Question 1 of 6: Convergent–Divergent Valve — Canister Blowdown
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — December 2018, 16-Mec-A6 Advanced Fluid Mechanics. Open book, 3 hours. Six questions of equal value (20 marks); any five constitute a complete paper. All six are solved here.
Given. Air blows down isentropically from a large canister (velocity inside negligible, so canister conditions are stagnation) through a fixed C–D nozzle to a fixed atmosphere.
$A_E = 3.0\ \text{mm}^2$; area ratio $A_E/A_T = 3.0$
Back pressure
$P_b = 100\ \text{kPa}$
Find. Exit flow rate, temperature and speed at $P_0 = 10$ and $5\ \text{MPa}$; the $P_0$ that puts a normal shock at the exit (with the post-exit state and flow rate); and the $P_0$ below which the throat is no longer sonic.
Figure 1. C–D valve. When choked the throat is sonic, so the divergent section accelerates the flow to the supersonic exit Mach set by the area ratio $A_E/A^{*}=3$.
Approach. For a choked C–D nozzle the throat is sonic ($A_T=A^{*}$), so the area ratio $A_E/A^{*}=3$ fixes two candidate exit Mach numbers (a subsonic and a supersonic branch). Comparing the fixed atmosphere $P_b$ with the three critical back pressures — first-critical (throat just sonic, subsonic exit), shock-at-exit, and design (isentropic supersonic exit) — classifies each operating point.
Exit Mach numbers for the area ratio. With $A_T=A^{*}$, the area–Mach relation
$$\frac{A_E}{A^{*}}=\frac{1}{M}\left[\frac{2}{\gamma+1}\Big(1+\tfrac{\gamma-1}{2}M^{2}\Big)\right]^{\frac{\gamma+1}{2(\gamma-1)}}=3.0$$
has a subsonic root $M_{sub}=0.197$ and a supersonic root $\boxed{M_{sup}=2.637}$.
The three critical back-pressure ratios. Using $P/P_0=(1+\tfrac{\gamma-1}{2}M^2)^{-\gamma/(\gamma-1)}$ on each branch, and the normal-shock rise $P_2/P_1=1+\tfrac{2\gamma}{\gamma+1}(M^2-1)=7.95$ at $M_{sup}$:
$$\Big(\tfrac{P_b}{P_0}\Big)_{\!first}=0.973,\quad \Big(\tfrac{P_b}{P_0}\Big)_{\!shock@exit}=0.0473\times7.95=0.376,\quad \Big(\tfrac{P_b}{P_0}\Big)_{\!design}=0.0473 .$$
The choked exit is design-supersonic whenever $P_b/P_0 \lt 0.0473$ (over-/under-expanded outside).
(a) $P_0=10\ \text{MPa}$. Here $P_b/P_0 = 0.01 \lt 0.0473$, so the nozzle is choked and under-expanded; the exit is the design supersonic state $M_E=2.637$. Then $T_E = T_0/(1+\tfrac{\gamma-1}{2}M_E^2)=125.5\ \text{K}$ and $V_E=M_E\sqrt{\gamma R T_E}$. The choked mass flow is
$$\dot m = A_T P_0\sqrt{\tfrac{\gamma}{RT_0}}\Big(\tfrac{2}{\gamma+1}\Big)^{\frac{\gamma+1}{2(\gamma-1)}} = (10^{-6})(10^{7})(4.03\times10^{-3})(0.5283).$$
$$\boxed{\dot m = 0.0233\ \text{kg/s},\quad T_E = 125.5\ \text{K},\quad V_E = 592\ \text{m/s}.}$$
(b) $P_0=5\ \text{MPa}$. Now $P_b/P_0=0.02$, still $\lt 0.0473$: choked and under-expanded, so the exit is the same design supersonic state — $T_E$ and $V_E$ are unchanged. Only the (choked) mass flow scales with $P_0$:
$$\boxed{\dot m = 0.0117\ \text{kg/s},\quad T_E = 125.5\ \text{K},\quad V_E = 592\ \text{m/s}.}$$
(c) Shock exactly at the exit. The flow is isentropic supersonic to the exit ($M_E=2.637$), then a normal shock returns the static pressure to $P_b$. This requires $P_b/P_0 = 0.376$, i.e.
$$\boxed{P_0 = 100/0.376 = 266\ \text{kPa}.}$$
The throat is still sonic, so $\dot m = 6.2\times10^{-4}\ \text{kg/s}$. Immediately after the exit (behind the shock) $M_2=0.501$, and since $T_0$ is conserved, $T_2=T_0/(1+\tfrac{\gamma-1}{2}M_2^2)$, $V_2=M_2\sqrt{\gamma R T_2}$:
$$\boxed{\dot m = 6.2\times10^{-4}\ \text{kg/s},\quad T_2 = 285.7\ \text{K},\quad V_2 = 170\ \text{m/s}.}$$
(d) Loss of choking. The throat stays sonic as long as $P_b$ is at or below the first-critical value $0.973\,P_0$. The valve is no longer choked once $P_b \gt 0.973\,P_0$, i.e. once
$$\boxed{P_0 \lt \frac{100\ \text{kPa}}{0.973}=103\ \text{kPa}.}$$
Below this the C–D passage acts as a subsonic venturi.
