Basics of Fluid Mechanics, 2014a

11.10. INTRODUCTION 473
or in simple terms as
Density Ratio
ρ 1
ρ 2
= U 2
U 1
= M 2
M 1
√
T2
T 1
(11.216)
or substituting equations (11.213) and (11.216) into equation (11.214) yields
T 2
= 1+kM 1 2
T
2
1 1+kM 2
M 2
M 1
√
T2
T 1
(11.217)
Transferring the temperature ratio to the left hand side and squaring the results gives
T 2
T 1
=
Temperature Ratio
[
2 1+kM1
1+kM 2
2
] 2 (
M2
M 1
) 2
(11.218)
The Rayleigh line exhibits two possible
maximums one for dT/ds = 0 and for
ds/dT =0. The second maximum can
be expressed as dT/ds = ∞. The second
law is used to find the expression for the
derivative.
s 1 − s 2
=ln T 2
− k − 1
C p T 1 k
ln P 2
P 1
(11.219)
T
M1
P=P ⋆
M=1
Fig. -11.38. The temperature entropy diagram
for Rayleigh line.
s
s 1 − s 2
C p
[
=2ln ( 1+kM 1 2 )
(1 + kM 2 2 )
]
M 2
M 1
+ k − 1
k
[ 1+kM21
2
]
ln
2
1+kM 1
(11.220)
Let the initial condition M 1 , and s 1 be constant and the variable parameters are M 2 ,
and s 2 . A derivative of equation (11.220) results in
1 ds
C p dM = 2(1− M 2 )
M (1 + kM 2 (11.221)
)
Taking the derivative of equation (11.221) and letting the variable parameters be T 2 ,
and M 2 results in
dT
dM = constant × 1 − kM2
(1 + kM 2 ) 3 (11.222)
Combining equations (11.221) and (11.222) by eliminating dM results in
dT
ds = constant × M(1 − kM 2 )
(1 − M 2 )(1 + kM 2 ) 2 (11.223)