A uniform flow of air has a Mach number of 3 3 The bottom of

A uniform flow of air has a Mach number of 3. 3. The bottom of the duct is bent upward at a 25degree angle. At the point where the shock intersects the upper wall, the boundary is bent 5degree upward as shown in Figure P7.17. Assume that the flow is supersonic through - out the system. Compute M_3, P_3/P_1, T_3/T_1, and beta.

Solution

solution:

1)here from point 1 to 2 deflection in flow is z=25 degree,hence shock wave angle is given as S1

where corelation for shock angle is

cotz=tanS((1+k)M^2/(2(M^2sin^2S-1))-1)

x^3+a1x^2+a2x+a3=0

a1=-1.433

a2=.0716

a3=-.00692

hence root are

sin^2S=1.3849 or .024

S=8.912 degree

Mn1=M1sinS1=.5111

Mn2^2=2+(k-1)Mn1^2/(2*k*Mn1^2-(k-1))

k=1.4

Mn2=2.52M2=Mn2/sin(Z-S1)=9.096

where

P2/p1=1+(2*k/k+1)(Mn1^2-1)=96.36

T2/T1=(P2/P1)(2+(k-1)Mn1^2/(k+1)Mn1^2)=323.58

4)here when flow is moving along 25 degree the it is again deflected bywall at 5 degree which resultant of 25-5=20

Z1=20

hence repeating as initial flow we get here

equatrionas

x^3-1.1879x^2+.1416x-.0001289=0

we get

x=sin^2S2

S2=21.69 degree

hence mn2=3.3618

Mn3=.4568

M3=15.49

P3/P2=13.01

T3/T2=3.1276

where resultant S2 with horizontal is

=21.69-25=-3.31 degree;above horizontal

P3/P1=P3/P2*P2/P1=1253.64

T3/T1=T3/T2*T2/T1=1012.02

final flow mach number is M3=15.49