Question 3 of 7: Question 3 (Part A, Question A3): Converging Nozzle Discharging Oxygen — Mass Flow at Three Back Pressures

Question 3 (Part A, Question A3): Converging Nozzle Discharging Oxygen — Mass Flow at Three Back Pressures (16 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The mass flow rate delivered at each of the three back pressures.

Approach. A converging nozzle can at most reach sonic conditions at its exit, so the whole question turns on one comparison: is the back pressure above or below the critical (choking) pressure? Compute that threshold first, then evaluate the aid-sheet mass-flow relation with the appropriate exit Mach number for each case.

1. Establish the critical pressure that separates the two regimes. Setting $M_e=1$ in the isentropic pressure ratio gives the choking threshold $$\frac{P^{*}}{P_0}=\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma}{\gamma-1}} =\left(\frac{1}{1.2}\right)^{3.5}=0.52828,$$ so $$\boxed{\ P^{*}=0.52828\times500=264.1\ \text{kPa}\ }$$ Any back pressure below 264.1 kPa chokes the nozzle; any back pressure above it leaves the exit subsonic with $P_e=P_b$.
2. Evaluate the choked (maximum) mass flow rate. The aid-sheet relation $$\dot m=\sqrt{\frac{\gamma}{R}}\;\frac{P_0}{\sqrt{T_0}}\,M \left(1+\frac{\gamma-1}{2}M^{2}\right)^{-\frac{\gamma+1}{2(\gamma-1)}}A$$ at $M=1$ collapses to $$\dot m_{\max}=\sqrt{\frac{1.4}{259.8}}\times\frac{500\,000}{\sqrt{1200}}\times (1.2)^{-3}\times1.0\times10^{-4} =0.073408\times14\,434\times0.57870\times10^{-4},$$ giving $$\boxed{\ \dot m_{\max}=0.0613\ \text{kg/s}=61.3\ \text{g/s}\ }$$ The corresponding throat state is $T^{*}=T_0/1.2=1000\ \text{K}$ and $V^{*}=\sqrt{\gamma RT^{*}}=603\ \text{m/s}$; multiplying by $\rho^{*}=P^{*}/RT^{*}=1.017\ \text{kg/m}^{3}$ and the area reproduces the same 0.0613 kg/s, which confirms the exponent arithmetic.
3. Case P_b = 0 (discharge to vacuum). Zero back pressure is far below the 264.1 kPa threshold, so the nozzle is choked. The exit plane holds $P_e=P^{*}=264.1\ \text{kPa}$, not zero — the remaining expansion to vacuum happens outside the nozzle in an under-expanded plume of oblique waves, which the nozzle cannot feel because no downstream signal can travel upstream through a sonic throat. Therefore $$\dot m=\dot m_{\max}=0.0613\ \text{kg/s}.$$
4. Case P_b = 250 kPa. Still below 264.1 kPa, so the nozzle is again choked and the answer is unchanged: $$\dot m=\dot m_{\max}=0.0613\ \text{kg/s}.$$ The nozzle is only 5 per cent under-expanded here, but "choked" is a yes-or-no condition; being marginally below the threshold delivers exactly the same flow as discharging to vacuum.
5. Case P_b = 400 kPa — find the subsonic exit Mach number. Now $P_b>P^{*}$, so the exit is subsonic and the exit static pressure equals the back pressure, $P_e=400\ \text{kPa}$. Inverting the isentropic relation, $$\begin{aligned} M_e&=\sqrt{\frac{2}{\gamma-1}\left[\left(\frac{P_0}{P_e}\right)^{\frac{\gamma-1}{\gamma}}-1\right]}\\ &=\sqrt{5\left[(1.25)^{0.28571}-1\right]}=\sqrt{5\times0.065831}=0.5737 \end{aligned}$$
6. Evaluate the mass flow for the subsonic case. Substituting $M_e=0.5737$ into the same aid-sheet relation, $$\dot m=0.073408\times14\,434\times0.5737\times(1.06583)^{-3}\times1.0\times10^{-4},$$ so $$\boxed{\ \dot m=0.0502\ \text{kg/s}=50.2\ \text{g/s}\ }$$ which is 82 per cent of the choked value. The exit conditions are $T_e=T_0/1.06583=1126\ \text{K}$ and $V_e=M_e\sqrt{\gamma RT_e}=367\ \text{m/s}$; a direct check gives $\rho_e V_e A_e=(400\,000/(259.8\times1126))\times367\times10^{-4}=0.0502\ \text{kg/s}$, matching.