Basics of Fluid Mechanics, 2014a

11.7. FANNO FLOW 449
M x M y
T y
T x
ρ y
ρ x
P y
P x
P 0y
P 0x
3.0000 0.47519 2.6790 3.8571 10.3333 0.32834
After the shock wave the flow is subsonic with “M 1 ”=0.47519. (Fanno flow
Table 11.6)
M
4fL
D
P
P ∗ P 0
P 0
∗
ρ
ρ ∗
U
U ∗
T
T ∗
0.47519 1.2919 2.2549 1.3904 1.9640 0.50917 1.1481
The stagnation values for M =0.47519 are
M
T
T 0
ρ
ρ 0
A
A ⋆
P
P 0
A×P
A ∗ ×P 0
F
F ∗
0.47519 0.95679 0.89545 1.3904 0.85676 1.1912 0.65326
The ratio of exit pressure to the chamber total pressure is
P 2
P 0
=
1
1
{( }} )( { )( )( ) {( }} ){
P2 P
∗ P1 P0y P0x
P ∗ P 1 P 0y P 0x P 0
= 1× 1 × 0.8568 × 0.32834 × 1
2.2549
= 0.12476
The actual pressure ratio 1/29.65 = 0.0338 is smaller than the case in which
shock occurs at the entrance. Thus, the shock is somewhere downstream. One
possible way to find the exit temperature, T 2 is by finding the location of the
shock. To find the location of the shock ratio of the pressure ratio, P 2
P 1
is needed.
With the location of shock, “claiming” upstream from the exit through shock
to the entrance. For example, calculate the parameters for shock location with
known 4fL
D
in the “y” side. Then either by utilizing shock table or the program,
to obtain the upstream Mach number.
The procedure for the calculations:
1)
Calculate the entrance Mach number assuming the shock occurs at the exit:
a) set M ′ 2 =1assume the flow in the entire tube is supersonic:
b) calculated M ′ 1
Note this Mach number is the high Value.