import math
# given data
G=39.0 # temperature gradient in K/km.
h2=10.0 # depth in km
rhor=2700.0 # kg/m^3
cr=820.0 # in J/kg-K
h1=120/G # T1-T0=120 K is given
h21=h2-h1 # in km
E0byA=(rhor*(1000**3)*G*cr*h21**2)/2 # in J/km^2 Heat content per square km
print" The Heat content per square km is ",E0byA,"J/km^2"
import math
# given data
G=39.0 # temperature gradient in K/km.
h2=10.0 # depth in km
rhor=2700.0 # kg/m^3
cr=820.0 # in J/kg-K
QbyA=0.5 #water flow rate in m^3/sec-km^2
rhow=1000.0 # density of water in kg/m^3
cw=4200.0 # specific heat of water in J/kg-K
h1=120.0/G # T1-T0=120 K is given
h21=h2-h1 # in km
t=25 # time in years
thetao=G*h21/2.0 # in degree K
print "Useful initial temp is",thetao,"degree K"
tau=rhor*cr*h21*(1000**3)/(QbyA*rhow*cw) # in seconds
tau=tau/(2*60*60*24*365) # in years
theta=thetao*math.exp(-t/tau) # in degree Kelvin
print "Useful average temp after 25 years is",round(theta,2),"degree K"
import math
# given data
G=39.0 # temperature gradient in K/km.
h2=10.0 # depth in km
rhor=2700.0 # kg/m^3
cr=820.0 # in J/kg-K
h1=120/G # T1-T0=120 K is given
h21=h2-h1 # in km
E0byA=(rhor*(1000**3)*G*cr*h21**2)/2 # in J/km^2 Heat content per square km
thetao=G*h21/2.0 # in degree K
tau=rhor*cr*h21*(1000**3)/(QbyA*rhow*cw) # in seconds
tau=tau/(2*60*60*24*365) # in years
theta=thetao*math.exp(-t/tau) # in degree Kelvin
Heatinitial=E0byA/(60*60*365*24*tau)/1000000 # intial heat extraction rate in MW /km^2
Heat25=Heatinitial*math.exp(-t/tau) # heat extraction rate after 25 years in MW /km^2
print "Initial Heat extraction rate is ",Heatinitial,"MW/km^2"
print "Final Heat extraction rate is ",round(Heat25,2),"MW/km^2"
import math
# given data
w=0.6 # in km
h2=2.5 # in km
p=5/100.0 # porosity
rhor=3000.0 # density of sediment in kg/m^3
cr=750.0 # specific heat of sediment in J/kg-K
rhow=1000.0 # density of water in kg/m^3
cw=4200.0 # specific heat of water in J/kg-K
G=35.0 # temperature gradient in degree C/km
T1=45.0 # temp 1 in degree celsius
T0=12.0 # temp 2 in degree celsius
Q=0.75 # water extraction rate in m^3/sec-km^2
T2=T0+G*h2 # initial temp in degree celsius
thetao=T2-T1 # in degree celsius
E0byA=(p*rhow*(1000**3)*cw+(1-p)*rhor*(1000**3)*cr)*w*thetao # in J/km^2
print "The heat content is",round(E0byA,-14),"J/km^2"
# the answer is different in textbook due to wrong thetao
import math
# given data
w=0.6 # in km
h2=2.5 # in km
p=5/100.0 # porosity
rhor=3000.0 # density of sediment in kg/m^3
cr=750.0 # specific heat of sediment in J/kg-K
rhow=1000.0 # density of water in kg/m^3
cw=4200.0 # specific heat of water in J/kg-K
G=35.0 # temperature gradient in degree C/km
T1=45.0 # temp 1 in degree celsius
T0=12.0 # temp 2 in degree celsius
Q=0.75 # water extraction rate in m^3/sec-km^2
tau=((p*rhow*cw+(1-p)*rhor*cr)*w*1000**3/(Q*rhow*cw))/(60*60*24*365) # in years
print "Time constant is ",round(tau,1),"years"
# the answer is different in textbook due to wrong calculations
import math
# given data
w=0.6 # in km
h2=2.5 # in km
p=5/100.0 # porosity
rhor=3000.0 # density of sediment in kg/m^3
cr=750.0 # specific heat of sediment in J/kg-K
rhow=1000.0 # density of water in kg/m^3
cw=4200.0 # specific heat of water in J/kg-K
G=35.0 # temperature gradient in degree C/km
T1=45.0 # temp 1 in degree celsius
T0=12.0 # temp 2 in degree celsius
Q=0.75 # water extraction rate in m^3/sec-km^2
T2=T0+G*h2 # initial temp in degree celsius
t=25 # time in years
thetao=T2-T1 # in degree celsius
E0byA=(p*rhow*(1000**3)*cw+(1-p)*rhor*(1000**3)*cr)*w*thetao # in J/km^2
tau=((p*rhow*cw+(1-p)*rhor*cr)*w*1000**3/(Q*rhow*cw)) # in seconds
Pperkm2=(E0byA)/(tau*1000000) # initial power per square km in MW/km^2
print "initial power per square km is",Pperkm2," MW/km^2"
Power20=Pperkm2*math.exp(-25*60*60*24*365/tau) # power per square km in MW/km^2 after 25 years
print "power per square km in MW/km^2 after 25 years is ",round(Power20,2),"MW/km^2"
# The answers are slightly different due to approximation in textbook