# Chapter09:Geothermal Energy¶

## Ex9.1.i:pg-302¶

In [30]:
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"

 The Heat content per square km is  2.06923846154e+18 J/km^2


## Ex9.1.ii:pg-302¶

In [27]:
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"

Useful initial temp  is 135.0 degree K
Useful average temp after 25 years is 108.77 degree K


## Ex9.1.iii:pg-302¶

In [39]:
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"

Initial Heat extraction rate is  567.0 MW/km^2
Final Heat extraction rate is  456.84 MW/km^2


## Ex9.2.i:pg-304¶

In [5]:
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

The heat content is 7.68e+16 J/km^2


## Ex9.2.ii:pg-304¶

In [58]:
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

Time constant is  14.2 years


## Ex9.2.iii:pg-304¶

In [68]:
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

initial power per square km is 171.675  MW/km^2
power per square km in MW/km^2 after 25 years is  29.44 MW/km^2