Page 209 - Physical Principles of Sedimentary Basin Analysis

6.22 Mantle adiabat 191

enthalpy h = e + pν. A different version of the left-hand side of the temperature
equation (6.322)is

DT Dp
c p − αT = rhs (6.332)
dt dt
Ds
T = rhs (6.333)
dt
Dh Dp
− = rhs (6.334)
dt dt
where rhs denotes the right-hand side of equation (6.322).

Exercise 6.33 Assume that the rate of change of temperature and pressure is equally
important in the temperature equation (6.322). How large is the pressure step dp that cor-
◦
responds to a temperature step dT = 1 C? Let c p = 1kJ kg −1 K −1 , α = 2.5 · 10 −5 K −1 ,
= 2500 kg m −3 and T = 1000 K.
Solution: The step in temperature and pressure are equally important when c p dT =
αTdp, which gives
c p
dp = dT = 100 MPa. (6.335)
αT
Exercise 6.34 A body is composed of parts of mass m k and volume V k . The bulk density

is = m/V where m = k m k and V = k V k . Show that
m
= = φ k k (6.336)
V
where φ k = V k /V is the volume fraction of component k and k = m k /V k is the density
of component k.

6.22 Mantle adiabat
There are convection cells in the mantle. Let us assume that the mantle upwells or down-
wells sufficiently fast for almost no heat to be lost from a small volume of mantle rock
to its surroundings, when it follows the flow. Such a process, without any loss of heat, is
called an adiabatic process. It is tempting to think that the temperature would remain con-
stant in a volume that does not exchange heat with its surroundings. That would also have
been the case if the mantle had been incompressible. To see that we look at the temperature
equation (6.322), which gives
DT Dp
c p − αT = 0 (6.337)
dt dt
when there is no conductive heat transfer. An adiabatic process in thermodynamics is a
process with constant entropy, and we see that equation (6.337) gives that ds = 0 when