Question 8.17: An aircraft engine employs a subsonic inlet diffuser of area......

Given: A_e / A_2=4 ; T_{01}=570  K ; p_{01}=1 \times 10^5  N / m ^2 ; M_1=2.2

Refer Figure 8.36. The Mach number of air downstream of the shock (inlet to the diffuser can be read from the normal shock tables. From shock tables (Appendix A4), for M_1 = 2.2.

\begin{aligned}M_2 &=0.547 \\\frac{p_{02}}{P_{01}}&=0.628 \\P_{02}&=1 \times 10^5 \times 0.628=6.28 \times 10^4  N / m ^2\end{aligned}

Since the flow in the diffuser region is isentropic, the stagnation pressure remains constant.

p_{0 e}=p_{02}=6.28 \times 10^4   N / m ^2

Also, the stagnation temperature is constant across the shock and along the diffuser

T_{01}=T_{02}=T_{0 e}=570  K

The pressure and temperature of air at inlet to the diffuser can be obtained using the isentropic stagnation pressure ratio at inlet. From isentropic tables (γ = 1.4) corresponding to M = 0.547.

\begin{aligned}\frac{p_2}{p_{02}}&=\left(\frac{p}{p_0}\right)_{M-0.547}=0.818 \\P_2 &=6.28 \times 10^4 \times 0.818=5.317 \times 10^4  N / m ^2 \\\frac{T_2}{T_{02}}&=\left(\frac{T}{T_0}\right)_{M-0.547}=0.9429 \\T_2 &=570 \times 0.9429=537.5  K\end{aligned}

For isentropic flow in the diffuser,

\frac{A_e}{A_2}=\frac{A_e / A^*}{A_2 / A^*}=\frac{\left(A / A^*\right)_{M_e}}{\left(A / A^*\right)_{M-0.547}}

From isentropic tables, corresponding to M = 0.547, A/A* = 1.26. Substituting the value, we get

\frac{A_e}{A^*}=\left(\frac{A}{A^*}\right)_{M_e}=1.26 \times 4=5.04

From isentropic tables, corresponding to A/A* = 5.04, the exit Mach number Me is read as:

M_e=0.116

The other flow parameters at the exit of the diffuser can be obtained from isentropic tables, corresponding to M_e = 0.116

\begin{aligned}\frac{T_e}{T_{0 e}}&=\left(\frac{T}{T_0}\right)_{M-0.116}=0.9975 \\T_e &=570 \times 0.9975=569  K \\\frac{p_e}{p_{0 e}}&=\left(\frac{p}{p_0}\right)_{M-0.116}=0.9951 \\P_e &=6.28 \times 10^4 \times 0.9951=6.25 \times 10^4  N / m ^2\end{aligned}

Stagnation pressure loss occurs only across the shock

\begin{aligned}p_{01}-p_{0 e}&=1 \times 10^5-6.25 \times 10^4 \\&=3.75 \times 10^4  N / m ^2\end{aligned}