Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Mec-B6 Fluid Machinery,
National Examinations May 2017 — three hours, closed book. Section A is
calculative (Questions 1–5) and Section B descriptive (Questions 6–8); the rubric asks for
four of Section A and two of Section B, six questions of ten marks each for a sixty-mark paper.
Reference data for individual questions are supplied as Attachments (pages 9–11) and a general
constants/equations sheet occupies pages 12–16. All eight questions are solved
here, because the set is a study resource rather than a timed attempt.
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.;
F. M. White, Fluid Mechanics, 8th ed.;
R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics, 9th ed.;
Y. A. Çengel & M. A. Boles, Thermodynamics: An Engineering Approach, 9th ed.
Constants are those printed on page 13 of the paper
(g = 9.81 m/s², ρwater = 1000 kg/m³,
ρair = 1.21 kg/m³ at 15 °C and 1.19 kg/m³
at 20 °C, cp = 1.005 kJ/kg·°C,
cv = 0.718 kJ/kg·°C, patm = 100 kPa,
pvapour = 2.34 kPa).
Subject note. Page 1 of the examination reads
16-MEC-B6 FLUID MACHINERY, and every question is a turbomachine question. The
solutions below answer the paper as printed.
Given. Two identical induced-draught fans in parallel on one exhaust
system:
Quantity
Symbol
Value
Fan characteristic
H
K1 − K2Q − K3Q² (kPa)
System characteristic
h
K4Q² (kPa)
Speed-dependent shut-off head
K1
4.5 × 10−6 N²
Linear fan term
K2
0.0
Quadratic fan term
K3
16.0 × 10−6
System resistance constant
K4
5.5 × 10−6
Full-load fan speed
N
1155 rev/min
Find. The two characteristics sketched with their operating points; the gas flow
with one fan and with two; the boiler load attainable on one fan as a percentage of the two-fan
maximum; and the reduced speed at which both fans together would deliver only that one-fan duty.
Figure 5.1 — head-flow characteristics at 1155 rev/min (part a). The single-fan curve, the parallel (two-fan) curve and the system resistance are shown; the marked intersections are the two operating points.
Approach. An operating point is simply where the fan characteristic meets
the system resistance. Two identical fans in parallel each pass half the total flow, so the combined
characteristic is the single-fan curve with Q replaced by Q/2 — which quarters
the quadratic term. Because K2 is zero, every algebraic step reduces to a single square
root, and the fan law relating head to speed is exact.
Evaluate the speed-dependent constant. At full load,
$$K_1=4.5\times 10^{-6}\,N^2=4.5\times 10^{-6}\times 1155^2=6.003\ \text{kPa}$$
This is the shut-off (zero-flow) head of one fan.
Part (a) — construct the two fan characteristics. One fan follows
$H=6.003-16.0\times 10^{-6}Q^2$. For two identical fans in parallel the head
is common and the flows add, so each machine handles Q/2 and the pair follows
$$H=K_1-K_3\left(\frac{Q}{2}\right)^{2}=6.003-4.0\times 10^{-6}\,Q^2$$
which is a visibly flatter curve reaching the same shut-off head. The system resistance
$h=5.5\times 10^{-6}Q^2$ rises from the origin. Figure 5.1 is the required
sketch with both intersections marked.
Part (b) — single-fan duty point. Equating fan head to system head,
$$K_1-K_3Q^2=K_4Q^2\qquad\Longrightarrow\qquad Q=\sqrt{\frac{K_1}{K_3+K_4}}
=\sqrt{\frac{6.003}{21.5\times 10^{-6}}}$$
$$\boxed{Q_{1\,fan}=528\ \text{m}^3/\text{s}}$$
at a system head of $K_4Q^2=1.54$ kPa.
Part (c) — two-fan duty point. Repeating with the parallel characteristic,
$$K_1-\frac{K_3}{4}Q^2=K_4Q^2\qquad\Longrightarrow\qquad
Q=\sqrt{\frac{K_1}{K_3/4+K_4}}=\sqrt{\frac{6.003}{9.5\times 10^{-6}}}$$
$$\boxed{Q_{2\,fans}=795\ \text{m}^3/\text{s}}$$
at a system head of 3.48 kPa. Note that doubling the fans has not doubled the flow —
it has raised it by only 50 %, because the steeply rising system resistance absorbs most of the extra
capability.
Part (d) — attainable boiler load on one fan. Combustion gas flow is
proportional to firing rate, so the load ratio is the flow ratio:
$$\text{load}=\frac{Q_{1\,fan}}{Q_{2\,fans}}\times 100=\frac{528.4}{794.9}\times 100$$
$$\boxed{\text{load}=66.5\ \%\ \text{of maximum}}$$
This is the practical value of the result: losing one ID fan does not halve the station output, it
costs about a third of it, so the unit can stay on line at two-thirds load while the failed fan is
repaired.
Part (e) — speed of both fans for the one-fan duty. Both fans running
together must now meet the system at $Q=528.4$ m³/s. Writing the parallel
characteristic at the unknown speed N′ and equating to the system head,
$$4.5\times 10^{-6}N'^2-\frac{K_3}{4}Q^2=K_4Q^2
\qquad\Longrightarrow\qquad N'=\sqrt{\frac{\left(K_4+K_3/4\right)Q^2}{4.5\times 10^{-6}}}$$
$$N'=\sqrt{\frac{9.5\times 10^{-6}\times 528.4^2}{4.5\times 10^{-6}}}$$
$$\boxed{N'=768\ \text{rev/min}}$$
that is 66.5 % of full speed.
Why the speed ratio equals the flow ratio here. Because K2 is zero,
both the fan and the system characteristics are pure quadratics in Q with heads proportional
to $N^2$, so the whole problem is homologous and
$N'/N=Q'/Q$ exactly — 768/1155 = 0.665, matching part (d) to three
figures. This coincidence is a free check on parts (d) and (e) together. The air power at the reduced
duty falls as the cube of speed, so running both fans slowly to hold two-thirds load costs about
$0.665^3=29\ \%$ of full-load fan power, against roughly half of it if one
fan alone were left at full speed.
