Basics of Fluid Mechanics, 2014a

426 CHAPTER 11. COMPRESSIBLE FLOW ONE DIMENSIONAL
Rearranging equation (11.143) is transformed into
U
U ∗ = √ kM (11.144)
Utilizing the continuity equation provides
ρU = ρ ∗ U ∗ ;=⇒ ρ ρ ∗ = √ 1
(11.145)
kM
Reusing the perfect–gas relationship
Pressure Ratio
P
P ∗ = ρ ρ ∗ = √ 1
kM
(11.146)
Utilizing the relation for stagnated isotropic pressure one can obtain
Substituting for P P ∗
P 0
P0
∗
= P P ∗ [
1+
k−1
2 M 2
1+ k−1
2k
] k
k−1
equation (11.146) and rearranging yields
(11.147)
P 0
P0
∗
= √ 1 ( 2 k
k 3 k − 1
Stagnation Pressure Ratio
) k
( k−1
1+ k − 1 ) k
M 2
2
k−1 1
M
(11.148)
And the stagnation temperature at the critical point can be expressed as
T 0
T0
∗
Stagnation Pressure Ratio
= T 1+ k − 1 M 2
2
T ∗ 1+ k − 1 = 2 k (
1+ k − 1 )
M 2
3 k − 1 2
2 k
These equations (11.144)-(11.149) are presented on in Figure (11.18).
(11.149)
11.6.3 The Entrance Limitation of Supersonic Branch
This setion deals with situations where the conditions at the tube exit have not arrived at
the critical condition. It is very useful to obtain the relationships between the entrance
and the exit conditions for this case. Denote 1 and 2 as the conditions at the inlet and
exit respectably. From equation (11.137)
4 fL
D
= 4 fL
D
∣ − 4 fL
max1
D ∣ = 1 − kM 1 2
2
− 1 − kM 2 2
2
max2 kM 1 kM 2
( ) 2 M1
+ln
M 2
(11.150)