Question 1 of 6: Convergent–Divergent Nozzle with an Exit Shock
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — May 2018, 16-Mec-A6 Advanced Fluid Mechanics. Open book, 3 hours. Six questions; any five (100 marks) constitute a complete paper, each worth 20 marks. All six are solved here.
Given. Air from a large reservoir through a C–D nozzle; the centre-line pressure trace shows a jump at the exit plane.
Given data
Ratio of specific heats
$\gamma = 1.4$
Gas constant
$R = 287\ \text{J kg}^{-1}\text{K}^{-1}$
Reservoir (stagnation) temperature
$T_0 = 300\ \text{K}$
Exit area
$A_E = 10\ \text{cm}^2$
Throat area
$A_T = 6.45\ \text{cm}^2$
Back pressure
$P_b = 100\ \text{kPa (abs)}$
Find. (a) $P_0$; (b) speed and temperature just downstream of the exit; (c) mass flow rate; (d) lowest $P_b$ for fully subsonic flow, and the corresponding mass flow.
Figure 1. Overexpanded C–D nozzle: the flow is isentropic and supersonic to the exit, where a normal shock raises the pressure discontinuously to the back pressure $P_b$.
Approach. The exit pressure jumps up to $P_b$, so a normal shock stands at the exit plane: the flow is choked, isentropic and supersonic from the (sonic) throat to the exit, then a normal shock returns it to $P_b$. Work exit Mach from the area ratio, chain the isentropic and normal-shock relations back to $P_0$, then get post-shock state, choked mass flow, and the subsonic first-critical back pressure.
Exit Mach from the area ratio (throat sonic). Because the flow is choked, $A_T = A^{*}$, and the supersonic solution of the area–Mach relation
$$\frac{A_E}{A^{*}}=\frac{1}{M_E}\left[\frac{2}{\gamma+1}\Big(1+\tfrac{\gamma-1}{2}M_E^{2}\Big)\right]^{\frac{\gamma+1}{2(\gamma-1)}}=\frac{10}{6.45}=1.550$$
gives $\boxed{M_E = 1.896}$ (supersonic branch).
Reservoir total pressure. Just upstream of the exit shock the flow is isentropic, so $P_E = P_0\,(1+\tfrac{\gamma-1}{2}M_E^{2})^{-\gamma/(\gamma-1)} = 0.1501\,P_0$. The normal shock raises the static pressure to $P_b$:
$$\frac{P_b}{P_E}=1+\frac{2\gamma}{\gamma+1}\big(M_E^{2}-1\big)=4.028 .$$
Combining, $P_0 = P_b/(0.1501\times 4.028)$, so
$$\boxed{P_0 = 165.4\ \text{kPa (abs)}} \qquad (P_E = 24.8\ \text{kPa}).$$
State directly downstream of the exit (behind the shock). The post-shock Mach number is
$$M_2=\sqrt{\frac{1+\tfrac{\gamma-1}{2}M_E^{2}}{\gamma M_E^{2}-\tfrac{\gamma-1}{2}}}=0.596 .$$
Since the flow is adiabatic, $T_0$ is conserved, so $T_2 = T_0/(1+\tfrac{\gamma-1}{2}M_2^{2})$ and $V_2 = M_2\sqrt{\gamma R T_2}$:
$$\boxed{T_2 = 280\ \text{K}\;(6.9\,{}^\circ\text{C}),\qquad V_2 = 200\ \text{m/s}.}$$
Mass flow rate (choked). With the throat sonic, the flow is the choked maximum
$$\dot m = A_T\,P_0\sqrt{\frac{\gamma}{R T_0}}\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}
=(6.45\times10^{-4})(165.4\times10^{3})\sqrt{\tfrac{1.4}{287\cdot300}}\,(0.5283) ,$$
$$\boxed{\dot m = 0.249\ \text{kg/s}.}$$
Lowest back pressure for subsonic flow throughout. The nozzle is fully subsonic only down to the first critical point, where the throat is just sonic and the diverging section follows the subsonic branch of the same area ratio, $M_{E,\text{sub}} = 0.413$. The matching back pressure is the isentropic subsonic exit pressure
$$P_{b,\min}=P_0\big(1+\tfrac{\gamma-1}{2}M_{E,\text{sub}}^{2}\big)^{-\gamma/(\gamma-1)}=\boxed{147\ \text{kPa}}.$$
At this limiting condition the throat is sonic, so the mass flow equals the choked value, $\boxed{\dot m = 0.249\ \text{kg/s}}$ — the same as part (c). For any $P_b \gt 147$ kPa the throat is subsonic and $\dot m$ falls below this.
