Mechanics of Fluids

490 Compressible flow of gases
lose, so the total entropy increases. There is no process by which the total
entropy can decrease: for a reversible process the total entropy is constant;
for all other it increases.
An important relation may be deduced for a frictionless adiabatic process
in a gas. Referring throughout to quantities divided by the mass of gas, we
may write for a frictionless (i.e. reversible) process: δqrev = Tδs and so specific
heat capacity = Tδs/δT. Hence cp = T(∂s/∂T)p and cv = T(∂s/∂T)v.
Now since entropy is a function of state (i.e. of absolute pressure p and
volume divided by mass v) we may write
� � � �
∂s ∂s
δs = δv + δp
∂v p ∂p v
where a suffix indicates the quantity held constant. Temperature does not
appear as a separate variable since it is uniquely related to p and v by the
equation of state. Since the assumed process is isentropic, however, δs = 0
and so, as δp and δv both tend to zero,
� �
∂p
=−
∂v
(∂s/∂v)p
=−
(∂s/∂p)v
(∂s/∂T)p
�
(∂s/∂T)v
(∂v/∂T)p (∂v/∂T)v
s
=−γ (∂p/∂T)v
(∂v/∂T)p
A, R (11.3)
where γ = cp/cv.
If the gas obeys Boyle’s Law pv = f (T), where f (T) represents any
function of T, then
� �
∂p
∂T
v
= 1 df
v dT and
� �
∂v
∂T p
These expressions substituted in eqn 11.3 give
� �
∂p γ p
=−
∂v v
s
= 1 df
p dT
A, R, B (11.4)
For a constant value of γ , integration then yields ln p =−γ ln v+ constant.
Thus
pv γ = constant = p/ρ γ
A, R, B γ const (11.5)
The equation of state for a perfect gas satisfies Boyle’s Law. For a perfect
gas p/(ρT) = constant and the isentropic relation 11.5 may be alternatively
expressed
p (γ −1)γ
= constant A, R, PG i.e., I, PG (11.6)
T
The equation of state which relates p, ρ and T, strictly refers to conditions
of thermodynamic equilibrium. Experience shows, however, that
this equation, and others derived from it, may in practice be used for
non-equilibrium conditions except when the departure from equilibrium is
extreme (as in explosions).