Table D3. Time for change in gradi-
The initial temperature gradient, at x = Xo, is
Ts = 13.69C; Ts′ = 11.0; Tf = 1.0
Tf - Tso
( Xo , 0) =
δo - δ
Note that this simplified form of δo gives virtually the same
heat flux as the value from eq D1, when x = Xo. Now
δo - δ Ts′ - Tso
1 + 2 ∑ (-1) e .
Tf - Tso
* Calculated without eq D14 approximation.
The time for a given change in the gradient can be closely ap-
= 2 ln
δo - δ Tf - Tso
δ Ts′ - Ts
Table D3 shows the results for Prudhoe Bay.
It would take about 490 years for the bottom growth to cease and 900 years for the bottom gradient to
change significantly. Thus, the approximations used in the derivation of eq D5 are acceptable.
The temperature in the frozen zone requires 1666 years for sensible heat adjustment, leaving ∆t = 15000
1666 = 13,334 years for bottom melt (the interglacial is 15,000 years long). The energy balance at the bottom
of the permafrost is
T - Ts′
- Aqg .
= kf f
Table D4. Per-
where A is the fraction of the geothermal energy that goes into melting; it can exceed 1.0.
The heat flow from the thawed zone is greater than the geothermal heat flow, as we
have seen. For the example discussed here, A = 1.179 at the beginning of thaw and will
decline towards 1 as thaw proceeds. The permafrost thickness Xf after ∆t years is given by
X + b
Xf - b ln f
= a + Xo
Xo + b
after surface tem-
kf (Tf - Ts′)
a = -A
qg = kuG.
The final permafrost thickness is strongly dependent upon the value of A. Table D4 shows values for the
Prudhoe Bay example.
The heat flow from the thawed zone varies continuously during the thaw, denoted in eq D15 as Α(t) qg.
The heat flow at equilibrium is such that A = Ae = 1.0. Thus, let A(t) be a linear function of X, given by
y - ye
A = ( Ao - Ae )
1 - ye