Question 1 of 8: Pump Power and Homologous Scaling
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examination 07-Mec-A6-1 Fluid Machinery (May 2014) — closed book, three hours, 60 marks. Section A is calculative (Q1–Q5) and Section B is descriptive (Q6–Q8); the rubric asks for four 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; H. Cohen, G. F. C. Rogers & H. I. H. Saravanamuttoo, Gas Turbine Theory (for the axial compressor stage); R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics (pump energy equation and affinity laws).
Question 1: Pump Power and Homologous Scaling (10 marks)
Given. Suction pipe D₁ = 150 mm and discharge pipe D₂ = 100 mm; suction gauge p₁ = −30 kPa located z₁ = −0.5 m; discharge gauge p₂ = +150 kPa located z₂ = +1.5 m; gasoline SG = 0.75 (ρ = 750 kg/m³); Q = 0.035 m³/s; pump efficiency η = 0.75.
Given data — Part I
Quantity
Suction (1)
Discharge (2)
Pipe diameter
150 mm
100 mm
Gauge pressure
−30 kPa
+150 kPa
Gauge elevation (rel. CL)
−0.5 m
+1.5 m
Find. The electrical (input) power drawn by the pump motor.
Pump installation: suction (150 mm, gauge 0.5 m below CL, 30 kPa vacuum) and discharge (100 mm, gauge 1.5 m above CL, 150 kPa). Flow 0.035 m³/s of gasoline (SG 0.75).
Approach. Apply the steady-flow energy (Bernoulli) equation between the two gauge stations to get the pump head, then multiply by ρgQ for water power and divide by efficiency for input power.
Pipe velocities from continuity. With $A=\tfrac{\pi}{4}D^2$,
$$V_1=\frac{Q}{A_1}=\frac{0.035}{\tfrac{\pi}{4}(0.150)^2}=1.98\ \text{m/s},\qquad V_2=\frac{Q}{A_2}=\frac{0.035}{\tfrac{\pi}{4}(0.100)^2}=4.46\ \text{m/s}.$$
Pump head from the energy equation. Between gauge 1 and gauge 2,
$$H=\frac{p_2-p_1}{\rho g}+(z_2-z_1)+\frac{V_2^2-V_1^2}{2g}.$$
Substituting (note the suction vacuum is a negative pressure, so $p_2-p_1=150-(-30)=180$ kPa and $z_2-z_1=2.0$ m):
$$H=\frac{180\,000}{750(9.81)}+2.0+\frac{4.46^2-1.98^2}{2(9.81)}=24.46+2.00+0.81.$$
$$\boxed{H=27.3\ \text{m of gasoline}}$$
Water (hydraulic) power. $P_{\text{fluid}}=\rho g Q H=750(9.81)(0.035)(27.3)=7.02\ \text{kW}.$
Electrical input power. Dividing by pump efficiency,
$$P_{\text{elec}}=\frac{P_{\text{fluid}}}{\eta}=\frac{7.02}{0.75}=9.37\ \text{kW}.$$
Part II — Homologous scaling
Given. Model: Dm = 188 mm, Nm = 3600 rev/min, Hm = 39.6 m, Qm = 0.085 m³/s, ηm = 84%. Prototype: Dp = 10 Dm = 1880 mm, Hp = 110 m, water in both.
Approach. Use the affinity (similarity) groups for geometrically similar machines — head coefficient $H/N^2D^2$ and flow coefficient $Q/ND^3$ constant — then the Moody formula for the efficiency step-up with size.
(a) Speed from the head coefficient. $\dfrac{H_m}{N_m^2 D_m^2}=\dfrac{H_p}{N_p^2 D_p^2}$ gives
$$N_p=N_m\sqrt{\frac{H_p}{H_m}}\left(\frac{D_m}{D_p}\right)=3600\sqrt{\frac{110}{39.6}}\left(\frac{1}{10}\right).$$
$$\boxed{N_p=600\ \text{rev/min}}$$
(b) Flow from the flow coefficient. $\dfrac{Q}{ND^3}$ constant:
$$Q_p=Q_m\frac{N_p}{N_m}\left(\frac{D_p}{D_m}\right)^3=0.085\left(\frac{600}{3600}\right)(10)^3=14.2\ \text{m}^3/\text{s}.$$
(c) Ideal power. With unit efficiency the input power equals the water power,
$$P_p=\rho g Q_p H_p=1000(9.81)(14.17)(110)=15.3\ \text{MW}.$$
(d) Efficiency by Moody's relationship. The full form allows for the different heads,
$$\eta_p=1-(1-\eta_m)\left(\frac{D_m}{D_p}\right)^{1/4}\left(\frac{H_m}{H_p}\right)^{1/10}=1-0.16(0.1)^{0.25}(0.36)^{0.1}=0.919.$$
The simpler diameter-only estimate $\;(1-\eta_p)=(1-\eta_m)(D_m/D_p)^{1/5}\;$ gives ηp ≈ 0.899, so the prototype efficiency lands near
$$\boxed{\eta_p\approx 90\text{–}92\%.}$$
