Question 8.18: Air enters a converging-diverging nozzle of throat area 10 c......

Given: A_t=10 \times 10^{-4}  m ^2 ; p_{0 i}=p_{01}=2.9 \times 10^6  Pa ; T_{0 i}=T_{01}=323  K; p_1 = 0.5 × 10^6   Pa

Refer Figure 8.37. Since the pressure upstream of the shock wave is given, the upstream Mach number can be found out using isentropic tables.

\frac{p_1}{P_{01}}=\frac{0.5 \times 10^6}{2.9 \times 10^6}=0.1724

Corresponding to this, from isentropic tables (γ = 1.4);

M_1=1.81

The Mach number and other parameters downstream of the normal shock can be obtained from normal shock tables. From shock tables (γ = 1.4), corresponding to M_1 = 1.81,

\begin{aligned}M_2 &=0.614 \\\frac{p_2}{p_1}&=3.6555 \\P_2 &=p_1 \times 3.6555=0.5 \times 3.6555 \\&=1.828   MPa\end{aligned}

Since the gas flowing is air (γ = 1.4), the mass flow rate can be calculated using the Fleigners equation,

\begin{aligned}\frac{\dot{m}}{A^*}\cdot \frac{\sqrt{T_0}}{P_0}&=0.0404 \\\dot{m}&=\frac{0.0404 \times 2.9 \times 10^6 \times 10 \times 10^{-4}}{\sqrt{323}}\\&=6.52  kg / s .\end{aligned}