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diff --git a/Non_Conventional_Energy_Resources/Chapter9.ipynb b/Non_Conventional_Energy_Resources/Chapter9.ipynb new file mode 100755 index 00000000..8736d9a8 --- /dev/null +++ b/Non_Conventional_Energy_Resources/Chapter9.ipynb @@ -0,0 +1,303 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:47ed2e07aa752d35a238880d941b7ff865358a9055d3a5d12934bca3f8fdda8e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "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": {}
+ }
+ ]
+}
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