{
"metadata": {
"name": "",
"signature": "sha256:47ed2e07aa752d35a238880d941b7ff865358a9055d3a5d12934bca3f8fdda8e"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter09:Geothermal Energy"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex9.1.i:pg-302"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# given data\n",
"G=39.0 # temperature gradient in K/km.\n",
"h2=10.0 # depth in km\n",
"rhor=2700.0 # kg/m^3\n",
"cr=820.0 # in J/kg-K\n",
"\n",
"h1=120/G # T1-T0=120 K is given\n",
"h21=h2-h1 # in km\n",
"E0byA=(rhor*(1000**3)*G*cr*h21**2)/2 # in J/km^2 Heat content per square km\n",
"print\" The Heat content per square km is \",E0byA,\"J/km^2\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The Heat content per square km is 2.06923846154e+18 J/km^2\n"
]
}
],
"prompt_number": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex9.1.ii:pg-302"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# given data\n",
"G=39.0 # temperature gradient in K/km.\n",
"h2=10.0 # depth in km\n",
"rhor=2700.0 # kg/m^3\n",
"cr=820.0 # in J/kg-K\n",
"QbyA=0.5 #water flow rate in m^3/sec-km^2 \n",
"rhow=1000.0 # density of water in kg/m^3\n",
"cw=4200.0 # specific heat of water in J/kg-K \n",
"h1=120.0/G # T1-T0=120 K is given\n",
"h21=h2-h1 # in km\n",
"t=25 # time in years\n",
"\n",
"thetao=G*h21/2.0 # in degree K\n",
"print \"Useful initial temp is\",thetao,\"degree K\"\n",
"tau=rhor*cr*h21*(1000**3)/(QbyA*rhow*cw) # in seconds\n",
"tau=tau/(2*60*60*24*365) # in years\n",
"theta=thetao*math.exp(-t/tau) # in degree Kelvin\n",
"print \"Useful average temp after 25 years is\",round(theta,2),\"degree K\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Useful initial temp is 135.0 degree K\n",
"Useful average temp after 25 years is 108.77 degree K\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex9.1.iii:pg-302"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# given data\n",
"G=39.0 # temperature gradient in K/km.\n",
"h2=10.0 # depth in km\n",
"rhor=2700.0 # kg/m^3\n",
"cr=820.0 # in J/kg-K\n",
"\n",
"h1=120/G # T1-T0=120 K is given\n",
"h21=h2-h1 # in km\n",
"E0byA=(rhor*(1000**3)*G*cr*h21**2)/2 # in J/km^2 Heat content per square km\n",
"\n",
"thetao=G*h21/2.0 # in degree K\n",
"tau=rhor*cr*h21*(1000**3)/(QbyA*rhow*cw) # in seconds\n",
"tau=tau/(2*60*60*24*365) # in years\n",
"theta=thetao*math.exp(-t/tau) # in degree Kelvin\n",
"\n",
"Heatinitial=E0byA/(60*60*365*24*tau)/1000000 # intial heat extraction rate in MW /km^2\n",
"\n",
"Heat25=Heatinitial*math.exp(-t/tau) # heat extraction rate after 25 years in MW /km^2\n",
"\n",
"print \"Initial Heat extraction rate is \",Heatinitial,\"MW/km^2\"\n",
"\n",
"print \"Final Heat extraction rate is \",round(Heat25,2),\"MW/km^2\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Initial Heat extraction rate is 567.0 MW/km^2\n",
"Final Heat extraction rate is 456.84 MW/km^2\n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex9.2.i:pg-304"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# given data\n",
"w=0.6 # in km \n",
"h2=2.5 # in km\n",
"p=5/100.0 # porosity\n",
"rhor=3000.0 # density of sediment in kg/m^3\n",
"cr=750.0 # specific heat of sediment in J/kg-K\n",
"rhow=1000.0 # density of water in kg/m^3\n",
"cw=4200.0 # specific heat of water in J/kg-K\n",
"G=35.0 # temperature gradient in degree C/km\n",
"T1=45.0 # temp 1 in degree celsius\n",
"T0=12.0 # temp 2 in degree celsius\n",
"Q=0.75 # water extraction rate in m^3/sec-km^2\n",
"\n",
"T2=T0+G*h2 # initial temp in degree celsius\n",
"\n",
"thetao=T2-T1 # in degree celsius\n",
"\n",
"E0byA=(p*rhow*(1000**3)*cw+(1-p)*rhor*(1000**3)*cr)*w*thetao # in J/km^2\n",
"\n",
"print \"The heat content is\",round(E0byA,-14),\"J/km^2\"\n",
"\n",
"# the answer is different in textbook due to wrong thetao\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The heat content is 7.68e+16 J/km^2\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex9.2.ii:pg-304"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# given data\n",
"w=0.6 # in km \n",
"h2=2.5 # in km\n",
"p=5/100.0 # porosity\n",
"rhor=3000.0 # density of sediment in kg/m^3\n",
"cr=750.0 # specific heat of sediment in J/kg-K\n",
"rhow=1000.0 # density of water in kg/m^3\n",
"cw=4200.0 # specific heat of water in J/kg-K\n",
"G=35.0 # temperature gradient in degree C/km\n",
"T1=45.0 # temp 1 in degree celsius\n",
"T0=12.0 # temp 2 in degree celsius\n",
"Q=0.75 # water extraction rate in m^3/sec-km^2\n",
"\n",
"tau=((p*rhow*cw+(1-p)*rhor*cr)*w*1000**3/(Q*rhow*cw))/(60*60*24*365) # in years\n",
"\n",
"print \"Time constant is \",round(tau,1),\"years\"\n",
"\n",
"# the answer is different in textbook due to wrong calculations\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Time constant is 14.2 years\n"
]
}
],
"prompt_number": 58
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex9.2.iii:pg-304"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"# given data\n",
"w=0.6 # in km \n",
"h2=2.5 # in km\n",
"p=5/100.0 # porosity\n",
"rhor=3000.0 # density of sediment in kg/m^3\n",
"cr=750.0 # specific heat of sediment in J/kg-K\n",
"rhow=1000.0 # density of water in kg/m^3\n",
"cw=4200.0 # specific heat of water in J/kg-K\n",
"G=35.0 # temperature gradient in degree C/km\n",
"T1=45.0 # temp 1 in degree celsius\n",
"T0=12.0 # temp 2 in degree celsius\n",
"Q=0.75 # water extraction rate in m^3/sec-km^2\n",
"T2=T0+G*h2 # initial temp in degree celsius\n",
"t=25 # time in years\n",
"thetao=T2-T1 # in degree celsius\n",
"\n",
"E0byA=(p*rhow*(1000**3)*cw+(1-p)*rhor*(1000**3)*cr)*w*thetao # in J/km^2\n",
"\n",
"tau=((p*rhow*cw+(1-p)*rhor*cr)*w*1000**3/(Q*rhow*cw)) # in seconds\n",
"Pperkm2=(E0byA)/(tau*1000000) # initial power per square km in MW/km^2\n",
"print \"initial power per square km is\",Pperkm2,\" MW/km^2\"\n",
"Power20=Pperkm2*math.exp(-25*60*60*24*365/tau) # power per square km in MW/km^2 after 25 years\n",
"print \"power per square km in MW/km^2 after 25 years is \",round(Power20,2),\"MW/km^2\"\n",
"\n",
"# The answers are slightly different due to approximation in textbook"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"initial power per square km is 171.675 MW/km^2\n",
"power per square km in MW/km^2 after 25 years is 29.44 MW/km^2\n"
]
}
],
"prompt_number": 68
}
],
"metadata": {}
}
]
}