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: National Examinations — 07-Mec-A6 Fluid Machinery, May 2013. Closed-book, three hours. Section A (Calculative) Q1–Q5, Section B (Descriptive) Q6–Q8; the candidate answers four from A and two from B, six questions of equal value (10 marks each). All eight questions are worked here as a complete study resource.
Reference texts: R. K. Turton, Principles of Turbomachinery; S. L. Dixon & C. A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery; R. W. Fox, A. T. McDonald, Introduction to Fluid Mechanics; Douglas, Gasiorek & Swaffield, Fluid Mechanics. Constants as printed on the exam attachment pages (g = 9.81 m/s², patm = 100 kPa, pvapour = 2.34 kPa at 20°C, ρwater = 1000 kg/m³).
Question 1: Pump Power and Homologous Scaling (10 marks)
Given. Suction pipe Ds = 0.150 m, discharge Dd = 0.100 m; pd = +150 kPa at zd = +1.5 m, ps = −30 kPa at zs = −0.5 m; ρ = 0.75 × 1000 = 750 kg/m³; Q = 0.035 m³/s; ηpump = 0.75.
Find. The electrical (motor input) power to drive the pump.
Pump installation: gauge locations, pipe sizes and elevations relative to the pump centre line.
Approach. Apply the steady-flow energy (Bernoulli) equation between the two gauge tappings to get the pump total head, convert to hydraulic power ρgQH, then divide by efficiency.
Pipe velocities from continuity. $V=\dfrac{Q}{A}=\dfrac{Q}{\tfrac{\pi}{4}D^{2}}$; $V_s=\dfrac{0.035}{\tfrac{\pi}{4}(0.150)^2}=1.98\ \text{m/s}$, $V_d=\dfrac{0.035}{\tfrac{\pi}{4}(0.100)^2}=4.46\ \text{m/s}$.
Total head across the pump. $H=\dfrac{p_d-p_s}{\rho g}+(z_d-z_s)+\dfrac{V_d^{2}-V_s^{2}}{2g}$. Substituting $\dfrac{(150-(-30))\times10^{3}}{750\times9.81}+2.0+\dfrac{4.46^{2}-1.98^{2}}{2\times9.81}=24.46+2.00+0.81$, so $\boxed{H=27.28\ \text{m}}$.
Hydraulic power delivered to the fluid. $P_{fluid}=\rho g Q H=750\times9.81\times0.035\times27.28=7.02\ \text{kW}$.
Electrical power at 75% efficiency. $P_{elec}=\dfrac{P_{fluid}}{\eta}=\dfrac{7.02}{0.75}$, giving $\boxed{P_{elec}=9.37\ \text{kW}}$.
Part II — Homologous prototype
Given. Model Dm = 0.188 m, Nm = 3600 rev/min, Hm = 39.6 m, Qm = 0.085 m³/s, ηm = 0.84; prototype Dp = 1.88 m (10×), Hp = 110 m, water.
Find. Prototype speed, flow, ideal power and efficiency under homologous (dynamically similar) operation.
Approach. Equate the dimensionless head, flow and power coefficients between model and prototype, and use the Moody step-up rule for efficiency.
(a) Speed from the head coefficient. $\dfrac{gH}{N^{2}D^{2}}=\text{const}\Rightarrow N_p=N_m\sqrt{\dfrac{H_p}{H_m}}\,\dfrac{D_m}{D_p}$. $N_p=3600\sqrt{\dfrac{110}{39.6}}\times\dfrac{1}{10}$, so $\boxed{N_p=600\ \text{rev/min}}$.
(b) Flow from the flow coefficient. $\dfrac{Q}{ND^{3}}=\text{const}\Rightarrow Q_p=Q_m\dfrac{N_p}{N_m}\left(\dfrac{D_p}{D_m}\right)^{3}=0.085\times\dfrac{600}{3600}\times10^{3}$, giving $\boxed{Q_p=14.17\ \text{m}^3/\text{s}}$.
(c) Ideal power. Assuming ideal (loss-free) conversion the water power is $P=\rho g Q_p H_p=1000\times9.81\times14.17\times110=15.3\ \text{MW}$.
(d) Efficiency by the Moody scale-up. $\dfrac{1-\eta_p}{1-\eta_m}=\left(\dfrac{D_m}{D_p}\right)^{1/5}$; $1-\eta_p=(1-0.84)(0.1)^{0.2}=0.101$, so $\boxed{\eta_p\approx90\%}$ — the larger machine is more efficient.
