
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.
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 PJ ≈ 86 Mtoe ≈ 1003 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?
Carnot 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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