diff options
Diffstat (limited to 'Hydraulics/Chapter_4.ipynb')
| -rw-r--r-- | Hydraulics/Chapter_4.ipynb | 778 |
1 files changed, 778 insertions, 0 deletions
diff --git a/Hydraulics/Chapter_4.ipynb b/Hydraulics/Chapter_4.ipynb new file mode 100644 index 00000000..9cf88603 --- /dev/null +++ b/Hydraulics/Chapter_4.ipynb @@ -0,0 +1,778 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4 : Flow Over Weirs Notches" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.1 page no : 103" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the length of the weir\n", + "import math\n", + "#initialisation of variables\n", + "p = 70. \t\t\t#per cent\n", + "Cd = 0.6\n", + "Q = 50. \t\t\t#million gallons\n", + "H = 2. \t\t\t#ft\n", + "w = 62.4 \t\t\t#lb/ft**3\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "#CALCULATIONS\n", + "Q1 = p*Q*10**6*10/(100*w*24*3600)\n", + "L = Q1*3/(2*Cd*math.sqrt(2*g)*H**1.5)\n", + "#RESULTS\n", + "print 'length of the weir = %.2f ft '%(L)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "length of the weir = 7.15 ft \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.2 page no : 104" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the theoretical horse power\n", + "\n", + "import math \n", + "\n", + "#initialisation of variables\n", + "L = 15. \t\t\t#ft\n", + "H = 1. \t\t\t#ft\n", + "Cd = 0.6\n", + "v = 80. \t\t\t#ft/min\n", + "g = 32.2 \t\t\t#ft/sec62\n", + "w = 62.4 \t\t\t#lb/ft**3\n", + "#CALCULATIONS\n", + "vo = v/60\n", + "Q = 2*Cd*math.sqrt(2*g)*L*((1+(vo**2/(2*g)))**1.5-(vo**2/(2*g))**1.5)*w*100/(3*550)\n", + "#RESULTS\n", + "print 'HP = %.f HP '%(Q)\n", + "\n", + "# This is accurate answer. Please calcualte manually." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "HP = 567 HP \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.3 pageno : 104" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the discharge per second\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "L = 11. \t\t\t#ft\n", + "H = 0.7 \t\t\t#ft\n", + "Cd = 0.6\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "h = 1.95 \t\t\t#ft\n", + "Q = 20.65 \t\t\t#cuses\n", + "Q1 = 21.2 \t\t\t#cfs\n", + "\n", + "#CALCULATIONS\n", + "Q = 2*Cd*math.sqrt(2*g)*L*H**1.5/3\n", + "vo = Q/(h*L)\n", + "h1 = vo**2/(2*g)\n", + "Q1 = 2*Cd*math.sqrt(2*g)*L*((H+(vo**2/(2*g)))**1.5-(vo**2/(2*g))**1.5)/3\n", + "v1 = Q1/(L*h)\n", + "Q2 = 2*Cd*math.sqrt(2*g)*L*((H+(v1**2/(2*g)))**1.5-(v1**2/(2*g))**1.5)/3\n", + "p = (Q2-Q1)*100/Q1\n", + "\n", + "#RESULTS\n", + "print \"Head to velocity approach = %.1f cu ft/sec\"%Q1\n", + "print \"Q2 = %.2f cu ft/sec\"%Q2\n", + "print 'discharge percent = %.3f per cent '%(p)\n", + "\n", + "# Note : answers may vary because of rounding error. Please calculate manually." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Head to velocity approach = 21.3 cu ft/sec\n", + "Q2 = 21.29 cu ft/sec\n", + "discharge percent = 0.148 per cent \n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.4 page no : 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find K and n\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "b = 3. \t\t\t#ft\n", + "H = 1 \t\t\t#ft\n", + "Q = 9 \t\t\t#cfs\n", + "Q1 = 1.105 # log Q from fig.\n", + "h = 0.1 \t\t# log H from fig. ft\n", + "#CALCULATIONS\n", + "K = Q/b\n", + "n = (Q1-math.log10(3*K))/h\n", + "#RESULTS\n", + "print 'K = %.f '%(K)\n", + "print 'n = %.1f '%(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "K = 3 \n", + "n = 1.5 \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.5 page no : 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "find the rate of discharge\n", + "'''\n", + "import math \n", + "#initialisation of variables\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "Cd = 0.62\n", + "L = 7.573 \t\t\t#ft\n", + "H = 1.2 \t\t\t#ft\n", + "S = 2.85 \t\t\t#ft\n", + "#CALCULATIONS\n", + "Q1 = 2*Cd*math.sqrt(2*g)*L*H**1.5/3\n", + "Q2 = 3.33*L*H**1.5\n", + "Q3 = math.sqrt(2*g)*L*H**1.5*(0.405+(0.00984/H))\n", + "He = H+0.004\n", + "Q4 = (3.227+0.435*(He/S))*L*He**1.5\n", + "#RESULTS\n", + "print 'Q = %.2f cuses '%(Q1)\n", + "print 'Q = %.2f cuses '%(Q2)\n", + "print 'Q = %.2f cuses '%(Q3)\n", + "print 'Q = %.2f cuses '%(Q4)\n", + "\n", + "# Note : answers may vari because of rounding error. Please check manually." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q = 33.02 cuses \n", + "Q = 33.15 cuses \n", + "Q = 33.01 cuses \n", + "Q = 34.12 cuses \n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.6 pageno : 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the length of the waste weir required.\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "H = 2.5 \t\t\t#ft\n", + "L = 10 \t\t\t#ft\n", + "A = 10 \t\t\t#miles\n", + "p = 30 \t\t\t#per cent\n", + "a = 2 \t\t\t#in/hr\n", + "w = 2 \t\t\t#ft\n", + "#CALCULATIONS\n", + "Q = L*1760**2*3**2*a*p/(60*60*12*100)\n", + "n = ((Q/(3.33*H**1.5))-(L-0.1*w*H))/(L-0.1*w*H)\n", + "#RESULTS\n", + "print 'n = %.f '%(n)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "n = 30 \n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.7 page no : 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the discharge of water \n", + "\n", + "import math \n", + "#initialisation of variables\n", + "L = 2.5 \t\t\t#ft\n", + "H = 1 \t\t\t#ft\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "Cd = 0.61\n", + "L1 = 1.75 \t\t\t#ft\n", + "L2 = 2.25 \t\t\t#ft\n", + "#CALCULATIONS\n", + "Q1 = 2*Cd*math.sqrt(2*g)*L*H/3\n", + "Q2 = 2*Cd*math.sqrt(2*g)*L1*(L1**1.5-1)/3\n", + "Q3 = 2*Cd*math.sqrt(2*g)*H*(L2**1.5-L1**1.5)/3\n", + "Q = Q1+Q2+Q3\n", + "#RESULTS\n", + "print 'Total discharge = %.1f cfs '%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total discharge = 19.1 cfs \n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.8 page no : 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the percentage error in estimating the discharge\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "h1 = 16.63 \t\t\t#cm\n", + "h2 = 10.18 \t\t\t#cm\n", + "h3 = 16.53 \t\t\t#cm\n", + "#CALCULATIONS\n", + "H1 = h1-h2\n", + "H2 = h3-h2\n", + "p = (H1**1.5-H2**1.5)*100/H1**1.5\n", + "#RESULTS\n", + "print 'Percent decrease in discharge = %.2f %% '%(p)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percent decrease in discharge = 2.32 % \n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.9 pageno : 111" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find time required to lower the level of a reservoir of superficial area\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "Cd = 0.6\n", + "a = 20000 \t\t\t#yd**2\n", + "H2 = 12 \t\t\t#in\n", + "L = 5 \t\t\t#ft\n", + "H1 = 2 \t\t\t#ft\n", + "g =32.2 \t\t\t#ft/s**2\n", + "#CALCULATIONS\n", + "t = 2*a*9*(L-H1)*((1/math.sqrt(H2/12))-(1/math.sqrt(H1)))/(2*60*Cd*math.sqrt(2*g)*L)\n", + "#RESULTS\n", + "print 'time required to lower level of reservoir = %.2f min '%(t)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "time required to lower level of reservoir = 109.49 min \n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.10 pageno : 113" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find depth of water through this notch.\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "L = 3. \t\t\t#ft\n", + "H = 6. \t\t\t#in\n", + "Cd = 0.62\n", + "Cd1 = 0.59\n", + "a = 45. \t\t\t#degrees\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "#CALCULATIONS\n", + "H = ((2./3)*Cd*math.sqrt(2*g)*L*(H/12)**1.5/((8./15)*Cd1*math.sqrt(2*g)))**0.4\n", + "#RESULTS\n", + "print 'depth of water = %.3f ft '%(H)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "depth of water = 1.142 ft \n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.11 page no : 114" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find in each case the percentage error \n", + "import math \n", + "#initialisation of variables\n", + "V = 20. \t\t\t#litres\n", + "g = 981. \t\t\t#cm/sec**2\n", + "Cd = 0.593\n", + "r = 2.5\n", + "r1 = 1.5\n", + "e = 2. \t\t\t#mm\n", + "Cd1 = 0.623\n", + "L = 30. \t\t\t#cm\n", + "#CALCULATIONS\n", + "H = (V*1000*15/(8*Cd*math.sqrt(2*g)))**0.4\n", + "dH1 = e/10.\n", + "p = r*dH1*100/H\n", + "H1 = (V*3*1000/(2*Cd1*math.sqrt(2*g)*L))**(2./3)\n", + "p1 = r1*dH1*100/H1\n", + "#RESULTS\n", + "print 'percentage error of discharge over the weir = %.2f %% '%(p)\n", + "print 'percentage error of discharge over the weir = %.2f %% '%(p1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "percentage error of discharge over the weir = 2.74 % \n", + "percentage error of discharge over the weir = 2.74 % \n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.12 page no : 116" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Estimate the discharge in the channel in gallons per minute\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "L = 16. \t\t\t#in\n", + "H = 9. \t\t\t#in\n", + "h = 18. \t\t\t#in\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "w = 2. \t\t\t#ft\n", + "Cd = 0.63\n", + "W = 62.4 \t\t\t#lbs/ft**3\n", + "#CALCULATIONS\n", + "Q = 2*Cd*math.sqrt(2*g)*(L/12)*(H/12)**1.5/3\n", + "v = Q/(w*(h/12))\n", + "H1 = v**2/(2*g)\n", + "Q1 = 2*Cd*math.sqrt(2*g)*(L/12)*(((H/12)+H1)**1.5-H1**1.5)*W*6/3.\n", + "#RESULTS\n", + "print 'Discharge = %.f gpm '%(Q1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Discharge = 1122 gpm \n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.13 pageno : 118" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Determine the discharge \n", + "\n", + "import math \n", + "#initialisation of variables\n", + "L = 100 \t\t\t#ft\n", + "H = 2.25 \t\t\t#ft\n", + "Cd = 0.95\n", + "w = 120 \t\t\t#ft\n", + "h = 2 \t\t\t#ft\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "#CALCULATIONS\n", + "Q = round(3.087*Cd*L*H**1.5)\n", + "v0 = round(Q/(w*(h+H)),2)\n", + "Q1 = 3.087*Cd*L*((H+(v0**2/(2*g)))**1.5-(v0**2/(2*g))**1.5)\n", + "#RESULTS\n", + "print 'Discharge = %.0f cuses '%(Q1)\n", + "\n", + "# Note: answer is slightly different because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Discharge = 1024 cuses \n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.14 pageno : 119" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find discharge over this weir in gallons per hour\n", + "import math \n", + "#initialisation of variables\n", + "L = 6 \t\t\t#ft\n", + "H1 = 0.5 \t\t\t#ft\n", + "H2 = 0.25 \t\t\t#ft\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "Cd1 = 0.58\n", + "Cd2 = 0.8\n", + "w = 6.24 \t\t\t#lb/ft**3\n", + "#CALCULATIONS\n", + "Q1 = 2*Cd1*math.sqrt(2*g)*L*(H1-H2)**1.5/3\n", + "Q2 = Cd2*L*H2*math.sqrt(2*g*(H1-H2))\n", + "Q = round((Q1+Q2)*w*3600,-3)\n", + "#RESULTS\n", + "print 'Discharge = %.f gph '%(Q)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Discharge = 160000 gph \n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.15 pageno : 120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the height of an anicut \n", + "import math \n", + "#initialisation of variables\n", + "W = 100 \t\t\t#ft\n", + "h = 10 \t\t\t#ft\n", + "v = 4 \t\t\t#ft/sec\n", + "h1 = 3 \t\t\t#ft\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "H = 5.4 \t\t\t#ft\n", + "Cd1 = 0.58\n", + "Cd2 = 0.8\n", + "#CALCULATIONS\n", + "v0 = (W*h*v)/(W*(h+h1))\n", + "h0 =v0**2/(2*g)\n", + "H2 = (W*h*v-(2*Cd1*W*math.sqrt(2*g)*((h1+h0)**1.5-h0**1.5)/3))/(Cd2*W*math.sqrt(2*g*(h1+h0)))\n", + "dh = h-H2\n", + "#RESULTS\n", + "print 'height of anicut which is drowned = %.f ft '%(dh)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "height of anicut which is drowned = 8 ft \n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.16 page no : 123" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find length\n", + "import math \n", + "#initialisation of variables\n", + "x = 6. \t\t\t#in\n", + "l = 200. \t\t\t#ft\n", + "d = 10. \t\t\t#ft\n", + "v = 4. \t\t\t#ft/sec\n", + "Ce = 0.95\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "#CALCULATIONS\n", + "l1 = math.sqrt(l**2/(Ce**2*(((x/12)*2*g/v**2)+(d**2/(d+(x/12))**2))))\n", + "#RESULTS\n", + "print 'length = %.f ft '%(l1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "length = 123 ft \n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.17 page no : 124" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the number of siphons required\n", + "\n", + "import math \n", + "#initialisation of variables\n", + "g = 32.2 \t\t\t#ft/sec**2\n", + "H = 25. \t\t\t#ft\n", + "l = 2.5 \t\t\t#ft\n", + "b = 5. \t\t\t#ft\n", + "Cd = 0.64\n", + "Q = 3200. \t\t\t#cuses\n", + "L =150. \t\t\t#ft\n", + "C =3.2\n", + "depth =0.5 \t\t\t#ft\n", + "A1 =5000000. \t\t\t#sq yards\n", + "#CALCULATIONS\n", + "Q1 = Cd*l*b*math.sqrt(2*g*H)\n", + "n = Q/Q1\n", + "h = (Q/(3.2*L))**(2./3)\n", + "hr =h-depth\n", + "Area =A1*9\n", + "V = round(Area*hr,-6)\n", + "#RESULTS\n", + "print 'number of spilways = %.f '%(n)\n", + "print \"Volume of extra water stored = %d cu ft\"%(V)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "number of spilways = 10 \n", + "Volume of extra water stored = 137000000 cu ft\n" + ] + } + ], + "prompt_number": 57 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file |