23-Chem-A2 Unit Operations and Separation Processes · December 2017
Question 2 of 6: Maximum Efficiency of a Centrifugal Pump
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 16-CHEM-A2 Unit Operations and Separation Processes, December 2017. 3 hours, open book (one text). Six problems — Part A Unit Operations (A1–A3) and Part B Separation Processes (B1–B3), each worth 25 marks; the rubric asks for at least two problems from each part (only the first two per part are marked). All six are worked below for completeness as a study resource.
Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Coulson & Richardson, Chemical Engineering Vol. 2 (particle technology, absorption, drying, crystallization); Treybal, Mass-Transfer Operations (3rd ed.); Perry's Chemical Engineers' Handbook (9th ed.).
Notes on chart / model reads. A1's minor-loss balance uses the Blasius smooth-pipe friction factor $f=0.079\,Re^{-0.25}$ (Fanning) — valid to $Re\approx10^5$, and all three test points fall in $3.8$–$8.9\times10^4$; the 1.5 m is taken as the head lost across the pipe-and-valve test section, because also charging an exit velocity head would make the measured open-valve flow physically impossible (see A1). B1(b) reads the Sherwood/Eckert flooding line supplied with the paper at abscissa $0.061$, giving flooding ordinate $\approx0.17$ (graphical, $0.16$–$0.18$), with the tabulated packing factor $F=160\ \text{ft}^{-1}$ converted to $525\ \text{m}^{-1}$ for the SI form of the ordinate. B3(a) sizes the vessel from its starting liquor volume, allowing for the water evaporated during the batch. Engineering choices are flagged in Check callouts.
Part A — Unit Operations
Question A2: Maximum Efficiency of a Centrifugal Pump (25 marks)
Find. The best-efficiency point: the maximum overall efficiency and the flow rate at which it occurs.
Figure A2 — Computed efficiency of every test point; the curve peaks (best-efficiency point) at $Q=0.91\times10^{-3}\ \text{m}^3/\text{s}$.
Approach. Efficiency is useful hydraulic power over shaft (brake) power; with the speed fixed the shaft power is just a constant times the arm force, so tabulate $\eta$ for all seven points and pick the maximum.
Fixed shaft-power coefficient. The angular speed is $\omega=\dfrac{2\pi N}{60}=\dfrac{2\pi(2500)}{60}=261.8\ \text{rad/s}$, and with the torque $T=Fr$ the brake power is $P_\text{shaft}=T\omega=Fr\omega$. Since $r\omega=0.179(261.8)=46.86$, $$P_\text{shaft}=46.86\,F\ \text{(W, }F\text{ in N)}.$$
Hydraulic power per point. The useful power delivered to the fluid is $P_\text{hyd}=Q\,\Delta P$ with $\Delta P$ in Pa, so the overall efficiency is $$\eta=\frac{P_\text{hyd}}{P_\text{shaft}}=\frac{Q\,\Delta P}{46.86\,F}.$$
Evaluate all seven rows.
$Q\,(\times10^{-3})$
$\Delta P$ (kPa)
$F$ (N)
$P_\text{hyd}$ (W)
$P_\text{shaft}$ (W)
$\eta$
1.47
18.8
4.6
27.6
215.6
12.8%
1.31
36.0
4.1
47.2
192.1
24.5%
1.18
44.8
4.0
52.9
187.4
28.2%
1.00
61.5
3.7
61.5
173.4
35.5%
0.91
63.6
3.3
57.9
154.6
37.4%
0.71
71.5
3.0
50.8
140.6
36.1%
0.48
79.7
2.4
38.3
112.5
34.0%
Locate the maximum. A representative point (row 5): $P_\text{hyd}=(0.91\times10^{-3})(63.6\times10^{3})=57.9\ \text{W}$, $P_\text{shaft}=3.3(46.86)=154.6\ \text{W}$, so $\eta=57.9/154.6=0.374$. Scanning the column, efficiency rises as flow falls, peaks, then declines: $$\boxed{\eta_\text{max}=37.4\%\ \text{at}\ Q=0.91\times10^{-3}\ \text{m}^3/\text{s}\ (\approx0.91\ \text{L/s}).}$$ The corresponding differential head is $\Delta P/\rho g=63.6\times10^{3}/(1000\cdot9.81)\approx6.5\ \text{m}$ of water.
