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diff --git a/sample_notebooks/GirishVora/GirishVora_version_backup/ch2.ipynb b/sample_notebooks/GirishVora/GirishVora_version_backup/ch2.ipynb new file mode 100755 index 00000000..9eb5c1ee --- /dev/null +++ b/sample_notebooks/GirishVora/GirishVora_version_backup/ch2.ipynb @@ -0,0 +1,428 @@ +{ + "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": {} + } + ] +}
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