Blowdown of a Pressurised Canister through a Convergent–Divergent Valve: 22-Mec-A6 Fluid Machinery, May 2017, Q1

Question 1: Blowdown of a Pressurised Canister through a Convergent–Divergent Valve (20 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. Air as an ideal gas discharging isentropically from a canister held at stagnation conditions through a convergent–divergent (C–D) nozzle to the atmosphere.

Given data
Quantity	Symbol	Value
Ratio of specific heats	$\gamma$	1.4
Gas constant	$R$	287 J·kg⁻¹·K⁻¹
Stagnation (canister) temperature	$T_0$	300 K (constant)
Throat / exit area	$A_T,\,A_E$	1 mm², 3.5 mm² → $A_E/A_T=3.5$
Back pressure	$P_b$	100 kPa
Initial canister pressure	$P_{0}$	10.0 MPa

Find. Mass flow rate, exit temperature and exit velocity at $P_0=$ 10 and 5 MPa; the canister pressure that places a normal shock at the exit plane (with its flow rate and post-shock temperature and speed); and the pressure below which the throat unchokes.

Figure 1: Convergent–divergent valve. Air (stagnation state in the canister) accelerates to sonic at the throat $A_T$ and supersonic in the diverging section to the exit $A_E$, discharging against back pressure $P_b$.

Approach. Because the flow is frictionless and adiabatic it is isentropic; the fixed area ratio $A_E/A_T$ fixes the exit Mach number, the choked throat fixes the mass flow, and the ratio of design exit pressure to $P_b$ decides the exit regime. A normal shock at the exit (part c) uses the standard normal-shock jump; unchoking (part d) is set by the subsonic isentropic branch.

Exit Mach number from the area ratio. The isentropic area–Mach relation $$\frac{A}{A^{*}}=\frac{1}{M}\left[\frac{1+\tfrac{\gamma-1}{2}M^{2}}{\tfrac{\gamma+1}{2}}\right]^{\frac{\gamma+1}{2(\gamma-1)}}=3.5$$ has two roots. The diverging section carries the supersonic one, $\boxed{M_E=2.80}$ (the subsonic root $M=0.168$ is used in part d).
Choked mass flow. With $P_0/P_b\gg\left(\tfrac{\gamma+1}{2}\right)^{\gamma/(\gamma-1)}=1.89$, the throat is sonic and $$\dot m=A_T\,P_0\sqrt{\frac{\gamma}{R\,T_0}}\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}.$$ The mass flow is simply proportional to $P_0$ while choked.
(a) $P_0=10$ MPa. The design exit pressure $P_E=P_0\left(1+\tfrac{\gamma-1}{2}M_E^{2}\right)^{-\gamma/(\gamma-1)}=368\ \text{kPa}$ exceeds $P_b=100$ kPa, so the nozzle is under-expanded: the flow reaches the exit plane at $M_E=2.80$ isentropically (matching to $P_b$ occurs outside in expansion fans). Hence $$T_E=\frac{T_0}{1+\tfrac{\gamma-1}{2}M_E^{2}}=116.8\ \text{K},\qquad V_E=M_E\sqrt{\gamma R T_E}=607\ \text{m/s},$$ $$\boxed{\dot m=0.0233\ \text{kg/s},\quad T_E=116.8\ \text{K},\quad V_E=607\ \text{m/s}.}$$
(b) $P_0=5$ MPa. The design exit pressure scales with $P_0$ to $184\ \text{kPa}$, still above $P_b$, so the nozzle is still under-expanded and the exit Mach number, temperature and speed are unchanged ($M_E=2.80$, $T_E=116.8$ K, $V_E=607$ m/s). Only the mass flow scales down with $P_0$: $$\boxed{\dot m=0.0117\ \text{kg/s},\quad T_E=116.8\ \text{K},\quad V_E=607\ \text{m/s}.}$$
(c) Shock exactly at the exit. Lowering $P_0$ drops the design exit pressure; a normal shock stands at the exit plane when the pressure just downstream of a shock at $M_E=2.80$ equals $P_b$. The static-pressure jump is $$\frac{P_2}{P_1}=1+\frac{2\gamma}{\gamma+1}\left(M_E^{2}-1\right)=8.98,$$ so the pre-shock exit pressure is $P_1=P_b/8.98=11.1\ \text{kPa}$, and since $P_1=P_0\,(P_E/P_0)$ with $P_E/P_0=1/27.1$, $$\boxed{P_0=302\ \text{kPa}.}$$ The throat is still choked ($P_0/P_b=3.0\gt1.89$), so $\dot m=7.05\times10^{-4}\ \text{kg/s}$. Behind the shock $M_2=0.488$, and since stagnation temperature is conserved, $$T_2=286\ \text{K},\qquad V_2=M_2\sqrt{\gamma R T_2}=166\ \text{m/s}.$$
(d) Unchoking pressure. The throat stays sonic for every back-pressure below the fully-subsonic (venturi) exit pressure. That subsonic branch has $M_E=0.168$, giving $P_E/P_0=\left(1+\tfrac{\gamma-1}{2}(0.168)^2\right)^{-\gamma/(\gamma-1)}=0.980$. The throat therefore unchokes once $P_b\ge0.980\,P_0$, i.e. $$\boxed{P_0\lt \frac{100\ \text{kPa}}{0.980}=102\ \text{kPa}.}$$

Question 1 — results
Case	$\dot m$ (kg/s)	Exit $T$ (K)	Exit $V$ (m/s)
(a) $P_0=10$ MPa (under-expanded, $M_E=2.80$)	0.0233	116.8	607
(b) $P_0=5$ MPa (under-expanded, $M_E=2.80$)	0.0117	116.8	607
(c) $P_0=302$ kPa (shock at exit)	$7.05\times10^{-4}$	286 (post-shock)	166 (post-shock)
(d) throat unchokes below	$P_0\approx102$ kPa

Check: the “exit” state in (a)/(b) is the isentropic supersonic exit-plane condition; the under-expanded jet matches to $P_b$ through expansion waves outside the nozzle, which do not change the exit-plane $M_E,\,T_E,\,V_E$. The blowdown is taken as isothermal in the canister ($T_0=300$ K held constant), as stated.

Topic: One-dimensional isentropic compressible flow in a C–D nozzle, choking, and normal shocks.

Key relations: area–Mach $A/A^{*}(M)$; isentropic $T_0/T=1+\tfrac{\gamma-1}{2}M^2$ and $P_0/P=(T_0/T)^{\gamma/(\gamma-1)}$; choked mass flux $\dot m=A_T P_0\sqrt{\gamma/RT_0}\,(2/(\gamma+1))^{(\gamma+1)/2(\gamma-1)}$; normal-shock static-pressure jump.

Why this works: for a fixed geometry the diverging-section area ratio pins a unique supersonic exit Mach number, independent of pressure level; the throat, once sonic, meters a mass flow that depends only on stagnation conditions and throat area, so $\dot m\propto P_0$ at fixed $T_0$. Whether the isentropic supersonic solution actually reaches the exit depends on how the design exit pressure compares with the back pressure — under-expanded (jet keeps expanding outside), over-expanded (shock train), or shock-in-nozzle. As $P_0$ falls the shock migrates from outside, to the exit plane (part c), into the diverging section, and finally the throat unchokes when even the subsonic branch cannot be sustained (part d).

Common pitfall: assuming the exit pressure equals $P_b$ for an under-expanded nozzle — it does not; the exit stays at the isentropic supersonic pressure and matches externally. A second trap is forgetting that stagnation temperature is preserved across a shock, which is what fixes the post-shock static temperature in (c).

Source: Anderson, Modern Compressible Flow, 3rd ed., Ch. 3–5; White, Fluid Mechanics, 8th ed., §9.4–9.7.

Question 1 of 6: Blowdown of a Pressurised Canister through a Convergent–Divergent Valve

Question 1: Blowdown of a Pressurised Canister through a Convergent–Divergent Valve (20 marks)