Respuesta :
Answer:
The back pressure for a normal shock to appear at the exit to the duct is 375.959 kPa
Explanation:
Here we have
A₁/A* = 3 = [tex]\frac{(1+0.2Ma^2_1)^3}{1.728Ma_1}}[/tex] Which gives
Mₐ₁ = 2.637
P₀₁ =[tex]P_1(1+0.2(2.637}^2)^{3.5}[/tex] = 1 MPa
∴ P₁ = 47298.69 Pa
P₂/P₁ shock = [tex]\frac{2.8(2.637)^2-0.4 }{2.4}[/tex] = 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 [tex]p_b[/tex] = 375959.457 Pa = 375.959 kPa