Net axial force on an insulated duct carrying choked adiabatic flow: 22-Mec-B6 Advanced Fluid Mechanics, May 2015, Q3

Question 3: Net axial force on an insulated duct carrying choked adiabatic flow (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
Inlet static pressure	$P_1$	700 kPa
Inlet static temperature	$T_1$	50 °C = 323.15 K
Inlet velocity	$V_1$	150 m/s
Duct diameter (constant)	$D$	0.15 m
Exit condition	$M_2$	1 (choked)
Air properties	$\gamma,\ R$	1.4, 287 J/(kg·K)

The duct is insulated (adiabatic), horizontal (no gravity term) and of constant area, so the flow follows a Fanno line: friction alone drives it from the inlet state to the sonic state at the exit.

Find. The net axial force that the fluid exerts on the pipe wall between the given inlet station and the choked exit plane.

Figure 3.1 — Control volume bounded by the inlet station, the choked exit plane and the pipe wall. The wall shear acts on the fluid in $-x$; by Newton’s third law the fluid acts on the pipe in $+x$.

Approach. Fix the inlet Mach number and mass flow, use the Fanno sonic-reference ratios to obtain the exit (starred) state, then apply the axial momentum balance to a control volume that spans the whole duct. No friction factor and no duct length are needed — the momentum balance already contains the friction as an unknown.

Compute the inlet Mach number. The speed of sound at the inlet is $$\begin{aligned} a_1&=\sqrt{\gamma R T_1}=\sqrt{(1.4)(287)(323.15)}=360.3\ \text{m/s}\cr M_1&=\frac{V_1}{a_1}=\frac{150}{360.3}=0.4163 \end{aligned}$$ The flow is subsonic, so friction accelerates it towards $M=1$, consistent with the stated choking at the exit.
Compute the mass flow rate. With $\rho_1=P_1/RT_1=700\,000/[(287)(323.15)]=7.548\ \text{kg/m}^{3}$ and $A=\pi D^{2}/4=\pi(0.15)^{2}/4=0.017671\ \text{m}^{2}$, $$\dot m=\rho_1 A V_1=(7.548)(0.017671)(150)=20.01\ \text{kg/s}.$$
Obtain the sonic reference state from the Fanno relations. For adiabatic constant-area flow the stagnation temperature is conserved, which gives $$\begin{aligned} \frac{T}{T^{*}}&=\frac{\gamma+1}{2+(\gamma-1)M^{2}}\cr \frac{P}{P^{*}}&=\frac{1}{M}\sqrt{\frac{\gamma+1}{2+(\gamma-1)M^{2}}} \end{aligned}$$ At $M_1=0.4163$ these evaluate to $T_1/T^{*}=1.1598$ and $P_1/P^{*}=2.587$, so $$\begin{aligned} T^{*}&=\frac{323.15}{1.1598}=278.6\ \text{K}\cr P^{*}&=\frac{700}{2.587}=270.6\ \text{kPa} \end{aligned}$$
Convert the sonic state to a velocity and check continuity. At the exit $M=1$, so $$V^{*}=a^{*}=\sqrt{\gamma R T^{*}}=\sqrt{(1.4)(287)(278.6)}=334.6\ \text{m/s}.$$ With $\rho^{*}=P^{*}/RT^{*}=3.384\ \text{kg/m}^{3}$, continuity closes independently: $\rho^{*}AV^{*}=(3.384)(0.017671)(334.6)=20.01$ kg/s, matching step 2 exactly. The stagnation temperature likewise checks, $T_{01}=T_1(1+0.2M_1^{2})=334.3$ K $=1.2\,T^{*}$.
Write the axial momentum balance for the control volume. Taking $x$ in the flow direction and letting $F$ be the force the fluid exerts on the pipe (so the pipe exerts $-F$ on the fluid), $$P_1A-P^{*}A-F=\dot m\,(V^{*}-V_1) \quad\Longrightarrow\quad F=(P_1-P^{*})A-\dot m\,(V^{*}-V_1).$$ Equivalently, $F$ is the drop in the impulse function $I=PA+\dot mV$ across the duct.
Substitute and evaluate. The pressure term is $(700\,000-270\,576)(0.017671)=7588\ \text{N}$ and the momentum term is $(20.01)(334.6-150)=3693\ \text{N}$, so $$F=7588-3693=3895\ \text{N}$$ $$\boxed{\;F=3.90\ \text{kN acting on the pipe in the flow direction}\;}$$ The sign is positive, as it must be: friction drags the pipe downstream, and the fluid loses far more pressure force than it gains momentum flux. The pipe restraints must therefore be designed to carry a 3.9 kN axial thrust, and an expansion joint anywhere in this run would transfer the whole of it to the nearest anchor.
Sanity-check the duct length implied by the answer. The Fanno length function at the inlet is $$\frac{fL^{*}}{D}=\frac{1-M^{2}}{\gamma M^{2}} +\frac{\gamma+1}{2\gamma}\ln\!\frac{(\gamma+1)M^{2}}{2+(\gamma-1)M^{2}}=2.032 ,$$ so for a commercial-steel value $f\approx0.020$ (Darcy convention) the duct is $L^{*}=(2.032)(0.15)/0.020=15.2$ m long. That is a physically sensible run for a 150 mm line, which supports the reading of the data; note the force answer itself is independent of $f$ and of $L$.

Quantity	Result
Inlet speed of sound, $a_1$	360.3 m/s
Inlet Mach number, $M_1$	0.4163
Inlet density, $\rho_1$	7.548 kg/m³
Mass flow rate, $\dot m$	20.01 kg/s
Sonic temperature, $T^{*}$	278.6 K
Sonic pressure, $P^{*}$	270.6 kPa
Exit velocity, $V^{}=a^{}$	334.6 m/s
Pressure term, $(P_1-P^{*})A$	7588 N
Momentum term, $\dot m(V^{*}-V_1)$	3693 N
Net force of the fluid on the pipe	3.90 kN in the flow direction
Implied duct length at $f=0.020$	15.2 m

Topic. Fanno flow — adiabatic compressible flow with wall friction in a constant-area duct — and the integral momentum balance that converts it into a structural load.

Key relations. $T_0=\text{const}$; $T/T^{*}=(\gamma+1)/[2+(\gamma-1)M^{2}]$; $P/P^{*}=(1/M)\sqrt{T/T^{*}}$; $fL^{*}/D=(1-M^{2})/(\gamma M^{2})+\tfrac{\gamma+1}{2\gamma}\ln\!\big[(\gamma+1)M^{2}/(2+(\gamma-1)M^{2})\big]$; impulse function $I=PA+\dot mV$ and $F=I_1-I_2$.

Why this works. The great convenience of Fanno flow is that every state on a given Fanno line can be referred to the same sonic point, because the mass flux and the stagnation temperature are both constants of the line. Choking at the exit therefore hands the exit state over without any information about the duct length or the friction factor: the exit is the reference point. The momentum balance then does the rest. Friction enters the balance as a single unknown — the integrated wall shear — so writing the balance across the whole duct and solving for that unknown short-circuits the entire integration of $\tau_w$ along the wall. The result is dominated by the pressure term because compressible friction flow loses static pressure very rapidly as $M\to1$: here the static pressure falls by 61 per cent while the velocity only rises by a factor 2.2. That imbalance is exactly why long choked gas lines impose large anchor loads, and why the axial thrust is computed from a control volume rather than from a friction-factor estimate.

Common pitfall. The commonest slip is to compute the friction force as $\tau_w\pi DL$ using an average shear stress, which requires both $f$ and $L$ and is far less accurate than the momentum balance. The second is a sign error: the momentum flux increases downstream, so it must be subtracted from the pressure term, and students who add it obtain 11.3 kN. A third is using stagnation pressures in the momentum balance; the balance is written on static pressures acting on the flow area, since stagnation pressure is not a force per unit area. Finally, remember that $T^{*}$ here is a Fanno reference temperature, not $T_0$ — the two differ by the factor 1.2.

Source. J. D. Anderson, Modern Compressible Flow, 3rd ed., §3.8 (Fanno flow); F. M. White, Fluid Mechanics, 8th ed., §9.7 and §3.4 (integral momentum).

Question 3 of 6: Net axial force on an insulated duct carrying choked adiabatic flow

Question 3: Net axial force on an insulated duct carrying choked adiabatic flow (20 marks)