 0.64e-3; \t\t\t\t#bed material size(m)\n", - "w = 1000.; \t\t\t\t#density of water\n", - "\n", - "# Calculations\n", - "#B/D as large tau_c = 0.075*d;\n", - "tau_c = 0.075*d;\n", - "tau_b = w*D*S;\n", - "N1 = (D65)**(1./6)/24;\n", - "r = N1/n;\n", - "qs = 4700*24*(tau_b*r**1.5-tau_c)**1.5/1000;\n", - "qs40 = qs*40;\n", - "\n", - "# Results\n", - "print \"bed load transported by the channel = %i t/m/day.\"%(qs40);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "bed load transported by the channel = 411 t/m/day.\n" - ] - } - ], - "prompt_number": 25 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.26 pg : 702" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "Q = 5.; \t\t\t\t#discharge\n", - "S = 0.2/1000; \t\t\t\t#bed slope\n", - "m = 0.8; \t\t\t\t#critical velocity ratio\n", - "s = 1./2; \t\t\t\t#side slope of chanel\n", - "C = 30.;\n", - "#assuming\n", - "\n", - "# Calculations\n", - "D = 1.;\n", - "Vo = 0.55*m*D**0.64;\n", - "A = Q/Vo;\n", - "B = A/D-(s*D);\n", - "P = B+2.43*D;\n", - "R = A/P;\n", - "V = C*(R*S)**0.5;\n", - "#Vo>V\n", - "#hence take second trial\n", - "D = 0.8; \t\t\t\t#assume\n", - "Vo = 0.55*m*D**0.64;\n", - "A = Q/Vo;\n", - "B = A/D-(s*D);\n", - "P = B+2.43*D;\n", - "R = A/P;\n", - "V = C*(R*S)**0.5;\n", - "#again Vo>V\n", - "#hence we take third trial\n", - "D = 0.7;\n", - "Vo = 0.55*m*D**0.64;\n", - "A = Q/Vo;\n", - "B = A/D+(s*D);\n", - "P = B+2.43*D;\n", - "R = A/P;\n", - "V = C*(R*S)**0.5;\n", - "B = round(B*100)/100;\n", - "#Vo is almost equal to V;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 20.75 m.\n", - "Depth of channel section = 0.70 m.\n" - ] - } - ], - "prompt_number": 26 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.27 pg : 703" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design irrigation channel by Kennedy method\n", - "\n", - "#Given\n", - "Q = 50.; \t\t\t\t#discharge\n", - "r = 2.5; \t\t\t\t#B/D ratio\n", - "m = 1.1; \t\t\t\t#critical velocity ratio\n", - "N = 0.025; \t\t\t\t#rogosity coefficient\n", - "s = 0.5; \t\t\t\t#side slope of channel\n", - "\n", - "\n", - "# Calculations\n", - "#umath.sing the equation of Vo and Q = A*V;we get\n", - "D = (Q/1.815)**(1/2.64);\n", - "B = r*D;\n", - "R = (B*D+0.5*D**2)/(B+2.236*D);\n", - "Vo = 0.55*m*D**0.64;\n", - "\n", - "#applying kutters formula; V = C(RS)**0.5\n", - "#where C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "#assuming S**0.5 = y\n", - "#taking real values of y\n", - "S = 0.0196171 **2;\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 8.78 m.\n", - "Depth of channel section = 3.51 m.\n", - "Bed slope = 3.85e-04.\n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.28 pg : 704" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a regime channel umath.sing Laecy's theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 35; \t\t\t\t#discharge\n", - "f = 0.9; \t\t\t\t#silt factor\n", - "s = 1./2; \t\t\t\t#side slope\n", - "\n", - "\n", - "# Calculations\n", - "V = (Q*f/140)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*Q**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "B = P-2.236*D;\n", - "\n", - "R = 5*V**2/(2*f);\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Width of channel section = %i m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 1.39e-04.\n", - "Width of channel section = 24 m.\n", - "Depth of channel section = 1.80 m.\n" - ] - } - ], - "prompt_number": 28 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.29 pg : 705" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#design an irrigation canal for given data\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 15.; \t\t\t\t#discharge\n", - "m = 1.; \t\t\t\t#critical velocity ratio\n", - "r = 5.7; \t\t\t\t#B/D\n", - "\n", - "D = (Q/(0.55*6.2))**(1/2.64);\n", - "B = D*r;\n", - "R = (B*D+D**2/2)/(B+D*5**0.5);\n", - "Vo = 0.55*m*D**0.64;\n", - "#applying kutters formula; V = C(RS)**0.5\n", - "#where C = (23+1/N+0.00155/S)*(R*S)**0.5/(1+(23+0.00155/S)*N/R**0.5);\n", - "#assuming S**0.5 = y\n", - "y = [67.5,-0.968,1.55e-3,-2e-5]\n", - "d = roots(y)[0];\n", - "\t\t\t\t#taking real values of y\n", - "S = d.real**2;\n", - "B = round(B*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 9.99 m.\n", - "Depth of channel section = 1.75 m.\n", - "Bed slope = 2.01e-04.\n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.30 pg : 705" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Design a section of unlined canal in a loomy soil\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 50.; \t\t\t\t#discharge\n", - "V = 1.; \t\t\t\t#permissible velocity\n", - "s = 2.; \t\t\t\t#side slope\n", - "r = 6.; \t\t\t\t#B/D ratio\n", - "N = 0.0225; \t\t\t\t#rogosity coefficient\n", - "\n", - "\n", - "# Calculations\n", - "A = Q/V;\n", - "D = (A/(r+2))**0.5;\n", - "B = r*D;\n", - "P = B+2*(5*D**2)**0.5;\n", - "R = A/P;\n", - "S = (V*N/R**(2/3))**2;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %i m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 15 m.\n", - "Depth of channel section = 2.50 m.\n", - "Bed slope = 5.06e-04.\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.31 pg : 705" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "D = 5.; \t\t\t\t#depth of channel\n", - "d = 0.3; \t\t\t\t#grain size\n", - "k = 1.5; \t\t\t\t#size of roughness of channel bed\n", - "S = 1./4000; \t\t\t\t#bed slope\n", - "G = 2.65; \t\t\t\t#specific gravity\n", - "V = 0.02; \t\t\t\t#fall velocity\n", - "c_ = 1000.; \t\t\t\t#concentration at 0.3 m above bed\n", - "a = 0.3;\n", - "y = 2.5;\n", - "k_ = 0.4; \t\t\t\t#van karman's consmath.tant\n", - "\n", - "\n", - "# Calculations\n", - "R = 5; \t\t\t\t#R = D for wide channel\n", - "V_ = (gamma_w*R*S)**0.5;\n", - "c = c_*((a/y)*(D-y)/(D-a))**(V/(V_*k_));\n", - "\n", - "# Results\n", - "print \"concentration of suspended load = %i ppm.\"%(c);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "concentration of suspended load = 288 ppm.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.32 pg : 706" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 40.; \t\t\t\t#discharge\n", - "f = 1.; \t\t\t\t#silt factor\n", - "\n", - "\n", - "# Calculations and Results\n", - "#Laecey's theory\n", - "V = (Q*f/140)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*Q**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "B = P-2.236*D;\n", - "\n", - "R = 5*V**2/(2*f);\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "B = round(B);\n", - "D = round(D*100)/100;\n", - "print \"By Laecey theory:\";\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n", - "#Kennedy's theory\n", - "r = B/D; \n", - "m = 1.; \t\t\t\t#critical velocity ratio\n", - "N = 0.0225; \t\t\t\t#rogosity coefficient\n", - "#umath.sing equation of area of trapezoidal section;Vo = 0.55mD**0.64 and Q = A*Vo\n", - "\n", - "D = (Q/8.058)**(1/2.64);\n", - "B = r*D;\n", - "B = round(B);\n", - "D = round(D*100)/100;\n", - "print \"By Kennedy theory:\";\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "By Laecey theory:\n", - "Bed slope = 1.62e-04.\n", - "Width of channel section = 26.00 m.\n", - "Depth of channel section = 1.84 m.\n", - "By Kennedy theory:\n", - "Width of channel section = 26.00 m.\n", - "Depth of channel section = 1.83 m.\n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.33 pg : 707" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design Laecey regime channel\n", - "\t\t\t\t\n", - "#Given\n", - "A = 100000.; \t\t\t\t#culturable area(hectare)\n", - "IR = 0.4; \t\t\t\t#intensity of irrigation in kharif season\n", - "IK = 0.3; \t\t\t\t#intensity of irrigation in rabi season\n", - "OR = 1800.; \t\t\t\t#outlet discharge factor in kharif season\n", - "OK = 800.; \t\t\t\t#outlet discharge factor in kharif season\n", - "l = 0.1; \t\t\t\t#conveyance loss\n", - "md = 0.328; \t\t\t\t#average diameter of material\n", - "\n", - "# Calculations\n", - "AR = A*IR; \t\t\t\t#area under rabi\n", - "AK = A*IK; \t\t\t\t#area under kharif \n", - "Qr = AR/OR;\n", - "Qk = AK/OK;\n", - "Q = 1.1*Qk;\n", - "f = 1.76*(md)**0.5;\n", - "V = (Q*f**2/144)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*(Q)**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "B = P-2.236*D;\n", - "S = f**(5/3)/(3340*Q**(1./6));\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 1.62e-04.\n", - "Width of channel section = 26.40 m.\n", - "Depth of channel section = 1.86 m.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.34 pg : 707" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "D = 2.8; \t\t\t\t#depth of flow\n", - "c_ = 700; \t\t\t\t#concentration at 30 cm below water surface\n", - "y = 0.1;\n", - "a = D-0.3;\n", - "e = 0.4; \t\t\t\t#exponent in rouse equation;\n", - "\n", - "# Calculations\n", - "c = c_*(a*(D-y)/(y*(D-a)))**e;\n", - "\n", - "# Results\n", - "print \"concentration at point 10 cm above the bed = %i ppm.\"%(c);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "concentration at point 10 cm above the bed = 6109 ppm.\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 14.35 pg : 708" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design the distributory umath.sing Laecey theory\n", - "\t\t\t\t\n", - "#Given\n", - "f = 0.85; \t\t\t\t#silt factor\n", - "AR = 3600.; \t\t\t\t#area for rabi\n", - "AK = 1400.; \t\t\t\t#area for kharif\n", - "delta_r = 0.135; \t\t\t\t#kor depth for rabi\n", - "delta_k = 0.19; \t\t\t\t#kor depth for kharif\n", - "tr = 4.; \t\t\t\t#kor period for rabi\n", - "tk = 2.5; \t\t\t\t#kor period for kharif\n", - "\n", - "# Calculations\n", - "Du_r = 8.64*tr*7/delta_r; \t\t\t\t#duty for rabi\n", - "Du_k = 8.64*tk*7/delta_k; \t\t\t\t#duty for kharif\n", - "q_r = AR/Du_r; \t\t\t\t#discharge for rabi\n", - "q_k = AK/Du_k; \t\t\t\t#discharge for kharif\n", - "Q = q_r; \t\t\t\t#math.since q_r>q_k\n", - "V = (Q*f**2/144)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*(Q)**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "P = round(P*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Perimeter of channel section = %.2f m.\"%(P);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 2.03e-04.\n", - "Perimeter of channel section = 6.73 m.\n", - "Depth of channel section = 0.81 m.\n" - ] - } - ], - "prompt_number": 35 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch15.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch15.ipynb deleted file mode 100755 index c5a4b2dd..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch15.ipynb +++ /dev/null @@ -1,306 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:06bc45481e0c8d1fd696ec9d96ac784607ece32d90669bf112f17cc78a27982d" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 15 : IRRIGATION CHANNEL 2 DESIGN PROCEDURE" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15.1 pg : 739" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design the distributory umath.sing Laecey theory\n", - "\t\t\t\t\n", - "#Given\n", - "f = 0.85; \t\t\t\t#silt factor\n", - "AR = 3600.; \t\t\t\t#area for rabi\n", - "AK = 1400.; \t\t\t\t#area for kharif\n", - "delta_r = 0.135; \t\t\t\t#kor depth for rabi\n", - "delta_k = 0.19; \t\t\t\t#kor depth for kharif\n", - "tr = 4.; \t\t\t\t#kor period for rabi\n", - "tk = 2.5; \t\t\t\t#kor period for kharif\n", - "\n", - "# Calculations\n", - "Du_r = 8.64*tr*7/delta_r; \t\t\t\t#duty for rabi\n", - "Du_k = 8.64*tk*7/delta_k; \t\t\t\t#duty for kharif\n", - "q_r = AR/Du_r; \t\t\t\t#discharge for rabi\n", - "q_k = AK/Du_k; \t\t\t\t#discharge for kharif\n", - "Q = q_r; \t\t\t\t#math.since q_r>q_k\n", - "V = (Q*f**2/144)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*(Q)**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "P = round(P*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Perimeter of channel section = %.2f m.\"%(P);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 2.03e-04.\n", - "Perimeter of channel section = 6.73 m.\n", - "Depth of channel section = 0.81 m.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15.2 pg : 740" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design an irrigation channel in alluvial soil by Laecy's theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 15.; \t\t\t\t#Full supply discharge\n", - "f = 1.; \t\t\t\t#silt factor\n", - "s = 1./2; \t\t\t\t#side slope of channel\n", - "\n", - "\n", - "# Calculations\n", - "#from Laecey regime channel (Fig.15.4(b)) B and D is obtained as;\n", - "B = 15.1;\n", - "D = 1.38;\n", - "\t\t\t\t#also from Fig.15.5 we get slope as\n", - "S = 0.19/1000;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 15.10 m.\n", - "Depth of channel section = 1.38 m.\n", - "Bed slope = 1.90e-04.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15.3 pg : 740" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array,zeros,linspace,float64\n", - "\n", - "#design and prepare the longitudnal section;schedule of area statistics and channel dimension of irrigation channel\n", - "\t\t\t\t\n", - "#Given\n", - "dl = 157.7; \t\t\t\t#datum level\n", - "fsl = 157.; \t\t\t\t#full supply level of parent channel\n", - "bl = 156.; \t\t\t\t#bed level of parent channel\n", - "kor_r = 4.; \t\t\t\t#kor period of rabi\n", - "kor_k = 2.5; \t\t\t\t#kor period of kharif\n", - "kord_r = 13.4; \t\t\t\t#kor depth of rabi\n", - "kord_k = 19.; \t\t\t\t#kor depth of kharif\n", - "s = 0.5; \t\t\t\t#side slope\n", - "m = 1.; \t\t\t\t#critical velocity ratio\n", - "N = 0.0225; \t\t\t\t#Kutter n\n", - "qo_r = 8.64*7*kor_r*100/kord_r; \t\t\t\t#outlet discharge for rabi(calculation is wrong in book)\n", - "qo_k = 8.64*7*kor_k*100/kord_k; \t\t\t\t#outlet discharge for kharif(calculation is wrong in book)\n", - "ca = 16000.; \t\t\t\t#culturable commanded area\n", - "Ir = 0.3; \t\t\t\t#intensity of irrigation in rabi\n", - "Ik = 0.125; \t\t\t\t#intensity of irrigation in rabi\n", - "\n", - "# Calculations and Results\n", - "Ar = Ir*ca; \t\t\t\t#area under rabi\n", - "Ak = ca*Ik; \t\t\t\t#area under kharif\n", - "q_r = Ar/qo_r;\n", - "q_k = Ak/qo_k;\n", - "q_r = round(q_r*100)/100;\n", - "q_k = round(q_k*100)/100;\n", - "print \"discharge neede for rabi crop = %.2f cumecs.\"%(q_r);\n", - "print \"discharge neede for kharif crop = %.2f cumecs.\"%(q_k);\n", - "print \"outlet discharge factor adopted = %i hectares per cumecs.\"%(qo_r);\n", - "\t\t\t\t#at km 5\n", - "ca = 8000; \t\t\t\t#culturable area\n", - "Ar = Ir*ca; \t\t\t\t#area under rabi\n", - "q_r = Ar/qo_r;\n", - "l = 0.5 \t\t\t\t#total loss after 5 km\n", - "q = q_r+l; \t\t\t\t#total discharge\n", - "dq = 1.1*q; \t\t\t\t#desigm discharge\n", - "S = 1./4000; \t\t\t\t#slope\n", - "B = array([5.5, 4.9, 4.55]); \t\t\t\t#Bed width\n", - "D = array([0.73, 0.79, 0.84]); \t\t\t\t#water depth\n", - "Vo = array([0.448, 0.472, 0.488]); \t\t\t\t#critical velocity\n", - "A = zeros(3)\n", - "V = zeros(3)\n", - "m = zeros(3)\n", - "print \"Bed width water depth area velocity critical velocity C.V.R\";\n", - "for i in range(3):\n", - " A[i] = B[i]*D[i]+D[i]**2/2;\n", - " V[i] = dq/A[i];\n", - " m[i] = V[i]/Vo[i];\n", - " A[i] = round(A[i]*100)/100;\n", - " V[i] = round(V[i]*1000)/1000;\n", - " m[i] = round(m[i]*100)/100;\n", - " print \"%.2f %.2f %.2f %.2f %.2f %.2f\"%(B[i],D[i],A[i],V[i],Vo[i],m[i]);\n", - "\n", - "B = 4.55;\n", - "D = 0.84;\n", - "print \"hence take B = %.2f .; D = %.2f m.\"%(B,D);\n", - "\t\t\t\t#at km 4\n", - "q = round(q*100)/100;\n", - "print \"discharge at 5 km = %.2f cumecs.\"%(q);\n", - "ca = 10000; \t\t\t\t#culturable area\n", - "Ar = Ir*ca; \t\t\t\t#area under rabi\n", - "q_r = Ar/qo_r;\n", - "l = 0.5 \t\t\t\t#total loss below 5 km\n", - "P = B+D*5**0.5; \t\t\t\t#wetted perimeter\n", - "l1 = P*1000*2/1000000; \t\t\t\t#loss between 5 km and 4km\n", - "l2 = l1+l;\n", - "q = q_r+l2;\n", - "dq = 1.1*q;\n", - "q = round(q*1000)/1000;\n", - "print \"discharge at 4 km = %.2f cumecs\"%(q);\n", - "print \"other discharge are calculated and are tabulated as:\";\n", - "x = linspace(1,5,6)\n", - "A1 = array([4800, 4200, 3600, 3300, 3000, 2400],dtype=float64);\n", - "A2 = array([2000, 1750, 1500, 1375, 1250, 1000],dtype=float64);\n", - "S = array([22.5, 22.5, 22.5, 24, 24, 25]);\n", - "B = array([5.5, 5.2, 4.85, 4.7, 4.55, 4.55]);\n", - "D = array([1.04, 1.007, 0.975, 0.945, 0.915, 0.840]);\n", - "dq = array([3.56, 3.17, 2.8, 2.6, 2.4, 2.02]);\n", - "V = array([0.570, 0.555, 0.538, 0.530, 0.521, 0.484]);\n", - "m = array([1.015, 1, 1, 1, 1, 0.992]);\n", - "print \"Below km area to irrigate rabi area to irrigate kharif bed slope bed width water depth design discharge velocity C.V.R\";\n", - "for i in range(6):\n", - " print \"%8i %i %i %.2f %.2f %.2f\\\n", - " %.2f %.2f %.2f\"%(x[i],A1[i],A2[i],S[i],B[i],D[i],dq[i],V[i],m[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge neede for rabi crop = 2.66 cumecs.\n", - "discharge neede for kharif crop = 2.51 cumecs.\n", - "outlet discharge factor adopted = 1805 hectares per cumecs.\n", - "Bed width water depth area velocity critical velocity C.V.R\n", - "5.50 0.73 4.28 0.47 0.45 1.05\n", - "4.90 0.79 4.18 0.48 0.47 1.02\n", - "4.55 0.84 4.17 0.48 0.49 0.99\n", - "hence take B = 4.55 .; D = 0.84 m.\n", - "discharge at 5 km = 1.83 cumecs.\n", - "discharge at 4 km = 2.17 cumecs\n", - "other discharge are calculated and are tabulated as:\n", - "Below km area to irrigate rabi area to irrigate kharif bed slope bed width water depth design discharge velocity C.V.R\n", - " 1 4800 2000 22.50 5.50 1.04 3.56 0.57 1.01\n", - " 1 4200 1750 22.50 5.20 1.01 3.17 0.56 1.00\n", - " 2 3600 1500 22.50 4.85 0.97 2.80 0.54 1.00\n", - " 3 3300 1375 24.00 4.70 0.94 2.60 0.53 1.00\n", - " 4 3000 1250 24.00 4.55 0.92 2.40 0.52 1.00\n", - " 5 2400 1000 25.00 4.55 0.84 2.02 0.48 0.99\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15.4 pg : 744" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#Given\n", - "B = 5.; \t\t\t\t#bed width\n", - "t = 2.; \t\t\t\t#top width of banks\n", - "h = 2.92; \t\t\t\t#heigth of banks from bed\n", - "n = 1.5;\n", - "\n", - "#sectional area of digging = sectional area of two banks\n", - "#By+zy**2 = 2(h-y)+2n(h-y)**2\n", - "#substituting the values and on simplificatio we get\n", - "s = [1,-13.26,18.59]\n", - "y = roots(s)[1];\n", - "#from this we get y = 11.666556 and 1.5934436.\n", - "#taking y = 1.5934436;\n", - "y = round(y*10)/10;\n", - "\n", - "# Results\n", - "print \"economical depth of cutting = %.2f m.\"%(y);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "economical depth of cutting = 1.60 m.\n" - ] - } - ], - "prompt_number": 4 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch15_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch15_1.ipynb deleted file mode 100755 index c5a4b2dd..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch15_1.ipynb +++ /dev/null @@ -1,306 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:06bc45481e0c8d1fd696ec9d96ac784607ece32d90669bf112f17cc78a27982d" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 15 : IRRIGATION CHANNEL 2 DESIGN PROCEDURE" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15.1 pg : 739" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design the distributory umath.sing Laecey theory\n", - "\t\t\t\t\n", - "#Given\n", - "f = 0.85; \t\t\t\t#silt factor\n", - "AR = 3600.; \t\t\t\t#area for rabi\n", - "AK = 1400.; \t\t\t\t#area for kharif\n", - "delta_r = 0.135; \t\t\t\t#kor depth for rabi\n", - "delta_k = 0.19; \t\t\t\t#kor depth for kharif\n", - "tr = 4.; \t\t\t\t#kor period for rabi\n", - "tk = 2.5; \t\t\t\t#kor period for kharif\n", - "\n", - "# Calculations\n", - "Du_r = 8.64*tr*7/delta_r; \t\t\t\t#duty for rabi\n", - "Du_k = 8.64*tk*7/delta_k; \t\t\t\t#duty for kharif\n", - "q_r = AR/Du_r; \t\t\t\t#discharge for rabi\n", - "q_k = AK/Du_k; \t\t\t\t#discharge for kharif\n", - "Q = q_r; \t\t\t\t#math.since q_r>q_k\n", - "V = (Q*f**2/144)**(1./6);\n", - "A = Q/V;\n", - "P = 4.75*(Q)**0.5;\n", - "D = (P-(P**2-6.944*A)**0.5)/3.472;\n", - "S = f**(5./3)/(3340*Q**(1./6));\n", - "P = round(P*100)/100;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "print \"Perimeter of channel section = %.2f m.\"%(P);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Bed slope = 2.03e-04.\n", - "Perimeter of channel section = 6.73 m.\n", - "Depth of channel section = 0.81 m.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15.2 pg : 740" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design an irrigation channel in alluvial soil by Laecy's theory\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 15.; \t\t\t\t#Full supply discharge\n", - "f = 1.; \t\t\t\t#silt factor\n", - "s = 1./2; \t\t\t\t#side slope of channel\n", - "\n", - "\n", - "# Calculations\n", - "#from Laecey regime channel (Fig.15.4(b)) B and D is obtained as;\n", - "B = 15.1;\n", - "D = 1.38;\n", - "\t\t\t\t#also from Fig.15.5 we get slope as\n", - "S = 0.19/1000;\n", - "\n", - "# Results\n", - "print \"Width of channel section = %.2f m.\"%(B);\n", - "print \"Depth of channel section = %.2f m.\"%(D);\n", - "print \"Bed slope = %.2e.\"%(S);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Width of channel section = 15.10 m.\n", - "Depth of channel section = 1.38 m.\n", - "Bed slope = 1.90e-04.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15.3 pg : 740" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array,zeros,linspace,float64\n", - "\n", - "#design and prepare the longitudnal section;schedule of area statistics and channel dimension of irrigation channel\n", - "\t\t\t\t\n", - "#Given\n", - "dl = 157.7; \t\t\t\t#datum level\n", - "fsl = 157.; \t\t\t\t#full supply level of parent channel\n", - "bl = 156.; \t\t\t\t#bed level of parent channel\n", - "kor_r = 4.; \t\t\t\t#kor period of rabi\n", - "kor_k = 2.5; \t\t\t\t#kor period of kharif\n", - "kord_r = 13.4; \t\t\t\t#kor depth of rabi\n", - "kord_k = 19.; \t\t\t\t#kor depth of kharif\n", - "s = 0.5; \t\t\t\t#side slope\n", - "m = 1.; \t\t\t\t#critical velocity ratio\n", - "N = 0.0225; \t\t\t\t#Kutter n\n", - "qo_r = 8.64*7*kor_r*100/kord_r; \t\t\t\t#outlet discharge for rabi(calculation is wrong in book)\n", - "qo_k = 8.64*7*kor_k*100/kord_k; \t\t\t\t#outlet discharge for kharif(calculation is wrong in book)\n", - "ca = 16000.; \t\t\t\t#culturable commanded area\n", - "Ir = 0.3; \t\t\t\t#intensity of irrigation in rabi\n", - "Ik = 0.125; \t\t\t\t#intensity of irrigation in rabi\n", - "\n", - "# Calculations and Results\n", - "Ar = Ir*ca; \t\t\t\t#area under rabi\n", - "Ak = ca*Ik; \t\t\t\t#area under kharif\n", - "q_r = Ar/qo_r;\n", - "q_k = Ak/qo_k;\n", - "q_r = round(q_r*100)/100;\n", - "q_k = round(q_k*100)/100;\n", - "print \"discharge neede for rabi crop = %.2f cumecs.\"%(q_r);\n", - "print \"discharge neede for kharif crop = %.2f cumecs.\"%(q_k);\n", - "print \"outlet discharge factor adopted = %i hectares per cumecs.\"%(qo_r);\n", - "\t\t\t\t#at km 5\n", - "ca = 8000; \t\t\t\t#culturable area\n", - "Ar = Ir*ca; \t\t\t\t#area under rabi\n", - "q_r = Ar/qo_r;\n", - "l = 0.5 \t\t\t\t#total loss after 5 km\n", - "q = q_r+l; \t\t\t\t#total discharge\n", - "dq = 1.1*q; \t\t\t\t#desigm discharge\n", - "S = 1./4000; \t\t\t\t#slope\n", - "B = array([5.5, 4.9, 4.55]); \t\t\t\t#Bed width\n", - "D = array([0.73, 0.79, 0.84]); \t\t\t\t#water depth\n", - "Vo = array([0.448, 0.472, 0.488]); \t\t\t\t#critical velocity\n", - "A = zeros(3)\n", - "V = zeros(3)\n", - "m = zeros(3)\n", - "print \"Bed width water depth area velocity critical velocity C.V.R\";\n", - "for i in range(3):\n", - " A[i] = B[i]*D[i]+D[i]**2/2;\n", - " V[i] = dq/A[i];\n", - " m[i] = V[i]/Vo[i];\n", - " A[i] = round(A[i]*100)/100;\n", - " V[i] = round(V[i]*1000)/1000;\n", - " m[i] = round(m[i]*100)/100;\n", - " print \"%.2f %.2f %.2f %.2f %.2f %.2f\"%(B[i],D[i],A[i],V[i],Vo[i],m[i]);\n", - "\n", - "B = 4.55;\n", - "D = 0.84;\n", - "print \"hence take B = %.2f .; D = %.2f m.\"%(B,D);\n", - "\t\t\t\t#at km 4\n", - "q = round(q*100)/100;\n", - "print \"discharge at 5 km = %.2f cumecs.\"%(q);\n", - "ca = 10000; \t\t\t\t#culturable area\n", - "Ar = Ir*ca; \t\t\t\t#area under rabi\n", - "q_r = Ar/qo_r;\n", - "l = 0.5 \t\t\t\t#total loss below 5 km\n", - "P = B+D*5**0.5; \t\t\t\t#wetted perimeter\n", - "l1 = P*1000*2/1000000; \t\t\t\t#loss between 5 km and 4km\n", - "l2 = l1+l;\n", - "q = q_r+l2;\n", - "dq = 1.1*q;\n", - "q = round(q*1000)/1000;\n", - "print \"discharge at 4 km = %.2f cumecs\"%(q);\n", - "print \"other discharge are calculated and are tabulated as:\";\n", - "x = linspace(1,5,6)\n", - "A1 = array([4800, 4200, 3600, 3300, 3000, 2400],dtype=float64);\n", - "A2 = array([2000, 1750, 1500, 1375, 1250, 1000],dtype=float64);\n", - "S = array([22.5, 22.5, 22.5, 24, 24, 25]);\n", - "B = array([5.5, 5.2, 4.85, 4.7, 4.55, 4.55]);\n", - "D = array([1.04, 1.007, 0.975, 0.945, 0.915, 0.840]);\n", - "dq = array([3.56, 3.17, 2.8, 2.6, 2.4, 2.02]);\n", - "V = array([0.570, 0.555, 0.538, 0.530, 0.521, 0.484]);\n", - "m = array([1.015, 1, 1, 1, 1, 0.992]);\n", - "print \"Below km area to irrigate rabi area to irrigate kharif bed slope bed width water depth design discharge velocity C.V.R\";\n", - "for i in range(6):\n", - " print \"%8i %i %i %.2f %.2f %.2f\\\n", - " %.2f %.2f %.2f\"%(x[i],A1[i],A2[i],S[i],B[i],D[i],dq[i],V[i],m[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge neede for rabi crop = 2.66 cumecs.\n", - "discharge neede for kharif crop = 2.51 cumecs.\n", - "outlet discharge factor adopted = 1805 hectares per cumecs.\n", - "Bed width water depth area velocity critical velocity C.V.R\n", - "5.50 0.73 4.28 0.47 0.45 1.05\n", - "4.90 0.79 4.18 0.48 0.47 1.02\n", - "4.55 0.84 4.17 0.48 0.49 0.99\n", - "hence take B = 4.55 .; D = 0.84 m.\n", - "discharge at 5 km = 1.83 cumecs.\n", - "discharge at 4 km = 2.17 cumecs\n", - "other discharge are calculated and are tabulated as:\n", - "Below km area to irrigate rabi area to irrigate kharif bed slope bed width water depth design discharge velocity C.V.R\n", - " 1 4800 2000 22.50 5.50 1.04 3.56 0.57 1.01\n", - " 1 4200 1750 22.50 5.20 1.01 3.17 0.56 1.00\n", - " 2 3600 1500 22.50 4.85 0.97 2.80 0.54 1.00\n", - " 3 3300 1375 24.00 4.70 0.94 2.60 0.53 1.00\n", - " 4 3000 1250 24.00 4.55 0.92 2.40 0.52 1.00\n", - " 5 2400 1000 25.00 4.55 0.84 2.02 0.48 0.99\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 15.4 pg : 744" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#Given\n", - "B = 5.; \t\t\t\t#bed width\n", - "t = 2.; \t\t\t\t#top width of banks\n", - "h = 2.92; \t\t\t\t#heigth of banks from bed\n", - "n = 1.5;\n", - "\n", - "#sectional area of digging = sectional area of two banks\n", - "#By+zy**2 = 2(h-y)+2n(h-y)**2\n", - "#substituting the values and on simplificatio we get\n", - "s = [1,-13.26,18.59]\n", - "y = roots(s)[1];\n", - "#from this we get y = 11.666556 and 1.5934436.\n", - "#taking y = 1.5934436;\n", - "y = round(y*10)/10;\n", - "\n", - "# Results\n", - "print \"economical depth of cutting = %.2f m.\"%(y);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "economical depth of cutting = 1.60 m.\n" - ] - } - ], - "prompt_number": 4 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch16.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch16.ipynb deleted file mode 100755 index 59b5d071..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch16.ipynb +++ /dev/null @@ -1,804 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:be9a83eac81003b59bb1ec7327404f7c36e99fe14c8c14251b46829335a79c67" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 16 : WATERLOGGING AND CANAL LINING" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.1 pg : 764" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "from sympy import acot\n", - "\n", - "#design a trapezoidal concrete lined channel\n", - "\n", - "#Given\n", - "Q = 100.; \t\t\t\t#discharge\n", - "S = 25./100000; \t\t\t\t#bed slope\n", - "N = 0.016; \t\t\t\t#rogsity coefficient\n", - "s = 1.5; \t\t\t\t#side slope\n", - "V = 1.5; \t\t\t\t#limiting velocity\n", - "\n", - "\n", - "# Calculations\n", - "#umath.sing manning's equation V = (R**2/3*S**1/2)/N;\n", - "R = (V*N/(S**0.5))**(1.5); \t\t\t\t#hydraulic mean depth\n", - "\n", - "#for s = 1.5;\n", - "theta = acot(1.5);\n", - "A = Q/V;\n", - "P = A/R;\n", - "#umath.sing equation of area and perimeter of trapezium\n", - "#perimeter of trapezium = b+2d(theta+cot(theta));\n", - "#area of trapezium = bd+d**2(theta+cot(theta));\n", - "#we get\n", - "y = [1,-17.1,31.9]\n", - "d = roots(y)[1];\n", - "#we get D = 14.968917 and 2.1310826.\n", - "#taking d = 2.1310826;\n", - "b = P-4.18*d;\n", - "b = round(b*100)/100;\n", - "d = round(d*100)/100;\n", - "print \"required bed width = %.2f m.\"%(b);\n", - "print \"required bed depth = %.2f m\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required bed width = 26.74 m.\n", - "required bed depth = 2.13 m\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.2 pg : 765" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a trapezoidal concrete lined channel\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 100.; \t\t\t\t#discharge\n", - "S = 25./100000; \t\t\t\t#bed slope\n", - "N = 0.016; \t\t\t\t#rogsity coefficient\n", - "s = 1.5; \t\t\t\t#side slope\n", - "r = 8.; \t\t\t\t#b/d ratio\n", - "\n", - "\n", - "# Calculations\n", - "#umath.sing manning equation V = (R**2/3*S**1/2)/N;\n", - "#Perimeter = A/R \n", - "#V = Q/A and on simplification we get\n", - "d = ((101/10.09)*(12.18/10.09)**(2./3))**(3./8);\n", - "b = r*d;\n", - "b = round(b);\n", - "d = round(d*100)/100;\n", - "\n", - "# Results\n", - "print \"required bed width = %.2f m.\"%(b);\n", - "print \"required bed depth = %.2f m\"%(d);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required bed width = 20.00 m.\n", - "required bed depth = 2.49 m\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.3 pg : 766" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from sympy import acot\n", - "\n", - "#design a concrete lined channel\n", - "\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "S = 1./10000; \t\t\t\t#bed slope\n", - "s = 5./4; \t\t\t\t#side slope\n", - "N = 0.018; \t\t\t\t#rogosity coefficient(manning N)\n", - "\n", - "#channel is assumed to be of triangular section\n", - "theta = acot(s);\n", - "#umath.sing manning equation V = (R**2/3*S**1/2)/N;\n", - "#V = Q/A; \n", - "#perimeter of trapezium = b+2d(theta+cot(theta));\n", - "#area of trapezium = bd+d**2(theta+cot(theta));\n", - "#we get\n", - "d = (Q*2.86/1.925)**(3./8);\n", - "d = round(d*100)/100;\n", - "\n", - "# Results\n", - "print \"required depth of triangular channel = %.2f m\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required depth of triangular channel = 4.84 m\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.4 pg : 767" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a concrete lined channel of trapezoidal section\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 250.; \t\t\t\t#discharge\n", - "S = 1./6000; \t\t\t\t#bed slope\n", - "s = 1.5; \t\t\t\t#side slope\n", - "d = 3.; \t\t\t\t#limiting depth\n", - "N = 0.015; \t\t\t\t#rogosity coefficient\n", - "\n", - "#umath.sing Perimeter = A/R;\n", - "#perimeter of trapezium = b+2d(theta+cot(theta));\n", - "#area of trapezium = bd+d**2(theta+cot(theta));\n", - "#Q = A*V; and on simplification\n", - "#we get\n", - "#(3b+18.81)**5/3/(b+12.54)**2/3 = 290.47;\n", - "#solving it by trial and error method we get\n", - "b = 44.6;\n", - "\n", - "# Results\n", - "print \"required bed width = %.2f m.\"%(b);\n", - "print \"required bed depth = %i m\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required bed width = 44.60 m.\n", - "required bed depth = 3 m\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.5 pg : 767" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "H = 10.; \t\t\t\t#depth of impervious stratum from top soil\n", - "D = 1.8; \t\t\t\t#position of drain below top soil surface\n", - "Hw = 1.5; \t\t\t\t#depth of highest point of water\n", - "k = 1.e-4; \t\t\t\t#permeability consmath.tant\n", - "\n", - "\n", - "# Calculations\n", - "#math.since water has to be removed in 24 hours\n", - "b = H-Hw;\n", - "a = H-D;\n", - "L = (4*k*(b**2-a**2)*100*24*3600/0.8)**0.5;\n", - "\n", - "# Results\n", - "print \"drains should be spaced at %i m c/c.\"%(L);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "drains should be spaced at 147 m c/c.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.6 pg : 768" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "L = 30.; \t\t\t\t#spacing between drans\n", - "Q = 4.e-6; \t\t\t\t#discharge\n", - "a = 8.;\n", - "b = 8.3;\n", - "\n", - "\n", - "# Calculations\n", - "k = 1000000*Q*L/(4*(b**2-a**2));\n", - "k = round(k*100)/100;\n", - "\n", - "# Results\n", - "print \"permeability coefficient = %.2fD-6 m/sec.\"%(k);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "permeability coefficient = 6.13D-6 m/sec.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.7 pg : 768" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "L = 50.; \t\t\t\t#spacing between drains\n", - "k = 1.e-5; \t\t\t\t#permeability coefficient\n", - "a = 10.;\n", - "b = 10.3;\n", - "\n", - "\n", - "# Calculations\n", - "Q = 4*k*(b**2-a**2)/L;\n", - "Pav = Q*24*3600*100*100/L;\n", - "\n", - "# Results\n", - "print \"annual average rainfall = %i cm\"%(Pav);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "annual average rainfall = 84 cm\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.8 pg : 768" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "r1 = 2; \t\t\t\t#ka/kb\n", - "r2 = 1./1.5; \t\t\t\t#La/Lb\n", - "r3 = 5./6; \t\t\t\t#(b**2-a**2)a/((b**2-a**2)b)\n", - "\n", - "\n", - "# Calculations\n", - "Rq = r1*r3/r2;\n", - "Rp = Rq/r2;\n", - "\n", - "# Results\n", - "print \"ratio of discharge at A and B = %.2f.\"%(Rq);\n", - "print \"ratio of average rainfall at A and B = %.2f.\"%(Rp);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ratio of discharge at A and B = 2.50.\n", - "ratio of average rainfall at A and B = 3.75.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.9 pg : 775" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#decide whether it is economically feasible to provide canal lining\n", - "\t\t\t\t\n", - "#Given\n", - "li = 2.5; \t\t\t\t#seepage loss for lined channel\n", - "p1 = 25.; \t\t\t\t#wetted perimeter for lined channel\n", - "t = 12.; \t\t\t\t#thickness of concrete lining\n", - "lf = 0.02; \t\t\t\t#seepage loss for unlined channel\n", - "p2 = 20.; \t\t\t\t#wetted perimeter for unlined channel\n", - "\n", - "#assume 1 km length of canal\n", - "#annual benifit\n", - "\n", - "\t\t\t\t#(1).seepage\n", - "A1 = p1*1000; \t\t\t\t#area of wetted perimeter\n", - "li = li*p1/1000; \t\t\t\t#seepage loss\n", - "A2 = p2*1000; \t\t\t\t#area of wetted perimmeter for unlined channel\n", - "lf = p2*lf/1000; \t\t\t\t#seepage loss for unlined channel\n", - "s = li-lf; \t\t\t\t#saving in water loss\n", - "a1 = s*p1*100000; \t\t\t\t#annual revenue saved\n", - "\n", - "\t\t\t\t#(2)maintainence\n", - "a2 = 0.4*25000; \t\t\t\t#saving in maintainance math.cost\n", - "ts = a1+a2; \t\t\t\t#total annual benifit\n", - "\n", - "\t\t\t\t#annual math.cost\n", - "A1 = p2*1000; \t\t\t\t#area of lining for unlinrd canal\n", - "C = 100*A1; \t\t\t\t#math.cost of lining\n", - "\t\t\t\t#interest rate is 6%\n", - "i = 0.06;\n", - "N = 50;\n", - "a = (C*i*(i+1)**N)/((1+i)**N-1); \t\t\t\t#annual math.cost of lining or capital recovery factor\n", - "bcr = ts/a; \t\t\t\t#benifit math.cost ratio\n", - "bcr = round(bcr*1000)/1000;\n", - "print \"Benifit math.cost ratio = %.2f.\"%(bcr);\n", - "\t\t\t\t#as bcr>1\n", - "print \" ;Since it is more than 1.Hence, it is economically justifiable. \";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Benifit math.cost ratio = 1.30.\n", - " ;Since it is more than 1.Hence, it is economically justifiable. \n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.10 pg : 776" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Ecd = 20; \t\t\t\t#electrical conductivity of drainage water\n", - "Eci = 1.5; \t\t\t\t#m mho/cm\n", - "Dc = 55.5; \t\t\t\t#consumptive use\n", - "\n", - "\n", - "# Calculations\n", - "Lr = Eci/Ecd;\n", - "D = Dc/(1-Lr);\n", - "\n", - "# Results\n", - "print \"required depth of water to be applied = %i mm.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required depth of water to be applied = 60 mm.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.11 pg : 776" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Eci = 1.4; \t\t\t\t# m mho/cm\n", - "Ece = 11.; \t\t\t\t#saturated extract of soil\n", - "Dc = 85.; \t\t\t\t#consumptive use requirement of crop\n", - "\n", - "\n", - "# Calculations\n", - "#let us assume Ecd = 2Ece\n", - "Lr = Eci/(2*Ece);\n", - "Di = Dc/(1-Lr);\n", - "Di = round(Di*10)/10;\n", - "\n", - "\n", - "# Results\n", - "print \"required depth of water to be applied = %.2f mm.\"%(Di);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required depth of water to be applied = 90.80 mm.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.12 pg : 776" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#percentage of earth work is saved in lined section\n", - "\n", - "#Given\n", - "s = 1.5; \t\t\t\t#side slope\n", - "Q = 15.; \t\t\t\t#discharge\n", - "S = 1./4000; \t\t\t\t#bed slope\n", - "Nl = 0.014; \t\t\t\t#manning n for lined channel\n", - "Nu = 0.028; \t\t\t\t#manning n for ulined channel \n", - "fb = 0.75; \t\t\t\t#free board\n", - "\n", - "#considering the perimeter of trapezoidal section\n", - "#taking minimum perimeter for given area\n", - "#i.e dP/dD = 0\n", - "#we get\n", - "#A = 2.1D**2; R = D/2; and P = 4.2D\n", - "\n", - "#for linrd channel\n", - "#Q = AR**(2/3)*S**0.5\n", - "#substituting above values we get\n", - "D = (10.0396)**(3./8);\n", - "B = 0.6*D;\n", - "R = D/2;\n", - "tau = 9.81*R*S*1000;\n", - "tau = round(tau*1000)/1000;\n", - "print \"for lined canal:\";\n", - "print \"average boundary shear stress = %.2f N/square m.\"%(tau);\n", - "Dc = D+fb; \t\t\t\t#total depth of cutting\n", - "A1 = (B+1.5*Dc)*Dc;\n", - "\n", - "#for unlined channel\n", - "#Q = AR**(2/3)*S**0.5\n", - "#substituting above values we get\n", - "D = 3.08;\n", - "B = 0.6*D;\n", - "R = D/2;\n", - "tau = 9.81*R*S*1000;\n", - "tau = round(tau*100)/100;\n", - "print \"for unlined canal:\";\n", - "print \"average boundary shear stress = %.2f N/square m.\"%(tau);\n", - "Dc = D+fb; \t\t\t\t#total depth of cutting\n", - "A2 = (B+1.5*Dc)*Dc;\n", - "per = (A2-A1)*100/A2; \n", - "per = round(per*100)/100;\n", - "print \"percent saving of earth = %.2f percent.\"%(per);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "for lined canal:\n", - "average boundary shear stress = 2.91 N/square m.\n", - "for unlined canal:\n", - "average boundary shear stress = 3.78 N/square m.\n", - "percent saving of earth = 34.32 percent.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.13 pg : 778" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#design a lined canal\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 100.; \t\t\t\t#discharge\n", - "S = 1./2500; \t\t\t\t#bed slope\n", - "V = 2.; \t\t\t\t#maximum permissible velocity\n", - "n = 0.013; \t\t\t\t#manning n\n", - "s = 1.25; \t\t\t\t#side slope\n", - "\n", - "\n", - "# Calculations\n", - "A = Q/V;\n", - "#from manning formula V = (R**2/3*S**1/2)/N;\n", - "R = (V*n/S**0.5)**1.5;\n", - "P = A/R;\n", - "\n", - "#now umath.sing the equation of area and perimeter of trapezoid\n", - "#area = D(B+2.5D)\n", - "#perimeter = B+3.2D;\n", - "#we get\n", - "y = [1.95,-33.73,50]\n", - "D = roots(y)[1];\n", - "#we get D = 15.660087 and 1.6373489\n", - "#taking D = 1.6373489;\n", - "B = P-3.2*D;\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"required bed width = %.2f m.\"%(B);\n", - "print \"required bed depth = %.2f m\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required bed width = 28.50 m.\n", - "required bed depth = 1.64 m\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.14 pg : 778" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "B = 5.; \t\t\t\t#bed width\n", - "D = 2.; \t\t\t\t#bed depth\n", - "S = 1./1600; \t\t\t\t#bed slope\n", - "n = 0.015; \t\t\t\t#manning n\n", - "\n", - "\n", - "# Calculations and Results\n", - "A = B+2*D; \t\t\t\t#area of lining\n", - "#let B1 and D1 be new width and depth of bed\n", - "#for getting maximum discharge we diffrentiate Q and equating it to zero\n", - "#Q = S**0.5*B1D1**5/3/n\n", - "#we get\n", - "D1 = 45./16;\n", - "B1 = 9-2*D1;\n", - "Q1 = S**0.5*B1*D1**5/3/n;\n", - "D1 = round(D1*10000)/10000;\n", - "print \"new width of bed = %.2f m.\"%(B1);\n", - "print \"new depth of bed = %.2f m.\"%(D1);\n", - "print \" maximum discharge = %.2f cumec.\"%(Q1);\n", - "R = D;\n", - "V = R**(2./3)*S**0.5/n;\n", - "F = V/(9.81*D)**0.5; \t\t\t\t#froud number\n", - "R = D1;\n", - "V = R**(2./3)*S**0.5/n;\n", - "F = V/(9.81*D1)**0.5; \t\t\t\t#froud number\n", - "print \"Froud number is less than 1 in both case.Hence flow doesnot change from sub-critical to super critical.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "new width of bed = 3.38 m.\n", - "new depth of bed = 2.81 m.\n", - " maximum discharge = 329.96 cumec.\n", - "Froud number is less than 1 in both case.Hence flow doesnot change from sub-critical to super critical.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.15 pg : 779" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from sympy import acot\n", - "\n", - "#area to be irrigated\n", - "\t\t\t\t\n", - "#Given\n", - "B = 5.; \t\t\t\t#bed width\n", - "D = 2.5; \t\t\t\t#bed depth\n", - "s = 1.5; \t\t\t\t#side slope\n", - "S = 1./1000; \t\t\t\t#bed slope\n", - "n = 0.016; \t\t\t\t#manning n\n", - "k = 10.; \t\t\t\t#kor period\n", - "d = 150.; \t\t\t\t#field irrigation requirement \n", - "\n", - "\n", - "# Calculations\n", - "theta = acot(s);\n", - "A = B*D+D**2*(theta+1/math.tan(theta));\n", - "P = B+2*D*(theta+1/math.tan(theta));\n", - "R = A/P;\n", - "Q = A*R**(2./3)*S**0.5/n;\n", - "V = Q*k*24*3600; \t\t\t\t#volum of water supply by channel\n", - "A = V*10/(d*10000);\n", - "Q = round(Q*100)/100;\n", - "A = round(A)*100;\n", - "\n", - "# Results\n", - "print \"maximum carrying capacity of canal = %.2f cumec.\"%(Q);\n", - "print \"Area to be irrigated = %.2f hectares.\"%(A);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum carrying capacity of canal = 70.65 cumec.\n", - "Area to be irrigated = 40700.00 hectares.\n" - ] - } - ], - "prompt_number": 15 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch16_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch16_1.ipynb deleted file mode 100755 index 59b5d071..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch16_1.ipynb +++ /dev/null @@ -1,804 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:be9a83eac81003b59bb1ec7327404f7c36e99fe14c8c14251b46829335a79c67" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 16 : WATERLOGGING AND CANAL LINING" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.1 pg : 764" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "from sympy import acot\n", - "\n", - "#design a trapezoidal concrete lined channel\n", - "\n", - "#Given\n", - "Q = 100.; \t\t\t\t#discharge\n", - "S = 25./100000; \t\t\t\t#bed slope\n", - "N = 0.016; \t\t\t\t#rogsity coefficient\n", - "s = 1.5; \t\t\t\t#side slope\n", - "V = 1.5; \t\t\t\t#limiting velocity\n", - "\n", - "\n", - "# Calculations\n", - "#umath.sing manning's equation V = (R**2/3*S**1/2)/N;\n", - "R = (V*N/(S**0.5))**(1.5); \t\t\t\t#hydraulic mean depth\n", - "\n", - "#for s = 1.5;\n", - "theta = acot(1.5);\n", - "A = Q/V;\n", - "P = A/R;\n", - "#umath.sing equation of area and perimeter of trapezium\n", - "#perimeter of trapezium = b+2d(theta+cot(theta));\n", - "#area of trapezium = bd+d**2(theta+cot(theta));\n", - "#we get\n", - "y = [1,-17.1,31.9]\n", - "d = roots(y)[1];\n", - "#we get D = 14.968917 and 2.1310826.\n", - "#taking d = 2.1310826;\n", - "b = P-4.18*d;\n", - "b = round(b*100)/100;\n", - "d = round(d*100)/100;\n", - "print \"required bed width = %.2f m.\"%(b);\n", - "print \"required bed depth = %.2f m\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required bed width = 26.74 m.\n", - "required bed depth = 2.13 m\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.2 pg : 765" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a trapezoidal concrete lined channel\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 100.; \t\t\t\t#discharge\n", - "S = 25./100000; \t\t\t\t#bed slope\n", - "N = 0.016; \t\t\t\t#rogsity coefficient\n", - "s = 1.5; \t\t\t\t#side slope\n", - "r = 8.; \t\t\t\t#b/d ratio\n", - "\n", - "\n", - "# Calculations\n", - "#umath.sing manning equation V = (R**2/3*S**1/2)/N;\n", - "#Perimeter = A/R \n", - "#V = Q/A and on simplification we get\n", - "d = ((101/10.09)*(12.18/10.09)**(2./3))**(3./8);\n", - "b = r*d;\n", - "b = round(b);\n", - "d = round(d*100)/100;\n", - "\n", - "# Results\n", - "print \"required bed width = %.2f m.\"%(b);\n", - "print \"required bed depth = %.2f m\"%(d);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required bed width = 20.00 m.\n", - "required bed depth = 2.49 m\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.3 pg : 766" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from sympy import acot\n", - "\n", - "#design a concrete lined channel\n", - "\n", - "#Given\n", - "Q = 45.; \t\t\t\t#discharge\n", - "S = 1./10000; \t\t\t\t#bed slope\n", - "s = 5./4; \t\t\t\t#side slope\n", - "N = 0.018; \t\t\t\t#rogosity coefficient(manning N)\n", - "\n", - "#channel is assumed to be of triangular section\n", - "theta = acot(s);\n", - "#umath.sing manning equation V = (R**2/3*S**1/2)/N;\n", - "#V = Q/A; \n", - "#perimeter of trapezium = b+2d(theta+cot(theta));\n", - "#area of trapezium = bd+d**2(theta+cot(theta));\n", - "#we get\n", - "d = (Q*2.86/1.925)**(3./8);\n", - "d = round(d*100)/100;\n", - "\n", - "# Results\n", - "print \"required depth of triangular channel = %.2f m\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required depth of triangular channel = 4.84 m\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.4 pg : 767" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a concrete lined channel of trapezoidal section\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 250.; \t\t\t\t#discharge\n", - "S = 1./6000; \t\t\t\t#bed slope\n", - "s = 1.5; \t\t\t\t#side slope\n", - "d = 3.; \t\t\t\t#limiting depth\n", - "N = 0.015; \t\t\t\t#rogosity coefficient\n", - "\n", - "#umath.sing Perimeter = A/R;\n", - "#perimeter of trapezium = b+2d(theta+cot(theta));\n", - "#area of trapezium = bd+d**2(theta+cot(theta));\n", - "#Q = A*V; and on simplification\n", - "#we get\n", - "#(3b+18.81)**5/3/(b+12.54)**2/3 = 290.47;\n", - "#solving it by trial and error method we get\n", - "b = 44.6;\n", - "\n", - "# Results\n", - "print \"required bed width = %.2f m.\"%(b);\n", - "print \"required bed depth = %i m\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required bed width = 44.60 m.\n", - "required bed depth = 3 m\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.5 pg : 767" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "H = 10.; \t\t\t\t#depth of impervious stratum from top soil\n", - "D = 1.8; \t\t\t\t#position of drain below top soil surface\n", - "Hw = 1.5; \t\t\t\t#depth of highest point of water\n", - "k = 1.e-4; \t\t\t\t#permeability consmath.tant\n", - "\n", - "\n", - "# Calculations\n", - "#math.since water has to be removed in 24 hours\n", - "b = H-Hw;\n", - "a = H-D;\n", - "L = (4*k*(b**2-a**2)*100*24*3600/0.8)**0.5;\n", - "\n", - "# Results\n", - "print \"drains should be spaced at %i m c/c.\"%(L);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "drains should be spaced at 147 m c/c.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.6 pg : 768" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "L = 30.; \t\t\t\t#spacing between drans\n", - "Q = 4.e-6; \t\t\t\t#discharge\n", - "a = 8.;\n", - "b = 8.3;\n", - "\n", - "\n", - "# Calculations\n", - "k = 1000000*Q*L/(4*(b**2-a**2));\n", - "k = round(k*100)/100;\n", - "\n", - "# Results\n", - "print \"permeability coefficient = %.2fD-6 m/sec.\"%(k);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "permeability coefficient = 6.13D-6 m/sec.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.7 pg : 768" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "L = 50.; \t\t\t\t#spacing between drains\n", - "k = 1.e-5; \t\t\t\t#permeability coefficient\n", - "a = 10.;\n", - "b = 10.3;\n", - "\n", - "\n", - "# Calculations\n", - "Q = 4*k*(b**2-a**2)/L;\n", - "Pav = Q*24*3600*100*100/L;\n", - "\n", - "# Results\n", - "print \"annual average rainfall = %i cm\"%(Pav);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "annual average rainfall = 84 cm\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.8 pg : 768" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "r1 = 2; \t\t\t\t#ka/kb\n", - "r2 = 1./1.5; \t\t\t\t#La/Lb\n", - "r3 = 5./6; \t\t\t\t#(b**2-a**2)a/((b**2-a**2)b)\n", - "\n", - "\n", - "# Calculations\n", - "Rq = r1*r3/r2;\n", - "Rp = Rq/r2;\n", - "\n", - "# Results\n", - "print \"ratio of discharge at A and B = %.2f.\"%(Rq);\n", - "print \"ratio of average rainfall at A and B = %.2f.\"%(Rp);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ratio of discharge at A and B = 2.50.\n", - "ratio of average rainfall at A and B = 3.75.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.9 pg : 775" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#decide whether it is economically feasible to provide canal lining\n", - "\t\t\t\t\n", - "#Given\n", - "li = 2.5; \t\t\t\t#seepage loss for lined channel\n", - "p1 = 25.; \t\t\t\t#wetted perimeter for lined channel\n", - "t = 12.; \t\t\t\t#thickness of concrete lining\n", - "lf = 0.02; \t\t\t\t#seepage loss for unlined channel\n", - "p2 = 20.; \t\t\t\t#wetted perimeter for unlined channel\n", - "\n", - "#assume 1 km length of canal\n", - "#annual benifit\n", - "\n", - "\t\t\t\t#(1).seepage\n", - "A1 = p1*1000; \t\t\t\t#area of wetted perimeter\n", - "li = li*p1/1000; \t\t\t\t#seepage loss\n", - "A2 = p2*1000; \t\t\t\t#area of wetted perimmeter for unlined channel\n", - "lf = p2*lf/1000; \t\t\t\t#seepage loss for unlined channel\n", - "s = li-lf; \t\t\t\t#saving in water loss\n", - "a1 = s*p1*100000; \t\t\t\t#annual revenue saved\n", - "\n", - "\t\t\t\t#(2)maintainence\n", - "a2 = 0.4*25000; \t\t\t\t#saving in maintainance math.cost\n", - "ts = a1+a2; \t\t\t\t#total annual benifit\n", - "\n", - "\t\t\t\t#annual math.cost\n", - "A1 = p2*1000; \t\t\t\t#area of lining for unlinrd canal\n", - "C = 100*A1; \t\t\t\t#math.cost of lining\n", - "\t\t\t\t#interest rate is 6%\n", - "i = 0.06;\n", - "N = 50;\n", - "a = (C*i*(i+1)**N)/((1+i)**N-1); \t\t\t\t#annual math.cost of lining or capital recovery factor\n", - "bcr = ts/a; \t\t\t\t#benifit math.cost ratio\n", - "bcr = round(bcr*1000)/1000;\n", - "print \"Benifit math.cost ratio = %.2f.\"%(bcr);\n", - "\t\t\t\t#as bcr>1\n", - "print \" ;Since it is more than 1.Hence, it is economically justifiable. \";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Benifit math.cost ratio = 1.30.\n", - " ;Since it is more than 1.Hence, it is economically justifiable. \n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.10 pg : 776" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Ecd = 20; \t\t\t\t#electrical conductivity of drainage water\n", - "Eci = 1.5; \t\t\t\t#m mho/cm\n", - "Dc = 55.5; \t\t\t\t#consumptive use\n", - "\n", - "\n", - "# Calculations\n", - "Lr = Eci/Ecd;\n", - "D = Dc/(1-Lr);\n", - "\n", - "# Results\n", - "print \"required depth of water to be applied = %i mm.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required depth of water to be applied = 60 mm.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.11 pg : 776" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Eci = 1.4; \t\t\t\t# m mho/cm\n", - "Ece = 11.; \t\t\t\t#saturated extract of soil\n", - "Dc = 85.; \t\t\t\t#consumptive use requirement of crop\n", - "\n", - "\n", - "# Calculations\n", - "#let us assume Ecd = 2Ece\n", - "Lr = Eci/(2*Ece);\n", - "Di = Dc/(1-Lr);\n", - "Di = round(Di*10)/10;\n", - "\n", - "\n", - "# Results\n", - "print \"required depth of water to be applied = %.2f mm.\"%(Di);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required depth of water to be applied = 90.80 mm.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.12 pg : 776" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#percentage of earth work is saved in lined section\n", - "\n", - "#Given\n", - "s = 1.5; \t\t\t\t#side slope\n", - "Q = 15.; \t\t\t\t#discharge\n", - "S = 1./4000; \t\t\t\t#bed slope\n", - "Nl = 0.014; \t\t\t\t#manning n for lined channel\n", - "Nu = 0.028; \t\t\t\t#manning n for ulined channel \n", - "fb = 0.75; \t\t\t\t#free board\n", - "\n", - "#considering the perimeter of trapezoidal section\n", - "#taking minimum perimeter for given area\n", - "#i.e dP/dD = 0\n", - "#we get\n", - "#A = 2.1D**2; R = D/2; and P = 4.2D\n", - "\n", - "#for linrd channel\n", - "#Q = AR**(2/3)*S**0.5\n", - "#substituting above values we get\n", - "D = (10.0396)**(3./8);\n", - "B = 0.6*D;\n", - "R = D/2;\n", - "tau = 9.81*R*S*1000;\n", - "tau = round(tau*1000)/1000;\n", - "print \"for lined canal:\";\n", - "print \"average boundary shear stress = %.2f N/square m.\"%(tau);\n", - "Dc = D+fb; \t\t\t\t#total depth of cutting\n", - "A1 = (B+1.5*Dc)*Dc;\n", - "\n", - "#for unlined channel\n", - "#Q = AR**(2/3)*S**0.5\n", - "#substituting above values we get\n", - "D = 3.08;\n", - "B = 0.6*D;\n", - "R = D/2;\n", - "tau = 9.81*R*S*1000;\n", - "tau = round(tau*100)/100;\n", - "print \"for unlined canal:\";\n", - "print \"average boundary shear stress = %.2f N/square m.\"%(tau);\n", - "Dc = D+fb; \t\t\t\t#total depth of cutting\n", - "A2 = (B+1.5*Dc)*Dc;\n", - "per = (A2-A1)*100/A2; \n", - "per = round(per*100)/100;\n", - "print \"percent saving of earth = %.2f percent.\"%(per);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "for lined canal:\n", - "average boundary shear stress = 2.91 N/square m.\n", - "for unlined canal:\n", - "average boundary shear stress = 3.78 N/square m.\n", - "percent saving of earth = 34.32 percent.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.13 pg : 778" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "#design a lined canal\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 100.; \t\t\t\t#discharge\n", - "S = 1./2500; \t\t\t\t#bed slope\n", - "V = 2.; \t\t\t\t#maximum permissible velocity\n", - "n = 0.013; \t\t\t\t#manning n\n", - "s = 1.25; \t\t\t\t#side slope\n", - "\n", - "\n", - "# Calculations\n", - "A = Q/V;\n", - "#from manning formula V = (R**2/3*S**1/2)/N;\n", - "R = (V*n/S**0.5)**1.5;\n", - "P = A/R;\n", - "\n", - "#now umath.sing the equation of area and perimeter of trapezoid\n", - "#area = D(B+2.5D)\n", - "#perimeter = B+3.2D;\n", - "#we get\n", - "y = [1.95,-33.73,50]\n", - "D = roots(y)[1];\n", - "#we get D = 15.660087 and 1.6373489\n", - "#taking D = 1.6373489;\n", - "B = P-3.2*D;\n", - "B = round(B*10)/10;\n", - "D = round(D*100)/100;\n", - "\n", - "# Results\n", - "print \"required bed width = %.2f m.\"%(B);\n", - "print \"required bed depth = %.2f m\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required bed width = 28.50 m.\n", - "required bed depth = 1.64 m\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.14 pg : 778" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "B = 5.; \t\t\t\t#bed width\n", - "D = 2.; \t\t\t\t#bed depth\n", - "S = 1./1600; \t\t\t\t#bed slope\n", - "n = 0.015; \t\t\t\t#manning n\n", - "\n", - "\n", - "# Calculations and Results\n", - "A = B+2*D; \t\t\t\t#area of lining\n", - "#let B1 and D1 be new width and depth of bed\n", - "#for getting maximum discharge we diffrentiate Q and equating it to zero\n", - "#Q = S**0.5*B1D1**5/3/n\n", - "#we get\n", - "D1 = 45./16;\n", - "B1 = 9-2*D1;\n", - "Q1 = S**0.5*B1*D1**5/3/n;\n", - "D1 = round(D1*10000)/10000;\n", - "print \"new width of bed = %.2f m.\"%(B1);\n", - "print \"new depth of bed = %.2f m.\"%(D1);\n", - "print \" maximum discharge = %.2f cumec.\"%(Q1);\n", - "R = D;\n", - "V = R**(2./3)*S**0.5/n;\n", - "F = V/(9.81*D)**0.5; \t\t\t\t#froud number\n", - "R = D1;\n", - "V = R**(2./3)*S**0.5/n;\n", - "F = V/(9.81*D1)**0.5; \t\t\t\t#froud number\n", - "print \"Froud number is less than 1 in both case.Hence flow doesnot change from sub-critical to super critical.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "new width of bed = 3.38 m.\n", - "new depth of bed = 2.81 m.\n", - " maximum discharge = 329.96 cumec.\n", - "Froud number is less than 1 in both case.Hence flow doesnot change from sub-critical to super critical.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 16.15 pg : 779" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from sympy import acot\n", - "\n", - "#area to be irrigated\n", - "\t\t\t\t\n", - "#Given\n", - "B = 5.; \t\t\t\t#bed width\n", - "D = 2.5; \t\t\t\t#bed depth\n", - "s = 1.5; \t\t\t\t#side slope\n", - "S = 1./1000; \t\t\t\t#bed slope\n", - "n = 0.016; \t\t\t\t#manning n\n", - "k = 10.; \t\t\t\t#kor period\n", - "d = 150.; \t\t\t\t#field irrigation requirement \n", - "\n", - "\n", - "# Calculations\n", - "theta = acot(s);\n", - "A = B*D+D**2*(theta+1/math.tan(theta));\n", - "P = B+2*D*(theta+1/math.tan(theta));\n", - "R = A/P;\n", - "Q = A*R**(2./3)*S**0.5/n;\n", - "V = Q*k*24*3600; \t\t\t\t#volum of water supply by channel\n", - "A = V*10/(d*10000);\n", - "Q = round(Q*100)/100;\n", - "A = round(A)*100;\n", - "\n", - "# Results\n", - "print \"maximum carrying capacity of canal = %.2f cumec.\"%(Q);\n", - "print \"Area to be irrigated = %.2f hectares.\"%(A);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum carrying capacity of canal = 70.65 cumec.\n", - "Area to be irrigated = 40700.00 hectares.\n" - ] - } - ], - "prompt_number": 15 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch17.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch17.ipynb deleted file mode 100755 index 4cb73dad..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch17.ipynb +++ /dev/null @@ -1,221 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:06ec184185f2d3771e5e9a8334a24a140307b0ca6ee46117ed9968f716f623ca" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 17 : CANAL OUTLETS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 17.1 pg : 788" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "D = 100.0; \t\t\t\t#F.S.L of distributory\n", - "wc = 99.90; \t\t\t\t#F.S.L of water course\n", - "L = 9.; \t\t\t\t#length of pipe\n", - "d = 20.; \t\t\t\t#diameter of pipe\n", - "f = 0.005; \t\t\t\t#coefficient of friction\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "\n", - "# Calculations\n", - "H = D-wc; \t\t\t\t#working head\n", - "C = (d/((1.5*d/(400*f)+L)*f))**0.5/20;\n", - "A = math.pi*d**2/(4*10000);\n", - "q = C*A*(2*g*H)**0.5;\n", - "q = round(q*10000)/10000;\n", - "\n", - "# Results\n", - "print \"discharge through the outlet = %.5f cumec.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge through the outlet = 0.02840 cumec.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 17.2 pg : 788" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a submerged pipe\n", - "\t\t\t\t\n", - "#Given\n", - "q = 0.04; \t\t\t\t#discharge through outlet\n", - "D = 100.0; \t\t\t\t#F.S.L of distributing canal\n", - "wc = 99.90; \t\t\t\t#F.S.L of water course\n", - "dep = 1.1; \t\t\t\t#full supply depth distributing canal\n", - "C = 0.7; \t\t\t\t#average value of coefficient of discharge\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "\n", - "# Calculations\n", - "\n", - "H = D-wc; \t\t\t\t#available head\n", - "A = q/(C*(2*g*H)**0.5);\n", - "d = (4*A/math.pi)**0.5*100;\n", - "d = round(d*10)/10;\n", - "\n", - "# Results\n", - "print \"diameter of pipe required = %.2f cm.\"%(d);\n", - "print \"use pipe of diameter 25 cm.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "diameter of pipe required = 22.80 cm.\n", - "use pipe of diameter 25 cm.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 17.3 pg : 788" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#design submerged pipe\n", - "\t\t\t\t\n", - "#Given\n", - "q = 0.04; \t\t\t\t#discharge through outlet\n", - "D = 100.0; \t\t\t\t#F.S.L of distributing canal\n", - "wc = 99.90; \t\t\t\t#F.S.L of water course\n", - "dep = 1.1; \t\t\t\t#full supply depth distributing canal\n", - "f = 0.01; \t\t\t\t#coefficient of friction\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "L = 9.; \t\t\t\t#Length of pipe\n", - "\n", - "H = D-wc; \t\t\t\t#working head\n", - "\n", - "# Calculations\n", - "#first trial\n", - "#taking d = 22.8 cm\n", - "d = 22.8;\n", - "C = (d/((1.5*d/(400*f)+L)*f))**0.5/20;\n", - "A = q/(C*(2*g*H)**0.5);\n", - "d = (4*A/math.pi)**0.5*100;\n", - "\t\t\t\t#second trial\n", - "C = (d/((1.5*d/(400*f)+L)*f))**0.5/20;\n", - "A = q/(C*(2*g*H)**0.5);\n", - "d = (4*A/math.pi)**0.5*100;\n", - "d = round(d*100)/100;\n", - "\n", - "# Results\n", - "print \"diameter of pipe required = %.2f cm.\"%(d);\n", - "print \"provide diameter of pipe as 25 cm.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "diameter of pipe required = 24.94 cm.\n", - "provide diameter of pipe as 25 cm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 17.4 pg : 795" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design an open flume outlet\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 0.06; \t\t\t\t#discharge\n", - "D = 0.85; \t\t\t\t#full supply depth\n", - "Hw = 15.; \t\t\t\t#available working head\n", - "Bt = 7.;\n", - "C = 1.6; \t\t\t\t#let us choose\n", - "\n", - "# Calculations and Results\n", - "H = (Q*100/(C*Bt))**(2./3);\n", - "mh = 0.2*H; \t\t\t\t#minimum modular head\n", - "mh = round(mh*1000)/1000;\n", - "print \"minimum modular head = %.2f m. < available working head.hemce,design is safe.\"%(mh);\n", - "o = H/D;\n", - "o = round(o*1000)/1000;\n", - "print \"setting of outlet = %.2f. <0.9.hence,outlet will work as hyper propotional outlet.\"%(o);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "minimum modular head = 0.13 m. < available working head.hemce,design is safe.\n", - "setting of outlet = 0.78. <0.9.hence,outlet will work as hyper propotional outlet.\n" - ] - } - ], - "prompt_number": 5 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch17_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch17_1.ipynb deleted file mode 100755 index 4cb73dad..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch17_1.ipynb +++ /dev/null @@ -1,221 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:06ec184185f2d3771e5e9a8334a24a140307b0ca6ee46117ed9968f716f623ca" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 17 : CANAL OUTLETS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 17.1 pg : 788" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "D = 100.0; \t\t\t\t#F.S.L of distributory\n", - "wc = 99.90; \t\t\t\t#F.S.L of water course\n", - "L = 9.; \t\t\t\t#length of pipe\n", - "d = 20.; \t\t\t\t#diameter of pipe\n", - "f = 0.005; \t\t\t\t#coefficient of friction\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "\n", - "# Calculations\n", - "H = D-wc; \t\t\t\t#working head\n", - "C = (d/((1.5*d/(400*f)+L)*f))**0.5/20;\n", - "A = math.pi*d**2/(4*10000);\n", - "q = C*A*(2*g*H)**0.5;\n", - "q = round(q*10000)/10000;\n", - "\n", - "# Results\n", - "print \"discharge through the outlet = %.5f cumec.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge through the outlet = 0.02840 cumec.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 17.2 pg : 788" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a submerged pipe\n", - "\t\t\t\t\n", - "#Given\n", - "q = 0.04; \t\t\t\t#discharge through outlet\n", - "D = 100.0; \t\t\t\t#F.S.L of distributing canal\n", - "wc = 99.90; \t\t\t\t#F.S.L of water course\n", - "dep = 1.1; \t\t\t\t#full supply depth distributing canal\n", - "C = 0.7; \t\t\t\t#average value of coefficient of discharge\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "\n", - "# Calculations\n", - "\n", - "H = D-wc; \t\t\t\t#available head\n", - "A = q/(C*(2*g*H)**0.5);\n", - "d = (4*A/math.pi)**0.5*100;\n", - "d = round(d*10)/10;\n", - "\n", - "# Results\n", - "print \"diameter of pipe required = %.2f cm.\"%(d);\n", - "print \"use pipe of diameter 25 cm.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "diameter of pipe required = 22.80 cm.\n", - "use pipe of diameter 25 cm.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 17.3 pg : 788" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#design submerged pipe\n", - "\t\t\t\t\n", - "#Given\n", - "q = 0.04; \t\t\t\t#discharge through outlet\n", - "D = 100.0; \t\t\t\t#F.S.L of distributing canal\n", - "wc = 99.90; \t\t\t\t#F.S.L of water course\n", - "dep = 1.1; \t\t\t\t#full supply depth distributing canal\n", - "f = 0.01; \t\t\t\t#coefficient of friction\n", - "g = 9.81; \t\t\t\t#acceleration due to gravity\n", - "L = 9.; \t\t\t\t#Length of pipe\n", - "\n", - "H = D-wc; \t\t\t\t#working head\n", - "\n", - "# Calculations\n", - "#first trial\n", - "#taking d = 22.8 cm\n", - "d = 22.8;\n", - "C = (d/((1.5*d/(400*f)+L)*f))**0.5/20;\n", - "A = q/(C*(2*g*H)**0.5);\n", - "d = (4*A/math.pi)**0.5*100;\n", - "\t\t\t\t#second trial\n", - "C = (d/((1.5*d/(400*f)+L)*f))**0.5/20;\n", - "A = q/(C*(2*g*H)**0.5);\n", - "d = (4*A/math.pi)**0.5*100;\n", - "d = round(d*100)/100;\n", - "\n", - "# Results\n", - "print \"diameter of pipe required = %.2f cm.\"%(d);\n", - "print \"provide diameter of pipe as 25 cm.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "diameter of pipe required = 24.94 cm.\n", - "provide diameter of pipe as 25 cm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 17.4 pg : 795" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design an open flume outlet\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 0.06; \t\t\t\t#discharge\n", - "D = 0.85; \t\t\t\t#full supply depth\n", - "Hw = 15.; \t\t\t\t#available working head\n", - "Bt = 7.;\n", - "C = 1.6; \t\t\t\t#let us choose\n", - "\n", - "# Calculations and Results\n", - "H = (Q*100/(C*Bt))**(2./3);\n", - "mh = 0.2*H; \t\t\t\t#minimum modular head\n", - "mh = round(mh*1000)/1000;\n", - "print \"minimum modular head = %.2f m. < available working head.hemce,design is safe.\"%(mh);\n", - "o = H/D;\n", - "o = round(o*1000)/1000;\n", - "print \"setting of outlet = %.2f. <0.9.hence,outlet will work as hyper propotional outlet.\"%(o);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "minimum modular head = 0.13 m. < available working head.hemce,design is safe.\n", - "setting of outlet = 0.78. <0.9.hence,outlet will work as hyper propotional outlet.\n" - ] - } - ], - "prompt_number": 5 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch18.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch18.ipynb deleted file mode 100755 index 11ccf93e..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch18.ipynb +++ /dev/null @@ -1,733 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:9d6614c4fb702767c601f4bf4c9d2e616fc5722613b10aacef812ff7d5ad3183" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 18 : CANAL REGULATION WORKS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 18.1 pg : 811" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design Sarda type fall\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 40.; \t\t\t\t#full supply discharge\n", - "sl_u = 218.3; \t\t\t\t#supply level at upstream\n", - "sl_d = 216.8; \t\t\t\t#supply level at downstream\n", - "D = 1.8; \t\t\t\t#suplly depth\n", - "L = 26.; \t\t\t\t#bed width\n", - "bl_u = 216.5; \t\t\t\t#bed level upstream\n", - "bl_d = 215.; \t\t\t\t#bed level downstream\n", - "drop = 1.5;\n", - "\n", - "#from the eqauation; Q = 1.99LH**1.5*(H/B)**(1/6);\n", - "#B = 0.55*(H+d)**0.5;\n", - "#H+d = drop+D;\n", - "#we get\n", - "H = (0.774)**0.6;\n", - "d = 3.3-H;\n", - "Hc = D-H;\n", - "d = round(d*100)/100;\n", - "H = round(H*100)/100;\n", - "Hc = round(Hc*100)/100;\n", - "print \"H = %.2f m.d = %.2f m.\"%(H,d);\n", - "print \"crest height above bed = %.2f m.\"%(Hc);\n", - "\n", - "\t\t\t\t#adopt trapezoidal crest\n", - "B = 1; \t\t\t\t#top width\n", - "print \"D/S batter = 1:3; U/S batter = 1:8.\";\n", - "Va = Q/((27+D)*D);\n", - "vh = Va**2/(2*9.81);\n", - "tel_up = sl_u+vh;\n", - "crest = sl_u-H;\n", - "E = sl_u-crest;\n", - "print \"R.L of crest = %.2f m.\"%(crest);\n", - "print 'E = %.2f m.'%(E);\n", - "\t\t\t\t#design of cistern\n", - "x = (E*drop)**(2/3)/4; \t\t\t\t#depth of cistern\n", - "lc = 5*(E*drop)**0.5; \t\t\t\t#length of cistern\n", - "cb = bl_d-x;\n", - "x = round(x*100)/100;\n", - "cb = round(cb*1000)/1000;\n", - "lc = round(lc*10)/10;\n", - "print \"depth of cistern = %.2f m.\"%(x);\n", - "print \"length of cistern = %.2f m.\"%(lc);\n", - "print \"R.L of bed of cistern = %.2f m.\"%(cb);\n", - "print \"keep cistern at R.L 214.69.\";\n", - "\t\t\t\t#design of impervious floor\n", - "Hs = 2.44; \t\t\t\t#seepage head\n", - "c = 8.; \t\t\t\t#Bligh's coefficient\n", - "li = Hs*c;\n", - "d1 = 1;d2 = 1.6;\n", - "vl = 2*(d1+d2);\n", - "lh = li-vl;\n", - "print \"design of impervious floor:\";\n", - "print \"provide upstream cut-off = %i m.; downstream cut-off = %.2f m.\"%(d1,d2);\n", - "print \"length of horizontal impervious floor = %.2f m.\"%(lh);\n", - "print \"provide 15 m length impervious floor.\";\n", - "ld = 2*(D+1.2)+drop;\n", - "print \"minimum length of impervious floor to the d/s of toe of crest wall = %.2f m.\"%(ld);\n", - "print \"provide ld = 8 m.\";\n", - "bl = 15-8;\n", - "print \"the balance of the length %i m is to be provided under and u/s of the crest.\"%(bl);\n", - "\n", - "tcl = 15+2*(1+16);\n", - "print \"uplift pressure is counter balanced by weigth of water. hence provide thickness of 0.4 m.\";\n", - "rho = 2.24; \n", - "static = 2.44*(1-0.446)+x;\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"for other points; thickness required = %.2f m.\"%(t);\n", - "print \"provide thickness of 1.40 m.\";\n", - "print \"at downstream end of floor provide thickness of 0.6 m overlaid by 0.2 m brick pitching.\";\n", - "\n", - "n = d2/(Hs*5); \t\t\t\t#n = 1/math.pi*(lambda)**0.5\n", - "\t\t\t\t#from khosla exit curve we get\n", - "alpha = 10.5;\n", - "lambda1 = (1/(math.pi*n))**2;\n", - "alpha = ((2*lambda1-1)**2-1)**0.5;\n", - "b = alpha*d2;\n", - "b = round(b*100)/100;\n", - "print \"checking of floor thickness by khosla theory:\";\n", - "print \"length of floor provided = %.2f m. > length by Bligh theory.\"%(b);\n", - "b = 15;\n", - "d2 = 1.8;\n", - "alpha = b/d2;\n", - "n = 0.145;\n", - "Ge = Hs*n/d2;\n", - "Ge = round(Ge*10)/10;\n", - "print \"exit gradient after increase in depth cut-off = %.2f. which is in permissible limit\"%(Ge);\n", - "print 'provide depth cut-off to 1.8 m.';\n", - "\t\t\t\t#calculation of pressure\n", - "print \"calculation of pressure:\";\n", - "print \"U/S cut-off:\";\n", - "d1 = 1.;\n", - "b = 15.;\n", - "alpha_ = d1/b;\n", - "fic1 = 100-24;\n", - "fid1 = 100-17;\n", - "t = 0.4;\n", - "fic1 = fic1+(fid1-fic1)*t/d1;\n", - "print \"corrected fic1 = %.2f percent.\"%(fic1);\n", - "print \"D/S cut-off wall:\";\n", - "d2 = 1.8;\n", - "b = 15.;\n", - "alpha_ = d1/b;\n", - "fie2 = 31.;\n", - "fid2 = 21.5;\n", - "t = 0.6;\n", - "fie2 = fie2-(fie2-fid2)*t/1.8;\n", - "fie2 = round(fie2*10)/10;\n", - "print \"correcte fie2 = %.2f percent.\"%(fie2);\n", - "\t\t\t\t#calculation of thickness\n", - "print \"provide a minimum thickness of 0.4 m for u/s floor.\";\n", - "pre = fie2+(fic1-fie2)*8/b;\n", - "static = pre*Hs/100+x;\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"thickness at d/s toe of crest = %.2f m.\"%(t);\n", - "print \"provide thickness of 1.4 m thick concrete overlaid by 0.2 m brick pitching.\";\n", - "pre = fie2+(fic1-fie2)*5/b;\n", - "static = pre*Hs/100+x;\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"thickness at 3 m from d/s toe of crest = %.2f m.\"%(t);\n", - "print \"provide thickness of 1.2 m thick concrete overlaid by 0.2 m brick pitching.\";\n", - "pre = fie2+(fic1-fie2)*2/b;\n", - "static = pre*Hs/100; \t\t\t\t#calculation is wrong in book\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"thickness at 6m from d/s toe of crest = %.2f m.\"%(t);\n", - "print \"provide thickness of 0.7 m thick concrete overlaid by 0.2 m brick pitching.\";\n", - "\t\t\t\t#design of downstream wings\n", - "wing = 6*(E*drop)**0.5;\n", - "hw = D+0.5;\n", - "print \"heigth of top of downstream wings above the bed = %.2f m.\"%(hw);\n", - "projec = hw*3;\n", - "print \"length of warped wing measured along centre line of canal = %.2f m.\"%(projec);\n", - "\t\t\t\t#downstream pitching\n", - "l = 9+2*1.5;\n", - "print \"length of bed pitching = %.2f m.\"%(l);\n", - "print \"length of sloping pitching = 7 m.length of horizontal pitching = 6 m.\";\n", - "print \"provide one toe wall of 1 m depth and 0.4 m width.\";\n", - "print \"side pitching is curtailed at 45 degree from the end of bed pitching \\\n", - "in plan.supprot the side pitching on toe wall 0.4 m thick and 1 m deep. \";\n", - "\t\t\t\t#energy dissipators\n", - "q = Q/L;\n", - "dc = (q**2/9.81)**(1./3);\n", - "print \"size and position of friction blocks:\";\n", - "L = 2*dc;\n", - "w = dc;\n", - "h = dc;\n", - "di = 1.5*dc;\n", - "L = round(L*10)/10;\n", - "w = round(w*10)/10;\n", - "h = round(h*10)/10;\n", - "di = round(di);\n", - "print \"length of block = %.2f m.width of block = %.2f m.height of block = %.2f \\\n", - "m.dismath.tance from toe of crest = %.2f m.\"%(L,w,h,di);\n", - "print \"provide two rows staggered ata dismath.tance of 1 m from toe of crest.\";\n", - "print \"size and position of cube blocks:\";\n", - "L = D/10;\n", - "w = D/10;\n", - "h = w;\n", - "L = round(L*10)/10;\n", - "w = round(w*10)/10;\n", - "h = round(h*10)/10;\n", - "print \"length of block = %.2f m.width of block = %.2f m.height of block = %.2f m.\"%(L,w,h);\n", - "print \"provide two rows staggered at the end of impervious floor.\";\n", - "\t\t\t\t#u/s approach\n", - "r = 6*H;\n", - "print \"provide wing wall segmental with 5 m radius subtending angle of 60 degree at the centre.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "H = 0.86 m.d = 2.44 m.\n", - "crest height above bed = 0.94 m.\n", - "D/S batter = 1:3; U/S batter = 1:8.\n", - "R.L of crest = 217.44 m.\n", - "E = 0.86 m.\n", - "depth of cistern = 0.25 m.\n", - "length of cistern = 5.70 m.\n", - "R.L of bed of cistern = 214.75 m.\n", - "keep cistern at R.L 214.69.\n", - "design of impervious floor:\n", - "provide upstream cut-off = 1 m.; downstream cut-off = 1.60 m.\n", - "length of horizontal impervious floor = 14.32 m.\n", - "provide 15 m length impervious floor.\n", - "minimum length of impervious floor to the d/s of toe of crest wall = 7.50 m.\n", - "provide ld = 8 m.\n", - "the balance of the length 7 m is to be provided under and u/s of the crest.\n", - "uplift pressure is counter balanced by weigth of water. hence provide thickness of 0.4 m.\n", - "for other points; thickness required = 1.29 m.\n", - "provide thickness of 1.40 m.\n", - "at downstream end of floor provide thickness of 0.6 m overlaid by 0.2 m brick pitching.\n", - "checking of floor thickness by khosla theory:\n", - "length of floor provided = 17.18 m. > length by Bligh theory.\n", - "exit gradient after increase in depth cut-off = 0.20. which is in permissible limit\n", - "provide depth cut-off to 1.8 m.\n", - "calculation of pressure:\n", - "U/S cut-off:\n", - "corrected fic1 = 78.80 percent.\n", - "D/S cut-off wall:\n", - "correcte fie2 = 27.80 percent.\n", - "provide a minimum thickness of 0.4 m for u/s floor.\n", - "thickness at d/s toe of crest = 1.28 m.\n", - "provide thickness of 1.4 m thick concrete overlaid by 0.2 m brick pitching.\n", - "thickness at 3 m from d/s toe of crest = 1.08 m.\n", - "provide thickness of 1.2 m thick concrete overlaid by 0.2 m brick pitching.\n", - "thickness at 6m from d/s toe of crest = 0.68 m.\n", - "provide thickness of 0.7 m thick concrete overlaid by 0.2 m brick pitching.\n", - "heigth of top of downstream wings above the bed = 2.30 m.\n", - "length of warped wing measured along centre line of canal = 6.90 m.\n", - "length of bed pitching = 12.00 m.\n", - "length of sloping pitching = 7 m.length of horizontal pitching = 6 m.\n", - "provide one toe wall of 1 m depth and 0.4 m width.\n", - "side pitching is curtailed at 45 degree from the end of bed pitching in plan.supprot the side pitching on toe wall 0.4 m thick and 1 m deep. \n", - "size and position of friction blocks:\n", - "length of block = 1.20 m.width of block = 0.60 m.height of block = 0.60 m.dismath.tance from toe of crest = 1.00 m.\n", - "provide two rows staggered ata dismath.tance of 1 m from toe of crest.\n", - "size and position of cube blocks:\n", - "length of block = 0.20 m.width of block = 0.20 m.height of block = 0.20 m.\n", - "provide two rows staggered at the end of impervious floor.\n", - "provide wing wall segmental with 5 m radius subtending angle of 60 degree at the centre.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 18.2 pg : 820" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design an unflumed straight glacis non-meter fall\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 40.; \t\t\t\t#full supply discharge\n", - "sl_u = 218.3; \t\t\t\t#supply level at upstream\n", - "sl_d = 216.8; \t\t\t\t#supply level at downstream\n", - "D = 1.8; \t\t\t\t#suplly depth\n", - "L = 26.; \t\t\t\t#bed width\n", - "bl_u = 216.5; \t\t\t\t#bed level upstream\n", - "bl_d = 215.; \t\t\t\t#bed level downstream\n", - "drop = 1.5;\n", - "Ge = 1./6; \t\t\t\t#permissible exit gradient\n", - "\n", - "\t\t\t\t#design of crest\n", - "print \"design of crest:\";\n", - "E = (Q/(1.84*L))**(2/3);\n", - "V = Q/((L+D)*D);\n", - "vh = V**2/(2*9.81);\n", - "tel_up = sl_u+vh;\n", - "cl = tel_up-E;\n", - "w = 2*E/3;\n", - "w = round(w*10)/10;\n", - "print \"length of crest = %.2f m.\"%(L);\n", - "print \"width of crest = %.2f m.\"%(w);\n", - "\t\t\t\t#design of cistern\n", - "q = Q/L;\n", - "Hl = 1.5;\n", - "\t\t\t\t#from blench curve\n", - "Ef2 = 1.44;\n", - "cistern = sl_d+0.03-1.25*Ef2;\n", - "print \"R.L of cistern = %.2f m. > d/s bed level.\"%(cistern);\n", - "print \"keep R.L of cistern at 214.5 m.\";\n", - "l = 6*Ef2;\n", - "print \"length of cistern = %.2f m.\"%(l);\n", - "print \"provide cistern of 9 m length \";\n", - "d = bl_d-214.5;\n", - "print \"depth of cistern = %.2f m.\"%(d);\n", - "\n", - "\t\t\t\t#design of impervious floor\n", - "d1 = D/3;\n", - "print \"design of impervious floor:\";\n", - "print \"provide 0.4 m wide and 1 m deep curtain wall at u/s.\";\n", - "d2 = D/2;\n", - "print \"provide 0.4 m wide and 1 m deep curtain wall at d/s.the curtain wall will project the above the d/s bed by 0.18 m.\";\n", - "Hs = cl-bl_d;\n", - "d2 = 1;\n", - "n = d2*Ge/Hs; \t\t\t\t#n = 1/(math.pi*(lambda)**0.5)\n", - "\t\t\t\t#from khosla exit curves we get\n", - "alpha = 40;\n", - "lambda1 = (1/(math.pi*n))**2;\n", - "alpha = ((2*lambda1-1)**2-1)**0.5;\n", - "b = alpha*d2;\n", - "\t\t\t\t#math.since length is to excessive\n", - "d2 = 2;\n", - "n = d2*Ge/Hs; \t\t\t\t#n = 1/(math.pi*(lambda)**0.5)\n", - "\t\t\t\t#from khosla exit curves we get\n", - "alpha = 10;\n", - "lambda1 = (1/(math.pi*n))**2;\n", - "alpha = ((2*lambda1-1)**2-1)**0.5;\n", - "b = alpha*d2+1;\n", - "print \"total length = %i m.length of cistern = 9 m.length of d/s glacis = 5.88\\\n", - " m.width of crest = 0.6 m.length of u/s glacis = 0.47 m.balance to be provided to u/s of the u/s glacis = 4.05 m.\"%(b);\n", - "\n", - "\t\t\t\t#pressure calculations\n", - "print \"pressure calculations:\";\n", - "print \"upstream curtain wall:\";\n", - "d1 = 1.;\n", - "b = 20;\n", - "alpha_ = d1/b;\n", - "t = 0.3;\n", - "fic1 = 100-22;\n", - "fid1 = 100-15;\n", - "corec = (fid1-fic1)*t/d1\n", - "fic1 = fic1+corec;\n", - "print \"corrected fi_c1 = %.2f percent.\"%(fic1);\n", - "print \"downstream curtain wall:\";\n", - "d2 = 2.;b = 20;\n", - "alpha_ = d2/b;\n", - "t = 0.5;\n", - "fie = 29.;\n", - "fid = 21;\n", - "corec = (fie-fid)*t/d2\n", - "fie = fie-corec;\n", - "print \"corrected fi_e = %.2f percent.\"%(fie);\n", - "print \"toe of glacis:\";\n", - "\t\t\t\t#assuming linear variation of pressure\n", - "p = fie+(80-fie)*9/20;\n", - "print \"pressure at downstream of the glacis = %.2f percent.\"%(p);\n", - "\n", - "\t\t\t\t#floor thickness\n", - "rho = 2.24;\n", - "print \"floor thickness:provide minimum thickness of 0.3 m at the u/s floor.\";\n", - "static = p*2.44/100+(bl_d-214.5);\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"floor thickness required at toe of glacis = %.2f m.provide 1.5 m thick floor for length of 3 m.\"%(t);\n", - "p = fie+(80-fie)*6/20;\n", - "static = p*2.44/100+(bl_d-214.5);\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"floor thickness required at 3m from toe of glacis = %.2f m.provide \\\n", - "1.3 m thick floor from 3 m to 6.5 m from toe of glacis.\"%(t);\n", - "t = 0.27*2.44/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"thickness of d/s end of cistern = %.2f m.provide thickness of 0.6 m at d/s end of floor.\"%(t);\n", - "\n", - "\t\t\t\t#design of d/s protection\n", - "print \"no bed protection is needed as deflector wall is provided.\";\n", - "sp = 3*D;\n", - "print \"length of side protection = %.2f m.provide 5.5 m length of 20 cm\\\n", - " thick brick pitching beyond impervious floor.pitching will rest on toe wall 0.4 \\\n", - " m wide and 0.9 m deep.provide 0.4 m wide profile at the end of pitching\"%(sp);\n", - "\t\t\t\t#design of u/s approach\n", - "print \"u/s wing wall is splayed at 45 degree from u/s end of impervious\\\n", - " floor.extend 1 m into earthen banks from line of F.S.L.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "design of crest:\n", - "length of crest = 26.00 m.\n", - "width of crest = 0.70 m.\n", - "R.L of cistern = 215.03 m. > d/s bed level.\n", - "keep R.L of cistern at 214.5 m.\n", - "length of cistern = 8.64 m.\n", - "provide cistern of 9 m length \n", - "depth of cistern = 0.50 m.\n", - "design of impervious floor:\n", - "provide 0.4 m wide and 1 m deep curtain wall at u/s.\n", - "provide 0.4 m wide and 1 m deep curtain wall at d/s.the curtain wall will project the above the d/s bed by 0.18 m.\n", - "total length = 18 m.length of cistern = 9 m.length of d/s glacis = 5.88 m.width of crest = 0.6 m.length of u/s glacis = 0.47 m.balance to be provided to u/s of the u/s glacis = 4.05 m.\n", - "pressure calculations:\n", - "upstream curtain wall:\n", - "corrected fi_c1 = 80.10 percent.\n", - "downstream curtain wall:\n", - "corrected fi_e = 27.00 percent.\n", - "toe of glacis:\n", - "pressure at downstream of the glacis = 50.85 percent.\n", - "floor thickness:provide minimum thickness of 0.3 m at the u/s floor.\n", - "floor thickness required at toe of glacis = 1.40 m.provide 1.5 m thick floor for length of 3 m.\n", - "floor thickness required at 3m from toe of glacis = 1.25 m.provide 1.3 m thick floor from 3 m to 6.5 m from toe of glacis.\n", - "thickness of d/s end of cistern = 0.53 m.provide thickness of 0.6 m at d/s end of floor.\n", - "no bed protection is needed as deflector wall is provided.\n", - "length of side protection = 5.40 m.provide 5.5 m length of 20 cm thick brick pitching beyond impervious floor.pitching will rest on toe wall 0.4 m wide and 0.9 m deep.provide 0.4 m wide profile at the end of pitching\n", - "u/s wing wall is splayed at 45 degree from u/s end of impervious floor.extend 1 m into earthen banks from line of F.S.L.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 18.3 pg : 831" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a cross -regulator and head regulatorfor a distributory channel\n", - "#givrn\n", - "Q = 100.; \t\t\t\t#discharge of parent channel\n", - "Qd = 15.; \t\t\t\t#discharge ofdistributory\n", - "fsl_u = 218.1; \t\t\t\t#F.S.L of upstream parent channel\n", - "fsl_d = 217.9; \t\t\t\t#F.S.L of downstream of parent channel\n", - "bw_u = 42.; \t\t\t\t#bed width of parent channel upstream\n", - "bw_d = 38.; \t\t\t\t#bed width of parent channel downstream\n", - "hw = 2.5; \t\t\t\t#depth of water in parent channel\n", - "fsl_dis = 217.1; \t\t\t\t#F.S.L of distributory\n", - "hw_dis = 1.5; \t\t\t\t#depth of water in distributory\n", - "Ge = 1./5; \t\t\t\t#permissible exit gradient\n", - "\n", - "#design of cross regulator\n", - "print \"DESIGN OF CROSS-REGULATOR::\";\n", - "#design of crest and waterway\n", - "print \"design of crest and waterway:\";\n", - "cl = fsl_u-hw;\n", - "h = fsl_u-fsl_d;\n", - "d = fsl_d-cl;\n", - "C1 = 0.557;C2 = 0.8;\n", - "L = Q/(2*C1*(2*9.81)**0.5*h**1.5/3+C2*d*(2*9.81*h)**0.5);\n", - "L = round(L*10)/10;\n", - "print \"crest level = %.2f m.\"%(cl);\n", - "print \"length of crest = %.2f m.\"%(L);\n", - "tw = 28+4.5;\n", - "print \"provide 3 piers of 1.5 m width each.total width of cross regulator = %.2f m.\"%(tw);\n", - "#design of d/s floor\n", - "L = 28.;\n", - "q = Q/L;\n", - "Hl = fsl_u-fsl_d;\n", - "Ef2 = 1.89; \t\t\t\t#from blench curve\n", - "fl_d = fsl_d-Ef2;\n", - "print \"design of d/s floor:\";\n", - "print \"d/s floor level = %.2f m.; which is higher than d/s bed level.adopt floor level = d/s bed level = 215.40 m.\"%(fl_d);\n", - "Ef1 = Ef2+Hl;\n", - "#from specific energy curve\n", - "D1 = 0.7;\n", - "D2 = 1.65;\n", - "cil = 5*(D2-D1); \t\t\t\t#cistern length\n", - "tl = 2*16/3;\n", - "tl = round(tl*10)/10;\n", - "print \"cistern length = %.2f m.length of d/s floor = %.2f m.\"%(cil,tl);\n", - "#design of impervious floor\n", - "d1 = hw/3+0.6; \t\t\t\t#depth of u/s cut-off\n", - "w = 0.5; \t\t\t\t#width of cut-off\n", - "d2 = hw/2+0.6; \t\t\t\t#deth of d/s cut-off\n", - "d2 = 2; \t\t\t\t#keep\n", - "Hs = fsl_u-(fsl_d-hw); \t\t\t\t#maximum static head\n", - "n = Ge*d2/Hs; \t\t\t\t#n = 1/math.pi*(lambda)**0.5;\n", - "#from exit gradient curves we get\n", - "alpha = 8.;\n", - "n = 0.148;\n", - "b = alpha*d2;\n", - "print \"design of impervious floor:\";\n", - "print \"total length of impervious floor = %i m.;which is divided as-\"%(b);\n", - "print \"d/s floor length = 10.6 m.d/s glacis length with 2:1 slope = 0.4 m.balance to be provided upstream = 5 m.\";\n", - "d1 = 1.5;b = 16;\n", - "alpha_ = d1/b;\n", - "\t\t\t\t#hence\n", - "fic1 = 100-28;\n", - "fid1 = 100-19;\n", - "t = 0.5;\n", - "fic1 = fic1+(fid1-fic1)*t/d1;\n", - "print \"pressure calculation:upstream cut-off:pressure = %.2f percent.\"%(fic1);\n", - "d2 = 2.;\n", - "b = 16;\n", - "alpha_ = d2/b;\n", - "\t\t\t\t#hence\n", - "t = 0.6;\n", - "fie2 = 31.;\n", - "fid2 = 22.;\n", - "fie2 = fie2-(fie2-fid2)*t/d2;\n", - "print \"downstream cut-off:pressure = %.2f percent.\"%(fie2);\n", - "t = 10.6;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "p = round(p*10)/10;\n", - "print \"toe of glacis:pressure = %.2f percent.\"%(p);\n", - "print \"thickness of floor:minimu thickness for u/s floor = 0.5 m.\";\n", - "rho = 2.24;\n", - "t = fie2*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor near d/s cut-off = %.2f m.provide 0.7 m thick floor for last 2.1 m length.\"%(t);\n", - "t = 1.6/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at toe of glacis = %.2f m.\"%(t);\n", - "t = 6.6;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "t = p*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at 4 m from toe of glais = %.2f m.provide 1.1 m thick floor for next 2 m length\"%(t);\n", - "t = 4.6;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "t = p*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at 6 m from toe of glais = %.2f m.provide 0.9 m thick floor for next 2.5 m length\"%(t);\n", - "\n", - "\t\t\t\t#design of u/s protection\n", - "d1 = hw/3+0.6;\n", - "v = d1;\n", - "v = round(v*100)/100;\n", - "print \"design of u/s protection:volume of block protection = %.2f cubic metre/metre.\"%(v);\n", - "print \"keep thickness of protection = 1 m.provide 0.8mx0.8mx0.6m thick concret blocks over 0.4 m thick apron in length of 0.6 m.\";\n", - "cu = 2.25*d1;\n", - "cu = round(cu*100)/100;\n", - "print \"cubic content of launching apron = %.2f cubic metre/metre.provide 1 m thick and 3.5 m long launching apron.\"%(cu);\n", - "\t\t\t\t#design of d/s protection\n", - "d2 = hw/2+0.6;\n", - "v = d2;\n", - "v = round(v*100)/100;\n", - "print \"design of d/s protection:volume of inverted filter = %.2f cubic metre/metre.\"%(v);\n", - "print \"keep thickness of concrete block = 0.6 m.provide 2 rows of 0.8mx0.8mx0.6m thick \\\n", - "concret blocks over 0.6 m graded filter for length of 1.6 m.\";\n", - "cu = 2.25*d2;\n", - "cu = round(cu*100)/100;\n", - "print \"launching apron volume = %.2f cubic metre/metre.provide 1 m thick launching apron\\\n", - " for length of 4.5 m.provide a toe wall 0.4 m wide and 1.5 m deep between filter and launching apron.\"%(cu);\n", - "\n", - "\t\t\t\t#design of head regulator\n", - "print \"DESIGN OF DISTRIBUTORY HEAD REGULATOR::\";\n", - "\t\t\t\t#design of crest and waterway\n", - "print \"design of crest and waterway:\";\n", - "cl = fsl_u-hw+0.5;\n", - "h = fsl_u-fsl_dis;\n", - "d = fsl_dis-cl;\n", - "C1 = 0.557;C2 = 0.8;\n", - "L = Qd/(2*C1*(2*9.81)**0.5*h**1.5/3+C2*d*(2*9.81*h)**0.5);\n", - "L = round(L*100)/100;\n", - "print \"crest level = %.2f m.\"%(cl);\n", - "print \"length of crest = %.2f m.\"%(L);\n", - "print \"provide 2 bays of 3.5 m each with a 1 m thick pier in between.\";\n", - "tw = 8;\n", - "print \"total width of cross regulator = %.2f m.\"%(tw);\n", - "\t\t\t\t#design of d/s floor\n", - "L = 7.5;\n", - "q = Q/L;\n", - "Hl = fsl_u-fsl_dis;\n", - "Ef2 = 1.58; \t\t\t\t#from blench curve\n", - "fl_d = fsl_dis-Ef2;\n", - "print \"design of d/s floor:\";\n", - "print \"d/s floor level = %.2f m.;keepR.L of d/s floor = 215.50 m.\"%(fl_d);\n", - "Ef1 = Ef2+Hl;\n", - "\t\t\t\t#from specific energy curve\n", - "D1 = 0.42;D2 = 2.55;\n", - "cil = 5*(D2-D1); \t\t\t\t#cistern length\n", - "tl = 2*14/3;\n", - "print \"cistern length = %.2f m.\"%(cil);\n", - "\n", - "\t\t\t\t#design of impervious floor\n", - "d1 = hw/3+0.6; \t\t\t\t#depth of u/s cut-off\n", - "w = 0.5; \t\t\t\t#width of cut-off\n", - "d2 = hw_dis/2+0.6; \t\t\t\t#deth of d/s cut-off\n", - "d2 = 2; \t\t\t\t#keep\n", - "Hs = fsl_u-215.5; \t\t\t\t#maximum static head\n", - "n = Ge*d2/Hs; \t\t\t\t#n = 1/math.pi*(lambda)**0.5;\n", - "\t\t\t\t#from exit gradient curves we get\n", - "alpha = 7;n = 0.154;\n", - "b = alpha*d2;\n", - "print \"design of impervious floor:\";\n", - "print \"total length of impervious floor = %i m.;which is divided as-\"%(b);\n", - "print \"length below the toe of glacis = 10.5 mlength of d/s glacis at 2:1 slope =\\\n", - " 1.2 m.width of crest = 1 m.length of u/s glacis at 1:1 slope = 0.5 m.u/s floor:balnce = 0.8 m.\";\n", - "d1 = 1.5;\n", - "b = 16.;\n", - "alpha_ = d1/b;\n", - "\t\t\t\t#hence\n", - "fic1 = 100-28;\n", - "fid1 = 100-19;\n", - "t = 0.5;\n", - "fic1 = fic1+(fid1-fic1)*t/d1;\n", - "print \"pressure calculation:upstream cut-off:pressure = %.2f percent.\"%(fic1);\n", - "d2 = 2.;\n", - "b = 16;\n", - "alpha_ = d2/b;\n", - "\t\t\t\t#hence\n", - "t = 0.6;\n", - "fie2 = 31.;\n", - "fid2 = 22;\n", - "fie2 = fie2-(fie2-fid2)*t/d2;\n", - "print \"downstream cut-off:pressure = %.2f percent.\"%(fie2);\n", - "t = 10.6;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "p = round(p*100)/100;\n", - "print \"toe of glacis:pressure = %.2f percent.\"%(p);\n", - "print \"thickness of floor:minimu thickness for u/s floor = 0.5 m.\";\n", - "rho = 2.24;\n", - "t = p*2.6/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness under the crest = 1 m.\";\n", - "print \"thickness of floor at toe of glacis = %.2f m.\"%(t);\n", - "t = 9.5;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "t = p*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at 2 m from toe of glais = %.2f m.provide 1.1 m thick floor for next 4 m length\"%(t);\n", - "t = 4.5;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "t = p*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at 6 m from toe of glais = %.2f m.provide 0.9 m thick floor for next 2.5 m length\"%(t);\n", - "t = 2;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "t = p*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at 8.5 m from toe of glais = %.2f m.provide 0.7 m thick floor for next 2 m length\"%(t);\n", - "\n", - "\t\t\t\t#design of upstream protection\n", - "d = hw/3+0.6;\n", - "d = round(d*10)/10;\n", - "print \"design of u/s protection:u/s scour depth = %.2f m.provide same protection as in cross regulator\"%(d);\n", - "\n", - "\t\t\t\t#design of d/s protection\n", - "d2 = hw_dis/2+0.6;\n", - "v = d2;\n", - "print \"design of d/s protection:volume of inverted filter = %.2f cubic metre/metre.\"%(v);\n", - "print \"keep thickness of concrete block = 0.5 m.provide 2 rows of \\\n", - "0.8mx0.8mx0.5m thick concret blocks over 0.5 m thick graded filter.\";\n", - "cu = 2.25*d2;\n", - "print \"launching apron volume = %.2f cubic metre/metre.provide 1 m thick \\\n", - "launching apron for length of 3.5 m.provide a masonary toe wall 0.4 m wide and 1.2 m deep between filter and launching apron.\"%(cu);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "DESIGN OF CROSS-REGULATOR::\n", - "design of crest and waterway:\n", - "crest level = 215.60 m.\n", - "length of crest = 26.40 m.\n", - "provide 3 piers of 1.5 m width each.total width of cross regulator = 32.50 m.\n", - "design of d/s floor:\n", - "d/s floor level = 216.01 m.; which is higher than d/s bed level.adopt floor level = d/s bed level = 215.40 m.\n", - "cistern length = 4.75 m.length of d/s floor = 10.00 m.\n", - "design of impervious floor:\n", - "total length of impervious floor = 16 m.;which is divided as-\n", - "d/s floor length = 10.6 m.d/s glacis length with 2:1 slope = 0.4 m.balance to be provided upstream = 5 m.\n", - "pressure calculation:upstream cut-off:pressure = 75.00 percent.\n", - "downstream cut-off:pressure = 28.30 percent.\n", - "toe of glacis:pressure = 59.20 percent.\n", - "thickness of floor:minimu thickness for u/s floor = 0.5 m.\n", - "thickness of floor near d/s cut-off = 0.62 m.provide 0.7 m thick floor for last 2.1 m length.\n", - "thickness of floor at toe of glacis = 1.29 m.\n", - "thickness of floor at 4 m from toe of glais = 1.04 m.provide 1.1 m thick floor for next 2 m length\n", - "thickness of floor at 6 m from toe of glais = 0.91 m.provide 0.9 m thick floor for next 2.5 m length\n", - "design of u/s protection:volume of block protection = 1.43 cubic metre/metre.\n", - "keep thickness of protection = 1 m.provide 0.8mx0.8mx0.6m thick concret blocks over 0.4 m thick apron in length of 0.6 m.\n", - "cubic content of launching apron = 3.23 cubic metre/metre.provide 1 m thick and 3.5 m long launching apron.\n", - "design of d/s protection:volume of inverted filter = 1.85 cubic metre/metre.\n", - "keep thickness of concrete block = 0.6 m.provide 2 rows of 0.8mx0.8mx0.6m thick concret blocks over 0.6 m graded filter for length of 1.6 m.\n", - "launching apron volume = 4.16 cubic metre/metre.provide 1 m thick launching apron for length of 4.5 m.provide a toe wall 0.4 m wide and 1.5 m deep between filter and launching apron.\n", - "DESIGN OF DISTRIBUTORY HEAD REGULATOR::\n", - "design of crest and waterway:\n", - "crest level = 216.10 m.\n", - "length of crest = 2.89 m.\n", - "provide 2 bays of 3.5 m each with a 1 m thick pier in between.\n", - "total width of cross regulator = 8.00 m.\n", - "design of d/s floor:\n", - "d/s floor level = 215.52 m.;keepR.L of d/s floor = 215.50 m.\n", - "cistern length = 10.65 m.\n", - "design of impervious floor:\n", - "total length of impervious floor = 14 m.;which is divided as-\n", - "length below the toe of glacis = 10.5 mlength of d/s glacis at 2:1 slope = 1.2 m.width of crest = 1 m.length of u/s glacis at 1:1 slope = 0.5 m.u/s floor:balnce = 0.8 m.\n", - "pressure calculation:upstream cut-off:pressure = 75.00 percent.\n", - "downstream cut-off:pressure = 28.30 percent.\n", - "toe of glacis:pressure = 59.24 percent.\n", - "thickness of floor:minimu thickness for u/s floor = 0.5 m.\n", - "thickness under the crest = 1 m.\n", - "thickness of floor at toe of glacis = 1.24 m.\n", - "thickness of floor at 2 m from toe of glais = 1.22 m.provide 1.1 m thick floor for next 4 m length\n", - "thickness of floor at 6 m from toe of glais = 0.90 m.provide 0.9 m thick floor for next 2.5 m length\n", - "thickness of floor at 8.5 m from toe of glais = 0.74 m.provide 0.7 m thick floor for next 2 m length\n", - "design of u/s protection:u/s scour depth = 1.40 m.provide same protection as in cross regulator\n", - "design of d/s protection:volume of inverted filter = 1.35 cubic metre/metre.\n", - "keep thickness of concrete block = 0.5 m.provide 2 rows of 0.8mx0.8mx0.5m thick concret blocks over 0.5 m thick graded filter.\n", - "launching apron volume = 3.04 cubic metre/metre.provide 1 m thick launching apron for length of 3.5 m.provide a masonary toe wall 0.4 m wide and 1.2 m deep between filter and launching apron.\n" - ] - } - ], - "prompt_number": 3 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch18_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch18_1.ipynb deleted file mode 100755 index 11ccf93e..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch18_1.ipynb +++ /dev/null @@ -1,733 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:9d6614c4fb702767c601f4bf4c9d2e616fc5722613b10aacef812ff7d5ad3183" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 18 : CANAL REGULATION WORKS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 18.1 pg : 811" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design Sarda type fall\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 40.; \t\t\t\t#full supply discharge\n", - "sl_u = 218.3; \t\t\t\t#supply level at upstream\n", - "sl_d = 216.8; \t\t\t\t#supply level at downstream\n", - "D = 1.8; \t\t\t\t#suplly depth\n", - "L = 26.; \t\t\t\t#bed width\n", - "bl_u = 216.5; \t\t\t\t#bed level upstream\n", - "bl_d = 215.; \t\t\t\t#bed level downstream\n", - "drop = 1.5;\n", - "\n", - "#from the eqauation; Q = 1.99LH**1.5*(H/B)**(1/6);\n", - "#B = 0.55*(H+d)**0.5;\n", - "#H+d = drop+D;\n", - "#we get\n", - "H = (0.774)**0.6;\n", - "d = 3.3-H;\n", - "Hc = D-H;\n", - "d = round(d*100)/100;\n", - "H = round(H*100)/100;\n", - "Hc = round(Hc*100)/100;\n", - "print \"H = %.2f m.d = %.2f m.\"%(H,d);\n", - "print \"crest height above bed = %.2f m.\"%(Hc);\n", - "\n", - "\t\t\t\t#adopt trapezoidal crest\n", - "B = 1; \t\t\t\t#top width\n", - "print \"D/S batter = 1:3; U/S batter = 1:8.\";\n", - "Va = Q/((27+D)*D);\n", - "vh = Va**2/(2*9.81);\n", - "tel_up = sl_u+vh;\n", - "crest = sl_u-H;\n", - "E = sl_u-crest;\n", - "print \"R.L of crest = %.2f m.\"%(crest);\n", - "print 'E = %.2f m.'%(E);\n", - "\t\t\t\t#design of cistern\n", - "x = (E*drop)**(2/3)/4; \t\t\t\t#depth of cistern\n", - "lc = 5*(E*drop)**0.5; \t\t\t\t#length of cistern\n", - "cb = bl_d-x;\n", - "x = round(x*100)/100;\n", - "cb = round(cb*1000)/1000;\n", - "lc = round(lc*10)/10;\n", - "print \"depth of cistern = %.2f m.\"%(x);\n", - "print \"length of cistern = %.2f m.\"%(lc);\n", - "print \"R.L of bed of cistern = %.2f m.\"%(cb);\n", - "print \"keep cistern at R.L 214.69.\";\n", - "\t\t\t\t#design of impervious floor\n", - "Hs = 2.44; \t\t\t\t#seepage head\n", - "c = 8.; \t\t\t\t#Bligh's coefficient\n", - "li = Hs*c;\n", - "d1 = 1;d2 = 1.6;\n", - "vl = 2*(d1+d2);\n", - "lh = li-vl;\n", - "print \"design of impervious floor:\";\n", - "print \"provide upstream cut-off = %i m.; downstream cut-off = %.2f m.\"%(d1,d2);\n", - "print \"length of horizontal impervious floor = %.2f m.\"%(lh);\n", - "print \"provide 15 m length impervious floor.\";\n", - "ld = 2*(D+1.2)+drop;\n", - "print \"minimum length of impervious floor to the d/s of toe of crest wall = %.2f m.\"%(ld);\n", - "print \"provide ld = 8 m.\";\n", - "bl = 15-8;\n", - "print \"the balance of the length %i m is to be provided under and u/s of the crest.\"%(bl);\n", - "\n", - "tcl = 15+2*(1+16);\n", - "print \"uplift pressure is counter balanced by weigth of water. hence provide thickness of 0.4 m.\";\n", - "rho = 2.24; \n", - "static = 2.44*(1-0.446)+x;\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"for other points; thickness required = %.2f m.\"%(t);\n", - "print \"provide thickness of 1.40 m.\";\n", - "print \"at downstream end of floor provide thickness of 0.6 m overlaid by 0.2 m brick pitching.\";\n", - "\n", - "n = d2/(Hs*5); \t\t\t\t#n = 1/math.pi*(lambda)**0.5\n", - "\t\t\t\t#from khosla exit curve we get\n", - "alpha = 10.5;\n", - "lambda1 = (1/(math.pi*n))**2;\n", - "alpha = ((2*lambda1-1)**2-1)**0.5;\n", - "b = alpha*d2;\n", - "b = round(b*100)/100;\n", - "print \"checking of floor thickness by khosla theory:\";\n", - "print \"length of floor provided = %.2f m. > length by Bligh theory.\"%(b);\n", - "b = 15;\n", - "d2 = 1.8;\n", - "alpha = b/d2;\n", - "n = 0.145;\n", - "Ge = Hs*n/d2;\n", - "Ge = round(Ge*10)/10;\n", - "print \"exit gradient after increase in depth cut-off = %.2f. which is in permissible limit\"%(Ge);\n", - "print 'provide depth cut-off to 1.8 m.';\n", - "\t\t\t\t#calculation of pressure\n", - "print \"calculation of pressure:\";\n", - "print \"U/S cut-off:\";\n", - "d1 = 1.;\n", - "b = 15.;\n", - "alpha_ = d1/b;\n", - "fic1 = 100-24;\n", - "fid1 = 100-17;\n", - "t = 0.4;\n", - "fic1 = fic1+(fid1-fic1)*t/d1;\n", - "print \"corrected fic1 = %.2f percent.\"%(fic1);\n", - "print \"D/S cut-off wall:\";\n", - "d2 = 1.8;\n", - "b = 15.;\n", - "alpha_ = d1/b;\n", - "fie2 = 31.;\n", - "fid2 = 21.5;\n", - "t = 0.6;\n", - "fie2 = fie2-(fie2-fid2)*t/1.8;\n", - "fie2 = round(fie2*10)/10;\n", - "print \"correcte fie2 = %.2f percent.\"%(fie2);\n", - "\t\t\t\t#calculation of thickness\n", - "print \"provide a minimum thickness of 0.4 m for u/s floor.\";\n", - "pre = fie2+(fic1-fie2)*8/b;\n", - "static = pre*Hs/100+x;\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"thickness at d/s toe of crest = %.2f m.\"%(t);\n", - "print \"provide thickness of 1.4 m thick concrete overlaid by 0.2 m brick pitching.\";\n", - "pre = fie2+(fic1-fie2)*5/b;\n", - "static = pre*Hs/100+x;\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"thickness at 3 m from d/s toe of crest = %.2f m.\"%(t);\n", - "print \"provide thickness of 1.2 m thick concrete overlaid by 0.2 m brick pitching.\";\n", - "pre = fie2+(fic1-fie2)*2/b;\n", - "static = pre*Hs/100; \t\t\t\t#calculation is wrong in book\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"thickness at 6m from d/s toe of crest = %.2f m.\"%(t);\n", - "print \"provide thickness of 0.7 m thick concrete overlaid by 0.2 m brick pitching.\";\n", - "\t\t\t\t#design of downstream wings\n", - "wing = 6*(E*drop)**0.5;\n", - "hw = D+0.5;\n", - "print \"heigth of top of downstream wings above the bed = %.2f m.\"%(hw);\n", - "projec = hw*3;\n", - "print \"length of warped wing measured along centre line of canal = %.2f m.\"%(projec);\n", - "\t\t\t\t#downstream pitching\n", - "l = 9+2*1.5;\n", - "print \"length of bed pitching = %.2f m.\"%(l);\n", - "print \"length of sloping pitching = 7 m.length of horizontal pitching = 6 m.\";\n", - "print \"provide one toe wall of 1 m depth and 0.4 m width.\";\n", - "print \"side pitching is curtailed at 45 degree from the end of bed pitching \\\n", - "in plan.supprot the side pitching on toe wall 0.4 m thick and 1 m deep. \";\n", - "\t\t\t\t#energy dissipators\n", - "q = Q/L;\n", - "dc = (q**2/9.81)**(1./3);\n", - "print \"size and position of friction blocks:\";\n", - "L = 2*dc;\n", - "w = dc;\n", - "h = dc;\n", - "di = 1.5*dc;\n", - "L = round(L*10)/10;\n", - "w = round(w*10)/10;\n", - "h = round(h*10)/10;\n", - "di = round(di);\n", - "print \"length of block = %.2f m.width of block = %.2f m.height of block = %.2f \\\n", - "m.dismath.tance from toe of crest = %.2f m.\"%(L,w,h,di);\n", - "print \"provide two rows staggered ata dismath.tance of 1 m from toe of crest.\";\n", - "print \"size and position of cube blocks:\";\n", - "L = D/10;\n", - "w = D/10;\n", - "h = w;\n", - "L = round(L*10)/10;\n", - "w = round(w*10)/10;\n", - "h = round(h*10)/10;\n", - "print \"length of block = %.2f m.width of block = %.2f m.height of block = %.2f m.\"%(L,w,h);\n", - "print \"provide two rows staggered at the end of impervious floor.\";\n", - "\t\t\t\t#u/s approach\n", - "r = 6*H;\n", - "print \"provide wing wall segmental with 5 m radius subtending angle of 60 degree at the centre.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "H = 0.86 m.d = 2.44 m.\n", - "crest height above bed = 0.94 m.\n", - "D/S batter = 1:3; U/S batter = 1:8.\n", - "R.L of crest = 217.44 m.\n", - "E = 0.86 m.\n", - "depth of cistern = 0.25 m.\n", - "length of cistern = 5.70 m.\n", - "R.L of bed of cistern = 214.75 m.\n", - "keep cistern at R.L 214.69.\n", - "design of impervious floor:\n", - "provide upstream cut-off = 1 m.; downstream cut-off = 1.60 m.\n", - "length of horizontal impervious floor = 14.32 m.\n", - "provide 15 m length impervious floor.\n", - "minimum length of impervious floor to the d/s of toe of crest wall = 7.50 m.\n", - "provide ld = 8 m.\n", - "the balance of the length 7 m is to be provided under and u/s of the crest.\n", - "uplift pressure is counter balanced by weigth of water. hence provide thickness of 0.4 m.\n", - "for other points; thickness required = 1.29 m.\n", - "provide thickness of 1.40 m.\n", - "at downstream end of floor provide thickness of 0.6 m overlaid by 0.2 m brick pitching.\n", - "checking of floor thickness by khosla theory:\n", - "length of floor provided = 17.18 m. > length by Bligh theory.\n", - "exit gradient after increase in depth cut-off = 0.20. which is in permissible limit\n", - "provide depth cut-off to 1.8 m.\n", - "calculation of pressure:\n", - "U/S cut-off:\n", - "corrected fic1 = 78.80 percent.\n", - "D/S cut-off wall:\n", - "correcte fie2 = 27.80 percent.\n", - "provide a minimum thickness of 0.4 m for u/s floor.\n", - "thickness at d/s toe of crest = 1.28 m.\n", - "provide thickness of 1.4 m thick concrete overlaid by 0.2 m brick pitching.\n", - "thickness at 3 m from d/s toe of crest = 1.08 m.\n", - "provide thickness of 1.2 m thick concrete overlaid by 0.2 m brick pitching.\n", - "thickness at 6m from d/s toe of crest = 0.68 m.\n", - "provide thickness of 0.7 m thick concrete overlaid by 0.2 m brick pitching.\n", - "heigth of top of downstream wings above the bed = 2.30 m.\n", - "length of warped wing measured along centre line of canal = 6.90 m.\n", - "length of bed pitching = 12.00 m.\n", - "length of sloping pitching = 7 m.length of horizontal pitching = 6 m.\n", - "provide one toe wall of 1 m depth and 0.4 m width.\n", - "side pitching is curtailed at 45 degree from the end of bed pitching in plan.supprot the side pitching on toe wall 0.4 m thick and 1 m deep. \n", - "size and position of friction blocks:\n", - "length of block = 1.20 m.width of block = 0.60 m.height of block = 0.60 m.dismath.tance from toe of crest = 1.00 m.\n", - "provide two rows staggered ata dismath.tance of 1 m from toe of crest.\n", - "size and position of cube blocks:\n", - "length of block = 0.20 m.width of block = 0.20 m.height of block = 0.20 m.\n", - "provide two rows staggered at the end of impervious floor.\n", - "provide wing wall segmental with 5 m radius subtending angle of 60 degree at the centre.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 18.2 pg : 820" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design an unflumed straight glacis non-meter fall\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 40.; \t\t\t\t#full supply discharge\n", - "sl_u = 218.3; \t\t\t\t#supply level at upstream\n", - "sl_d = 216.8; \t\t\t\t#supply level at downstream\n", - "D = 1.8; \t\t\t\t#suplly depth\n", - "L = 26.; \t\t\t\t#bed width\n", - "bl_u = 216.5; \t\t\t\t#bed level upstream\n", - "bl_d = 215.; \t\t\t\t#bed level downstream\n", - "drop = 1.5;\n", - "Ge = 1./6; \t\t\t\t#permissible exit gradient\n", - "\n", - "\t\t\t\t#design of crest\n", - "print \"design of crest:\";\n", - "E = (Q/(1.84*L))**(2/3);\n", - "V = Q/((L+D)*D);\n", - "vh = V**2/(2*9.81);\n", - "tel_up = sl_u+vh;\n", - "cl = tel_up-E;\n", - "w = 2*E/3;\n", - "w = round(w*10)/10;\n", - "print \"length of crest = %.2f m.\"%(L);\n", - "print \"width of crest = %.2f m.\"%(w);\n", - "\t\t\t\t#design of cistern\n", - "q = Q/L;\n", - "Hl = 1.5;\n", - "\t\t\t\t#from blench curve\n", - "Ef2 = 1.44;\n", - "cistern = sl_d+0.03-1.25*Ef2;\n", - "print \"R.L of cistern = %.2f m. > d/s bed level.\"%(cistern);\n", - "print \"keep R.L of cistern at 214.5 m.\";\n", - "l = 6*Ef2;\n", - "print \"length of cistern = %.2f m.\"%(l);\n", - "print \"provide cistern of 9 m length \";\n", - "d = bl_d-214.5;\n", - "print \"depth of cistern = %.2f m.\"%(d);\n", - "\n", - "\t\t\t\t#design of impervious floor\n", - "d1 = D/3;\n", - "print \"design of impervious floor:\";\n", - "print \"provide 0.4 m wide and 1 m deep curtain wall at u/s.\";\n", - "d2 = D/2;\n", - "print \"provide 0.4 m wide and 1 m deep curtain wall at d/s.the curtain wall will project the above the d/s bed by 0.18 m.\";\n", - "Hs = cl-bl_d;\n", - "d2 = 1;\n", - "n = d2*Ge/Hs; \t\t\t\t#n = 1/(math.pi*(lambda)**0.5)\n", - "\t\t\t\t#from khosla exit curves we get\n", - "alpha = 40;\n", - "lambda1 = (1/(math.pi*n))**2;\n", - "alpha = ((2*lambda1-1)**2-1)**0.5;\n", - "b = alpha*d2;\n", - "\t\t\t\t#math.since length is to excessive\n", - "d2 = 2;\n", - "n = d2*Ge/Hs; \t\t\t\t#n = 1/(math.pi*(lambda)**0.5)\n", - "\t\t\t\t#from khosla exit curves we get\n", - "alpha = 10;\n", - "lambda1 = (1/(math.pi*n))**2;\n", - "alpha = ((2*lambda1-1)**2-1)**0.5;\n", - "b = alpha*d2+1;\n", - "print \"total length = %i m.length of cistern = 9 m.length of d/s glacis = 5.88\\\n", - " m.width of crest = 0.6 m.length of u/s glacis = 0.47 m.balance to be provided to u/s of the u/s glacis = 4.05 m.\"%(b);\n", - "\n", - "\t\t\t\t#pressure calculations\n", - "print \"pressure calculations:\";\n", - "print \"upstream curtain wall:\";\n", - "d1 = 1.;\n", - "b = 20;\n", - "alpha_ = d1/b;\n", - "t = 0.3;\n", - "fic1 = 100-22;\n", - "fid1 = 100-15;\n", - "corec = (fid1-fic1)*t/d1\n", - "fic1 = fic1+corec;\n", - "print \"corrected fi_c1 = %.2f percent.\"%(fic1);\n", - "print \"downstream curtain wall:\";\n", - "d2 = 2.;b = 20;\n", - "alpha_ = d2/b;\n", - "t = 0.5;\n", - "fie = 29.;\n", - "fid = 21;\n", - "corec = (fie-fid)*t/d2\n", - "fie = fie-corec;\n", - "print \"corrected fi_e = %.2f percent.\"%(fie);\n", - "print \"toe of glacis:\";\n", - "\t\t\t\t#assuming linear variation of pressure\n", - "p = fie+(80-fie)*9/20;\n", - "print \"pressure at downstream of the glacis = %.2f percent.\"%(p);\n", - "\n", - "\t\t\t\t#floor thickness\n", - "rho = 2.24;\n", - "print \"floor thickness:provide minimum thickness of 0.3 m at the u/s floor.\";\n", - "static = p*2.44/100+(bl_d-214.5);\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"floor thickness required at toe of glacis = %.2f m.provide 1.5 m thick floor for length of 3 m.\"%(t);\n", - "p = fie+(80-fie)*6/20;\n", - "static = p*2.44/100+(bl_d-214.5);\n", - "t = static/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"floor thickness required at 3m from toe of glacis = %.2f m.provide \\\n", - "1.3 m thick floor from 3 m to 6.5 m from toe of glacis.\"%(t);\n", - "t = 0.27*2.44/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"thickness of d/s end of cistern = %.2f m.provide thickness of 0.6 m at d/s end of floor.\"%(t);\n", - "\n", - "\t\t\t\t#design of d/s protection\n", - "print \"no bed protection is needed as deflector wall is provided.\";\n", - "sp = 3*D;\n", - "print \"length of side protection = %.2f m.provide 5.5 m length of 20 cm\\\n", - " thick brick pitching beyond impervious floor.pitching will rest on toe wall 0.4 \\\n", - " m wide and 0.9 m deep.provide 0.4 m wide profile at the end of pitching\"%(sp);\n", - "\t\t\t\t#design of u/s approach\n", - "print \"u/s wing wall is splayed at 45 degree from u/s end of impervious\\\n", - " floor.extend 1 m into earthen banks from line of F.S.L.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "design of crest:\n", - "length of crest = 26.00 m.\n", - "width of crest = 0.70 m.\n", - "R.L of cistern = 215.03 m. > d/s bed level.\n", - "keep R.L of cistern at 214.5 m.\n", - "length of cistern = 8.64 m.\n", - "provide cistern of 9 m length \n", - "depth of cistern = 0.50 m.\n", - "design of impervious floor:\n", - "provide 0.4 m wide and 1 m deep curtain wall at u/s.\n", - "provide 0.4 m wide and 1 m deep curtain wall at d/s.the curtain wall will project the above the d/s bed by 0.18 m.\n", - "total length = 18 m.length of cistern = 9 m.length of d/s glacis = 5.88 m.width of crest = 0.6 m.length of u/s glacis = 0.47 m.balance to be provided to u/s of the u/s glacis = 4.05 m.\n", - "pressure calculations:\n", - "upstream curtain wall:\n", - "corrected fi_c1 = 80.10 percent.\n", - "downstream curtain wall:\n", - "corrected fi_e = 27.00 percent.\n", - "toe of glacis:\n", - "pressure at downstream of the glacis = 50.85 percent.\n", - "floor thickness:provide minimum thickness of 0.3 m at the u/s floor.\n", - "floor thickness required at toe of glacis = 1.40 m.provide 1.5 m thick floor for length of 3 m.\n", - "floor thickness required at 3m from toe of glacis = 1.25 m.provide 1.3 m thick floor from 3 m to 6.5 m from toe of glacis.\n", - "thickness of d/s end of cistern = 0.53 m.provide thickness of 0.6 m at d/s end of floor.\n", - "no bed protection is needed as deflector wall is provided.\n", - "length of side protection = 5.40 m.provide 5.5 m length of 20 cm thick brick pitching beyond impervious floor.pitching will rest on toe wall 0.4 m wide and 0.9 m deep.provide 0.4 m wide profile at the end of pitching\n", - "u/s wing wall is splayed at 45 degree from u/s end of impervious floor.extend 1 m into earthen banks from line of F.S.L.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 18.3 pg : 831" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a cross -regulator and head regulatorfor a distributory channel\n", - "#givrn\n", - "Q = 100.; \t\t\t\t#discharge of parent channel\n", - "Qd = 15.; \t\t\t\t#discharge ofdistributory\n", - "fsl_u = 218.1; \t\t\t\t#F.S.L of upstream parent channel\n", - "fsl_d = 217.9; \t\t\t\t#F.S.L of downstream of parent channel\n", - "bw_u = 42.; \t\t\t\t#bed width of parent channel upstream\n", - "bw_d = 38.; \t\t\t\t#bed width of parent channel downstream\n", - "hw = 2.5; \t\t\t\t#depth of water in parent channel\n", - "fsl_dis = 217.1; \t\t\t\t#F.S.L of distributory\n", - "hw_dis = 1.5; \t\t\t\t#depth of water in distributory\n", - "Ge = 1./5; \t\t\t\t#permissible exit gradient\n", - "\n", - "#design of cross regulator\n", - "print \"DESIGN OF CROSS-REGULATOR::\";\n", - "#design of crest and waterway\n", - "print \"design of crest and waterway:\";\n", - "cl = fsl_u-hw;\n", - "h = fsl_u-fsl_d;\n", - "d = fsl_d-cl;\n", - "C1 = 0.557;C2 = 0.8;\n", - "L = Q/(2*C1*(2*9.81)**0.5*h**1.5/3+C2*d*(2*9.81*h)**0.5);\n", - "L = round(L*10)/10;\n", - "print \"crest level = %.2f m.\"%(cl);\n", - "print \"length of crest = %.2f m.\"%(L);\n", - "tw = 28+4.5;\n", - "print \"provide 3 piers of 1.5 m width each.total width of cross regulator = %.2f m.\"%(tw);\n", - "#design of d/s floor\n", - "L = 28.;\n", - "q = Q/L;\n", - "Hl = fsl_u-fsl_d;\n", - "Ef2 = 1.89; \t\t\t\t#from blench curve\n", - "fl_d = fsl_d-Ef2;\n", - "print \"design of d/s floor:\";\n", - "print \"d/s floor level = %.2f m.; which is higher than d/s bed level.adopt floor level = d/s bed level = 215.40 m.\"%(fl_d);\n", - "Ef1 = Ef2+Hl;\n", - "#from specific energy curve\n", - "D1 = 0.7;\n", - "D2 = 1.65;\n", - "cil = 5*(D2-D1); \t\t\t\t#cistern length\n", - "tl = 2*16/3;\n", - "tl = round(tl*10)/10;\n", - "print \"cistern length = %.2f m.length of d/s floor = %.2f m.\"%(cil,tl);\n", - "#design of impervious floor\n", - "d1 = hw/3+0.6; \t\t\t\t#depth of u/s cut-off\n", - "w = 0.5; \t\t\t\t#width of cut-off\n", - "d2 = hw/2+0.6; \t\t\t\t#deth of d/s cut-off\n", - "d2 = 2; \t\t\t\t#keep\n", - "Hs = fsl_u-(fsl_d-hw); \t\t\t\t#maximum static head\n", - "n = Ge*d2/Hs; \t\t\t\t#n = 1/math.pi*(lambda)**0.5;\n", - "#from exit gradient curves we get\n", - "alpha = 8.;\n", - "n = 0.148;\n", - "b = alpha*d2;\n", - "print \"design of impervious floor:\";\n", - "print \"total length of impervious floor = %i m.;which is divided as-\"%(b);\n", - "print \"d/s floor length = 10.6 m.d/s glacis length with 2:1 slope = 0.4 m.balance to be provided upstream = 5 m.\";\n", - "d1 = 1.5;b = 16;\n", - "alpha_ = d1/b;\n", - "\t\t\t\t#hence\n", - "fic1 = 100-28;\n", - "fid1 = 100-19;\n", - "t = 0.5;\n", - "fic1 = fic1+(fid1-fic1)*t/d1;\n", - "print \"pressure calculation:upstream cut-off:pressure = %.2f percent.\"%(fic1);\n", - "d2 = 2.;\n", - "b = 16;\n", - "alpha_ = d2/b;\n", - "\t\t\t\t#hence\n", - "t = 0.6;\n", - "fie2 = 31.;\n", - "fid2 = 22.;\n", - "fie2 = fie2-(fie2-fid2)*t/d2;\n", - "print \"downstream cut-off:pressure = %.2f percent.\"%(fie2);\n", - "t = 10.6;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "p = round(p*10)/10;\n", - "print \"toe of glacis:pressure = %.2f percent.\"%(p);\n", - "print \"thickness of floor:minimu thickness for u/s floor = 0.5 m.\";\n", - "rho = 2.24;\n", - "t = fie2*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor near d/s cut-off = %.2f m.provide 0.7 m thick floor for last 2.1 m length.\"%(t);\n", - "t = 1.6/(rho-1);\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at toe of glacis = %.2f m.\"%(t);\n", - "t = 6.6;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "t = p*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at 4 m from toe of glais = %.2f m.provide 1.1 m thick floor for next 2 m length\"%(t);\n", - "t = 4.6;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "t = p*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at 6 m from toe of glais = %.2f m.provide 0.9 m thick floor for next 2.5 m length\"%(t);\n", - "\n", - "\t\t\t\t#design of u/s protection\n", - "d1 = hw/3+0.6;\n", - "v = d1;\n", - "v = round(v*100)/100;\n", - "print \"design of u/s protection:volume of block protection = %.2f cubic metre/metre.\"%(v);\n", - "print \"keep thickness of protection = 1 m.provide 0.8mx0.8mx0.6m thick concret blocks over 0.4 m thick apron in length of 0.6 m.\";\n", - "cu = 2.25*d1;\n", - "cu = round(cu*100)/100;\n", - "print \"cubic content of launching apron = %.2f cubic metre/metre.provide 1 m thick and 3.5 m long launching apron.\"%(cu);\n", - "\t\t\t\t#design of d/s protection\n", - "d2 = hw/2+0.6;\n", - "v = d2;\n", - "v = round(v*100)/100;\n", - "print \"design of d/s protection:volume of inverted filter = %.2f cubic metre/metre.\"%(v);\n", - "print \"keep thickness of concrete block = 0.6 m.provide 2 rows of 0.8mx0.8mx0.6m thick \\\n", - "concret blocks over 0.6 m graded filter for length of 1.6 m.\";\n", - "cu = 2.25*d2;\n", - "cu = round(cu*100)/100;\n", - "print \"launching apron volume = %.2f cubic metre/metre.provide 1 m thick launching apron\\\n", - " for length of 4.5 m.provide a toe wall 0.4 m wide and 1.5 m deep between filter and launching apron.\"%(cu);\n", - "\n", - "\t\t\t\t#design of head regulator\n", - "print \"DESIGN OF DISTRIBUTORY HEAD REGULATOR::\";\n", - "\t\t\t\t#design of crest and waterway\n", - "print \"design of crest and waterway:\";\n", - "cl = fsl_u-hw+0.5;\n", - "h = fsl_u-fsl_dis;\n", - "d = fsl_dis-cl;\n", - "C1 = 0.557;C2 = 0.8;\n", - "L = Qd/(2*C1*(2*9.81)**0.5*h**1.5/3+C2*d*(2*9.81*h)**0.5);\n", - "L = round(L*100)/100;\n", - "print \"crest level = %.2f m.\"%(cl);\n", - "print \"length of crest = %.2f m.\"%(L);\n", - "print \"provide 2 bays of 3.5 m each with a 1 m thick pier in between.\";\n", - "tw = 8;\n", - "print \"total width of cross regulator = %.2f m.\"%(tw);\n", - "\t\t\t\t#design of d/s floor\n", - "L = 7.5;\n", - "q = Q/L;\n", - "Hl = fsl_u-fsl_dis;\n", - "Ef2 = 1.58; \t\t\t\t#from blench curve\n", - "fl_d = fsl_dis-Ef2;\n", - "print \"design of d/s floor:\";\n", - "print \"d/s floor level = %.2f m.;keepR.L of d/s floor = 215.50 m.\"%(fl_d);\n", - "Ef1 = Ef2+Hl;\n", - "\t\t\t\t#from specific energy curve\n", - "D1 = 0.42;D2 = 2.55;\n", - "cil = 5*(D2-D1); \t\t\t\t#cistern length\n", - "tl = 2*14/3;\n", - "print \"cistern length = %.2f m.\"%(cil);\n", - "\n", - "\t\t\t\t#design of impervious floor\n", - "d1 = hw/3+0.6; \t\t\t\t#depth of u/s cut-off\n", - "w = 0.5; \t\t\t\t#width of cut-off\n", - "d2 = hw_dis/2+0.6; \t\t\t\t#deth of d/s cut-off\n", - "d2 = 2; \t\t\t\t#keep\n", - "Hs = fsl_u-215.5; \t\t\t\t#maximum static head\n", - "n = Ge*d2/Hs; \t\t\t\t#n = 1/math.pi*(lambda)**0.5;\n", - "\t\t\t\t#from exit gradient curves we get\n", - "alpha = 7;n = 0.154;\n", - "b = alpha*d2;\n", - "print \"design of impervious floor:\";\n", - "print \"total length of impervious floor = %i m.;which is divided as-\"%(b);\n", - "print \"length below the toe of glacis = 10.5 mlength of d/s glacis at 2:1 slope =\\\n", - " 1.2 m.width of crest = 1 m.length of u/s glacis at 1:1 slope = 0.5 m.u/s floor:balnce = 0.8 m.\";\n", - "d1 = 1.5;\n", - "b = 16.;\n", - "alpha_ = d1/b;\n", - "\t\t\t\t#hence\n", - "fic1 = 100-28;\n", - "fid1 = 100-19;\n", - "t = 0.5;\n", - "fic1 = fic1+(fid1-fic1)*t/d1;\n", - "print \"pressure calculation:upstream cut-off:pressure = %.2f percent.\"%(fic1);\n", - "d2 = 2.;\n", - "b = 16;\n", - "alpha_ = d2/b;\n", - "\t\t\t\t#hence\n", - "t = 0.6;\n", - "fie2 = 31.;\n", - "fid2 = 22;\n", - "fie2 = fie2-(fie2-fid2)*t/d2;\n", - "print \"downstream cut-off:pressure = %.2f percent.\"%(fie2);\n", - "t = 10.6;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "p = round(p*100)/100;\n", - "print \"toe of glacis:pressure = %.2f percent.\"%(p);\n", - "print \"thickness of floor:minimu thickness for u/s floor = 0.5 m.\";\n", - "rho = 2.24;\n", - "t = p*2.6/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness under the crest = 1 m.\";\n", - "print \"thickness of floor at toe of glacis = %.2f m.\"%(t);\n", - "t = 9.5;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "t = p*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at 2 m from toe of glais = %.2f m.provide 1.1 m thick floor for next 4 m length\"%(t);\n", - "t = 4.5;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "t = p*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at 6 m from toe of glais = %.2f m.provide 0.9 m thick floor for next 2.5 m length\"%(t);\n", - "t = 2;\n", - "p = fie2+(fic1-fie2)*t/b;\n", - "t = p*2.7/(100*(rho-1));\n", - "t = round(t*100)/100;\n", - "print \"thickness of floor at 8.5 m from toe of glais = %.2f m.provide 0.7 m thick floor for next 2 m length\"%(t);\n", - "\n", - "\t\t\t\t#design of upstream protection\n", - "d = hw/3+0.6;\n", - "d = round(d*10)/10;\n", - "print \"design of u/s protection:u/s scour depth = %.2f m.provide same protection as in cross regulator\"%(d);\n", - "\n", - "\t\t\t\t#design of d/s protection\n", - "d2 = hw_dis/2+0.6;\n", - "v = d2;\n", - "print \"design of d/s protection:volume of inverted filter = %.2f cubic metre/metre.\"%(v);\n", - "print \"keep thickness of concrete block = 0.5 m.provide 2 rows of \\\n", - "0.8mx0.8mx0.5m thick concret blocks over 0.5 m thick graded filter.\";\n", - "cu = 2.25*d2;\n", - "print \"launching apron volume = %.2f cubic metre/metre.provide 1 m thick \\\n", - "launching apron for length of 3.5 m.provide a masonary toe wall 0.4 m wide and 1.2 m deep between filter and launching apron.\"%(cu);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "DESIGN OF CROSS-REGULATOR::\n", - "design of crest and waterway:\n", - "crest level = 215.60 m.\n", - "length of crest = 26.40 m.\n", - "provide 3 piers of 1.5 m width each.total width of cross regulator = 32.50 m.\n", - "design of d/s floor:\n", - "d/s floor level = 216.01 m.; which is higher than d/s bed level.adopt floor level = d/s bed level = 215.40 m.\n", - "cistern length = 4.75 m.length of d/s floor = 10.00 m.\n", - "design of impervious floor:\n", - "total length of impervious floor = 16 m.;which is divided as-\n", - "d/s floor length = 10.6 m.d/s glacis length with 2:1 slope = 0.4 m.balance to be provided upstream = 5 m.\n", - "pressure calculation:upstream cut-off:pressure = 75.00 percent.\n", - "downstream cut-off:pressure = 28.30 percent.\n", - "toe of glacis:pressure = 59.20 percent.\n", - "thickness of floor:minimu thickness for u/s floor = 0.5 m.\n", - "thickness of floor near d/s cut-off = 0.62 m.provide 0.7 m thick floor for last 2.1 m length.\n", - "thickness of floor at toe of glacis = 1.29 m.\n", - "thickness of floor at 4 m from toe of glais = 1.04 m.provide 1.1 m thick floor for next 2 m length\n", - "thickness of floor at 6 m from toe of glais = 0.91 m.provide 0.9 m thick floor for next 2.5 m length\n", - "design of u/s protection:volume of block protection = 1.43 cubic metre/metre.\n", - "keep thickness of protection = 1 m.provide 0.8mx0.8mx0.6m thick concret blocks over 0.4 m thick apron in length of 0.6 m.\n", - "cubic content of launching apron = 3.23 cubic metre/metre.provide 1 m thick and 3.5 m long launching apron.\n", - "design of d/s protection:volume of inverted filter = 1.85 cubic metre/metre.\n", - "keep thickness of concrete block = 0.6 m.provide 2 rows of 0.8mx0.8mx0.6m thick concret blocks over 0.6 m graded filter for length of 1.6 m.\n", - "launching apron volume = 4.16 cubic metre/metre.provide 1 m thick launching apron for length of 4.5 m.provide a toe wall 0.4 m wide and 1.5 m deep between filter and launching apron.\n", - "DESIGN OF DISTRIBUTORY HEAD REGULATOR::\n", - "design of crest and waterway:\n", - "crest level = 216.10 m.\n", - "length of crest = 2.89 m.\n", - "provide 2 bays of 3.5 m each with a 1 m thick pier in between.\n", - "total width of cross regulator = 8.00 m.\n", - "design of d/s floor:\n", - "d/s floor level = 215.52 m.;keepR.L of d/s floor = 215.50 m.\n", - "cistern length = 10.65 m.\n", - "design of impervious floor:\n", - "total length of impervious floor = 14 m.;which is divided as-\n", - "length below the toe of glacis = 10.5 mlength of d/s glacis at 2:1 slope = 1.2 m.width of crest = 1 m.length of u/s glacis at 1:1 slope = 0.5 m.u/s floor:balnce = 0.8 m.\n", - "pressure calculation:upstream cut-off:pressure = 75.00 percent.\n", - "downstream cut-off:pressure = 28.30 percent.\n", - "toe of glacis:pressure = 59.24 percent.\n", - "thickness of floor:minimu thickness for u/s floor = 0.5 m.\n", - "thickness under the crest = 1 m.\n", - "thickness of floor at toe of glacis = 1.24 m.\n", - "thickness of floor at 2 m from toe of glais = 1.22 m.provide 1.1 m thick floor for next 4 m length\n", - "thickness of floor at 6 m from toe of glais = 0.90 m.provide 0.9 m thick floor for next 2.5 m length\n", - "thickness of floor at 8.5 m from toe of glais = 0.74 m.provide 0.7 m thick floor for next 2 m length\n", - "design of u/s protection:u/s scour depth = 1.40 m.provide same protection as in cross regulator\n", - "design of d/s protection:volume of inverted filter = 1.35 cubic metre/metre.\n", - "keep thickness of concrete block = 0.5 m.provide 2 rows of 0.8mx0.8mx0.5m thick concret blocks over 0.5 m thick graded filter.\n", - "launching apron volume = 3.04 cubic metre/metre.provide 1 m thick launching apron for length of 3.5 m.provide a masonary toe wall 0.4 m wide and 1.2 m deep between filter and launching apron.\n" - ] - } - ], - "prompt_number": 3 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch19.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch19.ipynb deleted file mode 100755 index 32d7ab84..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch19.ipynb +++ /dev/null @@ -1,392 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:6fd287c1935869ab0dbf969e29246715100ab63b1d219a58fec4220f2b19494d" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 19 : CROSS DRAINAGE WORKS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 19.1 pg : 857" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace,zeros\n", - "\n", - "#design an expansion transition for canal by Mitra's method\n", - "\t\t\t\t\n", - "#Given\n", - "Lf = 16.; \t\t\t\t#length of flume\n", - "Bf = 9.; \t\t\t\t#width of throat\n", - "Bo = 15.; \t\t\t\t#width of canal\n", - "\n", - "#width at any dismath.tance x from flumed section is given by\n", - "#Bx = Bo*Bf*Lf/(Lf*Bo-(Bo-Bf)x)\n", - "#on solving we get\n", - "#Bx = 2160/(240-6x)\n", - "\n", - "x = linspace(2,16,8) \t\t\t\t#dismath.tance\n", - "print \"width at any dismath.tance x from flumed section:\";\n", - "Bx = zeros(8)\n", - "for i in range(8):\n", - " Bx[i] = 2160./(240-6*x[i]);\n", - " Bx[i] = round(Bx[i]*100)/100;\n", - " print '%.2f'%(Bx[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "width at any dismath.tance x from flumed section:\n", - "9.47\n", - "10.00\n", - "10.59\n", - "11.25\n", - "12.00\n", - "12.86\n", - "13.85\n", - "15.00\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 19.2 pg : 857" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace,zeros\n", - "\n", - "#design an expansion transition for canal by Chaturvedi's method\n", - "\t\t\t\t\n", - "#Given;\n", - "Lf = 16.; \t\t\t\t#length of flume\n", - "Bf = 9.; \t\t\t\t#width of throat\n", - "Bo = 15.; \t\t\t\t#width of canal\n", - "\n", - "x = linspace(2,16,8); \t\t\t\t#dismath.tance\n", - "\n", - "#dismath.tance x is related as x = Lf*Bo**(2/3)(1-(Bf/Bx)**1.5)/(Bo**1.5-Bf**1.5)\n", - "#on solving we get\n", - "#(Bf/Bx)**1.5 = 1-(x/29.893) (relation is misprinted in book)\n", - "#let (Bf/Bx)**1.5 = r\n", - "r = zeros(8)\n", - "R = zeros(8)\n", - "Bx = zeros(8)\n", - "\n", - "print \"width at any dismath.tance x from flumed section:\";\n", - "for i in range(8):\n", - " r[i] = 1-(x[i]/29.893); \t\t\t\t#Bf/Bx**(1.5)\n", - " R[i] = r[i]**(2./3); \t\t\t\t#Bf/Bx\n", - " Bx[i] = Bf/R[i]; \n", - " Bx[i] = round(Bx[i]*100)/100; \n", - " print \"%.2f.\"%(Bx[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "width at any dismath.tance x from flumed section:\n", - "9.43.\n", - "9.90.\n", - "10.45.\n", - "11.08.\n", - "11.81.\n", - "12.67.\n", - "13.71.\n", - "15.00.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 19.3 pg : 860" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace,zeros,zeros_like\n", - "\n", - "#design a syphon aqueduct\n", - "\n", - "#Given\n", - "Q = 25.; \t\t\t\t#design discharge of canal\n", - "B = 20.; \t\t\t\t#bed width of canal\n", - "D = 1.5; \t\t\t\t#depth of water in canal\n", - "bl = 160.; \t\t\t\t#bed level of canal\n", - "hfq = 400.; \t\t\t\t#high flood discharge of drainage\n", - "hfl = 160.5; \t\t\t\t#high flood level of drainage\n", - "bl_drain = 158.; \t\t\t\t#bed level of drainage\n", - "gl = 160.; \t\t\t\t#general ground level\n", - "\n", - "#demath.sing of drainage water-way\n", - "P = 4.75*(hfq)**0.5; \t\t\t\t#laecey P-Q formula\n", - "print \"design of drainage water-way:wetted perimeter of river = %i m.provide 13 spans \\\n", - "of 6 m each,separated by 12 piers each of 1.25 m thick.\"%(P);\n", - "t = 78.+15;\n", - "print \"total length of water-way = %i m.\"%(t);\n", - "v = 2; \t\t\t\t#velocity through syphon\n", - "hb = hfq/(78*v);\n", - "ac = hfq/(6*2.5*1.3); \t\t\t\t#calculation is wrong in book\n", - "hb = round(hb*100)/100;\n", - "ac = round(ac*100)/100;\n", - "print \"height of barrels = %.2f m.provide recmath.tangular barrels 6 m wide and 2.5 m high.actual \\\n", - "velocity through barrels = %.2f m/sec.\"%(hb,ac);\n", - "\n", - "#design of canal waterway\n", - "print \"design of canal waterway:Type 3 aqueduct is adopted.\";\n", - "l1 = B-10;\n", - "l2 = (20-10)*3/2;\n", - "print \"providing a splay 2:1 in expansion,length of contraction transition = %i m.providing\\\n", - " a splay of 3:1 in expansion,length of expansion transition = %i m.\"%(l1,l2);\n", - "print 'In transition side slopes are warped from original slope of 1.5:1 to vertical.';\n", - "\n", - "#design of levels of different sectionn\n", - "print \"design of levels of different sectionn:at section 4-4:\";\n", - "A = (B+1.5*D); \t\t\t\t#area\n", - "V = Q/A; \t\t\t\t#velocity of flow\n", - "vh = V**2/(2*9.81); \t\t\t\t#velocity head\n", - "ws = gl+D; \t\t\t\t#R.L of water surface\n", - "tel = ws+vh;\n", - "tel = round(tel*1000)/1000;\n", - "print \"R.L of T.E.L = %.2f m. at section 3-3:\"%(tel);\n", - "A = 10*D; \t\t\t\t#area of trough\n", - "V = Q/A; \t\t\t\t#velocity\n", - "vh1 = V**2/(2*9.81); \t\t\t\t#velocity head\n", - "le = 0.3*(vh1-vh); \t\t\t\t#loss of head in expansion from section 3-3 to 4-4\n", - "tel = tel+le;\n", - "rlw = tel-vh1;\n", - "rlb = rlw-D;\n", - "tel = round(tel*1000)/1000;\n", - "rlb = round(rlb*1000)/1000;\n", - "print \"elevation of T.E.L = %.2f m.R.L of bed to maintain consmath.tant water depth = %.2f m.\"%(tel,rlb);\n", - "\n", - "#at section 2-2\n", - "R = A/P;\n", - "N = 0.016;\n", - "S = V**2*N**2/R**(4./3); \t\t\t\t#from manning's formula\n", - "L = 93; \t\t\t\t#length of trough\n", - "hl = L*S; \t\t\t\t#head loss\n", - "tel = tel+hl;\n", - "rlw = tel-vh1;\n", - "rlb = rlw-D;\n", - "tel = round(tel*1000)/1000;\n", - "rlb = round(rlb*1000)/1000;\n", - "print \"at section 2-2:R.L of T.E.L = %.2f m.R.L of bed to maintain consmath.tant water depth = %.2f m.\"%(tel,rlb);\n", - "\n", - "#at section 1-1\n", - "hl = 0.2*(vh1-vh); \t\t\t\t#loss of hed in contraction transition\n", - "tel = tel+hl;\n", - "rlw = tel-vh;\n", - "rlb = tel-D;\n", - "tel = round(tel*1000)/1000;\n", - "rlb = round(rlb*1000)/1000;\n", - "print \"at section 1-1:R.L of T.E.L = %.2f m.R.L of bed to maintain consmath.tant water depth = %.2f m.\"%(tel,rlb);\n", - "\n", - "#design of contraction transition\n", - "#it is designed on the basis of chaturvedi's formula\n", - "Bo = 20.;\n", - "Bf = 10.;\n", - "L = 10.;\n", - "#from chaturvedi formula we get relation between x and Bx as: x = 15.45(1-(10/Bx)**1.5);\n", - "Bx = linspace(10,20,11)\n", - "print \"design of contraction transition on the basis of chaturvedi formula:\\nBx x\";\n", - "x = zeros_like(Bx)\n", - "for i in range(11):\n", - " x[i] = 15.45*(1-(10/Bx[i])**1.5);\n", - " x[i] = round(x[i]*100)/100;\n", - " print \"%i %.2f\"%(Bx[i],x[i]);\n", - "\n", - "\n", - "#design of expansion transition on the basis of chaturvedi formula\n", - "L = 15.;\n", - "Bf = 10.;Bo = 20.;\n", - "#from chaturvedi formula we get relation between x and Bx as: x = 23.15(1-(10/Bx)**1.5);\n", - "print \"design of expansion transition on the basis of chaturvedi formula:\\nBx x\";\n", - "for i in range(11):\n", - " x[i] = 23.15*(1-(10/Bx[i])**1.5);\n", - " x[i] = round(x[i]*100)/100;\n", - " print \"%i %.2f\"%(Bx[i],x[i]);\n", - "\n", - "\n", - "#design of trough\n", - "print \"design of the trough:\";\n", - "print \"flumed water way of canal = 10 m.trough carrying canal will divide into two \\\n", - "compartments each 5 m wide an dseparated by 0.3 m thick partiions.heigth of trough will be \\\n", - " = 2 m.trough iss constructed umath.sing monolithic reinforced concrete.the outer and inner walls\\\n", - " ca be kept 0.4 m thick.thus,outer width of trough = 11.1 m.\";\n", - "\n", - "#head loss through syphon barrels\n", - "V = 2.05; \t\t\t\t#velocity through barrels\n", - "f1 = 0.505; \t\t\t\t#coefficient of loss of head at entry\n", - "a = 0.00316;\n", - "b = 0.030;\n", - "R = (6*2.5)/(2*(6+2.5));\n", - "f2 = a*(1+b/R);\n", - "L = 11.1; \t\t\t\t#length of barrel\n", - "h = (1+f1+f2*L/R)*V**2/(2*9.81);\n", - "hfl_up = hfl+h;\n", - "h = round(h*1000)/1000;\n", - "hfl_up = round(hfl_up*1000)/1000;\n", - "print \"head loss through syphon barrels = %.2f m.upstream H.F.L = %.2f m.\"%(h,hfl_up)\n", - "\n", - "#uplift pressure on the roof\n", - "bt = gl-0.4; \t\t\t\t#R.L of bottom of the trough\n", - "hl = 0.505*V**2/(2*9.81);\n", - "u = hfl_up-hl-159.6;\n", - "up = u*9.81;\n", - "print \"uplift pressure on the roof = %.2f kN/square m.trough slab is 0.4 m thick and exert a downward load of 9.42 kN.\"%(up);\n", - "print \"th ebalance of the uplift pressure has to be resisted by bending action of \\\n", - "trough slab.so,reinforcement has to be provided at the top of the slab.\";\n", - "\n", - "#uplift on the floor of the barrel and its design\n", - "#(a) static head\n", - "print \"uplift on the floor of the barrel and its design:a static head:\";\n", - "bf = bt-2.5; \t\t\t\t#R.L of barrel floor\n", - "t = 0.8; \t\t\t\t#tentative thickness of floor\n", - "bot = bf-t;\n", - "static = bl_drain-bot;\n", - "static = round(static*100)/100;\n", - "print \"static uplift on the floor = %.2f m.\"%(static);\n", - "\n", - "#(b) seepage head\n", - "L = 10.; \t\t\t\t#length of u/s transition\n", - "bs = 3.; \t\t\t\t#half the barrel span\n", - "df = 11.; \t\t\t\t#end drainage floor\n", - "tcl = 24.; \t\t\t\t#total creep length\n", - "tsh = 161.5-bl_drain; \t\t\t\t#total seepage head\n", - "rs = tsh*(1-13/tcl); \t\t\t\t#residual seepage at B\n", - "tu = (static+rs)*9.81;\n", - "tu = round(tu*100)/100;\n", - "print \"b) seepage head:total uplift = %.2f kN/square m.provide thickness of floor 0.8 m\"%(tu);\n", - "bending = tu-17.58;\n", - "bending = round(bending*100)/100;\n", - "print \"uplift to be resisted by bending action of floor = %.2f kN/square m.\"%(bending);\n", - "\n", - "#design of cut-off and protection works for drainage floor\n", - "print \"design of cut-off and protection works for drainage floor:\";\n", - "Q = 400;f = 1;\n", - "R = 0.47*(Q/f)**(1/3);\n", - "d_up = 1.5*R; \t\t\t\t#depth of u/s cut-off\n", - "bot_up = hfl_up-d_up; \t\t\t\t#R.L of bottom of u/s cut-off\n", - "d_down = 1.5*R; \t\t\t\t#depth of d/s cut-off\n", - "bot_down = hfl-d_down; \t\t\t\t#R.L of bottom of d/s cut-off\n", - "l_down = 2.5*(bl_drain-bot_down);\n", - "l_down1 = 2*(bl_drain-bot_up);\n", - "bot_up = round(bot_up*100)/100;\n", - "bot_down = round(bot_down*100)/100;\n", - "l_down = round(l_down);\n", - "l_down1 = round(l_down1);\n", - "print \"R.L of bottom of u/s cut-off = %.2f m.R.L of bottom of d/s cut-off = %.2f m.\"%(bot_up,bot_down);\n", - "print \"length of d/s protection consisting of 40 cm brick pritching = %.2f m.pitching is \\\n", - "supported by toe wall 0.4 m wide and 1.5 m deep at its d/s end.length of d/s protection consisting\\\n", - " of 0.4 cm brick pritching = %.2f m.pitching is supported by toe wall 0.4 m wide and\\\n", - " 1 m deep at its u/s end.\"%(l_down,l_down1);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "design of drainage water-way:wetted perimeter of river = 95 m.provide 13 spans of 6 m each,separated by 12 piers each of 1.25 m thick.\n", - "total length of water-way = 93 m.\n", - "height of barrels = 2.56 m.provide recmath.tangular barrels 6 m wide and 2.5 m high.actual velocity through barrels = 20.51 m/sec.\n", - "design of canal waterway:Type 3 aqueduct is adopted.\n", - "providing a splay 2:1 in expansion,length of contraction transition = 10 m.providing a splay of 3:1 in expansion,length of expansion transition = 15 m.\n", - "In transition side slopes are warped from original slope of 1.5:1 to vertical.\n", - "design of levels of different sectionn:at section 4-4:\n", - "R.L of T.E.L = 161.56 m. at section 3-3:\n", - "elevation of T.E.L = 161.59 m.R.L of bed to maintain consmath.tant water depth = 159.95 m.\n", - "at section 2-2:R.L of T.E.L = 162.36 m.R.L of bed to maintain consmath.tant water depth = 160.72 m.\n", - "at section 1-1:R.L of T.E.L = 162.38 m.R.L of bed to maintain consmath.tant water depth = 160.88 m.\n", - "design of contraction transition on the basis of chaturvedi formula:\n", - "Bx x\n", - "10 0.00\n", - "11 2.06\n", - "12 3.70\n", - "13 5.03\n", - "14 6.12\n", - "15 7.04\n", - "16 7.82\n", - "17 8.48\n", - "18 9.05\n", - "19 9.55\n", - "20 9.99\n", - "design of expansion transition on the basis of chaturvedi formula:\n", - "Bx x\n", - "10 0.00\n", - "11 3.08\n", - "12 5.54\n", - "13 7.53\n", - "14 9.17\n", - "15 10.55\n", - "16 11.71\n", - "17 12.71\n", - "18 13.56\n", - "19 14.31\n", - "20 14.97\n", - "design of the trough:\n", - "flumed water way of canal = 10 m.trough carrying canal will divide into two compartments each 5 m wide an dseparated by 0.3 m thick partiions.heigth of trough will be = 2 m.trough iss constructed umath.sing monolithic reinforced concrete.the outer and inner walls ca be kept 0.4 m thick.thus,outer width of trough = 11.1 m.\n", - "head loss through syphon barrels = 0.33 m.upstream H.F.L = 160.83 m.\n", - "uplift pressure on the roof = 11.01 kN/square m.trough slab is 0.4 m thick and exert a downward load of 9.42 kN.\n", - "th ebalance of the uplift pressure has to be resisted by bending action of trough slab.so,reinforcement has to be provided at the top of the slab.\n", - "uplift on the floor of the barrel and its design:a static head:\n", - "static uplift on the floor = 1.70 m.\n", - "b) seepage head:total uplift = 32.41 kN/square m.provide thickness of floor 0.8 m\n", - "uplift to be resisted by bending action of floor = 14.83 kN/square m.\n", - "design of cut-off and protection works for drainage floor:\n", - "R.L of bottom of u/s cut-off = 160.13 m.R.L of bottom of d/s cut-off = 159.79 m.\n", - "length of d/s protection consisting of 40 cm brick pritching = -4.00 m.pitching is supported by toe wall 0.4 m wide and 1.5 m deep at its d/s end.length of d/s protection consisting of 0.4 cm brick pritching = -4.00 m.pitching is supported by toe wall 0.4 m wide and 1 m deep at its u/s end.\n" - ] - } - ], - "prompt_number": 4 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch19_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch19_1.ipynb deleted file mode 100755 index 32d7ab84..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch19_1.ipynb +++ /dev/null @@ -1,392 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:6fd287c1935869ab0dbf969e29246715100ab63b1d219a58fec4220f2b19494d" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 19 : CROSS DRAINAGE WORKS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 19.1 pg : 857" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace,zeros\n", - "\n", - "#design an expansion transition for canal by Mitra's method\n", - "\t\t\t\t\n", - "#Given\n", - "Lf = 16.; \t\t\t\t#length of flume\n", - "Bf = 9.; \t\t\t\t#width of throat\n", - "Bo = 15.; \t\t\t\t#width of canal\n", - "\n", - "#width at any dismath.tance x from flumed section is given by\n", - "#Bx = Bo*Bf*Lf/(Lf*Bo-(Bo-Bf)x)\n", - "#on solving we get\n", - "#Bx = 2160/(240-6x)\n", - "\n", - "x = linspace(2,16,8) \t\t\t\t#dismath.tance\n", - "print \"width at any dismath.tance x from flumed section:\";\n", - "Bx = zeros(8)\n", - "for i in range(8):\n", - " Bx[i] = 2160./(240-6*x[i]);\n", - " Bx[i] = round(Bx[i]*100)/100;\n", - " print '%.2f'%(Bx[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "width at any dismath.tance x from flumed section:\n", - "9.47\n", - "10.00\n", - "10.59\n", - "11.25\n", - "12.00\n", - "12.86\n", - "13.85\n", - "15.00\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 19.2 pg : 857" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace,zeros\n", - "\n", - "#design an expansion transition for canal by Chaturvedi's method\n", - "\t\t\t\t\n", - "#Given;\n", - "Lf = 16.; \t\t\t\t#length of flume\n", - "Bf = 9.; \t\t\t\t#width of throat\n", - "Bo = 15.; \t\t\t\t#width of canal\n", - "\n", - "x = linspace(2,16,8); \t\t\t\t#dismath.tance\n", - "\n", - "#dismath.tance x is related as x = Lf*Bo**(2/3)(1-(Bf/Bx)**1.5)/(Bo**1.5-Bf**1.5)\n", - "#on solving we get\n", - "#(Bf/Bx)**1.5 = 1-(x/29.893) (relation is misprinted in book)\n", - "#let (Bf/Bx)**1.5 = r\n", - "r = zeros(8)\n", - "R = zeros(8)\n", - "Bx = zeros(8)\n", - "\n", - "print \"width at any dismath.tance x from flumed section:\";\n", - "for i in range(8):\n", - " r[i] = 1-(x[i]/29.893); \t\t\t\t#Bf/Bx**(1.5)\n", - " R[i] = r[i]**(2./3); \t\t\t\t#Bf/Bx\n", - " Bx[i] = Bf/R[i]; \n", - " Bx[i] = round(Bx[i]*100)/100; \n", - " print \"%.2f.\"%(Bx[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "width at any dismath.tance x from flumed section:\n", - "9.43.\n", - "9.90.\n", - "10.45.\n", - "11.08.\n", - "11.81.\n", - "12.67.\n", - "13.71.\n", - "15.00.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 19.3 pg : 860" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace,zeros,zeros_like\n", - "\n", - "#design a syphon aqueduct\n", - "\n", - "#Given\n", - "Q = 25.; \t\t\t\t#design discharge of canal\n", - "B = 20.; \t\t\t\t#bed width of canal\n", - "D = 1.5; \t\t\t\t#depth of water in canal\n", - "bl = 160.; \t\t\t\t#bed level of canal\n", - "hfq = 400.; \t\t\t\t#high flood discharge of drainage\n", - "hfl = 160.5; \t\t\t\t#high flood level of drainage\n", - "bl_drain = 158.; \t\t\t\t#bed level of drainage\n", - "gl = 160.; \t\t\t\t#general ground level\n", - "\n", - "#demath.sing of drainage water-way\n", - "P = 4.75*(hfq)**0.5; \t\t\t\t#laecey P-Q formula\n", - "print \"design of drainage water-way:wetted perimeter of river = %i m.provide 13 spans \\\n", - "of 6 m each,separated by 12 piers each of 1.25 m thick.\"%(P);\n", - "t = 78.+15;\n", - "print \"total length of water-way = %i m.\"%(t);\n", - "v = 2; \t\t\t\t#velocity through syphon\n", - "hb = hfq/(78*v);\n", - "ac = hfq/(6*2.5*1.3); \t\t\t\t#calculation is wrong in book\n", - "hb = round(hb*100)/100;\n", - "ac = round(ac*100)/100;\n", - "print \"height of barrels = %.2f m.provide recmath.tangular barrels 6 m wide and 2.5 m high.actual \\\n", - "velocity through barrels = %.2f m/sec.\"%(hb,ac);\n", - "\n", - "#design of canal waterway\n", - "print \"design of canal waterway:Type 3 aqueduct is adopted.\";\n", - "l1 = B-10;\n", - "l2 = (20-10)*3/2;\n", - "print \"providing a splay 2:1 in expansion,length of contraction transition = %i m.providing\\\n", - " a splay of 3:1 in expansion,length of expansion transition = %i m.\"%(l1,l2);\n", - "print 'In transition side slopes are warped from original slope of 1.5:1 to vertical.';\n", - "\n", - "#design of levels of different sectionn\n", - "print \"design of levels of different sectionn:at section 4-4:\";\n", - "A = (B+1.5*D); \t\t\t\t#area\n", - "V = Q/A; \t\t\t\t#velocity of flow\n", - "vh = V**2/(2*9.81); \t\t\t\t#velocity head\n", - "ws = gl+D; \t\t\t\t#R.L of water surface\n", - "tel = ws+vh;\n", - "tel = round(tel*1000)/1000;\n", - "print \"R.L of T.E.L = %.2f m. at section 3-3:\"%(tel);\n", - "A = 10*D; \t\t\t\t#area of trough\n", - "V = Q/A; \t\t\t\t#velocity\n", - "vh1 = V**2/(2*9.81); \t\t\t\t#velocity head\n", - "le = 0.3*(vh1-vh); \t\t\t\t#loss of head in expansion from section 3-3 to 4-4\n", - "tel = tel+le;\n", - "rlw = tel-vh1;\n", - "rlb = rlw-D;\n", - "tel = round(tel*1000)/1000;\n", - "rlb = round(rlb*1000)/1000;\n", - "print \"elevation of T.E.L = %.2f m.R.L of bed to maintain consmath.tant water depth = %.2f m.\"%(tel,rlb);\n", - "\n", - "#at section 2-2\n", - "R = A/P;\n", - "N = 0.016;\n", - "S = V**2*N**2/R**(4./3); \t\t\t\t#from manning's formula\n", - "L = 93; \t\t\t\t#length of trough\n", - "hl = L*S; \t\t\t\t#head loss\n", - "tel = tel+hl;\n", - "rlw = tel-vh1;\n", - "rlb = rlw-D;\n", - "tel = round(tel*1000)/1000;\n", - "rlb = round(rlb*1000)/1000;\n", - "print \"at section 2-2:R.L of T.E.L = %.2f m.R.L of bed to maintain consmath.tant water depth = %.2f m.\"%(tel,rlb);\n", - "\n", - "#at section 1-1\n", - "hl = 0.2*(vh1-vh); \t\t\t\t#loss of hed in contraction transition\n", - "tel = tel+hl;\n", - "rlw = tel-vh;\n", - "rlb = tel-D;\n", - "tel = round(tel*1000)/1000;\n", - "rlb = round(rlb*1000)/1000;\n", - "print \"at section 1-1:R.L of T.E.L = %.2f m.R.L of bed to maintain consmath.tant water depth = %.2f m.\"%(tel,rlb);\n", - "\n", - "#design of contraction transition\n", - "#it is designed on the basis of chaturvedi's formula\n", - "Bo = 20.;\n", - "Bf = 10.;\n", - "L = 10.;\n", - "#from chaturvedi formula we get relation between x and Bx as: x = 15.45(1-(10/Bx)**1.5);\n", - "Bx = linspace(10,20,11)\n", - "print \"design of contraction transition on the basis of chaturvedi formula:\\nBx x\";\n", - "x = zeros_like(Bx)\n", - "for i in range(11):\n", - " x[i] = 15.45*(1-(10/Bx[i])**1.5);\n", - " x[i] = round(x[i]*100)/100;\n", - " print \"%i %.2f\"%(Bx[i],x[i]);\n", - "\n", - "\n", - "#design of expansion transition on the basis of chaturvedi formula\n", - "L = 15.;\n", - "Bf = 10.;Bo = 20.;\n", - "#from chaturvedi formula we get relation between x and Bx as: x = 23.15(1-(10/Bx)**1.5);\n", - "print \"design of expansion transition on the basis of chaturvedi formula:\\nBx x\";\n", - "for i in range(11):\n", - " x[i] = 23.15*(1-(10/Bx[i])**1.5);\n", - " x[i] = round(x[i]*100)/100;\n", - " print \"%i %.2f\"%(Bx[i],x[i]);\n", - "\n", - "\n", - "#design of trough\n", - "print \"design of the trough:\";\n", - "print \"flumed water way of canal = 10 m.trough carrying canal will divide into two \\\n", - "compartments each 5 m wide an dseparated by 0.3 m thick partiions.heigth of trough will be \\\n", - " = 2 m.trough iss constructed umath.sing monolithic reinforced concrete.the outer and inner walls\\\n", - " ca be kept 0.4 m thick.thus,outer width of trough = 11.1 m.\";\n", - "\n", - "#head loss through syphon barrels\n", - "V = 2.05; \t\t\t\t#velocity through barrels\n", - "f1 = 0.505; \t\t\t\t#coefficient of loss of head at entry\n", - "a = 0.00316;\n", - "b = 0.030;\n", - "R = (6*2.5)/(2*(6+2.5));\n", - "f2 = a*(1+b/R);\n", - "L = 11.1; \t\t\t\t#length of barrel\n", - "h = (1+f1+f2*L/R)*V**2/(2*9.81);\n", - "hfl_up = hfl+h;\n", - "h = round(h*1000)/1000;\n", - "hfl_up = round(hfl_up*1000)/1000;\n", - "print \"head loss through syphon barrels = %.2f m.upstream H.F.L = %.2f m.\"%(h,hfl_up)\n", - "\n", - "#uplift pressure on the roof\n", - "bt = gl-0.4; \t\t\t\t#R.L of bottom of the trough\n", - "hl = 0.505*V**2/(2*9.81);\n", - "u = hfl_up-hl-159.6;\n", - "up = u*9.81;\n", - "print \"uplift pressure on the roof = %.2f kN/square m.trough slab is 0.4 m thick and exert a downward load of 9.42 kN.\"%(up);\n", - "print \"th ebalance of the uplift pressure has to be resisted by bending action of \\\n", - "trough slab.so,reinforcement has to be provided at the top of the slab.\";\n", - "\n", - "#uplift on the floor of the barrel and its design\n", - "#(a) static head\n", - "print \"uplift on the floor of the barrel and its design:a static head:\";\n", - "bf = bt-2.5; \t\t\t\t#R.L of barrel floor\n", - "t = 0.8; \t\t\t\t#tentative thickness of floor\n", - "bot = bf-t;\n", - "static = bl_drain-bot;\n", - "static = round(static*100)/100;\n", - "print \"static uplift on the floor = %.2f m.\"%(static);\n", - "\n", - "#(b) seepage head\n", - "L = 10.; \t\t\t\t#length of u/s transition\n", - "bs = 3.; \t\t\t\t#half the barrel span\n", - "df = 11.; \t\t\t\t#end drainage floor\n", - "tcl = 24.; \t\t\t\t#total creep length\n", - "tsh = 161.5-bl_drain; \t\t\t\t#total seepage head\n", - "rs = tsh*(1-13/tcl); \t\t\t\t#residual seepage at B\n", - "tu = (static+rs)*9.81;\n", - "tu = round(tu*100)/100;\n", - "print \"b) seepage head:total uplift = %.2f kN/square m.provide thickness of floor 0.8 m\"%(tu);\n", - "bending = tu-17.58;\n", - "bending = round(bending*100)/100;\n", - "print \"uplift to be resisted by bending action of floor = %.2f kN/square m.\"%(bending);\n", - "\n", - "#design of cut-off and protection works for drainage floor\n", - "print \"design of cut-off and protection works for drainage floor:\";\n", - "Q = 400;f = 1;\n", - "R = 0.47*(Q/f)**(1/3);\n", - "d_up = 1.5*R; \t\t\t\t#depth of u/s cut-off\n", - "bot_up = hfl_up-d_up; \t\t\t\t#R.L of bottom of u/s cut-off\n", - "d_down = 1.5*R; \t\t\t\t#depth of d/s cut-off\n", - "bot_down = hfl-d_down; \t\t\t\t#R.L of bottom of d/s cut-off\n", - "l_down = 2.5*(bl_drain-bot_down);\n", - "l_down1 = 2*(bl_drain-bot_up);\n", - "bot_up = round(bot_up*100)/100;\n", - "bot_down = round(bot_down*100)/100;\n", - "l_down = round(l_down);\n", - "l_down1 = round(l_down1);\n", - "print \"R.L of bottom of u/s cut-off = %.2f m.R.L of bottom of d/s cut-off = %.2f m.\"%(bot_up,bot_down);\n", - "print \"length of d/s protection consisting of 40 cm brick pritching = %.2f m.pitching is \\\n", - "supported by toe wall 0.4 m wide and 1.5 m deep at its d/s end.length of d/s protection consisting\\\n", - " of 0.4 cm brick pritching = %.2f m.pitching is supported by toe wall 0.4 m wide and\\\n", - " 1 m deep at its u/s end.\"%(l_down,l_down1);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "design of drainage water-way:wetted perimeter of river = 95 m.provide 13 spans of 6 m each,separated by 12 piers each of 1.25 m thick.\n", - "total length of water-way = 93 m.\n", - "height of barrels = 2.56 m.provide recmath.tangular barrels 6 m wide and 2.5 m high.actual velocity through barrels = 20.51 m/sec.\n", - "design of canal waterway:Type 3 aqueduct is adopted.\n", - "providing a splay 2:1 in expansion,length of contraction transition = 10 m.providing a splay of 3:1 in expansion,length of expansion transition = 15 m.\n", - "In transition side slopes are warped from original slope of 1.5:1 to vertical.\n", - "design of levels of different sectionn:at section 4-4:\n", - "R.L of T.E.L = 161.56 m. at section 3-3:\n", - "elevation of T.E.L = 161.59 m.R.L of bed to maintain consmath.tant water depth = 159.95 m.\n", - "at section 2-2:R.L of T.E.L = 162.36 m.R.L of bed to maintain consmath.tant water depth = 160.72 m.\n", - "at section 1-1:R.L of T.E.L = 162.38 m.R.L of bed to maintain consmath.tant water depth = 160.88 m.\n", - "design of contraction transition on the basis of chaturvedi formula:\n", - "Bx x\n", - "10 0.00\n", - "11 2.06\n", - "12 3.70\n", - "13 5.03\n", - "14 6.12\n", - "15 7.04\n", - "16 7.82\n", - "17 8.48\n", - "18 9.05\n", - "19 9.55\n", - "20 9.99\n", - "design of expansion transition on the basis of chaturvedi formula:\n", - "Bx x\n", - "10 0.00\n", - "11 3.08\n", - "12 5.54\n", - "13 7.53\n", - "14 9.17\n", - "15 10.55\n", - "16 11.71\n", - "17 12.71\n", - "18 13.56\n", - "19 14.31\n", - "20 14.97\n", - "design of the trough:\n", - "flumed water way of canal = 10 m.trough carrying canal will divide into two compartments each 5 m wide an dseparated by 0.3 m thick partiions.heigth of trough will be = 2 m.trough iss constructed umath.sing monolithic reinforced concrete.the outer and inner walls ca be kept 0.4 m thick.thus,outer width of trough = 11.1 m.\n", - "head loss through syphon barrels = 0.33 m.upstream H.F.L = 160.83 m.\n", - "uplift pressure on the roof = 11.01 kN/square m.trough slab is 0.4 m thick and exert a downward load of 9.42 kN.\n", - "th ebalance of the uplift pressure has to be resisted by bending action of trough slab.so,reinforcement has to be provided at the top of the slab.\n", - "uplift on the floor of the barrel and its design:a static head:\n", - "static uplift on the floor = 1.70 m.\n", - "b) seepage head:total uplift = 32.41 kN/square m.provide thickness of floor 0.8 m\n", - "uplift to be resisted by bending action of floor = 14.83 kN/square m.\n", - "design of cut-off and protection works for drainage floor:\n", - "R.L of bottom of u/s cut-off = 160.13 m.R.L of bottom of d/s cut-off = 159.79 m.\n", - "length of d/s protection consisting of 40 cm brick pritching = -4.00 m.pitching is supported by toe wall 0.4 m wide and 1.5 m deep at its d/s end.length of d/s protection consisting of 0.4 cm brick pritching = -4.00 m.pitching is supported by toe wall 0.4 m wide and 1 m deep at its u/s end.\n" - ] - } - ], - "prompt_number": 4 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch2.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch2.ipynb deleted file mode 100755 index 9eb5c1ee..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch2.ipynb +++ /dev/null @@ -1,428 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:3545b25396a8ddd5a6290547da9d278c18ef204a72b66bd3a11c0eea3ce41199" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 2 : METHODS OF IRRIGATION" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.1 pg : 21" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given\n", - "Q = 0.0108 #discharge through well\n", - "y = 0.075 #average depth of flow\n", - "I = 0.05 #average infiltration rate\n", - "A = 0.1 #area to cover\n", - "\n", - "# Calculations\n", - "t = (60*2.303*y*math.log10(Q/(Q-I*A)))/I\n", - "\n", - "# Results\n", - "print \"Time required to cover given area = %.f minutes.\"%(t)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Time required to cover given area = 56 minutes.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2 pg : 21" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "Q = 0.0108 #discharge through well\n", - "y = 0.075 #average depth of flow\n", - "I = 0.05 #average infiltration rate\n", - "A = 0.1 #area to cover\n", - "\n", - "# Calculations\n", - "Amax = Q/I;\n", - "\n", - "# Results\n", - "print \"Maximum area that can be irrigated = %.2f hectare.\"%(Amax);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Maximum area that can be irrigated = 0.22 hectare.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.3 pg : 22" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\n", - "#time of water application\n", - "#optimum length of each border strip\n", - "#dischrge for each border strip\n", - "\n", - "#Given\n", - "d = 0.05;\t\t\t\t\t\t\t\t#depth of root zone\n", - "I = 1.25E-5;\t\t\t\t\t\t\t\t#average infiltration rate\n", - "s = 0.0035\t\t\t\t\t\t\t\t#slope of border strip\n", - "t = d/(I*3600);\n", - "\n", - "# Calculations and Results\n", - "t = round(t*1000)/1000;\n", - "print \"Time of water application = %.2f hours.\"%(t);\n", - "\n", - "#Part (a)\n", - "q = 2E-3;\t\t\t\t\t\t\t\t#discharge entering water source\n", - "qdash = q*100**2*60;\n", - "n = 0.55425-(0.0001386*qdash);\n", - "yo = (n*q/(s**0.5))**0.6;\n", - "y = 0.665*yo;\n", - "L = (q/I)*(1-math.e**(-d/y));\n", - "L = round(10*L)/10;\n", - "print \"Part a:\";\n", - "print \"Optimum length of each border strip = %.2f m.\"%(L);\n", - "\n", - "#Part (b)\n", - "Lgiven = 150\t\t\t\t\t\t\t\t#given value of length\n", - "#First Trial\n", - "q = 3E-3;\n", - "qdash = q*100**2*60;\n", - "n = 0.55425-(0.0001386*qdash);\n", - "yo = (n*q/(s**0.5))**0.6;\n", - "y = 0.665*yo;\n", - "L = (q/I)*(1-math.e**(-d/y));\n", - "#second trial\n", - "q = 3.15E-3;\n", - "qdash = q*100**2*60;\n", - "n = 0.55425-(0.0001386*qdash);\n", - "yo = (n*q/(s**0.5))**0.6;\n", - "y = 0.665*yo;\n", - "L = (q/I)*(1-math.e**(-d/y));\n", - "q = 9*Lgiven*q*1000/L;\n", - "q = round(q*10)/10;\n", - "print \"Part b:\";\n", - "print \"Discharge for each border strip = %.2f lps.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Time of water application = 1.11 hours.\n", - "Part a:\n", - "Optimum length of each border strip = 101.90 m.\n", - "Part b:\n", - "Discharge for each border strip = 28.20 lps.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.4 pg : 26" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\n", - "\n", - "#deep percolation loss\n", - "#water application efficiency and time of irrigation.\n", - "\n", - "#Given\n", - "B = 12.;\t\t\t\t#breadth of bamath.sin\n", - "L = 36.\t\t\t\t#length of bamath.sin\n", - "d = 70.\t\t\t\t#depth of irrigation\n", - "Ic = 70.\t\t\t\t#cumulative infiltration\n", - "kdash = 9;\n", - "ndash = 0.42;\n", - "#Part (a) \n", - "a = 5;\n", - "b = 0.6;\n", - "q = 1.5;\t\t\t\t#stream size\n", - "\n", - "# Calculations and Results\n", - "Q = (q*B)/1000;\n", - "tl = (L/a)**(1/b);\n", - "td = (Ic/kdash)**(1/ndash);\n", - "T = tl+td;\n", - "p = (1-(td/T)**(ndash))*100;\n", - "eita = (1-p/100)*100;\n", - "Tdash = (d*L*B)/(10*eita*Q*60);\n", - "p = round(p*100)/100;\n", - "eita = round(eita*100)/100;\n", - "Tdash = round(Tdash*10)/10;\n", - "print \"Part a:\"\n", - "print \"Deep percolation loss = %.2f percent.\"%(p);\n", - "print \"Water application efficiency = %.2f percent.\"%(eita);\n", - "print \"Time of irrigation = %.2f minutes.\"%(Tdash);\n", - "#part (b)\n", - "a = 8;\n", - "b = 0.6;\n", - "q = 3;\n", - "Q = (q*B)/1000;\n", - "tl = (L/a)**(1/b);\n", - "td = (Ic/kdash)**(1/ndash);\n", - "T = tl+td;\n", - "p = (1-(td/T)**(ndash))*100;\n", - "eita = (1-p/100)*100;\n", - "Tdash = (d*L*B)/(10*eita*Q*60);\n", - "p = round(p*100)/100;\n", - "eita = round(eita*100)/100;\n", - "Tdash = round(Tdash*10)/10;\n", - "\n", - "print \"Part b:\"\n", - "print \"Deep percolation loss = %.2f percent.\"%(p);\n", - "print \"Water application efficiency = %.2f percent.\"%(eita);\n", - "print \"Time of irrigation = %.2f minutes.\"%(Tdash);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Part a:\n", - "Deep percolation loss = 7.47 percent.\n", - "Water application efficiency = 92.53 percent.\n", - "Time of irrigation = 30.30 minutes.\n", - "Part b:\n", - "Deep percolation loss = 3.66 percent.\n", - "Water application efficiency = 96.34 percent.\n", - "Time of irrigation = 14.50 minutes.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.5 pg : 31" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "\n", - "#given\n", - "d = 37.5\t\t\t\t#crop water requirement\n", - "W = 1.\t\t\t\t#furrow spacing\n", - "L = 120.\t\t\t\t#length of furrow\n", - "n = -0.49;\n", - "k = 38.;\n", - "Ttotal = 143.;\t\t\t\t#Total time of irrigation\n", - "A = [0 ,23, 52 ,88, 127]\t\t\t\t#given values of time of advance\n", - "B = zeros(5)\n", - "C = zeros(5)\n", - "D = zeros(5)\n", - "E = zeros(5)\n", - "\n", - "# Calculations\n", - "for i in range(5):\t\t\t\t#loop to find respective values of time of ponding\n", - " B[i] = 143-A[i] \n", - "\n", - "for j in range(5):\t\t\t\t#loop to find respective furrow infiltration\n", - " C[j] = B[j]**(n)*k;\n", - "\n", - "for K in range(4):\t\t\t\t#loop to find respective average infiltration\n", - " D[K] = (C[K]+C[K+1])/2;\n", - "\n", - "\n", - "E[0] = D[0];\n", - "for l in range(1,4):\t\t\t\t#loop to determine cumulative infiltration\n", - " E[l] = D[l]+E[l-1];\n", - "\n", - "I = E[3];\n", - "\n", - "T = (30*d*W*(n+1)/k)**(1/(n+1));\n", - "dav = ((24.5*Ttotal)+(I*(T-Ttotal)))/L;\n", - "q = ((120*37.5)-(24.5*143))/62;\n", - "T = round(T);\n", - "dav = round(dav*10)/10;\n", - "q = round(q*100)/100;\n", - "I = round(I*100)/100;\n", - "\n", - "# Results\n", - "print \"Maximum size of cut-back stream = %.2f lpm.\"%(I);\n", - "print \"Minimum size of cut-back stream = %.2f lpm.\"%(q);\n", - "print \"Time required for putting 37.5mm depth of water = %.2f minutes.\"%(T);\n", - "print \"Average depth of water required = %.2f mm.\"%(dav);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Maximum size of cut-back stream = 19.69 lpm.\n", - "Minimum size of cut-back stream = 16.07 lpm.\n", - "Time required for putting 37.5mm depth of water = 205.00 minutes.\n", - "Average depth of water required = 39.40 mm.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.6 pg : 32" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "L = 100.;\t\t\t\t#length of furrow\n", - "W = 1.;\t\t\t\t#furrow spacing\n", - "s = 0.3\t\t\t\t#longitudnal slope of furrow\n", - "t1 = 80.\t\t\t\t#initial time flow of stream\n", - "t2 = 35.\t\t\t\t#final time flow of stream\n", - "\n", - "# Calculations\n", - "qm = 0.6/s;\n", - "q = qm*0.4;\n", - "dav = ((q*t2*60)+(2*t1*60))/100;\n", - "\n", - "# Results\n", - "print \"Average depth of water applied = %.2f mm.\"%(dav);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Average depth of water applied = 112.80 mm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.7 pg : 40" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given\n", - "Q = 0.0072;\t\t\t\t#discharge through well\n", - "y = 0.1;\t\t\t\t#average depth of flow\n", - "I = 0.05\t\t\t\t#infiltration capacity of soil\n", - "A = 0.04\t\t\t\t#area of land\n", - "\n", - "# Calculations\n", - "t = (2.303*y*60/I)*math.log10(Q/(Q-I*A));\n", - "Amax = Q/I;\n", - "t = round(t*100)/100;\n", - "\n", - "# Results\n", - "print \"Time required to irrigate = %.2f minutes.\"%(t);\n", - "print \"Maximum area that can be irrigated = %.2f ha.\"%(Amax);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Time required to irrigate = 39.06 minutes.\n", - "Maximum area that can be irrigated = 0.14 ha.\n" - ] - } - ], - "prompt_number": 5 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch20.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch20.ipynb deleted file mode 100755 index 72e1d5c9..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch20.ipynb +++ /dev/null @@ -1,142 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:9fa9b8923853c5929c3c0b3b718ed4db74e8cca06a82acc75a09e2384344b2e1" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 20 : RIVER ENGINEERING" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 20.1 pg : 888" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a guide bank required for a bridge in a river\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 50000.; \t\t\t\t#discharge\n", - "f = 1.1; \t\t\t\t#silt factor\n", - "bl = 130.; \t\t\t\t#bed level of river\n", - "hfl = 140.; \t\t\t\t#high flood level\n", - "\n", - "# Calculations and Results\n", - "L = 4.75*(Q)**0.5;\n", - "L = L+212; \t\t\t\t#providing 20 percent more length\n", - "L_up = 5*L/4; \t\t\t\t#upstream length of guide bund\n", - "L_down = L/4; \t\t\t\t#downstream length of guide bund\n", - "r_up = 0.45*L; \t\t\t\t#radius of upstream curved head\n", - "print \"upstream length of guide bund = %i m.\"%(L_up);\n", - "print \"downstream length of guide bund = %i m.\"%(L_down);\n", - "print \"upstream radius of curved head = %i m.;it can be carved at 145 degrees.\"%(r_up);\n", - "print \"downstream radius of curved head = 287m.;it can be carved at 60 degrees.\";\n", - "\n", - "fb = 1.5; \t\t\t\t#free board\n", - "ltop = fb+hfl; \t\t\t\t#level of top of guide bund\n", - "print \"level of top of guide bund = %.2f m.\"%(ltop);\n", - "print \"adopt top level = 142 m.\";\n", - "ltop = 142;\n", - "Hr = ltop-bl;\n", - "print \"keep top width = 4 m. and side slope as 2:1.\";\n", - "T = 0.06*(Q)**(1./3); \t\t\t\t#thickness of stone pitching\n", - "T = round(T*100)/100;\n", - "print \"Thickness of stone pitching = %.2f m.\"%(T);\n", - "R = 0.47*(Q/f)**(1./3); \t\t\t\t#depth of scour\n", - "Rmax = 1.25*R; \t\t\t\t#maximum scour\n", - "rl = hfl-Rmax; \t\t\t\t#R.L at maximum anticipated cover\n", - "D = bl-rl; \t\t\t\t#depth of maximum scour\n", - "Lapron = 1.5*D;\n", - "R = round(R*100)/100;\n", - "Lapron = round(Lapron*100)/100;\n", - "print \"depth of scour = %.2f m.\"%(R);\n", - "print \"for straigtht reach of guide band:\";\n", - "print \"length of apron = %.2f m.\"%(Lapron);\n", - "Rmax = 1.5*R;\n", - "rl = hfl-Rmax;\n", - "D1 = bl-rl;\n", - "Lapron = 1.5*D1;\n", - "R = round(R*100)/100;\n", - "print \"for curvilinear transition portion of guide band:\";\n", - "print \"length of apron = %.2f m.\"%(Lapron);\n", - "T1 = 1.9*T;\n", - "T1 = round(T1*10)/10;\n", - "print \"thickness of apron = %.2f m.\"%(T1);\n", - "print \"volume of stones:\";\n", - "ss = 5**0.5*(141-130)*T;\n", - "as1 = 5**0.5*D*1.25*T;\n", - "ss = round(ss*100)/100;\n", - "as1 = round(as1*100)/100;\n", - "print \"at shank:\";\n", - "print \"on slope = %.2f cubic metre/m.\"%(ss);\n", - "print \"on apron with a slope 2:1 = %.2f cubic metre/m.\"%(as1);\n", - "\n", - "va = 5**0.5*D1*1.25*T;\n", - "vs = ss;\n", - "vs = round(vs*100)/100;\n", - "va = round(va*100)/100;\n", - "print \"U/S andD/S curved portion:\";\n", - "print \"on slope = %.2f cubic metre/m.\"%(vs);\n", - "print \"on apron = %.2f cubic metre/m.\"%(va);\n", - "\n", - "ta = va/(1.5*D1);\n", - "ta = round(ta*10)/10;\n", - "print \"thickness of launching apron = %.2f m.\"%(ta);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "upstream length of guide bund = 1592 m.\n", - "downstream length of guide bund = 318 m.\n", - "upstream radius of curved head = 573 m.;it can be carved at 145 degrees.\n", - "downstream radius of curved head = 287m.;it can be carved at 60 degrees.\n", - "level of top of guide bund = 141.50 m.\n", - "adopt top level = 142 m.\n", - "keep top width = 4 m. and side slope as 2:1.\n", - "Thickness of stone pitching = 2.21 m.\n", - "depth of scour = 16.77 m.\n", - "for straigtht reach of guide band:\n", - "length of apron = 16.45 m.\n", - "for curvilinear transition portion of guide band:\n", - "length of apron = 22.73 m.\n", - "thickness of apron = 4.20 m.\n", - "volume of stones:\n", - "at shank:\n", - "on slope = 54.36 cubic metre/m.\n", - "on apron with a slope 2:1 = 67.74 cubic metre/m.\n", - "U/S andD/S curved portion:\n", - "on slope = 54.36 cubic metre/m.\n", - "on apron = 93.61 cubic metre/m.\n", - "thickness of launching apron = 4.10 m.\n" - ] - } - ], - "prompt_number": 1 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch20_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch20_1.ipynb deleted file mode 100755 index 72e1d5c9..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch20_1.ipynb +++ /dev/null @@ -1,142 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:9fa9b8923853c5929c3c0b3b718ed4db74e8cca06a82acc75a09e2384344b2e1" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 20 : RIVER ENGINEERING" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 20.1 pg : 888" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#design a guide bank required for a bridge in a river\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 50000.; \t\t\t\t#discharge\n", - "f = 1.1; \t\t\t\t#silt factor\n", - "bl = 130.; \t\t\t\t#bed level of river\n", - "hfl = 140.; \t\t\t\t#high flood level\n", - "\n", - "# Calculations and Results\n", - "L = 4.75*(Q)**0.5;\n", - "L = L+212; \t\t\t\t#providing 20 percent more length\n", - "L_up = 5*L/4; \t\t\t\t#upstream length of guide bund\n", - "L_down = L/4; \t\t\t\t#downstream length of guide bund\n", - "r_up = 0.45*L; \t\t\t\t#radius of upstream curved head\n", - "print \"upstream length of guide bund = %i m.\"%(L_up);\n", - "print \"downstream length of guide bund = %i m.\"%(L_down);\n", - "print \"upstream radius of curved head = %i m.;it can be carved at 145 degrees.\"%(r_up);\n", - "print \"downstream radius of curved head = 287m.;it can be carved at 60 degrees.\";\n", - "\n", - "fb = 1.5; \t\t\t\t#free board\n", - "ltop = fb+hfl; \t\t\t\t#level of top of guide bund\n", - "print \"level of top of guide bund = %.2f m.\"%(ltop);\n", - "print \"adopt top level = 142 m.\";\n", - "ltop = 142;\n", - "Hr = ltop-bl;\n", - "print \"keep top width = 4 m. and side slope as 2:1.\";\n", - "T = 0.06*(Q)**(1./3); \t\t\t\t#thickness of stone pitching\n", - "T = round(T*100)/100;\n", - "print \"Thickness of stone pitching = %.2f m.\"%(T);\n", - "R = 0.47*(Q/f)**(1./3); \t\t\t\t#depth of scour\n", - "Rmax = 1.25*R; \t\t\t\t#maximum scour\n", - "rl = hfl-Rmax; \t\t\t\t#R.L at maximum anticipated cover\n", - "D = bl-rl; \t\t\t\t#depth of maximum scour\n", - "Lapron = 1.5*D;\n", - "R = round(R*100)/100;\n", - "Lapron = round(Lapron*100)/100;\n", - "print \"depth of scour = %.2f m.\"%(R);\n", - "print \"for straigtht reach of guide band:\";\n", - "print \"length of apron = %.2f m.\"%(Lapron);\n", - "Rmax = 1.5*R;\n", - "rl = hfl-Rmax;\n", - "D1 = bl-rl;\n", - "Lapron = 1.5*D1;\n", - "R = round(R*100)/100;\n", - "print \"for curvilinear transition portion of guide band:\";\n", - "print \"length of apron = %.2f m.\"%(Lapron);\n", - "T1 = 1.9*T;\n", - "T1 = round(T1*10)/10;\n", - "print \"thickness of apron = %.2f m.\"%(T1);\n", - "print \"volume of stones:\";\n", - "ss = 5**0.5*(141-130)*T;\n", - "as1 = 5**0.5*D*1.25*T;\n", - "ss = round(ss*100)/100;\n", - "as1 = round(as1*100)/100;\n", - "print \"at shank:\";\n", - "print \"on slope = %.2f cubic metre/m.\"%(ss);\n", - "print \"on apron with a slope 2:1 = %.2f cubic metre/m.\"%(as1);\n", - "\n", - "va = 5**0.5*D1*1.25*T;\n", - "vs = ss;\n", - "vs = round(vs*100)/100;\n", - "va = round(va*100)/100;\n", - "print \"U/S andD/S curved portion:\";\n", - "print \"on slope = %.2f cubic metre/m.\"%(vs);\n", - "print \"on apron = %.2f cubic metre/m.\"%(va);\n", - "\n", - "ta = va/(1.5*D1);\n", - "ta = round(ta*10)/10;\n", - "print \"thickness of launching apron = %.2f m.\"%(ta);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "upstream length of guide bund = 1592 m.\n", - "downstream length of guide bund = 318 m.\n", - "upstream radius of curved head = 573 m.;it can be carved at 145 degrees.\n", - "downstream radius of curved head = 287m.;it can be carved at 60 degrees.\n", - "level of top of guide bund = 141.50 m.\n", - "adopt top level = 142 m.\n", - "keep top width = 4 m. and side slope as 2:1.\n", - "Thickness of stone pitching = 2.21 m.\n", - "depth of scour = 16.77 m.\n", - "for straigtht reach of guide band:\n", - "length of apron = 16.45 m.\n", - "for curvilinear transition portion of guide band:\n", - "length of apron = 22.73 m.\n", - "thickness of apron = 4.20 m.\n", - "volume of stones:\n", - "at shank:\n", - "on slope = 54.36 cubic metre/m.\n", - "on apron with a slope 2:1 = 67.74 cubic metre/m.\n", - "U/S andD/S curved portion:\n", - "on slope = 54.36 cubic metre/m.\n", - "on apron = 93.61 cubic metre/m.\n", - "thickness of launching apron = 4.10 m.\n" - ] - } - ], - "prompt_number": 1 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch21.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch21.ipynb deleted file mode 100755 index a51714d4..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch21.ipynb +++ /dev/null @@ -1,180 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:cfc0d5f8100e77d08cba72409c2d6158116c0c87f0c1bc2797c7cdf67223be53" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 21 : WATER POWER ENGINEERING" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 21.1 pg : 906" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#total installed capacity\n", - "#load factor\n", - "#plant factor\n", - "#utilization factor\n", - "\t\t\t\t\n", - "#Given\n", - "c = 10000.; \t\t\t\t#capacity of each generator;\n", - "n = 3.; \t\t\t\t#number of generator\n", - "l1 = 12000.; \t\t\t\t#initial load on plant\n", - "l2 = 26000.; \t\t\t\t#final load on plant\n", - "\n", - "\n", - "# Calculations and Results\n", - "tc = n*c;\n", - "print \"Total installed capacity = %i kW.\"%(tc);\n", - "\n", - "avg = (l1+l2)/2; \t\t\t\t#average load\n", - "pk = l2; \t\t\t\t#peak load\n", - "lf = avg*100/pk;\n", - "lf = round(lf*10)/10;\n", - "print \"load factor = %.2f percent.\"%(lf);\n", - "\n", - "\t\t\t\t#take any time duration t hours\n", - "pf = avg*100/tc;\n", - "pf = round(pf*10)/10;\n", - "print \"plant factor = %.2f percent.\"%(pf);\n", - "\n", - "uf = pk*100/tc;\n", - "uf = round(uf*10)/10;\n", - "print \"utilization ratio = %.2f percent.\"%(uf);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Total installed capacity = 30000 kW.\n", - "load factor = 73.10 percent.\n", - "plant factor = 63.30 percent.\n", - "utilization ratio = 86.70 percent.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 21.2 pg : 906" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#pondage to be provided\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 40.; \t\t\t\t# minimum flow in river\n", - "H = 30.; \t\t\t\t#net head\n", - "lf = 0.73; \t\t\t\t#load factor\n", - "eita = 0.6; \t\t\t\t#plant efficiency\n", - "\n", - "# Calculations and Results\n", - "P = 9.81*Q*H*eita;\n", - "pk = P/lf;\n", - "pk = round(pk*10)/10;\n", - "print \"maximum generation capacity of generator = %.2f kW.\"%(pk);\n", - "\n", - "pp = pk-P; \t\t\t\t#power develop from pondage\n", - "Q = pp/(9.81*H*eita);\n", - "pr = Q*4*3600/10000;\n", - "pr = round(pr*10)/10;\n", - "print \"Pondage required = %.2fD+4 cubic metre.\"%(pr);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum generation capacity of generator = 9675.60 kW.\n", - "Pondage required = 21.30D+4 cubic metre.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 21.3 pg : 907" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#maximum load factor\n", - "\t\t\t\t\n", - "#Given\n", - "c = 15000.; \t\t\t\t#installed capacity of plant\n", - "lf = 0.3; \t\t\t\t#load factor\n", - "eita = 0.82; \t\t\t\t#plant efficiency\n", - "H = 25; \t\t\t\t#working head\n", - "\n", - "# Calculations and Results\n", - "avg = c*lf; \t\t\t\t#average power developed\n", - "Q = avg/(9.81*H*eita);\n", - "Q = round(Q*100)/100;\n", - "print \"minimum discharge required in the stream = %.2f cumecs.\"%(Q);\n", - "\n", - "Q = 32; \t\t\t\t#for second case\n", - "P = 9.81*H*Q*eita;\n", - "lf = P*100/c;\n", - "lf = round(lf*10)/10;\n", - "print \"maximum load factor = %.2f percent.\"%(lf);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "minimum discharge required in the stream = 22.38 cumecs.\n", - "maximum load factor = 42.90 percent.\n" - ] - } - ], - "prompt_number": 3 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch21_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch21_1.ipynb deleted file mode 100755 index a51714d4..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch21_1.ipynb +++ /dev/null @@ -1,180 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:cfc0d5f8100e77d08cba72409c2d6158116c0c87f0c1bc2797c7cdf67223be53" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 21 : WATER POWER ENGINEERING" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 21.1 pg : 906" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#total installed capacity\n", - "#load factor\n", - "#plant factor\n", - "#utilization factor\n", - "\t\t\t\t\n", - "#Given\n", - "c = 10000.; \t\t\t\t#capacity of each generator;\n", - "n = 3.; \t\t\t\t#number of generator\n", - "l1 = 12000.; \t\t\t\t#initial load on plant\n", - "l2 = 26000.; \t\t\t\t#final load on plant\n", - "\n", - "\n", - "# Calculations and Results\n", - "tc = n*c;\n", - "print \"Total installed capacity = %i kW.\"%(tc);\n", - "\n", - "avg = (l1+l2)/2; \t\t\t\t#average load\n", - "pk = l2; \t\t\t\t#peak load\n", - "lf = avg*100/pk;\n", - "lf = round(lf*10)/10;\n", - "print \"load factor = %.2f percent.\"%(lf);\n", - "\n", - "\t\t\t\t#take any time duration t hours\n", - "pf = avg*100/tc;\n", - "pf = round(pf*10)/10;\n", - "print \"plant factor = %.2f percent.\"%(pf);\n", - "\n", - "uf = pk*100/tc;\n", - "uf = round(uf*10)/10;\n", - "print \"utilization ratio = %.2f percent.\"%(uf);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Total installed capacity = 30000 kW.\n", - "load factor = 73.10 percent.\n", - "plant factor = 63.30 percent.\n", - "utilization ratio = 86.70 percent.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 21.2 pg : 906" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#pondage to be provided\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 40.; \t\t\t\t# minimum flow in river\n", - "H = 30.; \t\t\t\t#net head\n", - "lf = 0.73; \t\t\t\t#load factor\n", - "eita = 0.6; \t\t\t\t#plant efficiency\n", - "\n", - "# Calculations and Results\n", - "P = 9.81*Q*H*eita;\n", - "pk = P/lf;\n", - "pk = round(pk*10)/10;\n", - "print \"maximum generation capacity of generator = %.2f kW.\"%(pk);\n", - "\n", - "pp = pk-P; \t\t\t\t#power develop from pondage\n", - "Q = pp/(9.81*H*eita);\n", - "pr = Q*4*3600/10000;\n", - "pr = round(pr*10)/10;\n", - "print \"Pondage required = %.2fD+4 cubic metre.\"%(pr);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum generation capacity of generator = 9675.60 kW.\n", - "Pondage required = 21.30D+4 cubic metre.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 21.3 pg : 907" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#maximum load factor\n", - "\t\t\t\t\n", - "#Given\n", - "c = 15000.; \t\t\t\t#installed capacity of plant\n", - "lf = 0.3; \t\t\t\t#load factor\n", - "eita = 0.82; \t\t\t\t#plant efficiency\n", - "H = 25; \t\t\t\t#working head\n", - "\n", - "# Calculations and Results\n", - "avg = c*lf; \t\t\t\t#average power developed\n", - "Q = avg/(9.81*H*eita);\n", - "Q = round(Q*100)/100;\n", - "print \"minimum discharge required in the stream = %.2f cumecs.\"%(Q);\n", - "\n", - "Q = 32; \t\t\t\t#for second case\n", - "P = 9.81*H*Q*eita;\n", - "lf = P*100/c;\n", - "lf = round(lf*10)/10;\n", - "print \"maximum load factor = %.2f percent.\"%(lf);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "minimum discharge required in the stream = 22.38 cumecs.\n", - "maximum load factor = 42.90 percent.\n" - ] - } - ], - "prompt_number": 3 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch2_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch2_1.ipynb deleted file mode 100755 index 9eb5c1ee..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch2_1.ipynb +++ /dev/null @@ -1,428 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:3545b25396a8ddd5a6290547da9d278c18ef204a72b66bd3a11c0eea3ce41199" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 2 : METHODS OF IRRIGATION" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.1 pg : 21" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given\n", - "Q = 0.0108 #discharge through well\n", - "y = 0.075 #average depth of flow\n", - "I = 0.05 #average infiltration rate\n", - "A = 0.1 #area to cover\n", - "\n", - "# Calculations\n", - "t = (60*2.303*y*math.log10(Q/(Q-I*A)))/I\n", - "\n", - "# Results\n", - "print \"Time required to cover given area = %.f minutes.\"%(t)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Time required to cover given area = 56 minutes.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2 pg : 21" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "Q = 0.0108 #discharge through well\n", - "y = 0.075 #average depth of flow\n", - "I = 0.05 #average infiltration rate\n", - "A = 0.1 #area to cover\n", - "\n", - "# Calculations\n", - "Amax = Q/I;\n", - "\n", - "# Results\n", - "print \"Maximum area that can be irrigated = %.2f hectare.\"%(Amax);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Maximum area that can be irrigated = 0.22 hectare.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.3 pg : 22" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\n", - "#time of water application\n", - "#optimum length of each border strip\n", - "#dischrge for each border strip\n", - "\n", - "#Given\n", - "d = 0.05;\t\t\t\t\t\t\t\t#depth of root zone\n", - "I = 1.25E-5;\t\t\t\t\t\t\t\t#average infiltration rate\n", - "s = 0.0035\t\t\t\t\t\t\t\t#slope of border strip\n", - "t = d/(I*3600);\n", - "\n", - "# Calculations and Results\n", - "t = round(t*1000)/1000;\n", - "print \"Time of water application = %.2f hours.\"%(t);\n", - "\n", - "#Part (a)\n", - "q = 2E-3;\t\t\t\t\t\t\t\t#discharge entering water source\n", - "qdash = q*100**2*60;\n", - "n = 0.55425-(0.0001386*qdash);\n", - "yo = (n*q/(s**0.5))**0.6;\n", - "y = 0.665*yo;\n", - "L = (q/I)*(1-math.e**(-d/y));\n", - "L = round(10*L)/10;\n", - "print \"Part a:\";\n", - "print \"Optimum length of each border strip = %.2f m.\"%(L);\n", - "\n", - "#Part (b)\n", - "Lgiven = 150\t\t\t\t\t\t\t\t#given value of length\n", - "#First Trial\n", - "q = 3E-3;\n", - "qdash = q*100**2*60;\n", - "n = 0.55425-(0.0001386*qdash);\n", - "yo = (n*q/(s**0.5))**0.6;\n", - "y = 0.665*yo;\n", - "L = (q/I)*(1-math.e**(-d/y));\n", - "#second trial\n", - "q = 3.15E-3;\n", - "qdash = q*100**2*60;\n", - "n = 0.55425-(0.0001386*qdash);\n", - "yo = (n*q/(s**0.5))**0.6;\n", - "y = 0.665*yo;\n", - "L = (q/I)*(1-math.e**(-d/y));\n", - "q = 9*Lgiven*q*1000/L;\n", - "q = round(q*10)/10;\n", - "print \"Part b:\";\n", - "print \"Discharge for each border strip = %.2f lps.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Time of water application = 1.11 hours.\n", - "Part a:\n", - "Optimum length of each border strip = 101.90 m.\n", - "Part b:\n", - "Discharge for each border strip = 28.20 lps.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.4 pg : 26" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\n", - "\n", - "#deep percolation loss\n", - "#water application efficiency and time of irrigation.\n", - "\n", - "#Given\n", - "B = 12.;\t\t\t\t#breadth of bamath.sin\n", - "L = 36.\t\t\t\t#length of bamath.sin\n", - "d = 70.\t\t\t\t#depth of irrigation\n", - "Ic = 70.\t\t\t\t#cumulative infiltration\n", - "kdash = 9;\n", - "ndash = 0.42;\n", - "#Part (a) \n", - "a = 5;\n", - "b = 0.6;\n", - "q = 1.5;\t\t\t\t#stream size\n", - "\n", - "# Calculations and Results\n", - "Q = (q*B)/1000;\n", - "tl = (L/a)**(1/b);\n", - "td = (Ic/kdash)**(1/ndash);\n", - "T = tl+td;\n", - "p = (1-(td/T)**(ndash))*100;\n", - "eita = (1-p/100)*100;\n", - "Tdash = (d*L*B)/(10*eita*Q*60);\n", - "p = round(p*100)/100;\n", - "eita = round(eita*100)/100;\n", - "Tdash = round(Tdash*10)/10;\n", - "print \"Part a:\"\n", - "print \"Deep percolation loss = %.2f percent.\"%(p);\n", - "print \"Water application efficiency = %.2f percent.\"%(eita);\n", - "print \"Time of irrigation = %.2f minutes.\"%(Tdash);\n", - "#part (b)\n", - "a = 8;\n", - "b = 0.6;\n", - "q = 3;\n", - "Q = (q*B)/1000;\n", - "tl = (L/a)**(1/b);\n", - "td = (Ic/kdash)**(1/ndash);\n", - "T = tl+td;\n", - "p = (1-(td/T)**(ndash))*100;\n", - "eita = (1-p/100)*100;\n", - "Tdash = (d*L*B)/(10*eita*Q*60);\n", - "p = round(p*100)/100;\n", - "eita = round(eita*100)/100;\n", - "Tdash = round(Tdash*10)/10;\n", - "\n", - "print \"Part b:\"\n", - "print \"Deep percolation loss = %.2f percent.\"%(p);\n", - "print \"Water application efficiency = %.2f percent.\"%(eita);\n", - "print \"Time of irrigation = %.2f minutes.\"%(Tdash);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Part a:\n", - "Deep percolation loss = 7.47 percent.\n", - "Water application efficiency = 92.53 percent.\n", - "Time of irrigation = 30.30 minutes.\n", - "Part b:\n", - "Deep percolation loss = 3.66 percent.\n", - "Water application efficiency = 96.34 percent.\n", - "Time of irrigation = 14.50 minutes.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.5 pg : 31" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "\n", - "#given\n", - "d = 37.5\t\t\t\t#crop water requirement\n", - "W = 1.\t\t\t\t#furrow spacing\n", - "L = 120.\t\t\t\t#length of furrow\n", - "n = -0.49;\n", - "k = 38.;\n", - "Ttotal = 143.;\t\t\t\t#Total time of irrigation\n", - "A = [0 ,23, 52 ,88, 127]\t\t\t\t#given values of time of advance\n", - "B = zeros(5)\n", - "C = zeros(5)\n", - "D = zeros(5)\n", - "E = zeros(5)\n", - "\n", - "# Calculations\n", - "for i in range(5):\t\t\t\t#loop to find respective values of time of ponding\n", - " B[i] = 143-A[i] \n", - "\n", - "for j in range(5):\t\t\t\t#loop to find respective furrow infiltration\n", - " C[j] = B[j]**(n)*k;\n", - "\n", - "for K in range(4):\t\t\t\t#loop to find respective average infiltration\n", - " D[K] = (C[K]+C[K+1])/2;\n", - "\n", - "\n", - "E[0] = D[0];\n", - "for l in range(1,4):\t\t\t\t#loop to determine cumulative infiltration\n", - " E[l] = D[l]+E[l-1];\n", - "\n", - "I = E[3];\n", - "\n", - "T = (30*d*W*(n+1)/k)**(1/(n+1));\n", - "dav = ((24.5*Ttotal)+(I*(T-Ttotal)))/L;\n", - "q = ((120*37.5)-(24.5*143))/62;\n", - "T = round(T);\n", - "dav = round(dav*10)/10;\n", - "q = round(q*100)/100;\n", - "I = round(I*100)/100;\n", - "\n", - "# Results\n", - "print \"Maximum size of cut-back stream = %.2f lpm.\"%(I);\n", - "print \"Minimum size of cut-back stream = %.2f lpm.\"%(q);\n", - "print \"Time required for putting 37.5mm depth of water = %.2f minutes.\"%(T);\n", - "print \"Average depth of water required = %.2f mm.\"%(dav);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Maximum size of cut-back stream = 19.69 lpm.\n", - "Minimum size of cut-back stream = 16.07 lpm.\n", - "Time required for putting 37.5mm depth of water = 205.00 minutes.\n", - "Average depth of water required = 39.40 mm.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.6 pg : 32" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "L = 100.;\t\t\t\t#length of furrow\n", - "W = 1.;\t\t\t\t#furrow spacing\n", - "s = 0.3\t\t\t\t#longitudnal slope of furrow\n", - "t1 = 80.\t\t\t\t#initial time flow of stream\n", - "t2 = 35.\t\t\t\t#final time flow of stream\n", - "\n", - "# Calculations\n", - "qm = 0.6/s;\n", - "q = qm*0.4;\n", - "dav = ((q*t2*60)+(2*t1*60))/100;\n", - "\n", - "# Results\n", - "print \"Average depth of water applied = %.2f mm.\"%(dav);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Average depth of water applied = 112.80 mm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.7 pg : 40" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given\n", - "Q = 0.0072;\t\t\t\t#discharge through well\n", - "y = 0.1;\t\t\t\t#average depth of flow\n", - "I = 0.05\t\t\t\t#infiltration capacity of soil\n", - "A = 0.04\t\t\t\t#area of land\n", - "\n", - "# Calculations\n", - "t = (2.303*y*60/I)*math.log10(Q/(Q-I*A));\n", - "Amax = Q/I;\n", - "t = round(t*100)/100;\n", - "\n", - "# Results\n", - "print \"Time required to irrigate = %.2f minutes.\"%(t);\n", - "print \"Maximum area that can be irrigated = %.2f ha.\"%(Amax);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Time required to irrigate = 39.06 minutes.\n", - "Maximum area that can be irrigated = 0.14 ha.\n" - ] - } - ], - "prompt_number": 5 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch3.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch3.ipynb deleted file mode 100755 index 3e27009a..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch3.ipynb +++ /dev/null @@ -1,1440 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:fb46725fd841b00338dce4327765570824d962464729228190009d6ed499fe86" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 3 : WATER REQUIREMENTS OF CROPS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.1 pg : 44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Given\n", - "Na = 24.; \t\t\t\t#concentration of sodium ion\n", - "Ca = 3.6; \t\t\t\t#concentration of calcium ion\n", - "Mg = 2.; \t\t\t\t#concentration of magnesium ion\n", - "EC = 180.; \t\t\t\t#electrical conductivity\n", - "\n", - "# Calculations\n", - "SAR = Na/(((Ca+Mg)/2)**(0.5)); \t\t\t\t#Sodium absorption ratio\n", - "SAR = round(SAR*100)/100;\n", - "\n", - "# Results\n", - "print \"SAR = %.2f.\"%(SAR);\n", - "print \"Water falls under S2 class.\"; \t\t\t\t#from table 3.2\n", - "print \"For EC = 180\";\n", - "print \"water falls under C1 class.\"; \t\t\t\t#from table 3.1\n", - "print \"Water is medium sodium and low saline water.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "SAR = 14.34.\n", - "Water falls under S2 class.\n", - "For EC = 180\n", - "water falls under C1 class.\n", - "Water is medium sodium and low saline water.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.2 pg : 52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "gammad = 15.; \t\t\t\t#dry weigth of soil\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "Fc = 0.3; \t\t\t\t#field capacity\n", - "pwp = 0.08; \t\t\t\t#permanent wilting point\n", - "d = 0.8; \t\t\t\t#root zone depth\n", - "\n", - "# Calculations\n", - "d1 = gammad*Fc*1000/gammaw;\n", - "d2 = gammad*pwp*1000/gammaw;\n", - "d3 = gammad*d*(Fc-pwp)*1000/gammaw;\n", - "\n", - "# Results\n", - "print \"Depth of moisture in root zone at field capacity = %.f mm/m.\"%(d1);\n", - "print \"Depth of moisture in root zone at permanent wilting point = %.f mm/m.\"%(d2);\n", - "print \"Depth of moisture available in root zone = %.f mm/m.\"%(d3);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Depth of moisture in root zone at field capacity = 459 mm/m.\n", - "Depth of moisture in root zone at permanent wilting point = 122 mm/m.\n", - "Depth of moisture available in root zone = 269 mm/m.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.3 pg : 52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Given\n", - "gammad = 15.3; \t\t\t\t#dry weigth of soil\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "Fc = 0.15; \t\t\t\t#field capacity\n", - "Mc = 0.08; \t\t\t\t#moisture content before irrigation\n", - "D = 60.; \t\t\t\t#Depth of water applied\n", - "\n", - "# Calculations\n", - "d = (gammaw*D)/(gammad*(Fc-Mc));\n", - "\n", - "# Results\n", - "print \"Depth upto which soil profile is wetted = %.f mm.\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Depth upto which soil profile is wetted = 550 mm.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.4 pg :53" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Given\n", - "Sg = 1.6; \t\t\t\t#Apparent specific gravity\n", - "Fc = 0.2; \t\t\t\t#Field capacity\n", - "M1 = 150.; \t\t\t\t#mass of sample soil\n", - "M2 = 136.; \t\t\t\t#mass of sample after drying\n", - "d = 0.9; \t\t\t\t#depth of soil to be irrigated\n", - "\n", - "# Calculations\n", - "Mc = (M1-M2)/M2;\n", - "D = Sg*d*1000*(Fc-Mc);\n", - "\n", - "# Results\n", - "print \"Depth of water required to irrigate the soil = %.f mm.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Depth of water required to irrigate the soil = 140 mm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.5 pg : 53" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "\n", - "#Given\n", - "d = 2.; \t\t\t\t#root zone depth\n", - "Wc = 0.05; \t\t\t\t#existing water content\n", - "gammad = 15; \t\t\t\t#dry density of soil\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "Vw = 500. \t\t\t\t#water applied to the soil\n", - "Wl = 0.1; \t\t\t\t#water loss\n", - "A = 1000.; \t\t\t\t#area of plot\n", - "\n", - "# Calculations\n", - "Vu = Vw*0.9; \t\t\t\t#volume of water used in soil \n", - "Wu = Vu*gammaw; \t\t\t\t#weigth of water used in soil\n", - "Ws = A*d*gammad; \t\t\t\t#total dry weigth of soil\n", - "Wa = Wu*100/Ws; \t\t\t\t#percent water added\n", - "Fc = Wc*100+Wa; \n", - "Fc = round(Fc*100)/100;\n", - "\n", - "# Results\n", - "print \"The Field Capacity of soil is = %.2f percent.\"%(Fc);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The Field Capacity of soil is = 19.72 percent.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.6 pg : 53" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.22; \t\t\t\t#Field Capacity\n", - "wc = 0.1; \t\t\t\t#wilting coefficient\n", - "gammad = 15.; \t\t\t\t#dry unit weigth of soil\n", - "gammaw = 9.81; \t\t\t\t#unit wiegth of water\n", - "d = 0.7; \t\t\t\t#root zone depth\n", - "w = 0.14; \t\t\t\t#falled moisture content\n", - "E = 0.75; \t\t\t\t#water application efficiency\n", - "\n", - "# Calculations\n", - "SC = gammad*d*(Fc-wc)*100/gammaw;\n", - "D = gammad*d*(Fc-w)*1000/gammaw;\n", - "FIR = D/E; \t\t\t\t#Field irrigation requirement\n", - "SC = round(SC*10)/10;\n", - "D = round(D);\n", - "FIR = round(FIR)+1;\n", - "\n", - "# Results\n", - "print \"Maximum storage capacity of soil = %.2f cm.\"%(SC);\n", - "print \"Water depth required to be applied = %.2f mm\"%(D);\n", - "print \"Field Irrigation Requirement = %.2f mm\"%(FIR);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Maximum storage capacity of soil = 12.80 cm.\n", - "Water depth required to be applied = 86.00 mm\n", - "Field Irrigation Requirement = 115.00 mm\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.7 pg : 55" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.27; \t\t\t\t#Field capacity\n", - "pwp = 0.14; \t\t\t\t#permanent wilting point\n", - "gammad = 15.; \t\t\t\t#dry density of soil\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "d = 0.75; \t\t\t\t#effective depth of root zone\n", - "Du = 11.; \t\t\t\t#daily consumptive use of water\n", - "\n", - "# Calculations\n", - "Am = Fc-pwp; \t\t\t\t#Available moisture\n", - "#let readily available moisture be 80 percent of available moisture\n", - "RAm = 0.8*Am;\n", - "Mo = Fc-RAm;\n", - "D = gammad*d*(Fc-Mo)*100/gammaw;\n", - "WF = D*10/Du;\n", - "\n", - "# Results\n", - "print \"Watering Frequency = %i days.\"%(WF);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Watering Frequency = 10 days.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.8 pg : 56" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.22; \t\t\t\t#Field capacity\n", - "Sg = 1.56; \t\t\t\t#Apparent specific gravity\n", - "d = 0.6; \t\t\t\t#root zone depth\n", - "#irrigation is started when 70 percent of moisture is used\n", - "l = 250.; \t\t\t\t#length of field\n", - "b = 40.; \t\t\t\t#width of field\n", - "q = 20.; \t\t\t\t#Discharge\n", - "\n", - "# Calculations\n", - "m = (1-0.7)*Fc;\n", - "D = Sg*d*(Fc-m)*1000;\n", - "A = l*b;\n", - "t = A*D/(q*3600);\n", - "D = round(D);\n", - "t = round(t);\n", - "\n", - "# Results\n", - "print \"Net depth of irrigation water required = %.2f mm.\"%(D);\n", - "print \"Time required to irrigate field = %.2f hours.\"%(t);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Net depth of irrigation water required = 144.00 mm.\n", - "Time required to irrigate field = 20.00 hours.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.9 pg : 64" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "B = 110.; \t\t\t\t#Base period\n", - "D = 1400.; \t\t\t\t#Duty of water\n", - "\n", - "# Calculations\n", - "delta = 8.64*B*100/D;\n", - "delta = round(delta);\n", - "\n", - "# Results\n", - "print \"Delta for crop is = %.2f cm.\"%(delta);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Delta for crop is = 68.00 cm.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.10 pg : 64" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "B = 120.; \t\t\t\t#Base period\n", - "delta = 92.; \t\t\t\t#total depth requirement of crop\n", - "\n", - "# Calculations\n", - "D = 8.64*B*100/delta;\n", - "\n", - "# Results\n", - "print \"Duty of water = %.2f hectares/cumec.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Duty of water = 1126.96 hectares/cumec.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.11 pg : 65" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Cr = 2.; \t\t\t\t#crop ratio\n", - "A = 80000.; \t\t\t\t#Area of field\n", - "CI = 85.; \t\t\t\t#percent field culturable irrigable\n", - "IK = 30.; \t\t\t\t#irrigation intensity during kharif season\n", - "IR = 60.; \t\t\t\t#irrigation intensity for rabi season\n", - "DuK = 800.; \t\t\t\t#Duty of water for kharif season\n", - "DuR = 1700.; \t\t\t\t#Duty of water for rabi season\n", - "\n", - "# Calculations\n", - "CIA = A*CI/100; \t\t\t\t#Culturable irrigable area\n", - "AK = CIA*IK/100; \t\t\t\t#Area under kharif season\n", - "AR = CIA*IR/100; \t\t\t\t#Area under rabi season\n", - "DK = AK/DuK;\n", - "DR = AR/DuR;\n", - "\n", - "\n", - "# Results\n", - "print \"Dischage required at head of canal during Kharif season = %.2f cumecs.\"%(DK);\n", - "print \"Dischage required at head of canal during Rabi season = %.2f cumecs.\"%(DR);\n", - "print \"Water requirement during kharif is greater than during rabi season\";\n", - "print \"Hence,canal should be designed to carry discharge of %.2f cumecs.\"%(DK);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Dischage required at head of canal during Kharif season = 25.50 cumecs.\n", - "Dischage required at head of canal during Rabi season = 24.00 cumecs.\n", - "Water requirement during kharif is greater than during rabi season\n", - "Hence,canal should be designed to carry discharge of 25.50 cumecs.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.12 pg : 65" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "CA = 2600.; \t\t\t\t#culturable area\n", - "IS = 20.; \t\t\t\t#irrigation intensity for sugarcane\n", - "IR = 40.; \t\t\t\t#irrigation intensity for rice\n", - "DuS = 750.; \t\t\t\t#Duty of water for sugarcane\n", - "DuR = 1800.; \t\t\t\t#Duty of water for rice\n", - "PK = 1.2; \t\t\t\t#Peak demand\n", - "\n", - "# Calculations\n", - "AS = CA*IS/100; \t\t\t\t#Area under sugarcane \n", - "AR = CA*IR/100; \t\t\t\t#Area under rice\n", - "DS = AS/DuS;\n", - "DR = AR/DuR;\n", - "DT = DS+DR;\n", - "DD = PK*DT-0.005333+0.01;\n", - "DR = round(DR*1000)/1000;\n", - "DT = round(DT*1000)/1000;\n", - "\n", - "# Results\n", - "print \"Water required for Rice = %.2f cumecs.\"%(DR);\n", - "print \" Sugarcane is a perennial crop.\";\n", - "print \"Hence,Water required for Sugarcane = %.2f cumecs.\"%(DT);\n", - "print \"Design dischage to meet the peak demand = %.2f cumecs.\"%(DD);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Water required for Rice = 0.58 cumecs.\n", - " Sugarcane is a perennial crop.\n", - "Hence,Water required for Sugarcane = 1.27 cumecs.\n", - "Design dischage to meet the peak demand = 1.53 cumecs.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.13 pg : 66" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "ql = 20.; \t\t\t\t#discharge in left branch\n", - "Al = 20000.; \t\t\t\t#culturable area in left branch\n", - "Bl = 120.; \t\t\t\t#Base period in left branch\n", - "Il = 0.8; \t\t\t\t#intensity of rabi in left branch\n", - "qr = 8.; \t\t\t\t#discharge in rigth branch\n", - "Ar = 12000.; \t\t\t\t#culturable area in rigth branch\n", - "Br = 120.; \t\t\t\t#Base period in rigth branch\n", - "Ir = 0.5; \t\t\t\t#intensity of rabi in rigth branch\n", - "\n", - "# Calculations and Results\n", - "#for left canal\n", - "ARl = Al*Il;\n", - "Dl = ARl/ql;\n", - "print \"Duty for left canal is = %i hectares/cumecs.\"%(Dl);\n", - "\n", - "#for rigth canal\n", - "ARr = Ar*Ir;\n", - "Dr = ARr/qr;\n", - "print \"Duty for left canal is = %i hectares/cumecs.\"%(Dr);\n", - "print \"Since,left canal has higher duty,it is more efficient\";\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Duty for left canal is = 800 hectares/cumecs.\n", - "Duty for left canal is = 750 hectares/cumecs.\n", - "Since,left canal has higher duty,it is more efficient\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.14 pg : 66" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "CA = 1200.; \t\t\t\t#culturable area\n", - "IA = 0.4; \t\t\t\t#intensity of irrigation of crop A\n", - "IB = 0.35; \t\t\t\t#intensity of irrigation of crop B\n", - "bA = 20.; \t\t\t\t#kor period of crop A\n", - "bB = 15.; \t\t\t\t#kor period of crop B\n", - "deltaA = 0.1; \t\t\t\t#kor depth of crop A\n", - "deltaB = 0.16; \t\t\t\t#kor depth of crop B\n", - "\n", - "# Calculations and Results\n", - "#crop A\n", - "A = CA*IA;\n", - "Du = 8.64*bA/deltaA;\n", - "qA = A/Du;\n", - "qA = round(qA*1000)/1000;\n", - "print \"Discharge required for crop A = %.2f cumec.\"%(qA);\n", - "\n", - "#crop B\n", - "A = CA*IB;\n", - "Du = 8.64*bB/deltaB;\n", - "qB = A/Du;\n", - "qB = round(qB*1000)/1000;\n", - "print \"Discharge required for crop B = %.2f cumec.\"%(qB);\n", - "D = qA+qB;\n", - "D = round(D*10)/10;\n", - "print \"Design discharge of water course = %.2f cumec.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge required for crop A = 0.28 cumec.\n", - "Discharge required for crop B = 0.52 cumec.\n", - "Design discharge of water course = 0.80 cumec.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.15 pg : 67" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\n", - "#Given\n", - "B = 12.; \t\t\t\t#transplantaion period\n", - "D = 0.5; \t\t\t\t#total depth of water required by the crop\n", - "R = 0.1; \t\t\t\t#rain falling on field\n", - "L = 0.2; \t\t\t\t#loss of water\n", - "A = 600.; \t\t\t\t#irrigated area\n", - "I = 0.6; \t\t\t\t#intensity of irrigation\n", - "delta = D-R;\n", - "Dui = 8.64*B/delta;\n", - "\n", - "# Calculations and Results\n", - "#math.since water loss is 20 percent\n", - "Du = (1-L)*Dui;\n", - "print \"Duty of water required = %.2f hectares/cumec.\"%(Du);\n", - "\n", - "TA = I*A;\n", - "q = TA/Du;\n", - "q = round(q*100)/100;\n", - "print \"Discharge at head of water course = %.2f cumecs.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Duty of water required = 207.36 hectares/cumec.\n", - "Discharge at head of water course = 1.74 cumecs.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.16 pg : 67" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "CF = 0.8; \t\t\t\t#Capacity factory\n", - "Tf = 13./20; \t\t\t\t#time factor\n", - "A = [850., 120., 600., 500., 360.]; \t\t\t\t\n", - "#Given values of area\n", - "B = [320., 90., 120., 120., 120.]; \t\t\t\t\n", - "#Given values of Base period\n", - "D = [580., 580., 1600. ,2000., 600.]; \t\t\t\t\n", - "#Given values of duty at head canal\n", - "\n", - "# Calculations and Results\n", - "DS = A[0]/D[0]; \t\t\t\t#discharge for sugarcane \n", - "DOS = A[1]/D[1]; \t\t\t\t#discharge for overlap sugarcane\n", - "DW = A[2]/D[2]; \t\t\t\t#discharge for wheat\n", - "DB = A[3]/D[3]; \t\t\t\t#discharge for bajri\n", - "DV = A[4]/D[4]; \t\t\t\t#discharge for vegetables\n", - "DR = DS+DW;\n", - "DM = DS+DB;\n", - "DH = DS+DOS+DV;\n", - "print \"Maximum demand is in hot weather\";\n", - "q = DH/Tf;\n", - "D = q/CF;\n", - "q = round(1000*q)/1000;\n", - "D = round(100*D)/100;\n", - "print \"Full supply discharge at head = %.2f cumecs\"%(q);\n", - "print \"Design discharge = %.2f cumecs.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Maximum demand is in hot weather\n", - "Full supply discharge at head = 3.50 cumecs\n", - "Design discharge = 4.37 cumecs.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.17 pg : 68" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "#Given\n", - "CL = 0.2; \t\t\t\t#Canal loss\n", - "RL = 0.12; \t\t\t\t#Reservior loss\n", - "A = [4800., 5600., 2400., 3200., 1400]; \t\t\t\t\n", - "#Given values of area under crop\n", - "D = [1800., 800., 1400., 900., 700]; \t\t\t\t\n", - "#Given values of duty at field\n", - "B = [120., 360., 200., 120., 120]; \t\t\t\t\n", - "#Given values of base period\n", - "\n", - "# Calculations and Results\n", - "#(a) Wheat\n", - "d = A[0]/D[0];\n", - "V1 = d*B[0];\n", - "#(b) Sugarcane\n", - "d = A[1]/D[1];\n", - "V2 = d*B[1];\n", - "#(c) Cotton\n", - "d = A[2]/D[2];\n", - "V3 = round(d*B[2]);\n", - "#(d) Rice\n", - "d = A[3]/D[3];\n", - "V4 = round(d*B[3]);\n", - "#(e) vegetables\n", - "d = A[4]/D[4];\n", - "V5 = d*B[4];\n", - "\n", - "Vd = (V1+V2+V3+V4+V5)*8.64;\n", - "SC = Vd/((1-CL)*(1-RL));\n", - "print \"Reservior capacity = %.2f hectare-metres.\"%(SC);\n", - "\n", - "#Alternative method\n", - "delta = zeros(5)\n", - "for i in range(5):\n", - " delta[i] = 8.64*B[i]/D[i];\n", - "V = zeros(5)\n", - "for j in range(5):\n", - " V[j] = A[j]*delta[j];\n", - "\n", - "s = 0;\n", - "for k in range(5):\n", - " s = s+V[k];\n", - "\n", - "SC = s/((1-CL)*(1-RL));\n", - "\n", - "print \" By Alternative method.Storage capacity = %.f hectare-metres.\"%(SC);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Reservior capacity = 47250.00 hectare-metres.\n", - " By Alternative method.Storage capacity = 47244 hectare-metres.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.18 pg : 82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "eita = 0.7; \t\t\t\t#water application efficiency\n", - "k = 0.75; \t\t\t\t#crop factor\n", - "T = [19., 16., 12.,.5, 13.]; \t\t\t\t\n", - "#Given values of temperature\n", - "p = [7.19 ,7.15, 7.30, 7.03]; \t\t\t\t#daytime hours of the year\n", - "RD = 1.2; \t\t\t\t#rainfall in december\n", - "RJ = 0.8; \t\t\t\t#rainfall in january\n", - "\n", - "f = zeros(4)\n", - "for i in range(4):\n", - " f[i] = p[i]*(1.8*T[i]+32)/40;\n", - "\n", - "s = 0;\n", - "for i in range(4): \n", - " s = s+f[i];\n", - "\n", - "C = k*s;\n", - "R = RD+RJ;\n", - "CIR = C-R;\n", - "FIR = CIR/eita;\n", - "C = round(10*C)/10;\n", - "CIR = round(CIR*10)/10;\n", - "FIR = round(FIR*10)/10;\n", - "\n", - "# Results\n", - "print \"Consumptive use = %.2f cm.\"%(C);\n", - "print \"consumptive irrigatin requirement = %.2f cm.\"%(CIR);\n", - "print \"field irrigatio reqiurement = %.2f cm.\"%(FIR);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Consumptive use = 28.70 cm.\n", - "consumptive irrigatin requirement = 26.70 cm.\n", - "field irrigatio reqiurement = 38.20 cm.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.19 pg : 83" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "L = 20.; \t\t\t\t#latitude of place(degree North)\n", - "T = 15.; \t\t\t\t#mean monthly temperature(degree celcius)\n", - "RH = 0.5; \t\t\t\t#relative humidity\n", - "E = 250.; \t\t\t\t#elevation of area\n", - "V = 25.; \t\t\t\t#wind velocity at 2 m heigth\n", - "\n", - "#from table 3.10\n", - "VP = 12.79; \t\t\t\t#saturation vapour pressure\n", - "s = 0.8; \t\t\t\t#slope of curve between vapur pressure and temperature\n", - "#from table 3.11\n", - "R = 10.8;\n", - "#from table 3.12\n", - "N = 11.1;\n", - "#from table 3.9\n", - "n = 7.74;\n", - "\n", - "# Calculations\n", - "p = n/N;\n", - "e = VP*RH;\n", - "Ea = 0.002187*(160+V)*(VP-e);\n", - "r = 0.2;\n", - "alpha = 0.49;\n", - "sigma = 2.01E-9;\n", - "Ta = 293;\n", - "H = R*(1-r)*(0.29*math.cos(math.pi/9)+0.55*p)-sigma*Ta**4*(0.56-0.092*e**0.5)*(0.10+0.9*p);\n", - "Et = (s*H+alpha*Ea)*31/(s+alpha);\n", - "Et = round(Et*10)/10;\n", - "\n", - "# Results\n", - "print \"consumptive use of rice in january = %.2f mm of water.\"%(Et);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "consumptive use of rice in january = 71.60 mm of water.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.20 pg : 89" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.27; \t\t\t\t#Field capacity\n", - "pwp = 0.13; \t\t\t\t#permanent wilting point\n", - "d = 80.; \t\t\t\t#depth of soil(cm)\n", - "gammad = 1.5; \t\t\t\t#dry unit weigth of soil(g/cc)\n", - "gammaw = 1.; \t\t\t\t#unit weigth of water(g/cc)\n", - "M = 0.18; \t\t\t\t#avearge soil moisture\n", - "eita = 0.8; \t\t\t\t#field efficiency\n", - "FC = 0.15; \t\t\t\t#field channel\n", - "\n", - "# Calculations\n", - "SC = gammad*d*(Fc-pwp)/gammaw;\n", - "D = gammad*d*(Fc-M)/gammaw;\n", - "FIR = D/eita;\n", - "W = FIR/(1-FC);\n", - "W = round(W*10)/10;\n", - "\n", - "# Results\n", - "print \"maximum storage capacity = %.2f cm\"%(SC);\n", - "print \"depth of irrigation water = %.2f cm\"%(D);\n", - "print \"field irrigation requirement = %.2f cm\"%(FIR);\n", - "print \"water required at canal outlet = %.2f cm\"%(W);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum storage capacity = 16.80 cm\n", - "depth of irrigation water = 10.80 cm\n", - "field irrigation requirement = 13.50 cm\n", - "water required at canal outlet = 15.90 cm\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.21 pg : 89" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "W = 0.4; \t\t\t\t#amount of water available from precipitation\n", - "Cl = 0.15; \t\t\t\t#Channel loss\n", - "RL = 0.1; \t\t\t\t#reservior loss\n", - "B = [120., 320., 120., 200., 100]; \t\t\t\t#Base period\n", - "D = [1800., 800., 900., 1400., 1200];\t\t\t\t#Duty at field\n", - "A = [500., 600., 300., 1200., 500]; \t\t\t\t#Area under crop\n", - "\n", - "# Calculations\n", - "for i in range(5):\n", - " delta[i] = 8.64*B[i]/D[i];\n", - "\n", - "V = zeros(5)\n", - "for i in range(5):\n", - " V[i] = delta[i]*A[i];\n", - "\n", - "s = 0;\n", - "for i in range(5):\n", - " s = s+V[i];\n", - "\n", - "C = s*(1-W)/((1-Cl)*(1-RL));\n", - "\n", - "# Results\n", - "print \"Reservior capacity = %.f ha-m.\"%(C);\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Reservior capacity = 3567 ha-m.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.22 pg : 90" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "GCA = 10000.; \t\t\t\t#gross commanded area\n", - "CCA = 0.75*GCA; \t\t\t\t#Culturable commanded area\n", - "IR = 0.6; \t\t\t\t#intensity of irrigation during rabi season\n", - "IK = 0.3; \t\t\t\t#intensity of irrigation during kharif season \n", - "DuR = 2500.; \t\t\t\t#duty during rabi season\n", - "DuK = 1000.; \t\t\t\t#duty during kharif season\n", - "\n", - "# Calculations\n", - "AR = IR*CCA; \t\t\t\t#area under irrigation in rabi season\n", - "AK = IK*CCA; \t\t\t\t#area under irrigation in kharif season\n", - "DR = AR/DuR;\n", - "DK = AK/DuK;\n", - "\n", - "# Results\n", - "print \"discharge required at head of distributory = %.2f cumecs.\"%(DK);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge required at head of distributory = 2.25 cumecs.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.23 pg : 90" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.18; \t\t\t\t#field capacity\n", - "wc = 0.07; \t\t\t\t#wilting cofficient\n", - "Sg = 1.35; \t\t\t\t#bulk density of soil\n", - "d = 1.2; \t\t\t\t#root zone depth\n", - "\n", - "# Calculations and Results\n", - "m = Fc-wc;\n", - "mo = wc+m/3;\n", - "dw = 100*Sg*d*(Fc-mo);\n", - "print \"Depth of water required = %.2f cm\"%(dw);\n", - "ev1 = 1.1; \t\t\t\t#average evapotranspiration rates in 1 NOV-30 NOV\n", - "ev2 = 1.7; \t\t\t\t#average evapotranspiration rates in 1 DEC-31 DEC\n", - "ev3 = 2.4; \t\t\t\t#average evapotranspiration rates in 1 JAN-31 JAN\n", - "ev4 = 1.5; \t\t\t\t#average evapotranspiration rates in 1 FEB-28 FEB\n", - "ev5 = 3.5; \t\t\t\t#average evapotranspiration rates in 1 MAR-25 MAR\n", - "\t\t\t\t#irrigation requirement from 1 NOV to 3 JAn\n", - "dev = (ev1*30+ev2*31+ev3*3)/10;\n", - "print \"Water consumed by evapotranspiration = %.2f cm.\"%(dev);\n", - "print \"No water is required during 1 NOV-3 JAN\";\n", - "\n", - "\t\t\t\t#irrigation requirement after 3rd JAN\n", - "ws = (ev3-1.5)*16/10; \t\t\t\t#water consumed from soil from 4 JAN-19 JAN\n", - "ts = ws+dev; \t\t\t\t#water withdrawn from soil from 1 NOV-19 JAN\n", - "s = (dw-ts)*10;\n", - "day = s/ev3;\n", - "depth = ts+(4*ev3)/10+(2*ev3)/10;\n", - "print \"depth of water required in first irrigation = %.2f cm.\"%(depth);\n", - "\t\t\t\t#/irrigation requirement from 26 JAn to 25 MAR\n", - "w1 = ev3*6;\n", - "w2 = ev4*28;\n", - "w3 = ev5*25;\n", - "W = w1+w2+w3;\n", - "x = (dw*10-(14.4+42))/ev5;\n", - "print \"Hence second irrigation is required after %.2f days i.e on 18th March.\"%(x);\n", - "depth1 = (W-(dw*10))/10;\n", - "print \"required water depth = %.2f cm\"%(depth1);\n", - "print \"First Watering on 29 JAn and 30 JAN = %.2f cm.Second watering required on 18th March = %.2f cm.\"%(depth,depth1);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Depth of water required = 11.88 cm\n", - "Water consumed by evapotranspiration = 9.29 cm.\n", - "No water is required during 1 NOV-3 JAN\n", - "depth of water required in first irrigation = 12.17 cm.\n", - "Hence second irrigation is required after 17.83 days i.e on 18th March.\n", - "required water depth = 2.51 cm\n", - "First Watering on 29 JAn and 30 JAN = 12.17 cm.Second watering required on 18th March = 2.51 cm.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.24 pg : 91" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.26; \t\t\t\t#Field capacity of soil\n", - "A = 3000.; \t\t\t\t#Area of field\n", - "OM = 0.12; \t\t\t\t#optimum moisture \n", - "pwp = 0.1; \t\t\t\t#permanent wilting point\n", - "d = 80.; \t\t\t\t#depth of root zone\n", - "RD = 1.4; \t\t\t\t#relative density of soil\n", - "f = 10.; \t\t\t\t#frequency of irrigation\n", - "eita = 0.23; \t\t\t\t#overall efficiency\n", - "\n", - "# Calculations\n", - "D = RD*d*(Fc-OM);\n", - "U = D*10/f;\n", - "Wr = A*D*100;\n", - "q = Wr/(f*24*3600);\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"daily consumptive = %.2f mm.\"%(U);\n", - "print \"discharge in canal = %.2f q cumecs.\"%(q);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "daily consumptive = 15.68 mm.\n", - "discharge in canal = 5.44 q cumecs.\n" - ] - } - ], - "prompt_number": 25 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.25 pg : 92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "C1 = 0.2; \t\t\t\t#consumptive requirement of crop for 1 to 15 days\n", - "C2 = 0.3; \t\t\t\t#consumptive requirement of crop for 16 to 40 days\n", - "C3 = 0.5; \t\t\t\t#consumptive requirement of crop for 41 to 50 days\n", - "C4 = 0.1; \t\t\t\t#consumptive requirement of crop for 51 to 55 days\n", - "A = 50.; \t\t\t\t#area of land\n", - "wr = 5.; \t\t\t\t#presowing water requirement\n", - "R = 3.5; \t\t\t\t#rainfall during 36th and 45th day\n", - "\n", - "# Calculations\n", - "w1 = 15*C1*100;\n", - "w2 = 25*C2*100;\n", - "w3 = 10*C3*100;\n", - "w4 = 5*C4*100;\n", - "w5 = 5*100;\n", - "W = w1+w2+w3+w4+w5;\n", - "ER = 3.5*100;\n", - "q = (W-ER)*A;\n", - "\n", - "# Results\n", - "print \"total water to be delivered = %i cubic metre.\"%(q);\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "total water to be delivered = 87500 cubic metre.\n" - ] - } - ], - "prompt_number": 26 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.26 pg : 93" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.3; \t\t\t\t#field capacity\n", - "pwp = 0.11; \t\t\t\t#permanent wilting percent\n", - "gammad = 1300.; \t\t\t\t#density of soil\n", - "gammaw = 1000.; \t\t\t\t#density of water\n", - "d = 700.; \t\t\t\t#root zone depth\n", - "CW = 12.; \t\t\t\t#daily consumptive use of water\n", - "\n", - "# Calculations\n", - "WHC = Fc-pwp;\n", - "mo = Fc-(0.75*WHC);\n", - "D = gammad*d*(Fc-mo)/gammaw;\n", - "I = D/CW;\n", - "\n", - "# Results\n", - "print \" watering interval = %i days\"%(I); " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " watering interval = 10 days\n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.27 pg : 94" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "A = 1000.; \t\t\t\t#total area\n", - "AI = 0.7*A; \t\t\t\t#area under irrigation\n", - "B = 15.; \t\t\t\t#Base period\n", - "d = 500.; \t\t\t\t#depth of water required during transplantation\n", - "R = 120.; \t\t\t\t#useful rain falling\n", - "Wl = 0.2; \t\t\t\t#water loss\n", - "\n", - "# Calculations\n", - "delta = d-R;\n", - "Du = 8.64*B*1000/delta;\n", - "DuH = Du*(1-Wl);\n", - "q = AI/DuH;\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"Duty of water = %i hec/cumec.\"%(Du);\n", - "print \"discharge required in water course = %.2f cumecs.\"%(q);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Duty of water = 341 hec/cumec.\n", - "discharge required in water course = 2.57 cumecs.\n" - ] - } - ], - "prompt_number": 28 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.28 pg : 94" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "from numpy import zeros\n", - "\n", - "#Given\n", - "Ar = 4000.; \t\t\t\t#culturable commanded area\n", - "CL = 0.25; \t\t\t\t#canal loss\n", - "RL = 0.15; \t\t\t\t#reservior loss\n", - "B = [120., 360., 180., 120., 120.]; \t\t\t\t#base period\n", - "D = [1800., 1700., 1400., 800., 700.];\t\t\t\t#duty of water\n", - "I = [20., 20., 10., 15., 15.]; \t\t\t\t#intensity of irrigation\n", - "\n", - "A = zeros(5)\n", - "# Calculations\n", - "for i in range(5):\n", - " A[i] = Ar*I[i]/10; \t\t\t\t#area under crop\n", - "\n", - "Q = zeros(5)\n", - "for i in range(5):\n", - " Q[i] = A[i]/D[i]; \t\t\t\t#discharge required\n", - "\n", - "for i in range(5):\n", - " V[i] = 8.64E4*Q[i]*B[i]; \t\t\t\t#quantity of water\n", - "\n", - "s = 0;\n", - "for i in range(5):\n", - " s = s+V[i];\n", - "\n", - "SC = round(s/((1-CL)*(1-RL)*1000000));\n", - "\n", - "# Results\n", - "print \"Storage capacity = %iD+06 cubic metre.\"%(SC);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Storage capacity = 633D+06 cubic metre.\n" - ] - } - ], - "prompt_number": 29 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch3_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch3_1.ipynb deleted file mode 100755 index 3e27009a..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch3_1.ipynb +++ /dev/null @@ -1,1440 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:fb46725fd841b00338dce4327765570824d962464729228190009d6ed499fe86" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 3 : WATER REQUIREMENTS OF CROPS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.1 pg : 44" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Given\n", - "Na = 24.; \t\t\t\t#concentration of sodium ion\n", - "Ca = 3.6; \t\t\t\t#concentration of calcium ion\n", - "Mg = 2.; \t\t\t\t#concentration of magnesium ion\n", - "EC = 180.; \t\t\t\t#electrical conductivity\n", - "\n", - "# Calculations\n", - "SAR = Na/(((Ca+Mg)/2)**(0.5)); \t\t\t\t#Sodium absorption ratio\n", - "SAR = round(SAR*100)/100;\n", - "\n", - "# Results\n", - "print \"SAR = %.2f.\"%(SAR);\n", - "print \"Water falls under S2 class.\"; \t\t\t\t#from table 3.2\n", - "print \"For EC = 180\";\n", - "print \"water falls under C1 class.\"; \t\t\t\t#from table 3.1\n", - "print \"Water is medium sodium and low saline water.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "SAR = 14.34.\n", - "Water falls under S2 class.\n", - "For EC = 180\n", - "water falls under C1 class.\n", - "Water is medium sodium and low saline water.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.2 pg : 52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "gammad = 15.; \t\t\t\t#dry weigth of soil\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "Fc = 0.3; \t\t\t\t#field capacity\n", - "pwp = 0.08; \t\t\t\t#permanent wilting point\n", - "d = 0.8; \t\t\t\t#root zone depth\n", - "\n", - "# Calculations\n", - "d1 = gammad*Fc*1000/gammaw;\n", - "d2 = gammad*pwp*1000/gammaw;\n", - "d3 = gammad*d*(Fc-pwp)*1000/gammaw;\n", - "\n", - "# Results\n", - "print \"Depth of moisture in root zone at field capacity = %.f mm/m.\"%(d1);\n", - "print \"Depth of moisture in root zone at permanent wilting point = %.f mm/m.\"%(d2);\n", - "print \"Depth of moisture available in root zone = %.f mm/m.\"%(d3);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Depth of moisture in root zone at field capacity = 459 mm/m.\n", - "Depth of moisture in root zone at permanent wilting point = 122 mm/m.\n", - "Depth of moisture available in root zone = 269 mm/m.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.3 pg : 52" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Given\n", - "gammad = 15.3; \t\t\t\t#dry weigth of soil\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "Fc = 0.15; \t\t\t\t#field capacity\n", - "Mc = 0.08; \t\t\t\t#moisture content before irrigation\n", - "D = 60.; \t\t\t\t#Depth of water applied\n", - "\n", - "# Calculations\n", - "d = (gammaw*D)/(gammad*(Fc-Mc));\n", - "\n", - "# Results\n", - "print \"Depth upto which soil profile is wetted = %.f mm.\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Depth upto which soil profile is wetted = 550 mm.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.4 pg :53" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Given\n", - "Sg = 1.6; \t\t\t\t#Apparent specific gravity\n", - "Fc = 0.2; \t\t\t\t#Field capacity\n", - "M1 = 150.; \t\t\t\t#mass of sample soil\n", - "M2 = 136.; \t\t\t\t#mass of sample after drying\n", - "d = 0.9; \t\t\t\t#depth of soil to be irrigated\n", - "\n", - "# Calculations\n", - "Mc = (M1-M2)/M2;\n", - "D = Sg*d*1000*(Fc-Mc);\n", - "\n", - "# Results\n", - "print \"Depth of water required to irrigate the soil = %.f mm.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Depth of water required to irrigate the soil = 140 mm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.5 pg : 53" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "\n", - "#Given\n", - "d = 2.; \t\t\t\t#root zone depth\n", - "Wc = 0.05; \t\t\t\t#existing water content\n", - "gammad = 15; \t\t\t\t#dry density of soil\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "Vw = 500. \t\t\t\t#water applied to the soil\n", - "Wl = 0.1; \t\t\t\t#water loss\n", - "A = 1000.; \t\t\t\t#area of plot\n", - "\n", - "# Calculations\n", - "Vu = Vw*0.9; \t\t\t\t#volume of water used in soil \n", - "Wu = Vu*gammaw; \t\t\t\t#weigth of water used in soil\n", - "Ws = A*d*gammad; \t\t\t\t#total dry weigth of soil\n", - "Wa = Wu*100/Ws; \t\t\t\t#percent water added\n", - "Fc = Wc*100+Wa; \n", - "Fc = round(Fc*100)/100;\n", - "\n", - "# Results\n", - "print \"The Field Capacity of soil is = %.2f percent.\"%(Fc);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The Field Capacity of soil is = 19.72 percent.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.6 pg : 53" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.22; \t\t\t\t#Field Capacity\n", - "wc = 0.1; \t\t\t\t#wilting coefficient\n", - "gammad = 15.; \t\t\t\t#dry unit weigth of soil\n", - "gammaw = 9.81; \t\t\t\t#unit wiegth of water\n", - "d = 0.7; \t\t\t\t#root zone depth\n", - "w = 0.14; \t\t\t\t#falled moisture content\n", - "E = 0.75; \t\t\t\t#water application efficiency\n", - "\n", - "# Calculations\n", - "SC = gammad*d*(Fc-wc)*100/gammaw;\n", - "D = gammad*d*(Fc-w)*1000/gammaw;\n", - "FIR = D/E; \t\t\t\t#Field irrigation requirement\n", - "SC = round(SC*10)/10;\n", - "D = round(D);\n", - "FIR = round(FIR)+1;\n", - "\n", - "# Results\n", - "print \"Maximum storage capacity of soil = %.2f cm.\"%(SC);\n", - "print \"Water depth required to be applied = %.2f mm\"%(D);\n", - "print \"Field Irrigation Requirement = %.2f mm\"%(FIR);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Maximum storage capacity of soil = 12.80 cm.\n", - "Water depth required to be applied = 86.00 mm\n", - "Field Irrigation Requirement = 115.00 mm\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.7 pg : 55" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.27; \t\t\t\t#Field capacity\n", - "pwp = 0.14; \t\t\t\t#permanent wilting point\n", - "gammad = 15.; \t\t\t\t#dry density of soil\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "d = 0.75; \t\t\t\t#effective depth of root zone\n", - "Du = 11.; \t\t\t\t#daily consumptive use of water\n", - "\n", - "# Calculations\n", - "Am = Fc-pwp; \t\t\t\t#Available moisture\n", - "#let readily available moisture be 80 percent of available moisture\n", - "RAm = 0.8*Am;\n", - "Mo = Fc-RAm;\n", - "D = gammad*d*(Fc-Mo)*100/gammaw;\n", - "WF = D*10/Du;\n", - "\n", - "# Results\n", - "print \"Watering Frequency = %i days.\"%(WF);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Watering Frequency = 10 days.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.8 pg : 56" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.22; \t\t\t\t#Field capacity\n", - "Sg = 1.56; \t\t\t\t#Apparent specific gravity\n", - "d = 0.6; \t\t\t\t#root zone depth\n", - "#irrigation is started when 70 percent of moisture is used\n", - "l = 250.; \t\t\t\t#length of field\n", - "b = 40.; \t\t\t\t#width of field\n", - "q = 20.; \t\t\t\t#Discharge\n", - "\n", - "# Calculations\n", - "m = (1-0.7)*Fc;\n", - "D = Sg*d*(Fc-m)*1000;\n", - "A = l*b;\n", - "t = A*D/(q*3600);\n", - "D = round(D);\n", - "t = round(t);\n", - "\n", - "# Results\n", - "print \"Net depth of irrigation water required = %.2f mm.\"%(D);\n", - "print \"Time required to irrigate field = %.2f hours.\"%(t);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Net depth of irrigation water required = 144.00 mm.\n", - "Time required to irrigate field = 20.00 hours.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.9 pg : 64" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "B = 110.; \t\t\t\t#Base period\n", - "D = 1400.; \t\t\t\t#Duty of water\n", - "\n", - "# Calculations\n", - "delta = 8.64*B*100/D;\n", - "delta = round(delta);\n", - "\n", - "# Results\n", - "print \"Delta for crop is = %.2f cm.\"%(delta);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Delta for crop is = 68.00 cm.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.10 pg : 64" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "B = 120.; \t\t\t\t#Base period\n", - "delta = 92.; \t\t\t\t#total depth requirement of crop\n", - "\n", - "# Calculations\n", - "D = 8.64*B*100/delta;\n", - "\n", - "# Results\n", - "print \"Duty of water = %.2f hectares/cumec.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Duty of water = 1126.96 hectares/cumec.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.11 pg : 65" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Cr = 2.; \t\t\t\t#crop ratio\n", - "A = 80000.; \t\t\t\t#Area of field\n", - "CI = 85.; \t\t\t\t#percent field culturable irrigable\n", - "IK = 30.; \t\t\t\t#irrigation intensity during kharif season\n", - "IR = 60.; \t\t\t\t#irrigation intensity for rabi season\n", - "DuK = 800.; \t\t\t\t#Duty of water for kharif season\n", - "DuR = 1700.; \t\t\t\t#Duty of water for rabi season\n", - "\n", - "# Calculations\n", - "CIA = A*CI/100; \t\t\t\t#Culturable irrigable area\n", - "AK = CIA*IK/100; \t\t\t\t#Area under kharif season\n", - "AR = CIA*IR/100; \t\t\t\t#Area under rabi season\n", - "DK = AK/DuK;\n", - "DR = AR/DuR;\n", - "\n", - "\n", - "# Results\n", - "print \"Dischage required at head of canal during Kharif season = %.2f cumecs.\"%(DK);\n", - "print \"Dischage required at head of canal during Rabi season = %.2f cumecs.\"%(DR);\n", - "print \"Water requirement during kharif is greater than during rabi season\";\n", - "print \"Hence,canal should be designed to carry discharge of %.2f cumecs.\"%(DK);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Dischage required at head of canal during Kharif season = 25.50 cumecs.\n", - "Dischage required at head of canal during Rabi season = 24.00 cumecs.\n", - "Water requirement during kharif is greater than during rabi season\n", - "Hence,canal should be designed to carry discharge of 25.50 cumecs.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.12 pg : 65" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "CA = 2600.; \t\t\t\t#culturable area\n", - "IS = 20.; \t\t\t\t#irrigation intensity for sugarcane\n", - "IR = 40.; \t\t\t\t#irrigation intensity for rice\n", - "DuS = 750.; \t\t\t\t#Duty of water for sugarcane\n", - "DuR = 1800.; \t\t\t\t#Duty of water for rice\n", - "PK = 1.2; \t\t\t\t#Peak demand\n", - "\n", - "# Calculations\n", - "AS = CA*IS/100; \t\t\t\t#Area under sugarcane \n", - "AR = CA*IR/100; \t\t\t\t#Area under rice\n", - "DS = AS/DuS;\n", - "DR = AR/DuR;\n", - "DT = DS+DR;\n", - "DD = PK*DT-0.005333+0.01;\n", - "DR = round(DR*1000)/1000;\n", - "DT = round(DT*1000)/1000;\n", - "\n", - "# Results\n", - "print \"Water required for Rice = %.2f cumecs.\"%(DR);\n", - "print \" Sugarcane is a perennial crop.\";\n", - "print \"Hence,Water required for Sugarcane = %.2f cumecs.\"%(DT);\n", - "print \"Design dischage to meet the peak demand = %.2f cumecs.\"%(DD);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Water required for Rice = 0.58 cumecs.\n", - " Sugarcane is a perennial crop.\n", - "Hence,Water required for Sugarcane = 1.27 cumecs.\n", - "Design dischage to meet the peak demand = 1.53 cumecs.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.13 pg : 66" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "ql = 20.; \t\t\t\t#discharge in left branch\n", - "Al = 20000.; \t\t\t\t#culturable area in left branch\n", - "Bl = 120.; \t\t\t\t#Base period in left branch\n", - "Il = 0.8; \t\t\t\t#intensity of rabi in left branch\n", - "qr = 8.; \t\t\t\t#discharge in rigth branch\n", - "Ar = 12000.; \t\t\t\t#culturable area in rigth branch\n", - "Br = 120.; \t\t\t\t#Base period in rigth branch\n", - "Ir = 0.5; \t\t\t\t#intensity of rabi in rigth branch\n", - "\n", - "# Calculations and Results\n", - "#for left canal\n", - "ARl = Al*Il;\n", - "Dl = ARl/ql;\n", - "print \"Duty for left canal is = %i hectares/cumecs.\"%(Dl);\n", - "\n", - "#for rigth canal\n", - "ARr = Ar*Ir;\n", - "Dr = ARr/qr;\n", - "print \"Duty for left canal is = %i hectares/cumecs.\"%(Dr);\n", - "print \"Since,left canal has higher duty,it is more efficient\";\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Duty for left canal is = 800 hectares/cumecs.\n", - "Duty for left canal is = 750 hectares/cumecs.\n", - "Since,left canal has higher duty,it is more efficient\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.14 pg : 66" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "CA = 1200.; \t\t\t\t#culturable area\n", - "IA = 0.4; \t\t\t\t#intensity of irrigation of crop A\n", - "IB = 0.35; \t\t\t\t#intensity of irrigation of crop B\n", - "bA = 20.; \t\t\t\t#kor period of crop A\n", - "bB = 15.; \t\t\t\t#kor period of crop B\n", - "deltaA = 0.1; \t\t\t\t#kor depth of crop A\n", - "deltaB = 0.16; \t\t\t\t#kor depth of crop B\n", - "\n", - "# Calculations and Results\n", - "#crop A\n", - "A = CA*IA;\n", - "Du = 8.64*bA/deltaA;\n", - "qA = A/Du;\n", - "qA = round(qA*1000)/1000;\n", - "print \"Discharge required for crop A = %.2f cumec.\"%(qA);\n", - "\n", - "#crop B\n", - "A = CA*IB;\n", - "Du = 8.64*bB/deltaB;\n", - "qB = A/Du;\n", - "qB = round(qB*1000)/1000;\n", - "print \"Discharge required for crop B = %.2f cumec.\"%(qB);\n", - "D = qA+qB;\n", - "D = round(D*10)/10;\n", - "print \"Design discharge of water course = %.2f cumec.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge required for crop A = 0.28 cumec.\n", - "Discharge required for crop B = 0.52 cumec.\n", - "Design discharge of water course = 0.80 cumec.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.15 pg : 67" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\n", - "#Given\n", - "B = 12.; \t\t\t\t#transplantaion period\n", - "D = 0.5; \t\t\t\t#total depth of water required by the crop\n", - "R = 0.1; \t\t\t\t#rain falling on field\n", - "L = 0.2; \t\t\t\t#loss of water\n", - "A = 600.; \t\t\t\t#irrigated area\n", - "I = 0.6; \t\t\t\t#intensity of irrigation\n", - "delta = D-R;\n", - "Dui = 8.64*B/delta;\n", - "\n", - "# Calculations and Results\n", - "#math.since water loss is 20 percent\n", - "Du = (1-L)*Dui;\n", - "print \"Duty of water required = %.2f hectares/cumec.\"%(Du);\n", - "\n", - "TA = I*A;\n", - "q = TA/Du;\n", - "q = round(q*100)/100;\n", - "print \"Discharge at head of water course = %.2f cumecs.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Duty of water required = 207.36 hectares/cumec.\n", - "Discharge at head of water course = 1.74 cumecs.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.16 pg : 67" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "CF = 0.8; \t\t\t\t#Capacity factory\n", - "Tf = 13./20; \t\t\t\t#time factor\n", - "A = [850., 120., 600., 500., 360.]; \t\t\t\t\n", - "#Given values of area\n", - "B = [320., 90., 120., 120., 120.]; \t\t\t\t\n", - "#Given values of Base period\n", - "D = [580., 580., 1600. ,2000., 600.]; \t\t\t\t\n", - "#Given values of duty at head canal\n", - "\n", - "# Calculations and Results\n", - "DS = A[0]/D[0]; \t\t\t\t#discharge for sugarcane \n", - "DOS = A[1]/D[1]; \t\t\t\t#discharge for overlap sugarcane\n", - "DW = A[2]/D[2]; \t\t\t\t#discharge for wheat\n", - "DB = A[3]/D[3]; \t\t\t\t#discharge for bajri\n", - "DV = A[4]/D[4]; \t\t\t\t#discharge for vegetables\n", - "DR = DS+DW;\n", - "DM = DS+DB;\n", - "DH = DS+DOS+DV;\n", - "print \"Maximum demand is in hot weather\";\n", - "q = DH/Tf;\n", - "D = q/CF;\n", - "q = round(1000*q)/1000;\n", - "D = round(100*D)/100;\n", - "print \"Full supply discharge at head = %.2f cumecs\"%(q);\n", - "print \"Design discharge = %.2f cumecs.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Maximum demand is in hot weather\n", - "Full supply discharge at head = 3.50 cumecs\n", - "Design discharge = 4.37 cumecs.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.17 pg : 68" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "#Given\n", - "CL = 0.2; \t\t\t\t#Canal loss\n", - "RL = 0.12; \t\t\t\t#Reservior loss\n", - "A = [4800., 5600., 2400., 3200., 1400]; \t\t\t\t\n", - "#Given values of area under crop\n", - "D = [1800., 800., 1400., 900., 700]; \t\t\t\t\n", - "#Given values of duty at field\n", - "B = [120., 360., 200., 120., 120]; \t\t\t\t\n", - "#Given values of base period\n", - "\n", - "# Calculations and Results\n", - "#(a) Wheat\n", - "d = A[0]/D[0];\n", - "V1 = d*B[0];\n", - "#(b) Sugarcane\n", - "d = A[1]/D[1];\n", - "V2 = d*B[1];\n", - "#(c) Cotton\n", - "d = A[2]/D[2];\n", - "V3 = round(d*B[2]);\n", - "#(d) Rice\n", - "d = A[3]/D[3];\n", - "V4 = round(d*B[3]);\n", - "#(e) vegetables\n", - "d = A[4]/D[4];\n", - "V5 = d*B[4];\n", - "\n", - "Vd = (V1+V2+V3+V4+V5)*8.64;\n", - "SC = Vd/((1-CL)*(1-RL));\n", - "print \"Reservior capacity = %.2f hectare-metres.\"%(SC);\n", - "\n", - "#Alternative method\n", - "delta = zeros(5)\n", - "for i in range(5):\n", - " delta[i] = 8.64*B[i]/D[i];\n", - "V = zeros(5)\n", - "for j in range(5):\n", - " V[j] = A[j]*delta[j];\n", - "\n", - "s = 0;\n", - "for k in range(5):\n", - " s = s+V[k];\n", - "\n", - "SC = s/((1-CL)*(1-RL));\n", - "\n", - "print \" By Alternative method.Storage capacity = %.f hectare-metres.\"%(SC);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Reservior capacity = 47250.00 hectare-metres.\n", - " By Alternative method.Storage capacity = 47244 hectare-metres.\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.18 pg : 82" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "eita = 0.7; \t\t\t\t#water application efficiency\n", - "k = 0.75; \t\t\t\t#crop factor\n", - "T = [19., 16., 12.,.5, 13.]; \t\t\t\t\n", - "#Given values of temperature\n", - "p = [7.19 ,7.15, 7.30, 7.03]; \t\t\t\t#daytime hours of the year\n", - "RD = 1.2; \t\t\t\t#rainfall in december\n", - "RJ = 0.8; \t\t\t\t#rainfall in january\n", - "\n", - "f = zeros(4)\n", - "for i in range(4):\n", - " f[i] = p[i]*(1.8*T[i]+32)/40;\n", - "\n", - "s = 0;\n", - "for i in range(4): \n", - " s = s+f[i];\n", - "\n", - "C = k*s;\n", - "R = RD+RJ;\n", - "CIR = C-R;\n", - "FIR = CIR/eita;\n", - "C = round(10*C)/10;\n", - "CIR = round(CIR*10)/10;\n", - "FIR = round(FIR*10)/10;\n", - "\n", - "# Results\n", - "print \"Consumptive use = %.2f cm.\"%(C);\n", - "print \"consumptive irrigatin requirement = %.2f cm.\"%(CIR);\n", - "print \"field irrigatio reqiurement = %.2f cm.\"%(FIR);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Consumptive use = 28.70 cm.\n", - "consumptive irrigatin requirement = 26.70 cm.\n", - "field irrigatio reqiurement = 38.20 cm.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.19 pg : 83" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "L = 20.; \t\t\t\t#latitude of place(degree North)\n", - "T = 15.; \t\t\t\t#mean monthly temperature(degree celcius)\n", - "RH = 0.5; \t\t\t\t#relative humidity\n", - "E = 250.; \t\t\t\t#elevation of area\n", - "V = 25.; \t\t\t\t#wind velocity at 2 m heigth\n", - "\n", - "#from table 3.10\n", - "VP = 12.79; \t\t\t\t#saturation vapour pressure\n", - "s = 0.8; \t\t\t\t#slope of curve between vapur pressure and temperature\n", - "#from table 3.11\n", - "R = 10.8;\n", - "#from table 3.12\n", - "N = 11.1;\n", - "#from table 3.9\n", - "n = 7.74;\n", - "\n", - "# Calculations\n", - "p = n/N;\n", - "e = VP*RH;\n", - "Ea = 0.002187*(160+V)*(VP-e);\n", - "r = 0.2;\n", - "alpha = 0.49;\n", - "sigma = 2.01E-9;\n", - "Ta = 293;\n", - "H = R*(1-r)*(0.29*math.cos(math.pi/9)+0.55*p)-sigma*Ta**4*(0.56-0.092*e**0.5)*(0.10+0.9*p);\n", - "Et = (s*H+alpha*Ea)*31/(s+alpha);\n", - "Et = round(Et*10)/10;\n", - "\n", - "# Results\n", - "print \"consumptive use of rice in january = %.2f mm of water.\"%(Et);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "consumptive use of rice in january = 71.60 mm of water.\n" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.20 pg : 89" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.27; \t\t\t\t#Field capacity\n", - "pwp = 0.13; \t\t\t\t#permanent wilting point\n", - "d = 80.; \t\t\t\t#depth of soil(cm)\n", - "gammad = 1.5; \t\t\t\t#dry unit weigth of soil(g/cc)\n", - "gammaw = 1.; \t\t\t\t#unit weigth of water(g/cc)\n", - "M = 0.18; \t\t\t\t#avearge soil moisture\n", - "eita = 0.8; \t\t\t\t#field efficiency\n", - "FC = 0.15; \t\t\t\t#field channel\n", - "\n", - "# Calculations\n", - "SC = gammad*d*(Fc-pwp)/gammaw;\n", - "D = gammad*d*(Fc-M)/gammaw;\n", - "FIR = D/eita;\n", - "W = FIR/(1-FC);\n", - "W = round(W*10)/10;\n", - "\n", - "# Results\n", - "print \"maximum storage capacity = %.2f cm\"%(SC);\n", - "print \"depth of irrigation water = %.2f cm\"%(D);\n", - "print \"field irrigation requirement = %.2f cm\"%(FIR);\n", - "print \"water required at canal outlet = %.2f cm\"%(W);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum storage capacity = 16.80 cm\n", - "depth of irrigation water = 10.80 cm\n", - "field irrigation requirement = 13.50 cm\n", - "water required at canal outlet = 15.90 cm\n" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.21 pg : 89" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "W = 0.4; \t\t\t\t#amount of water available from precipitation\n", - "Cl = 0.15; \t\t\t\t#Channel loss\n", - "RL = 0.1; \t\t\t\t#reservior loss\n", - "B = [120., 320., 120., 200., 100]; \t\t\t\t#Base period\n", - "D = [1800., 800., 900., 1400., 1200];\t\t\t\t#Duty at field\n", - "A = [500., 600., 300., 1200., 500]; \t\t\t\t#Area under crop\n", - "\n", - "# Calculations\n", - "for i in range(5):\n", - " delta[i] = 8.64*B[i]/D[i];\n", - "\n", - "V = zeros(5)\n", - "for i in range(5):\n", - " V[i] = delta[i]*A[i];\n", - "\n", - "s = 0;\n", - "for i in range(5):\n", - " s = s+V[i];\n", - "\n", - "C = s*(1-W)/((1-Cl)*(1-RL));\n", - "\n", - "# Results\n", - "print \"Reservior capacity = %.f ha-m.\"%(C);\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Reservior capacity = 3567 ha-m.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.22 pg : 90" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "GCA = 10000.; \t\t\t\t#gross commanded area\n", - "CCA = 0.75*GCA; \t\t\t\t#Culturable commanded area\n", - "IR = 0.6; \t\t\t\t#intensity of irrigation during rabi season\n", - "IK = 0.3; \t\t\t\t#intensity of irrigation during kharif season \n", - "DuR = 2500.; \t\t\t\t#duty during rabi season\n", - "DuK = 1000.; \t\t\t\t#duty during kharif season\n", - "\n", - "# Calculations\n", - "AR = IR*CCA; \t\t\t\t#area under irrigation in rabi season\n", - "AK = IK*CCA; \t\t\t\t#area under irrigation in kharif season\n", - "DR = AR/DuR;\n", - "DK = AK/DuK;\n", - "\n", - "# Results\n", - "print \"discharge required at head of distributory = %.2f cumecs.\"%(DK);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge required at head of distributory = 2.25 cumecs.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.23 pg : 90" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.18; \t\t\t\t#field capacity\n", - "wc = 0.07; \t\t\t\t#wilting cofficient\n", - "Sg = 1.35; \t\t\t\t#bulk density of soil\n", - "d = 1.2; \t\t\t\t#root zone depth\n", - "\n", - "# Calculations and Results\n", - "m = Fc-wc;\n", - "mo = wc+m/3;\n", - "dw = 100*Sg*d*(Fc-mo);\n", - "print \"Depth of water required = %.2f cm\"%(dw);\n", - "ev1 = 1.1; \t\t\t\t#average evapotranspiration rates in 1 NOV-30 NOV\n", - "ev2 = 1.7; \t\t\t\t#average evapotranspiration rates in 1 DEC-31 DEC\n", - "ev3 = 2.4; \t\t\t\t#average evapotranspiration rates in 1 JAN-31 JAN\n", - "ev4 = 1.5; \t\t\t\t#average evapotranspiration rates in 1 FEB-28 FEB\n", - "ev5 = 3.5; \t\t\t\t#average evapotranspiration rates in 1 MAR-25 MAR\n", - "\t\t\t\t#irrigation requirement from 1 NOV to 3 JAn\n", - "dev = (ev1*30+ev2*31+ev3*3)/10;\n", - "print \"Water consumed by evapotranspiration = %.2f cm.\"%(dev);\n", - "print \"No water is required during 1 NOV-3 JAN\";\n", - "\n", - "\t\t\t\t#irrigation requirement after 3rd JAN\n", - "ws = (ev3-1.5)*16/10; \t\t\t\t#water consumed from soil from 4 JAN-19 JAN\n", - "ts = ws+dev; \t\t\t\t#water withdrawn from soil from 1 NOV-19 JAN\n", - "s = (dw-ts)*10;\n", - "day = s/ev3;\n", - "depth = ts+(4*ev3)/10+(2*ev3)/10;\n", - "print \"depth of water required in first irrigation = %.2f cm.\"%(depth);\n", - "\t\t\t\t#/irrigation requirement from 26 JAn to 25 MAR\n", - "w1 = ev3*6;\n", - "w2 = ev4*28;\n", - "w3 = ev5*25;\n", - "W = w1+w2+w3;\n", - "x = (dw*10-(14.4+42))/ev5;\n", - "print \"Hence second irrigation is required after %.2f days i.e on 18th March.\"%(x);\n", - "depth1 = (W-(dw*10))/10;\n", - "print \"required water depth = %.2f cm\"%(depth1);\n", - "print \"First Watering on 29 JAn and 30 JAN = %.2f cm.Second watering required on 18th March = %.2f cm.\"%(depth,depth1);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Depth of water required = 11.88 cm\n", - "Water consumed by evapotranspiration = 9.29 cm.\n", - "No water is required during 1 NOV-3 JAN\n", - "depth of water required in first irrigation = 12.17 cm.\n", - "Hence second irrigation is required after 17.83 days i.e on 18th March.\n", - "required water depth = 2.51 cm\n", - "First Watering on 29 JAn and 30 JAN = 12.17 cm.Second watering required on 18th March = 2.51 cm.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.24 pg : 91" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.26; \t\t\t\t#Field capacity of soil\n", - "A = 3000.; \t\t\t\t#Area of field\n", - "OM = 0.12; \t\t\t\t#optimum moisture \n", - "pwp = 0.1; \t\t\t\t#permanent wilting point\n", - "d = 80.; \t\t\t\t#depth of root zone\n", - "RD = 1.4; \t\t\t\t#relative density of soil\n", - "f = 10.; \t\t\t\t#frequency of irrigation\n", - "eita = 0.23; \t\t\t\t#overall efficiency\n", - "\n", - "# Calculations\n", - "D = RD*d*(Fc-OM);\n", - "U = D*10/f;\n", - "Wr = A*D*100;\n", - "q = Wr/(f*24*3600);\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"daily consumptive = %.2f mm.\"%(U);\n", - "print \"discharge in canal = %.2f q cumecs.\"%(q);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "daily consumptive = 15.68 mm.\n", - "discharge in canal = 5.44 q cumecs.\n" - ] - } - ], - "prompt_number": 25 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.25 pg : 92" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "C1 = 0.2; \t\t\t\t#consumptive requirement of crop for 1 to 15 days\n", - "C2 = 0.3; \t\t\t\t#consumptive requirement of crop for 16 to 40 days\n", - "C3 = 0.5; \t\t\t\t#consumptive requirement of crop for 41 to 50 days\n", - "C4 = 0.1; \t\t\t\t#consumptive requirement of crop for 51 to 55 days\n", - "A = 50.; \t\t\t\t#area of land\n", - "wr = 5.; \t\t\t\t#presowing water requirement\n", - "R = 3.5; \t\t\t\t#rainfall during 36th and 45th day\n", - "\n", - "# Calculations\n", - "w1 = 15*C1*100;\n", - "w2 = 25*C2*100;\n", - "w3 = 10*C3*100;\n", - "w4 = 5*C4*100;\n", - "w5 = 5*100;\n", - "W = w1+w2+w3+w4+w5;\n", - "ER = 3.5*100;\n", - "q = (W-ER)*A;\n", - "\n", - "# Results\n", - "print \"total water to be delivered = %i cubic metre.\"%(q);\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "total water to be delivered = 87500 cubic metre.\n" - ] - } - ], - "prompt_number": 26 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.26 pg : 93" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "Fc = 0.3; \t\t\t\t#field capacity\n", - "pwp = 0.11; \t\t\t\t#permanent wilting percent\n", - "gammad = 1300.; \t\t\t\t#density of soil\n", - "gammaw = 1000.; \t\t\t\t#density of water\n", - "d = 700.; \t\t\t\t#root zone depth\n", - "CW = 12.; \t\t\t\t#daily consumptive use of water\n", - "\n", - "# Calculations\n", - "WHC = Fc-pwp;\n", - "mo = Fc-(0.75*WHC);\n", - "D = gammad*d*(Fc-mo)/gammaw;\n", - "I = D/CW;\n", - "\n", - "# Results\n", - "print \" watering interval = %i days\"%(I); " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " watering interval = 10 days\n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.27 pg : 94" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "A = 1000.; \t\t\t\t#total area\n", - "AI = 0.7*A; \t\t\t\t#area under irrigation\n", - "B = 15.; \t\t\t\t#Base period\n", - "d = 500.; \t\t\t\t#depth of water required during transplantation\n", - "R = 120.; \t\t\t\t#useful rain falling\n", - "Wl = 0.2; \t\t\t\t#water loss\n", - "\n", - "# Calculations\n", - "delta = d-R;\n", - "Du = 8.64*B*1000/delta;\n", - "DuH = Du*(1-Wl);\n", - "q = AI/DuH;\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"Duty of water = %i hec/cumec.\"%(Du);\n", - "print \"discharge required in water course = %.2f cumecs.\"%(q);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Duty of water = 341 hec/cumec.\n", - "discharge required in water course = 2.57 cumecs.\n" - ] - } - ], - "prompt_number": 28 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 3.28 pg : 94" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "from numpy import zeros\n", - "\n", - "#Given\n", - "Ar = 4000.; \t\t\t\t#culturable commanded area\n", - "CL = 0.25; \t\t\t\t#canal loss\n", - "RL = 0.15; \t\t\t\t#reservior loss\n", - "B = [120., 360., 180., 120., 120.]; \t\t\t\t#base period\n", - "D = [1800., 1700., 1400., 800., 700.];\t\t\t\t#duty of water\n", - "I = [20., 20., 10., 15., 15.]; \t\t\t\t#intensity of irrigation\n", - "\n", - "A = zeros(5)\n", - "# Calculations\n", - "for i in range(5):\n", - " A[i] = Ar*I[i]/10; \t\t\t\t#area under crop\n", - "\n", - "Q = zeros(5)\n", - "for i in range(5):\n", - " Q[i] = A[i]/D[i]; \t\t\t\t#discharge required\n", - "\n", - "for i in range(5):\n", - " V[i] = 8.64E4*Q[i]*B[i]; \t\t\t\t#quantity of water\n", - "\n", - "s = 0;\n", - "for i in range(5):\n", - " s = s+V[i];\n", - "\n", - "SC = round(s/((1-CL)*(1-RL)*1000000));\n", - "\n", - "# Results\n", - "print \"Storage capacity = %iD+06 cubic metre.\"%(SC);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Storage capacity = 633D+06 cubic metre.\n" - ] - } - ], - "prompt_number": 29 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch4.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch4.ipynb deleted file mode 100755 index fac86470..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch4.ipynb +++ /dev/null @@ -1,4176 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:8233f932d93bf9a420e5546d5224a7f79d5530eb187360329fde39eb4d1d8811" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 4 : HYDROLOGY\n" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.1 pg : 113" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "p = [78.8, 90.2, 98.6, 102.4, 70.4]; \t\t\t\t#rain guage readings at respective stations\n", - "s = 0.;\n", - "\n", - "# Calculations and Results\n", - "for i in range(5):\n", - " s = s+p[i];\n", - "\n", - "pavg = s/5;\n", - "u = 0;\n", - "for i in range(5):\n", - " u = u+(p[i]-pavg)**2;\n", - "\n", - "sx = (u/4)**0.5;\n", - "Cv = sx*100/pavg;\n", - "N = (Cv/6)**2;\n", - "N = round(N*100)/100;\n", - "print \"mean rainfall = %.2f cm.\"%(pavg);\n", - "print \"total stations needed = %.2f.\"%(N);\n", - "#taking N = 7\n", - "N = 7;\n", - "n = N-5;\n", - "print \"additional guages needed = %i.\"%(n);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "mean rainfall = 88.08 cm.\n", - "total stations needed = 6.44.\n", - "additional guages needed = 2.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.2 pg : 115" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "pB = 48.; \t\t\t\t#precipitation at B\n", - "pC = 51.; \t\t\t\t#precpitation at C\n", - "pD = 45.; \t\t\t\t#precipitation at D\n", - "\n", - "# Calculations\n", - "pA = (pB+pC+pD)/3;\n", - "\n", - "# Results\n", - "print \"precipitation at A = %i mm.\"%(pA);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "precipitation at A = 48 mm.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.3 pg : 115" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "pA = 6.6; \t\t\t\t#precipitation at A\n", - "pB = 4.8; \t\t\t\t#precpitation at B\n", - "pC = 3.7; \t\t\t\t#precipitation at C\n", - "nA = 72.6; \t\t\t\t#normal precipitation at A\n", - "nB = 51.8; \t\t\t\t#normal precipitation at B\n", - "nC = 38.2; \t\t\t\t#normal precipitation at C\n", - "nX = 65.6; \t\t\t\t#normal precipitation at X\n", - "\n", - "# Calculations\n", - "pX = (nX*pA/nA+nX*pB/nB+nX*pC/nC)/3;\n", - "pX = round(pX*100)/100;\n", - "\n", - "# Results\n", - "print \"precipitation at x = %.2f cm.\"%(pX);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "precipitation at x = 6.13 cm.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.4 pg : 115" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "pB = 74.; \t\t\t\t#precipitation at B\n", - "pC = 88.; \t\t\t\t#precpitation at C\n", - "pD = 71.; \t\t\t\t#precipitation at D\n", - "pE = 80.; \t\t\t\t#precipitation at E\n", - "Bx = 9.;\n", - "By = 6.;\n", - "Cx = 12.;\n", - "Cy = -9.;\n", - "Dx = -11.;\n", - "Dy = -6.;\n", - "Ex = -7.;\n", - "Ey = 7.;\n", - "Ax = 0;\n", - "Ay = 0;\n", - "\n", - "# Calculations\n", - "Db = (Bx**2+By**2);\n", - "Dc = (Cx**2+Cy**2);\n", - "Dd = (Dx**2+Dy**2);\n", - "De = (Ex**2+Ey**2);\n", - "Wb = 1/Db;\n", - "Wc = 1/Dc;\n", - "Wd = 1/Dd;\n", - "We = 1/De;\n", - "s = pB*Wb+pC*Wc+pD*Wd+pE*We;\n", - "pA = s/(Wb+Wc+Wd+We);\n", - "pA = round(pA*10)/10;\n", - "\n", - "# Results\n", - "print \"precipitation at A = %.2f mm.\"%(pA);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "precipitation at A = 77.50 mm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.5 pg : 119" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "p = [58., 61., 69., 56., 84., 86., 69., 79., 71.]; \t\t\t\t#values of precipitation\n", - "s = 0;\n", - "\n", - "# Calculations and Results\n", - "for i in range(9):\n", - " s = s+p[i];\n", - "ar = s/9;\n", - "ar = round(ar*10)/10;\n", - "print \"umath.sing arithmatic average method:\"\n", - "print \"Average rainfall = %.2f cm.\"%(ar);\n", - "\n", - "I = [86., 85., 80., 75., 70., 65., 60., 55., 50.]; \t\t\t\t#isphytes\n", - "A = [0.43, 5.20, 4.0, 5.04, 5.85, 4.53, 4.09, 1.27]; \t\t\t\t#area between isohytes\n", - "\n", - "a = zeros(9)\n", - "for i in range(8):\n", - " a[i] = (I[i]+I[i+1])/2;\n", - "\n", - "P = zeros(8)\n", - "for i in range(8):\n", - " P[i] = A[i]*a[i];\n", - "\n", - "s = 0;\n", - "for i in range(8):\n", - " s = s+P[i];\n", - "\n", - "t = 0;\n", - "for i in range(8):\n", - " t = t+A[i];\n", - "\n", - "ar = s/t;\n", - "ar = round(ar*10)/10;\n", - "print \"isohytel method:\"\n", - "print \"Average rainfall = %.2f cm.\"%(ar);\n", - "\n", - "A = [3.26, 0.39, 1.61, 2.04, 2.46, 0.84, 3.91, 5.09, 0.41, 3.94, 2.06, 4.40]; \t\t\t\t#thiessen area\n", - "p = [58., 63., 71., 69., 86., 81., 84., 56., 53., 69., 61., 79.]; \t\t\t\t#observed precipitation\n", - "P = zeros(12)\n", - "for i in range(12):\n", - " P[i] = A[i]*p[i];\n", - "\n", - "s = 0;\n", - "for i in range(12):\n", - " s = s+P[i];\n", - "\n", - "t = 0;\n", - "for i in range(12):\n", - " t = t+A[i];\n", - "\n", - "ar = s/t;\n", - "ar = round(ar*10)/10;\n", - "print \"thiesson polygon method:\"\n", - "print \"Average rainfall = %.2f cm.\"%(ar);\n", - "\n", - "#mean rainfall obtained by thiesson polygon method is different from book as product(A*P) is round offed in book.\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "umath.sing arithmatic average method:\n", - "Average rainfall = 70.30 cm.\n", - "isohytel method:\n", - "Average rainfall = 69.70 cm.\n", - "thiesson polygon method:\n", - "Average rainfall = 70.00 cm.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.6 pg : 127" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "from matplotlib.pyplot import plot\n", - "from numpy import zeros\n", - "\n", - "\n", - "#Given\n", - "X = [69., 55., 62., 67., 87., 70., 65., 75., 90., 100., 90., 95., 85., 90., 75., 95.]; \t\t\t\t#annual rainfall at X\n", - "Y = [77., 62., 67., 68., 86., 90., 65., 75, 70, 70, 70 ,75 ,65 ,70, 55, 75]; \t\t\t\t#average rainfall at 10 base stations\n", - "cx = zeros(16)\n", - "cx[0] = 69; \t\t\t\t#accumulated annual values at station X \n", - "for i in range(1,16):\n", - " cx[i] = cx[i-1]+X[i];\n", - "\n", - "cy = zeros(16)\n", - "cy[0] = 77;\n", - "for i in range(1,16):\n", - " cy[i] = cy[i-1]+Y[i]; \t\t\t\t#accumulated annual values at ten stations\n", - " \n", - "\n", - "#since curve is not having unform slope\n", - "print \"Record at X is not consistent.\";\n", - "print \"From the curve regime is observed in the year 1978.\"\n", - "\n", - "Q = [1970., 1971., 1972., 1973., 1974., 1975., 1976., 1977.];\n", - "O = [95., 75., 90., 85., 95., 90., 100., 90.];\n", - "for i in range(8):\n", - " A[i] = 0.7051*O[i];\n", - "\n", - "print \"Year Observed rainfall Adjusted rainfall\";\n", - "for i in range(8):\n", - " print \"%i %i %i\"%(Q[i],O[i],A[i]);\n", - "\n", - "#graph is plotted between cx and cy\n", - "plot(cy,cx,cy,cx,\"ro\")" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "Record at X is not consistent.\n", - "From the curve regime is observed in the year 1978.\n", - "Year Observed rainfall Adjusted rainfall\n", - "1970 95 66\n", - "1971 75 52\n", - "1972 90 63\n", - "1973 85 59\n", - "1974 95 66\n", - "1975 90 63\n", - "1976 100 70\n", - "1977 90 63\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 9, - "text": [ - "[<matplotlib.lines.Line2D at 0x110983f50>,\n", - " <matplotlib.lines.Line2D at 0x110987210>]" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10e5863d0>" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.7 pg : 130" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "from numpy import zeros,linspace\n", - "\t\t\t\t\n", - "#Given\n", - "c = [0, 12.4, 22.1, 35.1, 52.7, 63.7, 81.9, 109.2, 123.5, 132.6, 143.3, 146., 146.]; \t\t\t\t#cumulative rainfall\n", - "T = linspace(0,13,len(c)) \t\t\t\t#Time\n", - "t = 15./60; \t\t\t\t#time interval\n", - "r = zeros(13)\n", - "I = zeros(13)\n", - "r[0] = 0;\n", - "print \"Rainfall intensity:\";\n", - "I[0] = 0;\n", - "for i in range(1,13):\n", - " r[i] = c[i]-c[i-1]; \n", - " I[i] = r[i]/t; \t\t\t\t#Rainfall intensity\n", - " print \"%.2f\"%(I[i]);\n", - "\n", - "\n", - "#graph is plotted between I and T\n", - "bar(T,I)\n", - "xlabel(\"Time hr\")\n", - "ylabel(\"Rain fall insentity\")" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "Rainfall intensity:\n", - "49.60\n", - "38.80\n", - "52.00\n", - "70.40\n", - "44.00\n", - "72.80\n", - "109.20\n", - "57.20\n", - "36.40\n", - "42.80\n", - "10.80\n", - "0.00\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 15, - "text": [ - "<matplotlib.text.Text at 0x110a8b610>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x11098cd50>" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.8 pg : 131" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "from numpy import *\n", - "\n", - "\n", - "#Given\n", - "CR = array([0, 12.4, 22.1 ,35.1, 52.7, 63.7, 81.9, 109.2, 123.5, 132.6, 143.3, 146.0, 146.0]); \t\t\t\t#cumulative rainfall\n", - "\n", - "c15 = zeros_like(CR)\n", - "c30 = zeros_like(CR)\n", - "c45 = zeros_like(CR)\n", - "c60 = zeros_like(CR)\n", - "c90 = zeros_like(CR)\n", - "c120 = zeros_like(CR)\n", - "\n", - "# Calculations and Results\n", - "c15[1] = 12.4;\n", - "c30[2] = 22.1;\n", - "c45[3] = 35.1;\n", - "c60[4] = 52.7;\n", - "c90[6] = 81.9;\n", - "c120[8] = 123.5;\n", - "for i in range(2,13):\n", - " c15[i] = CR[i]-CR[i-1];\n", - "\n", - "for i in range(3,13):\n", - " c30[i] = CR[i]-CR[i-2];\n", - "\n", - "for i in range(4,13):\n", - " c45[i] = CR[i]-CR[i-3];\n", - "\n", - "for i in range(5,13):\n", - " c60[i] = CR[i]-CR[i-4];\n", - "\n", - "for i in range(7,13):\n", - " c90[i] = CR[i]-CR[i-6];\n", - "\n", - "for i in range(9,13):\n", - " c120[i] = CR[i]-CR[i-8];\n", - "\n", - "print \"15min 30min 45min 60min 90min 120min\";\n", - "for i in range(13):\n", - " print \"%.2f %.2f %.2f %.2f %.2f %.2f\"%(c15[i],c30[i],c45[i],c60[i],c90[i],c120[i]);\n", - "\n", - "I = [109.2, 91, 79.7, 74.1, 67.6, 61.75]; \t\t\t\t#maximum intensity at respective durations\n", - "D = [15 ,30 ,45 ,60 ,90 ,120]; \t\t\t\t#durations\n", - "#greph is plotted between I and D\n", - "plot(D,I,D,I,\"ro\")\n", - "xlabel(\"Duration\")\n", - "ylabel(\"max rain fall intensity\")\n", - "show()" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "15min 30min 45min 60min 90min 120min\n", - "0.00 0.00 0.00 0.00 0.00 0.00\n", - "12.40 0.00 0.00 0.00 0.00 0.00\n", - "9.70 22.10 0.00 0.00 0.00 0.00\n", - "13.00 22.70 35.10 0.00 0.00 0.00\n", - "17.60 30.60 40.30 52.70 0.00 0.00\n", - "11.00 28.60 41.60 51.30 0.00 0.00\n", - "18.20 29.20 46.80 59.80 81.90 0.00\n", - "27.30 45.50 56.50 74.10 96.80 0.00\n", - "14.30 41.60 59.80 70.80 101.40 123.50\n", - "9.10 23.40 50.70 68.90 97.50 120.20\n", - "10.70 19.80 34.10 61.40 90.60 121.20\n", - "2.70 13.40 22.50 36.80 82.30 110.90\n", - "0.00 2.70 13.40 22.50 64.10 93.30\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x11083d250>" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.9 pg : 132" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "p = [475, 377, 731, 1066, 361, 305, 926, 628, 409, 236, 337, 853]; \t\t\t\t#precipitation value\n", - "N = 12.; \t\t\t\t#total number of years\n", - "T = 6; \t\t\t\t#recurrence interval\n", - "\n", - "# Calculations\n", - "m = N/T;\n", - "\n", - "# Results\n", - "print \"Ranking of storm = %i.\"%(m);\n", - "#hence pick 2nd severest storm\n", - "print \"preciptation value which has recurrence period of 6 years = %i mm.\"%(p[6]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Ranking of storm = 2.\n", - "preciptation value which has recurrence period of 6 years = 926 mm.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.10 pg : 132" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "from numpy import arange,array,zeros_like\n", - "\t\t\t\t\n", - "#Given\n", - "I = linspace(25,16,10) \t\t\t\t#isohytes\n", - "a = array([407 ,1008, 1522, 1909, 2216, 2460, 2651, 2782, 2910, 2936]); \t\t\t\t#enclosed area\n", - "ia = zeros_like(a)\n", - "ia[0] = 407.;\n", - "\n", - "# Calculations\n", - "for i in range(1,10):\n", - " ia[i] = a[i]-a[i-1];\n", - "r = linspace(25.5,16.5,10)\n", - "rv = r*ia\n", - "\n", - "cv = zeros_like(rv)\n", - "cv[0] = 10378;\n", - "for i in range(1,10):\n", - " cv[i] = cv[i-1]+rv[i];\n", - "\n", - "eud = cv/a\n", - "print \"From depth area curve we obtain average depth of precipitation = 24.1 mm for an area of 1800 sq. km.\";\n", - "#graph is plotted between eud and a.\n", - "# Results\n", - "plot(a,eud,a,eud,\"ro\")\n", - "xlabel(\"Area\")\n", - "ylabel(\"mean precipitation depth\")" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "From depth area curve we obtain average depth of precipitation = 24.1 mm for an area of 1800 sq. km.\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 27, - "text": [ - "<matplotlib.text.Text at 0x110b3b310>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x110c084d0>" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.11 pg :133" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "from numpy import array\n", - "\n", - "#24h max. rainfall with return period of 8,15 and 25.\n", - "#24h max rainfall with 40%,24% and 8% probability.\n", - "#probabilty of rainfall of magnitude equal to or exceeding 100 mm.\n", - "\t\t\t\t\n", - "#Given\n", - "N = 20.;\n", - "r = array([142, 126, 116, 108, 102, 95, 92, 88, 86, 82, 80, 78, 76, 73, 71, 69, 68, 66, 65, 64],dtype=float64); \t\t\t\t#rainfall in respective years\n", - "m = linspace(1,20,20) \t\t\t\t#ranking of storm\n", - "p = m*100/(N+1)\n", - "T = 100/p\n", - "\n", - "# Calculations and Results\n", - "#from frequency curve obtained we get\n", - "#Part (a)\n", - "T1 = array([8, 15, 25]);\n", - "r1 = array([119, 134, 149]);\n", - "print \"Tyears Rainfallmm\";\n", - "for i in range(3):\n", - " print \"%i %i\"%(T1[i],r1[i]);\n", - "\n", - "\n", - "#Part (b)\n", - "p1 = [40 ,24, 8];\n", - "r2 = [87, 101, 130];\n", - "print \"probabilitypercent Rainfallmm\";\n", - "for i in range(3):\n", - " print \"%i %i\"%(p1[i],r2[i]);\n", - "\n", - "print \"For rainfall = 100 m.T = 4 years.Probability = 25 percent.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "Tyears Rainfallmm\n", - "8 119\n", - "15 134\n", - "25 149\n", - "probabilitypercent Rainfallmm\n", - "40 87\n", - "24 101\n", - "8 130\n", - "For rainfall = 100 m.T = 4 years.Probability = 25 percent.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.12 pg : 135" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "\n", - "\n", - "#plot IDF curve for return period of 10,2 and 1 years umath.sing california formula\n", - "\t\t\t\t\n", - "#Given\n", - "t = array([5, 10, 20, 30, 60, 90, 120],dtype=float64); \t\t\t\t#duration\n", - "\t\t\t\t#value of P for respective return period is\n", - "p10 = array([10.6, 14.7, 19.3, 20.8, 25.5, 29, 34.7]); \t\t\t\t#rainfall for T = 10 years\n", - "p2 = array([8.2, 10.3, 13.2, 14.2, 16.6 ,19.4, 21.4]); \t\t\t\t#rainfall for T = 2 years\n", - "p1 = array([3.5, 6.2, 8.9, 10, 13.2, 15, 16.5]); \t\t\t\t#rainfall for T = 1 year\n", - "\n", - "\n", - "# Calculations\n", - "i1 = p10*60/t; \t\t\t\t#intensity of rainfall with return period of 10 years\n", - "i2 = p2*60/t; \t\t\t\t#intensity of rainfall with return period of 2 years\n", - "i3 = p1*60/t; \t\t\t\t#intensity of rainfall with return period of 1 year\n", - "\n", - "# Results\n", - "#graph is plotted between #t and i1 #t and i2 #t and i3\n", - "plot(t,i1)\n", - "plot(t,i2)\n", - "plot(t,i3)\n", - "xlabel(\"Duration\")\n", - "ylabel(\"Intensity\")\n", - "show()" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x110ba6310>" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.13 pg : 137" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "N = 20.;\n", - "m = linspace(1,20,20) \t\t\t\t#rank number\n", - "rd = array([82, 78, 75, 72, 70, 68, 65, 63, 61, 58, 56, 54, 52, 50, 46, 40, 36, 34, 32, 30]); \t\t\t\t#rainfall in decremath.sing order\n", - "\n", - "# Calculations\n", - "ri = rd[::-1]\n", - "T = N/(m-0.5);\n", - "\n", - "# Results\n", - "#from the curves\n", - "print \"maximum rainfall = 79cm for T = 15 years.\";\n", - "print \"minimum rainfall = 31 cm for T = 15 years.\";\n", - "#graph is plotted between rd and T;ri and T\n", - "subplot(121)\n", - "plot(T,rd)\n", - "xlabel(\"Reccurance interval\")\n", - "ylabel(\"rainfall cm\")\n", - "subplot(122)\n", - "plot(T,ri)\n", - "xlabel(\"Reccurance interval\")\n", - "ylabel(\"rainfall cm\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "maximum rainfall = 79cm for T = 15 years.\n", - "minimum rainfall = 31 cm for T = 15 years.\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 44, - "text": [ - "<matplotlib.text.Text at 0x110fc1410>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x11102ee50>" - ] - } - ], - "prompt_number": 44 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.14 pg : 144" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t#average evaporation for one week\n", - "\t\t\t\t\n", - "#Given\n", - "w = [12, 5, 2, -3, 1, 6, 11]; \t\t\t\t#water added or taken out\n", - "r = [0, 6, 8, 12, 9, 5, 0]; \t\t\t\t#rainfall\n", - "pan = zeros(7)\n", - "le = zeros(7)\n", - "for i in range(7):\n", - " pan[i] = w[i]+r[i]; \t\t\t\t#Pan evaporation\n", - " le[i] = 0.8*pan[i]; \t\t\t\t#lake evaporation\n", - "s = sum(le)\n", - "\n", - "print \"daily lake evaporationmm:\";\n", - "for i in range(7):\n", - " print \"%.2f\"%(le[i]);\n", - "\n", - "av = s/7;\n", - "av = round(av*100)/100;\n", - "print \"average evaporation for one week = %.2f mm.\"%(av);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "daily lake evaporationmm:\n", - "9.60\n", - "8.80\n", - "8.00\n", - "7.20\n", - "8.00\n", - "8.80\n", - "8.80\n", - "average evaporation for one week = 8.46 mm.\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.15 pg : 145" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#total depth and volume of evaporation loss\n", - "\t\t\t\t\n", - "#Given\n", - "Rh = 0.4; \t\t\t\t#relative humidity\n", - "A = 4.8; \t\t\t\t#average surface spread of reservior\n", - "v3 = 18.; \t\t\t\t#wind velocity at 3m above ground\n", - "es = 31.81; \t\t\t\t#saturated vapour pressure\n", - "Km = 0.36; \t\t\t\t#for large deep waters\n", - "\n", - "\n", - "# Calculations and Results\n", - "#umath.sing Meyer's formula\n", - "ea = es*Rh;\n", - "v9 = v3*(9./3)**(1./7);\n", - "E = Km*(es-ea)*(1+v9/16);\n", - "d = 7*E;\n", - "v = d*A*100/1000;\n", - "E = round(E*10)/10;\n", - "d = round(d*10)/10;\n", - "v = round(v*100)/100;\n", - "print \"umath.sing Meyers formula:\";\n", - "print \"average evaporation loss from reservior = %.2f mm/day.\"%(E);\n", - "print \"total depth = %.2f mm\"%(d);\n", - "print \"total volume = %.2f hectare-m.\"%(v);\n", - "\n", - "\t\t\t\t#umath.sing Rohwer's formula\n", - "Pa = 760.;\n", - "vdash = (0.6/2)**(1./7)*18;\n", - "E = 0.771*(1.465-0.000732*Pa)*(0.44+0.0733*vdash)*(es-ea);\n", - "d = 7*E;\n", - "v = d*A*100/1000;\n", - "E = round(E*10)/10;\n", - "d = round(d*10)/10;\n", - "v = round(v*10)/10;\n", - "print \"umath.sing Rohwers formula:\";\n", - "print \"average evaporation loss from reservior = %.2f mm/day.\"%(E);\n", - "print \"total depth = %.2f mm\"%(d);\n", - "print \"total volume = %.2f hectare-m.\"%(v);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "umath.sing Meyers formula:\n", - "average evaporation loss from reservior = 15.90 mm/day.\n", - "total depth = 111.40 mm\n", - "total volume = 53.47 hectare-m.\n", - "umath.sing Rohwers formula:\n", - "average evaporation loss from reservior = 20.70 mm/day.\n", - "total depth = 145.20 mm\n", - "total volume = 69.70 hectare-m.\n" - ] - } - ], - "prompt_number": 46 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.16 pg : 149" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "\n", - "#plot infiltration capacity curve\n", - "\t\t\t\t\n", - "#Given\n", - "D = 30.; \t\t\t\t#diameter of inside ring of infiltrometer\n", - "A = math.pi*D**2/4;\n", - "V = array([0, 200, 470, 840, 1405, 1840, 2245, 2510, 2745, 2885, 2990, 3130, 3270],dtype=float64); \t\t\t\t#cumulative volume;\n", - "t = array([0, 2, 5, 10, 20, 30, 45, 60, 80, 100, 120, 150, 180],dtype=float64); \t\t\t\t#Time(minutes)\n", - "\n", - "# Calculations and Results\n", - "dt = zeros_like(t)\n", - "\n", - "for i in range(1,13):\n", - " dt[i] = (t[i]-t[i-1])/60;\n", - "\n", - "F = V/A;\n", - "\n", - "Fd = zeros_like(F)\n", - "Fd[0] = F[0];\n", - "for i in range(1,13):\n", - " Fd[i] = F[i]-F[i-1];\n", - "\n", - "ft = Fd/dt \t\t\t\t#infirltration rate\n", - "\n", - "#from the graph\n", - "print \"constant rate of infiltration = 0.40 cm/hr.\";\n", - "avg10 = F[3]*60/10;\n", - "avg30 = F[5]*60/30;\n", - "avg10 = round(avg10*100)/100;\n", - "avg30 = round(avg30*100)/100;\n", - "print \"average rate of infiltration for first 10 min = %.2f cm/hr.\"%(avg10);\n", - "print \"average rate of infiltration for first 30 min = %.2f cm/hr.\"%(avg30);\n", - "#graph is plotted between ft and t\n", - "plot(t,ft)\n", - "xlabel(\"time in mins\")\n", - "ylabel(\"Infiltrtion rate\")\n", - "show()" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "constant rate of infiltration = 0.40 cm/hr.\n", - "average rate of infiltration for first 10 min = 7.13 cm/hr.\n", - "average rate of infiltration for first 30 min = 5.21 cm/hr.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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7ndNVTMXYt+yN2DeQSMAxJcaV+Jkaho1BrgOWY2OSQcclIiIiIiIiIiIiIiIi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\nIiIiIiIiIiIiIiIiIiIiIiIiIiIiYfZ/7AvpCyJ/2BkAAAAASUVORK5CYII=\n", - "text": [ - "<matplotlib.figure.Figure at 0x110b87a50>" - ] - } - ], - "prompt_number": 49 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.17 pg : 162" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#drainage desity\n", - "#form factor\n", - "#channel slope\n", - "#average overland flow length\n", - "\t\t\t\t\n", - "#Given\n", - "A = 82.; \t\t\t\t#area of watershed\n", - "d = 12.6; \t\t\t\t#dismath.tance between outlet and farther most point\n", - "l = 440.; \t\t\t\t#total length of channel\n", - "e = 656.; \t\t\t\t#elevation differnce between outlet and further most point\n", - "\n", - "\n", - "# Calculations\n", - "Dd = l/A;\n", - "ff = A/d**2;\n", - "cs = e/(d*1000);\n", - "lo = 1000/(2*Dd);\n", - "Dd = round(Dd*100)/100;\n", - "ff = round(ff*1000)/1000;\n", - "\n", - "# Results\n", - "print \"drainage desity = %.2f km/square.km.\"%(Dd);\n", - "print \"form factor = %.2f.\"%(ff);\n", - "print \"channel slope = %.2f.\"%(cs);\n", - "print \"average overland flow length = %i m.\"%(lo);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "drainage desity = 5.37 km/square.km.\n", - "form factor = 0.52.\n", - "channel slope = 0.05.\n", - "average overland flow length = 93 m.\n" - ] - } - ], - "prompt_number": 50 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.18 pg : 163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#compute fi and W index\n", - "\t\t\t\t\n", - "#Given\n", - "R = 3.6; \t\t\t\t#surface runoff\n", - "r = [0, 1.3, 2.8, 4.1, 3.9, 2.8, 2.0, 1.8, 0.9]; \t\t\t\t#rainfall at respective time\n", - "t = 4.; \t\t\t\t#total time\n", - "s = sum(r[2:]);\n", - "\n", - "# Calculations and Results\n", - "fi = (s-R*2)/6;\n", - "#math.since fi >1.3 and <1.8\n", - "print \"fi index = %.2f cm.\"%(fi);\n", - "print \"computations are correct.\";\n", - "\n", - "s = sum(r);\n", - "P = s/2;\n", - "Sr = 0.;\n", - "W = (P-R-Sr)/t;\n", - "print \"W index = %.2f cm/hr.\"%(W);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "fi index = 1.85 cm.\n", - "computations are correct.\n", - "W index = 1.55 cm/hr.\n" - ] - } - ], - "prompt_number": 51 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.19 pg : 163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import *\n", - "\t\t\t\t\n", - "#Given\n", - "T = linspace(1,9,9) \t\t\t\t#time from start\n", - "r = array([0.7, 1.4, 2.4, 3.7, 2.9, 2.6, 1.7, 0.8, 0.5]); \t\t\t\t#increamental rainfall\n", - "R = 9.3; \t\t\t\t#total run-off\n", - "s = sum(r)\n", - "\n", - "ti = s-R;\n", - "#first trial\n", - "tr = 9.; \t\t\t\t#assumed\n", - "fi1 = ti/tr;\n", - "#this makes 1st,8th and 9th hour ineffective\n", - "\n", - "#second trial\n", - "tr = 6.;\n", - "ti = s-R-r[0]-r[7]-r[8];\n", - "fi = ti/tr;\n", - "P = zeros_like(r)\n", - "for i in range(9):\n", - " P[i] = r[i]-fi;\n", - " if (P[i]<0):\n", - " P[i] = 0;\n", - " \n", - "print \"Timeh rainfall excess.\";\n", - "for i in range(9):\n", - " print \"%.2f %.2f\"%(T[i],P[i]);\n", - "\n", - "print \"fi index = %.2f cm/hr.\"%(fi);\n", - "print \"time of rainfall excess = %i hours..\"%(tr);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Timeh rainfall excess.\n", - "1.00 0.00\n", - "2.00 0.50\n", - "3.00 1.50\n", - "4.00 2.80\n", - "5.00 2.00\n", - "6.00 1.70\n", - "7.00 0.80\n", - "8.00 0.00\n", - "9.00 0.00\n", - "fi index = 0.90 cm/hr.\n", - "time of rainfall excess = 6 hours..\n" - ] - } - ], - "prompt_number": 52 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.20 pg : 164" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "P = array([72.2, 70.1, 73.3, 42.5, 81.3, 50.6, 52.9, 59.4, 60.3, 64.3, 68.8, 56.7, 77.2, 40.5, 44.1, 65.5]); \t\t\t\t#Precipitation\n", - "R = array([24.1, 22.7, 25.6, 11.3, 28.4, 12.7, 13.4, 15.7, 16.2, 17.7, 19.2, 14.9, 25.4, 10.6, 11.7, 17.9]); \t\t\t\t#runoff\n", - "\n", - "# Calculations\n", - "Ps = P**2\n", - "Rs = R**2;\n", - "PR = P*R;\n", - "\n", - "s = sum(Ps)\n", - "t = sum(Rs)\n", - "u = sum(PR)\n", - "q = sum(P)\n", - "w = sum(R)\n", - "N = 16.;\n", - "a = (N*u-q*w)/(N*s-q**2);\n", - "b = (w-a*q)/N;\n", - "b = round(b*1000)/1000;\n", - "a = round(a*10000)/10000;\n", - "\n", - "# Results\n", - "print \"Equation is:%.4f P %.3f.\"%(a,b);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Equation is:0.4375 P -8.823.\n" - ] - } - ], - "prompt_number": 61 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.21 pg : 165" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "\n", - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "A = 8.6; \t\t\t\t#catchment area\n", - "T = linspace(0,4,9) \t\t\t\t#time\n", - "r = array([0, 0.4, 1.1, 2.3, 3.8, 4.8, 5.6, 6.2, 6.7]); \t\t\t\t#accumulated rainfall\n", - "fi = 0.4; \t\t\t\t#fi index\n", - "dt = 0.5; \t\t\t\t#time interval\n", - "\n", - "# Calculations and Results\n", - "d = zeros_like(r)\n", - "\n", - "for i in range(1,9):\n", - " d[i] = r[i]-r[i-1]; \t\t\t\t#accumulated rainfall\n", - "\n", - "print \"Intensity of effective Rainfall:\";\n", - "I = zeros_like(r)\n", - "p = zeros_like(r)\n", - "s = 0;\n", - "for i in range(1,9):\n", - " p[i] = d[i]-fi; \t\t\t\t#effective rainfall\n", - " I[i] = p[i]/dt; \t\t\t\t#Intensity of effective Rainfall\n", - " s = s+I[i];\n", - " print \"%.2f\"%(I[i]);\n", - "\n", - "#graph is plotted between I and T\n", - "run = s*dt;\n", - "V = run*A*10000;\n", - "print \"Volume of direct run-off = %.2f cubic metre.\"%(V);\n", - "plot(T,I)\n", - "show()" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "Intensity of effective Rainfall:\n", - "0.00\n", - "0.60\n", - "1.60\n", - "2.20\n", - "1.20\n", - "0.80\n", - "0.40\n", - "0.20\n", - "Volume of direct run-off = 301000.00 cubic metre.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x11081b050>" - ] - } - ], - "prompt_number": 63 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.22 pg : 166" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "\n", - "#total rainfall\n", - "#total rainfall excess\n", - "#W index\n", - "\t\t\t\t\n", - "#Given\n", - "r = array([3.5, 6.5, 8.5 ,7.8, 6.4, 4, 4, 6]); \t\t\t\t#rainfall intensity\n", - "T = linspace(0,240,8) \t\t\t\t#time\n", - "dt = 30.; \t\t\t\t#time interval\n", - "\n", - "# Calculations\n", - "s = sum(r);\n", - "P = s*dt/60;\n", - "Pe = ((6.5-4.5)+(8.5-4.5)+(7.8-4.5)+(6.4-4.5)+(6-4.5))*dt/60; \t\t\t\t#area of graph above r = 4.5.\n", - "w = (P-Pe)/4;\n", - "\n", - "# Results\n", - "print \"total rainfall = %.2f cm.\"%(P);\n", - "print \"total rainfall excess = %.2f cm.\"%(Pe);\n", - "print \"W index = %.2f cm/hr.\"%(w);\n", - "plot(T,r)\n", - "show()" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "total rainfall = 23.35 cm.\n", - "total rainfall excess = 6.35 cm.\n", - "W index = 4.25 cm/hr.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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qcF4qDygiIgdXqgDPOQp4ESuxlABGAO+nMigREREREdlHc2AJsAx4wHEsLqwC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\no0m7AAAAAElFTkSuQmCC\n", - "text": [ - "<matplotlib.figure.Figure at 0x11118be90>" - ] - } - ], - "prompt_number": 65 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.23 pg : 166" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array,zeros_like\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "r = array([0, 8, 22, 74, 92, 105, 114, 120],dtype=float64); \t\t\t\t#raccumulated rainfall\n", - "T = array([0, 2, 4, 6, 8, 10, 12, 14],dtype=float64); \t\t\t\t#time for start of rainfall\n", - "V = 2e6; \t\t\t\t#volume of run-off\n", - "A = 40.; \t\t\t\t#catchment area\n", - "tr = 14.; \t\t\t\t#duration of rainfall\n", - "\n", - "# Calculations\n", - "d = V*1000/(40*1000000);\n", - "\n", - "l = r[7]-d;\n", - "W = l/tr;\n", - "I = zeros_like(r)\n", - "for i in range(1,8):\n", - " I[i] = r[i]-r[i-1]; \t\t\t\t#incremental rainfall\n", - "\n", - "\n", - "#rainfall excess is available in 4 time intervals of 2 hrs\n", - "tre = 8.;\n", - "fi = (l-I[1]-I[6]-I[7])/tre;\n", - "fi = round(fi*100)/100;\n", - "\n", - "# Results\n", - "print \"fi index = %.2f mm/hr.\"%(fi);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "fi index = 5.88 mm/hr.\n" - ] - } - ], - "prompt_number": 66 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.24 pg : 167" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array\n", - "\n", - "#Given\n", - "r = array([2.0 ,2.5, 7.6, 3.8, 10.6, 5.0, 7.0, 10.0, 6.4, 3.8, 1.4, 1.4]); \t\t\t\t#rainfall depths\n", - "R = 25.5;\n", - "\n", - "# Calculations\n", - "s = sum(r)\n", - "tf = s-R;\n", - "af = tf/12;\n", - "#rainfall is less than average infiltration in1st,2nd,11th and 12th hours\n", - "\n", - "f = (tf-r[0]-r[1]-r[10]-r[11])/8;\n", - "f = round(f*10)/10;\n", - "\n", - "# Results\n", - "print \"average infiltration index = %d cm/hour.\"%(f);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "average infiltration index = 3 cm/hour.\n" - ] - } - ], - "prompt_number": 68 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.25 pg : 167" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros,zeros_like,array\n", - "\t\t\t\t\n", - "#Given\n", - "r = array([2.0, 2.5, 7.6, 3.8, 10.6, 5.0, 7.0, 10.0, 6.4, 3.8, 1.4, 1.4]); \t\t\t\t#rainfall depths\n", - "A1 = 20.;\n", - "A2 = 40.;\n", - "A3 = 60.;\n", - "\n", - "# Calculations and Results\n", - "A = A1+A2+A3;\n", - "fi1 = 7.6;\n", - "fi2 = 3.8;\n", - "fi3 = 1.0;\n", - "R1 = zeros_like(r)\n", - "R2 = zeros_like(r)\n", - "R3 = zeros_like(r)\n", - "for i in range(12):\n", - " R1[i] = r[i]-fi1; \t\t\t\t#rainfall excess\n", - " R2[i] = r[i]-fi2;\n", - " R3[i] = r[i]-fi3;\n", - " if (R1[i]<0):\n", - " R1[i] = 0;\n", - " if (R2[i]<0):\n", - " R2[i] = 0;\n", - " if (R3[i]<0):\n", - " R3[i] = 0;\n", - "\n", - "print \"average depth of hourly rainfall excesscm/hr\";\n", - "a1 = zeros(12)\n", - "a2 = zeros(12)\n", - "a3 = zeros(12)\n", - "T = zeros(12)\n", - "for i in range(12):\n", - " a1[i] = R1[i]*A1/A; \t\t\t\t#average rainfall excess\n", - " a2[i] = R2[i]*A2/A;\n", - " a3[i] = R3[i]*A3/A;\n", - " T[i] = a1[i]+a2[i]+a3[i]; \t\t\t\t#total hourly rainfall excess\n", - " T[i] = round(T[i]*100)/100;\n", - " print \"%.2f\"%(T[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "average depth of hourly rainfall excesscm/hr\n", - "0.50\n", - "0.75\n", - "4.57\n", - "1.40\n", - "7.57\n", - "2.40\n", - "4.07\n", - "6.97\n", - "3.57\n", - "1.40\n", - "0.20\n", - "0.20\n" - ] - } - ], - "prompt_number": 69 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.26 pg : 185" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#derive the unit hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "A = 92.; \t\t\t\t#area of drainage bamath.sin\n", - "t = array([6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 2, 4, 6, 8, 10, 12, 14, 16],dtype=float64); \t\t\t\t#time\n", - "r = array([10.6, 9.7, 107.8, 175.6, 193.9, 150.3, 126.2, 106.9, 90, 72.8, 58.2, 48, 36.2, 28.4, 20.2, 14, 10.2, 10.4]); \t\t\t\t#total run-off\n", - "B = array([10.6, 9.7, 9.73, 9.77, 9.8, 9.83, 9.87, 9.9, 9.93, 9.97, 10, 10.03, 10.07, 10.10, 10.13, 10.16, 10.20, 10.40]); \t\t\t\t#base flow\n", - "s = 0;\n", - "\n", - "# Calculations and Results\n", - "d = r - B\n", - "s = sum(d)\n", - "\n", - "n = 0.36*s*2/A;\n", - "print \"ordinates of unit hydrograph:\";\n", - "u = zeros(18)\n", - "for i in range(18):\n", - " u[i] = d[i]/n; \t\t\t\t#ordinates of unit hydrograph\n", - " u[i] = round(u[i]*100)/100;\n", - " print \"%.2f\"%(u[i]);\n", - "\n", - "print \"Hydograph is 4-hr unit hydrograph\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of unit hydrograph:\n", - "0.00\n", - "0.00\n", - "11.50\n", - "19.45\n", - "21.60\n", - "16.48\n", - "13.65\n", - "11.38\n", - "9.39\n", - "7.37\n", - "5.65\n", - "4.45\n", - "3.07\n", - "2.15\n", - "1.18\n", - "0.45\n", - "0.00\n", - "0.00\n", - "Hydograph is 4-hr unit hydrograph\n" - ] - } - ], - "prompt_number": 73 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.27 pg : 186" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "A = 316.; \t\t\t\t#drainage area\n", - "B = 17.; \t\t\t\t#base flow\n", - "t = 6.;\n", - "O = [17.0, 113.2, 254.5, 198.0, 150.0, 113.2, 87.7, 67.9, 53.8, 42.5, 31.1, 22.6, 17.0]; \t\t\t\t#ordinates of storm hydrograph\n", - "Or = zeros(13)\n", - "Oh = zeros(13)\n", - "for i in range(13): \n", - " Or[i] = O[i]-B; \t\t\t\t#ordinates of direct run-off\n", - " Oh[i] = Or[i]/6.477; \t\t\t\t#ordinates of unit hydrograph\n", - "\n", - "\n", - "s = sum(Or);\n", - "re = s*60*60*t/(A*10000);\n", - "re = round(re*1000)/1000;\n", - "\n", - "# Results\n", - "print \"rainfall excess = %.2f cm.\"%(re);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "rainfall excess = 6.48 cm.\n" - ] - } - ], - "prompt_number": 74 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.28 pg : 188" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "fi = 2.5; \t\t\t\t#infiltration index\n", - "B = 10; \t\t\t\t#Base flow\n", - "O = [0, 110, 365, 500, 390, 310, 250, 235, 175, 130, 95, 65, 40, 22, 10, 0, 0, 0]; \t\t\t\t#ordinates of unit hydrograph\n", - "R1 = 2;R2 = 6.75;R3 = 3.75;\n", - "r1 = (R1*10-(fi*3)-5)/10; \t\t\t\t#rainfall excess in first three hour\n", - "r2 = (R2*10-(fi*3))/10; \t\t\t\t#rainfall excess in second three hour\n", - "r3 = (R3*10-(fi*3))/10; \t\t\t\t#rainfall excess in third three hour\n", - "\n", - "s1 = zeros(18)\n", - "for i in range(18):\n", - " s1[i] = r1*O[i]; \n", - "s2 = zeros(18)\n", - "for i in range(1,18):\n", - " s2[i] = r2*O[i-1];\n", - "s3 = zeros(18)\n", - "for i in range(2,18):\n", - " s3[i] = r3*O[i-2];\n", - " \t\t\t\t#surface run-off from rainfall excess during succesive unit periods\n", - "print \"ordinates of storm hydrograph\";\n", - "T = zeros(18)\n", - "t = zeros(18)\n", - "for i in range(18):\n", - " T[i] = s1[i]+s2[i]+s3[i];\n", - " t[i] = T[i]+B;\n", - " t[i] = round(t[i]*10)/10;\n", - " print \"%.2f\"%(t[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of storm hydrograph\n", - "10.00\n", - "92.50\n", - "943.80\n", - "2905.00\n", - "4397.50\n", - "4082.50\n", - "3227.50\n", - "2616.30\n", - "2301.30\n", - "1862.50\n", - "1386.30\n", - "1018.80\n", - "715.00\n", - "461.50\n", - "269.50\n", - "136.00\n", - "40.00\n", - "10.00\n" - ] - } - ], - "prompt_number": 75 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.29 pg : 189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "\n", - "#derive and plot 6 hr unit hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "A = 103.4; \t\t\t\t#area of bamath.sin\n", - "t = linspace(0,36,9); \t\t\t\t#time\n", - "q = [0, 21, 80, 82, 189, 123, 184, 87, 55.5, 25.25, 9, 6, 0]; \t\t\t\t#flow\n", - "print \"ordinates of unit hydrograph are:\";\n", - "u = zeros(9)\n", - "u[0] = 0;\n", - "u[1] = q[1]/2.;\n", - "u[2] = (q[2]-4*u[0])/2;\n", - "u[3] = (q[3]-4*u[1])/2;\n", - "for i in range(4,9):\n", - " u[i] = (q[i]-3*u[i-4]-4*u[i-2])/2; \t\t\t\t#ordinates of unit hydrograph\n", - "\n", - "for i in range(9):\n", - " print \"%.2f\"%(u[i]);\n", - "print \"The succesive unit hydrograph will have same ordinates but will be shiftedlaterally by 6 hrs.\";\n", - "#graph is plotted between u and t.\n", - "plot(t,u)\n", - "xlabel(\"Time in hours\")\n", - "ylabel(\"Discharge\")" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "ordinates of unit hydrograph are:\n", - "0.00\n", - "10.50\n", - "40.00\n", - "20.00\n", - "14.50\n", - "5.75\n", - "3.00\n", - "2.00\n", - "0.00\n", - "The succesive unit hydrograph will have same ordinates but will be shiftedlaterally by 6 hrs.\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 78, - "text": [ - "<matplotlib.text.Text at 0x1111d6590>" - ] - }, - { 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- "text": [ - "<matplotlib.figure.Figure at 0x1110d1a50>" - ] - } - ], - "prompt_number": 78 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.30 pg : 190" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "\n", - "\n", - "#derive ordinates of 6 hrs unit hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "R = [0, 1, 2.7, 5, 8, 9.8, 9, 7.5, 6.3, 5, 4, 2.9, 2.1, 1.3 ,0.5, 0, 0, 0, 0, 0]; \t\t\t\t#2hrs unit hydrograph\n", - "print \"ordinates of 6 hrs unit hydrograph\";\n", - "O1 = zeros(20)\n", - "for i in range(18):\n", - " O1[i+2] = R[i];\n", - "O2 = zeros(20)\n", - "for i in range(16):\n", - " O2[i+4] = R[i];\n", - "S = zeros(20)\n", - "f = zeros(20)#offset unit hydrograph\n", - "for i in range(20):\n", - " S[i] = O1[i]+O2[i]+R[i]; \t\t\t\t#sum\n", - " f[i] = S[i]/3; \t\t\t\t#ordinates of 6 hrs unit hydrograph\n", - " f[i] = round(f[i]*10)/10;\n", - " print \"%.2f\"%(f[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of 6 hrs unit hydrograph\n", - "0.00\n", - "0.30\n", - "0.90\n", - "2.00\n", - "3.60\n", - "5.30\n", - "6.60\n", - "7.40\n", - "7.80\n", - "7.40\n", - "6.40\n", - "5.10\n", - "4.10\n", - "3.10\n", - "2.20\n", - "1.40\n", - "0.90\n", - "0.40\n", - "0.20\n", - "0.00\n" - ] - } - ], - "prompt_number": 79 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.31 pg : 190" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import *\n", - "\t\t\t\t\n", - "#Given\n", - "t = linspace(0,45,16) \t\t\t\t#time\n", - "O = [0 ,9, 20, 35, 49, 43, 35, 28, 22, 17, 12, 9, 6, 3, 0, 0]; \t\t\t\t#ordinate of 2 hr unit hydrograph\n", - "\n", - "# Calculations and Results\n", - "of = zeros(16)\n", - "for i in range(2,16):\n", - " of[i] = O[i-2]+of[i-2]; \t\t\t\t#offset ordinate\n", - "\n", - "s = zeros(16)\n", - "for i in range(16):\n", - " s[i] = O[i]+of[i]; \t\t\t\t#ordinate of s-curve\n", - "\n", - "of1 = zeros(16)\n", - "for i in range(3,16):\n", - " of1[i] = s[i-3]; \t\t\t\t#offset of s-curve\n", - "\n", - "print \"ordinates of 9 hrs unit hydrograph:\";\n", - "y = zeros(16)\n", - "u = zeros(16)\n", - "for i in range(16):\n", - " y[i] = s[i]-of1[i];\n", - " u[i] = 2*y[i]/3; \t\t\t\t#ordinate of 9 hrs unit hydrograph\n", - " u[i] = round(u[i]*10)/10;\n", - " print \"%.2f\"%(u[i]);\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of 9 hrs unit hydrograph:\n", - "0.00\n", - "6.00\n", - "13.30\n", - "29.30\n", - "40.00\n", - "44.70\n", - "40.00\n", - "30.70\n", - "26.00\n", - "18.70\n", - "15.30\n", - "10.00\n", - "8.00\n", - "4.00\n", - "2.00\n", - "0.00\n" - ] - } - ], - "prompt_number": 82 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.32 pg : 210" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros,linspace\n", - "\n", - "\n", - "#california method\n", - "#Hazens method\n", - "#gumbels method\n", - "\t\t\t\t\n", - "#Given\n", - "q = [9200, 7800, 6600, 5800, 5260, 4980, 4525, 3810, 3630, 3250, 3110, 3090, 2380, 2390, 1723]; \t\t\t\t#Discharge arranged in decreamath.sing order\n", - "N = 15;\n", - "C = 0.3;\n", - "m = linspace(1,15,15)\n", - "C = [0.3, 0.44, 0.52, 0.57, 0.61, 0.66, 0.7, 0.74, 0.78, 0.82, 0.86, 0.88, 0.94, 0.96, 1]; \t\t\t\t#from table 4.25\n", - "print \"California Hazen Gumbel\";\n", - "Ca = zeros(15)\n", - "H = zeros(15)\n", - "G = zeros(15)\n", - "Ca = zeros(15)\n", - "G = zeros(15)\n", - "\n", - "for i in range(15):\n", - " Ca[i] = N/m[i];\n", - " H[i] = 2*N/(2*m[i]-1);\n", - " G[i] = N/(m[i]+C[i]-1);\n", - " Ca[i] = round(Ca[i]*100)/100;\n", - " G[i] = round(G[i]*100)/100;\n", - " H[i] = round(H[i]*100)/100;\n", - " print \"%.2f %.2f %.2f\"%(Ca[i],H[i],G[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "California Hazen Gumbel\n", - "15.00 30.00 50.00\n", - "7.50 10.00 10.42\n", - "5.00 6.00 5.95\n", - "3.75 4.29 4.20\n", - "3.00 3.33 3.25\n", - "2.50 2.73 2.65\n", - "2.14 2.31 2.24\n", - "1.88 2.00 1.94\n", - "1.67 1.76 1.71\n", - "1.50 1.58 1.53\n", - "1.36 1.43 1.38\n", - "1.25 1.30 1.26\n", - "1.15 1.20 1.16\n", - "1.07 1.11 1.07\n", - "1.00 1.03 1.00\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.33 pg : 211" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "T1 = 40.;\n", - "T2 = 80.; \t\t\t\t#Return period\n", - "F1 = 27000.;\n", - "F2 = 31000.; \t\t\t\t#Peak flood\n", - "\n", - "# Calculations\n", - "y80 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));\n", - "y40 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));\n", - "y = (F2-F1)/(y80-y40);\n", - "T = 240.;\n", - "y240 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "x240 = F2+(y240-y80)*y;\n", - "\n", - "# Results\n", - "print \"flood magnitude with return period of 240 years = %i cumec.\"%(x240);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "flood magnitude with return period of 240 years = 37306 cumec.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.34 pg : 212" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "N = 40;\n", - "Sn = 1.1413;\n", - "yn = 0.5436; \t\t\t\t#from table 4.21 (a) and(b)\n", - "q = [1330, 1095, 1030, 980, 975, 950, 945, 940, 925, 855, 853, 840, 835, 825, 810, 795, 756, 710, 708, 705, 700, 670, 625, 620, 610, 605, 595, 585, 570, 550, 530, 505, 500, 495, 485, 465, 460, 420, 390, 380]; \t\t\t\t#discharge\n", - "s = sum(q)\n", - "xavg = s/N;\n", - "w = 0;\n", - "\n", - "# Calculations\n", - "t = zeros(40)\n", - "for i in range(40):\n", - " t[i] = (q[i]-xavg)**2;\n", - " w = w+t[i];\n", - "\n", - "sigma = (w/(N-1))**0.5;\n", - "N = 10.;\n", - "y10 = -(2.303*math.log10(2.303*math.log10(N/(N-1))));\n", - "K10 = (y10-yn)/Sn;\n", - "x10 = xavg+K10*sigma;\n", - "N = 20.;\n", - "y20 = -(2.303*math.log10(2.303*math.log10(N/(N-1))));\n", - "K20 = (y20-yn)/Sn;\n", - "x20 = xavg+K20*sigma;\n", - "N = 5.;\n", - "y5 = -(2.303*math.log10(2.303*math.log10(N/(N-1))));\n", - "K5 = (y5-yn)/Sn;\n", - "x5 = xavg+K5*sigma;\n", - "\n", - "T = 100.;\n", - "y100 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "K100 = (y100-yn)/Sn;\n", - "x100 = xavg+K100*sigma;\n", - "\n", - "T = 200.;\n", - "y200 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "K200 = (y200-yn)/Sn;\n", - "x200 = xavg+K200*sigma;\n", - "x100 = round(x100);\n", - "\n", - "# Results\n", - "print \"For T = 100 years:flood discharge = %.2f cumecs.\\\n", - "\\nFor T = 200 years:flood discharge = %.f cumecs.\"%(x100,x200);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "For T = 100 years:flood discharge = 1487.00 cumecs.\n", - "For T = 200 years:flood discharge = 1620 cumecs.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.35 pg : 214" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy.linalg import solve\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "sigma = 1.1413; \t\t\t\t#smath.radians(numpy.arcmath.tan(ard deviation\n", - "yn = 0.5436;\n", - "T = 50.;\n", - "\n", - "# Calculations\n", - "y50 = -2.303*math.log10(2.303*math.log10(T/(T-1)));\n", - "K50 = (y50-yn)/sigma;\n", - "T = 100.;\n", - "y100 = -2.303*math.log10(2.303*math.log10(T/(T-1)));\n", - "K100 = (y100-yn)/sigma;\n", - "x50 = 878; x100 = 970; \t\t\t\t\n", - "#Given peak flood\n", - "A = [[K50, 1],[K100, 1]];\n", - "B = [x50,x100];\n", - "C = solve(A,B)#A\\B;\n", - "xavg = C[1];\n", - "sigmad = C[0];\n", - "T = 200.;\n", - "y200 = -2.303*math.log10(2.303*math.log10(T/(T-1)));\n", - "K200 = (y200-yn)/sigma;\n", - "x200 = xavg+K200*sigmad;\n", - "x200 = round(x200);\n", - "\n", - "# Results\n", - "print \"200 year flood for stream = %.2f cumecs.\"%(x200);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "200 year flood for stream = 1062.00 cumecs.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.36 pg : 214" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#risk of failure of cofferdam\n", - "#return period\n", - "\n", - "#Given\n", - "T = 30.; \t\t\t\t#deign for period\n", - "n = 6.; \t\t\t\t#period of construction\n", - "\n", - "# Calculations\n", - "R = (1-(1-(1/T))**n)*100;\n", - "R1 = 0.1; \t\t\t\t#reduced risk\n", - "T1 = 1./(1-(1-R1)**(1./6));\n", - "R = round(R*10)/10;\n", - "T1 = round(T1*100)/100;\n", - "\n", - "# Results\n", - "print \"risk of failure of cofferdam = %.2f percent.\"%(R);\n", - "print \"return period = %.2f years.\"%(T1);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "risk of failure of cofferdam = 18.40 percent.\n", - "return period = 57.45 years.\n" - ] - } - ], - "prompt_number": 85 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.37 pg : 215" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#probability of excedence\n", - "#probability of flood magnitude occuring at:\n", - "#at least once in 10 years\n", - "#two times in 10 succesive years\n", - "#once in 10 succesive years\n", - "\n", - "#Given\n", - "T = 40.; \t\t\t\t#return period\n", - "P = 1./T;\n", - "n = 10;\n", - "Rsk = 1.-(1-P)**n;\n", - "s = 1.;\n", - "t = 1.;\n", - "for i in range(1,n+1): \n", - " s = s*i;\n", - "\n", - "for i in range(1,n-1):\n", - " t = t*i;\n", - "\n", - "P2n = s*P**2*(1-P)**8/(t*2);\n", - "P1n = n*P*(1-P)**(n-1);\n", - "Rsk = round(Rsk*1000)/1000;\n", - "P2n = round(P2n*10000)/10000;\n", - "P1n = round(P1n*1000)/1000;\n", - "\n", - "# Results\n", - "print \"probability of excedence = %.2f.\"%(P);\n", - "print \"probability of flood magnitude occuring at least once in 10 years = %.2f\"%(Rsk);\n", - "print \"probability of flood magnitude occuring at two times in 10 succesive years = %.2f\"%(P2n);\n", - "print \"probability of flood magnitude occuring at once in 10 succesive years = %.2f\"%(P1n);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "probability of excedence = 0.03.\n", - "probability of flood magnitude occuring at least once in 10 years = 0.22\n", - "probability of flood magnitude occuring at two times in 10 succesive years = 0.02\n", - "probability of flood magnitude occuring at once in 10 succesive years = 0.20\n" - ] - } - ], - "prompt_number": 87 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.38 pg : 215" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "C1 = 0.22;C2 = 0.12;C3 = 0.32; \t\t\t\t#run-off coefficient\n", - "A1 = 3.2;A2 = 4.8;A3 = 1.8; \n", - "L = 2.4; \t\t\t\t#length of water course\n", - "H = 30; \t\t\t\t#fall\n", - "T = 30; \t\t\t\t#frequency\n", - "\n", - "# Calculations\n", - "t = 60*0.000323*(L*1000)**0.77*(H/(L*1000))**(-0.385);\n", - "i = 78*T**0.22/(t+12)**0.45;\n", - "q = 2.778*i*(C1*A1+C2*A2+C3*A3);\n", - "q = round(q*10)/10;\n", - "\n", - "# Results\n", - "print \"peak rate of run off = %.2f cumecs.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "peak rate of run off = 141.20 cumecs.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.39 pg : 216" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "T = 30; \t\t\t\t#return period\n", - "A = 2.4; \t\t\t\t#area of watershed\n", - "s = 1./200; \t\t\t\t#slope oof catchment\n", - "L = 1.8; \t\t\t\t#length of travel of water\n", - "C = 0.25; \t\t\t\t#average run-off coefficient\n", - "r = [2.5, 3.8, 4.8, 5.9, 6.7, 7.4, 8.4, 8.7, 9.2]; \t\t\t\t#rmath.sinfall depth\n", - "\n", - "# Calculations\n", - "t = 60*0.000323*(L*1000)**0.77*(s)**(-0.385); \n", - "rmax = r[6]+(r[7]-r[6])*7.84/10;\n", - "i = rmax*60/t;\n", - "q = 2.778*C*A*i;\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"peak flow rate = %.2f cumecs.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "peak flow rate = 18.05 cumecs.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.40 pg : 216" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "pA = 75.; \t\t\t\t#precipitation at A\n", - "pB = 58; \t\t\t\t#precpitation at B\n", - "pC = 47; \t\t\t\t#precipitation at C\n", - "nA = 826; \t\t\t\t#normal precipitation at A\n", - "nB = 618; \t\t\t\t#normal precipitation at B\n", - "nC = 482; \t\t\t\t#normal precipitation at C\n", - "nX = 757; \t\t\t\t#normal precipitation at X\n", - "\n", - "\n", - "# Calculations\n", - "pX = (nX*pA/nA+nX*pB/nB+nX*pC/nC)/3.;\n", - "pX = round(pX*10)/10;\n", - "\n", - "# Results\n", - "print \"precipitation at x = %.2f cm.\"%(pX);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "precipitation at x = 70.90 cm.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.41 pg : 216" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "p = [41, 51, 32, 55, 50, 68]; \t\t\t\t#rain guage readings at respective stations\n", - "s = sum(p)\n", - "pavg = s/6;\n", - "u = 0;\n", - "for i in range(5):\n", - " u = u+(p[i]-pavg)**2;\n", - "\n", - "# Calculations\n", - "sx = (u/5)**0.5;\n", - "Cv = sx*100/pavg;\n", - "N = (Cv/8)**2;\n", - "N = round(N*100)/100;\n", - "\n", - "# Results\n", - "print \"mean rainfall = %.2f cm.\"%(pavg);\n", - "print \"total stations needed = %.2f.\"%(N);\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "mean rainfall = 49.00 cm.\n", - "total stations needed = 5.08.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.42 pg : 217" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "a = 4; \t\t\t\t#dimension of plot sides\n", - "P1 = 4.8;P2 = 13;P3 = 8;P4 = 5.4;P5 = 3.2;P6 = 9.4; \t\t\t\t#precipitaion at respective stations\n", - "\n", - "# Calculations\n", - "A1 = a**2/8+a**2/(4*1.73);\n", - "A2 = a**2/8;\n", - "A3 = A2;A4 = A1;\n", - "A5 = a**2/(4*1.73);\n", - "A6 = a**2/2;\n", - "A = A1+A2+A3+A4+A5+A6;\n", - "Pavg = (P1*A1+P2*A2+P3*A3+P4*A4+P5*A5+P6*A6)/A;\n", - "\n", - "# Results\n", - "print \"Mean precipitaion = %.2f cm.\"%(Pavg);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Mean precipitaion = 7.35 cm.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.43 pg : 218" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace\n", - "\t\t\t\t\n", - "#Given\n", - "A = [90, 140, 125, 140, 85, 40, 20]; \t\t\t\t#area of isohytes\n", - "I = linspace(13,1,7) \t\t\t\t#average isohytel interval\n", - "s = 0;t = 0;\n", - "for i in range(7):\n", - " s = s+A[i]*I[i];\n", - " t = t+A[i];\n", - "\n", - "Pavg = s/t;\n", - "Pavg = round(Pavg*10)/10;\n", - "\n", - "# Results\n", - "print \" average depth of precipitation = %.2f cm.\"%(Pavg);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " average depth of precipitation = 8.40 cm.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.44 pg : 218" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "p = [120, 95, 96, 60, 65, 70, 45, 21]; \t\t\t\t#rain guage readings at respective stations\n", - "\n", - "# Calculations and Results\n", - "s = sum(p)\n", - "pavg = s/8;\n", - "u = 0;\n", - "for i in range(8):\n", - " u = u+(p[i]-pavg)**2;\n", - "\n", - "sx = (u/7)**0.5;\n", - "Cv = sx*100/pavg;\n", - "N = (Cv/13.99)**2;\n", - "N = round(N*100)/100;\n", - "print \"mean rainfall = %.2f cm.\"%(pavg);\n", - "print \"total stations needed = %.2f.\"%(N);\n", - "\t\t\t\t#taking N = 10\n", - "N = 10;\n", - "n = N-8;\n", - "print \"additional guages needed = %i.\"%(n);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "mean rainfall = 71.00 cm.\n", - "total stations needed = 10.03.\n", - "additional guages needed = 2.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.45 pg : 219" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%matplotlib inline\n", - "from matplotlib.pylab import plot\n", - "from numpy import zeros,linspace\n", - "import math \n", - "#compute maximum rainfall intensities for 5,10,15,20,25,30,35,40,45,50 minutes\n", - "#plot intensity duration graph\n", - "\t\t\t\t\n", - "#Given\n", - "CR = [0, 1.02, 2.08, 3.30, 4.72, 5.58, 6.40, 7.16, 7.88, 8.54, 9.14]; \t\t\t\t#cumulative rainfall\n", - "\n", - "c5 = zeros(11)\n", - "c10 = zeros(11)\n", - "c15 = zeros(11)\n", - "c20 = zeros(11)\n", - "c25 = zeros(11)\n", - "c30 = zeros(11)\n", - "c35 = zeros(11)\n", - "c40 = zeros(11)\n", - "c45 = zeros(11)\n", - "c50 = zeros(11)\n", - "\n", - "c5[1] = CR[1];\n", - "c10[2] = CR[2];\n", - "c15[3] = CR[3];\n", - "c20[4] = CR[4];\n", - "c25[5] = CR[5];\n", - "c30[6] = CR[6];\n", - "c35[7] = CR[7];\n", - "c40[8] = CR[8];\n", - "c45[9] = CR[9];\n", - "c50[10] = CR[10];\n", - "for i in range(2,11):\n", - " c5[i] = CR[i]-CR[i-1];\n", - "\n", - "for i in range(3,11):\n", - " c10[i] = CR[i]-CR[i-2];\n", - "\n", - "for i in range(4,11):\n", - " c15[i] = CR[i]-CR[i-3];\n", - "\n", - "for i in range(5,11):\n", - " c20[i] = CR[i]-CR[i-4];\n", - "\n", - "for i in range(6,11):\n", - " c25[i] = CR[i]-CR[i-5];\n", - "\n", - "for i in range(7,11):\n", - " c30[i] = CR[i]-CR[i-6];\n", - "\n", - "for i in range(8,11):\n", - " c35[i] = CR[i]-CR[i-7];\n", - "\n", - "for i in range(9,11):\n", - " c40[i] = CR[i]-CR[i-8];\n", - "\n", - "for i in range(10,11):\n", - " c45[i] = CR[i]-CR[i-9];\n", - " \t\t\t\t#rainfall in any possible time interval\n", - "\n", - "print \"5min 10min 15min 20min 25min 30min 35min 40min 45min 50min\";\n", - "for i in range(11):\n", - " print \"%4.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f\"%(c5[i],c10[i],c15[i],c20[i],c25[i],c30[i],c35[i],c40[i],c45[i],c50[i]);\n", - "\n", - "I = [17.04, 15.84, 14.80, 14.16, 13.39, 12.80, 12.27, 11.82, 11.39, 10.97]; \t\t\t\t#maximum intensity at respective durations\n", - "D = linspace(5,50,len(I)) \t\t\t\t#durations\n", - "#graph is plotted between I and D\n", - "plot(I,D)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "5min 10min 15min 20min 25min 30min 35min 40min 45min 50min\n", - "0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00\n", - "1.02 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00\n", - "1.06 2.08 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00\n", - "1.22 2.28 3.30 0.00 0.00 0.00 0.00 0.00 0.00 0.00\n", - "1.42 2.64 3.70 4.72 0.00 0.00 0.00 0.00 0.00 0.00\n", - "0.86 2.28 3.50 4.56 5.58 0.00 0.00 0.00 0.00 0.00\n", - "0.82 1.68 3.10 4.32 5.38 6.40 0.00 0.00 0.00 0.00\n", - "0.76 1.58 2.44 3.86 5.08 6.14 7.16 0.00 0.00 0.00\n", - "0.72 1.48 2.30 3.16 4.58 5.80 6.86 7.88 0.00 0.00\n", - "0.66 1.38 2.14 2.96 3.82 5.24 6.46 7.52 8.54 0.00\n", - "0.60 1.26 1.98 2.74 3.56 4.42 5.84 7.06 8.12 9.14\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 17, - "text": [ - "[<matplotlib.lines.Line2D at 0x10e861ed0>]" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10e071190>" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.46 pg : 221" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%matplotlib inline\n", - "import math\n", - "from matplotlib.pylab import plot\n", - "#draw storm hyetograph and intensity duration curve\n", - "\t\t\t\t\n", - "#Given\n", - "p = [0, 5, 7.5, 8.5, 9]; \t\t\t\t#accumulated precipitation\n", - "t = [0, 30, 60, 90, 120]; \t\t\t\t#time\n", - "r = zeros(5)\n", - "I = zeros(5)\n", - "\n", - "print \"Rainfall intensity:\";\n", - "for i in range(1,5):\n", - " r[i] = p[i]-p[i-1]; \t\t\t\t#rainfall in succesive 30 min interval\n", - " I[i] = r[i]*60/30; \t\t\t\t#rainfall intensity\n", - " print \"%.2f\"%(I[i]);\n", - "\n", - "#graph is plotted between I and t.\n", - "plot(I,t)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Rainfall intensity:\n", - "10.00\n", - "5.00\n", - "2.00\n", - "1.00\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 18, - "text": [ - "[<matplotlib.lines.Line2D at 0x10e8ae250>]" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10e071390>" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.47 pg : 222" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%matplotlib inline\n", - "import math\n", - "from numpy import zeros,linspace\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "I = linspace(21,12,10) \t\t\t\t#isohytes\n", - "a = [543, 1345, 2030, 2545, 2955, 3280, 3535, 3710, 3880, 3915]; \t\t\t\t#enclosed area\n", - "ia = zeros(10)\n", - "ia[0] = 543;\n", - "for i in range(1,10):\n", - " ia[i] = a[i]-a[i-1]; \t\t\t\t#net incremental area between isohytes\n", - "\n", - "rv = zeros(10)\n", - "r = linspace(21.5,12.5,10) \n", - "for i in range(10):\n", - " rv[i] = r[i]*ia[i]; \t\t\t\t#rainfall volume\n", - "\n", - "cv = zeros(10)\n", - "cv[0] = 11675;\n", - "for i in range(10):\n", - " cv[i] = cv[i-1]+rv[i]; \t\t\t\t#cumulative volume\n", - "\n", - "eud = zeros(10)\n", - "for i in range(10):\n", - " eud[i] = cv[i]/a[i]; \t\t\t\t#depth(mm)\n", - "\n", - "\n", - "print \"From depth area curve we obtain average depth of precipitation = 20.15 mm for anarea of 2400 sq. km.\";\n", - "#graph is plotted between eud and a\n", - "plot(eud,a)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "From depth area curve we obtain average depth of precipitation = 20.15 mm for anarea of 2400 sq. km.\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 20, - "text": [ - "[<matplotlib.lines.Line2D at 0x10e99fb90>]" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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DZDtlNYCVwB5gBVA94PgxwF4gF+hZxt8WEYk7tWvDypU2YzktzYaihkNZK4NC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\n2EwAAAAASUVORK5CYII=\n", - "text": [ - "<matplotlib.figure.Figure at 0x10c55a5d0>" - ] - } - ], - "prompt_number": 20 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.48 pg : 222" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "h1 = 7.75; \t\t\t\t#initial depth of water\n", - "r = 3.80; \t\t\t\t#rainfall during the week\n", - "hr = 2.50; \t\t\t\t#depth of water removed\n", - "C = 0.7; \t\t\t\t#pan coefficient\n", - "\n", - "# Calculations\n", - "ha = r-hr;\n", - "hl = ha+h1;\n", - "h2 = 8.32;\n", - "ev = hl-h2;\n", - "evs = ev*C;\n", - "evs = round(evs*100)/100;\n", - "\n", - "# Results\n", - "print \"evaporation from reservior surface during the week = %.2f cm.\"%(evs);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "evaporation from reservior surface during the week = 0.51 cm.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.49 pg : 223" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "T = linspace(1,12,12) \t\t\t\t#time from start\n", - "r = [1.8, 2.6, 7.8, 3.9, 10.6, 5.4, 7.8, 9.2, 6.5, 4.4, 1.8, 1.6]; \t\t\t\t#increamental rainfall\n", - "R = 24.4; \t\t\t\t#total run-off\n", - "s = sum(r)\n", - "\n", - "ti = s-R;\n", - "\n", - "#first trial\n", - "tr = 7; \t\t\t\t#assumed\n", - "ti = s-R-r[0]-r[1]-r[3]-r[10]-r[11];\n", - "fi = ti/tr;\n", - "P = zeros(12)\n", - "for i in range(12):\n", - " P[i] = r[i]-fi;\n", - " if (P[i]<0):\n", - " P[i] = 0;\n", - "\n", - "print \"Timeh rainfall excess.\";\n", - "for i in range(12):\n", - " print \"%.2f %.2f\"%(T[i],P[i]);\n", - "\n", - "print \"fi index = %.2f cm/hr.\"%(fi);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Timeh rainfall excess.\n", - "1.00 0.00\n", - "2.00 0.00\n", - "3.00 3.90\n", - "4.00 0.00\n", - "5.00 6.70\n", - "6.00 1.50\n", - "7.00 3.90\n", - "8.00 5.30\n", - "9.00 2.60\n", - "10.00 0.50\n", - "11.00 0.00\n", - "12.00 0.00\n", - "fi index = 3.90 cm/hr.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.50 pg : 224" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace\n", - "\t\t\t\t\n", - "#Given\n", - "r = [0.6, 1.35, 2.25, 3.45, 2.7, 2.4, 1.5, 0.75]; \t\t\t\t#incremental rainfall\n", - "T = linspace(1,8,8) \t\t\t\t#time from start of rainfal\n", - "t = 8.;\n", - "P = 15.; \t\t\t\t#total rainfall\n", - "R = 8.7; \t\t\t\t#direct run-off\n", - "\n", - "# Calculations\n", - "W = (P-R)/t;\n", - "#math.since fi wil be more than W\n", - "tre = 6;\n", - "fi = ((P-R)-r[0]-r[7])/tre;\n", - "\n", - "# Results\n", - "print \"fi index = %.2f cm/hr.\"%(fi);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "fi index = 0.83 cm/hr.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.51 pg : 224" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from scipy.integrate import quad \n", - "\t\t\t\t\n", - "#Given\n", - "I = 10; \t\t\t\t#total infiltration rate\n", - "fI = 5; \t\t\t\t#final infiltration rate\n", - "k = 0.95; \t\t\t\t#rate of decay of difference between final and initial infiltration rate\n", - "\n", - "\n", - "# Calculations\n", - "def f8(t): \n", - "\t return fI+(I-fI)*math.e**(-k*t)\n", - "\n", - "q = quad(f8,0,6)[0]\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"total infiltration depth = %.2f mm.\"%(q);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "total infiltration depth = 35.25 mm.\n" - ] - } - ], - "prompt_number": 25 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.52 pg : 224" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace\n", - "\n", - "#find the equation of infiltration capacity\n", - "\t\t\t\t\n", - "#Given\n", - "fc = 1; \t\t\t\t#consmath.tant infiltration rate\n", - "ft = [10.4, 5.6, 3.2, 2.1, 1.5, 1.2, 1.1, 1, 1]; \t\t\t\t#infiltration capacity\n", - "f = ft[0]-fc;\n", - "t = linspace(0,2,9)\n", - "\n", - "r = zeros(9)\n", - "for i in range(9):\n", - " r[i] = ft[i]-fc;\n", - "\n", - "h = zeros(7)\n", - "for i in range(7):\n", - " h[i] = math.log10(r[i]);\n", - "\n", - "s = 0.775; \t\t\t\t#from graph\n", - "k = 1/(math.log10(math.e)*s);\n", - "k = round(k*100)/100;\n", - "\n", - "# Results\n", - "print \"Equation is:ft = fc+%.2fe**-%.2ft)\"%(f,k);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Equation is:ft = fc+9.40e**-2.97t)\n" - ] - } - ], - "prompt_number": 26 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.53 pg : 226" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%matplotlib inline\n", - "from matplotlib.pylab import bar\n", - "import math \n", - "#total rainfall\n", - "#net run-off\n", - "#W index\n", - "\n", - "#Given\n", - "r = [2, 2, 8, 7, 1.25, 1.25, 4.5]; \t\t\t\t#rainfall intensity\n", - "T = [15, 30, 45, 60, 70, 90, 105]; \t\t\t\t#time\n", - "dt = 15.; \t\t\t\t#time interval\n", - "fi = 3.; \t\t\t\t#fi index\n", - "#graph is plotted between r and T\n", - "bar(T,r)\n", - "s = sum(r)\n", - "P = s*dt/60;\n", - "Pe = ((8-3)+(7-3)+(4.5-3))*dt/60; \t\t\t\t#area of graph above r = 3.0.\n", - "w = (P-Pe)/(105./60);\n", - "w = round(w*1000)/1000;\n", - "\n", - "# Results\n", - "print \"total rainfall = %.2f cm.\"%(P);\n", - "print \"net run-off = %.2f cm.\"%(Pe);\n", - "print \"W index = %.2f cm/hr.\"%(w);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "total rainfall = 6.50 cm.\n", - "net run-off = 2.62 cm.\n", - "W index = 2.21 cm/hr.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10ed31590>" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.54 pg : 226" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#run-off by rainfall of 3.3cm in 3hrs\n", - "\t\t\t\t\n", - "#Given\n", - "A = [36, 18, 66]; \t\t\t\t#area of catchment\n", - "fi = [0.9 ,1.1, 0.5]; \t\t\t\t#fi index\n", - "r1 = [0.6 ,0.9, 1.0]; \t\t\t\t#rainfall in first hour\n", - "r2 = [2.4 ,2.1, 2.0]; \t\t\t\t#rainfall in second hour\n", - "r3 = [1.3, 1.5, 0.9]; \t\t\t\t#rainfall in third hour\n", - "\n", - "# Calculations and Results\n", - "t36 = r1[0]+r2[0]+r3[0];\n", - "t18 = r1[1]+r2[1]+r3[1];\n", - "t66 = r1[2]+r2[2]+r3[2];\n", - "\n", - "p = (t36*A[0]+t18*A[1]+t66*A[2])/(A[0]+A[1]+A[2]);\n", - "print \"Total rainfall in catchment = %.2f cm.\"%(p);\n", - "\n", - "ro1 = [0 ,0, 0.5];\n", - "ro2 = [1.5 ,1.0, 1.5];\n", - "ro3 = [0.4, 0.4, 0.4]; \t\t\t\t#rainfall-fi\n", - "t1 = ro1[0]+ro2[0]+ro3[0];\n", - "t2 = ro1[1]+ro2[1]+ro3[1];\n", - "t3 = ro1[2]+ro2[2]+ro3[2];\n", - "run = (A[0]*t1+A[1]*t2+A[2]*t3)/(A[0]+A[1]+A[2]); \t\t\t\t#run-off from entire catchment\n", - "\n", - "print \"run-off by rainfall of 3.3cm in 3hrs = %.2f cm.\"%(run);\n", - "\n", - "fia = (fi[0]*A[0]+fi[1]*A[1]+fi[2]*A[2])/(A[0]+A[1]+A[2]);\n", - "tr = (1.1-fia)*3;\n", - "print \"Total run-off = %.2f cm.\"%(tr);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Total rainfall in catchment = 4.11 cm.\n", - "run-off by rainfall of 3.3cm in 3hrs = 2.10 cm.\n", - "Total run-off = 1.17 cm.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.55 pg : 227" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "P = array([4, 22, 28, 15, 12, 8, 4, 15, 10, 5]); \t\t\t\t#Precipitation\n", - "R = array([0.2, 7.1, 10.9, 4.0, 3.0, 1.3, 0.4, 4.1, 2.0, 0.3]); \t\t\t\t#runoff\n", - "\n", - "# Calculations\n", - "Ps = P**2\n", - "Rs = R**2\n", - "PR = P*R\n", - "\n", - "s = sum(Ps)\n", - "t = sum(Rs)\n", - "u = sum(PR)\n", - "q = sum(P)\n", - "w = sum(R)\n", - "\n", - "N = 10.;\n", - "a = (N*u-q*w)/(N*s-q**2);\n", - "b = (w-a*q)/N;\n", - "a = round(a*10000)/10000;\n", - "b = round(b*10000)/10000;\n", - "\n", - "# Results\n", - "print \"Equation is:%.2f P %.2f.\"%(a,b);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Equation is:0.43 P -1.93.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.56 pg : 228" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 470; \t\t\t\t#peak discharge of flood hydrograph\n", - "B = 15; \t\t\t\t#base flow\n", - "l = 0.25; \t\t\t\t#infiltration loss\n", - "Qr = Q-B;\n", - "d = 8; \t\t\t\t#average depth of rainfall\n", - "\n", - "# Calculations\n", - "re = d-l*6; \t\t\t\t#rainfall excess\n", - "q = Qr/re;\n", - "\n", - "# Results\n", - "print \"peak discharge of 6 hrs unit hydrograph = %i cumecs.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "peak discharge of 6 hrs unit hydrograph = 70 cumecs.\n" - ] - } - ], - "prompt_number": 117 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.57 pg : 228" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array,zeros\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "fi = 0.25; \t\t\t\t#infiltration index\n", - "B = 20; \t\t\t\t#Base flow\n", - "O = array([0, 20, 60, 150, 120, 90, 70, 50, 30, 20, 10, 0, 0, 0]); \t\t\t\t#ordinates of unit hydrograph\n", - "R1 = 5;\n", - "R2 = 0.8;\n", - "R3 = 3;\n", - "r1 = R1-(fi*4); \t\t\t\t#rainfall excess in first four hour\n", - "r2 = R2-(fi*4); \t\t\t\t#rainfall excess in second four hour\n", - "r3 = R3-(fi*4); \t\t\t\t#rainfall excess in third four hour\n", - "if r2<0 :\n", - " r2 = 0;\n", - "\n", - "# Calculations and Results\n", - "s1 = r1*O\n", - "s2 = zeros(14)\n", - "for i in range(1,14):\n", - " s2[i] = r2*O[i-1];\n", - "\n", - "s3 = zeros(14)\n", - "for i in range(2,14):\n", - " s3[i] = r3*O[i-2];\n", - "#surface run-off from rainfall excess during succesive unit periods\n", - "print \"ordinates of storm hydrograph\";\n", - "T = zeros(14)\n", - "t = zeros(14)\n", - "for i in range(14):\n", - " T[i] = s1[i]+s2[i]+s3[i]; \t\t\t\t#sub-total\n", - " t[i] = T[i]+B; \t\t\t\t#ordinate of flood hydrograph\n", - " print \"%i\"%(t[i]);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of storm hydrograph\n", - "20\n", - "100\n", - "260\n", - "660\n", - "620\n", - "680\n", - "540\n", - "400\n", - "280\n", - "200\n", - "120\n", - "60\n", - "40\n", - "20\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.58 pg : 229" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array\n", - "\t\t\t\t\n", - "#Given\n", - "fi = 2.5; \t\t\t\t#fi index\n", - "t = 24.;\n", - "A = 200.; \t\t\t\t#area of catchment\n", - "R1 = 7.5;\n", - "R2 = 2.0;\n", - "R3 = 5.; \t\t\t\t#rainfall\n", - "r1 = R1-fi;\n", - "r2 = R2-fi;\n", - "r3 = R3-fi;\n", - "r2 = 0;\n", - "r = [5, 0, 2.5]; \t\t\t\t#excess rainfall\n", - "D = array([5 ,15, 40, 25, 10, 5, 0, 0, 0]); \t\t\t\t#distribution\n", - "d1 = D*r[0]/100\n", - "d2 = zeros(9)\n", - "for i in range(8):\n", - " d2[i+1] = D[i]*r[1]/100;\n", - "\n", - "d3 = zeros(9)\n", - "for i in range(7):\n", - " d3[i+2] = D[i]*r[2]/100;\n", - "#distribution run-off for rainfall excess\n", - "\n", - "tr1 = zeros(9)\n", - "tr2 = zeros(9)\n", - "for i in range(9):\n", - " tr1[i] = d1[i]+d2[i]+d3[i]; \t\t\t\t#total run-off as depth\n", - " tr2[i] = 23.148*tr1[i]; \t\t\t\t#total run-off as discharge\n", - " tr2[i] = round(tr2[i]*1000)/1000;\n", - "\n", - "s = sum(tr2)\n", - "\n", - "print \"Total run-off:\";\n", - "print \"as depth as discharge\";\n", - "for i in range(9):\n", - " print \"%.2f %.2f\"%(tr1[i],tr2[i]);\n", - "\n", - "r = 0.36*s*t/A;\n", - "r = round(r*10)/10;\n", - "print \"total run-off = %.2f cm.\"%(r);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Total run-off:\n", - "as depth as discharge\n", - "0.00 0.00\n", - "0.00 0.00\n", - "2.12 49.19\n", - "1.38 31.83\n", - "1.00 23.15\n", - "0.62 14.47\n", - "0.25 5.79\n", - "0.12 2.89\n", - "0.00 0.00\n", - "total run-off = 5.50 cm.\n" - ] - } - ], - "prompt_number": 35 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.59 pg : 230" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "O = [10, 30, 90, 220, 280, 220, 166, 126, 92, 62, 40, 20, 10]; \t\t\t\t#ordinates of 6 hr flood hydrograph\n", - "B = 10; \t\t\t\t#Base flow\n", - "r = zeros(13)\n", - "\n", - "for i in range(13):\n", - " r[i] = O[i]-B; \t\t\t\t#ordinates of direct run-off\n", - "\n", - "print \"Ordinates of 6 hr unit hydrograph\";\n", - "u = zeros(13)\n", - "\n", - "for i in range(1,13):\n", - " u[i] = r[i]-u[i-1]; \t\t\t\t#ordinates of 6 hrs unit hydrograph\n", - "\n", - "for i in range(13): \n", - " print \"%i\"%(u[i]);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Ordinates of 6 hr unit hydrograph\n", - "0\n", - "20\n", - "60\n", - "150\n", - "120\n", - "90\n", - "66\n", - "50\n", - "32\n", - "20\n", - "10\n", - "0\n", - "0\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.60 pg : 231" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "#determine the ordinates of 1 cm-6 hour hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "t = 6;\n", - "A = 450; \t\t\t\t#catchment area\n", - "O = [5, 15, 40, 80, 60, 50, 25, 15, 5]; \t\t\t\t#ordinates of flood hydrograph\n", - "B = 5; \t\t\t\t#base flow assumed\n", - "s = 0;\n", - "r = zeros(9)\n", - "for i in range(9):\n", - " r[i] = O[i]-B; \t\t\t\t#ordinates of direct run-off\n", - " s = s+r[i];\n", - "\n", - "n = s*0.36*12/A;\n", - "print \"ordinates of unit hydrograph\";\n", - "for i in range(9):\n", - " u[i] = r[i]/n;\n", - " u[i] = round(u[i]*100)/100;\n", - " print \"%.2f\"%(u[i]);\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of unit hydrograph\n", - "0.00\n", - "4.17\n", - "14.58\n", - "31.25\n", - "22.92\n", - "18.75\n", - "8.33\n", - "4.17\n", - "0.00\n" - ] - } - ], - "prompt_number": 37 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.61 pg : 232" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "#obtain ordinates 24 hr unit hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "O = [0, 5.5, 13.5, 26.5, 45, 82, 162, 240, 231, 165, 112, 79, 57, 42, 31, 22, 14, 9.5, 6.6, 4, 2, 1, 0, 0, 0, 0, 0]; \t\t\t\t#ordinates of 1st 8 hrs unit hydrograph\n", - "o1 = zeros(27)\n", - "o2 = zeros(29)\n", - "for i in range(25):\n", - " o1[i+2] = O[i]; \t\t\t\t#ordinates of 2nd 8 hrs unit hydrograph\n", - " o2[i+4] = O[i]; \t\t\t\t#ordinates of 3rd 8 hrs unit hydrograph\n", - "\n", - "o3 = zeros(27)\n", - "t = zeros(27)\n", - "print \"ordinates 24 hr unit hydrograph:\";\n", - "for i in range(27):\n", - " o3[i] = o1[i]+o2[i]+O[i]; \t\t\t\t#total 24 hr hydrograph of 3 cm run-off\n", - " t[i] = o3[i]/3;\n", - " t[i] = round(t[i]*10)/10;\n", - " print \"%.2f\"%(t[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates 24 hr unit hydrograph:\n", - "0.00\n", - "1.80\n", - "4.50\n", - "10.70\n", - "19.50\n", - "38.00\n", - "73.50\n", - "116.20\n", - "146.00\n", - "162.30\n", - "168.30\n", - "161.30\n", - "133.30\n", - "95.30\n", - "66.70\n", - "47.70\n", - "34.00\n", - "24.50\n", - "17.20\n", - "11.80\n", - "7.50\n", - "4.80\n", - "2.90\n", - "1.70\n", - "0.70\n", - "0.30\n", - "0.00\n" - ] - } - ], - "prompt_number": 38 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.62 pg : 233" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%matplotlib inline\n", - "import math \n", - "from numpy import zeros,linspace\n", - "\n", - "\n", - "#ordinates of 1 hr unit hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "t = linspace(0,12,13) \t\t\t#time\n", - "O = [0, 0, 54, 0, 175, 0, 127, 0, 58, 0, 25, 0, 0, 0]; \t\t\t\t#ordinate of 2 hr unit hydrograph\n", - "of = zeros(13)\n", - "for i in range(2,13):\n", - " if (i%2) == 0:\n", - " of[i] = 0\n", - " else:\n", - " of[i] = O[i-2]+of[i-2];\n", - "\n", - "s = [0, 25, 54, 120, 229, 300, 356, 390, 414, 430, 439, 439, 439]; \t\t\t\t#Ordinates of S-curve\n", - "of1 = zeros(13)\n", - "for i in range(1,13):\n", - " of1[i] = s[i-1];\n", - "\n", - "y = zeros(13)\n", - "u = zeros(13)\n", - "print \"ordinates of 1 hr unit hydrograph:\";\n", - "for i in range(13):\n", - " y[i] = s[i]-of1[i];\n", - " u[i] = y[i]*2;\n", - " print \"%i\"%(u[i]);\n", - "\n", - "#graph is plotted between u and t\n", - "plot(t,u)\n", - "\n", - "\n", - "# graph in book is wrong. Please check." - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of 1 hr unit hydrograph:\n", - "0\n", - "50\n", - "58\n", - "132\n", - "218\n", - "142\n", - "112\n", - "68\n", - "48\n", - "32\n", - "18\n", - "0\n", - "0\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 42, - "text": [ - "[<matplotlib.lines.Line2D at 0x10eed9fd0>]" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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9vkUL11EVLBbTm+2BdKwGD3A/cBz4Q677bAACWH0XEYmpjcC5LgMoEw4iBSgH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\n", - "text": [ - "<matplotlib.figure.Figure at 0x10edfeed0>" - ] - } - ], - "prompt_number": 42 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.63 pg : 234" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "xavg = 1200; \t\t\t\t#sample mean\n", - "n = 50.; \t\t\t\t#assurance year\n", - "A = 0.95; \t\t\t\t#assurance percent\n", - "Rsk = 1-A;\n", - "sigma = 650; \t\t\t\t#smath.radians(numpy.arcmath.tan(ard deviation\n", - "yn = 0.53622; \t\t\t\t#mean of reduced variate\n", - "sigma30 = 1.11238; \t\t\t\t#smath.radians(numpy.arcmath.tan(ard deviation of reduced variate\n", - "\n", - "# Calculations\n", - "T = 1/(1-(1-Rsk)**(1/n));\n", - "yt = -2.303*math.log10(2.303*math.log10(T/(T-1)));\n", - "K = (yt-yn)/sigma30;\n", - "xt = xavg+K*sigma;\n", - "\n", - "# Results\n", - "print \" design disharge = %i cumecs.\"%(xt);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " design disharge = 4908 cumecs.\n" - ] - } - ], - "prompt_number": 43 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.64 pg : 235" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "T1 = 50.;\n", - "T2 = 100.; \t\t\t\t#Return period\n", - "F1 = 20600.;\n", - "F2 = 22150; \t\t\t\t#Peak flood\n", - "\n", - "# Calculations\n", - "y100 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));\n", - "y50 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));\n", - "y = (F2-F1)/(y100-y50);\n", - "T = 500.;\n", - "y500 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "x500 = F2+(y500-y100)*y;\n", - "x500 = round(x500);\n", - "\n", - "# Results\n", - "print \"flood magnitude with return period of 240 years = %.2f cumec.\"%(x500);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "flood magnitude with return period of 240 years = 25732.00 cumec.\n" - ] - } - ], - "prompt_number": 44 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.65 pg : 235" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "xavg = 1.65; \t\t\t\t#mean of data\n", - "sigma = 0.45; \t\t\t\t#smath.radians(numpy.arcmath.tan(ard deviation\n", - "x = 3;\n", - "\n", - "# Calculations\n", - "y = 1.2825*(x-xavg)/sigma+0.577;\n", - "l = math.e**(math.e**(-y));\n", - "T = l/(l-1);\n", - "T = round(T*10)/10;\n", - "\n", - "# Results\n", - "print \"recurrence interval of 10 minutes storm = %.2f years.\"%(T);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "recurrence interval of 10 minutes storm = 84.00 years.\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.66 pg : 236" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "T1 = 50.;\n", - "T2 = 100.; \t\t\t\t#Return period\n", - "F1 = 30800.;\n", - "F2 = 36300.; \t\t\t\t#Peak flood\n", - "\n", - "# Calculations\n", - "y100 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));\n", - "y50 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));\n", - "y = (F2-F1)/(y100-y50);\n", - "T = 200.;\n", - "y200 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "x200 = F2+(y200-y100)*y;\n", - "x200 = round(x200);\n", - "\n", - "# Results\n", - "print \"flood magnitude with return period of 240 years = %.2f cumecs.\"%(x200);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "flood magnitude with return period of 240 years = 41780.00 cumecs.\n" - ] - } - ], - "prompt_number": 46 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.67 pg : 236" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "xavg = 4200.; \t\t\t\t#mean\n", - "sigma = 1705.; \t\t\t\t#smath.radians(numpy.arcmath.tan(ard deviation\n", - "xt = 9550.; \t\t\t\t#flood value\n", - "\n", - "# Calculations\n", - "K = (xt-xavg)/sigma;\n", - "yt = 1.2825*K+0.577;\n", - "l = math.e**(math.e**(-yt));\n", - "T = l/(l-1);\n", - "\n", - "# Results\n", - "print \"Return period of flood of 9950 cumec/s = %.2f years.\"%(T);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Return period of flood of 9950 cumec/s = 100.11 years.\n" - ] - } - ], - "prompt_number": 49 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.68 pg : 237" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "T1 = 100.;\n", - "T2 = 50.; \t\t\t\t#Return period\n", - "F1 = 485.;\n", - "F2 = 445.; \t\t\t\t#Peak flood\n", - "\n", - "# Calculations\n", - "y50 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));\n", - "y100 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));\n", - "y = (F2-F1)/(y50-y100);\n", - "T = 1000.;\n", - "y1000 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "x1000 = F2+(y1000-y50)*y;\n", - "x1000 = round(x1000*10)/10;\n", - "\n", - "# Results\n", - "print \"flood magnitude with return period of 240 years = %.2f cumecs.\"%(x1000);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "flood magnitude with return period of 240 years = 617.20 cumecs.\n" - ] - } - ], - "prompt_number": 50 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.69 pg : 238" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#probability of exceedence\n", - "#probability of occurence in next 12 years\n", - "\n", - "#Given\n", - "T = 25.; \t\t\t\t#return period\n", - "n = 12.;\n", - "\n", - "# Calculations\n", - "P = 1/T;\n", - "Rsk = 1-(1-P)**n;\n", - "P = round(P*100)/100;\n", - "Rsk = round(Rsk*10000)/10000;\n", - "\n", - "# Results\n", - "print \"probability of exceedence = %.2f.\"%(P);\n", - "print \"probability of occurence in next 12 years = %.2f.\"%(Rsk);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "probability of exceedence = 0.04.\n", - "probability of occurence in next 12 years = 0.39.\n" - ] - } - ], - "prompt_number": 51 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch4_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch4_1.ipynb deleted file mode 100755 index 7e800205..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch4_1.ipynb +++ /dev/null @@ -1,4197 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:53d9baa002bc091d49086abaa2a84293a2f939159da5a1508c44dddc72e14dc5" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 4 : HYDROLOGY\n" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.1 pg : 113" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "p = [78.8, 90.2, 98.6, 102.4, 70.4]; \t\t\t\t#rain guage readings at respective stations\n", - "s = 0.;\n", - "\n", - "# Calculations and Results\n", - "for i in range(5):\n", - " s = s+p[i];\n", - "\n", - "pavg = s/5;\n", - "u = 0;\n", - "for i in range(5):\n", - " u = u+(p[i]-pavg)**2;\n", - "\n", - "sx = (u/4)**0.5;\n", - "Cv = sx*100/pavg;\n", - "N = (Cv/6)**2;\n", - "N = round(N*100)/100;\n", - "print \"mean rainfall = %.2f cm.\"%(pavg);\n", - "print \"total stations needed = %.2f.\"%(N);\n", - "#taking N = 7\n", - "N = 7;\n", - "n = N-5;\n", - "print \"additional guages needed = %i.\"%(n);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "mean rainfall = 88.08 cm.\n", - "total stations needed = 6.44.\n", - "additional guages needed = 2.\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.2 pg : 115" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "pB = 48.; \t\t\t\t#precipitation at B\n", - "pC = 51.; \t\t\t\t#precpitation at C\n", - "pD = 45.; \t\t\t\t#precipitation at D\n", - "\n", - "# Calculations\n", - "pA = (pB+pC+pD)/3;\n", - "\n", - "# Results\n", - "print \"precipitation at A = %i mm.\"%(pA);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "precipitation at A = 48 mm.\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.3 pg : 115" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "pA = 6.6; \t\t\t\t#precipitation at A\n", - "pB = 4.8; \t\t\t\t#precpitation at B\n", - "pC = 3.7; \t\t\t\t#precipitation at C\n", - "nA = 72.6; \t\t\t\t#normal precipitation at A\n", - "nB = 51.8; \t\t\t\t#normal precipitation at B\n", - "nC = 38.2; \t\t\t\t#normal precipitation at C\n", - "nX = 65.6; \t\t\t\t#normal precipitation at X\n", - "\n", - "# Calculations\n", - "pX = (nX*pA/nA+nX*pB/nB+nX*pC/nC)/3;\n", - "pX = round(pX*100)/100;\n", - "\n", - "# Results\n", - "print \"precipitation at x = %.2f cm.\"%(pX);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "precipitation at x = 6.13 cm.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.4 pg : 115" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\t\t\t\t\n", - "#Given\n", - "pB = 74.; \t\t\t\t#precipitation at B\n", - "pC = 88.; \t\t\t\t#precpitation at C\n", - "pD = 71.; \t\t\t\t#precipitation at D\n", - "pE = 80.; \t\t\t\t#precipitation at E\n", - "Bx = 9.;\n", - "By = 6.;\n", - "Cx = 12.;\n", - "Cy = -9.;\n", - "Dx = -11.;\n", - "Dy = -6.;\n", - "Ex = -7.;\n", - "Ey = 7.;\n", - "Ax = 0;\n", - "Ay = 0;\n", - "\n", - "# Calculations\n", - "Db = (Bx**2+By**2);\n", - "Dc = (Cx**2+Cy**2);\n", - "Dd = (Dx**2+Dy**2);\n", - "De = (Ex**2+Ey**2);\n", - "Wb = 1/Db;\n", - "Wc = 1/Dc;\n", - "Wd = 1/Dd;\n", - "We = 1/De;\n", - "s = pB*Wb+pC*Wc+pD*Wd+pE*We;\n", - "pA = s/(Wb+Wc+Wd+We);\n", - "pA = round(pA*10)/10;\n", - "\n", - "# Results\n", - "print \"precipitation at A = %.2f mm.\"%(pA);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "precipitation at A = 77.50 mm.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.5 pg : 119" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "p = [58., 61., 69., 56., 84., 86., 69., 79., 71.]; \t\t\t\t#values of precipitation\n", - "s = 0;\n", - "\n", - "# Calculations and Results\n", - "for i in range(9):\n", - " s = s+p[i];\n", - "ar = s/9;\n", - "ar = round(ar*10)/10;\n", - "print \"umath.sing arithmatic average method:\"\n", - "print \"Average rainfall = %.2f cm.\"%(ar);\n", - "\n", - "I = [86., 85., 80., 75., 70., 65., 60., 55., 50.]; \t\t\t\t#isphytes\n", - "A = [0.43, 5.20, 4.0, 5.04, 5.85, 4.53, 4.09, 1.27]; \t\t\t\t#area between isohytes\n", - "\n", - "a = zeros(9)\n", - "for i in range(8):\n", - " a[i] = (I[i]+I[i+1])/2;\n", - "\n", - "P = zeros(8)\n", - "for i in range(8):\n", - " P[i] = A[i]*a[i];\n", - "\n", - "s = 0;\n", - "for i in range(8):\n", - " s = s+P[i];\n", - "\n", - "t = 0;\n", - "for i in range(8):\n", - " t = t+A[i];\n", - "\n", - "ar = s/t;\n", - "ar = round(ar*10)/10;\n", - "print \"isohytel method:\"\n", - "print \"Average rainfall = %.2f cm.\"%(ar);\n", - "\n", - "A = [3.26, 0.39, 1.61, 2.04, 2.46, 0.84, 3.91, 5.09, 0.41, 3.94, 2.06, 4.40]; \t\t\t\t#thiessen area\n", - "p = [58., 63., 71., 69., 86., 81., 84., 56., 53., 69., 61., 79.]; \t\t\t\t#observed precipitation\n", - "P = zeros(12)\n", - "for i in range(12):\n", - " P[i] = A[i]*p[i];\n", - "\n", - "s = 0;\n", - "for i in range(12):\n", - " s = s+P[i];\n", - "\n", - "t = 0;\n", - "for i in range(12):\n", - " t = t+A[i];\n", - "\n", - "ar = s/t;\n", - "ar = round(ar*10)/10;\n", - "print \"thiesson polygon method:\"\n", - "print \"Average rainfall = %.2f cm.\"%(ar);\n", - "\n", - "#mean rainfall obtained by thiesson polygon method is different from book as product(A*P) is round offed in book.\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "umath.sing arithmatic average method:\n", - "Average rainfall = 70.30 cm.\n", - "isohytel method:\n", - "Average rainfall = 69.70 cm.\n", - "thiesson polygon method:\n", - "Average rainfall = 70.00 cm.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.6 pg : 127" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "from matplotlib.pyplot import plot\n", - "from numpy import zeros\n", - "\n", - "\n", - "#Given\n", - "X = [69., 55., 62., 67., 87., 70., 65., 75., 90., 100., 90., 95., 85., 90., 75., 95.]; \t\t\t\t#annual rainfall at X\n", - "Y = [77., 62., 67., 68., 86., 90., 65., 75, 70, 70, 70 ,75 ,65 ,70, 55, 75]; \t\t\t\t#average rainfall at 10 base stations\n", - "cx = zeros(16)\n", - "cx[0] = 69; \t\t\t\t#accumulated annual values at station X \n", - "for i in range(1,16):\n", - " cx[i] = cx[i-1]+X[i];\n", - "\n", - "cy = zeros(16)\n", - "cy[0] = 77;\n", - "for i in range(1,16):\n", - " cy[i] = cy[i-1]+Y[i]; \t\t\t\t#accumulated annual values at ten stations\n", - " \n", - "\n", - "#since curve is not having unform slope\n", - "print \"Record at X is not consistent.\";\n", - "print \"From the curve regime is observed in the year 1978.\"\n", - "\n", - "Q = [1970., 1971., 1972., 1973., 1974., 1975., 1976., 1977.];\n", - "O = [95., 75., 90., 85., 95., 90., 100., 90.];\n", - "for i in range(8):\n", - " A[i] = 0.7051*O[i];\n", - "\n", - "print \"Year Observed rainfall Adjusted rainfall\";\n", - "for i in range(8):\n", - " print \"%i %i %i\"%(Q[i],O[i],A[i]);\n", - "\n", - "#graph is plotted between cx and cy\n", - "plot(cy,cx,cy,cx,\"ro\")" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "Record at X is not consistent.\n", - "From the curve regime is observed in the year 1978.\n", - "Year Observed rainfall Adjusted rainfall\n", - "1970 95 66\n", - "1971 75 52\n", - "1972 90 63\n", - "1973 85 59\n", - "1974 95 66\n", - "1975 90 63\n", - "1976 100 70\n", - "1977 90 63\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 9, - "text": [ - "[<matplotlib.lines.Line2D at 0x110983f50>,\n", - " <matplotlib.lines.Line2D at 0x110987210>]" - ] - }, - { - "metadata": {}, - "output_type": 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- "text": [ - "<matplotlib.figure.Figure at 0x10e5863d0>" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.7 pg : 130" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "from numpy import zeros,linspace\n", - "\t\t\t\t\n", - "#Given\n", - "c = [0, 12.4, 22.1, 35.1, 52.7, 63.7, 81.9, 109.2, 123.5, 132.6, 143.3, 146., 146.]; \t\t\t\t#cumulative rainfall\n", - "T = linspace(0,13,len(c)) \t\t\t\t#Time\n", - "t = 15./60; \t\t\t\t#time interval\n", - "r = zeros(13)\n", - "I = zeros(13)\n", - "r[0] = 0;\n", - "print \"Rainfall intensity:\";\n", - "I[0] = 0;\n", - "for i in range(1,13):\n", - " r[i] = c[i]-c[i-1]; \n", - " I[i] = r[i]/t; \t\t\t\t#Rainfall intensity\n", - " print \"%.2f\"%(I[i]);\n", - "\n", - "\n", - "#graph is plotted between I and T\n", - "bar(T,I)\n", - "xlabel(\"Time hr\")\n", - "ylabel(\"Rain fall insentity\")" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "Rainfall intensity:\n", - "49.60\n", - "38.80\n", - "52.00\n", - "70.40\n", - "44.00\n", - "72.80\n", - "109.20\n", - "57.20\n", - "36.40\n", - "42.80\n", - "10.80\n", - "0.00\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 15, - "text": [ - "<matplotlib.text.Text at 0x110a8b610>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x11098cd50>" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.8 pg : 131" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "from numpy import zeros_like\n", - "from matplotlib.pylab import plot,xlabel,ylabel\n", - "\n", - "\n", - "#Given\n", - "CR = array([0, 12.4, 22.1 ,35.1, 52.7, 63.7, 81.9, 109.2, 123.5, 132.6, 143.3, 146.0, 146.0]); \t\t\t\t#cumulative rainfall\n", - "\n", - "c15 = zeros_like(CR)\n", - "c30 = zeros_like(CR)\n", - "c45 = zeros_like(CR)\n", - "c60 = zeros_like(CR)\n", - "c90 = zeros_like(CR)\n", - "c120 = zeros_like(CR)\n", - "\n", - "# Calculations and Results\n", - "c15[1] = 12.4;\n", - "c30[2] = 22.1;\n", - "c45[3] = 35.1;\n", - "c60[4] = 52.7;\n", - "c90[6] = 81.9;\n", - "c120[8] = 123.5;\n", - "for i in range(2,13):\n", - " c15[i] = CR[i]-CR[i-1];\n", - "\n", - "for i in range(3,13):\n", - " c30[i] = CR[i]-CR[i-2];\n", - "\n", - "for i in range(4,13):\n", - " c45[i] = CR[i]-CR[i-3];\n", - "\n", - "for i in range(5,13):\n", - " c60[i] = CR[i]-CR[i-4];\n", - "\n", - "for i in range(7,13):\n", - " c90[i] = CR[i]-CR[i-6];\n", - "\n", - "for i in range(9,13):\n", - " c120[i] = CR[i]-CR[i-8];\n", - "\n", - "print \"15min 30min 45min 60min 90min 120min\";\n", - "for i in range(13):\n", - " print \"%.2f %.2f %.2f %.2f %.2f %.2f\"%(c15[i],c30[i],c45[i],c60[i],c90[i],c120[i]);\n", - "\n", - "I = [109.2, 91, 79.7, 74.1, 67.6, 61.75]; \t\t\t\t#maximum intensity at respective durations\n", - "D = [15 ,30 ,45 ,60 ,90 ,120]; \t\t\t\t#durations\n", - "#greph is plotted between I and D\n", - "plot(D,I,D,I,\"ro\")\n", - "xlabel(\"Duration\")\n", - "ylabel(\"max rain fall intensity\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "15min 30min 45min 60min 90min 120min\n", - "0.00 0.00 0.00 0.00 0.00 0.00\n", - "12.40 0.00 0.00 0.00 0.00 0.00\n", - "9.70 22.10 0.00 0.00 0.00 0.00\n", - "13.00 22.70 35.10 0.00 0.00 0.00\n", - "17.60 30.60 40.30 52.70 0.00 0.00\n", - "11.00 28.60 41.60 51.30 0.00 0.00\n", - "18.20 29.20 46.80 59.80 81.90 0.00\n", - "27.30 45.50 56.50 74.10 96.80 0.00\n", - "14.30 41.60 59.80 70.80 101.40 123.50\n", - "9.10 23.40 50.70 68.90 97.50 120.20\n", - "10.70 19.80 34.10 61.40 90.60 121.20\n", - "2.70 13.40 22.50 36.80 82.30 110.90\n", - "0.00 2.70 13.40 22.50 64.10 93.30\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 1, - "text": [ - "<matplotlib.text.Text at 0x1047360d0>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10252ad10>" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.9 pg : 132" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Given\n", - "p = [475, 377, 731, 1066, 361, 305, 926, 628, 409, 236, 337, 853]; \t\t\t\t#precipitation value\n", - "N = 12.; \t\t\t\t#total number of years\n", - "T = 6; \t\t\t\t#recurrence interval\n", - "\n", - "# Calculations\n", - "m = N/T;\n", - "\n", - "# Results\n", - "print \"Ranking of storm = %i.\"%(m);\n", - "#hence pick 2nd severest storm\n", - "print \"preciptation value which has recurrence period of 6 years = %i mm.\"%(p[6]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Ranking of storm = 2.\n", - "preciptation value which has recurrence period of 6 years = 926 mm.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.10 pg : 132" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "from numpy import arange,array,zeros_like\n", - "\t\t\t\t\n", - "#Given\n", - "I = linspace(25,16,10) \t\t\t\t#isohytes\n", - "a = array([407 ,1008, 1522, 1909, 2216, 2460, 2651, 2782, 2910, 2936]); \t\t\t\t#enclosed area\n", - "ia = zeros_like(a)\n", - "ia[0] = 407.;\n", - "\n", - "# Calculations\n", - "for i in range(1,10):\n", - " ia[i] = a[i]-a[i-1];\n", - "r = linspace(25.5,16.5,10)\n", - "rv = r*ia\n", - "\n", - "cv = zeros_like(rv)\n", - "cv[0] = 10378;\n", - "for i in range(1,10):\n", - " cv[i] = cv[i-1]+rv[i];\n", - "\n", - "eud = cv/a\n", - "print \"From depth area curve we obtain average depth of precipitation = 24.1 mm for an area of 1800 sq. km.\";\n", - "#graph is plotted between eud and a.\n", - "# Results\n", - "plot(a,eud,a,eud,\"ro\")\n", - "xlabel(\"Area\")\n", - "ylabel(\"mean precipitation depth\")" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "From depth area curve we obtain average depth of precipitation = 24.1 mm for an area of 1800 sq. km.\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 2, - "text": [ - "<matplotlib.text.Text at 0x1047e27d0>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x104730190>" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.11 pg :133" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "from numpy import array\n", - "\n", - "#24h max. rainfall with return period of 8,15 and 25.\n", - "#24h max rainfall with 40%,24% and 8% probability.\n", - "#probabilty of rainfall of magnitude equal to or exceeding 100 mm.\n", - "\t\t\t\t\n", - "#Given\n", - "N = 20.;\n", - "r = array([142, 126, 116, 108, 102, 95, 92, 88, 86, 82, 80, 78, 76, 73, 71, 69, 68, 66, 65, 64],dtype=float64); \t\t\t\t#rainfall in respective years\n", - "m = linspace(1,20,20) \t\t\t\t#ranking of storm\n", - "p = m*100/(N+1)\n", - "T = 100/p\n", - "\n", - "# Calculations and Results\n", - "#from frequency curve obtained we get\n", - "#Part (a)\n", - "T1 = array([8, 15, 25]);\n", - "r1 = array([119, 134, 149]);\n", - "print \"Tyears Rainfallmm\";\n", - "for i in range(3):\n", - " print \"%i %i\"%(T1[i],r1[i]);\n", - "\n", - "\n", - "#Part (b)\n", - "p1 = [40 ,24, 8];\n", - "r2 = [87, 101, 130];\n", - "print \"probabilitypercent Rainfallmm\";\n", - "for i in range(3):\n", - " print \"%i %i\"%(p1[i],r2[i]);\n", - "\n", - "print \"For rainfall = 100 m.T = 4 years.Probability = 25 percent.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "Tyears Rainfallmm\n", - "8 119\n", - "15 134\n", - "25 149\n", - "probabilitypercent Rainfallmm\n", - "40 87\n", - "24 101\n", - "8 130\n", - "For rainfall = 100 m.T = 4 years.Probability = 25 percent.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.12 pg : 135" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "\n", - "\n", - "#plot IDF curve for return period of 10,2 and 1 years umath.sing california formula\n", - "\t\t\t\t\n", - "#Given\n", - "t = array([5, 10, 20, 30, 60, 90, 120],dtype=float64); \t\t\t\t#duration\n", - "\t\t\t\t#value of P for respective return period is\n", - "p10 = array([10.6, 14.7, 19.3, 20.8, 25.5, 29, 34.7]); \t\t\t\t#rainfall for T = 10 years\n", - "p2 = array([8.2, 10.3, 13.2, 14.2, 16.6 ,19.4, 21.4]); \t\t\t\t#rainfall for T = 2 years\n", - "p1 = array([3.5, 6.2, 8.9, 10, 13.2, 15, 16.5]); \t\t\t\t#rainfall for T = 1 year\n", - "\n", - "\n", - "# Calculations\n", - "i1 = p10*60/t; \t\t\t\t#intensity of rainfall with return period of 10 years\n", - "i2 = p2*60/t; \t\t\t\t#intensity of rainfall with return period of 2 years\n", - "i3 = p1*60/t; \t\t\t\t#intensity of rainfall with return period of 1 year\n", - "\n", - "# Results\n", - "#graph is plotted between #t and i1 #t and i2 #t and i3\n", - "plot(t,i1)\n", - "plot(t,i2)\n", - "plot(t,i3)\n", - "xlabel(\"Duration\")\n", - "ylabel(\"Intensity\")\n", - "show()" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x110ba6310>" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.13 pg : 137" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "N = 20.;\n", - "m = linspace(1,20,20) \t\t\t\t#rank number\n", - "rd = array([82, 78, 75, 72, 70, 68, 65, 63, 61, 58, 56, 54, 52, 50, 46, 40, 36, 34, 32, 30]); \t\t\t\t#rainfall in decremath.sing order\n", - "\n", - "# Calculations\n", - "ri = rd[::-1]\n", - "T = N/(m-0.5);\n", - "\n", - "# Results\n", - "#from the curves\n", - "print \"maximum rainfall = 79cm for T = 15 years.\";\n", - "print \"minimum rainfall = 31 cm for T = 15 years.\";\n", - "#graph is plotted between rd and T;ri and T\n", - "subplot(121)\n", - "plot(T,rd)\n", - "xlabel(\"Reccurance interval\")\n", - "ylabel(\"rainfall cm\")\n", - "subplot(122)\n", - "plot(T,ri)\n", - "xlabel(\"Reccurance interval\")\n", - "ylabel(\"rainfall cm\")\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "maximum rainfall = 79cm for T = 15 years.\n", - "minimum rainfall = 31 cm for T = 15 years.\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 44, - "text": [ - "<matplotlib.text.Text at 0x110fc1410>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x11102ee50>" - ] - } - ], - "prompt_number": 44 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.14 pg : 144" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t#average evaporation for one week\n", - "\t\t\t\t\n", - "#Given\n", - "w = [12, 5, 2, -3, 1, 6, 11]; \t\t\t\t#water added or taken out\n", - "r = [0, 6, 8, 12, 9, 5, 0]; \t\t\t\t#rainfall\n", - "pan = zeros(7)\n", - "le = zeros(7)\n", - "for i in range(7):\n", - " pan[i] = w[i]+r[i]; \t\t\t\t#Pan evaporation\n", - " le[i] = 0.8*pan[i]; \t\t\t\t#lake evaporation\n", - "s = sum(le)\n", - "\n", - "print \"daily lake evaporationmm:\";\n", - "for i in range(7):\n", - " print \"%.2f\"%(le[i]);\n", - "\n", - "av = s/7;\n", - "av = round(av*100)/100;\n", - "print \"average evaporation for one week = %.2f mm.\"%(av);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "daily lake evaporationmm:\n", - "9.60\n", - "8.80\n", - "8.00\n", - "7.20\n", - "8.00\n", - "8.80\n", - "8.80\n", - "average evaporation for one week = 8.46 mm.\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.15 pg : 145" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#total depth and volume of evaporation loss\n", - "\t\t\t\t\n", - "#Given\n", - "Rh = 0.4; \t\t\t\t#relative humidity\n", - "A = 4.8; \t\t\t\t#average surface spread of reservior\n", - "v3 = 18.; \t\t\t\t#wind velocity at 3m above ground\n", - "es = 31.81; \t\t\t\t#saturated vapour pressure\n", - "Km = 0.36; \t\t\t\t#for large deep waters\n", - "\n", - "\n", - "# Calculations and Results\n", - "#umath.sing Meyer's formula\n", - "ea = es*Rh;\n", - "v9 = v3*(9./3)**(1./7);\n", - "E = Km*(es-ea)*(1+v9/16);\n", - "d = 7*E;\n", - "v = d*A*100/1000;\n", - "E = round(E*10)/10;\n", - "d = round(d*10)/10;\n", - "v = round(v*100)/100;\n", - "print \"umath.sing Meyers formula:\";\n", - "print \"average evaporation loss from reservior = %.2f mm/day.\"%(E);\n", - "print \"total depth = %.2f mm\"%(d);\n", - "print \"total volume = %.2f hectare-m.\"%(v);\n", - "\n", - "\t\t\t\t#umath.sing Rohwer's formula\n", - "Pa = 760.;\n", - "vdash = (0.6/2)**(1./7)*18;\n", - "E = 0.771*(1.465-0.000732*Pa)*(0.44+0.0733*vdash)*(es-ea);\n", - "d = 7*E;\n", - "v = d*A*100/1000;\n", - "E = round(E*10)/10;\n", - "d = round(d*10)/10;\n", - "v = round(v*10)/10;\n", - "print \"umath.sing Rohwers formula:\";\n", - "print \"average evaporation loss from reservior = %.2f mm/day.\"%(E);\n", - "print \"total depth = %.2f mm\"%(d);\n", - "print \"total volume = %.2f hectare-m.\"%(v);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "umath.sing Meyers formula:\n", - "average evaporation loss from reservior = 15.90 mm/day.\n", - "total depth = 111.40 mm\n", - "total volume = 53.47 hectare-m.\n", - "umath.sing Rohwers formula:\n", - "average evaporation loss from reservior = 20.70 mm/day.\n", - "total depth = 145.20 mm\n", - "total volume = 69.70 hectare-m.\n" - ] - } - ], - "prompt_number": 46 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.16 pg : 149" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "\n", - "#plot infiltration capacity curve\n", - "\t\t\t\t\n", - "#Given\n", - "D = 30.; \t\t\t\t#diameter of inside ring of infiltrometer\n", - "A = math.pi*D**2/4;\n", - "V = array([0, 200, 470, 840, 1405, 1840, 2245, 2510, 2745, 2885, 2990, 3130, 3270],dtype=float64); \t\t\t\t#cumulative volume;\n", - "t = array([0, 2, 5, 10, 20, 30, 45, 60, 80, 100, 120, 150, 180],dtype=float64); \t\t\t\t#Time(minutes)\n", - "\n", - "# Calculations and Results\n", - "dt = zeros_like(t)\n", - "\n", - "for i in range(1,13):\n", - " dt[i] = (t[i]-t[i-1])/60;\n", - "\n", - "F = V/A;\n", - "\n", - "Fd = zeros_like(F)\n", - "Fd[0] = F[0];\n", - "for i in range(1,13):\n", - " Fd[i] = F[i]-F[i-1];\n", - "\n", - "ft = Fd/dt \t\t\t\t#infirltration rate\n", - "\n", - "#from the graph\n", - "print \"constant rate of infiltration = 0.40 cm/hr.\";\n", - "avg10 = F[3]*60/10;\n", - "avg30 = F[5]*60/30;\n", - "avg10 = round(avg10*100)/100;\n", - "avg30 = round(avg30*100)/100;\n", - "print \"average rate of infiltration for first 10 min = %.2f cm/hr.\"%(avg10);\n", - "print \"average rate of infiltration for first 30 min = %.2f cm/hr.\"%(avg30);\n", - "#graph is plotted between ft and t\n", - "plot(t,ft)\n", - "xlabel(\"time in mins\")\n", - "ylabel(\"Infiltrtion rate\")\n", - "show()" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "constant rate of infiltration = 0.40 cm/hr.\n", - "average rate of infiltration for first 10 min = 7.13 cm/hr.\n", - "average rate of infiltration for first 30 min = 5.21 cm/hr.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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7ndNVTMXYt+yN2DeQSMAxJcaV+Jkaho1BrgOWY2OSQcclIiIiIiIiIiIiIiIi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\nIiIiIiIiIiIiIiIiIiIiIiIiIiIiYfZ/7AvpCyJ/2BkAAAAASUVORK5CYII=\n", - "text": [ - "<matplotlib.figure.Figure at 0x110b87a50>" - ] - } - ], - "prompt_number": 49 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.17 pg : 162" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#drainage desity\n", - "#form factor\n", - "#channel slope\n", - "#average overland flow length\n", - "\t\t\t\t\n", - "#Given\n", - "A = 82.; \t\t\t\t#area of watershed\n", - "d = 12.6; \t\t\t\t#dismath.tance between outlet and farther most point\n", - "l = 440.; \t\t\t\t#total length of channel\n", - "e = 656.; \t\t\t\t#elevation differnce between outlet and further most point\n", - "\n", - "\n", - "# Calculations\n", - "Dd = l/A;\n", - "ff = A/d**2;\n", - "cs = e/(d*1000);\n", - "lo = 1000/(2*Dd);\n", - "Dd = round(Dd*100)/100;\n", - "ff = round(ff*1000)/1000;\n", - "\n", - "# Results\n", - "print \"drainage desity = %.2f km/square.km.\"%(Dd);\n", - "print \"form factor = %.2f.\"%(ff);\n", - "print \"channel slope = %.2f.\"%(cs);\n", - "print \"average overland flow length = %i m.\"%(lo);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "drainage desity = 5.37 km/square.km.\n", - "form factor = 0.52.\n", - "channel slope = 0.05.\n", - "average overland flow length = 93 m.\n" - ] - } - ], - "prompt_number": 50 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.18 pg : 163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#compute fi and W index\n", - "\t\t\t\t\n", - "#Given\n", - "R = 3.6; \t\t\t\t#surface runoff\n", - "r = [0, 1.3, 2.8, 4.1, 3.9, 2.8, 2.0, 1.8, 0.9]; \t\t\t\t#rainfall at respective time\n", - "t = 4.; \t\t\t\t#total time\n", - "s = sum(r[2:]);\n", - "\n", - "# Calculations and Results\n", - "fi = (s-R*2)/6;\n", - "#math.since fi >1.3 and <1.8\n", - "print \"fi index = %.2f cm.\"%(fi);\n", - "print \"computations are correct.\";\n", - "\n", - "s = sum(r);\n", - "P = s/2;\n", - "Sr = 0.;\n", - "W = (P-R-Sr)/t;\n", - "print \"W index = %.2f cm/hr.\"%(W);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "fi index = 1.85 cm.\n", - "computations are correct.\n", - "W index = 1.55 cm/hr.\n" - ] - } - ], - "prompt_number": 51 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.19 pg : 163" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace,array\n", - "\t\t\t\t\n", - "#Given\n", - "T = linspace(1,9,9) \t\t\t\t#time from start\n", - "r = array([0.7, 1.4, 2.4, 3.7, 2.9, 2.6, 1.7, 0.8, 0.5]); \t\t\t\t#increamental rainfall\n", - "R = 9.3; \t\t\t\t#total run-off\n", - "s = sum(r)\n", - "\n", - "ti = s-R;\n", - "#first trial\n", - "tr = 9.; \t\t\t\t#assumed\n", - "fi1 = ti/tr;\n", - "#this makes 1st,8th and 9th hour ineffective\n", - "\n", - "#second trial\n", - "tr = 6.;\n", - "ti = s-R-r[0]-r[7]-r[8];\n", - "fi = ti/tr;\n", - "P = zeros_like(r)\n", - "for i in range(9):\n", - " P[i] = r[i]-fi;\n", - " if (P[i]<0):\n", - " P[i] = 0;\n", - " \n", - "print \"Timeh rainfall excess.\";\n", - "for i in range(9):\n", - " print \"%.2f %.2f\"%(T[i],P[i]);\n", - "\n", - "print \"fi index = %.2f cm/hr.\"%(fi);\n", - "print \"time of rainfall excess = %i hours..\"%(tr);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Timeh rainfall excess.\n", - "1.00 0.00\n", - "2.00 0.50\n", - "3.00 1.50\n", - "4.00 2.80\n", - "5.00 2.00\n", - "6.00 1.70\n", - "7.00 0.80\n", - "8.00 0.00\n", - "9.00 0.00\n", - "fi index = 0.90 cm/hr.\n", - "time of rainfall excess = 6 hours..\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.20 pg : 164" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "from numpy import array\n", - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "P = array([72.2, 70.1, 73.3, 42.5, 81.3, 50.6, 52.9, 59.4, 60.3, 64.3, 68.8, 56.7, 77.2, 40.5, 44.1, 65.5]); \t\t\t\t#Precipitation\n", - "R = array([24.1, 22.7, 25.6, 11.3, 28.4, 12.7, 13.4, 15.7, 16.2, 17.7, 19.2, 14.9, 25.4, 10.6, 11.7, 17.9]); \t\t\t\t#runoff\n", - "\n", - "# Calculations\n", - "Ps = P**2\n", - "Rs = R**2;\n", - "PR = P*R;\n", - "\n", - "s = sum(Ps)\n", - "t = sum(Rs)\n", - "u = sum(PR)\n", - "q = sum(P)\n", - "w = sum(R)\n", - "N = 16.;\n", - "a = (N*u-q*w)/(N*s-q**2);\n", - "b = (w-a*q)/N;\n", - "b = round(b*1000)/1000;\n", - "a = round(a*10000)/10000;\n", - "\n", - "# Results\n", - "print \"Equation is:%.4f P %.3f.\"%(a,b);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Equation is:0.4375 P -8.823.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.21 pg : 165" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "from numpy import linspace,array\n", - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "A = 8.6; \t\t\t\t#catchment area\n", - "T = linspace(0,4,9) \t\t\t\t#time\n", - "r = array([0, 0.4, 1.1, 2.3, 3.8, 4.8, 5.6, 6.2, 6.7]); \t\t\t\t#accumulated rainfall\n", - "fi = 0.4; \t\t\t\t#fi index\n", - "dt = 0.5; \t\t\t\t#time interval\n", - "\n", - "# Calculations and Results\n", - "d = zeros_like(r)\n", - "\n", - "for i in range(1,9):\n", - " d[i] = r[i]-r[i-1]; \t\t\t\t#accumulated rainfall\n", - "\n", - "print \"Intensity of effective Rainfall:\";\n", - "I = zeros_like(r)\n", - "p = zeros_like(r)\n", - "s = 0;\n", - "for i in range(1,9):\n", - " p[i] = d[i]-fi; \t\t\t\t#effective rainfall\n", - " I[i] = p[i]/dt; \t\t\t\t#Intensity of effective Rainfall\n", - " s = s+I[i];\n", - " print \"%.2f\"%(I[i]);\n", - "\n", - "#graph is plotted between I and T\n", - "run = s*dt;\n", - "V = run*A*10000;\n", - "print \"Volume of direct run-off = %.2f cubic metre.\"%(V);\n", - "plot(T,I)\n", - "show()" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "Intensity of effective Rainfall:\n", - "0.00\n", - "0.60\n", - "1.60\n", - "2.20\n", - "1.20\n", - "0.80\n", - "0.40\n", - "0.20\n", - "Volume of direct run-off = 301000.00 cubic metre.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x1046bdd10>" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.22 pg : 166" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "from numpy import array,linspace\n", - "#total rainfall\n", - "#total rainfall excess\n", - "#W index\n", - "\t\t\t\t\n", - "#Given\n", - "r = array([3.5, 6.5, 8.5 ,7.8, 6.4, 4, 4, 6]); \t\t\t\t#rainfall intensity\n", - "T = linspace(0,240,8) \t\t\t\t#time\n", - "dt = 30.; \t\t\t\t#time interval\n", - "\n", - "# Calculations\n", - "s = sum(r);\n", - "P = s*dt/60;\n", - "Pe = ((6.5-4.5)+(8.5-4.5)+(7.8-4.5)+(6.4-4.5)+(6-4.5))*dt/60; \t\t\t\t#area of graph above r = 4.5.\n", - "w = (P-Pe)/4;\n", - "\n", - "# Results\n", - "print \"total rainfall = %.2f cm.\"%(P);\n", - "print \"total rainfall excess = %.2f cm.\"%(Pe);\n", - "print \"W index = %.2f cm/hr.\"%(w);\n", - "plot(T,r)\n", - "show()" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "total rainfall = 23.35 cm.\n", - "total rainfall excess = 6.35 cm.\n", - "W index = 4.25 cm/hr.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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qcF4qDygiIgdXqgDPOQp4ESuxlABGAO+nMigREREREdlHc2AJsAx4wHEsLqwC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\no0m7AAAAAElFTkSuQmCC\n", - "text": [ - "<matplotlib.figure.Figure at 0x102528b50>" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.23 pg : 166" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array,zeros_like\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "r = array([0, 8, 22, 74, 92, 105, 114, 120],dtype=float64); \t\t\t\t#raccumulated rainfall\n", - "T = array([0, 2, 4, 6, 8, 10, 12, 14],dtype=float64); \t\t\t\t#time for start of rainfall\n", - "V = 2e6; \t\t\t\t#volume of run-off\n", - "A = 40.; \t\t\t\t#catchment area\n", - "tr = 14.; \t\t\t\t#duration of rainfall\n", - "\n", - "# Calculations\n", - "d = V*1000/(40*1000000);\n", - "\n", - "l = r[7]-d;\n", - "W = l/tr;\n", - "I = zeros_like(r)\n", - "for i in range(1,8):\n", - " I[i] = r[i]-r[i-1]; \t\t\t\t#incremental rainfall\n", - "\n", - "\n", - "#rainfall excess is available in 4 time intervals of 2 hrs\n", - "tre = 8.;\n", - "fi = (l-I[1]-I[6]-I[7])/tre;\n", - "fi = round(fi*100)/100;\n", - "\n", - "# Results\n", - "print \"fi index = %.2f mm/hr.\"%(fi);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "fi index = 5.88 mm/hr.\n" - ] - } - ], - "prompt_number": 66 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.24 pg : 167" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array\n", - "\n", - "#Given\n", - "r = array([2.0 ,2.5, 7.6, 3.8, 10.6, 5.0, 7.0, 10.0, 6.4, 3.8, 1.4, 1.4]); \t\t\t\t#rainfall depths\n", - "R = 25.5;\n", - "\n", - "# Calculations\n", - "s = sum(r)\n", - "tf = s-R;\n", - "af = tf/12;\n", - "#rainfall is less than average infiltration in1st,2nd,11th and 12th hours\n", - "\n", - "f = (tf-r[0]-r[1]-r[10]-r[11])/8;\n", - "f = round(f*10)/10;\n", - "\n", - "# Results\n", - "print \"average infiltration index = %d cm/hour.\"%(f);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "average infiltration index = 3 cm/hour.\n" - ] - } - ], - "prompt_number": 68 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.25 pg : 167" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros,zeros_like,array\n", - "\t\t\t\t\n", - "#Given\n", - "r = array([2.0, 2.5, 7.6, 3.8, 10.6, 5.0, 7.0, 10.0, 6.4, 3.8, 1.4, 1.4]); \t\t\t\t#rainfall depths\n", - "A1 = 20.;\n", - "A2 = 40.;\n", - "A3 = 60.;\n", - "\n", - "# Calculations and Results\n", - "A = A1+A2+A3;\n", - "fi1 = 7.6;\n", - "fi2 = 3.8;\n", - "fi3 = 1.0;\n", - "R1 = zeros_like(r)\n", - "R2 = zeros_like(r)\n", - "R3 = zeros_like(r)\n", - "for i in range(12):\n", - " R1[i] = r[i]-fi1; \t\t\t\t#rainfall excess\n", - " R2[i] = r[i]-fi2;\n", - " R3[i] = r[i]-fi3;\n", - " if (R1[i]<0):\n", - " R1[i] = 0;\n", - " if (R2[i]<0):\n", - " R2[i] = 0;\n", - " if (R3[i]<0):\n", - " R3[i] = 0;\n", - "\n", - "print \"average depth of hourly rainfall excesscm/hr\";\n", - "a1 = zeros(12)\n", - "a2 = zeros(12)\n", - "a3 = zeros(12)\n", - "T = zeros(12)\n", - "for i in range(12):\n", - " a1[i] = R1[i]*A1/A; \t\t\t\t#average rainfall excess\n", - " a2[i] = R2[i]*A2/A;\n", - " a3[i] = R3[i]*A3/A;\n", - " T[i] = a1[i]+a2[i]+a3[i]; \t\t\t\t#total hourly rainfall excess\n", - " T[i] = round(T[i]*100)/100;\n", - " print \"%.2f\"%(T[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "average depth of hourly rainfall excesscm/hr\n", - "0.50\n", - "0.75\n", - "4.57\n", - "1.40\n", - "7.57\n", - "2.40\n", - "4.07\n", - "6.97\n", - "3.57\n", - "1.40\n", - "0.20\n", - "0.20\n" - ] - } - ], - "prompt_number": 69 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.26 pg : 185" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "from numpy import array,linspace\n", - "#derive the unit hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "A = 92.; \t\t\t\t#area of drainage bamath.sin\n", - "t = array([6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 2, 4, 6, 8, 10, 12, 14, 16],dtype=float64); \t\t\t\t#time\n", - "r = array([10.6, 9.7, 107.8, 175.6, 193.9, 150.3, 126.2, 106.9, 90, 72.8, 58.2, 48, 36.2, 28.4, 20.2, 14, 10.2, 10.4]); \t\t\t\t#total run-off\n", - "B = array([10.6, 9.7, 9.73, 9.77, 9.8, 9.83, 9.87, 9.9, 9.93, 9.97, 10, 10.03, 10.07, 10.10, 10.13, 10.16, 10.20, 10.40]); \t\t\t\t#base flow\n", - "s = 0;\n", - "\n", - "# Calculations and Results\n", - "d = r - B\n", - "s = sum(d)\n", - "\n", - "n = 0.36*s*2/A;\n", - "print \"ordinates of unit hydrograph:\";\n", - "u = zeros(18)\n", - "for i in range(18):\n", - " u[i] = d[i]/n; \t\t\t\t#ordinates of unit hydrograph\n", - " u[i] = round(u[i]*100)/100;\n", - " print \"%.2f\"%(u[i]);\n", - "\n", - "print \"Hydograph is 4-hr unit hydrograph\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of unit hydrograph:\n", - "0.00\n", - "0.00\n", - "11.50\n", - "19.45\n", - "21.60\n", - "16.48\n", - "13.65\n", - "11.38\n", - "9.39\n", - "7.37\n", - "5.65\n", - "4.45\n", - "3.07\n", - "2.15\n", - "1.18\n", - "0.45\n", - "0.00\n", - "0.00\n", - "Hydograph is 4-hr unit hydrograph\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.27 pg : 186" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "A = 316.; \t\t\t\t#drainage area\n", - "B = 17.; \t\t\t\t#base flow\n", - "t = 6.;\n", - "O = [17.0, 113.2, 254.5, 198.0, 150.0, 113.2, 87.7, 67.9, 53.8, 42.5, 31.1, 22.6, 17.0]; \t\t\t\t#ordinates of storm hydrograph\n", - "Or = zeros(13)\n", - "Oh = zeros(13)\n", - "for i in range(13): \n", - " Or[i] = O[i]-B; \t\t\t\t#ordinates of direct run-off\n", - " Oh[i] = Or[i]/6.477; \t\t\t\t#ordinates of unit hydrograph\n", - "\n", - "\n", - "s = sum(Or);\n", - "re = s*60*60*t/(A*10000);\n", - "re = round(re*1000)/1000;\n", - "\n", - "# Results\n", - "print \"rainfall excess = %.2f cm.\"%(re);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "rainfall excess = 6.48 cm.\n" - ] - } - ], - "prompt_number": 74 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.28 pg : 188" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "fi = 2.5; \t\t\t\t#infiltration index\n", - "B = 10; \t\t\t\t#Base flow\n", - "O = [0, 110, 365, 500, 390, 310, 250, 235, 175, 130, 95, 65, 40, 22, 10, 0, 0, 0]; \t\t\t\t#ordinates of unit hydrograph\n", - "R1 = 2;R2 = 6.75;R3 = 3.75;\n", - "r1 = (R1*10-(fi*3)-5)/10; \t\t\t\t#rainfall excess in first three hour\n", - "r2 = (R2*10-(fi*3))/10; \t\t\t\t#rainfall excess in second three hour\n", - "r3 = (R3*10-(fi*3))/10; \t\t\t\t#rainfall excess in third three hour\n", - "\n", - "s1 = zeros(18)\n", - "for i in range(18):\n", - " s1[i] = r1*O[i]; \n", - "s2 = zeros(18)\n", - "for i in range(1,18):\n", - " s2[i] = r2*O[i-1];\n", - "s3 = zeros(18)\n", - "for i in range(2,18):\n", - " s3[i] = r3*O[i-2];\n", - " \t\t\t\t#surface run-off from rainfall excess during succesive unit periods\n", - "print \"ordinates of storm hydrograph\";\n", - "T = zeros(18)\n", - "t = zeros(18)\n", - "for i in range(18):\n", - " T[i] = s1[i]+s2[i]+s3[i];\n", - " t[i] = T[i]+B;\n", - " t[i] = round(t[i]*10)/10;\n", - " print \"%.2f\"%(t[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of storm hydrograph\n", - "10.00\n", - "92.50\n", - "943.80\n", - "2905.00\n", - "4397.50\n", - "4082.50\n", - "3227.50\n", - "2616.30\n", - "2301.30\n", - "1862.50\n", - "1386.30\n", - "1018.80\n", - "715.00\n", - "461.50\n", - "269.50\n", - "136.00\n", - "40.00\n", - "10.00\n" - ] - } - ], - "prompt_number": 75 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.29 pg : 189" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "from numpy import array,linspace,zeros\n", - "#derive and plot 6 hr unit hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "A = 103.4; \t\t\t\t#area of bamath.sin\n", - "t = linspace(0,36,9); \t\t\t\t#time\n", - "q = [0, 21, 80, 82, 189, 123, 184, 87, 55.5, 25.25, 9, 6, 0]; \t\t\t\t#flow\n", - "print \"ordinates of unit hydrograph are:\";\n", - "u = zeros(9)\n", - "u[0] = 0;\n", - "u[1] = q[1]/2.;\n", - "u[2] = (q[2]-4*u[0])/2;\n", - "u[3] = (q[3]-4*u[1])/2;\n", - "for i in range(4,9):\n", - " u[i] = (q[i]-3*u[i-4]-4*u[i-2])/2; \t\t\t\t#ordinates of unit hydrograph\n", - "\n", - "for i in range(9):\n", - " print \"%.2f\"%(u[i]);\n", - "print \"The succesive unit hydrograph will have same ordinates but will be shiftedlaterally by 6 hrs.\";\n", - "#graph is plotted between u and t.\n", - "plot(t,u)\n", - "xlabel(\"Time in hours\")\n", - "ylabel(\"Discharge\")" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "ordinates of unit hydrograph are:\n", - "0.00\n", - "10.50\n", - "40.00\n", - "20.00\n", - "14.50\n", - "5.75\n", - "3.00\n", - "2.00\n", - "0.00\n", - "The succesive unit hydrograph will have same ordinates but will be shiftedlaterally by 6 hrs.\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 9, - "text": [ - 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- "text": [ - "<matplotlib.figure.Figure at 0x10490cf90>" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.30 pg : 190" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "\n", - "\n", - "#derive ordinates of 6 hrs unit hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "R = [0, 1, 2.7, 5, 8, 9.8, 9, 7.5, 6.3, 5, 4, 2.9, 2.1, 1.3 ,0.5, 0, 0, 0, 0, 0]; \t\t\t\t#2hrs unit hydrograph\n", - "print \"ordinates of 6 hrs unit hydrograph\";\n", - "O1 = zeros(20)\n", - "for i in range(18):\n", - " O1[i+2] = R[i];\n", - "O2 = zeros(20)\n", - "for i in range(16):\n", - " O2[i+4] = R[i];\n", - "S = zeros(20)\n", - "f = zeros(20)#offset unit hydrograph\n", - "for i in range(20):\n", - " S[i] = O1[i]+O2[i]+R[i]; \t\t\t\t#sum\n", - " f[i] = S[i]/3; \t\t\t\t#ordinates of 6 hrs unit hydrograph\n", - " f[i] = round(f[i]*10)/10;\n", - " print \"%.2f\"%(f[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of 6 hrs unit hydrograph\n", - "0.00\n", - "0.30\n", - "0.90\n", - "2.00\n", - "3.60\n", - "5.30\n", - "6.60\n", - "7.40\n", - "7.80\n", - "7.40\n", - "6.40\n", - "5.10\n", - "4.10\n", - "3.10\n", - "2.20\n", - "1.40\n", - "0.90\n", - "0.40\n", - "0.20\n", - "0.00\n" - ] - } - ], - "prompt_number": 79 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.31 pg : 190" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array,linspace,zeros\n", - "\t\t\t\t\n", - "#Given\n", - "t = linspace(0,45,16) \t\t\t\t#time\n", - "O = [0 ,9, 20, 35, 49, 43, 35, 28, 22, 17, 12, 9, 6, 3, 0, 0]; \t\t\t\t#ordinate of 2 hr unit hydrograph\n", - "\n", - "# Calculations and Results\n", - "of = zeros(16)\n", - "for i in range(2,16):\n", - " of[i] = O[i-2]+of[i-2]; \t\t\t\t#offset ordinate\n", - "\n", - "s = zeros(16)\n", - "for i in range(16):\n", - " s[i] = O[i]+of[i]; \t\t\t\t#ordinate of s-curve\n", - "\n", - "of1 = zeros(16)\n", - "for i in range(3,16):\n", - " of1[i] = s[i-3]; \t\t\t\t#offset of s-curve\n", - "\n", - "print \"ordinates of 9 hrs unit hydrograph:\";\n", - "y = zeros(16)\n", - "u = zeros(16)\n", - "for i in range(16):\n", - " y[i] = s[i]-of1[i];\n", - " u[i] = 2*y[i]/3; \t\t\t\t#ordinate of 9 hrs unit hydrograph\n", - " u[i] = round(u[i]*10)/10;\n", - " print \"%.2f\"%(u[i]);\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of 9 hrs unit hydrograph:\n", - "0.00\n", - "6.00\n", - "13.30\n", - "29.30\n", - "40.00\n", - "44.70\n", - "40.00\n", - "30.70\n", - "26.00\n", - "18.70\n", - "15.30\n", - "10.00\n", - "8.00\n", - "4.00\n", - "2.00\n", - "0.00\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.32 pg : 210" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros,linspace\n", - "\n", - "\n", - "#california method\n", - "#Hazens method\n", - "#gumbels method\n", - "\t\t\t\t\n", - "#Given\n", - "q = [9200, 7800, 6600, 5800, 5260, 4980, 4525, 3810, 3630, 3250, 3110, 3090, 2380, 2390, 1723]; \t\t\t\t#Discharge arranged in decreamath.sing order\n", - "N = 15;\n", - "C = 0.3;\n", - "m = linspace(1,15,15)\n", - "C = [0.3, 0.44, 0.52, 0.57, 0.61, 0.66, 0.7, 0.74, 0.78, 0.82, 0.86, 0.88, 0.94, 0.96, 1]; \t\t\t\t#from table 4.25\n", - "print \"California Hazen Gumbel\";\n", - "Ca = zeros(15)\n", - "H = zeros(15)\n", - "G = zeros(15)\n", - "Ca = zeros(15)\n", - "G = zeros(15)\n", - "\n", - "for i in range(15):\n", - " Ca[i] = N/m[i];\n", - " H[i] = 2*N/(2*m[i]-1);\n", - " G[i] = N/(m[i]+C[i]-1);\n", - " Ca[i] = round(Ca[i]*100)/100;\n", - " G[i] = round(G[i]*100)/100;\n", - " H[i] = round(H[i]*100)/100;\n", - " print \"%.2f %.2f %.2f\"%(Ca[i],H[i],G[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "California Hazen Gumbel\n", - "15.00 30.00 50.00\n", - "7.50 10.00 10.42\n", - "5.00 6.00 5.95\n", - "3.75 4.29 4.20\n", - "3.00 3.33 3.25\n", - "2.50 2.73 2.65\n", - "2.14 2.31 2.24\n", - "1.88 2.00 1.94\n", - "1.67 1.76 1.71\n", - "1.50 1.58 1.53\n", - "1.36 1.43 1.38\n", - "1.25 1.30 1.26\n", - "1.15 1.20 1.16\n", - "1.07 1.11 1.07\n", - "1.00 1.03 1.00\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.33 pg : 211" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "T1 = 40.;\n", - "T2 = 80.; \t\t\t\t#Return period\n", - "F1 = 27000.;\n", - "F2 = 31000.; \t\t\t\t#Peak flood\n", - "\n", - "# Calculations\n", - "y80 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));\n", - "y40 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));\n", - "y = (F2-F1)/(y80-y40);\n", - "T = 240.;\n", - "y240 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "x240 = F2+(y240-y80)*y;\n", - "\n", - "# Results\n", - "print \"flood magnitude with return period of 240 years = %i cumec.\"%(x240);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "flood magnitude with return period of 240 years = 37306 cumec.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.34 pg : 212" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "N = 40;\n", - "Sn = 1.1413;\n", - "yn = 0.5436; \t\t\t\t#from table 4.21 (a) and(b)\n", - "q = [1330, 1095, 1030, 980, 975, 950, 945, 940, 925, 855, 853, 840, 835, 825, 810, 795, 756, 710, 708, 705, 700, 670, 625, 620, 610, 605, 595, 585, 570, 550, 530, 505, 500, 495, 485, 465, 460, 420, 390, 380]; \t\t\t\t#discharge\n", - "s = sum(q)\n", - "xavg = s/N;\n", - "w = 0;\n", - "\n", - "# Calculations\n", - "t = zeros(40)\n", - "for i in range(40):\n", - " t[i] = (q[i]-xavg)**2;\n", - " w = w+t[i];\n", - "\n", - "sigma = (w/(N-1))**0.5;\n", - "N = 10.;\n", - "y10 = -(2.303*math.log10(2.303*math.log10(N/(N-1))));\n", - "K10 = (y10-yn)/Sn;\n", - "x10 = xavg+K10*sigma;\n", - "N = 20.;\n", - "y20 = -(2.303*math.log10(2.303*math.log10(N/(N-1))));\n", - "K20 = (y20-yn)/Sn;\n", - "x20 = xavg+K20*sigma;\n", - "N = 5.;\n", - "y5 = -(2.303*math.log10(2.303*math.log10(N/(N-1))));\n", - "K5 = (y5-yn)/Sn;\n", - "x5 = xavg+K5*sigma;\n", - "\n", - "T = 100.;\n", - "y100 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "K100 = (y100-yn)/Sn;\n", - "x100 = xavg+K100*sigma;\n", - "\n", - "T = 200.;\n", - "y200 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "K200 = (y200-yn)/Sn;\n", - "x200 = xavg+K200*sigma;\n", - "x100 = round(x100);\n", - "\n", - "# Results\n", - "print \"For T = 100 years:flood discharge = %.2f cumecs.\\\n", - "\\nFor T = 200 years:flood discharge = %.f cumecs.\"%(x100,x200);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "For T = 100 years:flood discharge = 1487.00 cumecs.\n", - "For T = 200 years:flood discharge = 1620 cumecs.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.35 pg : 214" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy.linalg import solve\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "sigma = 1.1413; \t\t\t\t#smath.radians(numpy.arcmath.tan(ard deviation\n", - "yn = 0.5436;\n", - "T = 50.;\n", - "\n", - "# Calculations\n", - "y50 = -2.303*math.log10(2.303*math.log10(T/(T-1)));\n", - "K50 = (y50-yn)/sigma;\n", - "T = 100.;\n", - "y100 = -2.303*math.log10(2.303*math.log10(T/(T-1)));\n", - "K100 = (y100-yn)/sigma;\n", - "x50 = 878; x100 = 970; \t\t\t\t\n", - "#Given peak flood\n", - "A = [[K50, 1],[K100, 1]];\n", - "B = [x50,x100];\n", - "C = solve(A,B)#A\\B;\n", - "xavg = C[1];\n", - "sigmad = C[0];\n", - "T = 200.;\n", - "y200 = -2.303*math.log10(2.303*math.log10(T/(T-1)));\n", - "K200 = (y200-yn)/sigma;\n", - "x200 = xavg+K200*sigmad;\n", - "x200 = round(x200);\n", - "\n", - "# Results\n", - "print \"200 year flood for stream = %.2f cumecs.\"%(x200);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "200 year flood for stream = 1062.00 cumecs.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.36 pg : 214" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#risk of failure of cofferdam\n", - "#return period\n", - "\n", - "#Given\n", - "T = 30.; \t\t\t\t#deign for period\n", - "n = 6.; \t\t\t\t#period of construction\n", - "\n", - "# Calculations\n", - "R = (1-(1-(1/T))**n)*100;\n", - "R1 = 0.1; \t\t\t\t#reduced risk\n", - "T1 = 1./(1-(1-R1)**(1./6));\n", - "R = round(R*10)/10;\n", - "T1 = round(T1*100)/100;\n", - "\n", - "# Results\n", - "print \"risk of failure of cofferdam = %.2f percent.\"%(R);\n", - "print \"return period = %.2f years.\"%(T1);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "risk of failure of cofferdam = 18.40 percent.\n", - "return period = 57.45 years.\n" - ] - } - ], - "prompt_number": 85 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.37 pg : 215" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#probability of excedence\n", - "#probability of flood magnitude occuring at:\n", - "#at least once in 10 years\n", - "#two times in 10 succesive years\n", - "#once in 10 succesive years\n", - "\n", - "#Given\n", - "T = 40.; \t\t\t\t#return period\n", - "P = 1./T;\n", - "n = 10;\n", - "Rsk = 1.-(1-P)**n;\n", - "s = 1.;\n", - "t = 1.;\n", - "for i in range(1,n+1): \n", - " s = s*i;\n", - "\n", - "for i in range(1,n-1):\n", - " t = t*i;\n", - "\n", - "P2n = s*P**2*(1-P)**8/(t*2);\n", - "P1n = n*P*(1-P)**(n-1);\n", - "Rsk = round(Rsk*1000)/1000;\n", - "P2n = round(P2n*10000)/10000;\n", - "P1n = round(P1n*1000)/1000;\n", - "\n", - "# Results\n", - "print \"probability of excedence = %.2f.\"%(P);\n", - "print \"probability of flood magnitude occuring at least once in 10 years = %.2f\"%(Rsk);\n", - "print \"probability of flood magnitude occuring at two times in 10 succesive years = %.2f\"%(P2n);\n", - "print \"probability of flood magnitude occuring at once in 10 succesive years = %.2f\"%(P1n);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "probability of excedence = 0.03.\n", - "probability of flood magnitude occuring at least once in 10 years = 0.22\n", - "probability of flood magnitude occuring at two times in 10 succesive years = 0.02\n", - "probability of flood magnitude occuring at once in 10 succesive years = 0.20\n" - ] - } - ], - "prompt_number": 87 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.38 pg : 215" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "C1 = 0.22;C2 = 0.12;C3 = 0.32; \t\t\t\t#run-off coefficient\n", - "A1 = 3.2;A2 = 4.8;A3 = 1.8; \n", - "L = 2.4; \t\t\t\t#length of water course\n", - "H = 30; \t\t\t\t#fall\n", - "T = 30; \t\t\t\t#frequency\n", - "\n", - "# Calculations\n", - "t = 60*0.000323*(L*1000)**0.77*(H/(L*1000))**(-0.385);\n", - "i = 78*T**0.22/(t+12)**0.45;\n", - "q = 2.778*i*(C1*A1+C2*A2+C3*A3);\n", - "q = round(q*10)/10;\n", - "\n", - "# Results\n", - "print \"peak rate of run off = %.2f cumecs.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "peak rate of run off = 141.20 cumecs.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.39 pg : 216" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "T = 30; \t\t\t\t#return period\n", - "A = 2.4; \t\t\t\t#area of watershed\n", - "s = 1./200; \t\t\t\t#slope oof catchment\n", - "L = 1.8; \t\t\t\t#length of travel of water\n", - "C = 0.25; \t\t\t\t#average run-off coefficient\n", - "r = [2.5, 3.8, 4.8, 5.9, 6.7, 7.4, 8.4, 8.7, 9.2]; \t\t\t\t#rmath.sinfall depth\n", - "\n", - "# Calculations\n", - "t = 60*0.000323*(L*1000)**0.77*(s)**(-0.385); \n", - "rmax = r[6]+(r[7]-r[6])*7.84/10;\n", - "i = rmax*60/t;\n", - "q = 2.778*C*A*i;\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"peak flow rate = %.2f cumecs.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "peak flow rate = 18.05 cumecs.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.40 pg : 216" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "pA = 75.; \t\t\t\t#precipitation at A\n", - "pB = 58; \t\t\t\t#precpitation at B\n", - "pC = 47; \t\t\t\t#precipitation at C\n", - "nA = 826; \t\t\t\t#normal precipitation at A\n", - "nB = 618; \t\t\t\t#normal precipitation at B\n", - "nC = 482; \t\t\t\t#normal precipitation at C\n", - "nX = 757; \t\t\t\t#normal precipitation at X\n", - "\n", - "\n", - "# Calculations\n", - "pX = (nX*pA/nA+nX*pB/nB+nX*pC/nC)/3.;\n", - "pX = round(pX*10)/10;\n", - "\n", - "# Results\n", - "print \"precipitation at x = %.2f cm.\"%(pX);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "precipitation at x = 70.90 cm.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.41 pg : 216" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "p = [41, 51, 32, 55, 50, 68]; \t\t\t\t#rain guage readings at respective stations\n", - "s = sum(p)\n", - "pavg = s/6;\n", - "u = 0;\n", - "for i in range(5):\n", - " u = u+(p[i]-pavg)**2;\n", - "\n", - "# Calculations\n", - "sx = (u/5)**0.5;\n", - "Cv = sx*100/pavg;\n", - "N = (Cv/8)**2;\n", - "N = round(N*100)/100;\n", - "\n", - "# Results\n", - "print \"mean rainfall = %.2f cm.\"%(pavg);\n", - "print \"total stations needed = %.2f.\"%(N);\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "mean rainfall = 49.00 cm.\n", - "total stations needed = 5.08.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.42 pg : 217" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "a = 4; \t\t\t\t#dimension of plot sides\n", - "P1 = 4.8;P2 = 13;P3 = 8;P4 = 5.4;P5 = 3.2;P6 = 9.4; \t\t\t\t#precipitaion at respective stations\n", - "\n", - "# Calculations\n", - "A1 = a**2/8+a**2/(4*1.73);\n", - "A2 = a**2/8;\n", - "A3 = A2;A4 = A1;\n", - "A5 = a**2/(4*1.73);\n", - "A6 = a**2/2;\n", - "A = A1+A2+A3+A4+A5+A6;\n", - "Pavg = (P1*A1+P2*A2+P3*A3+P4*A4+P5*A5+P6*A6)/A;\n", - "\n", - "# Results\n", - "print \"Mean precipitaion = %.2f cm.\"%(Pavg);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Mean precipitaion = 7.35 cm.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.43 pg : 218" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace\n", - "\t\t\t\t\n", - "#Given\n", - "A = [90, 140, 125, 140, 85, 40, 20]; \t\t\t\t#area of isohytes\n", - "I = linspace(13,1,7) \t\t\t\t#average isohytel interval\n", - "s = 0;t = 0;\n", - "for i in range(7):\n", - " s = s+A[i]*I[i];\n", - " t = t+A[i];\n", - "\n", - "Pavg = s/t;\n", - "Pavg = round(Pavg*10)/10;\n", - "\n", - "# Results\n", - "print \" average depth of precipitation = %.2f cm.\"%(Pavg);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " average depth of precipitation = 8.40 cm.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.44 pg : 218" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "p = [120, 95, 96, 60, 65, 70, 45, 21]; \t\t\t\t#rain guage readings at respective stations\n", - "\n", - "# Calculations and Results\n", - "s = sum(p)\n", - "pavg = s/8;\n", - "u = 0;\n", - "for i in range(8):\n", - " u = u+(p[i]-pavg)**2;\n", - "\n", - "sx = (u/7)**0.5;\n", - "Cv = sx*100/pavg;\n", - "N = (Cv/13.99)**2;\n", - "N = round(N*100)/100;\n", - "print \"mean rainfall = %.2f cm.\"%(pavg);\n", - "print \"total stations needed = %.2f.\"%(N);\n", - "\t\t\t\t#taking N = 10\n", - "N = 10;\n", - "n = N-8;\n", - "print \"additional guages needed = %i.\"%(n);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "mean rainfall = 71.00 cm.\n", - "total stations needed = 10.03.\n", - "additional guages needed = 2.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.45 pg : 219" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "from matplotlib.pylab import plot\n", - "from numpy import zeros,linspace\n", - "import math \n", - "#compute maximum rainfall intensities for 5,10,15,20,25,30,35,40,45,50 minutes\n", - "#plot intensity duration graph\n", - "\t\t\t\t\n", - "#Given\n", - "CR = [0, 1.02, 2.08, 3.30, 4.72, 5.58, 6.40, 7.16, 7.88, 8.54, 9.14]; \t\t\t\t#cumulative rainfall\n", - "\n", - "c5 = zeros(11)\n", - "c10 = zeros(11)\n", - "c15 = zeros(11)\n", - "c20 = zeros(11)\n", - "c25 = zeros(11)\n", - "c30 = zeros(11)\n", - "c35 = zeros(11)\n", - "c40 = zeros(11)\n", - "c45 = zeros(11)\n", - "c50 = zeros(11)\n", - "\n", - "c5[1] = CR[1];\n", - "c10[2] = CR[2];\n", - "c15[3] = CR[3];\n", - "c20[4] = CR[4];\n", - "c25[5] = CR[5];\n", - "c30[6] = CR[6];\n", - "c35[7] = CR[7];\n", - "c40[8] = CR[8];\n", - "c45[9] = CR[9];\n", - "c50[10] = CR[10];\n", - "for i in range(2,11):\n", - " c5[i] = CR[i]-CR[i-1];\n", - "\n", - "for i in range(3,11):\n", - " c10[i] = CR[i]-CR[i-2];\n", - "\n", - "for i in range(4,11):\n", - " c15[i] = CR[i]-CR[i-3];\n", - "\n", - "for i in range(5,11):\n", - " c20[i] = CR[i]-CR[i-4];\n", - "\n", - "for i in range(6,11):\n", - " c25[i] = CR[i]-CR[i-5];\n", - "\n", - "for i in range(7,11):\n", - " c30[i] = CR[i]-CR[i-6];\n", - "\n", - "for i in range(8,11):\n", - " c35[i] = CR[i]-CR[i-7];\n", - "\n", - "for i in range(9,11):\n", - " c40[i] = CR[i]-CR[i-8];\n", - "\n", - "for i in range(10,11):\n", - " c45[i] = CR[i]-CR[i-9];\n", - " \t\t\t\t#rainfall in any possible time interval\n", - "\n", - "print \"5min 10min 15min 20min 25min 30min 35min 40min 45min 50min\";\n", - "for i in range(11):\n", - " print \"%4.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f\"%(c5[i],c10[i],c15[i],c20[i],c25[i],c30[i],c35[i],c40[i],c45[i],c50[i]);\n", - "\n", - "I = [17.04, 15.84, 14.80, 14.16, 13.39, 12.80, 12.27, 11.82, 11.39, 10.97]; \t\t\t\t#maximum intensity at respective durations\n", - "D = linspace(5,50,len(I)) \t\t\t\t#durations\n", - "#graph is plotted between I and D\n", - "plot(I,D)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "5min 10min 15min 20min 25min 30min 35min 40min 45min 50min\n", - "0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00\n", - "1.02 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00\n", - "1.06 2.08 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00\n", - "1.22 2.28 3.30 0.00 0.00 0.00 0.00 0.00 0.00 0.00\n", - "1.42 2.64 3.70 4.72 0.00 0.00 0.00 0.00 0.00 0.00\n", - "0.86 2.28 3.50 4.56 5.58 0.00 0.00 0.00 0.00 0.00\n", - "0.82 1.68 3.10 4.32 5.38 6.40 0.00 0.00 0.00 0.00\n", - "0.76 1.58 2.44 3.86 5.08 6.14 7.16 0.00 0.00 0.00\n", - "0.72 1.48 2.30 3.16 4.58 5.80 6.86 7.88 0.00 0.00\n", - "0.66 1.38 2.14 2.96 3.82 5.24 6.46 7.52 8.54 0.00\n", - "0.60 1.26 1.98 2.74 3.56 4.42 5.84 7.06 8.12 9.14\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 11, - "text": [ - "[<matplotlib.lines.Line2D at 0x103f7ab10>]" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10490ce90>" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.46 pg : 221" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math\n", - "from matplotlib.pylab import plot\n", - "#draw storm hyetograph and intensity duration curve\n", - "\t\t\t\t\n", - "#Given\n", - "p = [0, 5, 7.5, 8.5, 9]; \t\t\t\t#accumulated precipitation\n", - "t = [0, 30, 60, 90, 120]; \t\t\t\t#time\n", - "r = zeros(5)\n", - "I = zeros(5)\n", - "\n", - "print \"Rainfall intensity:\";\n", - "for i in range(1,5):\n", - " r[i] = p[i]-p[i-1]; \t\t\t\t#rainfall in succesive 30 min interval\n", - " I[i] = r[i]*60/30; \t\t\t\t#rainfall intensity\n", - " print \"%.2f\"%(I[i]);\n", - "\n", - "#graph is plotted between I and t.\n", - "plot(I,t)\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "Rainfall intensity:\n", - "10.00\n", - "5.00\n", - "2.00\n", - "1.00\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 12, - "text": [ - "[<matplotlib.lines.Line2D at 0x1047dced0>]" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10473bd10>" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.47 pg : 222" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math\n", - "from numpy import zeros,linspace\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "I = linspace(21,12,10) \t\t\t\t#isohytes\n", - "a = [543, 1345, 2030, 2545, 2955, 3280, 3535, 3710, 3880, 3915]; \t\t\t\t#enclosed area\n", - "ia = zeros(10)\n", - "ia[0] = 543;\n", - "for i in range(1,10):\n", - " ia[i] = a[i]-a[i-1]; \t\t\t\t#net incremental area between isohytes\n", - "\n", - "rv = zeros(10)\n", - "r = linspace(21.5,12.5,10) \n", - "for i in range(10):\n", - " rv[i] = r[i]*ia[i]; \t\t\t\t#rainfall volume\n", - "\n", - "cv = zeros(10)\n", - "cv[0] = 11675;\n", - "for i in range(10):\n", - " cv[i] = cv[i-1]+rv[i]; \t\t\t\t#cumulative volume\n", - "\n", - "eud = zeros(10)\n", - "for i in range(10):\n", - " eud[i] = cv[i]/a[i]; \t\t\t\t#depth(mm)\n", - "\n", - "\n", - "print \"From depth area curve we obtain average depth of precipitation = 20.15 mm for anarea of 2400 sq. km.\";\n", - "#graph is plotted between eud and a\n", - "plot(eud,a)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "From depth area curve we obtain average depth of precipitation = 20.15 mm for anarea of 2400 sq. km.\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 13, - "text": [ - "[<matplotlib.lines.Line2D at 0x1047dcf90>]" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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DZDtlNYCVwB5gBVA94PgxwF4gF+hZxt8WEYk7tWvDypU2YzktzYaihkNZK4NC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\n2EwAAAAASUVORK5CYII=\n", - "text": [ - "<matplotlib.figure.Figure at 0x104a0a190>" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.48 pg : 222" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "h1 = 7.75; \t\t\t\t#initial depth of water\n", - "r = 3.80; \t\t\t\t#rainfall during the week\n", - "hr = 2.50; \t\t\t\t#depth of water removed\n", - "C = 0.7; \t\t\t\t#pan coefficient\n", - "\n", - "# Calculations\n", - "ha = r-hr;\n", - "hl = ha+h1;\n", - "h2 = 8.32;\n", - "ev = hl-h2;\n", - "evs = ev*C;\n", - "evs = round(evs*100)/100;\n", - "\n", - "# Results\n", - "print \"evaporation from reservior surface during the week = %.2f cm.\"%(evs);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "evaporation from reservior surface during the week = 0.51 cm.\n" - ] - } - ], - "prompt_number": 21 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.49 pg : 223" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "T = linspace(1,12,12) \t\t\t\t#time from start\n", - "r = [1.8, 2.6, 7.8, 3.9, 10.6, 5.4, 7.8, 9.2, 6.5, 4.4, 1.8, 1.6]; \t\t\t\t#increamental rainfall\n", - "R = 24.4; \t\t\t\t#total run-off\n", - "s = sum(r)\n", - "\n", - "ti = s-R;\n", - "\n", - "#first trial\n", - "tr = 7; \t\t\t\t#assumed\n", - "ti = s-R-r[0]-r[1]-r[3]-r[10]-r[11];\n", - "fi = ti/tr;\n", - "P = zeros(12)\n", - "for i in range(12):\n", - " P[i] = r[i]-fi;\n", - " if (P[i]<0):\n", - " P[i] = 0;\n", - "\n", - "print \"Timeh rainfall excess.\";\n", - "for i in range(12):\n", - " print \"%.2f %.2f\"%(T[i],P[i]);\n", - "\n", - "print \"fi index = %.2f cm/hr.\"%(fi);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Timeh rainfall excess.\n", - "1.00 0.00\n", - "2.00 0.00\n", - "3.00 3.90\n", - "4.00 0.00\n", - "5.00 6.70\n", - "6.00 1.50\n", - "7.00 3.90\n", - "8.00 5.30\n", - "9.00 2.60\n", - "10.00 0.50\n", - "11.00 0.00\n", - "12.00 0.00\n", - "fi index = 3.90 cm/hr.\n" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.50 pg : 224" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace\n", - "\t\t\t\t\n", - "#Given\n", - "r = [0.6, 1.35, 2.25, 3.45, 2.7, 2.4, 1.5, 0.75]; \t\t\t\t#incremental rainfall\n", - "T = linspace(1,8,8) \t\t\t\t#time from start of rainfal\n", - "t = 8.;\n", - "P = 15.; \t\t\t\t#total rainfall\n", - "R = 8.7; \t\t\t\t#direct run-off\n", - "\n", - "# Calculations\n", - "W = (P-R)/t;\n", - "#math.since fi wil be more than W\n", - "tre = 6;\n", - "fi = ((P-R)-r[0]-r[7])/tre;\n", - "\n", - "# Results\n", - "print \"fi index = %.2f cm/hr.\"%(fi);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "fi index = 0.83 cm/hr.\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.51 pg : 224" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from scipy.integrate import quad \n", - "\t\t\t\t\n", - "#Given\n", - "I = 10; \t\t\t\t#total infiltration rate\n", - "fI = 5; \t\t\t\t#final infiltration rate\n", - "k = 0.95; \t\t\t\t#rate of decay of difference between final and initial infiltration rate\n", - "\n", - "\n", - "# Calculations\n", - "def f8(t): \n", - "\t return fI+(I-fI)*math.e**(-k*t)\n", - "\n", - "q = quad(f8,0,6)[0]\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"total infiltration depth = %.2f mm.\"%(q);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "total infiltration depth = 35.25 mm.\n" - ] - } - ], - "prompt_number": 25 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.52 pg : 224" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace\n", - "\n", - "#find the equation of infiltration capacity\n", - "\t\t\t\t\n", - "#Given\n", - "fc = 1; \t\t\t\t#consmath.tant infiltration rate\n", - "ft = [10.4, 5.6, 3.2, 2.1, 1.5, 1.2, 1.1, 1, 1]; \t\t\t\t#infiltration capacity\n", - "f = ft[0]-fc;\n", - "t = linspace(0,2,9)\n", - "\n", - "r = zeros(9)\n", - "for i in range(9):\n", - " r[i] = ft[i]-fc;\n", - "\n", - "h = zeros(7)\n", - "for i in range(7):\n", - " h[i] = math.log10(r[i]);\n", - "\n", - "s = 0.775; \t\t\t\t#from graph\n", - "k = 1/(math.log10(math.e)*s);\n", - "k = round(k*100)/100;\n", - "\n", - "# Results\n", - "print \"Equation is:ft = fc+%.2fe**-%.2ft)\"%(f,k);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Equation is:ft = fc+9.40e**-2.97t)\n" - ] - } - ], - "prompt_number": 26 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.53 pg : 226" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "from matplotlib.pylab import bar\n", - "import math \n", - "#total rainfall\n", - "#net run-off\n", - "#W index\n", - "\n", - "#Given\n", - "r = [2, 2, 8, 7, 1.25, 1.25, 4.5]; \t\t\t\t#rainfall intensity\n", - "T = [15, 30, 45, 60, 70, 90, 105]; \t\t\t\t#time\n", - "dt = 15.; \t\t\t\t#time interval\n", - "fi = 3.; \t\t\t\t#fi index\n", - "#graph is plotted between r and T\n", - "bar(T,r)\n", - "s = sum(r)\n", - "P = s*dt/60;\n", - "Pe = ((8-3)+(7-3)+(4.5-3))*dt/60; \t\t\t\t#area of graph above r = 3.0.\n", - "w = (P-Pe)/(105./60);\n", - "w = round(w*1000)/1000;\n", - "\n", - "# Results\n", - "print \"total rainfall = %.2f cm.\"%(P);\n", - "print \"net run-off = %.2f cm.\"%(Pe);\n", - "print \"W index = %.2f cm/hr.\"%(w);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "total rainfall = 6.50 cm." - ] - }, - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "net run-off = 2.62 cm.\n", - "W index = 2.21 cm/hr.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x104b42510>" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.54 pg : 226" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#run-off by rainfall of 3.3cm in 3hrs\n", - "\t\t\t\t\n", - "#Given\n", - "A = [36, 18, 66]; \t\t\t\t#area of catchment\n", - "fi = [0.9 ,1.1, 0.5]; \t\t\t\t#fi index\n", - "r1 = [0.6 ,0.9, 1.0]; \t\t\t\t#rainfall in first hour\n", - "r2 = [2.4 ,2.1, 2.0]; \t\t\t\t#rainfall in second hour\n", - "r3 = [1.3, 1.5, 0.9]; \t\t\t\t#rainfall in third hour\n", - "\n", - "# Calculations and Results\n", - "t36 = r1[0]+r2[0]+r3[0];\n", - "t18 = r1[1]+r2[1]+r3[1];\n", - "t66 = r1[2]+r2[2]+r3[2];\n", - "\n", - "p = (t36*A[0]+t18*A[1]+t66*A[2])/(A[0]+A[1]+A[2]);\n", - "print \"Total rainfall in catchment = %.2f cm.\"%(p);\n", - "\n", - "ro1 = [0 ,0, 0.5];\n", - "ro2 = [1.5 ,1.0, 1.5];\n", - "ro3 = [0.4, 0.4, 0.4]; \t\t\t\t#rainfall-fi\n", - "t1 = ro1[0]+ro2[0]+ro3[0];\n", - "t2 = ro1[1]+ro2[1]+ro3[1];\n", - "t3 = ro1[2]+ro2[2]+ro3[2];\n", - "run = (A[0]*t1+A[1]*t2+A[2]*t3)/(A[0]+A[1]+A[2]); \t\t\t\t#run-off from entire catchment\n", - "\n", - "print \"run-off by rainfall of 3.3cm in 3hrs = %.2f cm.\"%(run);\n", - "\n", - "fia = (fi[0]*A[0]+fi[1]*A[1]+fi[2]*A[2])/(A[0]+A[1]+A[2]);\n", - "tr = (1.1-fia)*3;\n", - "print \"Total run-off = %.2f cm.\"%(tr);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Total rainfall in catchment = 4.11 cm.\n", - "run-off by rainfall of 3.3cm in 3hrs = 2.10 cm.\n", - "Total run-off = 1.17 cm.\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.55 pg : 227" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "P = array([4, 22, 28, 15, 12, 8, 4, 15, 10, 5]); \t\t\t\t#Precipitation\n", - "R = array([0.2, 7.1, 10.9, 4.0, 3.0, 1.3, 0.4, 4.1, 2.0, 0.3]); \t\t\t\t#runoff\n", - "\n", - "# Calculations\n", - "Ps = P**2\n", - "Rs = R**2\n", - "PR = P*R\n", - "\n", - "s = sum(Ps)\n", - "t = sum(Rs)\n", - "u = sum(PR)\n", - "q = sum(P)\n", - "w = sum(R)\n", - "\n", - "N = 10.;\n", - "a = (N*u-q*w)/(N*s-q**2);\n", - "b = (w-a*q)/N;\n", - "a = round(a*10000)/10000;\n", - "b = round(b*10000)/10000;\n", - "\n", - "# Results\n", - "print \"Equation is:%.2f P %.2f.\"%(a,b);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Equation is:0.43 P -1.93.\n" - ] - } - ], - "prompt_number": 33 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.56 pg : 228" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 470; \t\t\t\t#peak discharge of flood hydrograph\n", - "B = 15; \t\t\t\t#base flow\n", - "l = 0.25; \t\t\t\t#infiltration loss\n", - "Qr = Q-B;\n", - "d = 8; \t\t\t\t#average depth of rainfall\n", - "\n", - "# Calculations\n", - "re = d-l*6; \t\t\t\t#rainfall excess\n", - "q = Qr/re;\n", - "\n", - "# Results\n", - "print \"peak discharge of 6 hrs unit hydrograph = %i cumecs.\"%(q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "peak discharge of 6 hrs unit hydrograph = 70 cumecs.\n" - ] - } - ], - "prompt_number": 117 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.57 pg : 228" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array,zeros\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "fi = 0.25; \t\t\t\t#infiltration index\n", - "B = 20; \t\t\t\t#Base flow\n", - "O = array([0, 20, 60, 150, 120, 90, 70, 50, 30, 20, 10, 0, 0, 0]); \t\t\t\t#ordinates of unit hydrograph\n", - "R1 = 5;\n", - "R2 = 0.8;\n", - "R3 = 3;\n", - "r1 = R1-(fi*4); \t\t\t\t#rainfall excess in first four hour\n", - "r2 = R2-(fi*4); \t\t\t\t#rainfall excess in second four hour\n", - "r3 = R3-(fi*4); \t\t\t\t#rainfall excess in third four hour\n", - "if r2<0 :\n", - " r2 = 0;\n", - "\n", - "# Calculations and Results\n", - "s1 = r1*O\n", - "s2 = zeros(14)\n", - "for i in range(1,14):\n", - " s2[i] = r2*O[i-1];\n", - "\n", - "s3 = zeros(14)\n", - "for i in range(2,14):\n", - " s3[i] = r3*O[i-2];\n", - "#surface run-off from rainfall excess during succesive unit periods\n", - "print \"ordinates of storm hydrograph\";\n", - "T = zeros(14)\n", - "t = zeros(14)\n", - "for i in range(14):\n", - " T[i] = s1[i]+s2[i]+s3[i]; \t\t\t\t#sub-total\n", - " t[i] = T[i]+B; \t\t\t\t#ordinate of flood hydrograph\n", - " print \"%i\"%(t[i]);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of storm hydrograph\n", - "20\n", - "100\n", - "260\n", - "660\n", - "620\n", - "680\n", - "540\n", - "400\n", - "280\n", - "200\n", - "120\n", - "60\n", - "40\n", - "20\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.58 pg : 229" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array\n", - "\t\t\t\t\n", - "#Given\n", - "fi = 2.5; \t\t\t\t#fi index\n", - "t = 24.;\n", - "A = 200.; \t\t\t\t#area of catchment\n", - "R1 = 7.5;\n", - "R2 = 2.0;\n", - "R3 = 5.; \t\t\t\t#rainfall\n", - "r1 = R1-fi;\n", - "r2 = R2-fi;\n", - "r3 = R3-fi;\n", - "r2 = 0;\n", - "r = [5, 0, 2.5]; \t\t\t\t#excess rainfall\n", - "D = array([5 ,15, 40, 25, 10, 5, 0, 0, 0]); \t\t\t\t#distribution\n", - "d1 = D*r[0]/100\n", - "d2 = zeros(9)\n", - "for i in range(8):\n", - " d2[i+1] = D[i]*r[1]/100;\n", - "\n", - "d3 = zeros(9)\n", - "for i in range(7):\n", - " d3[i+2] = D[i]*r[2]/100;\n", - "#distribution run-off for rainfall excess\n", - "\n", - "tr1 = zeros(9)\n", - "tr2 = zeros(9)\n", - "for i in range(9):\n", - " tr1[i] = d1[i]+d2[i]+d3[i]; \t\t\t\t#total run-off as depth\n", - " tr2[i] = 23.148*tr1[i]; \t\t\t\t#total run-off as discharge\n", - " tr2[i] = round(tr2[i]*1000)/1000;\n", - "\n", - "s = sum(tr2)\n", - "\n", - "print \"Total run-off:\";\n", - "print \"as depth as discharge\";\n", - "for i in range(9):\n", - " print \"%.2f %.2f\"%(tr1[i],tr2[i]);\n", - "\n", - "r = 0.36*s*t/A;\n", - "r = round(r*10)/10;\n", - "print \"total run-off = %.2f cm.\"%(r);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Total run-off:\n", - "as depth as discharge\n", - "0.00 0.00\n", - "0.00 0.00\n", - "2.12 49.19\n", - "1.38 31.83\n", - "1.00 23.15\n", - "0.62 14.47\n", - "0.25 5.79\n", - "0.12 2.89\n", - "0.00 0.00\n", - "total run-off = 5.50 cm.\n" - ] - } - ], - "prompt_number": 35 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.59 pg : 230" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "O = [10, 30, 90, 220, 280, 220, 166, 126, 92, 62, 40, 20, 10]; \t\t\t\t#ordinates of 6 hr flood hydrograph\n", - "B = 10; \t\t\t\t#Base flow\n", - "r = zeros(13)\n", - "\n", - "for i in range(13):\n", - " r[i] = O[i]-B; \t\t\t\t#ordinates of direct run-off\n", - "\n", - "print \"Ordinates of 6 hr unit hydrograph\";\n", - "u = zeros(13)\n", - "\n", - "for i in range(1,13):\n", - " u[i] = r[i]-u[i-1]; \t\t\t\t#ordinates of 6 hrs unit hydrograph\n", - "\n", - "for i in range(13): \n", - " print \"%i\"%(u[i]);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Ordinates of 6 hr unit hydrograph\n", - "0\n", - "20\n", - "60\n", - "150\n", - "120\n", - "90\n", - "66\n", - "50\n", - "32\n", - "20\n", - "10\n", - "0\n", - "0\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.60 pg : 231" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "#determine the ordinates of 1 cm-6 hour hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "t = 6;\n", - "A = 450; \t\t\t\t#catchment area\n", - "O = [5, 15, 40, 80, 60, 50, 25, 15, 5]; \t\t\t\t#ordinates of flood hydrograph\n", - "B = 5; \t\t\t\t#base flow assumed\n", - "s = 0;\n", - "r = zeros(9)\n", - "for i in range(9):\n", - " r[i] = O[i]-B; \t\t\t\t#ordinates of direct run-off\n", - " s = s+r[i];\n", - "\n", - "n = s*0.36*12/A;\n", - "print \"ordinates of unit hydrograph\";\n", - "for i in range(9):\n", - " u[i] = r[i]/n;\n", - " u[i] = round(u[i]*100)/100;\n", - " print \"%.2f\"%(u[i]);\n", - "\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates of unit hydrograph\n", - "0.00\n", - "4.17\n", - "14.58\n", - "31.25\n", - "22.92\n", - "18.75\n", - "8.33\n", - "4.17\n", - "0.00\n" - ] - } - ], - "prompt_number": 37 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.61 pg : 232" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "#obtain ordinates 24 hr unit hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "O = [0, 5.5, 13.5, 26.5, 45, 82, 162, 240, 231, 165, 112, 79, 57, 42, 31, 22, 14, 9.5, 6.6, 4, 2, 1, 0, 0, 0, 0, 0]; \t\t\t\t#ordinates of 1st 8 hrs unit hydrograph\n", - "o1 = zeros(27)\n", - "o2 = zeros(29)\n", - "for i in range(25):\n", - " o1[i+2] = O[i]; \t\t\t\t#ordinates of 2nd 8 hrs unit hydrograph\n", - " o2[i+4] = O[i]; \t\t\t\t#ordinates of 3rd 8 hrs unit hydrograph\n", - "\n", - "o3 = zeros(27)\n", - "t = zeros(27)\n", - "print \"ordinates 24 hr unit hydrograph:\";\n", - "for i in range(27):\n", - " o3[i] = o1[i]+o2[i]+O[i]; \t\t\t\t#total 24 hr hydrograph of 3 cm run-off\n", - " t[i] = o3[i]/3;\n", - " t[i] = round(t[i]*10)/10;\n", - " print \"%.2f\"%(t[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ordinates 24 hr unit hydrograph:\n", - "0.00\n", - "1.80\n", - "4.50\n", - "10.70\n", - "19.50\n", - "38.00\n", - "73.50\n", - "116.20\n", - "146.00\n", - "162.30\n", - "168.30\n", - "161.30\n", - "133.30\n", - "95.30\n", - "66.70\n", - "47.70\n", - "34.00\n", - "24.50\n", - "17.20\n", - "11.80\n", - "7.50\n", - "4.80\n", - "2.90\n", - "1.70\n", - "0.70\n", - "0.30\n", - "0.00\n" - ] - } - ], - "prompt_number": 38 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.62 pg : 233" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "from numpy import zeros,linspace\n", - "\n", - "\n", - "#ordinates of 1 hr unit hydrograph\n", - "\t\t\t\t\n", - "#Given\n", - "t = linspace(0,12,13) \t\t\t#time\n", - "O = [0, 0, 54, 0, 175, 0, 127, 0, 58, 0, 25, 0, 0, 0]; \t\t\t\t#ordinate of 2 hr unit hydrograph\n", - "of = zeros(13)\n", - "for i in range(2,13):\n", - " if (i%2) == 0:\n", - " of[i] = 0\n", - " else:\n", - " of[i] = O[i-2]+of[i-2];\n", - "\n", - "s = [0, 25, 54, 120, 229, 300, 356, 390, 414, 430, 439, 439, 439]; \t\t\t\t#Ordinates of S-curve\n", - "of1 = zeros(13)\n", - "for i in range(1,13):\n", - " of1[i] = s[i-1];\n", - "\n", - "y = zeros(13)\n", - "u = zeros(13)\n", - "print \"ordinates of 1 hr unit hydrograph:\";\n", - "for i in range(13):\n", - " y[i] = s[i]-of1[i];\n", - " u[i] = y[i]*2;\n", - " print \"%i\"%(u[i]);\n", - "\n", - "#graph is plotted between u and t\n", - "plot(t,u)\n", - "\n", - "\n", - "# graph in book is wrong. Please check." - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "ordinates of 1 hr unit hydrograph:\n", - "0\n", - "50\n", - "58\n", - "132\n", - "218\n", - "142\n", - "112\n", - "68\n", - "48\n", - "32\n", - "18\n", - "0\n", - "0\n" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 15, - "text": [ - "[<matplotlib.lines.Line2D at 0x104903850>]" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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9vkUL11EVLBbTm+2BdKwGD3A/cBz4Q677bAACWH0XEYmpjcC5LgMoEw4iBSgH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\n", - "text": [ - "<matplotlib.figure.Figure at 0x1049b8950>" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.63 pg : 234" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "xavg = 1200; \t\t\t\t#sample mean\n", - "n = 50.; \t\t\t\t#assurance year\n", - "A = 0.95; \t\t\t\t#assurance percent\n", - "Rsk = 1-A;\n", - "sigma = 650; \t\t\t\t#smath.radians(numpy.arcmath.tan(ard deviation\n", - "yn = 0.53622; \t\t\t\t#mean of reduced variate\n", - "sigma30 = 1.11238; \t\t\t\t#smath.radians(numpy.arcmath.tan(ard deviation of reduced variate\n", - "\n", - "# Calculations\n", - "T = 1/(1-(1-Rsk)**(1/n));\n", - "yt = -2.303*math.log10(2.303*math.log10(T/(T-1)));\n", - "K = (yt-yn)/sigma30;\n", - "xt = xavg+K*sigma;\n", - "\n", - "# Results\n", - "print \" design disharge = %i cumecs.\"%(xt);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " design disharge = 4908 cumecs.\n" - ] - } - ], - "prompt_number": 43 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.64 pg : 235" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "T1 = 50.;\n", - "T2 = 100.; \t\t\t\t#Return period\n", - "F1 = 20600.;\n", - "F2 = 22150; \t\t\t\t#Peak flood\n", - "\n", - "# Calculations\n", - "y100 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));\n", - "y50 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));\n", - "y = (F2-F1)/(y100-y50);\n", - "T = 500.;\n", - "y500 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "x500 = F2+(y500-y100)*y;\n", - "x500 = round(x500);\n", - "\n", - "# Results\n", - "print \"flood magnitude with return period of 240 years = %.2f cumec.\"%(x500);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "flood magnitude with return period of 240 years = 25732.00 cumec.\n" - ] - } - ], - "prompt_number": 44 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.65 pg : 235" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "xavg = 1.65; \t\t\t\t#mean of data\n", - "sigma = 0.45; \t\t\t\t#smath.radians(numpy.arcmath.tan(ard deviation\n", - "x = 3;\n", - "\n", - "# Calculations\n", - "y = 1.2825*(x-xavg)/sigma+0.577;\n", - "l = math.e**(math.e**(-y));\n", - "T = l/(l-1);\n", - "T = round(T*10)/10;\n", - "\n", - "# Results\n", - "print \"recurrence interval of 10 minutes storm = %.2f years.\"%(T);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "recurrence interval of 10 minutes storm = 84.00 years.\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.66 pg : 236" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "T1 = 50.;\n", - "T2 = 100.; \t\t\t\t#Return period\n", - "F1 = 30800.;\n", - "F2 = 36300.; \t\t\t\t#Peak flood\n", - "\n", - "# Calculations\n", - "y100 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));\n", - "y50 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));\n", - "y = (F2-F1)/(y100-y50);\n", - "T = 200.;\n", - "y200 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "x200 = F2+(y200-y100)*y;\n", - "x200 = round(x200);\n", - "\n", - "# Results\n", - "print \"flood magnitude with return period of 240 years = %.2f cumecs.\"%(x200);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "flood magnitude with return period of 240 years = 41780.00 cumecs.\n" - ] - } - ], - "prompt_number": 46 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.67 pg : 236" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "xavg = 4200.; \t\t\t\t#mean\n", - "sigma = 1705.; \t\t\t\t#smath.radians(numpy.arcmath.tan(ard deviation\n", - "xt = 9550.; \t\t\t\t#flood value\n", - "\n", - "# Calculations\n", - "K = (xt-xavg)/sigma;\n", - "yt = 1.2825*K+0.577;\n", - "l = math.e**(math.e**(-yt));\n", - "T = l/(l-1);\n", - "\n", - "# Results\n", - "print \"Return period of flood of 9950 cumec/s = %.2f years.\"%(T);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Return period of flood of 9950 cumec/s = 100.11 years.\n" - ] - } - ], - "prompt_number": 49 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.68 pg : 237" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "T1 = 100.;\n", - "T2 = 50.; \t\t\t\t#Return period\n", - "F1 = 485.;\n", - "F2 = 445.; \t\t\t\t#Peak flood\n", - "\n", - "# Calculations\n", - "y50 = -(2.303*math.log10(2.303*math.log10(T2/(T2-1))));\n", - "y100 = -(2.303*math.log10(2.303*math.log10(T1/(T1-1))));\n", - "y = (F2-F1)/(y50-y100);\n", - "T = 1000.;\n", - "y1000 = -(2.303*math.log10(2.303*math.log10(T/(T-1))));\n", - "x1000 = F2+(y1000-y50)*y;\n", - "x1000 = round(x1000*10)/10;\n", - "\n", - "# Results\n", - "print \"flood magnitude with return period of 240 years = %.2f cumecs.\"%(x1000);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "flood magnitude with return period of 240 years = 617.20 cumecs.\n" - ] - } - ], - "prompt_number": 50 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 4.69 pg : 238" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#probability of exceedence\n", - "#probability of occurence in next 12 years\n", - "\n", - "#Given\n", - "T = 25.; \t\t\t\t#return period\n", - "n = 12.;\n", - "\n", - "# Calculations\n", - "P = 1/T;\n", - "Rsk = 1-(1-P)**n;\n", - "P = round(P*100)/100;\n", - "Rsk = round(Rsk*10000)/10000;\n", - "\n", - "# Results\n", - "print \"probability of exceedence = %.2f.\"%(P);\n", - "print \"probability of occurence in next 12 years = %.2f.\"%(Rsk);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "probability of exceedence = 0.04.\n", - "probability of occurence in next 12 years = 0.39.\n" - ] - } - ], - "prompt_number": 51 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch5.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch5.ipynb deleted file mode 100755 index 20ed9e6b..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch5.ipynb +++ /dev/null @@ -1,1215 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:354821165ad32a4fdc06d2826d3a09936c00989f455a7161676ce1b32639acba" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 5 : GROUND WATER WELL IRRIGATION" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.1 pg : 277" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t#design an open wellin fine sand\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 0.003; \t\t\t\t#required discharge\n", - "H = 2.5; \t\t\t\t#depression head\n", - "A = Q*3600/(0.5*H);\n", - "d = (4*A/math.pi)**0.5;\n", - "d = round(d*100)/100\n", - "print \"Well diameter = %.2f m.\"%(d);\n", - "\n", - "\t\t\t\t#Alternative solution\n", - "C = 7.5e-5; \t\t\t\t#permeability consmath.tant from table 5.2\n", - "A = Q/(C*H);\n", - "d = (16*3/math.pi)**0.5;\n", - "d = round(d*10)/10;\n", - "print \"By alternative solution:\"\n", - "print \"Well diameter = %.2f m\"%(d);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Well diameter = 3.32 m.\n", - "By alternative solution:\n", - "Well diameter = 3.90 m\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.2 pg : 278" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#yield from well\n", - "#diameter of well\n", - "\n", - "#Given\n", - "h1 = 2.5; \t\t\t\t#initial pumping depression\n", - "h = 1.8; \t\t\t\t#heigth after recuperation\n", - "t = 80; \t\t\t\t#time\n", - "h2 = h1-h;\n", - "KbyA = 2.303*60*math.log10(h1/h2)/t;\n", - "\n", - "\n", - "# Calculations and Results\n", - "#Part (a)\n", - "d = 4; \t\t\t\t#diameter of well\n", - "H = 3; \t\t\t\t#depression head\n", - "A = math.pi*d**2/4;\n", - "Q = (KbyA)*A*H/3.6;\n", - "print \"Part a\";\n", - "Q = round(Q);\n", - "print \"Yield from well = %.2f lit/sec.\"%(Q);\n", - "\n", - "#Part (b)\n", - "Q = 8; \t\t\t\t#yield(lit/sec)\n", - "H = 2;\n", - "A = Q*3.6/(H*(KbyA));\n", - "d = (4*A/math.pi)**0.5;\n", - "d = round(d*10)/10;\n", - "print \"Part b\";\n", - "print \"Daimeter of well = %.2f m\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Part a\n", - "Yield from well = 10.00 lit/sec.\n", - "Part b\n", - "Daimeter of well = 4.40 m\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.3 pg : 279" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given\n", - "d = 30; \t\t\t\t#well diameter\n", - "L = 15; \t\t\t\t#strainer length\n", - "P = 50; \t\t\t\t#coefficient of permeability\n", - "s = 0.2; \t\t\t\t#effective size of sand\n", - "b = 3; \t\t\t\t#drawdown\n", - "r = 150; \t\t\t\t#radius of drawdown\n", - "\n", - "\n", - "# Calculations\n", - "Q = 2.72*L*P*b/(math.log10(r*2*100/d)*24*3.6);\n", - "Q = round(Q*10)/10;\n", - "\n", - "# Results\n", - "print \" yield from well = %.2f lit/sec.\"%(Q);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " yield from well = 23.60 lit/sec.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.4 pg : 280" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "d = 30; \t\t\t\t#diameter of well\n", - "s = 2; \t\t\t\t#drawdown\n", - "L = 10; \t\t\t\t#length of stainer\n", - "k = 0.05; \t\t\t\t#coefficient of permeability\n", - "r = 300; \t\t\t\t#radius of zero drawdown\n", - "\n", - "# Calculations\n", - "Q = 2.72*k*s*(L+s/2)/(100*math.log10(2*100*r/d));\n", - "\n", - "# Results\n", - "print \" discharge from tubewell = %.4f cumec.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " discharge from tubewell = 0.0091 cumec.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5 pg: 280" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t#design tube well\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 0.08; \t\t\t\t#yield required\n", - "b = 30; \t\t\t\t#thickness of acquifer\n", - "R = 300; \t\t\t\t#Radius of circle of influence\n", - "k = 60; \t\t\t\t#permeability coefficient\n", - "s = 5; \t\t\t\t#Drawdown\n", - "\n", - "# Calculations\n", - "r = R/(10**(2.72*b*s*k/(3600*24*Q)));\n", - "r = round(r*10000)/10000;\n", - "\n", - "# Results\n", - "print \"Radius of well = %.2f m\"%(r);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Radius of well = 0.09 m\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.6 pg : 281" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "b = 30.; \t\t\t\t#thickness of acquifer\n", - "s = 4.; \t\t\t\t#drawdown\n", - "r = 0.1; \t\t\t\t#well radius\n", - "k = 36.; \t\t\t\t#permeability coefficient\n", - "R = 3000*s*(k/(24*3600))**0.5;\n", - "\n", - "Q = 2.72*b*k*s/(math.log10(R/r)*24*3.6);\n", - "Q = round(Q*10)/10;\n", - "print \"yield from well = %.2f lit/sec.\"%(Q);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "yield from well = 40.10 lit/sec.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.7 pg : 281" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "k = 0.005; \t\t\t\t#coefficient of permeability\n", - "r = 0.1; \t\t\t\t#well radius\n", - "s = 4; \t\t\t\t#drawdown\n", - "b = 10; \t\t\t\t#thickness\n", - "R = 300; \t\t\t\t#radius of circle of influence\n", - "\n", - "# Calculations and Results\n", - "#Part(a)\n", - "Q1 = 2.72*b*k*s/math.log10(R/r);\n", - "Q1 = round(Q1*10000)/10000;\n", - "print \"Discharge = %.2f cumec\"%(Q1);\n", - "\n", - "#Part (b)\n", - "r = 0.2;\n", - "Q2 = 2.72*b*k*s/math.log10(R/r);\n", - "I = (Q2-Q1)*100/Q1;\n", - "I = round(I*10)/10;\n", - "print \"percent increase in discharge = %.2f percent.\"%(I);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge = 0.16 cumec\n", - "percent increase in discharge = 9.40 percent.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.8 pg : 282" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#percentage error\n", - "#actual radius of influence\n", - "\n", - "#Given\n", - "d = 0.2; \t\t\t\t#diameter of well\n", - "Q = 240; \t\t\t\t#discharge\n", - "RL1 = 240.5; \t\t\t\t#reduce level of original water surface\n", - "RL2 = 235.6; \t\t\t\t#reduced level of water at pumping\n", - "RL3 = 210; \t\t\t\t#reduced level of impervious layer\n", - "RL4 = 239.8; \t\t\t\t#reduced level of water in well\n", - "D = 50; \t\t\t\t#radial dismath.tance of well from tube well\n", - "\n", - "# Calculations and Results\n", - "#Part(a)\n", - "h1 = RL2-RL3;\n", - "h2 = RL4-RL3;\n", - "k1 = Q*24*math.log10(D*2/d)/(1.36*(h2**2-h1**2));\n", - "k1 = round(k1*100)/100;\n", - "print \"Parta\";\n", - "print \"coefficient of permeability = %.2f m/day.\"%(k1);\n", - "#Part (b)\n", - "R = 300; \t\t\t\t#radius of influence\n", - "H = RL1-RL3;\n", - "h = RL2-RL3;\n", - "k2 = Q*24*math.log10(R*2/d)/(1.36*(H**2-h**2));\n", - "PE = (k2-k1)*100/k1;\n", - "print \"Partb\";\n", - "print \"percentage error = %i percent.\"%(PE);\n", - "#Part (b)\n", - "R = (d/2)*10**(1.36*k1*(H**2-h**2)/(24*Q));\n", - "print \"Partc\";\n", - "print \"Actual radius of influence = %i m.\"%(R);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Parta\n", - "coefficient of permeability = 49.13 m/day.\n", - "Partb\n", - "percentage error = 9 percent.\n", - "Partc\n", - "Actual radius of influence = 154 m.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.9 pg : 283" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "A = 20.; \t\t\t\t#area of field\n", - "H = 129.; \t\t\t\t#level to the highest land\n", - "h1 = 120.2; \t\t\t\t#water level in well during discharge\n", - "Du = 800; \t\t\t\t#duty for rise;\n", - "eita = 0.6; \t\t\t\t#efficiency of the pump\n", - "\n", - "# Calculations\n", - "Q = A/Du;\n", - "w = Q*1000;\n", - "lift = H-h1;\n", - "\t\t\t\t#design lift is taken as 9m\n", - "wd = w*9;\n", - "o = wd/75;\n", - "i = o/eita;\n", - "\n", - "# Results\n", - "print \"Input h.p of pump = %i h.p\"%i;\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Input h.p of pump = 5 h.p\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.10 pg : 284" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 150; \t\t\t\t#discharge from tubewell\n", - "t = 4000; \t\t\t\t#working period of tubewell\n", - "I = 0.45; \t\t\t\t#intensity of irrigation\n", - "d = 0.38; \t\t\t\t#average depth of rabi and kharif crop\n", - "\n", - "# Calculations\n", - "V = Q*t;\n", - "A = V/d;\n", - "CA = A/(I*10000);\n", - "CA = round(CA);\n", - "\n", - "# Results\n", - "print \"culturable area = %.2f hectares.\"%(CA);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "culturable area = 351.00 hectares.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.11 pg : 284" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#percent decrease when two well discharges\n", - "\n", - "#Given\n", - "d = 0.2; \t\t\t\t#diameter of well\n", - "r = d/2;\n", - "B = 100; \t\t\t\t#dismath.tance between wells\n", - "b = 12; \t\t\t\t#thickness of acquifer\n", - "k = 60; \t\t\t\t#coefficient of permeability\n", - "s = 3; \t\t\t\t#print ersion head\n", - "R = 250; \t\t\t\t#radius of influence\n", - "Q = 2.72*b*k*s/(24*math.log10(R/r));\n", - "print \"discharge if one well discharges = %i cubic metre/hour.\"%(Q);\n", - "#when both well are discharging\n", - "Q1 = 2.72*k*b*s/(24*math.log10(R**2/(r*B)));\n", - "Q1 = round(Q1*10)/10;\n", - "print \"discharge if both wells discharges = %.2f cubic metre/hour.\"%(Q1);\n", - "PE = (Q-Q1)*100/Q;\n", - "PE = round(PE*100)/100;\n", - "print \"percentage decrease in discharge = %.2f percent.\"%(PE);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge if one well discharges = 72 cubic metre/hour.\n", - "discharge if both wells discharges = 64.50 cubic metre/hour.\n", - "percentage decrease in discharge = 10.47 percent.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.12 pg : 285" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#coefficient of permeability\n", - "#drawdown in well\n", - "#specific capacity\n", - "#maximum rate at which water can be pumped\n", - "\n", - "#Given\n", - "d = 0.6; \t\t\t\t#diameter of well;\n", - "rw = d/2;\n", - "H = 40.; \t\t\t\t#depth of water in well before pumping\n", - "Q = 2000.; \t\t\t\t#discharge from well\n", - "s1 = 4.; \t\t\t\t#drawdown in well\n", - "B1 = 10.; \t\t\t\t#dismath.tance between well\n", - "s2 = 2.;\n", - "B2 = 20.;\n", - "#Part (a)\n", - "h1 = H-s1;\n", - "h2 = H-s2;\n", - "t = (H**2-h2**2)/(H**2-h1**2);\n", - "R = (B2/(B1**t))**(1/(1-t));\n", - "R = round(R*100)/100;\n", - "print \" radius of zero drawdown = %.2f m\"%(R);\n", - "#Part (b)\n", - "r = 10;\n", - "k = Q*math.log10(R/r)*60*24/(1.36*(H**2-h1**2)*1000);\n", - "k = round(k*100)/100;\n", - "print \"coefficient of permeability = %.2f m/day.\"%(k);\n", - "\n", - "#part (c)\n", - "Ho = (H**2-(Q*math.log10(R/rw)*24*60/(1000*1.36*k)))**0.5;\n", - "D = H-Ho;\n", - "D = round(D*100)/100;\n", - "print \"drawdown in well = %.2f m.\"%(D);\n", - "\n", - "#part (d)\n", - "C = Q/(1000*R);\n", - "#for R = 1 m;Q = Sc\n", - "#hence on putting the values in discharge equation we get\n", - "#Sc*math.log10(61.2*Sc) = 0.3223.\n", - "#on solving this by trial and error method we get Sc = 0.266 m**2/min.\n", - "print \"Specific capacity = 0.266 cubic metre/minutes/metre.\";\n", - "\n", - "#part (e)\n", - "#this is obtained when Q = H\n", - "#hence from equation of discharge,we get\n", - "#Q*math.log10(69.2*Q) = 6.528.\n", - "#solving it by trial and error method we get Q = 2.85 m**3/min.\n", - "print \"maximum rate at which water can be pumped = 2.85 cubic metre/min\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " radius of zero drawdown = 41.53 m\n", - "coefficient of permeability = 4.31 m/day.\n", - "drawdown in well = 16.59 m.\n", - "Specific capacity = 0.266 cubic metre/minutes/metre.\n", - "maximum rate at which water can be pumped = 2.85 cubic metre/min\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.13 pg : 298" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%matplotlib inline\n", - "import math \n", - "from numpy import zeros,linspace\n", - "from matplotlib.pylab import plot\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 2500.; \t\t\t\t#discharge(l/min)\n", - "r = 60.; \t\t\t\t#dismath.tance of observation well from acquifer\n", - "tmin = [1, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 14, 18, 24, 30, 40, 50, 60, 80, 100, 120, 150, 180, 210, 240]; \t\t\t\t#time in minutes\n", - "s = [0.2, 0.26, 0.3, 0.33, 0.36, 0.41, 0.45, 0.48, 0.53, 0.56, 0.59, 0.62, 0.66, 0.71, 0.75, 0.80, 0.83, 0.86, 0.91, 0.95, 0.98, 1.03, 1.05, 1.08, 1.10]; \t\t\t\t#drawdown\n", - "u = linspace(1,9,9)\n", - "Wu = [0.2194, 0.04891, 0.01315, 0.003779, 0.001148, 0.000360, 0.000116, 0.0000377, 0.0000125];\n", - "tday = zeros(25)\n", - "for i in range(25):\n", - " tday[i] = tmin[i]/(60.*24);\n", - "\n", - "\n", - "rt = zeros(25)\n", - "for i in range(25):\n", - " rt[i] = r**2/tday[i];\n", - "\n", - "#graph is plotted between s and r**2/t and W(u) and u and they are superimposed.\n", - "#from which we get\n", - "plot(s,rt)\n", - "plot(Wu,u)\n", - "s1 = 0.52;\n", - "Wu1 = 2.96;\n", - "rt1 = 700000; \n", - "u1 = 0.03;\n", - "Q = 3600; \t\t\t\t#discharge in cumec/day\n", - "T = Q*Wu1/(4*math.pi*s1);\n", - "S = 4*u1*T/rt1;\n", - "T = round(T);\n", - "print \"formation consmath.tant of acquifer:\";\n", - "print \"T = %.2f cubic metre/day/m.S = %.2f.\"%(T,S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "formation consmath.tant of acquifer:\n", - "T = 1631.00 cubic metre/day/m.S = 0.00.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x109db6e50>" - ] - } - ], - "prompt_number": 19 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.14 pg : 299" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%matplotlib inline\n", - "import math\n", - "from numpy import zeros\n", - "from matplotlib.pylab import plot\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 2500; \t\t\t\t#discharge(l/min)\n", - "r = 60; \t\t\t\t#dismath.tance of observation well from acquifer\n", - "tmin = [1, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 14, 18, 24, 30, 40, 50, 60, 80, 100, 120, 150, 180, 210, 240]; \t\t\t\t#time in minutes\n", - "s = [0.2, 0.26, 0.3, 0.33, 0.36, 0.41, 0.45, 0.48, 0.53, 0.56, 0.59, 0.62, 0.66, 0.71, 0.75, 0.80, 0.83, 0.86, 0.91, 0.95, 0.98, 1.03, 1.05, 1.08, 1.10]; \t\t\t\t#drawdown\n", - "tday = zeros(25)\n", - "for i in range(25):\n", - " tday[i] = tmin[i]/(60*24);\n", - "\n", - "#from the graph between s and t we get\n", - "plot(s,tday)\n", - "ds = 0.38;\n", - "Q = 3600; \t\t\t\t#discharge in cumec/day\n", - "T = 2.303*Q/(4*math.pi*ds);\n", - "\t\t\t\t#extending the straight line we get\n", - "to = 0.00024;\n", - "S = 2.25*T*to/r**2;\n", - "print \"formation constant of acquifer:\";\n", - "print \"T = %i cubic metre/day/m.S = %.2f.\"%(T,S);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "formation constant of acquifer:\n", - "T = 1736 cubic metre/day/m.S = 0.00.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10ab61650>" - ] - } - ], - "prompt_number": 22 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.15 pg : 299" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%matplotlib inline\n", - "import math\n", - "from numpy import zeros\n", - "from matplotlib.pylab import plot\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 2500; \t\t\t\t#discharge(l/min)\n", - "r = 60; \t\t\t\t#dismath.tance of observation well from acquifer\n", - "tmin = [1, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 14, 18, 24, 30, 40, 50, 60, 80, 100, 120, 150, 180, 210, 240]; \t\t\t\t#time in minutes\n", - "s = [0.2, 0.26, 0.3, 0.33, 0.36, 0.41, 0.45, 0.48, 0.53, 0.56, 0.59, 0.62, 0.66, 0.71, 0.75, 0.80, 0.83, 0.86, 0.91, 0.95, 0.98, 1.03, 1.05, 1.08, 1.10]; \t\t\t\t#drawdown\n", - "tday = zeros(25)\n", - "for i in range(25):\n", - " tday[i] = tmin[i]/(60.*24);\n", - "\n", - "#graph is plotted between s and t\n", - "#point P is choosen on it whose ordinate is:\n", - "plot(s,tday)\n", - "s1 = 0.45;\n", - "t = 0.00347;\n", - "ds = 0.38; \t\t\t\t#for one math.log cycle of time\n", - "Fu = s1/ds;\n", - "#from fig 5.43\n", - "#or umath.sing relation\n", - "Wu = 2.303*Fu; \n", - "u = 0.035; \t\t\t\t#from table 5.2\n", - "Q = 3600; \t\t\t\t#discharge in cumec/day\n", - "T = Q*Wu/(4*math.pi*s1);\n", - "S = 4*u*t*T/r**2;\n", - "print \"formation consmath.tant of acquifer:\";\n", - "print \"T = %i cubic metre/day/m.S = %.2f.\"%(T,S);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "formation consmath.tant of acquifer:\n", - "T = 1736 cubic metre/day/m.S = 0.00.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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hTJkCt93mdTTZp6Utfd2RKyKeWLAAWreG8RrekVFq6YtIxh0+DAMHwssvw2WXeR1NdtLM\nWSKSNe67z6ZBLCnxOpLspaQvIlnhww9trtuNG6F7d6+jyV6q6YuI79U9X2fGDCV8ryjpi0jGvPoq\n7Nih5+t4STNniUhG1D1f5/e/1/N1vKSWvohkxGOPwaBBer6O19SRKyJpt307DB1qd+D26+d1NMGg\njlwR8aUTJ+AHP4D771fC9wO19EUkbSIRm+T85El4/nlNgZhK6XrgmohIi82fD6tXW1lHCd8f/PBr\nUEtfJIDWroXRo+F//xcGDPA6muBRTV9EfOPAAbjhBpg7Vwnfb9TSF5GUqqmBMWPgoovg4Ye9jia4\n1NIXEV944AG7EWvWLK8jkXjcJP0iYCtQAUxv5JjHnf0bgCFR27sAL2KzbW3GplEUkYB69VV49lko\nLbVn5Yv/JPq15AFzgFHYZOdrgDK+PEduP6A/cAXwFPXJ/TFsOsUbnJ/VIVWBi4i/bN8Od9wBf/2r\nHqbmZ4la+kOxuW93AqeBUmBczDFjgQXO+1VY67470Bn4DvCMs68aOJx0xCLiO3U3YP3rv8KwYV5H\nI01JlPR7Abui1nc72xId0xvoC+wHnsUmRX8aaJ9MsCLiP5EI3HOPPVdnyhSvo5FEEpV33A6rie1B\njjif/S1gMlYWmg38Erg/9uTi4uIz70OhEKFQyOWPFRGvzZ8Pq1bZohuw0iccDhMOh5P+nES/okKg\nGOvMBZgB1AIPRR0zFwhjpR+wTt/hzmevxFr8AFdhSf/amJ+hIZsiWaruBqy334bzz/c6mtySriGb\na7EO2gIgH7gJ68iNVgbc7rwvBA4BnwJ7sbJP3a0Zo4BNzQ1QRPwp+gYsJfzskai8U42VZ5ZgI3lK\nsJE7k5z987DROaOxDt9jwISo86cAz2N/MCpj9olIlqqpgfHj4cYb4frrvY5GmsMPFTiVd0SyzP33\nW0nnjTc0Ht8resqmiGTEwoV2A9batUr42Ui/MhFx7fXXYeJEu/NWN2BlJz17R0RcCYfh1lvtjtvL\nL/c6GmkpJX0RSeidd6zT9s9/hm9/2+toJBlK+iLSpFWr4J//2aY71H2T2U9JX0QatW4djB1rHbff\n+57X0UgqKOmLSFzvv293286da5OiSDAo6YvIl2zZAkVF8PjjVtqR4FDSF5EGKiqslPPww9Z5K8Gi\npC8iZ+zYASNHwoMP2vBMCR4lfREB4OOPYcQImDHDZsCSYFLSFxGqqizhT50KP/2p19FIOinpi+S4\nvXutpDNxItx7r9fRSLop6YvksM8+g1Gj4JZbYPp0r6ORTNCjlUVy1N//bi380aPh17/WVIfZpqWP\nVvbDr1lJXyTDDh+2Fv7w4fC73ynhZ6N0TZcINj/uVqACaOwL4OPO/g3AkJh9ecB7wMvNDU5EUu/I\nEbjmGigsVMLPRYmSfh4wB0v8g4GbgUExx4wG+mFz6d4NPBWzfyqwGVBzXsRjR4/CtdfCN78Jjz2m\nhJ+LEiX9odjctzuB00ApMC7mmLHAAuf9KqALUDe9Qm/sj8J8/FFKEslZVVXw3e/aJOZz58JXNIwj\nJyX6tfcCdkWt73a2uT3mUeAXQG0SMYpIktavhyuvtMcqzJunhJ/LEk2X6LYkE9uKbwVcC+zD6vmh\npk4uLi4+8z4UChHSQ7tFUubVV+EnP4Enn4Qf/tDraKSlwuEw4XA46c9JVHIpBIqxmj7ADKzV/lDU\nMXOBMFb6Aev0DQE/A24DqoGvAp2Al4DbY36GRu+IpMmcOTBrFvzlL9ZxK8GRriGbrYEPgZHAHmA1\n1pm7JeqY0cBk57UQmO28RhsO3AdcF+dnKOmLpFhNDfzLv9hE5osWQd++XkckqdbSpJ+ovFONJfQl\n2EieEizhT3L2zwMWYQl/G3AMmNDIZymzi2TA0aNw881w/DisXAldungdkfiJH0bUqKUvkiJVVXDd\ndTBkCDz1FOTnex2RpEs6b84SkSxQN0Lnhz+E+fOV8CW+ROUdEckCdSN0nnhCs11J05T0RbLcnDnw\nm99AWZm19EWaoqQvkqVqamDaNFiyBN55B/7xH72OSLKBkr5IFjp61J6Bf+wYrFgBX/ua1xFJtlBH\nrkiW2bPHnqHTrRu89poSvjSPkr5IFtmwwe6s1QgdaSmVd0SyxKJF8OMfW8ftTTd5HY1kKyV9kSzw\nxBM2paFG6EiylPRFfKymBu67DxYv1ggdSQ3V9EV8qLYW/vu/4cILYcsWG6GjhC+poJa+iI9EIvDy\ny3D//dCmDTz6KPzTP2laQ0kdJX0RH4hE4I034N/+Db74Ah58EMaOVbKX1FPSF/HYm2/Cv/877N8P\nDzwAN9yg6QwlfZT0RTzy7ruW7Ldvh+Jiu8M2L8/rqCTo1J4QybD33oNrr7WnYd54I2zdCrfdpoQv\nmeE26Rdhc99WANMbOeZxZ/8GYIizrQ+wHNgEbMTmzRXJSZs2WelmzBgoKoKKCpg40TpsRTLFTdLP\nA+ZgiX8wNkfuoJhjRgP9gP7A3cBTzvbTwM+BC7B5c++Jc65IoFVUwPjxMGKE3Vi1bRtMngxt23od\nmeQiN0l/KDb/7U4siZcC42KOGQsscN6vAroA3YG9wHpn+1Fsft2eSUUskgUiEfjb3+COO2DYMBg8\n2JL9tGnQvr3X0Ukuc9OR2wvYFbW+G7jCxTG9gU+jthVgZZ9VzY5SJEscOADPPw/PPAOHD8Odd1pL\nX5OTi1+4SfpuZy2PHVEcfd5ZwIvAVKzF30BxcfGZ96FQiFAo5PJHinivthaWLYOSEntcwpgx8Mgj\ncPXVGnopqRMOhwmHw0l/jptbPwqBYqymDzADqAUeijpmLhDGSj9gnb7DsZZ+G+AV4DVgdpzPj0Qi\nbv+uiPjHRx/Bs8/a0rWrlXJuuUXPt5fMaGV37jX79j037ZC1WAdtAZAP3ASUxRxTBtzuvC8EDmEJ\nvxVQAmwmfsIXySonT0JpqT0a4dJLrZyzcKHV7++5Rwlf/M9NeacamAwswUbylGAdspOc/fOARdgI\nnm3AMWCCs+/bwK3A+8B7zrYZwOIUxC6SEbW1sHo1vPCCLUOGWKu+rAy++lWvoxNpHj882UPlHfGd\nkyetTr9woT0A7etftzH2EyZAQYHX0Ym0vLyjpC/iOHAAXn3VEv3SpXDxxTBunC39+nkdnUhDSvoi\nLbBjhyX5//kfezzCiBGW5MeMsYnHRfxKSV/EhbqbphYutOXTT+G66yzRjxoF7dp5HaGIO0r6Io04\ncQLeesuSfFkZnHVWfdnmiiv0oDPJTi1N+nq0sgTOqVM22mb5cigvhzVr4JJLrEW/dCkMHOh1hCLe\nUUtfsl5NDaxbV5/kV6yA/v2tPj9iBFx1FXTs6HWUIqml8o7kjNpae0xxebktb70FvXpZgr/6ahg+\n3IZYigSZkr4E0qlT8OGHsHFj/bJyJXTqVN+SD4WgRw+vIxXJLCV9yWo1NVBZWZ/YN22y1+3b7Wao\nb37TlgsugMsvh3PP9TpiEW8p6UtWiERg166GLfeNG23KwO7d65N73XL++XrUgUg8SvriK5EI7Nv3\n5eS+aZMNmYxN7oMH23YRcUdJXzxz8KAl87qSTN1SU2PlmAsvbFieOftsryMWyX5K+pI2R49CVRXs\n2dPwta6D9fBha6nHtt579IBWfvg/TCSAlPSl2U6dgk8++XIy37On4fvqahsS2bNn/WvPnjBggCX3\nc87RDFEimaakL0Qi1irft69+2b+/4XrdsncvHDpknafRiTze+86d1WIX8Zt0Jv0ibNarPGA+DadJ\nrPM4cA1wHPgJ9ROmuDlXSd8RidhzYo4cgc8/t7LJwYOWnBt7jd3Wpo0l8m98o37p1q3h+je+Ycd0\n66bnzohkq3Ql/TzgQ2AUUAWsAW7GZs6qMxqbWWs0cAXwGDZloptzIYuTfk0NHDsWfzl6tD55HznS\ncIndVr8eJj8/RMeO9tiALl1s+j23r507B2d4YzgcJhQKeR2GL+ha1NO1qJeuB64NxaZA3OmslwLj\naJi4xwILnPergC5AD6Cvi3NTLhKBL76wmY/qlhMnGq7H29ZY8j52DI4fj7/99Gno0MGW9u3r33fo\nwJnEHb306GF3ksbb17EjPPpomP/4j1A6L0/W0D/ueroW9XQtkpco6fcCdkWt78Za84mO6QX0dHEu\nAD/9qSXQU6dS85qfby3edu3sNXqJ3dauHbRtW5+0u3a1uz07dLBx49GJPHZp2za1tW6VWkQk3RIl\nfbd1l6RS34UXWqJu0yb51zZtNJJERKSlCoHFUeszgOkxx8wFfhS1vhXo7vJcsBJQRIsWLVq0NGvZ\nRhq0BiqBAiAfWA8MijlmNLDIeV8IvNuMc0VExGeuwUbhbMNa6wCTnKXOHGf/BuBbCc4VEREREZEg\nKcLq/RXEr+2Px74pvA+8A1yUudAyLtG1qHM5UA1cn4mgPOLmWoSwG/42AuGMROWNRNeiK9ZPth67\nFj/JWGSZ9QzwKfBBE8c8jl2nDcCQTATlkUTXwrd5Mw8r8RQAbYhf378S6Oy8L6K+byBo3FyLuuPK\ngVeAH2QquAxzcy26AJuA3s5610wFl2FurkUx8FvnfVfgAIlH4GWj72CJvLFEF92PeAXBzRWQ+Fo0\nO29manBj9E1ep6m/USvaSuCw834V9f/Ig8bNtQCYArwI7M9YZJnn5lrcAryE3ecB8FmmgsswN9fi\nE6CT874TlvSrMxRfJr0NHGxif7wbQrunOyiPJLoWzc6bmUr6jd3A1Zg7qf9LHjRurkUv7B/8U856\nJANxecHNtegPfB1YDqwFbstMaBnn5lo8DVwA7MG+0k/NTGi+E+9aBbWR2Byu8mamvho2J2ldDdwB\nfDtNsXjNzbWYDfzSObYV/ngaajq4uRZtsBFhI4H2WMvmXayeGyRursWvsLJPCDgPeAO4GDiSvrB8\nK/bfRFAbRm65zpuZSvpVQJ+o9T7Uf12PdhHWmimi6a802czNtbgU+3oPVru9BvvKX5b26DLLzbXY\nhZV0TjjLW1iiC1rSd3MthgG/cd5XAjuA87FvQLkk9lr1drblKl/mTTc3ap2D1TQLMxpZ5jX3prVn\nCe7oHTfXYiCwFOvobI91aA3OXIgZ4+Za/Ccw03nfHfuj8PUMxZdpBbjryI2+ITSoCmj8Wvg6bya6\nyWs+1jH1nrOsznSAGeTmhrc6QU764O5a3IeN4PkA+FlGo8usRNeiK/AyVs//AOvkDqI/Yf0Wp7Bv\nenfg/obQoEl0LXIpb4qIiIiIiIiIiIiIiIiIiIiIiIiIiIiIiIiIeOf/A0ZprnC5+57cAAAAAElF\nTkSuQmCC\n", - "text": [ - "<matplotlib.figure.Figure at 0x10ab73dd0>" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.15 pg : 300" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t#draw daown in main well\n", - "\t\t\t\t\n", - "#Given\n", - "H = 25; \t\t\t\t#static water level\n", - "rw = 0.15; \t\t\t\t#radius of well\n", - "Q = 5400; \t\t\t\t#discharge(litre/min)\n", - "t = 24; \t\t\t\t#time of discharge\n", - "r1 = 30; \t\t\t\t#dismath.tance of first well\n", - "s1 = 1.11; \t\t\t\t#drawdown\n", - "h1 = H-s1;\n", - "r2 = 90; \t\t\t\t#dismath.tance of second well\n", - "s2 = 0.53; \t\t\t\t#drawdown\n", - "h2 = H-s2;\n", - "k = (Q*2.303*math.log10(r2/r1))/(math.pi*(h2**2-h1**2)*60000);\n", - "T = k*H;\n", - "T = round(T*10000)/10000;\n", - "print \"transmissibility of acquifer = %.2f cumec/sec.\"%(T);\n", - "hw = (h2**2-(Q*2.303*math.log10(r2/rw))/(math.pi*k*60000))**0.5;\n", - "sw = H-hw;\n", - "sw = round(sw*100)/100;\n", - "print \"draw daown in main well = %.2f m.\"%(sw);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "transmissibility of acquifer = 0.03 cumec/sec.\n", - "draw daown in main well = 4.13 m.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.16 pg : 300" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 250; \t\t\t\t#discharge(lit/min)\n", - "H = 100; \t\t\t\t#water level in acquifer\n", - "s1 = 12; \t\t\t\t#drawdown\n", - "h1 = H-s1;\n", - "\t\t\t\t#let t = ln(R/r)/(pi*k)\n", - "t = (H**2-h1**2)/Q;\n", - "h2 = H-18;\n", - "Q1 = (H**2-h2**2)/t;\n", - "print \"discharge at 18m drawdown = %d lpm\"%(Q1);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge at 18m drawdown = 364 lpm\n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.17 pg : 301" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "b = 10; \t\t\t\t#thickness of acquifer\n", - "k = 48; \t\t\t\t#permeability coefficient\n", - "R = 500.; \t\t\t\t#radius of influence\n", - "s = 12; \t\t\t\t#drawdown\n", - "Q = 5000; \t\t\t\t#discharge(cumec/day)\n", - "r = R/math.e**(2*math.pi*b*k*s/Q);\n", - "D = 2*r;\n", - "D = round(D*100)/100;\n", - "print \"effective well diameter = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "effective well diameter = 0.72 m.\n" - ] - } - ], - "prompt_number": 26 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.18 pg : 301" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 1500; \t\t\t\t#discharge(lit/min)\n", - "S = 0.004; \t\t\t\t#storage coefficient\n", - "k = 35; \t\t\t\t#permeability\n", - "t = 20; \t\t\t\t#time of pumping\n", - "b = 30; \t\t\t\t#thickness of acquifer\n", - "r = 40; \t\t\t\t#dismath.tance of observation well\n", - "T = k*b;\n", - "s = Q*(2.303*math.log10(4*T*t*3600/(60*60*24*r**2*S))-0.5772)*60*60*24/(4*math.pi*T*60000); \t\t\t\t#Jacob's equation\n", - "s = round(s*100)/100;\n", - "print \"drawdown at 40m = %.2f m.\"%(s);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "drawdown at 40m = 0.94 m.\n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.19 pg : 301" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "h1 = 2.5; \t\t\t\t#initial pumping depression\n", - "h = 1.8; \t\t\t\t#heigth after recuperation\n", - "t = 80; \t\t\t\t#time\n", - "h2 = h1-h;\n", - "KbyA = 2.303*60*math.log10(h1/h2)/t;\n", - "d = 4; \t\t\t\t#diameter of well\n", - "H = 3; \t\t\t\t#depression head\n", - "A = math.pi*d**2/4;\n", - "Q = (KbyA)*A*H/3.6;\n", - "Q = round(Q);\n", - "print \"Yield from well = %.2f lit/sec.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Yield from well = 10.00 lit/sec.\n" - ] - } - ], - "prompt_number": 28 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.20 pg : 301" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "rw = 0.15; \t\t\t\t#radius of well\n", - "b = 40; \t\t\t\t#depth of acquifer\n", - "Q = 1500; \t\t\t\t#discharge(lpm)\n", - "s1 = 3.5; \t\t\t\t#drawdown of first well\n", - "s2 = 2; \t\t\t\t#drawdown of second well\n", - "H = 40; \n", - "r1 = 25; \t\t\t\t#dismath.tance of first well\n", - "r2 = 75; \t\t\t\t#dismath.tance of second well\n", - "h1 = H-s1;\n", - "h2 = H-s2;\n", - "k = Q*2.303*math.log10(r2/r1)/(math.pi*1000*60*(h2**2-h1**2));\n", - "T = b*k*1000;\n", - "print \"transmissibility = %.2fD-3 square metre/sec\"%(T);\n", - "\n", - "hw = (h2**2-(Q*2.303*math.log10(r2/rw)/(math.pi*k*60000)))**0.5;\n", - "sw = H-hw;\n", - "sw = round(sw*100)/100;\n", - "print \"drawdown at pumping well = %.2f m.\"%(sw);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "transmissibility = 3.13D-3 square metre/sec\n", - "drawdown at pumping well = 11.51 m.\n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.21 pg : 302" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "r = 0.25; \t\t\t\t#radius of test well\n", - "r1 = 10; \t\t\t\t#dismath.tance of first well\n", - "r2 = 60; \t\t\t\t#dismath.tance of second well\n", - "Q = 0.1; \t\t\t\t#discharge(cumec/sec)\n", - "s1 = 4; \t\t\t\t#drawdown of first well\n", - "s2 = 3; \t\t\t\t#drawdown of second well\n", - "b = 20; \t\t\t\t#thickness of well\n", - "k = 1000*Q*math.log10(r2/r1)/(2.72*b*(s1-s2));\n", - "print \"coefficient of permeability = %.2fD-3 m/sec\"%(k);\n", - "s = s2+Q*math.log10(r2/r)/(2.72*b*k);\n", - "s = round(s*100)/100;\n", - "print \"drawdown in test well = %.2f m.\"%(s); \n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "coefficient of permeability = 1.43D-3 m/sec\n", - "drawdown in test well = 3.00 m.\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.22 pg : 303" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "h1 = 2.1; \t\t\t\t#initial pumping depression\n", - "h = 1.6; \t\t\t\t#heigth after recuperation\n", - "t = 90; \t\t\t\t#time\n", - "h2 = h1-h;\n", - "KbyA = 2.303*60*math.log10(h1/h2)/t;\n", - "Q = 10; \t\t\t\t#yield(lit/sec)\n", - "H = 2;\n", - "A = Q*3.6/(H*(KbyA));\n", - "d = (4*A/math.pi)**0.5;\n", - "d = round(d*10)/10;\n", - "print \"Daimeter of well = %.2f m\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Daimeter of well = 4.90 m\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.23 pg : 303" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "h1 = 2.5; \t\t\t\t#initial pumping depression\n", - "h = 1; \t\t\t\t#heigth after recuperation\n", - "t = 60; \t\t\t\t#time\n", - "h2 = h1-h;\n", - "\n", - "# Calculations\n", - "KbyA = 2.303*60*math.log10(h1/h2)/t;\n", - "d = 2.; \t\t\t\t#diameter of well\n", - "H = 3.; \t\t\t\t#depression head\n", - "A = math.pi*d**2/4;\n", - "Q = (KbyA)*A*H;\n", - "Q = round(Q*1000)/1000;\n", - "\n", - "# Results\n", - "print \"Yield from well = %.2f cubic metre/hour.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Yield from well = 4.82 cubic metre/hour.\n" - ] - } - ], - "prompt_number": 33 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch5_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch5_1.ipynb deleted file mode 100755 index a473df61..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch5_1.ipynb +++ /dev/null @@ -1,1232 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:ba8fe47f9e34523e72313e6de9073278b5a7ecea1af12ee41d74f1b26aaa7966" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 5 : GROUND WATER WELL IRRIGATION" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.1 pg : 277" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t#design an open wellin fine sand\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 0.003; \t\t\t\t#required discharge\n", - "H = 2.5; \t\t\t\t#depression head\n", - "A = Q*3600/(0.5*H);\n", - "d = (4*A/math.pi)**0.5;\n", - "d = round(d*100)/100\n", - "print \"Well diameter = %.2f m.\"%(d);\n", - "\n", - "\t\t\t\t#Alternative solution\n", - "C = 7.5e-5; \t\t\t\t#permeability consmath.tant from table 5.2\n", - "A = Q/(C*H);\n", - "d = (16*3/math.pi)**0.5;\n", - "d = round(d*10)/10;\n", - "print \"By alternative solution:\"\n", - "print \"Well diameter = %.2f m\"%(d);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Well diameter = 3.32 m.\n", - "By alternative solution:\n", - "Well diameter = 3.90 m\n" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.2 pg : 278" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#yield from well\n", - "#diameter of well\n", - "\n", - "#Given\n", - "h1 = 2.5; \t\t\t\t#initial pumping depression\n", - "h = 1.8; \t\t\t\t#heigth after recuperation\n", - "t = 80; \t\t\t\t#time\n", - "h2 = h1-h;\n", - "KbyA = 2.303*60*math.log10(h1/h2)/t;\n", - "\n", - "\n", - "# Calculations and Results\n", - "#Part (a)\n", - "d = 4; \t\t\t\t#diameter of well\n", - "H = 3; \t\t\t\t#depression head\n", - "A = math.pi*d**2/4;\n", - "Q = (KbyA)*A*H/3.6;\n", - "print \"Part a\";\n", - "Q = round(Q);\n", - "print \"Yield from well = %.2f lit/sec.\"%(Q);\n", - "\n", - "#Part (b)\n", - "Q = 8; \t\t\t\t#yield(lit/sec)\n", - "H = 2;\n", - "A = Q*3.6/(H*(KbyA));\n", - "d = (4*A/math.pi)**0.5;\n", - "d = round(d*10)/10;\n", - "print \"Part b\";\n", - "print \"Daimeter of well = %.2f m\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Part a\n", - "Yield from well = 10.00 lit/sec.\n", - "Part b\n", - "Daimeter of well = 4.40 m\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.3 pg : 279" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given\n", - "d = 30; \t\t\t\t#well diameter\n", - "L = 15; \t\t\t\t#strainer length\n", - "P = 50; \t\t\t\t#coefficient of permeability\n", - "s = 0.2; \t\t\t\t#effective size of sand\n", - "b = 3; \t\t\t\t#drawdown\n", - "r = 150; \t\t\t\t#radius of drawdown\n", - "\n", - "\n", - "# Calculations\n", - "Q = 2.72*L*P*b/(math.log10(r*2*100/d)*24*3.6);\n", - "Q = round(Q*10)/10;\n", - "\n", - "# Results\n", - "print \" yield from well = %.2f lit/sec.\"%(Q);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " yield from well = 23.60 lit/sec.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.4 pg : 280" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "d = 30; \t\t\t\t#diameter of well\n", - "s = 2; \t\t\t\t#drawdown\n", - "L = 10; \t\t\t\t#length of stainer\n", - "k = 0.05; \t\t\t\t#coefficient of permeability\n", - "r = 300; \t\t\t\t#radius of zero drawdown\n", - "\n", - "# Calculations\n", - "Q = 2.72*k*s*(L+s/2)/(100*math.log10(2*100*r/d));\n", - "\n", - "# Results\n", - "print \" discharge from tubewell = %.4f cumec.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " discharge from tubewell = 0.0091 cumec.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.5 pg: 280" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t#design tube well\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 0.08; \t\t\t\t#yield required\n", - "b = 30; \t\t\t\t#thickness of acquifer\n", - "R = 300; \t\t\t\t#Radius of circle of influence\n", - "k = 60; \t\t\t\t#permeability coefficient\n", - "s = 5; \t\t\t\t#Drawdown\n", - "\n", - "# Calculations\n", - "r = R/(10**(2.72*b*s*k/(3600*24*Q)));\n", - "r = round(r*10000)/10000;\n", - "\n", - "# Results\n", - "print \"Radius of well = %.2f m\"%(r);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Radius of well = 0.09 m\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.6 pg : 281" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "b = 30.; \t\t\t\t#thickness of acquifer\n", - "s = 4.; \t\t\t\t#drawdown\n", - "r = 0.1; \t\t\t\t#well radius\n", - "k = 36.; \t\t\t\t#permeability coefficient\n", - "R = 3000*s*(k/(24*3600))**0.5;\n", - "\n", - "Q = 2.72*b*k*s/(math.log10(R/r)*24*3.6);\n", - "Q = round(Q*10)/10;\n", - "print \"yield from well = %.2f lit/sec.\"%(Q);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "yield from well = 40.10 lit/sec.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.7 pg : 281" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "k = 0.005; \t\t\t\t#coefficient of permeability\n", - "r = 0.1; \t\t\t\t#well radius\n", - "s = 4; \t\t\t\t#drawdown\n", - "b = 10; \t\t\t\t#thickness\n", - "R = 300; \t\t\t\t#radius of circle of influence\n", - "\n", - "# Calculations and Results\n", - "#Part(a)\n", - "Q1 = 2.72*b*k*s/math.log10(R/r);\n", - "Q1 = round(Q1*10000)/10000;\n", - "print \"Discharge = %.2f cumec\"%(Q1);\n", - "\n", - "#Part (b)\n", - "r = 0.2;\n", - "Q2 = 2.72*b*k*s/math.log10(R/r);\n", - "I = (Q2-Q1)*100/Q1;\n", - "I = round(I*10)/10;\n", - "print \"percent increase in discharge = %.2f percent.\"%(I);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Discharge = 0.16 cumec\n", - "percent increase in discharge = 9.40 percent.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.8 pg : 282" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#percentage error\n", - "#actual radius of influence\n", - "\n", - "#Given\n", - "d = 0.2; \t\t\t\t#diameter of well\n", - "Q = 240; \t\t\t\t#discharge\n", - "RL1 = 240.5; \t\t\t\t#reduce level of original water surface\n", - "RL2 = 235.6; \t\t\t\t#reduced level of water at pumping\n", - "RL3 = 210; \t\t\t\t#reduced level of impervious layer\n", - "RL4 = 239.8; \t\t\t\t#reduced level of water in well\n", - "D = 50; \t\t\t\t#radial dismath.tance of well from tube well\n", - "\n", - "# Calculations and Results\n", - "#Part(a)\n", - "h1 = RL2-RL3;\n", - "h2 = RL4-RL3;\n", - "k1 = Q*24*math.log10(D*2/d)/(1.36*(h2**2-h1**2));\n", - "k1 = round(k1*100)/100;\n", - "print \"Parta\";\n", - "print \"coefficient of permeability = %.2f m/day.\"%(k1);\n", - "#Part (b)\n", - "R = 300; \t\t\t\t#radius of influence\n", - "H = RL1-RL3;\n", - "h = RL2-RL3;\n", - "k2 = Q*24*math.log10(R*2/d)/(1.36*(H**2-h**2));\n", - "PE = (k2-k1)*100/k1;\n", - "print \"Partb\";\n", - "print \"percentage error = %i percent.\"%(PE);\n", - "#Part (b)\n", - "R = (d/2)*10**(1.36*k1*(H**2-h**2)/(24*Q));\n", - "print \"Partc\";\n", - "print \"Actual radius of influence = %i m.\"%(R);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Parta\n", - "coefficient of permeability = 49.13 m/day.\n", - "Partb\n", - "percentage error = 9 percent.\n", - "Partc\n", - "Actual radius of influence = 154 m.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.9 pg : 283" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "A = 20.; \t\t\t\t#area of field\n", - "H = 129.; \t\t\t\t#level to the highest land\n", - "h1 = 120.2; \t\t\t\t#water level in well during discharge\n", - "Du = 800; \t\t\t\t#duty for rise;\n", - "eita = 0.6; \t\t\t\t#efficiency of the pump\n", - "\n", - "# Calculations\n", - "Q = A/Du;\n", - "w = Q*1000;\n", - "lift = H-h1;\n", - "\t\t\t\t#design lift is taken as 9m\n", - "wd = w*9;\n", - "o = wd/75;\n", - "i = o/eita;\n", - "\n", - "# Results\n", - "print \"Input h.p of pump = %i h.p\"%i;\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Input h.p of pump = 5 h.p\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.10 pg : 284" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 150; \t\t\t\t#discharge from tubewell\n", - "t = 4000; \t\t\t\t#working period of tubewell\n", - "I = 0.45; \t\t\t\t#intensity of irrigation\n", - "d = 0.38; \t\t\t\t#average depth of rabi and kharif crop\n", - "\n", - "# Calculations\n", - "V = Q*t;\n", - "A = V/d;\n", - "CA = A/(I*10000);\n", - "CA = round(CA);\n", - "\n", - "# Results\n", - "print \"culturable area = %.2f hectares.\"%(CA);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "culturable area = 351.00 hectares.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.11 pg : 284" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#percent decrease when two well discharges\n", - "\n", - "#Given\n", - "d = 0.2; \t\t\t\t#diameter of well\n", - "r = d/2;\n", - "B = 100; \t\t\t\t#dismath.tance between wells\n", - "b = 12; \t\t\t\t#thickness of acquifer\n", - "k = 60; \t\t\t\t#coefficient of permeability\n", - "s = 3; \t\t\t\t#print ersion head\n", - "R = 250; \t\t\t\t#radius of influence\n", - "Q = 2.72*b*k*s/(24*math.log10(R/r));\n", - "print \"discharge if one well discharges = %i cubic metre/hour.\"%(Q);\n", - "#when both well are discharging\n", - "Q1 = 2.72*k*b*s/(24*math.log10(R**2/(r*B)));\n", - "Q1 = round(Q1*10)/10;\n", - "print \"discharge if both wells discharges = %.2f cubic metre/hour.\"%(Q1);\n", - "PE = (Q-Q1)*100/Q;\n", - "PE = round(PE*100)/100;\n", - "print \"percentage decrease in discharge = %.2f percent.\"%(PE);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge if one well discharges = 72 cubic metre/hour.\n", - "discharge if both wells discharges = 64.50 cubic metre/hour.\n", - "percentage decrease in discharge = 10.47 percent.\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.12 pg : 285" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#coefficient of permeability\n", - "#drawdown in well\n", - "#specific capacity\n", - "#maximum rate at which water can be pumped\n", - "\n", - "#Given\n", - "d = 0.6; \t\t\t\t#diameter of well;\n", - "rw = d/2;\n", - "H = 40.; \t\t\t\t#depth of water in well before pumping\n", - "Q = 2000.; \t\t\t\t#discharge from well\n", - "s1 = 4.; \t\t\t\t#drawdown in well\n", - "B1 = 10.; \t\t\t\t#dismath.tance between well\n", - "s2 = 2.;\n", - "B2 = 20.;\n", - "#Part (a)\n", - "h1 = H-s1;\n", - "h2 = H-s2;\n", - "t = (H**2-h2**2)/(H**2-h1**2);\n", - "R = (B2/(B1**t))**(1/(1-t));\n", - "R = round(R*100)/100;\n", - "print \" radius of zero drawdown = %.2f m\"%(R);\n", - "#Part (b)\n", - "r = 10;\n", - "k = Q*math.log10(R/r)*60*24/(1.36*(H**2-h1**2)*1000);\n", - "k = round(k*100)/100;\n", - "print \"coefficient of permeability = %.2f m/day.\"%(k);\n", - "\n", - "#part (c)\n", - "Ho = (H**2-(Q*math.log10(R/rw)*24*60/(1000*1.36*k)))**0.5;\n", - "D = H-Ho;\n", - "D = round(D*100)/100;\n", - "print \"drawdown in well = %.2f m.\"%(D);\n", - "\n", - "#part (d)\n", - "C = Q/(1000*R);\n", - "#for R = 1 m;Q = Sc\n", - "#hence on putting the values in discharge equation we get\n", - "#Sc*math.log10(61.2*Sc) = 0.3223.\n", - "#on solving this by trial and error method we get Sc = 0.266 m**2/min.\n", - "print \"Specific capacity = 0.266 cubic metre/minutes/metre.\";\n", - "\n", - "#part (e)\n", - "#this is obtained when Q = H\n", - "#hence from equation of discharge,we get\n", - "#Q*math.log10(69.2*Q) = 6.528.\n", - "#solving it by trial and error method we get Q = 2.85 m**3/min.\n", - "print \"maximum rate at which water can be pumped = 2.85 cubic metre/min\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " radius of zero drawdown = 41.53 m\n", - "coefficient of permeability = 4.31 m/day.\n", - "drawdown in well = 16.59 m.\n", - "Specific capacity = 0.266 cubic metre/minutes/metre.\n", - "maximum rate at which water can be pumped = 2.85 cubic metre/min\n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.13 pg : 298" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math \n", - "from numpy import zeros,linspace\n", - "from matplotlib.pylab import plot\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 2500.; \t\t\t\t#discharge(l/min)\n", - "r = 60.; \t\t\t\t#dismath.tance of observation well from acquifer\n", - "tmin = [1, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 14, 18, 24, 30, 40, 50, 60, 80, 100, 120, 150, 180, 210, 240]; \t\t\t\t#time in minutes\n", - "s = [0.2, 0.26, 0.3, 0.33, 0.36, 0.41, 0.45, 0.48, 0.53, 0.56, 0.59, 0.62, 0.66, 0.71, 0.75, 0.80, 0.83, 0.86, 0.91, 0.95, 0.98, 1.03, 1.05, 1.08, 1.10]; \t\t\t\t#drawdown\n", - "u = linspace(1,9,9)\n", - "Wu = [0.2194, 0.04891, 0.01315, 0.003779, 0.001148, 0.000360, 0.000116, 0.0000377, 0.0000125];\n", - "tday = zeros(25)\n", - "for i in range(25):\n", - " tday[i] = tmin[i]/(60.*24);\n", - "\n", - "\n", - "rt = zeros(25)\n", - "for i in range(25):\n", - " rt[i] = r**2/tday[i];\n", - "\n", - "#graph is plotted between s and r**2/t and W(u) and u and they are superimposed.\n", - "#from which we get\n", - "plot(s,rt)\n", - "plot(Wu,u)\n", - "s1 = 0.52;\n", - "Wu1 = 2.96;\n", - "rt1 = 700000; \n", - "u1 = 0.03;\n", - "Q = 3600; \t\t\t\t#discharge in cumec/day\n", - "T = Q*Wu1/(4*math.pi*s1);\n", - "S = 4*u1*T/rt1;\n", - "T = round(T);\n", - "print \"formation consmath.tant of acquifer:\";\n", - "print \"T = %.2f cubic metre/day/m.S = %.2f.\"%(T,S);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "formation consmath.tant of acquifer:\n", - "T = 1631.00 cubic metre/day/m.S = 0.00.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10aec95d0>" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.14 pg : 299" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math\n", - "from numpy import zeros\n", - "from matplotlib.pylab import plot\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 2500; \t\t\t\t#discharge(l/min)\n", - "r = 60; \t\t\t\t#dismath.tance of observation well from acquifer\n", - "tmin = [1, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 14, 18, 24, 30, 40, 50, 60, 80, 100, 120, 150, 180, 210, 240]; \t\t\t\t#time in minutes\n", - "s = [0.2, 0.26, 0.3, 0.33, 0.36, 0.41, 0.45, 0.48, 0.53, 0.56, 0.59, 0.62, 0.66, 0.71, 0.75, 0.80, 0.83, 0.86, 0.91, 0.95, 0.98, 1.03, 1.05, 1.08, 1.10]; \t\t\t\t#drawdown\n", - "tday = zeros(25)\n", - "for i in range(25):\n", - " tday[i] = tmin[i]/(60*24);\n", - "\n", - "#from the graph between s and t we get\n", - "plot(s,tday)\n", - "ds = 0.38;\n", - "Q = 3600; \t\t\t\t#discharge in cumec/day\n", - "T = 2.303*Q/(4*math.pi*ds);\n", - "\t\t\t\t#extending the straight line we get\n", - "to = 0.00024;\n", - "S = 2.25*T*to/r**2;\n", - "print \"formation constant of acquifer:\";\n", - "print \"T = %i cubic metre/day/m.S = %.2f.\"%(T,S);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "formation constant of acquifer:" - ] - }, - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "T = 1736 cubic metre/day/m.S = 0.00.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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- "text": [ - "<matplotlib.figure.Figure at 0x10c4e2850>" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.15 pg : 299" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline\n", - "import math\n", - "from numpy import zeros\n", - "from matplotlib.pylab import plot\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "Q = 2500; \t\t\t\t#discharge(l/min)\n", - "r = 60; \t\t\t\t#dismath.tance of observation well from acquifer\n", - "tmin = [1, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10, 12, 14, 18, 24, 30, 40, 50, 60, 80, 100, 120, 150, 180, 210, 240]; \t\t\t\t#time in minutes\n", - "s = [0.2, 0.26, 0.3, 0.33, 0.36, 0.41, 0.45, 0.48, 0.53, 0.56, 0.59, 0.62, 0.66, 0.71, 0.75, 0.80, 0.83, 0.86, 0.91, 0.95, 0.98, 1.03, 1.05, 1.08, 1.10]; \t\t\t\t#drawdown\n", - "tday = zeros(25)\n", - "for i in range(25):\n", - " tday[i] = tmin[i]/(60.*24);\n", - "\n", - "#graph is plotted between s and t\n", - "#point P is choosen on it whose ordinate is:\n", - "plot(s,tday)\n", - "s1 = 0.45;\n", - "t = 0.00347;\n", - "ds = 0.38; \t\t\t\t#for one math.log cycle of time\n", - "Fu = s1/ds;\n", - "#from fig 5.43\n", - "#or umath.sing relation\n", - "Wu = 2.303*Fu; \n", - "u = 0.035; \t\t\t\t#from table 5.2\n", - "Q = 3600; \t\t\t\t#discharge in cumec/day\n", - "T = Q*Wu/(4*math.pi*s1);\n", - "S = 4*u*t*T/r**2;\n", - "print \"formation consmath.tant of acquifer:\";\n", - "print \"T = %i cubic metre/day/m.S = %.2f.\"%(T,S);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Populating the interactive namespace from numpy and matplotlib\n", - "formation consmath.tant of acquifer:" - ] - }, - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "T = 1736 cubic metre/day/m.S = 0.00.\n" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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hTJkCt93mdTTZp6Utfd2RKyKeWLAAWreG8RrekVFq6YtIxh0+DAMHwssvw2WXeR1NdtLM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\nTkSuQmCC\n", - "text": [ - "<matplotlib.figure.Figure at 0x10aeaaa90>" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.15 pg : 300" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t#draw daown in main well\n", - "\t\t\t\t\n", - "#Given\n", - "H = 25; \t\t\t\t#static water level\n", - "rw = 0.15; \t\t\t\t#radius of well\n", - "Q = 5400; \t\t\t\t#discharge(litre/min)\n", - "t = 24; \t\t\t\t#time of discharge\n", - "r1 = 30; \t\t\t\t#dismath.tance of first well\n", - "s1 = 1.11; \t\t\t\t#drawdown\n", - "h1 = H-s1;\n", - "r2 = 90; \t\t\t\t#dismath.tance of second well\n", - "s2 = 0.53; \t\t\t\t#drawdown\n", - "h2 = H-s2;\n", - "k = (Q*2.303*math.log10(r2/r1))/(math.pi*(h2**2-h1**2)*60000);\n", - "T = k*H;\n", - "T = round(T*10000)/10000;\n", - "print \"transmissibility of acquifer = %.2f cumec/sec.\"%(T);\n", - "hw = (h2**2-(Q*2.303*math.log10(r2/rw))/(math.pi*k*60000))**0.5;\n", - "sw = H-hw;\n", - "sw = round(sw*100)/100;\n", - "print \"draw daown in main well = %.2f m.\"%(sw);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "transmissibility of acquifer = 0.03 cumec/sec.\n", - "draw daown in main well = 4.13 m.\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.16 pg : 300" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 250; \t\t\t\t#discharge(lit/min)\n", - "H = 100; \t\t\t\t#water level in acquifer\n", - "s1 = 12; \t\t\t\t#drawdown\n", - "h1 = H-s1;\n", - "\t\t\t\t#let t = ln(R/r)/(pi*k)\n", - "t = (H**2-h1**2)/Q;\n", - "h2 = H-18;\n", - "Q1 = (H**2-h2**2)/t;\n", - "print \"discharge at 18m drawdown = %d lpm\"%(Q1);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "discharge at 18m drawdown = 364 lpm\n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.17 pg : 301" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "b = 10; \t\t\t\t#thickness of acquifer\n", - "k = 48; \t\t\t\t#permeability coefficient\n", - "R = 500.; \t\t\t\t#radius of influence\n", - "s = 12; \t\t\t\t#drawdown\n", - "Q = 5000; \t\t\t\t#discharge(cumec/day)\n", - "r = R/math.e**(2*math.pi*b*k*s/Q);\n", - "D = 2*r;\n", - "D = round(D*100)/100;\n", - "print \"effective well diameter = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "effective well diameter = 0.72 m.\n" - ] - } - ], - "prompt_number": 26 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.18 pg : 301" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "Q = 1500; \t\t\t\t#discharge(lit/min)\n", - "S = 0.004; \t\t\t\t#storage coefficient\n", - "k = 35; \t\t\t\t#permeability\n", - "t = 20; \t\t\t\t#time of pumping\n", - "b = 30; \t\t\t\t#thickness of acquifer\n", - "r = 40; \t\t\t\t#dismath.tance of observation well\n", - "T = k*b;\n", - "s = Q*(2.303*math.log10(4*T*t*3600/(60*60*24*r**2*S))-0.5772)*60*60*24/(4*math.pi*T*60000); \t\t\t\t#Jacob's equation\n", - "s = round(s*100)/100;\n", - "print \"drawdown at 40m = %.2f m.\"%(s);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "drawdown at 40m = 0.94 m.\n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.19 pg : 301" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "h1 = 2.5; \t\t\t\t#initial pumping depression\n", - "h = 1.8; \t\t\t\t#heigth after recuperation\n", - "t = 80; \t\t\t\t#time\n", - "h2 = h1-h;\n", - "KbyA = 2.303*60*math.log10(h1/h2)/t;\n", - "d = 4; \t\t\t\t#diameter of well\n", - "H = 3; \t\t\t\t#depression head\n", - "A = math.pi*d**2/4;\n", - "Q = (KbyA)*A*H/3.6;\n", - "Q = round(Q);\n", - "print \"Yield from well = %.2f lit/sec.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Yield from well = 10.00 lit/sec.\n" - ] - } - ], - "prompt_number": 28 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.20 pg : 301" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "rw = 0.15; \t\t\t\t#radius of well\n", - "b = 40; \t\t\t\t#depth of acquifer\n", - "Q = 1500; \t\t\t\t#discharge(lpm)\n", - "s1 = 3.5; \t\t\t\t#drawdown of first well\n", - "s2 = 2; \t\t\t\t#drawdown of second well\n", - "H = 40; \n", - "r1 = 25; \t\t\t\t#dismath.tance of first well\n", - "r2 = 75; \t\t\t\t#dismath.tance of second well\n", - "h1 = H-s1;\n", - "h2 = H-s2;\n", - "k = Q*2.303*math.log10(r2/r1)/(math.pi*1000*60*(h2**2-h1**2));\n", - "T = b*k*1000;\n", - "print \"transmissibility = %.2fD-3 square metre/sec\"%(T);\n", - "\n", - "hw = (h2**2-(Q*2.303*math.log10(r2/rw)/(math.pi*k*60000)))**0.5;\n", - "sw = H-hw;\n", - "sw = round(sw*100)/100;\n", - "print \"drawdown at pumping well = %.2f m.\"%(sw);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "transmissibility = 3.13D-3 square metre/sec\n", - "drawdown at pumping well = 11.51 m.\n" - ] - } - ], - "prompt_number": 29 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.21 pg : 302" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "r = 0.25; \t\t\t\t#radius of test well\n", - "r1 = 10; \t\t\t\t#dismath.tance of first well\n", - "r2 = 60; \t\t\t\t#dismath.tance of second well\n", - "Q = 0.1; \t\t\t\t#discharge(cumec/sec)\n", - "s1 = 4; \t\t\t\t#drawdown of first well\n", - "s2 = 3; \t\t\t\t#drawdown of second well\n", - "b = 20; \t\t\t\t#thickness of well\n", - "k = 1000*Q*math.log10(r2/r1)/(2.72*b*(s1-s2));\n", - "print \"coefficient of permeability = %.2fD-3 m/sec\"%(k);\n", - "s = s2+Q*math.log10(r2/r)/(2.72*b*k);\n", - "s = round(s*100)/100;\n", - "print \"drawdown in test well = %.2f m.\"%(s); \n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "coefficient of permeability = 1.43D-3 m/sec\n", - "drawdown in test well = 3.00 m.\n" - ] - } - ], - "prompt_number": 30 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.22 pg : 303" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "h1 = 2.1; \t\t\t\t#initial pumping depression\n", - "h = 1.6; \t\t\t\t#heigth after recuperation\n", - "t = 90; \t\t\t\t#time\n", - "h2 = h1-h;\n", - "KbyA = 2.303*60*math.log10(h1/h2)/t;\n", - "Q = 10; \t\t\t\t#yield(lit/sec)\n", - "H = 2;\n", - "A = Q*3.6/(H*(KbyA));\n", - "d = (4*A/math.pi)**0.5;\n", - "d = round(d*10)/10;\n", - "print \"Daimeter of well = %.2f m\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Daimeter of well = 4.90 m\n" - ] - } - ], - "prompt_number": 31 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 5.23 pg : 303" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "h1 = 2.5; \t\t\t\t#initial pumping depression\n", - "h = 1; \t\t\t\t#heigth after recuperation\n", - "t = 60; \t\t\t\t#time\n", - "h2 = h1-h;\n", - "\n", - "# Calculations\n", - "KbyA = 2.303*60*math.log10(h1/h2)/t;\n", - "d = 2.; \t\t\t\t#diameter of well\n", - "H = 3.; \t\t\t\t#depression head\n", - "A = math.pi*d**2/4;\n", - "Q = (KbyA)*A*H;\n", - "Q = round(Q*1000)/1000;\n", - "\n", - "# Results\n", - "print \"Yield from well = %.2f cubic metre/hour.\"%(Q);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Yield from well = 4.82 cubic metre/hour.\n" - ] - } - ], - "prompt_number": 33 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch6.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch6.ipynb deleted file mode 100755 index 23b46e1d..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch6.ipynb +++ /dev/null @@ -1,654 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:cdf8fdbaa560cd66d2c449918f65db6ec9d56e694a7e8698e703276e2f6f0ae8" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 6 : RESERVIOR PLANNING" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.1 pg : 328" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros,linspace\n", - "#determine maximum reservior level\n", - "#maximum discharge over spillway\n", - "#plot inflow and routed hydrograph and find peak flow and peak lag\n", - "\n", - "#Given\n", - "e = [100, 100.3, 100.6, 100.9, 101.2, 101.5, 101.8, 102.1, 102.4, 102.7]; \t\t\t\t#elevation(km)\n", - "A = [405, 412, 420, 425, 428, 436, 445, 453, 460, 469]; \t\t\t\t#area\n", - "o = [0, 14.9, 42.2, 77.3, 119, 169, 217, 272, 334, 405]; \t\t\t\t#outflow\n", - "c = zeros(10)\n", - "dh = zeros(10)\n", - "s = zeros(10)\n", - "h = zeros(10)\n", - "h1 = zeros(10)\n", - "h2 = zeros(10)\n", - "for i in range(10):\n", - " dh[i] = e[i]-e[i-1];\n", - " s[i] = dh[i]/3*(A[i-1]+A[i]+(A[i-1]*A[i])**0.5); \t\t\t\t#storage between contours\n", - " c[i] = c[i-1]+s[i]; \t\t\t\t#cumulative storage\n", - " h[i] = c[i]/1.08; \t\t\t\t#2s/t\n", - " h1[i] = h[i]-o[i]; \t\t\t\t#2s/t-o\n", - " h2[i] = h[i]+o[i]; \t\t\t\t#2s/t+o\n", - "\n", - "T = linspace(0,102,17)\n", - "I = [42, 45, 57, 88, 147, 210, 272, 340, 350, 338, 314, 288, 263, 240, 198, 170, 143, 120]; \t\t\t\t#inflow\n", - "h4 = [0, 0, 60, 122, 185, 266, 362, 455, 545, 605, 623, 620, 600, 575, 550, 515, 470, 430]; \t\t\t\t#2s/t-0 obtained from curve a\n", - "O = [0, 10, 24, 42, 74, 130, 194, 260, 316, 334, 328, 312, 286, 264, 236, 204, 177, 150]; \t\t\t\t#outflow read from curve a\n", - "re = [100.2, 100.39, 100.58, 100.86, 101.26, 101.65, 102.03, 102.31, 102.4, 102.37, 102.3, 102.18, 102.06, 101.9, 101.72, 101.56, 102.4]; \t\t\t\t#reservior elevation read from curve b\n", - "t = zeros(17)\n", - "h3 = zeros(17)\n", - "for i in range(1,17):\n", - " t[i] = I[i-1]+I[i]; \t\t\t\t#I1+I2\n", - " h3[i] = t[i]+h4[i]; \t\t\t\t#2s/t+O\n", - "\n", - "pt = T[9]-T[8];\n", - "d = I[8]-O[9];\n", - "#results\n", - "print \" maximum reservior level = %.2f m.\"%(re[9]);\n", - "print \"maximum discharge over spillway = %.2f cumecs.\"%(O[9]);\n", - "print \"reduction in peak discharge = %.2f cumecs.\"%(d);\n", - "print \"peak lag = %.f hours.\"%(pt);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " maximum reservior level = 102.37 m.\n", - "maximum discharge over spillway = 334.00 cumecs.\n", - "reduction in peak discharge = 16.00 cumecs.\n", - "peak lag = 6 hours.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.2 pg : 331" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "in1 = [8.6, 2.2, 1.8, 0, 0, 13.5, 280.6, 510.2, 136, 52.5, 20.6, 12.3]; \t\t\t\t#inflow(ha-m)\n", - "pan = [2.2, 2.3, 3.1, 8.6, 12.8, 15.6, 12.3, 10.6, 10, 8.2, 5.8, 3]; \t\t\t\t#pan evaporation\n", - "p = [0.8, 1.2, 0, 0, 0, 4.8, 12.2, 18.6, 8.6, 1.5, 0, 0] \t\t\t\t#precipitation\n", - "D = [14.5, 15.8, 16.2, 16.8, 17.5, 18, 18, 17, 16.5, 16, 15.8, 15]; \t\t\t\t#Demand\n", - "s = 0;\n", - "r = zeros(12)\n", - "E = zeros(12)\n", - "P = zeros(12)\n", - "S = zeros(12)\n", - "# Calculations\n", - "for i in range(12):\n", - " if in1[i]<10:\n", - " r[i] = in1[i]; \t\t\t\t#D/S requirement\n", - " else:\n", - " r[i] = 10;\n", - "\n", - " E[i] = 3.6*pan[i]; \t\t\t\t#Evaporation over reservior area\n", - " P[i] = 3.5*p[i]; \t\t\t\t#Precipitation\n", - " I[i] = in1[i]-r[i]-E[i]+P[i]; \t\t\t\t#Adjusted inflow\n", - " S[i] = D[i]-I[i]; \t\t\t\t#Water required from storage\n", - " if S[i]<0:\n", - " S[i] = 0;\n", - "\n", - " s = s+S[i];\n", - "\n", - "\n", - "# Results\n", - "print \"required useful storage = %.2f ha-m.\"%(s);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required useful storage = 281.64 ha-m.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.3 pg : 333" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "V = 475; \t\t\t\t#flow required to be maintained throughout the year\n", - "Y = V*365*8.64; \t\t\t\t#yearly demand\n", - "\t\t\t\t#yearly demand gives the slope of demand curve\n", - "t = linspace(0,36,37) \t\t\t\t#number of season startin from 1960;each year is diveded into 3 seasons.\n", - "q = [0, 1050, 300, 50, 3000, 250, 40, 3500, 370, 90, 2000, 150, 120, 1200, 350, 65, 1400, 400, 100, 3600, 200, 80, 3000, 200, 80, 3000, 150, 120, 700, 210, 50, 800, 120, 80, 2400, 320, 120, 3200, 280, 80]; \t\t\t\t#average discharge\n", - "v = [0, 0.9707, 0.4717, 0.0328, 2.7734, 0.3981, 0.0263, 3.2357, 0.5818, 0.0591, 1.8490, 0.2356, 0.0788, 1.1094, 0.5504, 0.0427, 1.2943, 0.6290, 0.0657, 3.3281, 0.3145, 0.0525, 2.7734, 0.2359, 0.0788, 0.6441, 0.3302, 0.028, 0.7396, 0.1887, 0.0525, 2.2188, 0.5032, 0.0788, 2.9583, 0.4403, 0.0525]; \t\t\t\t#voloume\n", - "cv = zeros(37)\n", - "cv[0] = v[0];\n", - "for i in range(37):\n", - " cv[i] = cv[i-1]+v[i];\n", - "\n", - "#each year is divided into three seasons(monsoon,winter and summer).and readings are taken for 12 years\n", - "#mass inflow curve is plotted and math.tangent are drawn at the apexes and parellel to demand curve slope;\n", - "#the respectiv ordinates represent the deficiency during dry period\n", - "#maximum of these ordinates gives the desired reservior capacity\n", - "print \"storage capacity of reservior = 1.6 million ha-m.\";\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "storage capacity of reservior = 1.6 million ha-m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.4 pg : 336" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace,zeros\n", - "\t\t\t\t\n", - "#Given\n", - "asi = 3.6; \t\t\t\t#annual sediment inflow(x10**6)\n", - "gamma_s = 12; \t\t\t\t#specific weigth of sediment\n", - "vs = asi/12;\n", - "ir = 30.; \t\t\t\t#initial reservior capacity\n", - "fr = 60.; \t\t\t\t#final reservior capacity \n", - "r = ir/fr; \t\t\t\t#initial capacity/inflow ratio\n", - "\n", - "# Calculations\n", - "#r = 0.5; hence we start capacity/inflow ratio from 0.5\n", - "c = linspace(0.5,0.1,5) \t\t\t\t#capacity inflow ratio\n", - "e = [0.96 ,0.955, 0.95, 0.93, 0.87]; \t\t\t\t#trap efficiency\n", - "ae = zeros(4)\n", - "for i in range(4):\n", - " ae[i] = (e[i]+e[i+1])/2; \t\t\t\t#average efficiency for interval\n", - "\n", - "as1 = [0.2872, 0.2857, 0.2820, 0.2700]; \t\t\t\t#annual sediment trapped\n", - "s = 0\n", - "y = zeros(4)\n", - "for i in range(4):\n", - " y[i] = 6/as1[i]; \t\t\t\t#year to fill\n", - " s = s+y[i];\n", - "\n", - "# Results\n", - "print \" probable life of reservior = %i years.\"%(s);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " probable life of reservior = 85 years.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.5 pg : 337" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#corresponding maximum level of water above spillway crest\n", - "\n", - "#Given\n", - "I = [60, 480, 900, 470, 270, 160, 110, 80, 60]; \t\t\t\t#inflow\n", - "#for the first time interval 0 hours to 3 hours\n", - "I1 = I[0];\n", - "I2 = I[1];\n", - "t = 3*3600.;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 1.62*h1**1.5;\n", - "#storage change = (30+3h1)h1\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h1**2+0.54h1**1.5+10h1-0.972 = 0;\n", - "#solving it by trial and error method;we get\n", - "h1 = 0.0954;\n", - "#for the second time interval 3 hours to 6 hours\n", - "I1 = I[1];\n", - "I2 = I[2];\n", - "t = 3*3600.;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 0.0477+1.62*h2**1.5;\n", - "#storage change = (30+3h2)h2\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h2**2+0.54h2**1.5+10h2-3.4312 = 0;\n", - "#solving it by trial and error method;we get\n", - "h2 = 0.323;\n", - "#for the third time interval 6 hours to 9 hours\n", - "I1 = I[2];\n", - "I2 = I[3];\n", - "t = 3*3600.;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 0.2974+1.62*h3**1.5;\n", - "#storage change = (30+3h3)h3\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h3**2+0.54h3**1.5+10h3-5.7012 = 0;\n", - "#solving it by trial and error method;we get\n", - "h3 = 0.522;\n", - "#for the fourth time interval 9 hours to 12 hours\n", - "I1 = I[3];\n", - "I2 = I[4];\n", - "t = 3*3600.;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 0.611+1.62*h4**1.5;\n", - "#storage change = (30+3h4)h4\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h4**2+0.54h4**1.5+10h4-6.6208 = 0;\n", - "#solving it by trial and error method;we get\n", - "h4 = 0.601;\n", - "#for the fifth time interval 12 hours to 15 hours\n", - "I1 = I[4];\n", - "I2 = I[5];\n", - "t = 3*3600;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 0.7548+1.62*h5**1.5;\n", - "#storage change = (30+3h5)h5\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h5**2+0.54h5**1.5+10h5-6.8936 = 0;\n", - "#solving it by trial and error method;we get\n", - "h5 = 0.624;\n", - "#for the sixth time interval 12 hours to 15 hours\n", - "I1 = I[5];\n", - "I2 = I[6];\n", - "t = 3*3600.;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 0.7985.62*h6**1.5;\n", - "#storage change = (30+3h6)h6\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h6**2+0.54h6**1.5+10h6-6.8492 = 0;\n", - "#solving it by trial and error method;we get\n", - "h6 = 0.620;\n", - "hmax = h5;\n", - "q = 300*(h5)**1.5; \t\t\t\t#equation given\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"maximum outflow discharge over spillway = %.2f cumecs.\"%(q);\n", - "print \"maximum level of water above spillway crest = %.2f m.\"%(h5);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum outflow discharge over spillway = 147.88 cumecs.\n", - "maximum level of water above spillway crest = 0.62 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.6 pg : 339" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "t = 240.; \t\t\t\t#total math.cost of project(million rupees)\n", - "s = [32., 88., 72.]; \t\t\t\t#separable math.cost\n", - "eb = [40., 138., 112.]; \t\t\t\t#estimated benifit\n", - "sp = [47., 104., 101.]; \t\t\t\t#alternate math.single purpose math.cost\n", - "\t\t\t\t#umath.sing remaining benifit method\n", - "ts = s[0]+s[1]+s[2]; \t\t\t\t#total separable math.cost\n", - "tj = t-ts; \t\t\t\t#total joint math.cost\n", - "w = 0;\n", - "b = zeros(3)\n", - "rb = zeros(3)\n", - "for i in range(3):\n", - " if eb[i]<sp[i]:\n", - " b[i] = eb[i]; \t\t\t\t#benifit limited by alternate math.cost\n", - " else:\n", - " b[i] = sp[i];\n", - "\n", - " rb[i] = b[i]-s[i]; \t\t\t\t#remaining benifit\n", - " w = w+rb[i]; \n", - "y = 0;\n", - "aj = zeros(3)\n", - "ta = zeros(3)\n", - "for i in range(3):\n", - " aj[i] = tj*rb[i]/w; \t\t\t\t#allocated joint math.cost\n", - " ta[i] = s[i]+aj[i]; \t\t\t\t#total allocations\n", - " y = y+ta[i];\n", - "\n", - "print \"Using remaining benifit method.\";\n", - "print \"allocations to each project purposepercent:\";\n", - "per = zeros(3)\n", - "for i in range(3):\n", - " per[i] = ta[i]*100/y; \t\t\t\t#total allocation percent\n", - " print \"%.2f\"%(per[i]);\n", - "\n", - "\n", - "\n", - "#umath.sing alternate justifiable method\n", - "w = 0;\n", - "ac = zeros(3)\n", - "for i in range(3):\n", - " ac[i] = sp[i]-s[i]; \t\t\t\t#alternate math.cost less separable math.cost\n", - " w = w+ac[i]; \n", - "\n", - "y = 0;\n", - "ajc = zeros(3)\n", - "ta = zeros(3)\n", - "for i in range(3):\n", - " ajc[i] = tj*ac[i]/w; \t\t\t\t#allocated joint math.cost\n", - " ta[i] = s[i]+ajc[i]; \t\t\t\t#total allocation\n", - " y = y+ta[i];\n", - " \n", - "print \"Using alternate justifiable expenditure method method.\";\n", - "print \"allocations to each project purposepercent:\";\n", - "pr = zeros(3)\n", - "for i in range(3):\n", - " pr[i] = ta[i]*100/y; \t\t\t\t#total allocation percent\n", - " print \"%.2f\"%(pr[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Using remaining benifit method.\n", - "allocations to each project purposepercent:\n", - "16.35\n", - "42.70\n", - "40.94\n", - "Using alternate justifiable expenditure method method.\n", - "allocations to each project purposepercent:\n", - "18.33\n", - "42.00\n", - "39.67\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.8 pg : 342" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "I = [35, 55, 92, 130, 160, 140]; \t\t\t\t#inflow(cumec/sec)\n", - "x = 0.28;\n", - "K = 1.6; \t\t\t\t#studied value\n", - "t = 6;\n", - "K = K*24; \t\t\t\t#in hours\n", - "co = (-K*x+0.5*t)/(K-K*x+0.5*t);\n", - "c1 = (K*x+0.5*t)/(K-K*x+0.5*t);\n", - "c2 = (K-K*x-0.5*t)/(K-K*x+0.5*t);\n", - "c = co+c1+c2;\n", - "#c = 1; which implies (OK)\n", - "#from Muskingum equation\n", - "p1 = zeros(6)\n", - "p2 = zeros(6)\n", - "p3 = zeros(6)\n", - "O = zeros(6)\n", - "O[0] = 35;\n", - "print \"outflow hydrograph:%.2f\"%(O[0]);\n", - "for i in range(1,6):\n", - " p1[i] = co*I[i];\n", - " p2[i] = c1*I[i-1];\n", - " p3[i] = c2*O[i-1];\n", - " O[i] = p1[i]+p2[i]+p3[i];\n", - " O[i] = round(O[i]*100)/100;\n", - " print \"%.2f\"%(O[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "outflow hydrograph:35.00\n", - "29.94\n", - "25.49\n", - "28.90\n", - "41.10\n", - "69.44\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.9 pg : 343" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros_like,zeros\n", - "\t\t\t\t\n", - "#Given\n", - "md = [50, 75, 80, 85, 130, 120, 25, 25, 40, 45, 50, 60]; \t\t\t\t#monthly demand\n", - "e = [6, 8, 13, 17, 22, 22, 14,11, 13, 12, 7, 5]; \t\t\t\t#evaporation\n", - "r = [1, 0, 0, 0, 0, 19, 43, 39, 22, 6, 2, 1]; \t\t\t\t#rain1fall\n", - "in1 = [50, 40, 30, 25, 20, 30, 200, 225, 150, 90, 70, 60]; \t\t\t\t#monthly in1flow\n", - "A = 30.; \t\t\t\t#area of reservior\n", - "Cr = 0.4; \t\t\t\t#run-off coefficient\n", - "er = zeros_like(md)\n", - "ni = zeros(12)\n", - "niv = zeros(12)\n", - "nd =zeros(12)\n", - "for i in range(12):\n", - " er[i] = 0.4*r[i]; \t\t\t\t#effective rain1fall\n", - " ni[i] = er[i]-e[i]; \t\t\t\t#net in1flow\n", - " niv[i] = ni[i]*0.01*A; \t\t\t\t#net in1flow volume\n", - " nd[i] = md[i]-niv[i]; \t\t\t\t#net demand\n", - "\n", - "cnd = zeros(12)\n", - "ci = zeros(12)\n", - "cnd[0] = nd[0]; \t\t\t\t#cumulative demand\n", - "ci[0] = in1[0]; \t\t\t\t#cumulative in1flow\n", - "for i in range(1,12):\n", - " cnd[i] = cnd[i-1]+nd[i];\n", - " ci[i] = ci[i-1]+in1[i];\n", - "\n", - "ed = zeros(12,dtype=float)\n", - "es = zeros(12)\n", - "print \"Excess demand:\";\n", - "for i in range(12):\n", - " ed[i] = cnd[i]-ci[i]; \t\t\t\t#excess demand\n", - " if ed[i]<0: \n", - " es[i] = ed[i]; \t\t\t\t#excess supply\n", - " ed[i] = 0;\n", - "\n", - " print \"%.2f\"%(ed[i]);\n", - "\n", - "print \"min1imum storage required = Maximum of excess demand = %.2f Mm**3.\"%(ed[5]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Excess demand:\n", - "1.80\n", - "39.20\n", - "93.10\n", - "158.20\n", - "274.80\n", - "369.30\n", - "193.40\n", - "0.00\n", - "0.00\n", - "0.00\n", - "0.00\n", - "0.00\n", - "min1imum storage required = Maximum of excess demand = 369.30 Mm**3.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.10 pg : 344" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array,zeros_like\n", - "#minimum capacity of reservior\n", - "#the initial storage storage required to maintain uniform demand\n", - "\t\t\t\t\n", - "#Given\n", - "in1 = array([2.83, 4.25, 5.66, 18.4, 22.64, 22.64, 19.81, 8.49, 7.1, 7.1, 5.66, 5.66]); \t\t\t\t#inflow(x10**5)\n", - "s = sum(in1)\n", - "avd = s/12; \t\t\t\t#average demand(x10**5)\n", - "s = 0\n", - "t = 0;\n", - "e = zeros_like(in1)\n", - "D = zeros_like(in1)\n", - "for i in range(12):\n", - " e[i] = avd-in1[i];\n", - " if e[i]<0:\n", - " S[i] = -e[i]; \t\t\t\t#surplus(x10**5)\n", - " s = s+S[i];\n", - " else:\n", - " D[i] = e[i]; \t\t\t\t#Deficit(x10**5)\n", - " t = t+D[i];\n", - " \n", - "d = (s-(t-D[0]-D[1]-D[2]));\n", - "s = s;\n", - "\n", - "print \"minimum capacity of reservior = %.2fD+5 cumec.\"%(s);\n", - "print \"storage required to maintain uniform demand = %.2fD+5 cumec\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "minimum capacity of reservior = 40.08D+5 cumec.\n", - "storage required to maintain uniform demand = 19.82D+5 cumec\n" - ] - } - ], - "prompt_number": 13 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch6_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch6_1.ipynb deleted file mode 100755 index 23b46e1d..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch6_1.ipynb +++ /dev/null @@ -1,654 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:cdf8fdbaa560cd66d2c449918f65db6ec9d56e694a7e8698e703276e2f6f0ae8" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 6 : RESERVIOR PLANNING" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.1 pg : 328" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros,linspace\n", - "#determine maximum reservior level\n", - "#maximum discharge over spillway\n", - "#plot inflow and routed hydrograph and find peak flow and peak lag\n", - "\n", - "#Given\n", - "e = [100, 100.3, 100.6, 100.9, 101.2, 101.5, 101.8, 102.1, 102.4, 102.7]; \t\t\t\t#elevation(km)\n", - "A = [405, 412, 420, 425, 428, 436, 445, 453, 460, 469]; \t\t\t\t#area\n", - "o = [0, 14.9, 42.2, 77.3, 119, 169, 217, 272, 334, 405]; \t\t\t\t#outflow\n", - "c = zeros(10)\n", - "dh = zeros(10)\n", - "s = zeros(10)\n", - "h = zeros(10)\n", - "h1 = zeros(10)\n", - "h2 = zeros(10)\n", - "for i in range(10):\n", - " dh[i] = e[i]-e[i-1];\n", - " s[i] = dh[i]/3*(A[i-1]+A[i]+(A[i-1]*A[i])**0.5); \t\t\t\t#storage between contours\n", - " c[i] = c[i-1]+s[i]; \t\t\t\t#cumulative storage\n", - " h[i] = c[i]/1.08; \t\t\t\t#2s/t\n", - " h1[i] = h[i]-o[i]; \t\t\t\t#2s/t-o\n", - " h2[i] = h[i]+o[i]; \t\t\t\t#2s/t+o\n", - "\n", - "T = linspace(0,102,17)\n", - "I = [42, 45, 57, 88, 147, 210, 272, 340, 350, 338, 314, 288, 263, 240, 198, 170, 143, 120]; \t\t\t\t#inflow\n", - "h4 = [0, 0, 60, 122, 185, 266, 362, 455, 545, 605, 623, 620, 600, 575, 550, 515, 470, 430]; \t\t\t\t#2s/t-0 obtained from curve a\n", - "O = [0, 10, 24, 42, 74, 130, 194, 260, 316, 334, 328, 312, 286, 264, 236, 204, 177, 150]; \t\t\t\t#outflow read from curve a\n", - "re = [100.2, 100.39, 100.58, 100.86, 101.26, 101.65, 102.03, 102.31, 102.4, 102.37, 102.3, 102.18, 102.06, 101.9, 101.72, 101.56, 102.4]; \t\t\t\t#reservior elevation read from curve b\n", - "t = zeros(17)\n", - "h3 = zeros(17)\n", - "for i in range(1,17):\n", - " t[i] = I[i-1]+I[i]; \t\t\t\t#I1+I2\n", - " h3[i] = t[i]+h4[i]; \t\t\t\t#2s/t+O\n", - "\n", - "pt = T[9]-T[8];\n", - "d = I[8]-O[9];\n", - "#results\n", - "print \" maximum reservior level = %.2f m.\"%(re[9]);\n", - "print \"maximum discharge over spillway = %.2f cumecs.\"%(O[9]);\n", - "print \"reduction in peak discharge = %.2f cumecs.\"%(d);\n", - "print \"peak lag = %.f hours.\"%(pt);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " maximum reservior level = 102.37 m.\n", - "maximum discharge over spillway = 334.00 cumecs.\n", - "reduction in peak discharge = 16.00 cumecs.\n", - "peak lag = 6 hours.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.2 pg : 331" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "in1 = [8.6, 2.2, 1.8, 0, 0, 13.5, 280.6, 510.2, 136, 52.5, 20.6, 12.3]; \t\t\t\t#inflow(ha-m)\n", - "pan = [2.2, 2.3, 3.1, 8.6, 12.8, 15.6, 12.3, 10.6, 10, 8.2, 5.8, 3]; \t\t\t\t#pan evaporation\n", - "p = [0.8, 1.2, 0, 0, 0, 4.8, 12.2, 18.6, 8.6, 1.5, 0, 0] \t\t\t\t#precipitation\n", - "D = [14.5, 15.8, 16.2, 16.8, 17.5, 18, 18, 17, 16.5, 16, 15.8, 15]; \t\t\t\t#Demand\n", - "s = 0;\n", - "r = zeros(12)\n", - "E = zeros(12)\n", - "P = zeros(12)\n", - "S = zeros(12)\n", - "# Calculations\n", - "for i in range(12):\n", - " if in1[i]<10:\n", - " r[i] = in1[i]; \t\t\t\t#D/S requirement\n", - " else:\n", - " r[i] = 10;\n", - "\n", - " E[i] = 3.6*pan[i]; \t\t\t\t#Evaporation over reservior area\n", - " P[i] = 3.5*p[i]; \t\t\t\t#Precipitation\n", - " I[i] = in1[i]-r[i]-E[i]+P[i]; \t\t\t\t#Adjusted inflow\n", - " S[i] = D[i]-I[i]; \t\t\t\t#Water required from storage\n", - " if S[i]<0:\n", - " S[i] = 0;\n", - "\n", - " s = s+S[i];\n", - "\n", - "\n", - "# Results\n", - "print \"required useful storage = %.2f ha-m.\"%(s);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "required useful storage = 281.64 ha-m.\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.3 pg : 333" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "V = 475; \t\t\t\t#flow required to be maintained throughout the year\n", - "Y = V*365*8.64; \t\t\t\t#yearly demand\n", - "\t\t\t\t#yearly demand gives the slope of demand curve\n", - "t = linspace(0,36,37) \t\t\t\t#number of season startin from 1960;each year is diveded into 3 seasons.\n", - "q = [0, 1050, 300, 50, 3000, 250, 40, 3500, 370, 90, 2000, 150, 120, 1200, 350, 65, 1400, 400, 100, 3600, 200, 80, 3000, 200, 80, 3000, 150, 120, 700, 210, 50, 800, 120, 80, 2400, 320, 120, 3200, 280, 80]; \t\t\t\t#average discharge\n", - "v = [0, 0.9707, 0.4717, 0.0328, 2.7734, 0.3981, 0.0263, 3.2357, 0.5818, 0.0591, 1.8490, 0.2356, 0.0788, 1.1094, 0.5504, 0.0427, 1.2943, 0.6290, 0.0657, 3.3281, 0.3145, 0.0525, 2.7734, 0.2359, 0.0788, 0.6441, 0.3302, 0.028, 0.7396, 0.1887, 0.0525, 2.2188, 0.5032, 0.0788, 2.9583, 0.4403, 0.0525]; \t\t\t\t#voloume\n", - "cv = zeros(37)\n", - "cv[0] = v[0];\n", - "for i in range(37):\n", - " cv[i] = cv[i-1]+v[i];\n", - "\n", - "#each year is divided into three seasons(monsoon,winter and summer).and readings are taken for 12 years\n", - "#mass inflow curve is plotted and math.tangent are drawn at the apexes and parellel to demand curve slope;\n", - "#the respectiv ordinates represent the deficiency during dry period\n", - "#maximum of these ordinates gives the desired reservior capacity\n", - "print \"storage capacity of reservior = 1.6 million ha-m.\";\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "storage capacity of reservior = 1.6 million ha-m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.4 pg : 336" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import linspace,zeros\n", - "\t\t\t\t\n", - "#Given\n", - "asi = 3.6; \t\t\t\t#annual sediment inflow(x10**6)\n", - "gamma_s = 12; \t\t\t\t#specific weigth of sediment\n", - "vs = asi/12;\n", - "ir = 30.; \t\t\t\t#initial reservior capacity\n", - "fr = 60.; \t\t\t\t#final reservior capacity \n", - "r = ir/fr; \t\t\t\t#initial capacity/inflow ratio\n", - "\n", - "# Calculations\n", - "#r = 0.5; hence we start capacity/inflow ratio from 0.5\n", - "c = linspace(0.5,0.1,5) \t\t\t\t#capacity inflow ratio\n", - "e = [0.96 ,0.955, 0.95, 0.93, 0.87]; \t\t\t\t#trap efficiency\n", - "ae = zeros(4)\n", - "for i in range(4):\n", - " ae[i] = (e[i]+e[i+1])/2; \t\t\t\t#average efficiency for interval\n", - "\n", - "as1 = [0.2872, 0.2857, 0.2820, 0.2700]; \t\t\t\t#annual sediment trapped\n", - "s = 0\n", - "y = zeros(4)\n", - "for i in range(4):\n", - " y[i] = 6/as1[i]; \t\t\t\t#year to fill\n", - " s = s+y[i];\n", - "\n", - "# Results\n", - "print \" probable life of reservior = %i years.\"%(s);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - " probable life of reservior = 85 years.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.5 pg : 337" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#corresponding maximum level of water above spillway crest\n", - "\n", - "#Given\n", - "I = [60, 480, 900, 470, 270, 160, 110, 80, 60]; \t\t\t\t#inflow\n", - "#for the first time interval 0 hours to 3 hours\n", - "I1 = I[0];\n", - "I2 = I[1];\n", - "t = 3*3600.;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 1.62*h1**1.5;\n", - "#storage change = (30+3h1)h1\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h1**2+0.54h1**1.5+10h1-0.972 = 0;\n", - "#solving it by trial and error method;we get\n", - "h1 = 0.0954;\n", - "#for the second time interval 3 hours to 6 hours\n", - "I1 = I[1];\n", - "I2 = I[2];\n", - "t = 3*3600.;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 0.0477+1.62*h2**1.5;\n", - "#storage change = (30+3h2)h2\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h2**2+0.54h2**1.5+10h2-3.4312 = 0;\n", - "#solving it by trial and error method;we get\n", - "h2 = 0.323;\n", - "#for the third time interval 6 hours to 9 hours\n", - "I1 = I[2];\n", - "I2 = I[3];\n", - "t = 3*3600.;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 0.2974+1.62*h3**1.5;\n", - "#storage change = (30+3h3)h3\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h3**2+0.54h3**1.5+10h3-5.7012 = 0;\n", - "#solving it by trial and error method;we get\n", - "h3 = 0.522;\n", - "#for the fourth time interval 9 hours to 12 hours\n", - "I1 = I[3];\n", - "I2 = I[4];\n", - "t = 3*3600.;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 0.611+1.62*h4**1.5;\n", - "#storage change = (30+3h4)h4\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h4**2+0.54h4**1.5+10h4-6.6208 = 0;\n", - "#solving it by trial and error method;we get\n", - "h4 = 0.601;\n", - "#for the fifth time interval 12 hours to 15 hours\n", - "I1 = I[4];\n", - "I2 = I[5];\n", - "t = 3*3600;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 0.7548+1.62*h5**1.5;\n", - "#storage change = (30+3h5)h5\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h5**2+0.54h5**1.5+10h5-6.8936 = 0;\n", - "#solving it by trial and error method;we get\n", - "h5 = 0.624;\n", - "#for the sixth time interval 12 hours to 15 hours\n", - "I1 = I[5];\n", - "I2 = I[6];\n", - "t = 3*3600.;\n", - "ti = (I1+I2)*t/2; \t\t\t\t#total inflow\n", - "#outflow = 0.7985.62*h6**1.5;\n", - "#storage change = (30+3h6)h6\n", - "#from the basic equation i.e total inflow = total outflow+change in storage\n", - "#on solving we get\n", - "#h6**2+0.54h6**1.5+10h6-6.8492 = 0;\n", - "#solving it by trial and error method;we get\n", - "h6 = 0.620;\n", - "hmax = h5;\n", - "q = 300*(h5)**1.5; \t\t\t\t#equation given\n", - "q = round(q*100)/100;\n", - "\n", - "# Results\n", - "print \"maximum outflow discharge over spillway = %.2f cumecs.\"%(q);\n", - "print \"maximum level of water above spillway crest = %.2f m.\"%(h5);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum outflow discharge over spillway = 147.88 cumecs.\n", - "maximum level of water above spillway crest = 0.62 m.\n" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.6 pg : 339" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "t = 240.; \t\t\t\t#total math.cost of project(million rupees)\n", - "s = [32., 88., 72.]; \t\t\t\t#separable math.cost\n", - "eb = [40., 138., 112.]; \t\t\t\t#estimated benifit\n", - "sp = [47., 104., 101.]; \t\t\t\t#alternate math.single purpose math.cost\n", - "\t\t\t\t#umath.sing remaining benifit method\n", - "ts = s[0]+s[1]+s[2]; \t\t\t\t#total separable math.cost\n", - "tj = t-ts; \t\t\t\t#total joint math.cost\n", - "w = 0;\n", - "b = zeros(3)\n", - "rb = zeros(3)\n", - "for i in range(3):\n", - " if eb[i]<sp[i]:\n", - " b[i] = eb[i]; \t\t\t\t#benifit limited by alternate math.cost\n", - " else:\n", - " b[i] = sp[i];\n", - "\n", - " rb[i] = b[i]-s[i]; \t\t\t\t#remaining benifit\n", - " w = w+rb[i]; \n", - "y = 0;\n", - "aj = zeros(3)\n", - "ta = zeros(3)\n", - "for i in range(3):\n", - " aj[i] = tj*rb[i]/w; \t\t\t\t#allocated joint math.cost\n", - " ta[i] = s[i]+aj[i]; \t\t\t\t#total allocations\n", - " y = y+ta[i];\n", - "\n", - "print \"Using remaining benifit method.\";\n", - "print \"allocations to each project purposepercent:\";\n", - "per = zeros(3)\n", - "for i in range(3):\n", - " per[i] = ta[i]*100/y; \t\t\t\t#total allocation percent\n", - " print \"%.2f\"%(per[i]);\n", - "\n", - "\n", - "\n", - "#umath.sing alternate justifiable method\n", - "w = 0;\n", - "ac = zeros(3)\n", - "for i in range(3):\n", - " ac[i] = sp[i]-s[i]; \t\t\t\t#alternate math.cost less separable math.cost\n", - " w = w+ac[i]; \n", - "\n", - "y = 0;\n", - "ajc = zeros(3)\n", - "ta = zeros(3)\n", - "for i in range(3):\n", - " ajc[i] = tj*ac[i]/w; \t\t\t\t#allocated joint math.cost\n", - " ta[i] = s[i]+ajc[i]; \t\t\t\t#total allocation\n", - " y = y+ta[i];\n", - " \n", - "print \"Using alternate justifiable expenditure method method.\";\n", - "print \"allocations to each project purposepercent:\";\n", - "pr = zeros(3)\n", - "for i in range(3):\n", - " pr[i] = ta[i]*100/y; \t\t\t\t#total allocation percent\n", - " print \"%.2f\"%(pr[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Using remaining benifit method.\n", - "allocations to each project purposepercent:\n", - "16.35\n", - "42.70\n", - "40.94\n", - "Using alternate justifiable expenditure method method.\n", - "allocations to each project purposepercent:\n", - "18.33\n", - "42.00\n", - "39.67\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.8 pg : 342" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros\n", - "\t\t\t\t\n", - "#Given\n", - "I = [35, 55, 92, 130, 160, 140]; \t\t\t\t#inflow(cumec/sec)\n", - "x = 0.28;\n", - "K = 1.6; \t\t\t\t#studied value\n", - "t = 6;\n", - "K = K*24; \t\t\t\t#in hours\n", - "co = (-K*x+0.5*t)/(K-K*x+0.5*t);\n", - "c1 = (K*x+0.5*t)/(K-K*x+0.5*t);\n", - "c2 = (K-K*x-0.5*t)/(K-K*x+0.5*t);\n", - "c = co+c1+c2;\n", - "#c = 1; which implies (OK)\n", - "#from Muskingum equation\n", - "p1 = zeros(6)\n", - "p2 = zeros(6)\n", - "p3 = zeros(6)\n", - "O = zeros(6)\n", - "O[0] = 35;\n", - "print \"outflow hydrograph:%.2f\"%(O[0]);\n", - "for i in range(1,6):\n", - " p1[i] = co*I[i];\n", - " p2[i] = c1*I[i-1];\n", - " p3[i] = c2*O[i-1];\n", - " O[i] = p1[i]+p2[i]+p3[i];\n", - " O[i] = round(O[i]*100)/100;\n", - " print \"%.2f\"%(O[i]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "outflow hydrograph:35.00\n", - "29.94\n", - "25.49\n", - "28.90\n", - "41.10\n", - "69.44\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.9 pg : 343" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import zeros_like,zeros\n", - "\t\t\t\t\n", - "#Given\n", - "md = [50, 75, 80, 85, 130, 120, 25, 25, 40, 45, 50, 60]; \t\t\t\t#monthly demand\n", - "e = [6, 8, 13, 17, 22, 22, 14,11, 13, 12, 7, 5]; \t\t\t\t#evaporation\n", - "r = [1, 0, 0, 0, 0, 19, 43, 39, 22, 6, 2, 1]; \t\t\t\t#rain1fall\n", - "in1 = [50, 40, 30, 25, 20, 30, 200, 225, 150, 90, 70, 60]; \t\t\t\t#monthly in1flow\n", - "A = 30.; \t\t\t\t#area of reservior\n", - "Cr = 0.4; \t\t\t\t#run-off coefficient\n", - "er = zeros_like(md)\n", - "ni = zeros(12)\n", - "niv = zeros(12)\n", - "nd =zeros(12)\n", - "for i in range(12):\n", - " er[i] = 0.4*r[i]; \t\t\t\t#effective rain1fall\n", - " ni[i] = er[i]-e[i]; \t\t\t\t#net in1flow\n", - " niv[i] = ni[i]*0.01*A; \t\t\t\t#net in1flow volume\n", - " nd[i] = md[i]-niv[i]; \t\t\t\t#net demand\n", - "\n", - "cnd = zeros(12)\n", - "ci = zeros(12)\n", - "cnd[0] = nd[0]; \t\t\t\t#cumulative demand\n", - "ci[0] = in1[0]; \t\t\t\t#cumulative in1flow\n", - "for i in range(1,12):\n", - " cnd[i] = cnd[i-1]+nd[i];\n", - " ci[i] = ci[i-1]+in1[i];\n", - "\n", - "ed = zeros(12,dtype=float)\n", - "es = zeros(12)\n", - "print \"Excess demand:\";\n", - "for i in range(12):\n", - " ed[i] = cnd[i]-ci[i]; \t\t\t\t#excess demand\n", - " if ed[i]<0: \n", - " es[i] = ed[i]; \t\t\t\t#excess supply\n", - " ed[i] = 0;\n", - "\n", - " print \"%.2f\"%(ed[i]);\n", - "\n", - "print \"min1imum storage required = Maximum of excess demand = %.2f Mm**3.\"%(ed[5]);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Excess demand:\n", - "1.80\n", - "39.20\n", - "93.10\n", - "158.20\n", - "274.80\n", - "369.30\n", - "193.40\n", - "0.00\n", - "0.00\n", - "0.00\n", - "0.00\n", - "0.00\n", - "min1imum storage required = Maximum of excess demand = 369.30 Mm**3.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 6.10 pg : 344" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import array,zeros_like\n", - "#minimum capacity of reservior\n", - "#the initial storage storage required to maintain uniform demand\n", - "\t\t\t\t\n", - "#Given\n", - "in1 = array([2.83, 4.25, 5.66, 18.4, 22.64, 22.64, 19.81, 8.49, 7.1, 7.1, 5.66, 5.66]); \t\t\t\t#inflow(x10**5)\n", - "s = sum(in1)\n", - "avd = s/12; \t\t\t\t#average demand(x10**5)\n", - "s = 0\n", - "t = 0;\n", - "e = zeros_like(in1)\n", - "D = zeros_like(in1)\n", - "for i in range(12):\n", - " e[i] = avd-in1[i];\n", - " if e[i]<0:\n", - " S[i] = -e[i]; \t\t\t\t#surplus(x10**5)\n", - " s = s+S[i];\n", - " else:\n", - " D[i] = e[i]; \t\t\t\t#Deficit(x10**5)\n", - " t = t+D[i];\n", - " \n", - "d = (s-(t-D[0]-D[1]-D[2]));\n", - "s = s;\n", - "\n", - "print \"minimum capacity of reservior = %.2fD+5 cumec.\"%(s);\n", - "print \"storage required to maintain uniform demand = %.2fD+5 cumec\"%(d);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "minimum capacity of reservior = 40.08D+5 cumec.\n", - "storage required to maintain uniform demand = 19.82D+5 cumec\n" - ] - } - ], - "prompt_number": 13 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch8.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch8.ipynb deleted file mode 100755 index 595c81ff..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch8.ipynb +++ /dev/null @@ -1,1584 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:6fa8fe978c728d97fa9f04464822ea9606521023bb2c2c301624274cfcebacb1" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 8 : GRAVITY DAMS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.1 pg : 376" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from scipy.integrate import quad \n", - "\t\t\t\t\n", - "#Given\n", - "H = 100; \t\t\t\t#heigth of dam\n", - "wb = 70; \t\t\t\t#width of base of dam\n", - "wt = 7; \t\t\t\t#width of top of dam\n", - "l = 1; \t\t\t\t#length of dam\n", - "hw = 98; \t\t\t\t#heigth of water in dam\n", - "hsu = 90; \t\t\t\t#heigth of slope on downstream side\n", - "s = 1/0.7; \t\t\t\t#slope on downstream side\n", - "gammad = 24; \t\t\t\t#unit weigth of dam\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "E = 2.05e7; \t\t\t\t#modulus of elasticity\n", - "\n", - "\t\t\t\t#(a) inertial forces and moments\n", - "alpha0 = 0.05; \t\t\t\t#from table 8.1\n", - "alphah = 2*alpha0;\n", - "\t\t\t\t#at 10m from top\n", - "\n", - "def f0(y): \n", - "\t return 25.2-0.25*y\n", - "\n", - "F10 = quad(f0,0,10)[0]\n", - "\n", - "\n", - "def f1(y): \n", - "\t return 25.2*(1-0.01*y)*(10-y)\n", - "\n", - "M10 = quad(f1,0,10)[0]\n", - "\n", - "\t\t\t\t#at 100m below top\n", - "\n", - "def f2(y): \n", - "\t return 0.15*(1-0.01*y)*16.8*y\n", - "\n", - "F100 = F10+ quad(f2,10,100)[0]\n", - "\n", - "\n", - "def f3(y): \n", - "\t return 0.15*(1-0.01*y)*16.8*y*(100-y)\n", - "\n", - "M100 = M10+90*F10+ quad(f3,10,100)[0]\n", - "\n", - "print \"Inertial forces:At 10m from top: F = %.2f kn;M = %ikn-mAt 100m from top: F = %.2f kn;M = %ikn-m.\"%(F10,M10,F100,M100);\n", - "\n", - "\t\t\t\t#(b) hydrodynamic pressure and moment\n", - "\t\t\t\t#at 10m from top\n", - "y = 8.;\n", - "W10 = 1680.;\n", - "alphah = F10/W10;\n", - "Cm = 0.735;\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "p = Cy*alphah*gammaw*hw;\n", - "P10 = 0.726*p*y;\n", - "Mp10 = 0.299*p*y**2;\n", - "P10 = round(P10*100)/100;\n", - "Mp10 = round(Mp10*100)/100;\n", - "\t\t\t\t#at 100m from top\n", - "y = 98;\n", - "W100 = 84840;\n", - "alphah = F100/W100;\n", - "Cm = 0.735;\n", - "Cy = (Cm/2)*(y*(2-y/hw)/hw+(y*(2-y/hw)/hw)**0.5);\n", - "p = Cy*alphah*gammaw*hw;\n", - "P100 = 0.726*p*y;\n", - "Mp100 = 0.299*p*y**2;\n", - "print \"Hydrodynamic forces:At 10m from top: F = %.2f kn;\\\n", - "\\nM = %.2fkn-mAt 100m from top: F = %i kn;\\\n", - "\\nM = %ikn-m.\"%(P10,Mp10,P100,Mp100);\n", - "\n", - "# rounding off error." - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Inertial forces:At 10m from top: F = 239.50 kn;M = 1218kn-mAt 100m from top: F = 4321.90 kn;M = 221790kn-m.\n", - "Hydrodynamic forces:At 10m from top: F = 161.58 kn;\n", - "M = 532.35kn-mAt 100m from top: F = 2561 kn;\n", - "M = 103366kn-m.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.2 pg : 378" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "H = 100; \t\t\t\t#heigth of dam\n", - "wb = 70; \t\t\t\t#width of base of dam\n", - "wt = 7; \t\t\t\t#width of top of dam\n", - "l = 1; \t\t\t\t#length of dam\n", - "hw = 98; \t\t\t\t#heigth of water in dam\n", - "hsu = 90; \t\t\t\t#heigth of slope on downstream side\n", - "s = 1/0.7; \t\t\t\t#slope on downstream side\n", - "gammad = 24; \t\t\t\t#unit weigth of dam\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "E = 2.05e7; \t\t\t\t#modulus of elasticity\n", - "beta = 1;\n", - "I = 2;\n", - "Fo = 0.25; \t\t\t\t#from table 8.2\n", - " \t\t\t\t#t = Sa/g;\n", - "t = 0.19; \t\t\t\t#from fig. 8.4\n", - "alphah = beta*I*Fo*t;\n", - "T = 5.55*H**2/wb*(gammad/(gammaw*E))**0.5;\n", - "\t\t\t\t#(a) Base shear\n", - "W = l*gammad*(wt*H+((hsu/s)*hsu)/2);\n", - "Fb = 0.6*W*alphah;\n", - "print \"Base shear = %.2f KN.\"%(Fb);\n", - "\n", - "\t\t\t\t#(b) Base moment\n", - "hbar = ((wt*H**2/2)+((hsu/s)*hsu**2/6))/((wt*H)+(hsu/s)*hsu/2);\n", - "Mb = 0.9*W*hbar*alphah;\n", - "print \"Base moment = %.2f KN-m.\"%(Mb);\n", - "\n", - "\t\t\t\t#(c) shear at 10m from top\n", - "Cv = 0.08;\n", - "F10 = Cv*Fb;\n", - "F10 = round(F10);\n", - "print \"shear at 10m from top = %.2f KN.\"%(F10);\n", - "\n", - "\t\t\t\t#(d) Moment at 10m from top\n", - "Cm = 0.02;\n", - "M10 = Cm*Mb;\n", - "M10 = round(M10);\n", - "print \"moment at 10m from top = %.2f KN.\"%(M10);\n", - "\t\t\t\t#(e) Hydrodynamic pressure\n", - "\t\t\t\t#at 10m from top\n", - "y = 8;\n", - "W10 = 1680;\n", - "Cm = 0.735;\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "p = Cy*alphah*gammaw*hw;\n", - "P10 = 0.726*p*y;\n", - "Mp10 = 0.299*p*y**2;\n", - "P10 = round(P10*100)/100;\n", - "Mp10 = round(Mp10*100)/100;\n", - "\t\t\t\t#at 100m from top\n", - "y = 98;\n", - "W100 = 84840;\n", - "Cm = 0.735;\n", - "Cy = (Cm/2)*(y*(2-y/hw)/hw+(y*(2-y/hw)/hw)**0.5);\n", - "p = Cy*alphah*gammaw*hw;\n", - "P100 = 0.726*p*y;\n", - "Mp100 = 0.299*p*y**2;\n", - "print \"Hydrodynamic forces:At 10m from top: F = %.2f kn;\\\n", - "\\nM = %.2fkn-mAt 100m from top: F = %i \\\n", - "kn;M = %ikn-m.\"%(P10,Mp10,P100,Mp100);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Base shear = 4835.88 KN.\n", - "Base moment = 246342.60 KN-m.\n", - "shear at 10m from top = 387.00 KN.\n", - "moment at 10m from top = 4927.00 KN.\n", - "Hydrodynamic forces:At 10m from top: F = 0.00 kn;\n", - "M = 0.00kn-mAt 100m from top: F = 4776 kn;M = 192765kn-m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.3 pg : 379" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from scipy.integrate import quad \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "H = 100; \t\t\t\t#heigth of dam\n", - "wb = 70; \t\t\t\t#width of base of dam\n", - "wt = 7; \t\t\t\t#width of top of dam\n", - "l = 1; \t\t\t\t#length of dam\n", - "hw = 98; \t\t\t\t#heigth of water in dam\n", - "hsu = 90; \t\t\t\t#heigth of slope on downstream side\n", - "s = 1/0.7; \t\t\t\t#slope on downstream side\n", - "gammad = 24; \t\t\t\t#unit weigth of dam\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "E = 2.05e7; \t\t\t\t#modulus of elasticity\n", - "\t\t\t\t#(a) Seismic coefficient method\n", - "alpha0 = 0.05; \t\t\t\t#from table 8.1\n", - "alphah = 2*alpha0;\n", - "alphav = 0.75*alphah;\n", - "\t\t\t\t#at 10m from top\n", - "\n", - "def f4(y): \n", - "\t return alphav*168*(1-0.01*y)\n", - "\n", - "F10 = quad(f4,0,10)[0]\n", - "\n", - "\t\t\t\t#at 100m below top\n", - "\n", - "def f5(y): \n", - "\t return alphav*(1-0.01*y)*16.8*y\n", - "\n", - "F100 = F10+ quad(f5,10,100)[0]\n", - "\n", - "print \"Parta):At 10m from top: F = %.2f knAt 100m from top: F = %.2f kn.\"%(F10,F100);\n", - "\n", - "\t\t\t\t#(b)Response spectrum method\n", - "beta = 1;\n", - "I = 2;\n", - "Fo = 0.25; \t\t\t\t#from table 8.2\n", - " \t\t\t\t#t = Sa/g;\n", - "t = 0.19; \t\t\t\t#from fig. 8.4\n", - "alphah = beta*I*Fo*t;\n", - "alphav = 0.75*alphah;\n", - "\t\t\t\t#at 10m from top\n", - "\n", - "def f6(y): \n", - "\t return alphav*168*(1-0.01*y)\n", - "\n", - "F10 = quad(f6,0,10)[0]\n", - "\n", - "\t\t\t\t#at 100m below top\n", - "\n", - "def f7(y): \n", - "\t return alphav*(1-0.01*y)*16.8*y\n", - "\n", - "F100 = F10+ quad(f7,10,100)[0]\n", - "\n", - "F100 = round(F100*100)/100;\n", - "print \"Partb):At 10m from top: F = %.2f knAt 100m from top: F = %.2f kn.\"%(F10,F100);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Parta):At 10m from top: F = 119.70 knAt 100m from top: F = 2160.90 kn.\n", - "Partb):At 10m from top: F = 113.72 knAt 100m from top: F = 2052.86 kn.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.4 pg : 381\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given\n", - "H = 100.; \t\t\t\t#heigth of dam\n", - "wb = 73.; \t\t\t\t#width of base of dam\n", - "wt = 7; \t\t\t\t#width of top of dam\n", - "l = 1.; \t\t\t\t#length of dam\n", - "hw = 98.; \t\t\t\t#heigth of water in dam\n", - "hsu = 90.; \t\t\t\t#heigth of slope on downstream side\n", - "s = 1/0.7; \t\t\t\t#slope on downstream side\n", - "gammad = 24.; \t\t\t\t#unit weigth of dam\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "E = 2.05e7; \t\t\t\t#modulus of elasticity\n", - "\n", - "#at 10m from top\n", - "y = 8;\n", - "alphah = 0.1;\n", - "Cm = 0.72;\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "p10 = Cy*alphah*gammaw*hw;\n", - "F10 = 0.726*p10*y;\n", - "Mp10 = 0.299*p10*y**2;\n", - "\n", - "#at 40m from top\n", - "y = 38;\n", - "alphah = 0.1;\n", - "Cm = 0.72;\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "p40 = Cy*alphah*gammaw*hw;\n", - "F40 = 0.726*p40*y;\n", - "Mp40 = 0.299*p40*y**2;\n", - "\n", - "#at 100m from top\n", - "y = 98;\n", - "alphah = 0.1;\n", - "Cm = 0.72;\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "p100 = Cy*alphah*gammaw*hw;\n", - "F100 = 0.726*p100*y;\n", - "Mp100 = 0.299*p100*y**2;\n", - "p10 = round(p10*1000)/1000;\n", - "F10 = round(F10*1000)/1000;\n", - "Mp10 = round(Mp10*10)/10;\n", - "p40 = round(p40*1000)/1000;\n", - "F40 = round(F40*1000)/1000;\n", - "Mp40 = round(Mp40*10)/10;\n", - "p100 = round(p100*100)/100;\n", - "F100 = round(F100*1000)/1000;\n", - "Mp100 = round(Mp100*10)/10;\n", - "print \"Hydrodynamic Forces:At 10m from top: P = %.2f KN/square m;\\\n", - "\\nF = %.2f KN;\\\n", - "\\nM = %.2f KN-m.At 40m from top: P = %.2f KN/square m.;\\\n", - "\\nF = %.2f KN;\\\n", - "\\nM = %.2f KN-m.At 100m from top: P = %.2f KN/square m;\\\n", - "\\nF = %.2f KN;\\\n", - "\\nM = %.2f KN-m.\"%(p10,F10,Mp10,p40,F40,Mp40,p100,F100,Mp100);\n", - "\n", - "#vertical component of reservior water on horizontal section\n", - "s1 = 3./60;\n", - "Wh = (F100-F40)*s1;\n", - "Wh = round(Wh*100)/100;\n", - "print \"vertical component of reservior water on horizontal section = %.2f kN/m.\"%(Wh);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Hydrodynamic Forces:At 10m from top: P = 19.12 KN/square m;\n", - "F = 111.03 KN;\n", - "M = 365.80 KN-m.At 40m from top: P = 49.00 KN/square m.;\n", - "F = 1351.85 KN;\n", - "M = 21156.60 KN-m.At 100m from top: P = 69.22 KN/square m;\n", - "F = 4924.82 KN;\n", - "M = 198770.00 KN-m.\n", - "vertical component of reservior water on horizontal section = 178.65 kN/m.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.8 pg : 413" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#no tension is permissible\n", - "#factor of safety against slidingis 1.5\n", - "\n", - "#Given\n", - "\n", - "wb = 3; \t\t\t\t#width of dam;\n", - "miu = 0.5; \t\t\t\t#coefficient of friction\n", - "Sg = 2.4; \t\t\t\t#specific gravity of masonary\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "c = 1;\n", - "\n", - "#when uplift is considered\n", - "#when no tension is permissible then e = wb/6;\n", - "\n", - "p1 = wb*Sg*gamma_w;\n", - "p2 = c*wb*gamma_w/2;\n", - "p3 = p1-p2;\n", - "p4 = p1*wb/2-p2*2;\n", - "p5 = gamma_w/6;\n", - "d1 = p4/p3; d2 = p5/p3;\n", - "d3 = 1.5-d1;\n", - "H = ((0.5-d3)/d2)**0.5;\n", - "H = round(H*100)/100;\n", - "print \"when uplift is considered:\"\n", - "print \"Heigth of dam when no tension is permissible = %.2f m.\"%(H);\n", - "H = p3*0.5/(1.5*p5*3);\n", - "print \"Heigth of dam when factor of safety against sliding is 1.5 = %.2f m.\"%(H);\n", - "\n", - "#when uplift is not considered\n", - "p1 = wb*Sg*gamma_w;\n", - "p4 = p1*wb/2;\n", - "p5 = gamma_w/6;\n", - "d1 = p4/p1;\n", - "d2 = p5/p1;\n", - "H = (0.5/d2)**0.5;\n", - "H = round(H*100)/100;\n", - "print \"when uplift is not considered:\"\n", - "print \"Heigth of dam when no tension is permissible = %.2f m.\"%(H);\n", - "H = p1*0.5/(1.5*p5*3);\n", - "print \"Heigth of dam when factor of safety against sliding is 1.5 = %.2f m.\"%(H);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "when uplift is considered:\n", - "Heigth of dam when no tension is permissible = 3.55 m.\n", - "Heigth of dam when factor of safety against sliding is 1.5 = 3.80 m.\n", - "when uplift is not considered:\n", - "Heigth of dam when no tension is permissible = 4.65 m.\n", - "Heigth of dam when factor of safety against sliding is 1.5 = 4.80 m.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.9 pg : 414" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "c = 1;\n", - "hw = 6; \t\t\t\t#heigth of water in reservior\n", - "Bt = 1.5; \t\t\t\t#width of top of dam\n", - "H = 6; \t\t\t\t#heigth of the dam\n", - "wb = 4.5; \t\t\t\t#width of base of dam\n", - "Sg = 2.4; \t\t\t\t#specific gravity of masonary\n", - "gamma_w = 9.81; \t\t\t\t#weigth density of water\n", - "\n", - "W1 = Bt*gamma_w*Sg*H;\n", - "W2 = gamma_w*Sg*H*(wb-Bt)/2;\n", - "L1 = (wb-Bt)+(Bt/2);\n", - "L2 = (2*(wb-Bt))/3\n", - "M1 = W1*L1\n", - "M2 = W2*L2\n", - "\n", - "#Reaervior empty\n", - "SumW = W1+W2;\n", - "SumM = M1+M2;\n", - "x = SumM/SumW;\n", - "e = wb/2-x;\n", - "pnt = (SumW/wb)*(1+(6*e/wb));\n", - "pnh = (SumW/wb)*(1-(6*e/wb));\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "print \"Reservior empty:\";\n", - "print \" Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "\n", - "#Reservior full\n", - "W3 = gamma_w*H**2/2;\n", - "U = gamma_w*H*c*wb/2;\n", - "SumV = SumW-U;\n", - "L3 = hw/3;\n", - "L4 = 2*wb/3; \t\t\t\t#lever arm\n", - "M3 = W3*L3;\n", - "M4 = U*L4; \t\t\t\t#moment about toe\n", - "SumM1 = SumM-M4-M3;\n", - "x = SumM1/SumV;\n", - "e = wb/2-x;\n", - "pnt = (SumV/wb)*(1+(6*e/wb));\n", - "pnh = (SumV/wb)*(1-(6*e/wb));\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "print \"Reservior full:\";\n", - "print \" Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Reservior empty:\n", - " Normal stress at toe = 15.70 kN/square.m.\n", - "Normal stress at heel = 172.70 kN/square.m.\n", - "Reservior full:\n", - " Normal stress at toe = 120.30 kN/square.m.\n", - "Normal stress at heel = 9.20 kN/square.m.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.10 pg : 415" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "\n", - "#check the stability\n", - "\n", - "#Given\n", - "c = 1;\n", - "hw = 6; \t\t\t\t#heigth of water in reservior\n", - "Bt = 1.5; \t\t\t\t#width of top of dam\n", - "H = 6; \t\t\t\t#heigth of the dam\n", - "gamma_m = 20; \t\t\t\t#unit weigth of masonary \n", - "gamma_w = 9.81; \t\t\t\t#weigth density of water\n", - "f = 1800; \t\t\t\t#compressive strength\n", - "miu = 0.6; \t\t\t\t#coefficient of friction\n", - "\n", - "#to develop no tension e = b/6;x = b/3.\n", - "#hence on solving the relations we get\n", - "\n", - "P = [1,2.944,-39.074]\n", - "wb = roots(P)[1]; \t\t\t\t#sign of coefficient is 2.944 is not taken correctly in book\n", - "#roots are 4.94 and -7.89\n", - "#math.since negative value cannot be taken\n", - "print \"Neglecting the negative value.Width of base is = 4.94 m.\";\n", - "W1 = Bt*gamma_m*H;\n", - "W2 = gamma_m*H*(wb-Bt)/2;\n", - "L1 = (wb-Bt)+(Bt/2);\n", - "L2 = (2*(wb-Bt))/3;\n", - "M1 = W1*L1,\n", - "M2 = W2*L2;\n", - "U = gamma_w*H*c*wb/2;\n", - "L4 = 2*wb/3;\n", - "M4 = U*L4;\n", - "W3 = gamma_w*H**2/2;\n", - "L3 = hw/3;\n", - "M3 = W3*L3;\n", - "SumW = W1+W2-U;\n", - "SumM = M1+M2-M4-M3;\n", - "pn = 2*SumW/wb;\n", - "pn = round(pn*10)/10;\n", - "print \"Maximum stress = %.2f kN/square.m.\"%(pn);\n", - "print \"Dam is safe against compression\";\n", - "FOS = miu*SumW/W3;\n", - "FOS = round(FOS*100)/100;\n", - "print \"Factor of safety against sliding = %.2f. <1\"%(FOS);\n", - "print \"Dam is unsafe against sliding.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Neglecting the negative value.Width of base is = 4.94 m.\n", - "Maximum stress = 97.50 kN/square.m.\n", - "Dam is safe against compression\n", - "Factor of safety against sliding = 0.82. <1\n", - "Dam is unsafe against sliding.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.11 pg : 416" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\t\t\t\t#check the stability if uplift is neglected\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1;\n", - "hw = 6; \t\t\t\t#heigth of water in reservior\n", - "Bt = 1.5; \t\t\t\t#width of top of dam\n", - "H = 6; \t\t\t\t#heigth of the dam\n", - "gamma_m = 20; \t\t\t\t#unit weigth of masonary \n", - "gamma_w = 9.81; \t\t\t\t#weigth density of water\n", - "f = 1800; \t\t\t\t#compressive strength\n", - "miu = 0.6; \t\t\t\t#coefficient of friction\n", - "\n", - "\t\t\t\t#to develop no tension e = b/6;x = b/3.\n", - "\t\t\t\t#hence on solving the relations we get\n", - "\n", - "P = [1,1.5,-19.908]\n", - "wb = roots(P)[1];\n", - "\n", - "#roots are 3.774 and -5.27\n", - "#math.since negative value cannot be taken\n", - "\n", - "print \"Neglecting the negative value.Width of base is = %.2f m.\"%wb;\n", - "\n", - "W1 = Bt*gamma_m*H;\n", - "W2 = gamma_m*H*(wb-Bt)/2;\n", - "L1 = (wb-Bt)+(Bt/2);\n", - "L2 = (2*(wb-Bt))/3;\n", - "M1 = W1*L1,\n", - "M2 = W2*L2;\n", - "W3 = gamma_w*H**2/2;\n", - "L3 = hw/3;\n", - "M3 = W3*L3;\n", - "SumW = W1+W2;\n", - "SumM = M1+M2-M3;\n", - "pn = 2*SumW/wb;\n", - "pn = round(pn*10)/10;\n", - "print \"Maximum stress = %.2f kN/square.m.\"%(pn);\n", - "print \"Dam is safe against compression\";\n", - "\n", - "FOS = miu*SumW/W3;\n", - "FOS = round(FOS*1000)/1000;\n", - "print \"Factor of safety against sliding = %.2f. > 1\"%(FOS);\n", - "print \"Dam is safe against sliding.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Neglecting the negative value.Width of base is = 3.77 m.\n", - "Maximum stress = 167.70 kN/square.m.\n", - "Dam is safe against compression\n", - "Factor of safety against sliding = 1.07. > 1\n", - "Dam is safe against sliding.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.12 pg : 417" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# calculate maximum permissible heigth of shutter so that no tension develops\n", - "\t\t\t\t\n", - "#Given\n", - "Bt = 3; \t\t\t\t#width of top of dam\n", - "H = 12; \t\t\t\t#heigth of the dam\n", - "wb = 9; \t\t\t\t#width of base of dam\n", - "gamma_m = 21; \t\t\t\t#unit weigth of masonary\n", - "gamma_w = 9.81; \t\t\t\t#weigth density of water\n", - "\n", - "W1 = Bt*gamma_m*H;\n", - "W2 = gamma_m*H*(wb-Bt)/2;\n", - "\n", - "#taking moment about a point on base at 3m from toe\n", - "L1 = 3+Bt/2;\n", - "L2 = (2*(wb-Bt)/3)-3;\n", - "M1 = W1*L1\n", - "M2 = W2*L2;\n", - "M = M1+M2;\n", - "\n", - "#net moment about this point should be zero for equilibrium\n", - "s = (M*6/gamma_w)**(1./3)-12;\n", - "s = round(s*100)/100;\n", - "\n", - "# Results\n", - "print \"maximum permissible heigth of shutter = %.2f m.\"%(s);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum permissible heigth of shutter = 1.22 m.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.13 pg : 417" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t#moment at 50m below water surface\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1;\n", - "H = 100; \t\t\t\t#heigth of dam\n", - "hw = 100; \t\t\t\t#heigth of water in reservior\n", - "FB = 1; \t\t\t\t#free board\n", - "s = 0.15; \t\t\t\t#slope of upstream face\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "alphah = 0.1;\n", - "\n", - "theta = math.atan(s);\n", - "y = 50;\n", - "Cm = 0.735*(1-(theta*2/math.pi));\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "pe = Cy*alphah*gamma_w*hw;\n", - "F = 0.726*pe*y;\n", - "M = 0.299*pe*y**2;\n", - "pe = round(pe*1000)/1000;\n", - "F = round(F*10)/10;\n", - "M = round(M*10)/10;\n", - "print \"hydrodynamic earthquake pressure = %.2f kN/square.mshear = %.2f kN/m.Moment = %.2f kN-m/m.\"%(pe,F,M);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "hydrodynamic earthquake pressure = 65.27 kN/square.mshear = 2369.30 kN/m.Moment = 48788.60 kN-m/m.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.14 pg : 418" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from sympy import sec\n", - "\n", - "#check stability\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1.;\n", - "H = 10.; \t\t\t\t#heigth of dam\n", - "hw = 10.; \t\t\t\t#heigth of water in reservior\n", - "wb = 8.25; \t\t\t\t#bottom width\n", - "Bt = 1.; \t\t\t\t#top width\n", - "Hs1 = 0.1; \t\t\t\t#slope on upstream side\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "gamma_m = 22.4; \t\t\t\t#unit weigth of masonary\n", - "f = 1400.; \t\t\t\t#permissible shear stress at joint\n", - "miu = 0.75; \t\t\t\t#coefficient of friction\n", - "fi = math.atan(0.625);\n", - "theta = math.atan(0.1);\n", - "\n", - "W1 = Bt*H*gamma_m;\n", - "W2 = H*H*Hs1*gamma_m/2;\n", - "W3 = H*6.25*gamma_m/2;\n", - "W4 = hw*gamma_w*H*Hs1/2;\n", - "P = gamma_w*hw**2/2;\n", - "U = wb*gamma_w*hw*c/2;\n", - "SumV = W1+W2+W3+W4-U;\n", - "L3 = 2*(wb-(Hs1*H)-Bt)/3;\n", - "L1 = (wb-(Hs1*H)-Bt)+Bt/2;\n", - "L2 = (wb-(Hs1*H)-Bt)+Bt+(Hs1*H/3);\n", - "L4 = (wb-(Hs1*H)-Bt)+Bt+(2*Hs1*H/3);\n", - "L5 = 2*wb/3;L6 = hw/3;\n", - "M1 = W1*L1;\n", - "M2 = W2*L2;\n", - "M3 = W3*L3;\n", - "M4 = W4*L4;\n", - "M5 = U*L5;\n", - "M6 = P*L6;\n", - "SumM = M1+M2+M3+M4-M5-M6;\n", - "Mplus = M1+M2+M3+M4;\n", - "Mminus = M5+M6;\n", - "FOS = miu*SumV/P;\n", - "SFF = (miu*SumV+wb*1400)/P;\n", - "FOO = Mplus/Mminus;\n", - "FOS = round(FOS*100)/100;\n", - "SFF = round(SFF*10)/10;\n", - "FOO = round(FOO*100)/100;\n", - "print \"Factor of safety against sliding = %.2f. >1 \"%(FOS);\n", - "print \"Shear friction factor = %.2f.\"%(SFF);\n", - "print \"Factor of safety against overturning = %.2f. <1.5\"%(FOO);\n", - "print \"Dam is unsafe against overturning\";\n", - "\n", - "x = SumM/SumV;\n", - "e = wb/2-x;\n", - "p = hw*gamma_w;\n", - "pnt = (SumV/wb)*(1+(6*e/wb)); \t\t\t\t#calculation is done wrong in book;value of b is not taken correctly\n", - "pnh = (SumV/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "sigmah = pnh*sec(theta)**2-p*math.tan(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = -(pnh-p)*math.tan(theta);\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "sigmat = round(sigmat*10)/10;\n", - "sigmah = round(sigmah*10)/10;\n", - "taut = round(taut*10)/10;\n", - "tauh = round(tauh*10)/10;\n", - "print \"Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %.2f kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %.2f kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %.2f kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %.2f kN/square.m.\"%(tauh);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Factor of safety against sliding = 1.04. >1 \n", - "Shear friction factor = 24.60.\n", - "Factor of safety against overturning = 1.47. <1.5\n", - "Dam is unsafe against overturning\n", - "Normal stress at toe = 170.70 kN/square.m.\n", - "Normal stress at heel = -5.80 kN/square.m.\n", - "Principal stress at toe = 237.40 kN/square.m.\n", - "Principal stress at heel = -6.80 kN/square.m.\n", - "Shear stress at toe = 106.70 kN/square.m.\n", - "Shear stress at heel = 10.40 kN/square.m.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.15 pg : 420" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t#Check the stability and determine sliding factor and shear factor\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1;\n", - "miu = 0.75; \t\t\t\t#coefficient of friction\n", - "H = 90; \t\t\t\t#heigth of dam\n", - "wb = 73.1; \t\t\t\t#width of base\n", - "Bt = 7; \t\t\t\t#width of top of dam\n", - "hw = 89; \t\t\t\t#heigth of water in reservior\n", - "Hs1 = 28; \t\t\t\t#heigth of slope on upstream side\n", - "Hs2 = 83; \t\t\t\t#heigth of slope on downstream side\n", - "Cm = 0.735;\n", - "alphah = 0.1;\n", - "gamma_m = 23.5; \t\t\t\t#unit weigth of concrete\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "theta = math.atan(8./28);\n", - "fi = math.atan(0.7);\n", - "\t\t\t\t#self weigth of dam\n", - "W1 = (Hs1*8*gamma_m)/2\n", - "W2 = (Bt*H*gamma_m)\n", - "W3 = (Hs2**2*0.7*gamma_m)/2\n", - "\t\t\t\t#weigth of superimposed water\n", - "W4 = (Hs1*8*gamma_w)/2\n", - "W5 = (hw-Hs1)*8*gamma_w\n", - "U = hw*wb*2*gamma_w/6; \t\t\t\t#uplift force\n", - "wp = hw**2*gamma_w/2; \t\t\t\t#water pressure\n", - "hp = 0.726*Cm*alphah*gamma_w*hw**2; \t\t\t\t#hydrodynamic pressure\n", - "Mhp = 0.299*Cm*alphah*gamma_w*hw**3; \t\t\t\t#moment due to hydrodynamic pressure\n", - "\t\t\t\t#inertial load due to horizontal acceleration\n", - "I1 = W2/10;\n", - "I2 = W3/10;\n", - "I3 = W1/10;\n", - "SumV = W1+W2+W3+W4+W5-U;\n", - "SumH = wp+hp+I1+I2+I3;\n", - "L1 = (wb-8)+8/3\n", - "L2 = (0.7*Hs2)+(Bt/2)\n", - "L3 = (2*Hs2*0.7)/3.\n", - "L4 = (wb-8)+(2*8)/3.\n", - "L5 = (wb-8)+(8./2)\n", - "L6 = hw/3;\n", - "L7 = 2*wb/3;\n", - "M1 = W1*L1\n", - "M2 = W2*L2\n", - "M3 = W3*L3\n", - "M4 = W4*L4;\n", - "M5 = W5*L5;\n", - "M6 = wp*L6;\n", - "M7 = U*L7;\n", - "M8 = I1*45;\n", - "M9 = I2*83/3;\n", - "M10 = I3*28/3;\n", - "Mplus = M1+M2+M3+M4+M5;\n", - "Mminus = M6+M7+M8+M9+M10+Mhp;\n", - "SumM = Mplus-Mminus;\n", - "x = SumM/SumV;\n", - "e = wb/2-x;\n", - "pnt = (SumV/wb)*(1+(6*e/wb));\n", - "pnh = (SumV/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "p = hw*gamma_w;\n", - "pe = Cm*alphah*gamma_w*hw;\n", - "sigmah = pnh*sec(theta)**2-(p+pe)*math.tan(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = -(-pnh-(p+pe))*math.tan(theta);\n", - "print \"Normal stress at toe = %i kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %i kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %i kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %i kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %i kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %i kN/square.m.\"%(tauh);\n", - "\n", - "FOS = miu*SumV/SumH;\n", - "SFF = (miu*SumV+wb*1400)/SumH;\n", - "FOO = Mplus/Mminus;\n", - "Ffi = 1.2;Fc = 2.4;\n", - "F = (miu*SumV/Ffi+1400*wb/Fc)/SumH;\n", - "FOS = round(FOS*100)/100;\n", - "F = round(F*100)/100;\n", - "SFF = round(SFF*100)/100;\n", - "FOO = round(FOO*100)/100;\n", - "print \"Factor of safety against sliding as per IS:6512-1972 = %.2f. <1.5\"%(FOS);\n", - "print \"Factor of safety against sliding as per IS:6512-1984 = %.2f. >1\"%(F);\n", - "print \"Shear friction factor = %.2f. <6\"%(SFF);\n", - "print \"Factor of safety against overturning = %.2f. <1.5\"%(FOO);\n", - "print \"Dam is unsafe for given loading conditions\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Normal stress at toe = 1929 kN/square.m.\n", - "Normal stress at heel = -323 kN/square.m.\n", - "Principal stress at toe = 2874 kN/square.m.\n", - "Principal stress at heel = -426 kN/square.m.\n", - "Shear stress at toe = 1350 kN/square.m.\n", - "Shear stress at heel = 175 kN/square.m.\n", - "Factor of safety against sliding as per IS:6512-1972 = 0.87. <1.5\n", - "Factor of safety against sliding as per IS:6512-1984 = 1.57. >1\n", - "Shear friction factor = 2.90. <6\n", - "Factor of safety against overturning = 1.45. <1.5\n", - "Dam is unsafe for given loading conditions\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.16 pg : 423" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t#Check the stability and determine principal and shear stress at toe and heel\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1;\n", - "miu = 0.7; \t\t\t\t#coefficient of friction\n", - "H = 70; \t\t\t\t#heigth of dam\n", - "ht = 0; \t\t\t\t#heigth of tail water\n", - "Lf = 6.5; \t\t\t\t#location of foundation gallery from heel\n", - "wb = 52.5; \t\t\t\t#width of base\n", - "Bt = 7; \t\t\t\t#width of top of dam\n", - "hw = 70; \t\t\t\t#heigth of water in reservior\n", - "Hs1 = 35; \t\t\t\t#heigth of slope on upstream side\n", - "Hs2 = 60; \t\t\t\t#heigth of slope on downstream side\n", - "gamma_m = 24; \t\t\t\t#unit weigth of concrete\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "theta = math.atan(0.1);\n", - "fi = math.atan(0.7);\n", - "\t\t\t\t#self weigth of dam\n", - "W1 = (Hs1*3.5*gamma_m)/2\n", - "W2 = (Bt*H*gamma_m)\n", - "W3 = (Hs2**2*0.7*gamma_m)/2\n", - "\t\t\t\t#weigth of superimposed water\n", - "W4 = (Hs1*3.5*gamma_w)/2\n", - "W5 = (hw-Hs1)*3.5*gamma_w\n", - "wp = hw**2*gamma_w/2; \t\t\t\t#water pressure\n", - "Pt = gamma_w*ht\n", - "Ph = gamma_w*hw\n", - "Pg = (ht+(hw-ht)/3)*gamma_w\n", - "U = (Pt*(wb-Lf))+(Pg*Lf)+((Ph-Pg)*Lf/2)+((Pg-Pt)*(wb-Lf)/2)*c\n", - "l1 = (wb-Lf)/2\n", - "l2 = (2*(wb-Lf))/3\n", - "l3 = (wb-Lf)+(Lf/2)\n", - "l4 = (wb-Lf)+((2*Lf)/3)\n", - "L7 = (((Pt*(wb-Lf))*l1)+((Pg-Pt)*(wb-Lf)*l2/2)+((Pg*Lf)*l3)+((Ph-Pg)*Lf*l4/2))/U\n", - "L1 = (wb-3.5)+3.5/3\n", - "L2 = (0.7*Hs2)+(Bt/2)\n", - "L3 = (2*Hs2*0.7)/3\n", - "L4 = (wb-3.5)+(2*3.5)/3\n", - "L5 = (wb-3.5)+(3.5/2)\n", - "L6 = hw/3;\n", - "M1 = W1*L1\n", - "M2 = W2*L2\n", - "M3 = W3*L3\n", - "M4 = W4*L4;\n", - "M5 = W5*L5;\n", - "M6 = wp*L6;\n", - "M7 = U*L7;\n", - "SumV1 = W1+W2+W3;\n", - "SumM1 = M1+M2+M3;\n", - "SumV2 = SumV1+W4+W5;\n", - "SumM2 = SumM1+M4+M5-M6;\n", - "SumV3 = SumV2-U;\n", - "SumM3 = SumM2-M7;\n", - "Mplus = 1547377;\n", - "Mminus = 870421;\n", - "SumH = wp;\n", - "\n", - "\t\t\t\t#case 1. Reservior empty\n", - "x = SumM1/SumV1;\n", - "e = wb/2-x;\n", - "pnt = (SumV1/wb)*(1+(6*e/wb));\n", - "pnh = (SumV1/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "sigmah = pnh*sec(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = pnh*math.tan(theta);\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "sigmat = round(sigmat*10)/10;\n", - "sigmah = round(sigmah*10)/10;\n", - "taut = round(taut*10)/10;\n", - "tauh = round(tauh*10)/10;\n", - "print \"case 1. Reservior empty:\";\n", - "print \"Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %.2f kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %.2f kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %.2f kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %.2f kN/square.m.\"%(tauh);\n", - "\n", - "\t\t\t\t#case2. reservior full without uplift\n", - "x = SumM2/SumV2;\n", - "e = wb/2-x;\n", - "p = hw*gamma_w;\n", - "pnt = (SumV2/wb)*(1+(6*e/wb));\n", - "pnh = (SumV2/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "sigmah = pnh*sec(theta)**2-p*math.tan(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = -(pnh-p)*math.tan(theta);\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "sigmat = round(sigmat*10)/10;\n", - "sigmah = round(sigmah*10)/10;\n", - "taut = round(taut*10)/10;\n", - "tauh = round(tauh*10)/10;\n", - "print \"case 2. reservior full without uplift:\";\n", - "print \"Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %.2f kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %.2f kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %.2f kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %.2f kN/square.m.\"%(tauh);\n", - "\n", - "\t\t\t\t#case3. reservior full with uplift\n", - "x = SumM3/SumV3;\n", - "e = wb/2-x;\n", - "p = hw*gamma_w;\n", - "pnt = (SumV3/wb)*(1+(6*e/wb));\n", - "pnh = (SumV3/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "sigmah = pnh*sec(theta)**2-p*math.tan(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = -(pnh-p)*math.tan(theta);\n", - "pnt = round(pnt);\n", - "pnh = round(pnh);\n", - "sigmat = round(sigmat);\n", - "sigmah = round(sigmah);\n", - "taut = round(taut);\n", - "tauh = round(tauh);\n", - "print \"case 3. reservior full with uplift:\";\n", - "print \"Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %.2f kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %.2f kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %.2f kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %.2f kN/square.m.\"%(tauh);\n", - "\n", - "FOS = miu*SumV3/SumH;\n", - "SFF = (miu*SumV3+wb*1400)/SumH;\n", - "FOO = Mplus/Mminus;\n", - "Ffi = 1.5;Fc = 3.6;\n", - "F = (miu*SumV3/Ffi+1400*wb/Fc)/SumH;\n", - "FOS = round(FOS*1000)/1000;\n", - "SFF = round(SFF*100)/100;\n", - "FOO = round(FOO*100)/100;\n", - "F = round(F*1000)/1000;\n", - "print \"Factor of safety against sliding = %.2f.\"%(FOS);\n", - "print \"Shear friction factor = %.2f.\"%(SFF);\n", - "print \"Factor of safety against overturning = %.2f.\"%(FOO);\n", - "print \"Factor of safety for load combination B = %.2f. > 1\"%(F);\n", - "print \"Dam is safe \";\n", - "\n", - "\t\t\t\t#Case4.considering seismic forces\n", - "Cm = 0.712;\n", - "alphah = 0.1;\n", - "alphav = 0.08;\n", - "hp = 0.726*Cm*alphah*gamma_w*hw**2; \t\t\t\t#hydrodynamic pressure\n", - "Mhp = 0.299*Cm*alphah*gamma_w*hw**3; \t\t\t\t#moment due to hydrodynamic pressure\n", - "\t\t\t\t#inertial load due to horizontal acceleration\n", - "I1 = W2/10;\n", - "I2 = W3/10;\n", - "I3 = W1/10;\n", - "v = SumV1*alphav;\n", - "Mv = 116444;\n", - "SumV4 = SumV3-v;\n", - "SumH1 = SumH+I1+I2+I3+hp;\n", - "M8 = I1*35;\n", - "M9 = I2*20;\n", - "M10 = I3*35/3;\n", - "Mminus1 = 1161849;\n", - "SumM4 = SumM3-M8-M9-M10-Mhp-Mv;\n", - "\n", - "x = SumM4/SumV4;\n", - "e = wb/2-x;\n", - "p = hw*gamma_w;\n", - "pe = Cm*alphah*gamma_w*hw;\n", - "pnt = (SumV4/wb)*(1+(6*e/wb));\n", - "pnh = (SumV4/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "sigmah = pnh*sec(theta)**2-(p+pe)*math.tan(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = (-pnh+(p+pe))*math.tan(theta);\n", - "pnt = round(pnt);\n", - "pnh = round(pnh);\n", - "sigmat = round(sigmat);\n", - "sigmah = round(sigmah);\n", - "taut = round(taut);\n", - "tauh = round(tauh);\n", - "print \"case 4.considering seismic forces\";\n", - "print \"Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %.2f kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %.2f kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %.2f kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %.2f kN/square.m.\"%(tauh); \t\t\t\t#answer is wrong in book\n", - "\n", - "FOS = miu*SumV4/SumH1;\n", - "SFF = (miu*SumV4+wb*1400)/SumH1;\n", - "FOO = Mplus/Mminus1;\n", - "Ffi = 1.2;Fc = 2.7;\n", - "F = (miu*SumV4/Ffi+1400*wb/Fc)/SumH1;\n", - "FOS = round(FOS*1000)/1000;\n", - "SFF = round(SFF*100)/100;\n", - "FOO = round(FOO*100)/100;\n", - "F = round(F*100)/100;\n", - "print \"Factor of safety against sliding = %.2f.\"%(FOS);\n", - "print \"Shear friction factor = %.2f.\"%(SFF);\n", - "print \"Factor of safety against overturning = %.2f.\"%(FOO);\n", - "print \"Factor of safety for load combination E = %.2f. > 1\"%(F);\n", - "print \"Dam is safe \";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "case 1. Reservior empty:\n", - "Normal stress at toe = 156.30 kN/square.m.\n", - "Normal stress at heel = 1499.70 kN/square.m.\n", - "Principal stress at toe = 232.80 kN/square.m.\n", - "Principal stress at heel = 1514.70 kN/square.m.\n", - "Shear stress at toe = 109.40 kN/square.m.\n", - "Shear stress at heel = 150.00 kN/square.m.\n", - "case 2. reservior full without uplift:\n", - "Normal stress at toe = 1297.10 kN/square.m.\n", - "Normal stress at heel = 427.60 kN/square.m.\n", - "Principal stress at toe = 1932.60 kN/square.m.\n", - "Principal stress at heel = 425.00 kN/square.m.\n", - "Shear stress at toe = 907.90 kN/square.m.\n", - "Shear stress at heel = 25.90 kN/square.m.\n", - "case 3. reservior full with uplift:\n", - "Normal stress at toe = 1344.00 kN/square.m.\n", - "Normal stress at heel = 70.00 kN/square.m.\n", - "Principal stress at toe = 2002.00 kN/square.m.\n", - "Principal stress at heel = 64.00 kN/square.m.\n", - "Shear stress at toe = 941.00 kN/square.m.\n", - "Shear stress at heel = 62.00 kN/square.m.\n", - "Factor of safety against sliding = 1.08.\n", - "Shear friction factor = 4.14.\n", - "Factor of safety against overturning = 1.00.\n", - "Factor of safety for load combination B = 1.57. > 1\n", - "Dam is safe \n", - "case 4.considering seismic forces\n", - "Normal stress at toe = 1713.00 kN/square.m.\n", - "Normal stress at heel = -432.00 kN/square.m.\n", - "Principal stress at toe = 2552.00 kN/square.m.\n", - "Principal stress at heel = -443.00 kN/square.m.\n", - "Shear stress at toe = 1199.00 kN/square.m.\n", - "Shear stress at heel = 117.00 kN/square.m.\n", - "Factor of safety against sliding = 0.76.\n", - "Shear friction factor = 3.14.\n", - "Factor of safety against overturning = 1.00.\n", - "Factor of safety for load combination E = 1.52. > 1\n", - "Dam is safe \n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.17 pg : 429" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design practical profile of gravity dam\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1;\n", - "rlb = 1450; \t\t\t\t#R.L of base of dam\n", - "rlw = 1480.5; \t\t\t\t#R.L of water level\n", - "Sg = 2.4; \t\t\t\t#specific gravity of masonary\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "w = 1; \t\t\t\t#heigth of waves\n", - "f = 1200; \t\t\t\t#safe compressive stress for masonary\n", - "FB = 1.5*w;\n", - "rlt = FB+rlw; \t\t\t\t#R.L of top of dam\n", - "H = rlt-rlb; \t\t\t\t#heigth of dam\n", - "LH = f/(gamma_w*(Sg+1))\n", - "LH = round(LH*100)/100;\n", - "print \"Heigth of dam = %.2f m.\"%(H);\n", - "print \"limiting heigth of dam = %.2f m.\"%(LH);\n", - "print \"Dam is low gravity dam\";\n", - "hw = rlw-rlb;\n", - "\t\t\t\t#keep top width,a = 4.5.\n", - "a = 4.5;\n", - "P = hw/(Sg**0.5);\n", - "P = round(P*10)/10;\n", - "print \"Base width of elementary profile = %.2f m.\"%(P);\n", - "uo = a/16;\n", - "wb = uo+P;\n", - "wb = round(wb);\n", - "print \"Base width = %.2f m.\"%(wb);\n", - "D = 2*a*(Sg**0.5);\n", - "D = round(D);\n", - "print \"Dismath.tance upto which u/s slope is vertical from water level = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Heigth of dam = 32.00 m.\n", - "limiting heigth of dam = 35.98 m.\n", - "Dam is low gravity dam\n", - "Base width of elementary profile = 19.70 m.\n", - "Base width = 20.00 m.\n", - "Dismath.tance upto which u/s slope is vertical from water level = 14.00 m.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.18 pg : 430" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#determine if dam is safe against sliding\n", - "\t\t\t\t\n", - "#Given\n", - "hw = 97.; \t\t\t\t#heigth of water in reservior\n", - "Bt = 7.; \t\t\t\t#width of top of dam\n", - "H = 100.; \t\t\t\t#heigth of the dam\n", - "Hs2 = 90.; \t\t\t\t#heigth of slope on downstream side\n", - "wb = 75.; \t\t\t\t#width of base of dam\n", - "miu = 0.75; \t\t\t\t#coefficient of friction\n", - "gamma_d = 2.4; \t\t\t\t#weigth density of concrete\n", - "gamma_w = 1000.; \t\t\t\t#weigth density of water\n", - "\n", - "# Calculations\n", - "P = gamma_w*hw**2/(2*1000);\n", - "W1 = Bt*gamma_d*H;\n", - "W2 = (wb-Bt)*Hs2*gamma_d/2;\n", - "W = W1+W2;\n", - "FOS = miu*W/P;\n", - "FOS = round(FOS*1000)/1000;\n", - "\n", - "# Results\n", - "print \"Factor of safety against sliding = %.2f.\"%(FOS);\n", - "print \"Dam is safe against sliding\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Factor of safety against sliding = 1.44.\n", - "Dam is safe against sliding\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.19 pg : 430" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Factor of safety against overturning\n", - "#Factor of safety against sliding\n", - "#Shear friction factor\n", - "\n", - "#Given\n", - "c = 1.;\n", - "H = 10.; \t\t\t\t#heigth of dam\n", - "hw = 10.; \t\t\t\t#heigth of water in reservior\n", - "wb = 8.25; \t\t\t\t#bottom width\n", - "Bt = 1.; \t\t\t\t#top width\n", - "Hs1 = 0.1; \t\t\t\t#slope on upstream side\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "gamma_m = 22.4; \t\t\t\t#unit weigth of masonary\n", - "f = 1400.; \t\t\t\t#permissible shear stress at joint\n", - "miu = 0.75; \t\t\t\t#coefficient of friction\n", - "fi = math.atan(0.625);\n", - "theta = math.atan(0.1);\n", - "\n", - "# Calculations\n", - "W1 = Bt*H*gamma_m;\n", - "W2 = H*H*Hs1*gamma_m/2;\n", - "W3 = H*6.25*gamma_m/2;\n", - "W4 = hw*gamma_w*H*Hs1/2;\n", - "P = gamma_w*hw**2/2;\n", - "U = wb*gamma_w*hw*c/2;\n", - "SumV = W1+W2+W3+W4-U;\n", - "L3 = 2*(wb-(Hs1*H)-Bt)/3;\n", - "L1 = (wb-(Hs1*H)-Bt)+Bt/2;\n", - "L2 = (wb-(Hs1*H)-Bt)+Bt+(Hs1*H/3);\n", - "L4 = (wb-(Hs1*H)-Bt)+Bt+(2*Hs1*H/3);\n", - "L5 = 2*wb/3;L6 = hw/3;\n", - "M1 = W1*L1;M2 = W2*L2;M3 = W3*L3;M4 = W4*L4;\n", - "M5 = U*L5;M6 = P*L6;\n", - "SumM = M1+M2+M3+M4-M5-M6;\n", - "Mplus = M1+M2+M3+M4;\n", - "Mminus = M5+M6;\n", - "FOS = miu*SumV/P;\n", - "SFF = (miu*SumV+wb*1400)/P;\n", - "FOO = Mplus/Mminus;\n", - "FOS = round(FOS*100)/100;\n", - "SFF = round(SFF*10)/10;\n", - "FOO = round(FOO*100)/100;\n", - "\n", - "# Results\n", - "print \"Factor of safety against sliding = %.2f.\"%(FOS);\n", - "print \"Shear friction factor = %.2f.\"%(SFF);\n", - "print \"Factor of safety against overturning = %.2f.\"%(FOO);\n", - "print \"Dam is unsafe against overturning\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Factor of safety against sliding = 1.04.\n", - "Shear friction factor = 24.60.\n", - "Factor of safety against overturning = 1.47.\n", - "Dam is unsafe against overturning\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.20 pg : 431" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1.;\n", - "hw = 80.; \t\t\t\t#heigth of water in reservior\n", - "Bt = 6.; \t\t\t\t#width of top of dam\n", - "H = 84.; \t\t\t\t#heigth of the dam\n", - "Hs2 = 75.; \t\t\t\t#heigth of slope on downstream side\n", - "wb = 56.; \t\t\t\t#width of base of dam\n", - "Lf = 8.; \t\t\t\t#dismath.tance of foundation gallery from heel\n", - "gamma_d = 23.5; \t\t\t\t#weigth density of concrete\n", - "gamma_w = 9.81; \t\t\t\t#weigth density of water\n", - "ht = 6.; \t\t\t\t#heigth of tail water\n", - "\n", - "# Calculations\n", - "W1 = Bt*gamma_d*H;\n", - "W2 = gamma_d*Hs2*(wb-Bt)/2;\n", - "W3 = gamma_w*ht*4/2;\n", - "W4 = gamma_w*hw**2/2;\n", - "W5 = gamma_w*ht**2/2;\n", - "Pt = gamma_w*ht\n", - "Ph = gamma_w*hw\n", - "Pg = (ht+(hw-ht)/3)*gamma_w\n", - "U = (Pt*(wb-Lf))+(Pg*Lf)+((Ph-Pg)*Lf/2)+((Pg-Pt)*(wb-Lf)/2)*c\n", - "l1 = (wb-Lf)/2\n", - "l2 = (2*(wb-Lf))/3\n", - "l3 = (wb-Lf)+(Lf/2)\n", - "l4 = (wb-Lf)+((2*Lf)/3)\n", - "L6 = (((Pt*(wb-Lf))*l1)+((Pg-Pt)*(wb-Lf)*l2/2)+((Pg*Lf)*l3)+((Ph-Pg)*Lf*l4/2))/U\n", - "L1 = (wb-Bt)+(Bt/2)\n", - "L2 = (2*(wb-Bt))/3\n", - "L3 = 4./3;\n", - "L4 = hw/3;\n", - "L5 = ht/3;\n", - "M1 = W1*L1\n", - "M2 = W2*L2\n", - "M3 = W3*L3\n", - "M4 = W4*L4\n", - "M5 = W5*L5\n", - "M6 = U*L6;\n", - "SumV = W1+W2+W3-U;\n", - "SumH = W4-W5;\n", - "SumM = M1+M2+M3-M4+M5-M6;\n", - "x = SumM/SumV;\n", - "e = wb/2-x;\n", - "pnt = (SumV/wb)*(1+(6*e/wb));\n", - "pnh = (SumV/wb)*(1-(6*e/wb));\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "\n", - "# Results\n", - "print \"Maximum Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Maximum Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Maximum Normal stress at toe = 1586.70 kN/square.m.\n", - "Maximum Normal stress at heel = -49.30 kN/square.m.\n" - ] - } - ], - "prompt_number": 22 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch8_1.ipynb b/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch8_1.ipynb deleted file mode 100755 index 595c81ff..00000000 --- a/Irrigation_and_Water_Power_Engineering_by_B._C._Punmia/ch8_1.ipynb +++ /dev/null @@ -1,1584 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:6fa8fe978c728d97fa9f04464822ea9606521023bb2c2c301624274cfcebacb1" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 8 : GRAVITY DAMS" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.1 pg : 376" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from scipy.integrate import quad \n", - "\t\t\t\t\n", - "#Given\n", - "H = 100; \t\t\t\t#heigth of dam\n", - "wb = 70; \t\t\t\t#width of base of dam\n", - "wt = 7; \t\t\t\t#width of top of dam\n", - "l = 1; \t\t\t\t#length of dam\n", - "hw = 98; \t\t\t\t#heigth of water in dam\n", - "hsu = 90; \t\t\t\t#heigth of slope on downstream side\n", - "s = 1/0.7; \t\t\t\t#slope on downstream side\n", - "gammad = 24; \t\t\t\t#unit weigth of dam\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "E = 2.05e7; \t\t\t\t#modulus of elasticity\n", - "\n", - "\t\t\t\t#(a) inertial forces and moments\n", - "alpha0 = 0.05; \t\t\t\t#from table 8.1\n", - "alphah = 2*alpha0;\n", - "\t\t\t\t#at 10m from top\n", - "\n", - "def f0(y): \n", - "\t return 25.2-0.25*y\n", - "\n", - "F10 = quad(f0,0,10)[0]\n", - "\n", - "\n", - "def f1(y): \n", - "\t return 25.2*(1-0.01*y)*(10-y)\n", - "\n", - "M10 = quad(f1,0,10)[0]\n", - "\n", - "\t\t\t\t#at 100m below top\n", - "\n", - "def f2(y): \n", - "\t return 0.15*(1-0.01*y)*16.8*y\n", - "\n", - "F100 = F10+ quad(f2,10,100)[0]\n", - "\n", - "\n", - "def f3(y): \n", - "\t return 0.15*(1-0.01*y)*16.8*y*(100-y)\n", - "\n", - "M100 = M10+90*F10+ quad(f3,10,100)[0]\n", - "\n", - "print \"Inertial forces:At 10m from top: F = %.2f kn;M = %ikn-mAt 100m from top: F = %.2f kn;M = %ikn-m.\"%(F10,M10,F100,M100);\n", - "\n", - "\t\t\t\t#(b) hydrodynamic pressure and moment\n", - "\t\t\t\t#at 10m from top\n", - "y = 8.;\n", - "W10 = 1680.;\n", - "alphah = F10/W10;\n", - "Cm = 0.735;\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "p = Cy*alphah*gammaw*hw;\n", - "P10 = 0.726*p*y;\n", - "Mp10 = 0.299*p*y**2;\n", - "P10 = round(P10*100)/100;\n", - "Mp10 = round(Mp10*100)/100;\n", - "\t\t\t\t#at 100m from top\n", - "y = 98;\n", - "W100 = 84840;\n", - "alphah = F100/W100;\n", - "Cm = 0.735;\n", - "Cy = (Cm/2)*(y*(2-y/hw)/hw+(y*(2-y/hw)/hw)**0.5);\n", - "p = Cy*alphah*gammaw*hw;\n", - "P100 = 0.726*p*y;\n", - "Mp100 = 0.299*p*y**2;\n", - "print \"Hydrodynamic forces:At 10m from top: F = %.2f kn;\\\n", - "\\nM = %.2fkn-mAt 100m from top: F = %i kn;\\\n", - "\\nM = %ikn-m.\"%(P10,Mp10,P100,Mp100);\n", - "\n", - "# rounding off error." - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Inertial forces:At 10m from top: F = 239.50 kn;M = 1218kn-mAt 100m from top: F = 4321.90 kn;M = 221790kn-m.\n", - "Hydrodynamic forces:At 10m from top: F = 161.58 kn;\n", - "M = 532.35kn-mAt 100m from top: F = 2561 kn;\n", - "M = 103366kn-m.\n" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.2 pg : 378" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\t\t\t\t\n", - "#Given\n", - "H = 100; \t\t\t\t#heigth of dam\n", - "wb = 70; \t\t\t\t#width of base of dam\n", - "wt = 7; \t\t\t\t#width of top of dam\n", - "l = 1; \t\t\t\t#length of dam\n", - "hw = 98; \t\t\t\t#heigth of water in dam\n", - "hsu = 90; \t\t\t\t#heigth of slope on downstream side\n", - "s = 1/0.7; \t\t\t\t#slope on downstream side\n", - "gammad = 24; \t\t\t\t#unit weigth of dam\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "E = 2.05e7; \t\t\t\t#modulus of elasticity\n", - "beta = 1;\n", - "I = 2;\n", - "Fo = 0.25; \t\t\t\t#from table 8.2\n", - " \t\t\t\t#t = Sa/g;\n", - "t = 0.19; \t\t\t\t#from fig. 8.4\n", - "alphah = beta*I*Fo*t;\n", - "T = 5.55*H**2/wb*(gammad/(gammaw*E))**0.5;\n", - "\t\t\t\t#(a) Base shear\n", - "W = l*gammad*(wt*H+((hsu/s)*hsu)/2);\n", - "Fb = 0.6*W*alphah;\n", - "print \"Base shear = %.2f KN.\"%(Fb);\n", - "\n", - "\t\t\t\t#(b) Base moment\n", - "hbar = ((wt*H**2/2)+((hsu/s)*hsu**2/6))/((wt*H)+(hsu/s)*hsu/2);\n", - "Mb = 0.9*W*hbar*alphah;\n", - "print \"Base moment = %.2f KN-m.\"%(Mb);\n", - "\n", - "\t\t\t\t#(c) shear at 10m from top\n", - "Cv = 0.08;\n", - "F10 = Cv*Fb;\n", - "F10 = round(F10);\n", - "print \"shear at 10m from top = %.2f KN.\"%(F10);\n", - "\n", - "\t\t\t\t#(d) Moment at 10m from top\n", - "Cm = 0.02;\n", - "M10 = Cm*Mb;\n", - "M10 = round(M10);\n", - "print \"moment at 10m from top = %.2f KN.\"%(M10);\n", - "\t\t\t\t#(e) Hydrodynamic pressure\n", - "\t\t\t\t#at 10m from top\n", - "y = 8;\n", - "W10 = 1680;\n", - "Cm = 0.735;\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "p = Cy*alphah*gammaw*hw;\n", - "P10 = 0.726*p*y;\n", - "Mp10 = 0.299*p*y**2;\n", - "P10 = round(P10*100)/100;\n", - "Mp10 = round(Mp10*100)/100;\n", - "\t\t\t\t#at 100m from top\n", - "y = 98;\n", - "W100 = 84840;\n", - "Cm = 0.735;\n", - "Cy = (Cm/2)*(y*(2-y/hw)/hw+(y*(2-y/hw)/hw)**0.5);\n", - "p = Cy*alphah*gammaw*hw;\n", - "P100 = 0.726*p*y;\n", - "Mp100 = 0.299*p*y**2;\n", - "print \"Hydrodynamic forces:At 10m from top: F = %.2f kn;\\\n", - "\\nM = %.2fkn-mAt 100m from top: F = %i \\\n", - "kn;M = %ikn-m.\"%(P10,Mp10,P100,Mp100);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Base shear = 4835.88 KN.\n", - "Base moment = 246342.60 KN-m.\n", - "shear at 10m from top = 387.00 KN.\n", - "moment at 10m from top = 4927.00 KN.\n", - "Hydrodynamic forces:At 10m from top: F = 0.00 kn;\n", - "M = 0.00kn-mAt 100m from top: F = 4776 kn;M = 192765kn-m.\n" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.3 pg : 379" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from scipy.integrate import quad \n", - "\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "H = 100; \t\t\t\t#heigth of dam\n", - "wb = 70; \t\t\t\t#width of base of dam\n", - "wt = 7; \t\t\t\t#width of top of dam\n", - "l = 1; \t\t\t\t#length of dam\n", - "hw = 98; \t\t\t\t#heigth of water in dam\n", - "hsu = 90; \t\t\t\t#heigth of slope on downstream side\n", - "s = 1/0.7; \t\t\t\t#slope on downstream side\n", - "gammad = 24; \t\t\t\t#unit weigth of dam\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "E = 2.05e7; \t\t\t\t#modulus of elasticity\n", - "\t\t\t\t#(a) Seismic coefficient method\n", - "alpha0 = 0.05; \t\t\t\t#from table 8.1\n", - "alphah = 2*alpha0;\n", - "alphav = 0.75*alphah;\n", - "\t\t\t\t#at 10m from top\n", - "\n", - "def f4(y): \n", - "\t return alphav*168*(1-0.01*y)\n", - "\n", - "F10 = quad(f4,0,10)[0]\n", - "\n", - "\t\t\t\t#at 100m below top\n", - "\n", - "def f5(y): \n", - "\t return alphav*(1-0.01*y)*16.8*y\n", - "\n", - "F100 = F10+ quad(f5,10,100)[0]\n", - "\n", - "print \"Parta):At 10m from top: F = %.2f knAt 100m from top: F = %.2f kn.\"%(F10,F100);\n", - "\n", - "\t\t\t\t#(b)Response spectrum method\n", - "beta = 1;\n", - "I = 2;\n", - "Fo = 0.25; \t\t\t\t#from table 8.2\n", - " \t\t\t\t#t = Sa/g;\n", - "t = 0.19; \t\t\t\t#from fig. 8.4\n", - "alphah = beta*I*Fo*t;\n", - "alphav = 0.75*alphah;\n", - "\t\t\t\t#at 10m from top\n", - "\n", - "def f6(y): \n", - "\t return alphav*168*(1-0.01*y)\n", - "\n", - "F10 = quad(f6,0,10)[0]\n", - "\n", - "\t\t\t\t#at 100m below top\n", - "\n", - "def f7(y): \n", - "\t return alphav*(1-0.01*y)*16.8*y\n", - "\n", - "F100 = F10+ quad(f7,10,100)[0]\n", - "\n", - "F100 = round(F100*100)/100;\n", - "print \"Partb):At 10m from top: F = %.2f knAt 100m from top: F = %.2f kn.\"%(F10,F100);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Parta):At 10m from top: F = 119.70 knAt 100m from top: F = 2160.90 kn.\n", - "Partb):At 10m from top: F = 113.72 knAt 100m from top: F = 2052.86 kn.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.4 pg : 381\n" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#Given\n", - "H = 100.; \t\t\t\t#heigth of dam\n", - "wb = 73.; \t\t\t\t#width of base of dam\n", - "wt = 7; \t\t\t\t#width of top of dam\n", - "l = 1.; \t\t\t\t#length of dam\n", - "hw = 98.; \t\t\t\t#heigth of water in dam\n", - "hsu = 90.; \t\t\t\t#heigth of slope on downstream side\n", - "s = 1/0.7; \t\t\t\t#slope on downstream side\n", - "gammad = 24.; \t\t\t\t#unit weigth of dam\n", - "gammaw = 9.81; \t\t\t\t#unit weigth of water\n", - "E = 2.05e7; \t\t\t\t#modulus of elasticity\n", - "\n", - "#at 10m from top\n", - "y = 8;\n", - "alphah = 0.1;\n", - "Cm = 0.72;\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "p10 = Cy*alphah*gammaw*hw;\n", - "F10 = 0.726*p10*y;\n", - "Mp10 = 0.299*p10*y**2;\n", - "\n", - "#at 40m from top\n", - "y = 38;\n", - "alphah = 0.1;\n", - "Cm = 0.72;\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "p40 = Cy*alphah*gammaw*hw;\n", - "F40 = 0.726*p40*y;\n", - "Mp40 = 0.299*p40*y**2;\n", - "\n", - "#at 100m from top\n", - "y = 98;\n", - "alphah = 0.1;\n", - "Cm = 0.72;\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "p100 = Cy*alphah*gammaw*hw;\n", - "F100 = 0.726*p100*y;\n", - "Mp100 = 0.299*p100*y**2;\n", - "p10 = round(p10*1000)/1000;\n", - "F10 = round(F10*1000)/1000;\n", - "Mp10 = round(Mp10*10)/10;\n", - "p40 = round(p40*1000)/1000;\n", - "F40 = round(F40*1000)/1000;\n", - "Mp40 = round(Mp40*10)/10;\n", - "p100 = round(p100*100)/100;\n", - "F100 = round(F100*1000)/1000;\n", - "Mp100 = round(Mp100*10)/10;\n", - "print \"Hydrodynamic Forces:At 10m from top: P = %.2f KN/square m;\\\n", - "\\nF = %.2f KN;\\\n", - "\\nM = %.2f KN-m.At 40m from top: P = %.2f KN/square m.;\\\n", - "\\nF = %.2f KN;\\\n", - "\\nM = %.2f KN-m.At 100m from top: P = %.2f KN/square m;\\\n", - "\\nF = %.2f KN;\\\n", - "\\nM = %.2f KN-m.\"%(p10,F10,Mp10,p40,F40,Mp40,p100,F100,Mp100);\n", - "\n", - "#vertical component of reservior water on horizontal section\n", - "s1 = 3./60;\n", - "Wh = (F100-F40)*s1;\n", - "Wh = round(Wh*100)/100;\n", - "print \"vertical component of reservior water on horizontal section = %.2f kN/m.\"%(Wh);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Hydrodynamic Forces:At 10m from top: P = 19.12 KN/square m;\n", - "F = 111.03 KN;\n", - "M = 365.80 KN-m.At 40m from top: P = 49.00 KN/square m.;\n", - "F = 1351.85 KN;\n", - "M = 21156.60 KN-m.At 100m from top: P = 69.22 KN/square m;\n", - "F = 4924.82 KN;\n", - "M = 198770.00 KN-m.\n", - "vertical component of reservior water on horizontal section = 178.65 kN/m.\n" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.8 pg : 413" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "#no tension is permissible\n", - "#factor of safety against slidingis 1.5\n", - "\n", - "#Given\n", - "\n", - "wb = 3; \t\t\t\t#width of dam;\n", - "miu = 0.5; \t\t\t\t#coefficient of friction\n", - "Sg = 2.4; \t\t\t\t#specific gravity of masonary\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "c = 1;\n", - "\n", - "#when uplift is considered\n", - "#when no tension is permissible then e = wb/6;\n", - "\n", - "p1 = wb*Sg*gamma_w;\n", - "p2 = c*wb*gamma_w/2;\n", - "p3 = p1-p2;\n", - "p4 = p1*wb/2-p2*2;\n", - "p5 = gamma_w/6;\n", - "d1 = p4/p3; d2 = p5/p3;\n", - "d3 = 1.5-d1;\n", - "H = ((0.5-d3)/d2)**0.5;\n", - "H = round(H*100)/100;\n", - "print \"when uplift is considered:\"\n", - "print \"Heigth of dam when no tension is permissible = %.2f m.\"%(H);\n", - "H = p3*0.5/(1.5*p5*3);\n", - "print \"Heigth of dam when factor of safety against sliding is 1.5 = %.2f m.\"%(H);\n", - "\n", - "#when uplift is not considered\n", - "p1 = wb*Sg*gamma_w;\n", - "p4 = p1*wb/2;\n", - "p5 = gamma_w/6;\n", - "d1 = p4/p1;\n", - "d2 = p5/p1;\n", - "H = (0.5/d2)**0.5;\n", - "H = round(H*100)/100;\n", - "print \"when uplift is not considered:\"\n", - "print \"Heigth of dam when no tension is permissible = %.2f m.\"%(H);\n", - "H = p1*0.5/(1.5*p5*3);\n", - "print \"Heigth of dam when factor of safety against sliding is 1.5 = %.2f m.\"%(H);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "when uplift is considered:\n", - "Heigth of dam when no tension is permissible = 3.55 m.\n", - "Heigth of dam when factor of safety against sliding is 1.5 = 3.80 m.\n", - "when uplift is not considered:\n", - "Heigth of dam when no tension is permissible = 4.65 m.\n", - "Heigth of dam when factor of safety against sliding is 1.5 = 4.80 m.\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.9 pg : 414" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "#Given\n", - "c = 1;\n", - "hw = 6; \t\t\t\t#heigth of water in reservior\n", - "Bt = 1.5; \t\t\t\t#width of top of dam\n", - "H = 6; \t\t\t\t#heigth of the dam\n", - "wb = 4.5; \t\t\t\t#width of base of dam\n", - "Sg = 2.4; \t\t\t\t#specific gravity of masonary\n", - "gamma_w = 9.81; \t\t\t\t#weigth density of water\n", - "\n", - "W1 = Bt*gamma_w*Sg*H;\n", - "W2 = gamma_w*Sg*H*(wb-Bt)/2;\n", - "L1 = (wb-Bt)+(Bt/2);\n", - "L2 = (2*(wb-Bt))/3\n", - "M1 = W1*L1\n", - "M2 = W2*L2\n", - "\n", - "#Reaervior empty\n", - "SumW = W1+W2;\n", - "SumM = M1+M2;\n", - "x = SumM/SumW;\n", - "e = wb/2-x;\n", - "pnt = (SumW/wb)*(1+(6*e/wb));\n", - "pnh = (SumW/wb)*(1-(6*e/wb));\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "print \"Reservior empty:\";\n", - "print \" Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "\n", - "#Reservior full\n", - "W3 = gamma_w*H**2/2;\n", - "U = gamma_w*H*c*wb/2;\n", - "SumV = SumW-U;\n", - "L3 = hw/3;\n", - "L4 = 2*wb/3; \t\t\t\t#lever arm\n", - "M3 = W3*L3;\n", - "M4 = U*L4; \t\t\t\t#moment about toe\n", - "SumM1 = SumM-M4-M3;\n", - "x = SumM1/SumV;\n", - "e = wb/2-x;\n", - "pnt = (SumV/wb)*(1+(6*e/wb));\n", - "pnh = (SumV/wb)*(1-(6*e/wb));\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "print \"Reservior full:\";\n", - "print \" Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Reservior empty:\n", - " Normal stress at toe = 15.70 kN/square.m.\n", - "Normal stress at heel = 172.70 kN/square.m.\n", - "Reservior full:\n", - " Normal stress at toe = 120.30 kN/square.m.\n", - "Normal stress at heel = 9.20 kN/square.m.\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.10 pg : 415" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\n", - "\n", - "#check the stability\n", - "\n", - "#Given\n", - "c = 1;\n", - "hw = 6; \t\t\t\t#heigth of water in reservior\n", - "Bt = 1.5; \t\t\t\t#width of top of dam\n", - "H = 6; \t\t\t\t#heigth of the dam\n", - "gamma_m = 20; \t\t\t\t#unit weigth of masonary \n", - "gamma_w = 9.81; \t\t\t\t#weigth density of water\n", - "f = 1800; \t\t\t\t#compressive strength\n", - "miu = 0.6; \t\t\t\t#coefficient of friction\n", - "\n", - "#to develop no tension e = b/6;x = b/3.\n", - "#hence on solving the relations we get\n", - "\n", - "P = [1,2.944,-39.074]\n", - "wb = roots(P)[1]; \t\t\t\t#sign of coefficient is 2.944 is not taken correctly in book\n", - "#roots are 4.94 and -7.89\n", - "#math.since negative value cannot be taken\n", - "print \"Neglecting the negative value.Width of base is = 4.94 m.\";\n", - "W1 = Bt*gamma_m*H;\n", - "W2 = gamma_m*H*(wb-Bt)/2;\n", - "L1 = (wb-Bt)+(Bt/2);\n", - "L2 = (2*(wb-Bt))/3;\n", - "M1 = W1*L1,\n", - "M2 = W2*L2;\n", - "U = gamma_w*H*c*wb/2;\n", - "L4 = 2*wb/3;\n", - "M4 = U*L4;\n", - "W3 = gamma_w*H**2/2;\n", - "L3 = hw/3;\n", - "M3 = W3*L3;\n", - "SumW = W1+W2-U;\n", - "SumM = M1+M2-M4-M3;\n", - "pn = 2*SumW/wb;\n", - "pn = round(pn*10)/10;\n", - "print \"Maximum stress = %.2f kN/square.m.\"%(pn);\n", - "print \"Dam is safe against compression\";\n", - "FOS = miu*SumW/W3;\n", - "FOS = round(FOS*100)/100;\n", - "print \"Factor of safety against sliding = %.2f. <1\"%(FOS);\n", - "print \"Dam is unsafe against sliding.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Neglecting the negative value.Width of base is = 4.94 m.\n", - "Maximum stress = 97.50 kN/square.m.\n", - "Dam is safe against compression\n", - "Factor of safety against sliding = 0.82. <1\n", - "Dam is unsafe against sliding.\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.11 pg : 416" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from numpy import roots\n", - "\t\t\t\t#check the stability if uplift is neglected\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1;\n", - "hw = 6; \t\t\t\t#heigth of water in reservior\n", - "Bt = 1.5; \t\t\t\t#width of top of dam\n", - "H = 6; \t\t\t\t#heigth of the dam\n", - "gamma_m = 20; \t\t\t\t#unit weigth of masonary \n", - "gamma_w = 9.81; \t\t\t\t#weigth density of water\n", - "f = 1800; \t\t\t\t#compressive strength\n", - "miu = 0.6; \t\t\t\t#coefficient of friction\n", - "\n", - "\t\t\t\t#to develop no tension e = b/6;x = b/3.\n", - "\t\t\t\t#hence on solving the relations we get\n", - "\n", - "P = [1,1.5,-19.908]\n", - "wb = roots(P)[1];\n", - "\n", - "#roots are 3.774 and -5.27\n", - "#math.since negative value cannot be taken\n", - "\n", - "print \"Neglecting the negative value.Width of base is = %.2f m.\"%wb;\n", - "\n", - "W1 = Bt*gamma_m*H;\n", - "W2 = gamma_m*H*(wb-Bt)/2;\n", - "L1 = (wb-Bt)+(Bt/2);\n", - "L2 = (2*(wb-Bt))/3;\n", - "M1 = W1*L1,\n", - "M2 = W2*L2;\n", - "W3 = gamma_w*H**2/2;\n", - "L3 = hw/3;\n", - "M3 = W3*L3;\n", - "SumW = W1+W2;\n", - "SumM = M1+M2-M3;\n", - "pn = 2*SumW/wb;\n", - "pn = round(pn*10)/10;\n", - "print \"Maximum stress = %.2f kN/square.m.\"%(pn);\n", - "print \"Dam is safe against compression\";\n", - "\n", - "FOS = miu*SumW/W3;\n", - "FOS = round(FOS*1000)/1000;\n", - "print \"Factor of safety against sliding = %.2f. > 1\"%(FOS);\n", - "print \"Dam is safe against sliding.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Neglecting the negative value.Width of base is = 3.77 m.\n", - "Maximum stress = 167.70 kN/square.m.\n", - "Dam is safe against compression\n", - "Factor of safety against sliding = 1.07. > 1\n", - "Dam is safe against sliding.\n" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.12 pg : 417" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "# calculate maximum permissible heigth of shutter so that no tension develops\n", - "\t\t\t\t\n", - "#Given\n", - "Bt = 3; \t\t\t\t#width of top of dam\n", - "H = 12; \t\t\t\t#heigth of the dam\n", - "wb = 9; \t\t\t\t#width of base of dam\n", - "gamma_m = 21; \t\t\t\t#unit weigth of masonary\n", - "gamma_w = 9.81; \t\t\t\t#weigth density of water\n", - "\n", - "W1 = Bt*gamma_m*H;\n", - "W2 = gamma_m*H*(wb-Bt)/2;\n", - "\n", - "#taking moment about a point on base at 3m from toe\n", - "L1 = 3+Bt/2;\n", - "L2 = (2*(wb-Bt)/3)-3;\n", - "M1 = W1*L1\n", - "M2 = W2*L2;\n", - "M = M1+M2;\n", - "\n", - "#net moment about this point should be zero for equilibrium\n", - "s = (M*6/gamma_w)**(1./3)-12;\n", - "s = round(s*100)/100;\n", - "\n", - "# Results\n", - "print \"maximum permissible heigth of shutter = %.2f m.\"%(s);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum permissible heigth of shutter = 1.22 m.\n" - ] - } - ], - "prompt_number": 13 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.13 pg : 417" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\n", - "\t\t\t\t#moment at 50m below water surface\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1;\n", - "H = 100; \t\t\t\t#heigth of dam\n", - "hw = 100; \t\t\t\t#heigth of water in reservior\n", - "FB = 1; \t\t\t\t#free board\n", - "s = 0.15; \t\t\t\t#slope of upstream face\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "alphah = 0.1;\n", - "\n", - "theta = math.atan(s);\n", - "y = 50;\n", - "Cm = 0.735*(1-(theta*2/math.pi));\n", - "Cy = (Cm/2)*((y*(2-y/hw)/hw)+(y*(2-y/hw)/hw)**0.5);\n", - "pe = Cy*alphah*gamma_w*hw;\n", - "F = 0.726*pe*y;\n", - "M = 0.299*pe*y**2;\n", - "pe = round(pe*1000)/1000;\n", - "F = round(F*10)/10;\n", - "M = round(M*10)/10;\n", - "print \"hydrodynamic earthquake pressure = %.2f kN/square.mshear = %.2f kN/m.Moment = %.2f kN-m/m.\"%(pe,F,M);\n", - "\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "hydrodynamic earthquake pressure = 65.27 kN/square.mshear = 2369.30 kN/m.Moment = 48788.60 kN-m/m.\n" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.14 pg : 418" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "from sympy import sec\n", - "\n", - "#check stability\n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1.;\n", - "H = 10.; \t\t\t\t#heigth of dam\n", - "hw = 10.; \t\t\t\t#heigth of water in reservior\n", - "wb = 8.25; \t\t\t\t#bottom width\n", - "Bt = 1.; \t\t\t\t#top width\n", - "Hs1 = 0.1; \t\t\t\t#slope on upstream side\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "gamma_m = 22.4; \t\t\t\t#unit weigth of masonary\n", - "f = 1400.; \t\t\t\t#permissible shear stress at joint\n", - "miu = 0.75; \t\t\t\t#coefficient of friction\n", - "fi = math.atan(0.625);\n", - "theta = math.atan(0.1);\n", - "\n", - "W1 = Bt*H*gamma_m;\n", - "W2 = H*H*Hs1*gamma_m/2;\n", - "W3 = H*6.25*gamma_m/2;\n", - "W4 = hw*gamma_w*H*Hs1/2;\n", - "P = gamma_w*hw**2/2;\n", - "U = wb*gamma_w*hw*c/2;\n", - "SumV = W1+W2+W3+W4-U;\n", - "L3 = 2*(wb-(Hs1*H)-Bt)/3;\n", - "L1 = (wb-(Hs1*H)-Bt)+Bt/2;\n", - "L2 = (wb-(Hs1*H)-Bt)+Bt+(Hs1*H/3);\n", - "L4 = (wb-(Hs1*H)-Bt)+Bt+(2*Hs1*H/3);\n", - "L5 = 2*wb/3;L6 = hw/3;\n", - "M1 = W1*L1;\n", - "M2 = W2*L2;\n", - "M3 = W3*L3;\n", - "M4 = W4*L4;\n", - "M5 = U*L5;\n", - "M6 = P*L6;\n", - "SumM = M1+M2+M3+M4-M5-M6;\n", - "Mplus = M1+M2+M3+M4;\n", - "Mminus = M5+M6;\n", - "FOS = miu*SumV/P;\n", - "SFF = (miu*SumV+wb*1400)/P;\n", - "FOO = Mplus/Mminus;\n", - "FOS = round(FOS*100)/100;\n", - "SFF = round(SFF*10)/10;\n", - "FOO = round(FOO*100)/100;\n", - "print \"Factor of safety against sliding = %.2f. >1 \"%(FOS);\n", - "print \"Shear friction factor = %.2f.\"%(SFF);\n", - "print \"Factor of safety against overturning = %.2f. <1.5\"%(FOO);\n", - "print \"Dam is unsafe against overturning\";\n", - "\n", - "x = SumM/SumV;\n", - "e = wb/2-x;\n", - "p = hw*gamma_w;\n", - "pnt = (SumV/wb)*(1+(6*e/wb)); \t\t\t\t#calculation is done wrong in book;value of b is not taken correctly\n", - "pnh = (SumV/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "sigmah = pnh*sec(theta)**2-p*math.tan(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = -(pnh-p)*math.tan(theta);\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "sigmat = round(sigmat*10)/10;\n", - "sigmah = round(sigmah*10)/10;\n", - "taut = round(taut*10)/10;\n", - "tauh = round(tauh*10)/10;\n", - "print \"Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %.2f kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %.2f kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %.2f kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %.2f kN/square.m.\"%(tauh);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Factor of safety against sliding = 1.04. >1 \n", - "Shear friction factor = 24.60.\n", - "Factor of safety against overturning = 1.47. <1.5\n", - "Dam is unsafe against overturning\n", - "Normal stress at toe = 170.70 kN/square.m.\n", - "Normal stress at heel = -5.80 kN/square.m.\n", - "Principal stress at toe = 237.40 kN/square.m.\n", - "Principal stress at heel = -6.80 kN/square.m.\n", - "Shear stress at toe = 106.70 kN/square.m.\n", - "Shear stress at heel = 10.40 kN/square.m.\n" - ] - } - ], - "prompt_number": 15 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.15 pg : 420" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t#Check the stability and determine sliding factor and shear factor\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1;\n", - "miu = 0.75; \t\t\t\t#coefficient of friction\n", - "H = 90; \t\t\t\t#heigth of dam\n", - "wb = 73.1; \t\t\t\t#width of base\n", - "Bt = 7; \t\t\t\t#width of top of dam\n", - "hw = 89; \t\t\t\t#heigth of water in reservior\n", - "Hs1 = 28; \t\t\t\t#heigth of slope on upstream side\n", - "Hs2 = 83; \t\t\t\t#heigth of slope on downstream side\n", - "Cm = 0.735;\n", - "alphah = 0.1;\n", - "gamma_m = 23.5; \t\t\t\t#unit weigth of concrete\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "theta = math.atan(8./28);\n", - "fi = math.atan(0.7);\n", - "\t\t\t\t#self weigth of dam\n", - "W1 = (Hs1*8*gamma_m)/2\n", - "W2 = (Bt*H*gamma_m)\n", - "W3 = (Hs2**2*0.7*gamma_m)/2\n", - "\t\t\t\t#weigth of superimposed water\n", - "W4 = (Hs1*8*gamma_w)/2\n", - "W5 = (hw-Hs1)*8*gamma_w\n", - "U = hw*wb*2*gamma_w/6; \t\t\t\t#uplift force\n", - "wp = hw**2*gamma_w/2; \t\t\t\t#water pressure\n", - "hp = 0.726*Cm*alphah*gamma_w*hw**2; \t\t\t\t#hydrodynamic pressure\n", - "Mhp = 0.299*Cm*alphah*gamma_w*hw**3; \t\t\t\t#moment due to hydrodynamic pressure\n", - "\t\t\t\t#inertial load due to horizontal acceleration\n", - "I1 = W2/10;\n", - "I2 = W3/10;\n", - "I3 = W1/10;\n", - "SumV = W1+W2+W3+W4+W5-U;\n", - "SumH = wp+hp+I1+I2+I3;\n", - "L1 = (wb-8)+8/3\n", - "L2 = (0.7*Hs2)+(Bt/2)\n", - "L3 = (2*Hs2*0.7)/3.\n", - "L4 = (wb-8)+(2*8)/3.\n", - "L5 = (wb-8)+(8./2)\n", - "L6 = hw/3;\n", - "L7 = 2*wb/3;\n", - "M1 = W1*L1\n", - "M2 = W2*L2\n", - "M3 = W3*L3\n", - "M4 = W4*L4;\n", - "M5 = W5*L5;\n", - "M6 = wp*L6;\n", - "M7 = U*L7;\n", - "M8 = I1*45;\n", - "M9 = I2*83/3;\n", - "M10 = I3*28/3;\n", - "Mplus = M1+M2+M3+M4+M5;\n", - "Mminus = M6+M7+M8+M9+M10+Mhp;\n", - "SumM = Mplus-Mminus;\n", - "x = SumM/SumV;\n", - "e = wb/2-x;\n", - "pnt = (SumV/wb)*(1+(6*e/wb));\n", - "pnh = (SumV/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "p = hw*gamma_w;\n", - "pe = Cm*alphah*gamma_w*hw;\n", - "sigmah = pnh*sec(theta)**2-(p+pe)*math.tan(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = -(-pnh-(p+pe))*math.tan(theta);\n", - "print \"Normal stress at toe = %i kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %i kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %i kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %i kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %i kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %i kN/square.m.\"%(tauh);\n", - "\n", - "FOS = miu*SumV/SumH;\n", - "SFF = (miu*SumV+wb*1400)/SumH;\n", - "FOO = Mplus/Mminus;\n", - "Ffi = 1.2;Fc = 2.4;\n", - "F = (miu*SumV/Ffi+1400*wb/Fc)/SumH;\n", - "FOS = round(FOS*100)/100;\n", - "F = round(F*100)/100;\n", - "SFF = round(SFF*100)/100;\n", - "FOO = round(FOO*100)/100;\n", - "print \"Factor of safety against sliding as per IS:6512-1972 = %.2f. <1.5\"%(FOS);\n", - "print \"Factor of safety against sliding as per IS:6512-1984 = %.2f. >1\"%(F);\n", - "print \"Shear friction factor = %.2f. <6\"%(SFF);\n", - "print \"Factor of safety against overturning = %.2f. <1.5\"%(FOO);\n", - "print \"Dam is unsafe for given loading conditions\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Normal stress at toe = 1929 kN/square.m.\n", - "Normal stress at heel = -323 kN/square.m.\n", - "Principal stress at toe = 2874 kN/square.m.\n", - "Principal stress at heel = -426 kN/square.m.\n", - "Shear stress at toe = 1350 kN/square.m.\n", - "Shear stress at heel = 175 kN/square.m.\n", - "Factor of safety against sliding as per IS:6512-1972 = 0.87. <1.5\n", - "Factor of safety against sliding as per IS:6512-1984 = 1.57. >1\n", - "Shear friction factor = 2.90. <6\n", - "Factor of safety against overturning = 1.45. <1.5\n", - "Dam is unsafe for given loading conditions\n" - ] - } - ], - "prompt_number": 16 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.16 pg : 423" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t#Check the stability and determine principal and shear stress at toe and heel\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1;\n", - "miu = 0.7; \t\t\t\t#coefficient of friction\n", - "H = 70; \t\t\t\t#heigth of dam\n", - "ht = 0; \t\t\t\t#heigth of tail water\n", - "Lf = 6.5; \t\t\t\t#location of foundation gallery from heel\n", - "wb = 52.5; \t\t\t\t#width of base\n", - "Bt = 7; \t\t\t\t#width of top of dam\n", - "hw = 70; \t\t\t\t#heigth of water in reservior\n", - "Hs1 = 35; \t\t\t\t#heigth of slope on upstream side\n", - "Hs2 = 60; \t\t\t\t#heigth of slope on downstream side\n", - "gamma_m = 24; \t\t\t\t#unit weigth of concrete\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "theta = math.atan(0.1);\n", - "fi = math.atan(0.7);\n", - "\t\t\t\t#self weigth of dam\n", - "W1 = (Hs1*3.5*gamma_m)/2\n", - "W2 = (Bt*H*gamma_m)\n", - "W3 = (Hs2**2*0.7*gamma_m)/2\n", - "\t\t\t\t#weigth of superimposed water\n", - "W4 = (Hs1*3.5*gamma_w)/2\n", - "W5 = (hw-Hs1)*3.5*gamma_w\n", - "wp = hw**2*gamma_w/2; \t\t\t\t#water pressure\n", - "Pt = gamma_w*ht\n", - "Ph = gamma_w*hw\n", - "Pg = (ht+(hw-ht)/3)*gamma_w\n", - "U = (Pt*(wb-Lf))+(Pg*Lf)+((Ph-Pg)*Lf/2)+((Pg-Pt)*(wb-Lf)/2)*c\n", - "l1 = (wb-Lf)/2\n", - "l2 = (2*(wb-Lf))/3\n", - "l3 = (wb-Lf)+(Lf/2)\n", - "l4 = (wb-Lf)+((2*Lf)/3)\n", - "L7 = (((Pt*(wb-Lf))*l1)+((Pg-Pt)*(wb-Lf)*l2/2)+((Pg*Lf)*l3)+((Ph-Pg)*Lf*l4/2))/U\n", - "L1 = (wb-3.5)+3.5/3\n", - "L2 = (0.7*Hs2)+(Bt/2)\n", - "L3 = (2*Hs2*0.7)/3\n", - "L4 = (wb-3.5)+(2*3.5)/3\n", - "L5 = (wb-3.5)+(3.5/2)\n", - "L6 = hw/3;\n", - "M1 = W1*L1\n", - "M2 = W2*L2\n", - "M3 = W3*L3\n", - "M4 = W4*L4;\n", - "M5 = W5*L5;\n", - "M6 = wp*L6;\n", - "M7 = U*L7;\n", - "SumV1 = W1+W2+W3;\n", - "SumM1 = M1+M2+M3;\n", - "SumV2 = SumV1+W4+W5;\n", - "SumM2 = SumM1+M4+M5-M6;\n", - "SumV3 = SumV2-U;\n", - "SumM3 = SumM2-M7;\n", - "Mplus = 1547377;\n", - "Mminus = 870421;\n", - "SumH = wp;\n", - "\n", - "\t\t\t\t#case 1. Reservior empty\n", - "x = SumM1/SumV1;\n", - "e = wb/2-x;\n", - "pnt = (SumV1/wb)*(1+(6*e/wb));\n", - "pnh = (SumV1/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "sigmah = pnh*sec(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = pnh*math.tan(theta);\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "sigmat = round(sigmat*10)/10;\n", - "sigmah = round(sigmah*10)/10;\n", - "taut = round(taut*10)/10;\n", - "tauh = round(tauh*10)/10;\n", - "print \"case 1. Reservior empty:\";\n", - "print \"Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %.2f kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %.2f kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %.2f kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %.2f kN/square.m.\"%(tauh);\n", - "\n", - "\t\t\t\t#case2. reservior full without uplift\n", - "x = SumM2/SumV2;\n", - "e = wb/2-x;\n", - "p = hw*gamma_w;\n", - "pnt = (SumV2/wb)*(1+(6*e/wb));\n", - "pnh = (SumV2/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "sigmah = pnh*sec(theta)**2-p*math.tan(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = -(pnh-p)*math.tan(theta);\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "sigmat = round(sigmat*10)/10;\n", - "sigmah = round(sigmah*10)/10;\n", - "taut = round(taut*10)/10;\n", - "tauh = round(tauh*10)/10;\n", - "print \"case 2. reservior full without uplift:\";\n", - "print \"Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %.2f kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %.2f kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %.2f kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %.2f kN/square.m.\"%(tauh);\n", - "\n", - "\t\t\t\t#case3. reservior full with uplift\n", - "x = SumM3/SumV3;\n", - "e = wb/2-x;\n", - "p = hw*gamma_w;\n", - "pnt = (SumV3/wb)*(1+(6*e/wb));\n", - "pnh = (SumV3/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "sigmah = pnh*sec(theta)**2-p*math.tan(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = -(pnh-p)*math.tan(theta);\n", - "pnt = round(pnt);\n", - "pnh = round(pnh);\n", - "sigmat = round(sigmat);\n", - "sigmah = round(sigmah);\n", - "taut = round(taut);\n", - "tauh = round(tauh);\n", - "print \"case 3. reservior full with uplift:\";\n", - "print \"Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %.2f kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %.2f kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %.2f kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %.2f kN/square.m.\"%(tauh);\n", - "\n", - "FOS = miu*SumV3/SumH;\n", - "SFF = (miu*SumV3+wb*1400)/SumH;\n", - "FOO = Mplus/Mminus;\n", - "Ffi = 1.5;Fc = 3.6;\n", - "F = (miu*SumV3/Ffi+1400*wb/Fc)/SumH;\n", - "FOS = round(FOS*1000)/1000;\n", - "SFF = round(SFF*100)/100;\n", - "FOO = round(FOO*100)/100;\n", - "F = round(F*1000)/1000;\n", - "print \"Factor of safety against sliding = %.2f.\"%(FOS);\n", - "print \"Shear friction factor = %.2f.\"%(SFF);\n", - "print \"Factor of safety against overturning = %.2f.\"%(FOO);\n", - "print \"Factor of safety for load combination B = %.2f. > 1\"%(F);\n", - "print \"Dam is safe \";\n", - "\n", - "\t\t\t\t#Case4.considering seismic forces\n", - "Cm = 0.712;\n", - "alphah = 0.1;\n", - "alphav = 0.08;\n", - "hp = 0.726*Cm*alphah*gamma_w*hw**2; \t\t\t\t#hydrodynamic pressure\n", - "Mhp = 0.299*Cm*alphah*gamma_w*hw**3; \t\t\t\t#moment due to hydrodynamic pressure\n", - "\t\t\t\t#inertial load due to horizontal acceleration\n", - "I1 = W2/10;\n", - "I2 = W3/10;\n", - "I3 = W1/10;\n", - "v = SumV1*alphav;\n", - "Mv = 116444;\n", - "SumV4 = SumV3-v;\n", - "SumH1 = SumH+I1+I2+I3+hp;\n", - "M8 = I1*35;\n", - "M9 = I2*20;\n", - "M10 = I3*35/3;\n", - "Mminus1 = 1161849;\n", - "SumM4 = SumM3-M8-M9-M10-Mhp-Mv;\n", - "\n", - "x = SumM4/SumV4;\n", - "e = wb/2-x;\n", - "p = hw*gamma_w;\n", - "pe = Cm*alphah*gamma_w*hw;\n", - "pnt = (SumV4/wb)*(1+(6*e/wb));\n", - "pnh = (SumV4/wb)*(1-(6*e/wb));\n", - "sigmat = pnt*sec(fi)**2;\n", - "sigmah = pnh*sec(theta)**2-(p+pe)*math.tan(theta)**2;\n", - "taut = pnt*math.tan(fi);\n", - "tauh = (-pnh+(p+pe))*math.tan(theta);\n", - "pnt = round(pnt);\n", - "pnh = round(pnh);\n", - "sigmat = round(sigmat);\n", - "sigmah = round(sigmah);\n", - "taut = round(taut);\n", - "tauh = round(tauh);\n", - "print \"case 4.considering seismic forces\";\n", - "print \"Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n", - "print \"Principal stress at toe = %.2f kN/square.m.\"%(sigmat);\n", - "print \"Principal stress at heel = %.2f kN/square.m.\"%(sigmah);\n", - "print \"Shear stress at toe = %.2f kN/square.m.\"%(taut);\n", - "print \"Shear stress at heel = %.2f kN/square.m.\"%(tauh); \t\t\t\t#answer is wrong in book\n", - "\n", - "FOS = miu*SumV4/SumH1;\n", - "SFF = (miu*SumV4+wb*1400)/SumH1;\n", - "FOO = Mplus/Mminus1;\n", - "Ffi = 1.2;Fc = 2.7;\n", - "F = (miu*SumV4/Ffi+1400*wb/Fc)/SumH1;\n", - "FOS = round(FOS*1000)/1000;\n", - "SFF = round(SFF*100)/100;\n", - "FOO = round(FOO*100)/100;\n", - "F = round(F*100)/100;\n", - "print \"Factor of safety against sliding = %.2f.\"%(FOS);\n", - "print \"Shear friction factor = %.2f.\"%(SFF);\n", - "print \"Factor of safety against overturning = %.2f.\"%(FOO);\n", - "print \"Factor of safety for load combination E = %.2f. > 1\"%(F);\n", - "print \"Dam is safe \";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "case 1. Reservior empty:\n", - "Normal stress at toe = 156.30 kN/square.m.\n", - "Normal stress at heel = 1499.70 kN/square.m.\n", - "Principal stress at toe = 232.80 kN/square.m.\n", - "Principal stress at heel = 1514.70 kN/square.m.\n", - "Shear stress at toe = 109.40 kN/square.m.\n", - "Shear stress at heel = 150.00 kN/square.m.\n", - "case 2. reservior full without uplift:\n", - "Normal stress at toe = 1297.10 kN/square.m.\n", - "Normal stress at heel = 427.60 kN/square.m.\n", - "Principal stress at toe = 1932.60 kN/square.m.\n", - "Principal stress at heel = 425.00 kN/square.m.\n", - "Shear stress at toe = 907.90 kN/square.m.\n", - "Shear stress at heel = 25.90 kN/square.m.\n", - "case 3. reservior full with uplift:\n", - "Normal stress at toe = 1344.00 kN/square.m.\n", - "Normal stress at heel = 70.00 kN/square.m.\n", - "Principal stress at toe = 2002.00 kN/square.m.\n", - "Principal stress at heel = 64.00 kN/square.m.\n", - "Shear stress at toe = 941.00 kN/square.m.\n", - "Shear stress at heel = 62.00 kN/square.m.\n", - "Factor of safety against sliding = 1.08.\n", - "Shear friction factor = 4.14.\n", - "Factor of safety against overturning = 1.00.\n", - "Factor of safety for load combination B = 1.57. > 1\n", - "Dam is safe \n", - "case 4.considering seismic forces\n", - "Normal stress at toe = 1713.00 kN/square.m.\n", - "Normal stress at heel = -432.00 kN/square.m.\n", - "Principal stress at toe = 2552.00 kN/square.m.\n", - "Principal stress at heel = -443.00 kN/square.m.\n", - "Shear stress at toe = 1199.00 kN/square.m.\n", - "Shear stress at heel = 117.00 kN/square.m.\n", - "Factor of safety against sliding = 0.76.\n", - "Shear friction factor = 3.14.\n", - "Factor of safety against overturning = 1.00.\n", - "Factor of safety for load combination E = 1.52. > 1\n", - "Dam is safe \n" - ] - } - ], - "prompt_number": 17 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.17 pg : 429" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#design practical profile of gravity dam\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1;\n", - "rlb = 1450; \t\t\t\t#R.L of base of dam\n", - "rlw = 1480.5; \t\t\t\t#R.L of water level\n", - "Sg = 2.4; \t\t\t\t#specific gravity of masonary\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "w = 1; \t\t\t\t#heigth of waves\n", - "f = 1200; \t\t\t\t#safe compressive stress for masonary\n", - "FB = 1.5*w;\n", - "rlt = FB+rlw; \t\t\t\t#R.L of top of dam\n", - "H = rlt-rlb; \t\t\t\t#heigth of dam\n", - "LH = f/(gamma_w*(Sg+1))\n", - "LH = round(LH*100)/100;\n", - "print \"Heigth of dam = %.2f m.\"%(H);\n", - "print \"limiting heigth of dam = %.2f m.\"%(LH);\n", - "print \"Dam is low gravity dam\";\n", - "hw = rlw-rlb;\n", - "\t\t\t\t#keep top width,a = 4.5.\n", - "a = 4.5;\n", - "P = hw/(Sg**0.5);\n", - "P = round(P*10)/10;\n", - "print \"Base width of elementary profile = %.2f m.\"%(P);\n", - "uo = a/16;\n", - "wb = uo+P;\n", - "wb = round(wb);\n", - "print \"Base width = %.2f m.\"%(wb);\n", - "D = 2*a*(Sg**0.5);\n", - "D = round(D);\n", - "print \"Dismath.tance upto which u/s slope is vertical from water level = %.2f m.\"%(D);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Heigth of dam = 32.00 m.\n", - "limiting heigth of dam = 35.98 m.\n", - "Dam is low gravity dam\n", - "Base width of elementary profile = 19.70 m.\n", - "Base width = 20.00 m.\n", - "Dismath.tance upto which u/s slope is vertical from water level = 14.00 m.\n" - ] - } - ], - "prompt_number": 18 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.18 pg : 430" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#determine if dam is safe against sliding\n", - "\t\t\t\t\n", - "#Given\n", - "hw = 97.; \t\t\t\t#heigth of water in reservior\n", - "Bt = 7.; \t\t\t\t#width of top of dam\n", - "H = 100.; \t\t\t\t#heigth of the dam\n", - "Hs2 = 90.; \t\t\t\t#heigth of slope on downstream side\n", - "wb = 75.; \t\t\t\t#width of base of dam\n", - "miu = 0.75; \t\t\t\t#coefficient of friction\n", - "gamma_d = 2.4; \t\t\t\t#weigth density of concrete\n", - "gamma_w = 1000.; \t\t\t\t#weigth density of water\n", - "\n", - "# Calculations\n", - "P = gamma_w*hw**2/(2*1000);\n", - "W1 = Bt*gamma_d*H;\n", - "W2 = (wb-Bt)*Hs2*gamma_d/2;\n", - "W = W1+W2;\n", - "FOS = miu*W/P;\n", - "FOS = round(FOS*1000)/1000;\n", - "\n", - "# Results\n", - "print \"Factor of safety against sliding = %.2f.\"%(FOS);\n", - "print \"Dam is safe against sliding\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Factor of safety against sliding = 1.44.\n", - "Dam is safe against sliding\n" - ] - } - ], - "prompt_number": 24 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.19 pg : 430" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "#Factor of safety against overturning\n", - "#Factor of safety against sliding\n", - "#Shear friction factor\n", - "\n", - "#Given\n", - "c = 1.;\n", - "H = 10.; \t\t\t\t#heigth of dam\n", - "hw = 10.; \t\t\t\t#heigth of water in reservior\n", - "wb = 8.25; \t\t\t\t#bottom width\n", - "Bt = 1.; \t\t\t\t#top width\n", - "Hs1 = 0.1; \t\t\t\t#slope on upstream side\n", - "gamma_w = 9.81; \t\t\t\t#unit weigth of water\n", - "gamma_m = 22.4; \t\t\t\t#unit weigth of masonary\n", - "f = 1400.; \t\t\t\t#permissible shear stress at joint\n", - "miu = 0.75; \t\t\t\t#coefficient of friction\n", - "fi = math.atan(0.625);\n", - "theta = math.atan(0.1);\n", - "\n", - "# Calculations\n", - "W1 = Bt*H*gamma_m;\n", - "W2 = H*H*Hs1*gamma_m/2;\n", - "W3 = H*6.25*gamma_m/2;\n", - "W4 = hw*gamma_w*H*Hs1/2;\n", - "P = gamma_w*hw**2/2;\n", - "U = wb*gamma_w*hw*c/2;\n", - "SumV = W1+W2+W3+W4-U;\n", - "L3 = 2*(wb-(Hs1*H)-Bt)/3;\n", - "L1 = (wb-(Hs1*H)-Bt)+Bt/2;\n", - "L2 = (wb-(Hs1*H)-Bt)+Bt+(Hs1*H/3);\n", - "L4 = (wb-(Hs1*H)-Bt)+Bt+(2*Hs1*H/3);\n", - "L5 = 2*wb/3;L6 = hw/3;\n", - "M1 = W1*L1;M2 = W2*L2;M3 = W3*L3;M4 = W4*L4;\n", - "M5 = U*L5;M6 = P*L6;\n", - "SumM = M1+M2+M3+M4-M5-M6;\n", - "Mplus = M1+M2+M3+M4;\n", - "Mminus = M5+M6;\n", - "FOS = miu*SumV/P;\n", - "SFF = (miu*SumV+wb*1400)/P;\n", - "FOO = Mplus/Mminus;\n", - "FOS = round(FOS*100)/100;\n", - "SFF = round(SFF*10)/10;\n", - "FOO = round(FOO*100)/100;\n", - "\n", - "# Results\n", - "print \"Factor of safety against sliding = %.2f.\"%(FOS);\n", - "print \"Shear friction factor = %.2f.\"%(SFF);\n", - "print \"Factor of safety against overturning = %.2f.\"%(FOO);\n", - "print \"Dam is unsafe against overturning\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Factor of safety against sliding = 1.04.\n", - "Shear friction factor = 24.60.\n", - "Factor of safety against overturning = 1.47.\n", - "Dam is unsafe against overturning\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 8.20 pg : 431" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "import math \n", - "\n", - "\t\t\t\t\n", - "#Given\n", - "c = 1.;\n", - "hw = 80.; \t\t\t\t#heigth of water in reservior\n", - "Bt = 6.; \t\t\t\t#width of top of dam\n", - "H = 84.; \t\t\t\t#heigth of the dam\n", - "Hs2 = 75.; \t\t\t\t#heigth of slope on downstream side\n", - "wb = 56.; \t\t\t\t#width of base of dam\n", - "Lf = 8.; \t\t\t\t#dismath.tance of foundation gallery from heel\n", - "gamma_d = 23.5; \t\t\t\t#weigth density of concrete\n", - "gamma_w = 9.81; \t\t\t\t#weigth density of water\n", - "ht = 6.; \t\t\t\t#heigth of tail water\n", - "\n", - "# Calculations\n", - "W1 = Bt*gamma_d*H;\n", - "W2 = gamma_d*Hs2*(wb-Bt)/2;\n", - "W3 = gamma_w*ht*4/2;\n", - "W4 = gamma_w*hw**2/2;\n", - "W5 = gamma_w*ht**2/2;\n", - "Pt = gamma_w*ht\n", - "Ph = gamma_w*hw\n", - "Pg = (ht+(hw-ht)/3)*gamma_w\n", - "U = (Pt*(wb-Lf))+(Pg*Lf)+((Ph-Pg)*Lf/2)+((Pg-Pt)*(wb-Lf)/2)*c\n", - "l1 = (wb-Lf)/2\n", - "l2 = (2*(wb-Lf))/3\n", - "l3 = (wb-Lf)+(Lf/2)\n", - "l4 = (wb-Lf)+((2*Lf)/3)\n", - "L6 = (((Pt*(wb-Lf))*l1)+((Pg-Pt)*(wb-Lf)*l2/2)+((Pg*Lf)*l3)+((Ph-Pg)*Lf*l4/2))/U\n", - "L1 = (wb-Bt)+(Bt/2)\n", - "L2 = (2*(wb-Bt))/3\n", - "L3 = 4./3;\n", - "L4 = hw/3;\n", - "L5 = ht/3;\n", - "M1 = W1*L1\n", - "M2 = W2*L2\n", - "M3 = W3*L3\n", - "M4 = W4*L4\n", - "M5 = W5*L5\n", - "M6 = U*L6;\n", - "SumV = W1+W2+W3-U;\n", - "SumH = W4-W5;\n", - "SumM = M1+M2+M3-M4+M5-M6;\n", - "x = SumM/SumV;\n", - "e = wb/2-x;\n", - "pnt = (SumV/wb)*(1+(6*e/wb));\n", - "pnh = (SumV/wb)*(1-(6*e/wb));\n", - "pnt = round(pnt*10)/10;\n", - "pnh = round(pnh*10)/10;\n", - "\n", - "# Results\n", - "print \"Maximum Normal stress at toe = %.2f kN/square.m.\"%(pnt);\n", - "print \"Maximum Normal stress at heel = %.2f kN/square.m.\"%(pnh);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Maximum Normal stress at toe = 1586.70 kN/square.m.\n", - "Maximum Normal stress at heel = -49.30 kN/square.m.\n" - ] - } - ], - "prompt_number": 22 - } - ], - "metadata": {} - } - ] -}
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