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.
Given. Fan curve $H=K_1-K_3Q^2$ (K2=0); system curve $h=K_4Q^2$; $K_1=4.5\times10^{-6}N^2$; $K_3=16.0\times10^{-6}$; $K_4=5.5\times10^{-6}$; $N=1155\ \text{rev/min}$.
Find. Q with one fan; Q with both fans in parallel; the one-fan load as a % of the two-fan maximum; and the speed both fans must run at to reproduce the one-fan flow.
Approach. First evaluate the constant K1 at the given speed. For a single fan, equate its H–Q curve to the system curve. For two identical fans in parallel, each carries half the total flow at the shared head, so replace Q by Q/2 in the fan-curve term before equating to the system curve. The load ratio follows directly from the two flows, and the required new speed follows by re-scaling K1 (which is the only speed-dependent constant, per the fan affinity laws built into this curve family) so that the two-fan operating point reproduces the target flow.
(a) Fan shut-off head at N = 1155 rev/min (the sketch itself, part (a), is given as Figs. Q5(i)/(ii) below, with the operating points marked). $$K_1=4.5\times10^{-6}(1155)^2=6.00\ \text{kPa}.$$
(b) One fan operating. Equating fan and system curves, $K_1-K_3Q_b^2=K_4Q_b^2$, so$$Q_b=\sqrt{\frac{K_1}{K_3+K_4}}=\sqrt{\frac{6.00}{(16.0+5.5)\times10^{-6}}}.$$$$\boxed{Q_b=528\ \text{m}^3/\text{s}}\quad(H_b=K_4Q_b^2=1.54\ \text{kPa})$$
(c) Both fans in parallel. Each fan supplies Qc/2, so the combined characteristic is $H=K_1-K_3(Q_c/2)^2$; equating to the system curve,$$K_1-K_3\left(\frac{Q_c}{2}\right)^2=K_4Q_c^2\ \Rightarrow\ Q_c=\sqrt{\frac{K_1}{K_4+K_3/4}}.$$$$\boxed{Q_c=795\ \text{m}^3/\text{s}}\quad(H_c=K_4Q_c^2=3.48\ \text{kPa})$$
Fig. Q5(i) - One fan operating: fan curve H=K1-K3Q^2 vs. system curve h=K4Q^2, operating point at Q~528 m3/s.
Fig. Q5(ii) - Both fans in parallel: combined fan curve H=K1-K3(Q/2)^2 vs. the same system curve, operating point at Q~795 m3/s.
(d) Load with one fan, as a % of the two-fan maximum. Boiler (gas) load scales directly with exhaust flow rate, so$$\%\text{load}=\frac{Q_b}{Q_c}\times100=\frac{528}{795}\times100.$$$$\boxed{66.5\%}$$
(e) Speed for both fans to match the one-fan flow. Target flow is $Q_b=528\ \text{m}^3/\text{s}$ with both fans running at a new speed $N'$ (new $K_1'=4.5\times10^{-6}N'^2$, since $K_3$ and $K_4$ are geometry/system constants independent of speed). Two fans in parallel at $Q_b$ require$$K_1'=K_4Q_b^2+K_3\left(\frac{Q_b}{2}\right)^2=(5.5\times10^{-6})(528)^2+(16.0\times10^{-6})(264)^2=2.653\ \text{kPa},$$$$N'=\sqrt{\frac{K_1'}{4.5\times10^{-6}}}.$$$$\boxed{N'=768\ \text{rev/min}}$$
Check: The fan curve family $H=K_1(N)-K_3Q^2$ with $K_1\propto N^2$ and $K_3$ speed-independent is exactly the form implied by the fan affinity laws ($Q\propto N$, $H\propto N^2$) for a fixed-geometry machine, which is why step 5 can re-scale $K_1$ alone to find the new speed. Boiler load is taken as directly proportional to exhaust gas volume flow (combustion air/gas flow scales with fuel burned); the problem gives no separate load-vs-flow calibration.
