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23-Chem-A2 Unit Operations and Separation Processes · May 2014

Question 1 of 6: Elevation Difference Between Two Reservoirs

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 2014 — open-book, 3 hours. Two sections: Section A (Mechanical Operations, A1–A3) and Section B (Thermal Operations, B1–B3); all problems are worth 25 marks. The rubric asks candidates to attempt two problems per section; all six are solved in full below as a study resource.

Reference texts: McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed., McGraw-Hill) — pipe-flow friction and loss coefficients, agitator power correlations (Fig. 9.13, Table 9.3), and filtration theory; de Nevers, Fluid Mechanics for Chemical Engineers (3rd ed.) and Brodkey & Hershey, Transport Phenomena — mechanical-energy balance, fitting equivalent lengths, and the appended Fanning chart; Coulson & Richardson, Chemical Engineering Vol. 2 — rotary-drum filtration; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (7th ed., Wiley) and Lienhard, A Heat Transfer Textbook — composite-wall resistance networks, the Colburn (Chilton–Colburn) analogy, and LMTD/ε–NTU cross-flow exchangers.

Property note. Water properties are as printed in each question; the appended Table A.2 loss coefficients ($k$) and Table A3 turbine constants ($K_T$) are used directly, and pipe-friction factors are taken from Colebrook (identical to the appended Fanning chart, Fig. A1) so the numbers are reproducible without reading a graph.

Question A1: Elevation Difference Between Two Reservoirs (25 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. Water at 20 °C flowing through a single 10 cm commercial wrought-iron line connecting two open reservoirs.

QuantityValue
Volumetric flow $Q$$0.04\ \mathrm{m^3/s}$
Inside diameter $D$$0.10\ \mathrm{m}$
Density $\rho$ / viscosity $\mu$$998.2\ \mathrm{kg/m^3}$ / $1.002\times10^{-3}\ \mathrm{Pa\,s}$
Roughness $\varepsilon$ (commercial steel / wrought iron)$0.0457\ \mathrm{mm}$ (Table A1)
Straight length $L=20+10+20$$50\ \mathrm{m}$
Fittings (Table A2, $k$)entrance 0.5, open globe 6.0, 2×90° elbow 0.75, exit 1.0

Find. the elevation difference $h$ between the free surfaces of the two reservoirs that drives $0.04\ \mathrm{m^3/s}$ through the line.

upper reservoir lower reservoir 20 m horizontal globe valve (open) 90° elbow 10 m 90° elbow 20 m horizontal h
Figure A1 — Pipe routing of Fig. 1. A single 10 cm line carries $0.04\ \mathrm{m^3/s}$ from the upper to the lower reservoir; the elevation drop $h$ supplies the exit kinetic energy plus all friction and fitting losses.

Approach. Apply the mechanical-energy balance between the two free surfaces; both are at atmospheric pressure with negligible surface velocity, so $h$ equals the total head loss $=(4f\,L/D+\Sigma k)\,v^2/2g$, evaluated at the single pipe velocity.

  1. Pipe velocity from continuity. With $A=\tfrac{\pi}{4}D^2=7.854\times10^{-3}\ \mathrm{m^2}$, $$v=\frac{Q}{A}=\frac{0.04}{7.854\times10^{-3}}=5.093\ \mathrm{m/s},\qquad \frac{v^2}{2g}=\frac{5.093^2}{2(9.81)}=1.322\ \mathrm{m}.$$ The velocity head $1.322\ \mathrm{m}$ is the common multiplier for every loss term.
  2. Reynolds number and Fanning factor. $$\mathrm{Re}=\frac{\rho v D}{\mu}=\frac{998.2(5.093)(0.10)}{1.002\times10^{-3}}=5.07\times10^{5}.$$ With relative roughness $\varepsilon/D=0.0457/100=4.57\times10^{-4}$, the Colebrook equation gives $f_{\text{Fanning}}=0.00434$ (equivalently $f_{\text{Darcy}}=4f=0.0174$), matching the appended Fanning chart Fig. A1.
  3. Sum the fitting loss coefficients (Table A2). $$\Sigma k = \underbrace{0.5}_{\text{entrance}}+\underbrace{6.0}_{\text{open globe}}+\underbrace{2(0.75)}_{\text{two }90^\circ\text{ elbows}}+\underbrace{1.0}_{\text{exit}}=9.0.$$ The wide-open globe valve alone accounts for $6.0$ of this — the single largest minor loss.
  4. Major (pipe friction) loss. Using $h_f=4f\dfrac{L}{D}\dfrac{v^2}{2g}$ with $L=50\ \mathrm{m}$: $$h_f=4(0.00434)\frac{50}{0.10}(1.322)=11.5\ \mathrm{m}.$$
  5. Minor (fitting) loss. $$h_m=\Sigma k\,\frac{v^2}{2g}=9.0(1.322)=11.9\ \mathrm{m}.$$ The open globe valve by itself contributes $6.0(1.322)=7.9\ \mathrm{m}$ — two-thirds of the fitting loss.
  6. Elevation difference. The energy balance $h=h_f+h_m$ gives $$h=11.5+11.9=23.4\ \mathrm{m}.$$ $h\approx 23.4\ \mathrm{m}$
QuantityResult
Pipe velocity $v$ / velocity head5.09 m/s / 1.32 m
$\mathrm{Re}$ / $f_{\text{Fanning}}$$5.07\times10^{5}$ / 0.00434
Friction loss $h_f$ / fitting loss $h_m$11.5 m / 11.9 m
Elevation difference $h$≈ 23.4 m
Check — routing assumption Fig. 1 is a schematic; the 20/10/20 m split and the two standard 90° elbows plus one open globe valve are read from the figure as printed. Because every loss term shares the same $v^2/2g$, the answer is insensitive to how the 50 m is distributed — only the totals $L=50$ m and $\Sigma k=9.0$ matter.
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