Question 1 of 6: Convergent–Divergent Nozzle — Normal Shock at the Exit Plane
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — December 2019, 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 as a study resource.
Given. Air discharges from a large reservoir (interior velocity negligible, so reservoir conditions are stagnation) through a fixed C–D nozzle; a pressure jump at the exit signals a normal shock standing exactly at the exit plane.
$A_E = 10\ \text{cm}^2$; area ratio $A_E/A_T = 1.439$
Back pressure
$P_b = 100\ \text{kPa}$ (absolute)
Find. The reservoir total pressure $P_0$; the temperature and speed just downstream of the exit (behind the shock); the mass flow rate; and the lowest $P_b$ giving entirely subsonic flow, with its mass flow rate.
Figure 1. Supersonic (isentropic) expansion through the divergent section drops $P/P_0$ monotonically; a normal shock standing at the exit plane raises the static pressure discontinuously up to $P_b$.
Approach. A jump up in static pressure at the exit is a normal shock at the exit plane: the diverging section runs the isentropic supersonic branch (throat choked, $A_T=A^{*}$), so $A_E/A^{*}$ fixes the pre-shock exit Mach number. The shock relations then tie the exit static pressure to $P_b$, which back-solves $P_0$; the post-shock state gives the downstream $T$ and $V$; the choked relation gives $\dot m$; and the first-critical condition gives the lowest fully-subsonic $P_b$.
Exit Mach number from the area ratio. With the throat sonic ($A_T=A^{*}$), the area–Mach relation
$$\frac{A_E}{A^{*}}=\frac{1}{M}\left[\frac{2}{\gamma+1}\Big(1+\tfrac{\gamma-1}{2}M^{2}\Big)\right]^{\frac{\gamma+1}{2(\gamma-1)}}=1.439$$
has a supersonic root $\boxed{M_E=1.800}$ (and a subsonic root $M_{sub}=0.454$ used in part d).
(a) Reservoir total pressure. The static pressure just upstream of the shock is the isentropic supersonic value $P_E=P_0\,(1+\tfrac{\gamma-1}{2}M_E^{2})^{-\gamma/(\gamma-1)}=0.1740\,P_0$. The normal shock at the exit raises it to $P_b$ through $P_2/P_1=1+\tfrac{2\gamma}{\gamma+1}(M_E^{2}-1)=3.613$:
$$P_b=P_E\,(P_2/P_1)=0.1740\times3.613\,P_0=0.6287\,P_0 .$$
Solving with $P_b=100\ \text{kPa}$,
$$\boxed{P_0=\frac{100\ \text{kPa}}{0.6287}=159\ \text{kPa}.}$$
(b) State directly downstream of the exit. Behind the shock the flow is subsonic with $M_2=\sqrt{\dfrac{1+\tfrac{\gamma-1}{2}M_E^{2}}{\gamma M_E^{2}-\tfrac{\gamma-1}{2}}}=0.617$. The stagnation temperature is conserved across the shock, so $T_2=T_0/(1+\tfrac{\gamma-1}{2}M_2^{2})$ and $V_2=M_2\sqrt{\gamma R T_2}$:
$$\boxed{T_2=279\ \text{K},\qquad V_2=206\ \text{m/s}.}$$
(c) Mass flow rate. The throat is sonic, so the flow is choked and
$$\dot m = A_T\,P_0\sqrt{\frac{\gamma}{R T_0}}\Big(\frac{2}{\gamma+1}\Big)^{\frac{\gamma+1}{2(\gamma-1)}} =(6.95\times10^{-4})(1.59\times10^{5})(4.03\times10^{-3})(0.5283).$$
$$\boxed{\dot m = 0.258\ \text{kg/s}.}$$
(d) Lowest back pressure for subsonic flow throughout. As $P_b$ rises the shock moves upstream; the flow is subsonic everywhere only once the throat is just sonic with the diverging section on the subsonic (decelerating) branch — the first-critical condition. Then the exit static pressure is $P_b=P_0\,(1+\tfrac{\gamma-1}{2}M_{sub}^{2})^{-\gamma/(\gamma-1)}=0.868\,P_0$:
$$\boxed{P_{b,\min}=0.868\times159\ \text{kPa}=138\ \text{kPa}.}$$
At first critical the throat is still exactly sonic, so the mass flow is unchanged from part (c): $\dot m = 0.258\ \text{kg/s}$. (For any $P_b$ above this the throat is subsonic and $\dot m$ is smaller.)
