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 2017 — 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.); 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.
Check: This paper, although listed under “Marine Control Systems”, is printed and headed “98-Mar-B5, Fluid Machinery” throughout, with zero marine-control-systems content (no PID loops, governors, or automation). It is solved here exactly as printed. Question 2's design method (annulus sizing from mass flow, 50% reaction symmetric velocity triangles) follows the standard preliminary axial-compressor design procedure of Dixon & Hall ch. 3–5; since no stage/compressor efficiency is supplied, the number of stages is obtained from the ideal (isentropic) overall temperature rise divided by the actual work done per stage — the standard simplification for this class of preliminary-design problem. Question 3 assumes zero exit whirl at the Francis runner outlet (design/best-efficiency condition), the standard assumption when no exit-blade data is given.
Find. Rotational speed N, shaft power P, number of nozzles, and dimensionless specific speed.
Approach. Get the jet velocity from the head and nozzle coefficient, the blade speed from the given speed ratio, and hence the rotational speed from the wheel diameter. Use ideal (frictionless, 180°-turning) bucket theory with the mechanical efficiency to get shaft power, size one jet's flow from its diameter and divide into the total flow for the nozzle count, then substitute into the reference-sheet specific-speed formula.
(a) Jet velocity and wheel speed. $V_{jet}=K\sqrt{2gH}=0.98\sqrt{2(9.81)(200)}=61.39\ \text{m/s}$. The blade (bucket) speed is $U=0.47\,V_{jet}=28.85\ \text{m/s}$, and with $U=\pi DN/60$,$$N=\frac{60U}{\pi D}=\frac{60(28.85)}{\pi(1.47)}.$$$$\boxed{N=375\ \text{rev/min}}$$
(b) Power output. For an ideal frictionless bucket turning the jet through 180°, the force on the wheel is $F=2\rho Q(V_{jet}-U)$ and the runner power is $P_{bucket}=FU=2\rho QU(V_{jet}-U)$. Equivalently, the ideal bucket hydraulic efficiency is $\eta_h=4(U/V_{jet})(1-U/V_{jet})=4(0.47)(0.53)=0.996$ of the jet's kinetic power $\tfrac12\rho QV_{jet}^2$. Applying only the mechanical efficiency beyond this ideal bucket result,$$P=\eta_{mech}\times2\rho QU(V_{jet}-U)=0.88\times2(1000)(4)(28.85)(61.39-28.85).$$$$\boxed{P=6.61\ \text{MW}}$$
(c) Number of nozzles. Each jet's diameter is $d=0.113D=0.113(1.47)=0.1661\ \text{m}$, so its area is $A_j=\tfrac{\pi}{4}d^2=0.02167\ \text{m}^2$ and its flow is $Q_j=A_jV_{jet}=0.02167(61.39)=1.330\ \text{m}^3/\text{s}$. Dividing the total flow among identical jets,$$n=\frac{Q}{Q_j}=\frac{4}{1.330}=3.01\ \rightarrow\ \boxed{n=3\ \text{nozzles}}$$
(d) Specific speed. $\omega=2\pi N/60=2\pi(374.9)/60=39.26\ \text{rad/s}$; $gH=9.81(200)=1962\ \text{m}^2/\text{s}^2$. Using the reference-sheet turbine specific speed,$$N_s=\frac{\omega\sqrt{P}}{\rho^{1/2}(gH)^{5/4}}=\frac{39.26\sqrt{6.61\times10^6}}{(1000)^{1/2}(1962)^{1.25}}.$$$$\boxed{N_s=0.244}$$(a low value, consistent with the low-specific-speed, high-head Pelton design region of the attached efficiency chart.)
Check: Power is computed from ideal (frictionless, full 180° deflection) bucket momentum theory, since no bucket friction coefficient or exit deflection angle is supplied — only the mechanical efficiency is applied beyond this ideal hydraulic result. A real bucket (finite friction, ~165° deflection) would give a slightly lower power for the same flow and speed; the sizing method for N, nozzle count and specific speed is unaffected.
