Convergent–divergent nozzle with a shock in the exit plane: 22-Mec-B6 Advanced Fluid Mechanics, December 2016, Q1

Question 1: Convergent–divergent nozzle with a shock in the exit plane (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.

Quantity	Symbol	Value
Ratio of specific heats	γ	1.4
Gas constant of air	R	287 J kg⁻¹ K⁻¹
Reservoir stagnation temperature	T₀	300 K
Throat area	A_T	6.45 cm² = 6.45 × 10⁻⁴ m²
Exit area	A_E	10 cm² = 10 × 10⁻⁴ m²
Back pressure	P_b	100 kPa absolute
Observed feature	—	pressure jump located exactly at the exit plane

Find. The reservoir total pressure, the temperature and speed of the air immediately downstream of the exit plane, the mass flow rate, and the lowest back pressure at which the channel remains subsonic throughout together with the mass flow at that condition.

Figure 1 — The nozzle and its centre-line pressure distribution. The measured jump sits exactly at the exit plane, which is the signature of a normal shock standing in that plane: everything upstream of it is isentropic from the reservoir, and the 100 kPa back pressure is the pressure behind the shock.

Approach. A discontinuity located exactly at the exit means a normal shock stands in the exit plane, so the whole nozzle is isentropic with a choked throat, the area ratio alone fixes the supersonic Mach number just ahead of the shock, and the normal-shock pressure ratio then converts the measured back pressure into the reservoir total pressure.

Establish that the throat is choked and identify the sonic reference area. Because the pressure falls continuously from the reservoir to the exit plane and only jumps there, the divergent section carries supersonic flow, which is possible only if the throat has reached M = 1. The sonic reference area is therefore the throat itself, $$\frac{A_E}{A^{*}}=\frac{A_E}{A_T}=\frac{10}{6.45}=1.5504 .$$
Invert the area–Mach relation on the supersonic branch. The quasi-one-dimensional area relation is $$\frac{A}{A^{*}}=\frac{1}{M}\left[\frac{2}{\gamma+1}\left(1+\frac{\gamma-1}{2}M^{2}\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}.$$ Setting the left-hand side to 1.5504 and taking the root with M > 1 gives the Mach number immediately ahead of the shock, M₁ = 1.896.
Work back through the shock to the static pressure at the nozzle exit. Across a normal shock the static pressures are related by $$\frac{p_2}{p_1}=\frac{2\gamma M_1^{2}-(\gamma-1)}{\gamma+1} =\frac{2(1.4)(1.896)^{2}-0.4}{2.4}=4.028 .$$ The measured back pressure is the pressure behind the shock, so p₂ = 100 kPa and $$p_1=\frac{100\ \text{kPa}}{4.028}=24.83\ \text{kPa}.$$
Recover the reservoir total pressure isentropically. Upstream of the shock the flow is isentropic from the reservoir, so $$P_0=p_1\left(1+\frac{\gamma-1}{2}M_1^{2}\right)^{\frac{\gamma}{\gamma-1}} =24.83\left(1+0.2(1.896)^{2}\right)^{3.5}=24.83\times 6.660,$$ $$\boxed{P_0=165.4\ \text{kPa (absolute)}}$$ which answers part (a).
Find the Mach number behind the shock. The normal-shock Mach relation $$M_2^{2}=\frac{M_1^{2}+\dfrac{2}{\gamma-1}}{\dfrac{2\gamma}{\gamma-1}M_1^{2}-1} =\frac{(1.896)^{2}+5}{7(1.896)^{2}-1}$$ gives M₂ = 0.5964, comfortably subsonic as every normal shock demands.
Convert to temperature and speed downstream of the exit. A shock is adiabatic, so the stagnation temperature is unchanged at 300 K on both sides and $$T_2=\frac{T_0}{1+\frac{\gamma-1}{2}M_2^{2}}=\frac{300}{1+0.2(0.5964)^{2}}=280.1\ \text{K},$$ $$V_2=M_2\sqrt{\gamma R T_2}=0.5964\sqrt{1.4(287)(280.1)}=0.5964(335.5),$$ $$\boxed{T_2=280.1\ \text{K},\qquad V_2=200.1\ \text{m/s}}$$ which answers part (b). For contrast the flow arriving at the shock is at T₁ = 174.5 K and V₁ = 502.1 m/s, so the shock removes about 60 per cent of the speed in a distance of a few mean free paths.
Compute the mass flow from the choked throat. With the throat sonic the mass flow depends only on the reservoir state and the throat area, $$\dot m=\frac{A_T P_0}{\sqrt{T_0}}\sqrt{\frac{\gamma}{R}} \left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}} =\frac{(6.45\times10^{-4})(165\,361)}{\sqrt{300}}(0.04042),$$ $$\boxed{\dot m=0.2489\ \text{kg/s}}$$ answering part (c). Evaluating ρ₁A_EV₁ at the exit plane returns the same 0.2489 kg/s, which is a free check that steps 2 to 4 are mutually consistent.
Locate the first-choking back pressure for part (d). The channel is subsonic everywhere only while the exit sits on the subsonic branch of the same area ratio. Solving A/A^* = 1.5504 for M < 1 gives M_E = 0.4127, so $$p_b=\frac{P_0}{\left(1+0.2M_E^{2}\right)^{3.5}}=\frac{165.4}{1.1244},$$ $$\boxed{p_{b,\min}=147.1\ \text{kPa (absolute)}}$$ Lower the receiver below this and the throat cannot stay subsonic; raise it above and the whole channel is subsonic with a throat Mach number below unity.
State the mass flow at that back pressure. At exactly 147.1 kPa the throat is on the point of reaching M = 1, so the nozzle is passing its choked maximum and the mass flow is unchanged at 0.2489 kg/s. This is the whole content of choking: every back pressure from 147.1 kPa down to hard vacuum delivers the same 0.2489 kg/s, and only above 147.1 kPa does the flow rate begin to fall.

Check: the solution assumes the throat is choked, which the answer then confirms rather than presumes — the given back pressure of 100 kPa lies well below the 147.1 kPa first-choking value computed in step 8, so a sonic throat is the only possibility. The 100 kPa back pressure also lies above the design (fully expanded) exit pressure of 24.8 kPa, which is exactly the window in which a shock can stand inside the divergent section or in its exit plane.

Quantity	Symbol	Result
Exit Mach number ahead of the shock	M₁	1.896
(a) Reservoir total pressure	P₀	165.4 kPa absolute
(b) Temperature downstream of the exit	T₂	280.1 K
(b) Speed downstream of the exit	V₂	200.1 m/s
(c) Mass flow rate	ṁ	0.2489 kg/s
(d) Lowest back pressure for subsonic flow	p_b,min	147.1 kPa absolute
(d) Mass flow at that back pressure	ṁ	0.2489 kg/s (choked maximum)

Topic. Quasi-one-dimensional nozzle flow, choking, and the normal shock as a back-pressure diagnostic.

Key relations. Area–Mach relation $A/A^{*}=f(M,\gamma)$; isentropic ratios $P_0/p=(1+\tfrac{\gamma-1}{2}M^{2})^{\gamma/(\gamma-1)}$ and $T_0/T=1+\tfrac{\gamma-1}{2}M^{2}$; normal-shock relations $p_2/p_1=[2\gamma M_1^{2}-(\gamma-1)]/(\gamma+1)$ and $M_2^{2}=(M_1^{2}+\tfrac{2}{\gamma-1})/(\tfrac{2\gamma}{\gamma-1}M_1^{2}-1)$; choked mass flow $\dot m=A^{*}P_0\sqrt{\gamma/R}\,(2/(\gamma+1))^{(\gamma+1)/2(\gamma-1)}/\sqrt{T_0}$.

Why this works. Once the throat is sonic the nozzle stops communicating with the receiver: no downstream disturbance can travel upstream past M = 1, so the reservoir sees a fixed sonic throat and delivers a fixed mass flow. Everything downstream then rearranges itself to meet the back pressure, and a normal shock is the only mechanism available for making the static pressure jump discontinuously. Where the shock sits is set entirely by how much pressure recovery the receiver demands: at the design pressure the flow leaves supersonically with no shock at all, at a slightly higher back pressure the shock appears in the exit plane, and as the back pressure rises further the shock marches upstream into the divergent section where the local Mach number, and therefore the strength of the jump, is smaller. Reading the problem backwards — from the position of the jump to the reservoir pressure — is possible precisely because that mapping is one-to-one.

Common pitfall. Two mistakes recur. The first is treating the 100 kPa back pressure as the pressure at the nozzle exit before the shock; it is the pressure behind it, and using it as p₁ inflates P₀ by a factor of four. The second is recomputing the mass flow in part (d): candidates who evaluate ρVA at the subsonic exit state get a smaller answer and miss that a just-choked throat passes exactly the same flow as a fully choked one.

Source. Anderson, Modern Compressible Flow, 3rd ed., §3.3 (isentropic nozzle flow) and §3.6 (normal shocks); White, Fluid Mechanics, 8th ed., §9.4–9.6.

Question 1 of 6: Convergent–divergent nozzle with a shock in the exit plane

Question 1: Convergent–divergent nozzle with a shock in the exit plane (20 marks)