23-Chem-A2 Unit Operations and Separation Processes · May 2013
Question 2 of 6: Adiabatic Choked Relief Flow of Nitrogen
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exam 04-Chem-A2 Mechanical and Thermal Operations, May 2013 — open-book, 3 hours. Two sections: Section A (Mechanical Operations, A1–A3) and Section B (Thermal Operations, B1–B3); all problems 25 marks. The rubric asks candidates to attempt two problems per section; all six are solved in full below.
Reference texts: McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed., McGraw-Hill) — pipe-flow friction, loss coefficients, sphericity and the Ergun equation (Tables 7.1, 5.1); de Nevers, Fluid Mechanics for Chemical Engineers (3rd ed.) and Brodkey & Hershey, Transport Phenomena — mechanical-energy balance and sudden expansion/contraction losses; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (7th ed., Wiley) — composite-wall resistance networks, LMTD/ε–NTU cross-flow exchangers and lumped radiative cooling; Lienhard, A Heat Transfer Textbook and Özişik, Radiative Transfer for the appended correction-factor and emissivity charts.
Compressible-flow note. Water properties at 180 °F are taken as $\rho=60.55\ \mathrm{lb/ft^3}=970\ \mathrm{kg/m^3}$ and $\mu=2.32\times10^{-4}\ \mathrm{lb/(ft\cdot s)}=3.45\times10^{-4}\ \mathrm{Pa\cdot s}$; commercial-steel roughness $\varepsilon=0.0457$ mm (Table A2).
Section A — Mechanical Operations
Question A2: Adiabatic Choked Relief Flow of Nitrogen (25 marks)
Given. Reservoir (source) stagnation state $P_0=1.4\ \mathrm{MPa(g)}+101=1501$ kPa abs, $T_0=299.15$ K; 1-in Sch-40 pipe, $D=1.049\ \mathrm{in}=0.02664$ m, $A=5.576\times10^{-4}\ \mathrm{m^2}$, $L=10$ m; $\gamma=1.4$, $R_s=R/M=8314/28=296.9\ \mathrm{J/(kg\,K)}$; commercial-steel $\varepsilon/D=1.72\times10^{-3}$.
Find. the choked (maximum) mass flow rate the relief must handle when the pipe discharges adiabatically with wall friction (Fanno flow) and chokes at the exit.
Figure A2 — On regulator failure the source drives the 1-in line to sonic (choked) conditions at the exit. Adiabatic pipe flow with wall friction is Fanno flow; the pipe’s $4fL/D=8.48$ throttles the flow far below the frictionless-nozzle value.
Approach. Model the 10 m line as adiabatic Fanno flow choking at the exit ($M=1$); the pipe friction parameter $4fL/D$ fixes the inlet Mach number, and the stagnation state then sets the mass flux $G=\rho_1 v_1$ and hence $\dot m=GA$.
Confirm the flow is choked. The critical pressure ratio is $$\frac{P^\*}{P_0}=\left(\frac{2}{\gamma+1}\right)^{\gamma/(\gamma-1)}=0.528\;\Rightarrow\;P^\*=793\ \mathrm{kPa}.$$ The tank back-pressure (701 kPa abs) is below $P^\*$, so the line is choked and the flow is the maximum the pipe can pass.
Friction parameter. For commercial steel at high $\mathrm{Re}$, $f_{\text{Fanning}}=0.0056$, so $$\frac{4fL}{D}=\frac{4(0.0056)(10)}{0.02664}=8.48.$$
Inlet Mach number from the Fanno function. With the exit choked, $\dfrac{4fL^\*}{D}\big|_{M_1}=8.48$, where $$\frac{4fL^\*}{D}=\frac{1-M^2}{\gamma M^2}+\frac{\gamma+1}{2\gamma}\ln\!\frac{(\gamma+1)M^2}{2+(\gamma-1)M^2}.$$ Solving gives $M_1=0.250$.
Static state at the pipe inlet. From the isentropic reservoir relations at $M_1=0.250$: $T_1=T_0/(1+0.2M_1^2)=295.5$ K, $P_1=P_0(T_1/T_0)^{3.5}=1437$ kPa, hence $$\rho_1=\frac{P_1}{R_sT_1}=16.4\ \mathrm{kg/m^3},\qquad v_1=M_1\sqrt{\gamma R_sT_1}=87.6\ \mathrm{m/s}.$$
Mass flow. The mass flux is $G=\rho_1v_1=1436\ \mathrm{kg/(m^2 s)}$, so $$\dot m=GA=1436\times5.576\times10^{-4}=\boxed{0.80\ \mathrm{kg/s}}.$$ The relief device must vent at least this rate. (A frictionless choked-nozzle estimate would give 1.9 kg/s; wall friction more than halves it.)
