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 5: Boiler Draught Fans in Parallel (10 marks)
Given. Shutoff head K1 = 4.5×10−6(1155)² = 6.003 kPa; fan H = 6.003 − 16.0×10−6Q²; system h = 5.5×10−6Q². Parallel fans add flow at equal head.
Find. Operating flows for one and two fans, the load ratio, and the reduced fan speed.
Head–flow characteristics: one fan (blue), two fans in parallel (red), and the system resistance (green). Operating points are the intersections with the system curve.
Approach. Intersect each fan characteristic with the system curve; for two fans in parallel, replace Q by Q/2 in the single-fan head equation (each fan carries half the total flow at the common head).
(b) One fan. Set fan = system: $6.003-16\times10^{-6}Q^2=5.5\times10^{-6}Q^2\Rightarrow Q^2=\dfrac{6.003}{21.5\times10^{-6}}$, so $\boxed{Q_1=528\ \text{m}^3/\text{s}}$ at h = 1.54 kPa.
(c) Both fans in parallel. The combined characteristic is $H=6.003-16\times10^{-6}(Q/2)^2=6.003-4\times10^{-6}Q^2$. Set equal to the system: $6.003=9.5\times10^{-6}Q^2$, so $\boxed{Q_2=795\ \text{m}^3/\text{s}}$ at h = 3.48 kPa.
(d) One-fan load as a percentage. $\dfrac{Q_1}{Q_2}=\dfrac{528}{795}=66.5\%$ — one fan alone still delivers two-thirds of maximum boiler load because the system resistance rises steeply with flow.
(e) Speed of both fans to match the one-fan load. Both fans share Q1 = 528 m³/s, so each carries 264 m³/s against the system head at Q1 (1.54 kPa). With $4.5\times10^{-6}N'^{2}-16\times10^{-6}(264)^2=1.54$, solve $N'=\sqrt{\dfrac{1.54+1.117}{4.5\times10^{-6}}}$, so $\boxed{N'\approx768\ \text{rev/min}}$.
