Mechanics of Fluids

Compressible flow in pipes of constant cross-section 541
The reason why an outlet Mach number of � (1/γ ) corresponds to the
maximum mass flow rate is seen from eqn 11.82. This gives
dp
dl
�
f ρu2
=
2A/P
�
ρu2 �
− 1 =
p f ρu2
�
(γ M
2A/P
2 − 1) (11.86)
Thus when M = � (1/γ ) (= 0.845 for air) the pressure gradient is infinite;
that is, there is a discontinuity of pressure (and also, by eqn 11.81, of density
and velocity).
For a given initial Mach number M1, there is thus a limiting length for
continuous isothermal flow, and this is given by setting M2 = � (1/γ ) in
eqn 11.85:
flmax
A/P
1
=
γ M2 − 1 + ln(γ M
1
2 1 )
If the actual length were made greater than lmax the rate of flow would adjust
itself so that M = � (1/γ ) was not reached until the end of the pipe. Thus
it is seen that the phenomenon of choking may occur in isothermal flow but
that the limiting value of the Mach number is (theoretically) � (1/γ ) instead
of unity as in adiabatic flow.
In practice, however, isothermal flow at Mach numbers close to � (1/γ )
cannot be obtained. For a perfect gas under isothermal conditions both p/ρ
and the internal energy divided by mass e are constant, and so the steady-flow
energy equation (3.13) reduces to
q = 1
2 u2 1
2 −
2 u2 1
(11.87)
the gravity terms again being neglected. Hence, if points 1 and 2 are separated
by a distance δl, the heat transfer gradient dq/dl = udu/dl. Substitution from
eqns 11.81 and 11.86 then gives
dq
dl
= udu
dl =−u2
dp
p dl =
f γ M2u2 2(1 − γ M2 )A/P
For conditions near to M � (1/γ ), and for large velocities generally, the
required high values of dq/dl are difficult to achieve and the flow becomes
more nearly adiabatic. Indeed, the limiting condition M = � (1/γ ) cannot
be achieved at all because dq/dl would then have to be infinite.
Except for high Mach numbers, results for isothermal and adiabatic
flow do not in fact differ widely. This is because in adiabatic flow at low
Mach numbers there is little variation of temperature (see Table A3.3).
In isothermal flow
p2
p1
= ρ2
=
ρ1
u1
=
u2
M1
M2
(11.88)