Basics of Fluid Mechanics, 2014a

440 CHAPTER 11. COMPRESSIBLE FLOW ONE DIMENSIONAL
Rearranging equation (11.177) results in
(
dM 2 kM 2 1+ k − 1 )
M 2
M 2 = 2 4 fdx
1 − M 2 (11.178)
D
After similar mathematical manipulation one can get the relationship for the velocity
to read
dU
U = kM 2 4 fdx
2(1− M 2 (11.179)
) D
and the relationship for the temperature is
dT
T = 1 4
dc (k − 1) M 4 fdx
= −k
2 c 2(1− M 2 (11.180)
) D
density is obtained by utilizing equations (11.179) and (11.170) to obtain
dρ
ρ = − kM2
2(1− M 2 )
The stagnation pressure is similarly obtained as
dP 0
P 0
= − kM2
2
The second law reads
( dT
ds = C p ln
T
4 fdx
D
4 fdx
D
) ( dP
− R ln
P
)
(11.181)
(11.182)
(11.183)
The stagnation temperature expresses as T 0 = T (1+(1− k)/2M 2 ). Taking derivative
of this expression when M remains constant yields dT 0 = dT (1 + (1 − k)/2M 2 ) and
thus when these equations are divided they yield
dT/T = dT 0 /T 0 (11.184)
In similar fashion the relationship between the stagnation pressure and the pressure can
be substituted into the entropy equation and result in
( ) ( )
dT0
dP0
ds = C p ln − R ln
(11.185)
T 0 P 0
The first law requires that the stagnation temperature remains constant, (dT 0 =0).
Therefore the entropy change is
ds (k − 1) dP 0
= − (11.186)
C p k P 0
Using the equation for stagnation pressure the entropy equation yields
2
ds (k − 1) M
=
C p 2
4 fdx
D
(11.187)