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Analysis

[2]

Consider a large mass of uniform rock, with specific heat capacity cr density ρr and cross-section A, and ignore heat convection.  

With depth z increasing downward,  T = To + (dT/dz) ∙z = To + Gz where G is the temperature gradient.

GEROM - Analysis

Let the minimum useful temperature be T1 at depth z1; let the temperature at the maximum depth z2 be T2, and let the average available temperature greater than T1 be θ:

GEROM - AnalysisT1 = To + Gz1 and T2 = To + Gz2  

θ = (T2 - T1)/2 = G(z2 - z1 )/2

The total useful heat content of the rock segment between z1 and z2 : Eo = Cr θ, where Cr = thermal capacity = A ρrcr(z2 - z1 )
Therefore Eo = ½ AρrcrG(z2 - z1)2 (eq.1)

In an ideal heat exchange process, water with volume flow rate V´, density ρw and specific heat capacity cw is heated uniformly by the hot rocks through a temperature difference θ, thus
V´ρwcw θ = - Cr (dθ /dt)

(dθ/θ) = - V´ρwcw/Cr∙ dt = - d t /τ  ,
where the time constant τ is given by Cr/ V´ρwcw 

But Cr = A ρrcr (z2 - z1 )
therefore the time constant
τ = A ρrcr(z2 - z1 ) / V´ρwcw  (eq.2)

Hence θ = θo e –t/τ

The useful heat content is E = Cr θ = Eo e –t/τ  and the rate of change of heat extraction
dE/dt = (Eo / τ )∙e –t/τ  (eq.3)


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