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23-Chem-A2 Unit Operations and Separation Processes · May 2016

Question 1 of 6: Affinity-Law Scale-Up of a Centrifugal Pump

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 2016. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt at least two from each section (only the first two per section are marked). All six are worked below for completeness.

Reference texts. Coulson & Richardson, Chemical Engineering Vol. 2 (particle technology, sedimentation, fluidization, filtration, crystallization, evaporation) and Vol. 1 (heat transfer); McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Perry's Chemical Engineers' Handbook (9th ed.).

Note on saturation data. Sections B2 and B3 need water saturation temperatures and latent heats that the exam expects from an open-book steam table. These are taken from standard tables (and, for reproducibility). The B1 latent heat of evaporation is not printed in the exam and is taken as 2370 kJ/kg at the cooling range; this and other engineering choices are flagged in Check callouts.

Question A1: Affinity-Law Scale-Up of a Centrifugal Pump (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. A geometrically similar (homologous) model and prototype centrifugal pump.

QuantityModel (1)Prototype (2)
Capacity $Q$108 m³/hr?
Head $H$107 m?
NPSH required3.05 m?
Power $P$41 kW?
Speed $N$3500 rpm1170 rpm
Impeller $D$10.5 in20 in

Find. (a) $Q$, $H$, $P$ and NPSH of the prototype; (b) the (dimensional) specific speed $N_s$ and suction specific speed $N_{ss}$ of both machines.

Homologous pumps — affinity (fan) laws Model (1) D₁=10.5″, N₁=3500 rpm Q₁=108, H₁=107, P₁=41 Prototype (2) D₂=20″, N₂=1170 rpm Q₂,H₂,P₂ = ? scale D, N
Figure A1 — Model-to-prototype scaling. Because the machines are geometrically similar, the affinity laws in $N$ and $D$ transfer every performance point.

Approach. Apply the affinity (fan) laws for homologous pumps — $Q\propto ND^3$, $H\propto N^2D^2$, $P\propto N^3D^5$, and NPSH (a head) $\propto N^2D^2$ — then compute the dimensional specific speed and suction specific speed, which are invariant along a homologous family.

  1. Form the speed and size ratios. $$r_N=\frac{N_2}{N_1}=\frac{1170}{3500}=0.3343,\qquad r_D=\frac{D_2}{D_1}=\frac{20}{10.5}=1.905.$$ The prototype turns slower but is roughly twice the diameter, so the size effect dominates capacity.
  2. Scale the capacity. Since $Q\propto ND^3$, $$Q_2=Q_1\,r_N\,r_D^{3}=108(0.3343)(1.905)^3=108(0.3343)(6.912)=\;\boxed{249.5\ \mathrm{m^3/hr}}.$$
  3. Scale the head. With $H\propto N^2D^2$, $$H_2=H_1\,r_N^2\,r_D^2=107(0.1118)(3.628)=43.4\ \mathrm{m}.$$ The lower speed pulls the head down even though the impeller grew.
  4. Scale the required NPSH. NPSH is a head and follows the same group: $$\mathrm{NPSH}_2=3.05\,r_N^2\,r_D^2=3.05(0.1118)(3.628)=1.24\ \mathrm{m}.$$
  5. Scale the power. With $P\propto \rho N^3D^5$ (same fluid), $$P_2=P_1\,r_N^3\,r_D^5=41(0.03736)(25.04)=38.4\ \mathrm{kW}.$$ Prototype: $Q_2=249.5$ m³/hr, $H_2=43.4$ m, $P_2=38.4$ kW, NPSH$_2=1.24$ m
  6. Specific speed (dimensional, rpm·m³/s·m). Using $N_s=\dfrac{N\sqrt{Q}}{H^{3/4}}$ with $Q$ in m³/s ($108/3600=0.0300$): $$N_{s,1}=\frac{3500\sqrt{0.0300}}{107^{0.75}}=\frac{606.2}{33.28}=18.2.$$ For the prototype, $Q_2=0.0693$ m³/s and $H_2=43.4$ m give $N_{s,2}=\dfrac{1170\sqrt{0.0693}}{43.4^{0.75}}=18.2$ — identical, as it must be for a homologous pair.
  7. Suction specific speed. Replace $H$ with NPSH: $$N_{ss}=\frac{N\sqrt{Q}}{\mathrm{NPSH}^{3/4}}\;\Rightarrow\; N_{ss,1}=\frac{606.2}{3.05^{0.75}}=263,\quad N_{ss,2}=\frac{1170\sqrt{0.0693}}{1.24^{0.75}}=263.$$ $N_s=18.2$ and $N_{ss}=263$ for both machines
QuantityModel (1)Prototype (2)
Capacity $Q$108 m³/hr249.5 m³/hr
Head $H$107 m43.4 m
NPSH required3.05 m1.24 m
Power $P$41 kW38.4 kW
Specific speed $N_s$18.218.2
Suction specific speed $N_{ss}$263263
Check (convention): $N_s$ and $N_{ss}$ are reported here in the dimensional SI form (rpm, m³/s, m). US-customary practice uses rpm, US-gpm and ft, which multiplies these figures by a constant (~51.6); the invariance across the homologous pair is what the question tests, not the unit system.
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