Answer:
The back pressure for a normal shock to appear at the exit to the duct is 375.959 kPa
Step-by-step explanation:
Here we have
A₁/A* = 3 =
Which gives
Mₐ₁ = 2.637
P₀₁ =
= 1 MPa
∴ P₁ = 47298.69 Pa
P₂/P₁ shock =
= 7.949
∴ P₂ = 47298.69 Pa× 7.949 = 375959.457 Pa
Therefore, for a normal shock to appear at the exit to the duct we have the back pressure
= 375959.457 Pa = 375.959 kPa