Question 3 of 6: Adiabatic Constant-Area Duct with Friction (Fanno Flow)
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. Insulated (adiabatic) constant-area duct, air as an ideal gas, choked exit ($M_2=1$).
Given data
Inlet pressure
$P_1 = 680\ \text{kPa}$
Inlet temperature
$T_1 = 60\,{}^\circ\text{C}=333.15\ \text{K}$
Inlet velocity
$V_1 = 110\ \text{m/s}$
Diameter
$D = 0.13\ \text{m}\;(A=0.01327\ \text{m}^2)$
Exit
choked, $M_2=1$
Find. The net axial force exerted by the pipe on the fluid.
Figure 2 (control volume). Constant-area adiabatic duct; wall friction retards the flow (acts upstream), and the flow reaches $M=1$ at the exit.
Approach. This is Fanno flow (adiabatic, constant area, friction). Fix the inlet Mach number, use the Fanno reference ($^{*}$) relations to get the choked exit state, then apply the axial momentum equation on the control volume to extract the wall (friction) force — the only force the parallel walls can exert on the fluid.
Inlet Mach number and mass flow. $a_1=\sqrt{\gamma R T_1}=365.9$ m/s, so $M_1=V_1/a_1=0.301$. With $\rho_1=P_1/(RT_1)=7.11\ \text{kg/m}^3$ and $A=\pi D^2/4=0.01327\ \text{m}^2$,
$$\dot m=\rho_1 A V_1=\boxed{10.38\ \text{kg/s}.}$$
Choked exit state via Fanno $^{*}$ relations. At $M_2=1$ the exit is the sonic reference state:
$$T_2=T^{*}=T_1\frac{2+(\gamma-1)M_1^2}{\gamma+1}=282.6\ \text{K},\quad
P_2=P^{*}=P_1 M_1\sqrt{\frac{2+(\gamma-1)M_1^2}{\gamma+1}}=188.3\ \text{kPa},$$
$$V_2=V^{*}=\sqrt{\gamma R T^{*}}=337.0\ \text{m/s}.$$
(Continuity check: $\rho_2 A V_2=\dot m$.)
Axial momentum on the control volume. Taking $x$ positive in the flow direction, the pressures act on the end faces and the wall exerts an axial force $R_x$ on the fluid (constant area ⇒ the only wall force is friction):
$$P_1 A - P_2 A + R_x=\dot m\,(V_2-V_1).$$
Solving,
$$R_x=\dot m(V_2-V_1)-(P_1-P_2)A=2357-6526=\boxed{-4.17\ \text{kN}.}$$
Interpretation. The negative sign means the pipe pushes the fluid upstream: the net force of the pipe on the fluid is a friction (shear) drag of magnitude $4.17\ \text{kN}$ opposing the flow. Equivalently, the fluid exerts $4.17\ \text{kN}$ on the pipe in the downstream direction.
