{
"metadata": {
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 9 : Groundwater Flow"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.1 Page No : 9-11"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\n",
"#initialisation of variables\n",
"t = 10\t#C\n",
"s = 74.2\t#days\n",
"c = 0.01\t#mm\n",
"d = 245\t#mm\n",
"\t\n",
"#CALCULATIONS\n",
"h = s/(d*c)\t#cm\n",
"\t\n",
"#RESULTS\n",
"print 'the high will water at a temperature = %.1f cm'%(h)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the high will water at a temperature = 30.3 cm\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.2 Page No : 9-13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\n",
"#initialisation of variables\n",
"p1 = 1000\t#ft\n",
"p2 = 50\t#ft\n",
"g = 20\t#ft/mile\n",
"v = 5280.\t#ft\n",
"q = 7.5*10**-6\t#ft\n",
"t = 60\t#F\n",
"k = 2835\t#ft/days\n",
"p = 7.5\t#ft\n",
"\t\n",
"#CALCULATIONS\n",
"S = g/v\t#ft\n",
"W = k*(g/v)\t#ft/day\n",
"Q = W*p1*p2*q\t#mgd\n",
"P = k*p\t#ft\n",
"P1 = P*p2\t#mgd\n",
"\t\n",
"#RESULTS\n",
"print 'the velocity of flow = %.2f mgd'%(Q)\n",
"print 'the standard coefficient pf permeability = %.2e mgd'%(P1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the velocity of flow = 4.03 mgd\n",
"the standard coefficient pf permeability = 1.06e+06 mgd\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.3 Page No : 9-19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\n",
"#initialisation of variables\n",
"p = 40. \t#ft\n",
"d = 56. \t#ft\n",
"d1 = 140. \t#ft\n",
"p1 = 30. \t#ft\n",
"w = 3.28*10**-4\t#fps\n",
"\t\n",
"#CALCULATIONS\n",
"Q = w*(p/d1)*2*d*p\t#cfs\n",
"q = Q/p\t#cfs\n",
"K = w*(p/d1)\t#fps\n",
"x0 = q/(2*math.pi*K)\t#ft\n",
"Z = 2*math.pi*x0\t#ft\n",
"\t\n",
"#RESULTS\n",
"print 'the yield of the well if the coefficient of permeability = %.1f ft'%(x0)\n",
"print 'the distance of the point of stagnation = %.0f ft'%(Z)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the yield of the well if the coefficient of permeability = 17.8 ft\n",
"the distance of the point of stagnation = 112 ft\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.4 Page No : 9-23"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\n",
"#initialisation of variables\n",
"p = 5.*10**6\t #ft\n",
"Q = 350. \t#gpm\n",
"x = 225. \t#ft\n",
"u = 10.**-2\t#ft\n",
"g = 1.87\t#ft\n",
"p2 = 7.*10**2\t#ft\n",
"p3 = 10.9\t#ft\n",
"w = 4.0\t#ft\n",
"t = 114.6\t #ft\n",
"d = 10. \t#ft\n",
"p1 = 5. \t#ft\n",
"w1 = 3.2*10**4\t#ft\n",
"W = 21.75\t#ft\n",
"\t\n",
"#CALCULATIONS\n",
"T = t*Q*4/p1\t#gpd/ft\n",
"S = u*(w1)/(g*(p))\t#ft\n",
"U = g*((S)/(T))*(x**2/d)\t#ft\n",
"P = t*(p2)*p3/(T)\t#ft\n",
"U1 = g*((S)/(T))*(1./d)\t#ft\n",
"P1 = t*(p2)*W/(T)\t#ft\n",
"\t\n",
"#RESULTS\n",
"print 'the type curve as if a transparency of the observed data had moved into place over the type = %.0f ft'%(P1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the type curve as if a transparency of the observed data had moved into place over the type = 54 ft\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.5 Page No : 9-24"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\t\n",
"#initialisation of variables\n",
"Q = 350. \t#gpm\n",
"x = 225. \t#ft\n",
"t = 1. \t#min\n",
"p = 1.6\t #ft\n",
"t2 = 10. \t#min\n",
"p2 = 4.5\t#ft\n",
"p3 = 700. \t#gpm\n",
"T = 3.2*10**4\t#gpd/ft\n",
"t0 = 0.3 \t#min\n",
"u = 1.15*10**-5\n",
"\t\n",
"#CALCULATIONS\n",
"S = t0*(T)*t0/((x)**2*1440)\t#ft\n",
"P = ((114.6*p3)/(T))*(-0.5772*2.3*math.log(u))\t#ft\n",
"\n",
"#RESULTS\n",
"print 'A straight line being drawn through the ppints for the higher = %.0f ft'%(P)\n",
"\n",
"# note : book answer is wrong. please check."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"A straight line being drawn through the ppints for the higher = 38 ft\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 9.6 Page No : 9-31"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\n",
"#initialisation of variables\n",
"h = 4.8\t #ft\n",
"m = 13.4\t #ft\n",
"k = 10.**-1\t #cm/sec\n",
"k1 = 3.28*10**-3\t#fps\n",
"n = 7. \t#ft\n",
"n1 = 11. \t#ft\n",
"q = 1.0*10**-2\n",
"\t\n",
"#CALCULATIONS\n",
"Q = k1*h*n/n1\t#cfs/ft\n",
"Q1 = 2*q*10**2\t#cfs\n",
"\t\n",
"#RESULTS\n",
"print 'A satisfactory orthogonal system the flow of into the collector = %.0f cfs'%(Q1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"A satisfactory orthogonal system the flow of into the collector = 2 cfs\n"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}