Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examination 07-Mec-A6-1 Fluid Machinery (May 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 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; H. Cohen, G. F. C. Rogers & H. I. H. Saravanamuttoo, Gas Turbine Theory (for the axial compressor stage); R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics (pump energy equation and affinity laws).
Question 5: Boiler Draught Fans in Parallel (10 marks)
Given. Single-fan curve $H=K_1-K_3Q^2$ (K₂ = 0); system curve $h=K_4Q^2$; K₁ = 4.5×10⁻⁶N² = 6.003 kPa at N = 1155; K₃ = 16.0×10⁻⁶; K₄ = 5.5×10⁻⁶.
Given data — Q5
Constant
Value
Constant
Value
K₁ (at 1155 rpm)
6.003 kPa
K₃
16.0×10⁻⁶
K₂
0
K₄ (system)
5.5×10⁻⁶
Find. (b) one-fan flow, (c) two-fan flow, (d) one-fan load as % of two-fan, (e) reduced speed of both fans matching one-fan load.
Head–flow characteristics. One-fan and two-fan (parallel: flow doubled at each head) fan curves meet the rising system curve at operating points A (528 m³/s) and B (795 m³/s).
Approach. The operating point is where the fan curve meets the system curve. For two identical fans in parallel the combined curve is the single-fan curve with Q replaced by Q/2 (each fan supplies half the total flow at the common head).
Fan constant at full speed. $K_1=4.5\times10^{-6}(1155)^2=6.003$ kPa, so one fan gives $H=6.003-16.0\times10^{-6}\,Q^2$.
(b) One fan operating. Set fan = system:
$$6.003-16.0\times10^{-6}Q^2=5.5\times10^{-6}Q^2\ \Rightarrow\ Q^2=\frac{6.003}{21.5\times10^{-6}},$$
$$\boxed{Q_{1}=528\ \text{m}^3/\text{s}.}$$
(c) Both fans in parallel. Replace $Q\to Q/2$ in the fan curve, giving $H=6.003-4.0\times10^{-6}Q^2$; set equal to the system:
$$6.003=(4.0+5.5)\times10^{-6}\,Q^2\ \Rightarrow\ Q_{2}=795\ \text{m}^3/\text{s}.$$
(d) One-fan load fraction. Boiler load scales with the extracted gas flow:
$$\frac{Q_1}{Q_2}=\frac{528}{795}=66.5\%.$$
So a single fan still carries about two-thirds of full boiler load — the steep system curve blunts the benefit of the second fan.
(e) Reduced speed for both fans. Both fans must together pass $Q_1=528$ m³/s, at which the system head is $h=5.5\times10^{-6}(528)^2=1.536$ kPa. The combined two-fan curve at speed N′ is $H=4.5\times10^{-6}N'^2-4.0\times10^{-6}Q^2$; setting it to 1.536 kPa at Q = 528:
$$4.5\times10^{-6}N'^2=1.536+4.0\times10^{-6}(528)^2=2.652\ \Rightarrow\ N'=768\ \text{rev/min}.$$
