Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 98-Mar-B5 Fluid Machinery, December 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 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. A. Sabersky, A. J. Acosta, E. G. Hauptmann & E. M. Gates, Fluid Flow: A First Course in Fluid Mechanics (4th ed.) — source of the pump-selection charts Figs 15.11/15.12; H. Cohen, G. F. C. Rogers & H. I. H. Saravanamuttoo, Gas Turbine Theory; R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics.
the same misfiling pattern already documented for other 98-Mar-coded papers in this discipline. It is solved here exactly as printed. Separately, Questions 3 relies on chart reads from the examination attachments (Figs 15.11 & 15.12); values read from those charts (φe, σc, η) are quoted to the precision the graphs allow and are flagged where used. The induction-motor pole count is not stated in Q3, so a standard 4-pole (1800 rev/min synchronous) machine is assumed — the governing method is unaffected.
Given. Turbine (dimensionless) specific speed Ns = 0.20; effective head H = 120 m; nozzle coefficient Cv = 0.985; runner speed N = 880 rev/min; blade/jet speed ratio U/Vjet = 0.47; overall efficiency ηo = 0.88.
Given data — Pelton wheel
Quantity
Symbol
Value
Specific speed
Ns
0.20
Effective head
H
120 m
Nozzle coefficient
Cv
0.985
Runner speed
N
880 rev/min
Speed ratio
U/Vjet
0.47
Overall efficiency
ηo
0.88
Find. Shaft power P, volume flow rate Q, jet area Ajet, and wheel-to-jet diameter ratio D/d.
Approach. Invert the dimensionless turbine specific speed to get shaft power, then use the overall efficiency to back out flow, the nozzle equation for jet speed and area, and the speed ratio plus rotational speed for the two diameters.
Shaft power from specific speed. The dimensionless turbine specific speed is $N_s=\dfrac{\omega\sqrt{P}}{\rho^{1/2}(gH)^{5/4}}$ with $\omega=\dfrac{2\pi N}{60}=\dfrac{2\pi(880)}{60}=92.15\ \text{rad/s}$. Solving for power,$$P=\left[\frac{N_s\,\rho^{1/2}(gH)^{5/4}}{\omega}\right]^2=\left[\frac{0.20\,(1000)^{1/2}(9.81\times120)^{5/4}}{92.15}\right]^2.$$$$\boxed{P=224\ \text{kW}}$$
Volume flow rate from overall efficiency. The overall efficiency relates shaft power to the available water power, $\eta_o=\dfrac{P}{\rho g Q H}$, so$$Q=\frac{P}{\eta_o\,\rho g H}=\frac{224\,000}{0.88(1000)(9.81)(120)}.$$$$\boxed{Q=0.216\ \text{m}^3/\text{s}=216\ \text{L/s}}$$
Jet velocity and jet area. The nozzle converts head to jet velocity with the velocity coefficient, $V_{jet}=C_v\sqrt{2gH}=0.985\sqrt{2(9.81)(120)}=47.79\ \text{m/s}$. Continuity then gives the jet area,$$A_{jet}=\frac{Q}{V_{jet}}=\frac{0.216}{47.79}=4.52\times10^{-3}\ \text{m}^2.$$$$\boxed{A_{jet}=4.52\times10^{3}\ \text{mm}^2}$$
Wheel and jet diameters. The blade (bucket) speed is $U=0.47\,V_{jet}=22.46\ \text{m/s}$; with $U=\pi D N/60$ the wheel pitch diameter is $D=\dfrac{60U}{\pi N}=\dfrac{60(22.46)}{\pi(880)}=0.488\ \text{m}$. The jet diameter follows from $A_{jet}=\tfrac{\pi}{4}d^2$, giving $d=\sqrt{4A_{jet}/\pi}=0.0759\ \text{m}$. Hence$$\frac{D}{d}=\frac{0.488}{0.0759}=6.42.$$$$\boxed{D/d=6.4}$$
