Mechanics of Fluids

534 Compressible flow of gases
Successively using p = ρRT, ρu = constant (by continuity) and eqns 11.69
and 11.70, we have
dp
p
= dρ
ρ
dT
+
T =−du
dT du
+ = − 2dM
u T u M
�
(γ − 1)M
=−
2 + 1
1
2 (γ − 1)M2 �
dM
+ 1 M
(11.71)
If we divide the momentum equation 11.67 by u2 = M2γ p/ρ and substitute
for du/u and dp/p from eqns 11.70 and 11.71, we obtain
�
M2 − 1
1
2 (γ − 1)M2 �
dM f dl
+ = 0 (11.72)
+ 1 γ M3 2A/P
We now have only two main variables, M and l, and these are separated.
But the friction factor f is a function of Reynolds number and this is not
constant, changing with distance along the pipe (see Table 11.2). Changes of
temperature are appreciable only at high velocities, and therefore in general
only at high Reynolds numbers. Fortunately, the higher the Reynolds number
the less f depends on it, especially for rough pipes. For fully developed flow
(i.e., at a distance greater than say 50 times the diameter from the pipe
entrance) the value of f is apparently uninfluenced by Mach number. The
boundary layer, however, is greatly affected by the oblique shock waves that
in supersonic flow form near the entrance, and so the apparent value of f is
then notably reduced below the value for subsonic flow at the same Reynolds
number. However, if we consider a mean value
f = 1
l
� 1
we may integrate eqn 11.72 between points 1 and 2 a distance l apart:
�
1 1
2γ M2 −
2
1
M2 �
+
1
γ + 1
��M2 �2 (γ − 1)M2 1 + 2
ln
4γ M1 (γ − 1)M2 �
+
2 + 2
fl
= 0
2A/P
(11.73)
We may note in passing that if dM/M is eliminated between eqns 11.71 and
11.72 and ρu2 /γ M2 substituted for p we obtain
dp f
=−
dl 2A/P ρu2
�
1 + (γ − 1)M2 1 − M2 �
This expression reduces to Darcy’s formula as M → 0, so, if the Mach
number remains low, the fluid may, with small error, be treated as incompressible.
For air at M = 0.1, for example, the error in dp/dl is 1.41%.
When M is not small, however, solutions to problems require the use of
eqn 11.73 or its equivalent.
0
f dl