Consider a formation of sedimentary rock with a temperature gradient G = 0.018 °C m^{1}
Let the surface temperature T_{o} = 15 °C
Let the minimum useful temperature T_{1}= 120 °C
Since T_{1} = T_{o} + Gz_{1} it follows that z_{1} = (120 15)/ 0.018 ≈ 5830 m
Let the maximum reachable well depth z_{2} = 10,280 m
Then T_{2} = T_{o} + Gz_{2} = 15 + (0.018x10280) = 200 °C
A system of injection and return bore holes is heatmining an engineered reservoir with a projected area A = 9 km^{2 }and with a seawater flow rate V´= 10 m^{3} s^{1} extracting heat from the rocks.
The average Temperature is 160 °C, and the rock formation to be heat mined is deep (z_{2}  z_{2} ) = 10280 m – 5830 m = 4450 m.
Given that the density for carbonate (e.g. limestone) formations, ρ_{r}= 2483 kg m^{3},
And the specific heat capacity, c_{r} = 907 J kg^{1}K
Given that the density of sea water at 10Km head, ρ_{w}= 1073 kg m^{3}
And the specific heat capacity of sea water at 180 °C and 10 km head, c_{w} = 4400 J kg^{1}K
(eq.1)
E_{o} = ½ Aρ_{r}c_{r}G(z_{2}  z_{1})^{2}
E_{o} = ½ 9x10^{6} x 2483 x 907 x 0.018 x 4450^{2} = 3.61x10^{18}J
Total useful heat content E_{o} = 3610 PJ ≈ 86 Mtoe ≈ 1003 TWh_{th}
(eq.2)
τ = A ρ_{r}c_{r}(z_{2}  z_{1} ) / V´ρ_{w}c_{w}
= (9x10^{6} x 2483 x 907 x 4450) / (10 x 1073 x 4400)
= 1.91 x 10^{9} s = 60.5 years
The time constant for useful heat extraction using 10 m^{3}s^{1} of sea water τ = 1.91 x10^{9}s = 60.5 y
(eq.3)
dE/dt = (E_{o} / τ )∙e ^{–t/τ}
dE/dt _{(t=0)} = (E_{o} / τ )∙1= 3.61x10^{18}J / 1.91 x10^{9} s = 1.89 x 10^{9} W = 1.89 GW
dE/dt _{(t=25y = 7.89E8 s)} = (E_{o} / τ )∙ e ^{–t/τ} = 1.89 x 10^{9}. e^{(7.89E8/1.91E9)}
= 1.89 x 10^{9} .e^{0.413} = 1.25 GW
The useful heat extraction rate P_{o} (initially) and P_{25y}(after 25 years) = 1.89GW_{th} and 1.25 GW_{th} respectively
Other Flow rates/ Areas
For the same example, but different areas and flow rates, we have

Area
km^{2} 
Flow
m^{3}s^{1} 
E_{o}
TWh 
τ
years 
P_{o}
GW 
P_{25y}
GW 
(a) 
9 
10 
1003 
60 
1.89 
1.25 
(b) 
9 
3 
1003 
202 
0.57 
0.50 
(c) 
6 
3 
668 
134 
0.57 
0.47 
(d)

4 
3 
446 
90 
0.57 
0.43 
(e)

3 
2 
334 
101 
0.38 
0.30 
Efficiency
How much of this heat energy can be converted into electricity
and how much of it can be used directly for heating and cooling?
Carnot efficiency, η = 1 T _{L}/ T_{H}
Taking condenser T _{L}= 15 °C + 273 = 288 K,(assuming condenser is water cooled), and T_{H} = θ = ½(120+200)=160 °C = 433 K
η = 1 (288/433) = 0.335
Actually only a fraction of this η is attained = 15% efficiency at best
Assuming 97% generator efficiency, the amount that can be converted to electricity with today’s technology is about 14.5 % (Small plants are usually assumed to yield only 10 % efficiency)
