Page 204 - Physical Principles of Sedimentary Basin Analysis

186 Heat flow

deposited sediments becomes unreasonable. To assure that the last condition is fulfilled we
require that vt 1/2
l 0 , which is the same as

Pe ln2
f (Pe) = 1
1. (6.304)
2
π + Pe 2
4
The function f (Pe) has a maximum at Pe = 2π where it is ln2/π ≈ 0.22. Since the
function f is rather large in the interval from Pe = 0.5toPe = 100 we cannot expect
the reduction in the temperature to reach a steady state for moderate deposition rates. The
Fourier series solution (6.166) of the convection–diffusion equation was used by Wangen
(1995) in a study of thermal blanketing.

6.21 Conservation of energy once more

A more general temperature equation than equation (6.15) can be derived by starting with
a more complete expression for energy conservation than what was used in Section 6.1.
The following derivation is not only more complete, but also more formal. It makes use of
Reynolds transport theorem, the continuity equation, Newton’s second law and some ther-
modynamics. The final result is the same temperature equation except for two terms – one
term that accounts for the work done on the system and a new term for the rate of change of
internal energy. We could have just added the first term to the previous temperature equa-
tion, since it has a simple and direct interpretation. However, the formal derivation shows
important parts of continuum mechanics, and also provides better insight into the foun-
dations of the temperature equation. The first law of thermodynamics expresses energy
conservation as
E = W + Q (6.305)

where the energy E of the system is equal to the work W done on the system and the heat
Q added to the system. The energy E is the internal energy E I plus the kinetic energy E K
and the first law becomes
d d
(E K + E I ) = (W + Q) (6.306)
dt dt
when written in terms of rates. The kinetic energy is
1 2
E K = v dV (6.307)
2 V
2
where the velocity squared is v = v · v = v i v i , and the volume of the system is V .The
internal energy inside the same volume V is

E I = edV (6.308)
V