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Large scale example

Consider a formation of sedimentary rock with a temperature gradient G = 0.018 °C m-1

Let the surface temperature To = 15 °C

Let the minimum useful temperature T1= 120 °C

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

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

Then T2 = To + 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.

GEROM

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, ρr= 2483 kg m-3,
And the specific heat capacity, cr = 907 J kg-1K
Given that the density of sea water at 10Km head, ρw= 1073 kg m-3
And the specific heat capacity of sea water at 180 °C  and 10 km head, cw = 4400 J kg-1K

(eq.1)
Eo = ½ AρrcrG(z2 - z1)2

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

Total useful heat content Eo = 3610 PJ86 Mtoe1003 TWhth

(eq.2)
τ = A ρrcr(z2 - z1 ) / V´ρwcw 

= (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 = (Eo / τ )∙e –t/τ

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

dE/dt (t=25y = 7.89E8 s) = (Eo / 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 Po (initially) and P25y(after 25 years) = 1.89GWth and 1.25 GWth respectively

Other Flow rates/ Areas

For the same example, but different areas and flow rates, we have

Area
km2

Flow
m3s-1

Eo
TWh

τ
years

Po
GW

P25y
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?

GEROMCarnot efficiency,  η = 1- T L/ TH

Taking condenser T L= 15 °C + 273 = 288 K,(assuming condenser is water cooled), and TH = 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)


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