Question 2 of 8: Pump Application — Preliminary Design
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 98-Mar-B5 Fluid Machinery, May 2016 — closed book, three hours, 60 marks. Section A is calculative (Q1–Q5) and Section B is descriptive (Q6–Q8); the rubric asks for four questions of Section A plus two of Section B (six questions, each of equal value, 10 marks). All eight questions are solved in full as a study resource. General constants supplied with the paper: g = 9.81 m/s², patm = 100 kPa, pvapour = 2.34 kPa (20 °C), ρwater = 1000 kg/m³, ρair = 1.21 kg/m³ (15 °C), cp,air = 1.005, cv,air = 0.718 kJ/kg·K.
Reference texts. S. L. Dixon & C. A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed.; R. K. Turton, Principles of Turbomachinery, 2nd ed.; H. Cohen, G. F. C. Rogers & H. I. H. Saravanamuttoo, Gas Turbine Theory, 6th ed.; R. H. Sabersky, A. J. Acosta, E. G. Hauptmann & E. M. Gates, Fluid Flow: A First Course in Fluid Mechanics, 4th ed. — pump-characteristic and cavitation-parameter charts; R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics, 8th ed.; Y. A. Çengel & J. M. Cimbala, Fluid Mechanics: Fundamentals and Applications.
Given. Q = 85 L/s = 0.085 m³/s; H = 30 m; 60 Hz induction drive; slip 3%; electrical losses 4%; inlet head loss hL = 1.0 m; pvapour = 2 kPa; patm = 100 kPa.
Given data
Quantity
Value
Duty flow Q
0.085 m³/s
Duty head H
30 m
Supply frequency / slip
60 Hz / 3%
Electrical losses
4%
Inlet head loss / vapour pressure
1.0 m / 2 kPa
Find. The eight preliminary-design quantities (a)–(h).
Check: the motor pole count is not stated. A 4-pole 60 Hz machine (synchronous 1800 rev/min) at 3% slip runs at N = 1746 rev/min, which places this duty near the top of the radial-flow efficiency curve; the design is carried out at that speed and the assumption is flagged.
Approach. Compute the dimensionless specific speed to fix the pump type and read optimum efficiency, peripheral-velocity factor and critical cavitation parameter from the supplied charts; then size the impeller, NPSH and setting elevation, and finally the electrical draw.
(a) Running speed and specific speed. $N=1800(1-0.03)=1746\ \text{rev/min}$, $\omega=182.8\ \text{rad/s}$. $$N_s=\frac{\omega\,Q^{1/2}}{(gH)^{3/4}}=\frac{182.8\,(0.085)^{1/2}}{(9.81\times30)^{3/4}}=\boxed{0.75}.$$
(b) Pump type. A dimensionless specific speed near 0.75 falls in the radial-flow band on the Page 11 chart, so a single-stage centrifugal (radial-flow) pump with a closed impeller is appropriate (approaching the mixed-flow transition). The impeller is a backward-curved, shrouded radial rotor.
(c) Impeller diameter. At $N_s=0.75$ (equivalently $N_s\approx2050$ on the gpm scale) the $\phi_e$ curve on Page 11 reads $\phi_e\approx1.1$, so the tip speed is $V_{B2}=\phi_e\sqrt{2gH}=1.1\sqrt{2(9.81)(30)}=1.1(24.26)=26.69\ \text{m/s}$, hence $D=\dfrac{60\,V_{B2}}{\pi N}=\dfrac{60(26.69)}{\pi(1746)}=\boxed{0.292\ \text{m}}$.
(d) Critical cavitation parameter. From the Page 12 chart at $N_s\approx0.75$ (about 2050 on the gpm scale, just right of the 2000 gridline) the curve reads $\boxed{\sigma_c\approx0.15}$.
(f) Maximum pump elevation above supply. $$\Delta z_{max}=\frac{p_{atm}-p_{vap}}{\rho g}-h_L-NPSH=\frac{100-2\ \text{kPa}}{9.81}-1.0-4.5=\boxed{+4.49\ \text{m}}.$$ Of the 9.99 m of atmospheric-minus-vapour head, 1.0 m is lost in the inlet piping and 4.5 m must remain as NPSH at the impeller eye, so the pump centre line may be set up to about 4.5 m above the supply water level (a suction lift). A practical design would keep a margin below this limit.
(g) Pump efficiency. The Page 11 optimum-efficiency curve at $N_s\approx0.75$ reads $\boxed{\eta_p\approx92\%}$ (the curve peaks at about 92.5% near $N_s\approx0.9$).
(h) Electric power. Water power $P_w=\rho gQH=25.0\ \text{kW}$; shaft power $P_{shaft}=P_w/\eta_p=25.0/0.92=27.2\ \text{kW}$; with motor efficiency $\eta_m=(1-0.03)(1-0.04)=0.931$, $P_{elec}=27.2/0.931=\boxed{29.2\ \text{kW}}$.
