{ "metadata": { "name": "", "signature": "sha256:bcdc954bb2d82ee9797b3a585d54241820072bea3f7c78ca9c6a68f4f6eefc10" }, "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": [ "" ] } ], "prompt_number": 1 }, { "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": [ "" ] } ], "prompt_number": 2 }, { "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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\nTkSuQmCC\n", "text": [ "" ] } ], "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": {} } ] }