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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 T₁= 120 °C

Since T₁ = T_o + Gz₁ it follows that z₁ = (120 -15)/ 0.018 ≈ 5830 m

Let the maximum reachable well depth z₂ = 10,280 m

Then T₂ = T_o + Gz₂ = 15 + (0.018x10280) = 200 °C

A system of injection and return bore holes is heat-mining an engineered reservoir with a projected area A = 9 km² and with a sea-water flow rate V´= 10 m³ s^-1 extracting heat from the rocks.

The average Temperature is 160 °C, and the rock formation to be heat mined is deep (z₂ - z₂ ) = 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^-1K
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^-1K

(eq.1)
E_o = ½ Aρ_rc_rG(z₂ - z₁)²

E_o = ½ 9x10⁶ x 2483 x 907 x 0.018 x 4450² = 3.61x10¹⁸J

Total useful heat content E_o = 3610 PJ ≈ 86 Mtoe ≈ 1003 TWh_th

(eq.2)
τ = A ρ_rc_r(z₂ - z₁ ) / V´ρ_wc_w 

= (9x10⁶ x 2483 x 907 x 4450) / (10 x 1073 x 4400)

= 1.91 x 10⁹ s = 60.5 years

The time constant for useful heat extraction using 10 m³s^-1 of sea water τ = 1.91 x10⁹s = 60.5 y

(eq.3)
dE/dt = (E_o / τ )∙e ^–t/τ

dE/dt _(t=0) = (E_o / τ )∙1= 3.61x10¹⁸J / 1.91 x10⁹ s = 1.89 x 10⁹ W = 1.89 GW

dE/dt _{(t=25y = 7.89E8 s)} = (E_o / τ )∙ e ^–t/τ = 1.89 x 10⁹. e^{(-7.89E8/1.91E9)}

= 1.89 x 10⁹ .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

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)