Convergent nozzle: choking and mass flow

Critical pressure ratio for air:

$$\frac{p^*}{p_0} = \left(\frac{2}{\gamma + 1}\right)^{\gamma/(\gamma-1)} = 0.5283 \;\Rightarrow\; p^* = 0.5283 \times 500 = 264.1\ \text{kPa}$$

(a) Back pressure 350 kPa

Since $350 > 264.1$ kPa, the nozzle is not choked and the exit pressure equals the back pressure, $p_e = 350$ kPa.

$$T_e = T_0\left(\frac{p_e}{p_0}\right)^{0.2857} = 500 \times 0.7^{0.2857} = 451.6\ \text{K}$$ $$C_e = \sqrt{2c_p(T_0 - T_e)} = \sqrt{2 \times 1005 \times 48.4} = 312.0\ \text{m/s}$$ $$\rho_e = \frac{p_e}{RT_e} = \frac{350}{0.287 \times 451.6} = 2.701\ \text{kg/m}^3$$ $$\dot m = \rho_e C_e A_e = 2.701 \times 312.0 \times 0.0020 = 1.685\ \text{kg/s}$$

Exit Mach number $= 312.0/\sqrt{1.4 \times 287 \times 451.6} = 0.73$.

(b) Back pressure 200 kPa

Since $200 < 264.1$ kPa, the nozzle is choked. The exit pressure is $p^* = 264.1$ kPa (the remaining drop to 200 kPa takes place outside the nozzle) and $Ma_e = 1$.

$$T^* = \frac{2T_0}{\gamma + 1} = \frac{2 \times 500}{2.4} = 416.7\ \text{K}$$ $$C^* = \sqrt{\gamma R T^*} = \sqrt{1.4 \times 287 \times 416.7} = 409.2\ \text{m/s}$$ $$\rho^* = \frac{p^*}{RT^*} = \frac{264.1}{0.287 \times 416.7} = 2.209\ \text{kg/m}^3$$ $$\dot m_{max} = \rho^* C^* A = 2.209 \times 409.2 \times 0.0020 = 1.808\ \text{kg/s}$$

Lowering the back pressure further does not change the mass flow; it can only be raised by increasing $p_0$ or $A$.

Answer: (a) $C_e$ = 312.0 m/s, $\dot m$ = 1.685 kg/s; (b) choked, $p_e$ = 264.1 kPa, $C_e$ = 409.2 m/s, $\dot m$ = 1.808 kg/s.