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 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 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; 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 (pump energy equation and affinity laws); Y. A. Çengel & J. M. Cimbala, Fluid Mechanics: Fundamentals and Applications.
Question 1: Pump Power and Homologous Scaling (10 marks)
Given. The pumped fluid is gasoline (SG = 0.75, ρ = 750 kg/m³) at Q = 0.035 m³/s; the two gauges straddle the pump as tabulated below.
Given data — Part I
Quantity
Suction (1)
Discharge (2)
Pipe diameter
150 mm
100 mm
Gauge pressure
−30 kPa
+150 kPa
Gauge elevation (rel. centre line)
−0.5 m
+1.5 m
Find. The electrical input power at a pump efficiency of 75%.
Pump installation: the suction gauge reads a vacuum below the centre line, the discharge gauge a positive pressure above it; the flow accelerates from the 150 mm suction into the 100 mm discharge.
Approach. Apply the steady-flow energy (Bernoulli) equation between the two gauge sections to obtain the pump head, then convert to fluid power and divide by efficiency.
Pipe velocities from continuity. $V=\dfrac{Q}{\tfrac{\pi}{4}D^{2}}$ gives $V_1=\dfrac{0.035}{\tfrac{\pi}{4}(0.150)^2}=1.98\ \text{m/s}$ and $V_2=\dfrac{0.035}{\tfrac{\pi}{4}(0.100)^2}=4.46\ \text{m/s}$.
Pump head between the gauges. With $\rho g = 750\times9.81 = 7358\ \text{Pa/m}$, $$H=\frac{p_2-p_1}{\rho g}+(z_2-z_1)+\frac{V_2^{2}-V_1^{2}}{2g}=\frac{150-(-30)\ \text{kPa}}{7.358}+2.0+\frac{4.46^{2}-1.98^{2}}{2(9.81)}.$$ The three contributions are 24.46 m (pressure) + 2.0 m (elevation) + 0.81 m (velocity head) $=\boxed{H=27.28\ \text{m}}$.
Fluid power. $P_{f}=\rho g Q H = 750(9.81)(0.035)(27.28)=7.02\ \text{kW}$.
Approach. Homologous machines share the head and flow coefficients, so use $\dfrac{gH}{N^{2}D^{2}}$ and $\dfrac{Q}{ND^{3}}$ constant, then the Moody efficiency step-up for the size change.
Prototype speed. From $\dfrac{H}{N^{2}D^{2}}=\text{const}$, $N_p=N_m\sqrt{\dfrac{H_p}{H_m}}\dfrac{D_m}{D_p}=3600\sqrt{\dfrac{110}{39.6}}\left(\dfrac{1}{10}\right)=\boxed{600\ \text{rev/min}}$.
Prototype flow. From $\dfrac{Q}{ND^{3}}=\text{const}$, $Q_p=Q_m\dfrac{N_p}{N_m}\left(\dfrac{D_p}{D_m}\right)^{3}=0.085\left(\dfrac{600}{3600}\right)(10)^{3}=\boxed{14.17\ \text{m}^3/\text{s}}$.
Ideal power. $P_p=\rho g Q_p H_p=1000(9.81)(14.17)(110)=\boxed{15.29\ \text{MW}}$.
Anticipated efficiency (Moody). The full Moody relation $1-\eta_p=(1-\eta_m)\left(\dfrac{D_m}{D_p}\right)^{1/4}\left(\dfrac{H_m}{H_p}\right)^{1/10}$ gives $\eta_p=91.9\%$; the diameter-only approximation $1-\eta_p=(1-\eta_m)\left(\dfrac{D_m}{D_p}\right)^{1/5}$ gives $\eta_p=89.9\%$. Report $\boxed{\eta_p\approx 92\%}$.
