Consider a large mass of uniform rock, with specific heat capacity cr density ρ_{r} and crosssection A, and ignore heat convection.
With depth z increasing downward, T = T_{o} + (dT/dz) ∙z = T_{o} + Gz where G is the temperature gradient.
Let the minimum useful temperature be T_{1} at depth z_{1}; let the temperature at the maximum depth z_{2} be T_{2}, and let the average available temperature greater than T_{1} be θ:
T_{1} = T_{o} + Gz1 and T_{2} = T_{o} + Gz_{2}
θ = (T_{2}  T_{1})/2 = G(z_{2}  z_{1} )/2
The total useful heat content of the rock segment between z_{1} and z_{2} :
E_{o} = C_{r} θ, where C_{r} = thermal capacity = A ρ_{r}c_{r}(z_{2}  z_{1} )
Therefore E_{o} = ½ Aρ_{r}c_{r}G(z_{2}  z_{1})^{2} (eq.1)
In an ideal heat exchange process, water with volume flow rate V´, density ρ_{w} and specific heat capacity c_{w} is heated uniformly by the hot rocks through a temperature difference θ, thus
V´ρ_{w}c_{w} θ =  C_{r} (dθ /dt)
(dθ/θ) =  V´ρ_{w}c_{w}/C_{r}∙ dt =  d t /τ ,
where the time constant τ is given by C_{r}/ V´ρ_{w}c_{w}
But C_{r} = A ρ_{r}c_{r} (z_{2}  z_{1} )
therefore the time constant
τ = A ρ_{r}c_{r}(z_{2}  z_{1} ) / V´ρ_{w}c_{w} (eq.2)
Hence θ = θ_{o} e ^{–t/τ}
The useful heat content is E = C_{r} θ = E_{o} e ^{–t/τ}
and the rate of change of heat extraction
dE/dt = (E_{o} / τ )∙e ^{–t/τ} (eq.3)
