Convergent–Divergent Nozzle Between Two Reservoirs: 22-Mec-A6 Fluid Machinery, December 2017, Q1

Question 1: Convergent–Divergent Nozzle Between Two Reservoirs (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. Stagnation air state in (a): $P_0=270$ kPa, $T_0=373.15\ \text{K}$. Nozzle area ratio $A_E/A_T=36/9=4$. Manometer between throat and reservoir (b): $h=18\ \text{cm}$ of mercury, $\rho_{\text{Hg}}=13{,}550\ \text{kg/m}^3$. Air $\gamma=1.4$, $R=287\ \text{J/(kg K)}$.

Find. The back pressure $P_b$, whether/where a normal shock stands, and the manometer reading at design.

Figure 1 (reconstructed schematic): choked C–D nozzle; a normal shock stands in the divergent section (M > 1 upstream, M < 1 downstream of it). The throat leg of the manometer is depressed, so $P_{\text{throat}}\gt P_b$.

Approach. Assume the nozzle is choked (throat sonic), fix the throat static pressure from isentropic relations, read $P_b$ off the manometer, then compare $P_b$ against the critical back pressures (first critical, shock-at-exit, design) to place the shock.

Exit Mach numbers for the fixed area ratio. With the throat choked, $A^\ast=A_T$ and $A_E/A^\ast=4$. Solving the area–Mach relation $$\frac{A}{A^\ast}=\frac{1}{M}\left[\frac{1+\tfrac{\gamma-1}{2}M^2}{\tfrac{\gamma+1}{2}}\right]^{\frac{\gamma+1}{2(\gamma-1)}}=4$$ gives a supersonic branch $M_{e,\text{sup}}=2.94$ and a subsonic branch $M_{e,\text{sub}}=0.147$.
Throat (sonic) static pressure. A choked throat is at $M=1$, so $$P^\ast=\frac{P_0}{\left(1+\tfrac{\gamma-1}{2}\right)^{\gamma/(\gamma-1)}}=\frac{270}{1.893}=142.6\ \text{kPa}.$$
Read the back pressure from the manometer. The mercury column difference is $$\Delta P=\rho_{\text{Hg}}\,g\,h=13{,}550\times 9.81\times 0.18=23.9\ \text{kPa}.$$ The figure shows the throat leg depressed, i.e. $P_{\text{throat}}\gt P_b$, so $$\boxed{P_b=P^\ast-\Delta P=142.6-23.9=118.7\ \text{kPa}.}$$
Critical back pressures (part b). The subsonic branch fixes the first critical (highest back pressure that still chokes): $P_{b,1}=P_0\left(P/P_0\right)_{M_{e,\text{sub}}}=266\ \text{kPa}$. Full supersonic expansion gives the design exit pressure $P_{e,\text{des}}=P_0\left(P/P_0\right)_{2.94}=8.04\ \text{kPa}$. A normal shock exactly at the exit plane raises this to $$P_{b,\text{exit}}=P_{e,\text{des}}\left[1+\frac{2\gamma}{\gamma+1}\left(M_{e,\text{sup}}^2-1\right)\right]=8.04\times 9.92=79.8\ \text{kPa}.$$ Since $79.8\ \text{kPa}\lt P_b=118.7\ \text{kPa}\lt 266\ \text{kPa}$, the flow is over-expanded with a shock inside the divergent section: $\boxed{\text{yes, a normal shock stands in the nozzle.}}$
Locate the shock (part c). Because $P_b=118.7$ kPa is well above the shock-at-exit value (79.8 kPa), the shock sits upstream of the exit plane — a higher back pressure pushes the shock toward the throat. Matching the post-shock isentropic recompression to $P_b$ (find the pre-shock Mach $M_x$ whose downstream subsonic branch delivers exit static $=P_b$) gives $$M_x=2.57,\qquad \boxed{\frac{A_s}{A_T}=2.80}$$ i.e. about 60% of the way from throat ($A/A_T=1$) to exit ($A/A_T=4$). The shock therefore stands well inside the divergent section, not at the exit plane.
Manometer reading at design (part d). At the supersonic design condition the flow expands isentropically to $M_{e,\text{sup}}=2.94$ with $P_b=P_{e,\text{des}}=8.04$ kPa, while the throat remains sonic at 142.6 kPa. The mercury difference is $$\Delta P_{\text{des}}=142.6-8.04=134.6\ \text{kPa}\ \Rightarrow\ \boxed{h=\frac{\Delta P_{\text{des}}}{\rho_{\text{Hg}}\,g}=1.01\ \text{m}=101\ \text{cm}.}$$

Question 1 — results
Quantity	Result
Downstream reservoir pressure $P_b$	118.7 kPa
Normal shock present?	Yes — inside the divergent section
Shock location $A_s/A_T$ ($M_x$)	2.80 ($M_x=2.57$), upstream of exit
Manometer reading at design	$h\approx 101$ cm

Topic. Quasi-one-dimensional compressible flow in a converging–diverging nozzle and normal-shock placement.

Key relations. Area–Mach relation $A/A^\ast(M)$; isentropic $P_0/P=(1+\tfrac{\gamma-1}{2}M^2)^{\gamma/(\gamma-1)}$; normal-shock static jump $p_2/p_1=1+\tfrac{2\gamma}{\gamma+1}(M_1^2-1)$.

Why this works. Once the throat is choked the mass flow and the throat static pressure are frozen, so the entire internal solution is set by geometry until the exit; the back pressure only decides where the flow recompresses. Comparing $P_b$ with the three critical back pressures (first critical, shock-at-exit, design) sorts the flow into subsonic / shock-in-nozzle / over-expanded / design / under-expanded regimes. The manometer simply converts a pressure difference into a mercury height.

Common pitfall. Treating the throat pressure as unknown, or using stagnation instead of static pressure at the throat. Another is assuming the shock sits at the exit; the manometer here gives a back pressure above the shock-at-exit value, so it must be upstream.

Source. Anderson, Modern Compressible Flow, Ch. 5 (nozzle flows & shocks); White, Fluid Mechanics, §9.4–9.7.

Question 1 of 6: Convergent–Divergent Nozzle Between Two Reservoirs

Question 1: Convergent–Divergent Nozzle Between Two Reservoirs (20 marks)