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 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 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. K. Turton, Principles of Turbomachinery, 2nd ed.; H. Cohen, G. F. C. Rogers & H. I. H. Saravanamuttoo, Gas Turbine Theory, 6th ed.; 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, 8th ed.; Y. A. Çengel & J. M. Cimbala, Fluid Mechanics: Fundamentals and Applications.
It is solved here exactly as printed.
Question 5: Boiler Draught Fans in Parallel (10 marks)
Given. Fan curve $H=K_1-K_3Q^{2}$ (K2 = 0); system curve $h=K_4Q^{2}$; $K_1=4.5\times10^{-6}(1155)^2=6.003\ \text{kPa}$; $K_3=16.0\times10^{-6}$; $K_4=5.5\times10^{-6}$; N = 1155 rev/min.
Find. (a) the characteristic sketch and operating points; (b) one-fan flow; (c) two-fan flow; (d) one-fan load as % of two-fan; (e) the fan speed for two fans to match the one-fan load.
Operating points are the intersections of the system resistance (dashed) with the one-fan and two-fan characteristics. Two fans in parallel double the flow at a given head, shifting the operating point out to B.
Approach. At each operating point the fan head equals the system head; for two identical fans in parallel each carries half the total flow, so replace Q by Q/2 in the single-fan curve to build the combined characteristic.
(b) One fan. Setting fan = system: $K_1-K_3Q^{2}=K_4Q^{2}\Rightarrow Q=\sqrt{\dfrac{K_1}{K_3+K_4}}=\sqrt{\dfrac{6.003}{21.5\times10^{-6}}}=\boxed{528\ \text{m}^3/\text{s}}$.
(c) Both fans. Each fan passes Q/2, so $K_1-K_3(Q/2)^{2}=K_4Q^{2}\Rightarrow Q=\sqrt{\dfrac{K_1}{K_3/4+K_4}}=\sqrt{\dfrac{6.003}{9.5\times10^{-6}}}=\boxed{795\ \text{m}^3/\text{s}}$.
(d) One-fan load fraction. Boiler load scales with the exhaust-gas flow, so one fan gives $\dfrac{528}{795}=\boxed{66.5\%}$ of the two-fan maximum. Doubling the fans yields only ~1.5× the flow because the steep system resistance rises quickly with Q.
(e) Speed for two fans to match one-fan load. Require the two-fan flow to equal 528 m³/s: $4.5\times10^{-6}N'^{2}-\dfrac{K_3}{4}(528)^2=K_4(528)^2$, giving $N'=\sqrt{\dfrac{(K_4+K_3/4)(528)^2}{4.5\times10^{-6}}}=\boxed{768\ \text{rev/min}}$.
