Question 2 of 5: Blowdown of a pressurised air canister through a convergent–divergent valve

Question 2: Blowdown of a pressurised air canister through a convergent–divergent valve (25 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. Mass flow, exit temperature and exit speed at 10.0 MPa and at 5.0 MPa; the canister pressure that places a normal shock exactly at the exit plane with the mass flow and post-shock state there; and the canister pressure below which the throat is no longer choked.

Approach. The area ratio alone fixes the exit Mach number whenever the nozzle runs full and supersonic, so the whole question reduces to one isentropic area-Mach solution, the aid-sheet mass-flow formula evaluated at the choked throat, and the normal-shock relations applied at the exit plane. The canister is treated as quasi-steady: the flow through the valve responds instantly to the instantaneous stagnation state inside.

1. Fix the design exit Mach number from the area ratio. The aid-sheet mass-flow relation, written once at the throat (M = 1) and once at the exit and divided, gives the standard area–Mach function $$\frac{A}{A^{*}}=\frac{1}{M}\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M^{2}\right) \right]^{\frac{\gamma+1}{2(\gamma-1)}}$$ With $ A_E/A_T=3.5/1.0=3.5 $ and γ = 1.4 this has two roots. Solving numerically, $$\boxed{\;M_{E,\text{sup}}=2.800,\qquad M_{E,\text{sub}}=0.1682\;}$$ The supersonic root governs parts (a) to (c); the subsonic root is the one that matters in part (d). Note how cleanly the paper is designed: the area ratio was chosen to land on M = 2.80 to three figures.
2. Record the isentropic exit state that goes with M = 2.800. From the adiabatic and isentropic relations, $$\frac{T_o}{T_E}=1+\frac{\gamma-1}{2}M_E^{2}=1+0.2(2.800)^{2}=2.568, \qquad \frac{P_o}{P_E}=(2.568)^{3.5}=27.14$$ so, with $ T_o = 300 $ K held constant by the canister, $$\begin{aligned} T_E&=\frac{300}{2.568}=116.8\ \text{K}\\ a_E&=\sqrt{\gamma R T_E}=\sqrt{1.4(287)(116.8)}=216.7\ \text{m/s}\\ V_E&=M_E\,a_E=2.800(216.7)=606.6\ \text{m/s} \end{aligned}$$ Because $ T_o $ never changes and $ M_E $ is set by geometry, the exit temperature and speed are the same for every choked, fully supersonic condition — only the mass flow falls as the canister empties.
3. Part (a) — check the nozzle really runs full at 10.0 MPa, then size the flow. The design exit pressure is $ P_E = P_o/27.14 = 10.0\times10^{6}/27.14 = 368\ \text{kPa} $, which exceeds the 100 kPa back pressure. The nozzle is therefore under-expanded: the adjustment to ambient happens outside in a plume of expansion waves, and the internal solution is the fully supersonic isentropic one. Evaluating the aid-sheet mass flow at the choked throat, $$\begin{aligned} \dot m&=\sqrt{\frac{\gamma}{R}}\,\frac{P_o}{\sqrt{T_o}}\,M\left(1+\frac{\gamma-1}{2}M^{2} \right)^{-\frac{\gamma+1}{2(\gamma-1)}}A\bigg|_{M=1,\ A=A_T}\\ &=\sqrt{\frac{1.4}{287}}\,\frac{10.0\times10^{6}}{\sqrt{300}}\,(1.2)^{-3}\,(1.0\times10^{-6}) \end{aligned}$$ $$\boxed{\;\dot m=0.02334\ \text{kg/s}=23.3\ \text{g/s},\qquad T_E=116.8\ \text{K}, \qquad V_E=606.6\ \text{m/s}\;}$$
4. Part (b) — halve the canister pressure and use the linearity of choked flow. At $ P_o = 5.0 $ MPa the design exit pressure is still $ 5.0\times10^{6}/27.14=184\ \text{kPa} > 100\ \text{kPa} $, so the nozzle is still under-expanded and still runs full. Choked mass flow is directly proportional to $ P_o $ at fixed $ T_o $, so it simply halves while the exit state is untouched: $$\boxed{\;\dot m=0.01167\ \text{kg/s}=11.7\ \text{g/s},\qquad T_E=116.8\ \text{K}, \qquad V_E=606.6\ \text{m/s}\;}$$ The physical picture is worth stating: for the whole first decade of the blowdown the jet looks identical — same Mach number, same temperature, same speed — and only its density, and hence its thrust and mass flow, decay.
5. Part (c) — place a normal shock exactly on the exit plane. A shock standing at the exit is fed by the undisturbed supersonic flow, $ M_1 = M_E = 2.800 $, and must discharge directly into the ambient, so the post-shock pressure equals the back pressure. The aid-sheet shock relation gives the static pressure jump $$\frac{P_2}{P_1}=\frac{2\gamma M_1^{2}-(\gamma-1)}{\gamma+1} =\frac{2(1.4)(2.800)^{2}-0.4}{2.4}=\frac{21.55}{2.4}=8.980$$ Requiring $ P_2 = P_b = 100 $ kPa fixes the pre-shock static pressure at $ P_1 = 100/8.980 = 11.14 $ kPa, and running that back up the isentropic nozzle to the canister, $$\boxed{\;P_o=27.14\,P_1=27.14(11.14)=302\ \text{kPa}\;}$$
6. Scale the mass flow to that canister pressure. The throat is still choked at 302 kPa (the shock is downstream and cannot signal upstream), so mass flow again scales linearly with $ P_o $: $$\dot m=0.02334\times\frac{302.2\times10^{3}}{10.0\times10^{6}} =7.05\times10^{-4}\ \text{kg/s}=0.705\ \text{g/s}$$ a factor of 33 below the opening flow — the canister is nearly spent by the time the shock reaches the lip.
7. Get the state immediately after the exit shock. The shock is adiabatic, so $ T_o = 300 $ K carries across it unchanged; only the Mach number and the entropy change. From the aid sheet, $$M_2^{2}=\frac{M_1^{2}+\dfrac{2}{\gamma-1}}{\dfrac{2\gamma}{\gamma-1}M_1^{2}-1} =\frac{(2.800)^{2}+5}{7(2.800)^{2}-1}=\frac{12.84}{53.88}=0.2383 \quad\Longrightarrow\quad M_2=0.4882$$ $$\begin{aligned} T_2&=\frac{T_o}{1+0.2M_2^{2}}=\frac{300}{1.0477}=286.4\ \text{K}\\ V_2&=M_2\sqrt{\gamma R T_2}=0.4882\sqrt{1.4(287)(286.4)}=165.6\ \text{m/s} \end{aligned}$$ $$\boxed{\;M_2=0.488,\qquad T_2=286.4\ \text{K},\qquad V_2=165.6\ \text{m/s}\;}$$ The shock converts nearly all of the 607 m/s kinetic energy back into thermal energy: the jet leaves at little more than half the speed of the earlier plume, and 170 K hotter.
8. Part (d) — find the pressure at which the throat unchokes. The throat stops being sonic once the nozzle can pass the required flow entirely subsonically, which happens when the exit runs on the subsonic branch of the same area ratio, $ M_{E,\text{sub}} = 0.1682 $, with the exit pressure equal to the ambient. That requires $$\frac{P_o}{P_b}=\left(1+0.2M_{E,\text{sub}}^{2}\right)^{3.5}=(1.005659)^{3.5}=1.0199$$ $$\boxed{\;P_o=1.0199\,(100\ \text{kPa})=102.0\ \text{kPa}\;}$$ Above 102.0 kPa the throat remains sonic; below it the entire valve is subsonic, the mass flow starts to depend on the back pressure as well as on $ P_o $, and the blowdown decays asymptotically instead of linearly. The choked window therefore covers the canister's life from 10 MPa down to about 1.02 bar — essentially the whole discharge.