Question 2 (Part A, Question A2): Supersonic Nozzle Feeding an Insulated Pipe — Fanno Flow with a Shock at the Exit: 22-Mec-B6 Advanced Fluid Mechanics, December 2013, Q2

Question 2 (Part A, Question A2): Supersonic Nozzle Feeding an Insulated Pipe — Fanno Flow with a Shock at the Exit (16 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
Reservoir stagnation pressure	P₀	500 kPa
Reservoir stagnation temperature	T₀	77 °C = 350.15 K
Nozzle area ratio (exit / throat)	A₂/A_T	1.688
Pipe inner diameter	D	5 mm
Pipe length	L	5.63 cm
Friction factor	f	0.015 (Darcy)
Gas	air	γ = 1.4, R = 287 J/kg·K
Duct	well insulated ⇒ adiabatic, so T₀ is constant everywhere

Find. The back pressure P_b that places a normal shock exactly at the pipe exit, and the mass flow rate the system passes.

Figure A2 — Reservoir, convergent-divergent nozzle (isentropic, A to B) and insulated constant-area pipe (adiabatic with friction, B to C), with the normal shock standing at the pipe exit plane C.

Approach. Work downstream in three isolated pieces: the area ratio fixes the supersonic Mach number leaving the nozzle, the Fanno relation for an adiabatic constant-area duct carries that Mach number to the pipe exit, and the normal-shock pressure jump then converts the exit static pressure into the back pressure. The mass flow is fixed independently at the choked throat.

Find the Mach number at the nozzle exit from the area ratio. With the throat choked, the aid-sheet mass-flow relation evaluated at $M$ and at $M=1$ gives the isentropic area relation $$\frac{A}{A^{*}}=\frac{1}{M}\left[\frac{2}{\gamma+1} \left(1+\frac{\gamma-1}{2}M^{2}\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}.$$ Because the nozzle is designed to run supersonically, we take the supersonic root of $A_2/A^{*}=1.688$. Substituting $M=2$ gives $\tfrac{1}{2}\left[(1+0.8)/1.2\right]^{3}=\tfrac{1}{2}(1.5)^{3}=1.6875$, which reproduces the stated ratio to four figures, so $$\boxed{\ M_B=2.00\ }$$ The paper has chosen the area ratio so that this comes out exactly; there is no interpolation to do.
Get the static state at the pipe inlet B. Section A–B is isentropic, so the reservoir stagnation values apply: $$\begin{aligned} P_B&=\frac{P_0}{\left(1+\tfrac{\gamma-1}{2}M_B^{2}\right)^{\gamma/(\gamma-1)}} =\frac{500}{(1.8)^{3.5}}=\frac{500}{7.824}=63.9\ \text{kPa}\\ T_B&=\frac{T_0}{1+\tfrac{\gamma-1}{2}M_B^{2}}=\frac{350.15}{1.8}=194.5\ \text{K} \end{aligned}$$ The pipe now receives air at Mach 2, 63.9 kPa and 194.5 K.
Evaluate the friction parameter for the pipe. The Fanno duct parameter is $$\frac{fL}{D}=\frac{0.015\times0.0563}{0.005}=0.1689 .$$ Reading $f$ as the Darcy friction factor is what makes the problem well posed. The Fanno function tabulated as $4f_{\text{Fanning}}L^{*}/D$ is numerically the same as $f_{\text{Darcy}}L^{*}/D$, and its value at $M=2$ is 0.3051. Had $f$ instead been the Fanning factor, the duct parameter would be $4\times0.1689=0.6756$, far beyond the 0.3051 available at Mach 2; the supersonic branch could not survive the pipe at all and a shock would be driven back inside the nozzle, contradicting the question. The Darcy reading is therefore the one the paper intends.
Carry the Mach number to the pipe exit along the Fanno line. With $$\frac{4f L^{*}}{D}=\frac{1-M^{2}}{\gamma M^{2}} +\frac{\gamma+1}{2\gamma}\ln\!\left[\frac{(\gamma+1)M^{2}} {2\left(1+\tfrac{\gamma-1}{2}M^{2}\right)}\right],$$ supersonic Fanno flow decelerates toward Mach 1, so the duct consumes friction length: $$\left(\frac{4fL^{*}}{D}\right)_{C}=\left(\frac{4fL^{*}}{D}\right)_{B}-\frac{fL}{D} =0.3051-0.1689=0.1362 .$$ Solving the Fanno function for that value gives 0.1360 at $M=1.5$, so $$\boxed{\ M_C=1.50\ }$$ The flow is still supersonic at the exit, which is the precondition for a normal shock to stand there.
Convert the Mach numbers into the exit static state. Fanno ratios referenced to the sonic point are $$\begin{aligned} \frac{P}{P^{*}}&=\frac{1}{M}\sqrt{\frac{\gamma+1}{2+(\gamma-1)M^{2}}}\\ \frac{T}{T^{*}}&=\frac{\gamma+1}{2+(\gamma-1)M^{2}} \end{aligned}$$ which give 0.4082 and 0.6667 at $M=2$, and 0.6065 and 0.8276 at $M=1.5$. Taking ratios so that the sonic reference cancels, $$P_C=P_B\frac{(P/P^{*})_{1.5}}{(P/P^{*})_{2.0}}=63.9\times1.4856=94.9\ \text{kPa},\qquad T_C=194.5\times1.2414=241.4\ \text{K}.$$ As a check on the adiabatic assumption, $T_{0C}=241.4\times(1+0.2\times1.5^{2})=350.1\ \text{K}$, which recovers the reservoir stagnation temperature exactly — friction destroys stagnation pressure, never stagnation temperature.
Apply the normal shock at the exit plane to obtain the back pressure. Downstream of a normal shock the static pressure is that of the surroundings, so the back pressure is the post-shock pressure. From the aid-sheet shock relation with $M_1=M_C=1.5$, $$\frac{P_b}{P_C}=\frac{2\gamma M_C^{2}-(\gamma-1)}{\gamma+1} =\frac{2.8\times2.25-0.4}{2.4}=2.4598,$$ so $$\boxed{\ P_b=94.9\times2.4598=233\ \text{kPa}\ }$$ The shock also drops the flow to $M=0.7011$ and raises the temperature to 319 K, but only the pressure is asked for.
Compute the mass flow rate at the choked throat. Because the throat is sonic, the mass flow is set there and is independent of everything downstream. The pipe area is $A_2=\pi D^{2}/4=19.63\ \text{mm}^{2}$, so the throat area is $A_T=A_2/1.688=11.63\ \text{mm}^{2}$. The aid-sheet mass-flow relation at $M=1$ reads $$\dot m=\sqrt{\frac{\gamma}{R}}\;\frac{P_0}{\sqrt{T_0}} \left(\frac{\gamma+1}{2}\right)^{-\frac{\gamma+1}{2(\gamma-1)}}A_T =0.069843\times\frac{500\,000}{18.712}\times0.57870\times1.1632\times10^{-5}.$$ Hence $$\boxed{\ \dot m=0.01256\ \text{kg/s}=12.6\ \text{g/s}\ }$$ Recomputing from the pipe-exit state as an independent check, $\rho_C=P_C/RT_C=1.369\ \text{kg/m}^{3}$ and $V_C=M_C\sqrt{\gamma RT_C}=467\ \text{m/s}$, so $\rho_C V_C A_2=0.01256\ \text{kg/s}$ — the same value, as continuity demands.

Station / quantity	Symbol	Value
Nozzle exit Mach number	M_B	2.00
Nozzle exit static pressure / temperature	P_B, T_B	63.9 kPa, 194.5 K
Fanno duct parameter	fL/D	0.1689
Pipe exit Mach number	M_C	1.50
Pipe exit static pressure / temperature	P_C, T_C	94.9 kPa, 241.4 K
Back pressure with the shock at the exit	P_b	233 kPa
Mass flow rate	ṁ	0.01256 kg/s (12.6 g/s)
Throat area (for reference)	A_T	11.63 mm²

Check: friction-factor convention. The question quotes "a friction factor of f = 0.015" without saying which. The solution above reads it as the Darcy–Weisbach factor, i.e. the duct parameter is fL/D. That choice is not arbitrary: it is the only one under which the stated pipe length is short enough for the flow to stay supersonic to the exit, which is what the question presupposes. It also makes the numbers land on exactly Mach 2.00 and Mach 1.50, a signature of a deliberately constructed problem. A candidate who reads f as the Fanning factor should state the assumption and note that the duct would then be over-long and the shock would move upstream into the divergent section.

Topic. Adiabatic constant-area duct flow with friction (Fanno flow) downstream of a supersonic nozzle, terminated by a normal shock.

Key relations. Isentropic area–Mach relation $A/A^{*}=f(M)$; isentropic stagnation ratios for $P$ and $T$; the Fanno friction-length function $4fL^{*}/D=f(M)$ and the Fanno property ratios $P/P^{*}$, $T/T^{*}$; the normal-shock static-pressure ratio $P_2/P_1=\left[2\gamma M_1^{2}-(\gamma-1)\right]/(\gamma+1)$; choked mass flow $\dot m=\sqrt{\gamma/R}\,(P_0/\sqrt{T_0})\left[(\gamma+1)/2\right]^{-(\gamma+1)/2(\gamma-1)}A_T$.

Why this works. Each element of the system changes the flow in one way only, and the aid sheet gives a closed relation for each. The nozzle is isentropic, so it conserves both stagnation properties and merely trades area for Mach number. The pipe is insulated and of constant area, so it conserves stagnation temperature and mass flux while friction drives the Mach number toward unity from whichever side it starts — supersonic flow decelerates, subsonic flow accelerates. Because the starred sonic reference state is common to every point on one Fanno line, ratios of the tabulated functions let you move between two stations without ever computing the reference state itself. Finally, a normal shock is a discontinuity of zero thickness: it changes the static pressure abruptly but cannot change the mass flow, which was already locked in at the choked throat.

Common pitfall. Adding the duct's friction length to the value at the inlet instead of subtracting it. The tabulated $4fL^{*}/D$ is the length still remaining before the flow would choke, so travelling downstream always consumes it. A second trap is assuming the shock changes the mass flow or the back pressure independently — the mass flow is set at the throat, and once the shock sits at the exit plane the back pressure is determined, not free.

Source. Anderson, Modern Compressible Flow, 3rd ed., §3.2 and §3.9 (normal shocks and Fanno flow); White, Fluid Mechanics, 8th ed., §9.7 (adiabatic flow with friction).

Question 2 of 7: Question 2 (Part A, Question A2): Supersonic Nozzle Feeding an Insulated Pipe — Fanno Flow with a Shock at the Exit

Question 2 (Part A, Question A2): Supersonic Nozzle Feeding an Insulated Pipe — Fanno Flow with a Shock at the Exit (16 marks)