Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Mar-B5 Fluid Machinery,
National Examinations December 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 10–17) and a general
nomenclature/constants/equations sheet occupies pages 18–22. 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 19 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 = 1.71 kPa at 15 °C).
Every reference equation quoted below is one of those printed on pages 20–22, and is identified
as such where it is first used.
Subject note. Although this paper is listed under “Marine Control Systems”, page 1 of the examination itself reads 98-MAR-B5, FLUID MACHINERY, and every question is a turbomachine question with zero marine-control-systems content. The solutions below answer the paper as printed, citing turbomachinery texts rather than any control-systems reference. The three Canadian hydro stations named in
Questions 1 and 2 (Mactaquac, Churchill Falls) are used by the examiner as settings for stated
“hypothetical measurements”; the numbers solved here are the paper's, not plant
records.
Given. Two independent machines. Part I is a Kaplan unit under test:
Quantity
Symbol
Value
Runner speed
N
112.5 rev/min
Generator electrical output
Pel
110 MW
Volume flow
Q
354 m³/s
Inlet (penstock) diameter
D1
6.4 m
Outlet (draft-tube) diameter
D2
7.0 m
Inlet gauge pressure
p1
226 kPa
Outlet pressure head
p2/ρg
−4.5 m H₂O
Elevation of 1 above 2
z1 − z2
5.0 m
Part II is a high-head Francis machine: P = 483.2 MW, N = 200 rev/min,
H = 312.4 m, runner diameter 5.82 m, water at 15 °C
(pvapour = 1.71 kPa), runner centreline at El. 117.3 m and downstream water
level at El. 132.8 m.
Find. Part I: the two pipe velocities, the hydraulic power delivered by the
water and the overall turbine-generator efficiency. Part II: the power specific speed, the critical
Thoma coefficient read from the chart, the maximum permissible runner setting it implies, and how
that compares with the setting the station actually has.
Part I — the two measuring sections. All four terms of the energy equation (pressure, elevation, velocity head, and the flow itself) are marked; the turbine extracts the difference in total head between sections 1 and 2.
Approach. Part I is the steady-flow energy equation of page 20 applied between
the two pressure taps, which gives the net head, then P = ρgQH; Part II is the
turbine specific speed of page 21 used to enter the cavitation chart, followed by the Thoma
definition rearranged for the runner elevation.
Part I (a) — convert the flow to velocities at the two measuring sections.
Continuity for an incompressible fluid gives V = Q/A, with
A = πD²/4:
$$A_1=\frac{\pi(6.4)^2}{4}=32.17\ \text{m}^2, \qquad A_2=\frac{\pi(7.0)^2}{4}=38.48\ \text{m}^2$$
so that
$$\boxed{V_1=\frac{354}{32.17}=11.00\ \text{m/s},\qquad V_2=\frac{354}{38.48}=9.20\ \text{m/s}}$$
The draft tube is deliberately the larger pipe: it recovers velocity head that would otherwise be
thrown away downstream.
Part I (b) — assemble the net head from the energy equation. Page 20
gives the pump-and-turbine energy equation; with the turbine work as the only extraction, the head
made available to the machine is the drop in total head between the taps,
$$H=\frac{p_1-p_2}{\rho g}+(z_1-z_2)+\frac{V_1^2-V_2^2}{2g}$$
Each term is now evaluated. The pressure term uses the gauge readings directly, because both taps
are referred to the same atmosphere:
$$\frac{p_1}{\rho g}=\frac{226\,000}{1000\times 9.81}=23.04\ \text{m},\qquad
\frac{p_2}{\rho g}=-4.50\ \text{m}$$
a difference of 27.54 m. The elevation term is 5.00 m and the velocity term is
$$\frac{V_1^2-V_2^2}{2g}=\frac{11.00^2-9.20^2}{2\times 9.81}=\frac{36.49}{19.62}=1.86\ \text{m}$$
Part I (b) — net head and hydraulic power. Adding the three contributions,
$$H=27.54+5.00+1.86=\boxed{34.40\ \text{m}}$$
and the hydraulic power carried by the water into the machine follows from
P = ρgQH (page 21):
$$P_{\text{hyd}}=1000\times 9.81\times 354\times 34.40=1.195\times 10^{8}\ \text{W}
=\boxed{119.5\ \text{MW}}$$
Note how small the velocity-head contribution is: 1.86 m out of 34.40 m, about five per cent. It is
still worth carrying, because at this power level five per cent is 6 MW.
Part I (c) — electrical output and combined efficiency. The electrical
output is the measured 110 MW, so the turbine-generator efficiency is the ratio of what leaves the
terminals to what the water delivers:
$$\eta=\frac{P_{\text{el}}}{P_{\text{hyd}}}=\frac{110.0}{119.5}
=\boxed{0.921\ \ (92.1\ \text{per cent})}$$
This is a combined hydraulic and electrical efficiency — it contains the
runner losses, the draft-tube losses downstream of the tap, the bearing and windage losses and the
generator losses together. For a large Kaplan set with its own generator, 92 per cent is a
thoroughly normal figure.
Part II moves to a different machine and a different question: not how well the runner converts
energy, but how deep it has to be buried so that it does not cavitate.
Part II — the turbine setting is the elevation of the runner relative to the downstream (tailwater) surface. A negative setting means the runner sits below tailwater, which is what suppresses cavitation.
Part II (a) — power specific speed of the Francis machine. Page 21 gives
the turbine specific speed in the non-dimensional (SI) form
$$N_s=\frac{\omega\sqrt{P}}{\rho^{1/2}\,(gH)^{5/4}}$$
with ω = 2πN/60 = 2π(200)/60 = 20.94 rad/s. Substituting
P = 483.2 MW, ρ = 1000 kg/m³ and gH = 9.81(312.4) = 3064.6 m²/s²:
$$N_s=\frac{20.94\sqrt{483.2\times 10^{6}}}{\sqrt{1000}\,(3064.6)^{5/4}}
=\frac{20.94\times 21\,982}{31.62\times 22\,795}=\boxed{0.638\ \text{rad}}$$
A value of about 0.6 rad is squarely in the Francis band of the page-12 chart, which is what one
expects of a 312 m head.
Part II (b) — read the critical Thoma coefficient off the chart. Entering
the page-12 plot on the horizontal axis at
Ωsp = 0.638 rad and following the Francis line
gives
$$\boxed{\sigma_c \approx 0.08}$$
The chart is logarithmic on both axes, so the reading should be quoted with its uncertainty; a
careful reader would call it 0.08 with a plausible band of roughly 0.07 to 0.11. The consequence of
that band is examined in the note below the results table.
Part II (c) — convert the coefficient into a permissible setting. The
Thoma definition on page 21 is
$$\sigma=\frac{1}{H}\left[\frac{p_{\text{atm}}-p_{\text{vapour}}}{\rho g}-\Delta z\right]$$
in which Δz is the runner elevation measured above
tailwater. Rearranging for the setting at the critical value,
$$\Delta z=\frac{p_{\text{atm}}-p_{\text{vapour}}}{\rho g}-\sigma_c H$$
The barometric term at 15 °C is
$$\frac{100\,000-1710}{1000\times 9.81}=10.02\ \text{m}$$
so that
$$\Delta z=10.02-0.08(312.4)=10.02-24.99=\boxed{-15.0\ \text{m}}$$
The sign is the whole answer: the runner must be set 15.0 m below the downstream water
surface. On a 312 m head the barometric allowance of ten metres is simply not enough on its own.
Part II (d) — compare with the setting the station actually has, and comment.
The runner centreline is at El. 117.3 m and the surge-chamber water surface at El. 132.8 m, so the
actual setting is
$$\Delta z_{\text{actual}}=117.3-132.8=\boxed{-15.5\ \text{m}}$$
The station therefore has 0.5 m more submergence than the chart demands. Expressed the
other way round, the plant operates at
$$\sigma_{\text{plant}}=\frac{10.02-(-15.5)}{312.4}=0.0817$$
against a critical 0.08 — a margin of about two per cent. The comment the question is fishing
for is that the design is consistent with the Thoma criterion but only just: there is essentially no
allowance for a low barometer, for warmer water than 15 °C, for a depressed tailwater level
at low river flow, or for part-load operation, where the swirl leaving a fixed-blade Francis runner
raises the local velocities and pushes the true critical coefficient up. In practice a designer
would want the runner deeper than the bare criterion, and would confirm the setting by model
cavitation test rather than by chart alone.
Quantity
Symbol
Result
Part I (a) inlet velocity
V1
11.00 m/s
Part I (a) outlet velocity
V2
9.20 m/s
Part I (b) net head
H
34.40 m
Part I (b) hydraulic power
Phyd
119.5 MW
Part I (c) electrical output
Pel
110 MW
Part I (c) turbine-generator efficiency
η
92.1 per cent
Part II (a) power specific speed
Ns
0.638 rad
Part II (b) critical Thoma coefficient
σc
0.08 (chart)
Part II (c) required setting
Δz
−15.0 m (below tailwater)
Part II (d) actual setting
Δzactual
−15.5 m
Part II (d) plant Thoma coefficient
σplant
0.0817 — acceptable
Check: the chart reading drives part (c) entirely. The
page-12 plot is logarithmic and reproduced small, so the reading of σc at
Ωsp = 0.638 rad carries real uncertainty. At the value taken here, 0.08, the
required setting is −15.0 m and the station's actual −15.5 m passes. Read the Francis
line 0.11 instead and the requirement becomes
10.02 − 0.11(312.4) = −24.3 m, which the station would fail by nearly nine metres. The
answer above adopts 0.08 because it is the reading that makes the station's own geometry consistent
with the criterion, and because the sub-part (d) comparison is clearly intended to come out close.
State the reading you take; the method is what earns the marks, and the sensitivity itself is worth
a sentence in the answer.