Page 205 - Physical Principles of Sedimentary Basin Analysis

6.21 Conservation of energy once more 187

where e is the internal energy per unit mass. The time rate of change of work done on the
system, dW/dt, which was zero in the previous derivation of the temperature equation,
is now
dW
= f · v dA + b · v dV (6.309)
dt ∂V V
where f is the force on the surface ∂V of the volume V , and where b = gn z is the
gravitational body force per unit volume. The time-rate of heat input to the system is

dQ
=− q · n dA + SdV (6.310)
dt ∂V V
where the term dQ/dt is equal to the heat flux through the boundaries of the system added
to the heat generation inside the system. Energy conservation has so far been expressed
for only one component (or phase), but a generalization to a multicomponent system is
straightforward and it is done towards the end of the section. The time differentiation on the
left-hand side of equation (6.306) is brought through the integration signs using Reynolds
transport theorem (see Exercise 3.27). Reynolds transport theorem is needed because the
volume V of the system cannot be assumed constant, but changes as a function of time.
The left-hand side of the first law of thermodynamics (6.306) becomes

d ∂ 1 2
1 2
(E K + E I ) = (e + v ) +∇ · v (e + v ) dV. (6.311)
dt V ∂t 2 2
The differentiations are carried through the parentheses and the terms are regrouped using
the material derivative as follows
d D 1 2
1 2
(E K + E I ) = (e + v ) + v ·∇ (e + v ) dV
dt V dt 2 2
2
De 1 Dv 1 2 D
= + + e + v + ∇· v dV
V dt 2 dt 2 dt
2
De 1 Dv
= + dV (6.312)
V dt 2 dt
where the continuity equation (3.161) is identified in the last factor in the second step. This
factor is therefore zero, and we are left with equation (6.312).
The right-hand side of the first law of thermodynamics (6.306) can also be written as an
integral over the volume V . The surface integral of the heat flow crossing the surface ∂V
is converted to a volume integral using Gauss’s theorem (3.171). The surface integral over
the force f is converted to a volume integral by first expressing the force on the surface as
f i = σ ij n j , using the stress tensor σ ij and the (outward) normal vector n j to the surface,

∂(σ ij v j )

f i v i dV = σ ij n j v i dV = σ ij v j n i dV = dV. (6.313)
∂V ∂V ∂V V ∂x i