23-Chem-A2 Unit Operations and Separation Processes · May 2015
Question 1 of 6: Pump Power for a Sanitary Apple-Juice Line
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 04-CHEM-A2 Mechanical and Thermal Operations, May 2015. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt two from each section. All six are worked below for completeness.
Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 1 (fluid flow, heat transfer) and Vol. 2 (particle technology, evaporation); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Perry's Chemical Engineers' Handbook (9th ed.).
Note on the charts and tables. Page 3 (Fanning friction-factor chart), page 5 (LMTD correction-factor plot) and page 6 (Table B.2, saturated water by pressure) were read from the printed figure. The smooth-pipe friction factor is computed from the Colebrook equation, which the chart plots; the B3 condenser is isothermal on the steam side, so the correction factor is exactly $F=1$; and the B2 saturation temperatures are read from the supplied Table B.2.
Question A1: Pump Power for a Sanitary Apple-Juice Line (25 marks)
Given. Incompressible flow between two open free surfaces (both velocities negligible, both at atmospheric pressure), lifted from 3 m to 12 m.
Quantity
Value
Density $\rho$
997.1 kg/m³
Viscosity $\mu$
$2.1\times10^{-3}$ Pa·s
Inside diameter $D$
2.291 cm = 0.02291 m
Mass flow $\dot m$
1 kg/s
Straight length $L$
30 m
Fittings
2 × 90° std elbow, 1 angle valve (open)
Static lift $\Delta z$
$12-3 = 9$ m
Pump efficiency $\eta$
0.60
Find. The shaft power the pump must deliver to raise 1 kg/s of apple juice over this piping network at 60% efficiency.
Figure A1 — Sanitary line lifting apple juice from the 3 m supply surface to the 12 m discharge surface; pump P supplies the static lift plus friction.
Approach. Apply the mechanical-energy (extended Bernoulli) balance between the two free surfaces; the pump work equals the static lift plus the total friction loss (straight pipe + fittings), then divide by efficiency.
Pipe velocity and Reynolds number. With $A=\tfrac{\pi}{4}D^2 = \tfrac{\pi}{4}(0.02291)^2 = 4.12\times10^{-4}\ \mathrm{m^2}$, the mean velocity is $$v=\frac{\dot m/\rho}{A}=\frac{1/997.1}{4.12\times10^{-4}}=2.43\ \mathrm{m/s},\qquad \mathrm{Re}=\frac{\rho v D}{\mu}=\frac{997.1(2.43)(0.02291)}{2.1\times10^{-3}}=2.65\times10^{4}.$$ The flow is turbulent.
Fanning friction factor (smooth pipe). The Colebrook equation with $\varepsilon/D=0$ gives $f=0.00605$ (Darcy $4f=0.0242$). $$\boxed{f = 0.00605}$$
Fitting losses. Using equivalent loss coefficients $K$ (entrance 0.5, each std elbow 0.75, open angle valve 2.0, exit 1.0): $$\textstyle\sum K = 0.5 + 2(0.75) + 2.0 + 1.0 = 5.0.$$
Total friction head (as specific energy). Combining straight-pipe and fitting terms with the velocity head $v^2/2 = (2.43)^2/2 = 2.96\ \mathrm{J/kg}$: $$h_f=\left(\frac{4fL}{D}+\sum K\right)\frac{v^2}{2}=\left(\frac{4(0.00605)(30)}{0.02291}+5.0\right)(2.96)=108.5\ \mathrm{J/kg}.$$ The straight run contributes $\tfrac{4fL}{D}=31.7$ velocity heads; the fittings only 5.
Mechanical-energy balance. Both surfaces are open and quiescent, so the pump specific work covers lift plus friction: $$W_s=g\,\Delta z + h_f = 9.81(9) + 108.5 = 88.3 + 108.5 = 196.8\ \mathrm{J/kg}.$$ Friction and lift are comparable here (friction slightly larger).
