22-Mec-B6 Advanced Fluid Mechanics · December 2019

Question 5 of 8: Hydro Turbines

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Examinations, December 2019 — 16-Mec-B6 Fluid Machinery. Closed book, three hours, 22 pages. Section A is calculative (Questions 1–5) and Section B descriptive (Questions 6–8); a candidate answers four questions from Section A and two from Section B — six questions of ten marks each, 60 marks in all. Reference data for individual questions are supplied in the Attachments on pages 12–17, and the nomenclature, general constants and reference equations on pages 18–22. All eight questions are solved below, because this set is a study resource rather than a three-hour sitting.

Check — angle conventions are taken from the paper's own attachments, and on this sitting they are uniform. The gas-turbine velocity diagram on page 13 strikes $\theta$, $\phi$, $\gamma$ and $\delta$ off the tangential direction (the plane of rotation); the pump diagram on page 14 strikes $\alpha_1$, $\alpha_2$, $\beta_1$ and $\beta_2$ off the tangential direction as well; and the Francis diagram on page 15 defines $\alpha_1$ between $V_1$ and $u_1$, again tangential. Every angle below is therefore measured from the direction of blade motion. Every constant used is the paper's own page-19 value: $g = 9.81\ \text{m}\,\text{s}^{-2}$, $\rho_{\text{water}} = 1000\ \text{kg}\,\text{m}^{-3}$, $p_{\text{atm}} = 100\ \text{kPa}$ and $p_{\text{vapour}} = 1.71\ \text{kPa}$ at $15^{\circ}\text{C}$.

Reference texts

Dixon, S. L. and Hall, C. A., Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed. — Ch. 1 (specific speed), Ch. 4 (axial turbines), Ch. 7 (centrifugal pumps), Ch. 9 (hydraulic turbines, Thoma cavitation parameter). The efficiency chart on page 15 and the cavitation chart on page 17 of the examination paper are reproduced from this text.
Cohen, H., Rogers, G. F. C. and Saravanamuttoo, H. I. H., Gas Turbine Theory, 7th ed. — Ch. 5 (axial compressors, stage loading and stall), Ch. 7 (axial turbines, stage work and blade angles).
Turton, R. K., Principles of Turbomachinery, 2nd ed. — Ch. 2 (the Euler equation), Ch. 5 (radial machines, vane number and slip), Ch. 6 (hydraulic turbines).
White, F. M., Fluid Mechanics, 8th ed. — Ch. 11 (turbomachinery: pump characteristics, specific speed, Pelton, Francis and Kaplan turbines, cavitation). The pump characteristic curves reproduced on page 11 of the examination paper come from this text.
Fox, R. W., McDonald, A. T. and Pritchard, P. J., Introduction to Fluid Mechanics, 10th ed. — Ch. 10 (fluid machinery, Euler turbomachine equation, cavitation and NPSH).
Douglas, J. F., Gasiorek, J. M. and Swaffield, J. A., Fluid Mechanics, 6th ed. — Ch. 23 (rotodynamic machines, impeller velocity triangles).

Question 5 — Hydro Turbines (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two real machines. Part I is the six-nozzle vertical-shaft Pelton wheel at Bridge River in British Columbia, quoted in imperial units; Part II is a 120 MW Francis unit at Vanderkloof, quoted in SI.

Quantity	Symbol	Part I — Bridge River	Part II — Vanderkloof
Head on the turbine	$H$	gross $1226\ \text{ft} = 373.7\ \text{m}$; net $1118\ \text{ft} = 340.8\ \text{m}$	$65\ \text{m}$
Power output	$P$	$62\,000\ \text{HP} = 46.25\ \text{MW}$	$120\ \text{MW}$
Rotational speed	$N$	$300\ \text{rev}\,\text{min}^{-1}$	$125\ \text{rev}\,\text{min}^{-1}$
Runner / pitch diameter	$D$	$95\ \text{in} = 2.413\ \text{m}$	$5.462\ \text{m}$
Other data	—	6 nozzles	$Q = 217\ \text{m}^3\,\text{s}^{-1}$, spiral-casing inlet 7 m, 11 kV

Find. For Part I, the blade-speed ratio, its percentage departure from the ideal with a physical reason, and the flow the plant must pass. For Part II, the power specific speed, the critical Thoma coefficient read from the page-17 chart, and the maximum elevation of the runner above the tailrace.

Approach. Part I is a units exercise wrapped around one dimensionless group: convert everything to SI, form $U$ and $\sqrt{2gH_{\text{net}}}$, and compare their ratio with the theoretical optimum of one half. Part II forms the same power specific speed used in Question 4, reads the critical cavitation parameter off the Francis curve, and rearranges the page-21 Thoma definition for the elevation.

Part I — Pelton wheel

Convert the data to SI. $$H_{\text{net}} = 1118 \times 0.3048 = 340.8\ \text{m},\qquad H_{\text{gross}} = 1226 \times 0.3048 = 373.7\ \text{m}$$ $$P = 62\,000 \times 746 = 46.25\ \text{MW},\qquad D = 95 \times 0.0254 = 2.413\ \text{m}$$ The 33 m difference between gross and net head is the penstock friction loss — nearly 9 per cent, which is normal for a long high-head penstock and is exactly why the net head is the one that drives the jet.
Actual blade velocity and anticipated jet velocity (part a). The bucket centre moves at the pitch-circle speed, and the jet is anticipated from the net head as a free jet: $$U = \frac{\pi D N}{60} = \frac{\pi \times 2.413 \times 300}{60} = 37.90\ \text{m}\,\text{s}^{-1}$$ $$V_{\text{jet}} = \sqrt{2 g H_{\text{net}}} = \sqrt{2 \times 9.81 \times 340.8} = 81.77\ \text{m}\,\text{s}^{-1}$$ so the ratio asked for is $$\boxed{\varphi = \frac{U}{V_{\text{jet}}} = \frac{37.90}{81.77} = 0.4635}$$

The Bridge River machine: a vertical-shaft Pelton wheel with six nozzles arranged around a single runner. Multiple nozzles let one wheel pass six times the flow of a single jet without increasing the bucket size, which is how a Pelton unit reaches 46 MW at a head of 341 m.

Deviation from the ideal ratio, and why (part b). The theoretical optimum for a Pelton wheel is $\varphi = 0.5$, so $$\boxed{\text{deviation} = \frac{0.4635 - 0.5}{0.5} \times 100 = -7.3\ \text{per cent}}$$ There are three good reasons, and they all push the same way. The first, and the strongest, is that the wheel is directly coupled to a synchronous generator: at 300 rev/min on a 60 Hz system the machine must have 24 poles, and the speed is therefore imposed by the grid, not chosen to suit the hydraulics. The second is that the real jet is slower than the anticipated free jet, because the nozzle has a velocity coefficient of about 0.98; measured against the actual jet the ratio is $0.4635/0.98 = 0.473$, much closer to the ideal. The third is that windage, bearing friction and bucket friction shift the true efficiency peak of a real wheel slightly below $\varphi = 0.5$, so operating a little slow costs almost nothing — the penalty at $\varphi = 0.4635$ is only $1 - 4\varphi(1-\varphi) = 0.5$ per cent of the ideal peak.
Volume flow rate required (part c). The flow needed follows from the water power at the net head: $$\boxed{Q = \frac{P}{\rho g H_{\text{net}}} = \frac{46.25\times10^{6}}{1000 \times 9.81 \times 340.8} = 13.84\ \text{m}^3\,\text{s}^{-1}}$$ which is $13.84/6 = 2.31\ \text{m}^3\,\text{s}^{-1}$ through each of the six nozzles.

Check — no efficiency is stated for Part I, so the flow above is the theoretical minimum. The 13.84 m³/s figure assumes every joule of the net head reaches the shaft. That is very nearly what the blade-speed ratio alone predicts — an ideal bucket at $\varphi = 0.4635$ recovers $4\varphi(1-\varphi) = 99.5$ per cent of the jet energy — but a real machine also loses in the nozzle, the buckets and the bearings. At a realistic overall efficiency of 88 per cent the plant would actually need $46.25\times10^{6}/(0.88 \times 1000 \times 9.81 \times 340.8) = 15.7\ \text{m}^3\,\text{s}^{-1}$, or 2.62 m³/s per nozzle. Both figures are quoted so the reader can see which assumption is doing the work; the boxed answer is the one the question's wording (no efficiency given) supports.

Part II — turbine setting at Vanderkloof

Power specific speed (part a). With $\omega = 2\pi \times 125/60 = 13.09\ \text{rad}\,\text{s}^{-1}$, $gH = 9.81 \times 65 = 637.65$ and $(gH)^{5/4} = 3204.9$, $$\boxed{\Omega_{sp} = \frac{\omega\sqrt{P}}{\rho^{1/2}(gH)^{5/4}} = \frac{13.09 \times \sqrt{120\times10^{6}}}{31.623 \times 3204.9} = 1.415\ \text{rad}}$$ The page-15 efficiency chart puts $\Omega_{sp} = 1.4$ squarely in the Francis band, which is a useful confirmation that the machine type stated is the right one for this head and speed.
Read the Thoma coefficient off the page-17 chart (part b). Entering the chart at $\Omega_{sp} = 1.415$ and rising to the Francis curve gives $$\boxed{\sigma_c \approx 0.23}$$ The chart is logarithmic on both axes; digitising the printed Francis curve shows it follows $\sigma_c \approx 0.132\,\Omega_{sp}^{1.60}$ across the range 0.3 to 2.2, which returns 0.231 at this specific speed and is quoted here as the reading rather than an eyeballed value.

[Figure not reproduced: The page-17 chart redrawn, with the reading marked. Above the Francis curve the machine is safe; below it, the pressure at the runner exit falls to vapour pressure somewhere on the blade and cavitation sets in. The higher the specific speed, the deeper the runner must be buried. See the official exam paper.]

Maximum runner elevation above tailwater (part c). The page-21 definition of the critical cavitation parameter is $$\sigma = \frac{\dfrac{p_{\text{atm}} - p_{\text{vapour}}}{\rho g} - \Delta z}{H}$$ The barometric head available, using the page-19 constants at $15^{\circ}\text{C}$, is $$\frac{p_{\text{atm}} - p_{\text{vapour}}}{\rho g} = \frac{100\,000 - 1710}{1000 \times 9.81} = 10.02\ \text{m}$$ Rearranging for the setting and substituting the chart reading, $$\Delta z = 10.02 - \sigma_c H = 10.02 - 0.23 \times 65 = 10.02 - 14.95$$ $$\boxed{\Delta z = -4.9\ \text{m}}$$ The negative sign is the answer's whole meaning: the runner centreline must sit about 4.9 m below the tailrace water level, not above it. At 65 m head and this specific speed the atmosphere simply cannot supply enough suction head, so the machine must be drowned.
Sanity-check the stated duty. The plant data are internally consistent: $\eta = P/(\rho g Q H) = 120\times10^{6}/(1000 \times 9.81 \times 217 \times 65) = 0.867$, a wholly believable overall efficiency for a large Francis unit, and the runner peripheral speed $U = \pi \times 5.462 \times 125/60 = 35.74\ \text{m}\,\text{s}^{-1}$ gives a velocity factor $U/\sqrt{2gH} = 1.00$, high but typical of a high-specific-speed Francis wheel.

Check — the setting is sensitive to the chart reading, so the band matters more than the point value. Reading $\sigma_c = 0.20$ instead of 0.23 gives $\Delta z = -3.0\ \text{m}$; reading 0.26 gives $-6.9\ \text{m}$. A designer would take the conservative end and add a margin of a metre or two on top, because the consequence of getting this wrong is blade erosion rather than a lost efficiency point. The reading also assumes the tailrace is at sea-level atmospheric pressure and the water is at $15^{\circ}\text{C}$; a plant on the South African highveld at 1 200 m would lose about 1.3 m of barometric head and would need to be set correspondingly deeper.

Question 5 — final results
Quantity	Symbol	Result
Part I — net head, power, pitch diameter in SI	$H$, $P$, $D$	$340.8\ \text{m}$, $46.25\ \text{MW}$, $2.413\ \text{m}$
Part I (a) — blade speed / anticipated jet speed	$U$, $V_{\text{jet}}$	$37.90$ / $81.77\ \text{m}\,\text{s}^{-1}$
Part I (a) — ratio	$\varphi$	$0.4635$
Part I (b) — deviation from the ideal 0.5	—	$-7.3\ \text{per cent}$ (synchronous speed, nozzle coefficient, friction)
Part I (c) — flow required (ideal) / per nozzle	$Q$	$13.84$ / $2.31\ \text{m}^3\,\text{s}^{-1}$
Part II (a) — power specific speed	$\Omega_{sp}$	$1.415\ \text{rad}$
Part II (b) — critical Thoma coefficient	$\sigma_c$	$0.23$
Part II (c) — maximum runner setting	$\Delta z$	$-4.9\ \text{m}$ (below tailwater)

Topic. Impulse-wheel speed ratio, and cavitation-limited setting of a reaction turbine.

Key relations. Blade speed, $U = \pi D N/60$; free jet, $V_{\text{jet}} = \sqrt{2gH_{\text{net}}}$; Pelton efficiency, $\eta = 2\varphi(1-\varphi)(1 + k\cos\beta)$, maximum at $\varphi = 0.5$; water power, $P = \rho g Q H$; power specific speed, $\Omega_{sp} = \omega P^{1/2}/\left[\rho^{1/2}(gH)^{5/4}\right]$; Thoma parameter, $\sigma = \left[(p_{\text{atm}} - p_v)/\rho g - \Delta z\right]/H$.

Why this works. The two halves of this question are the two ways a hydraulic machine is constrained. An impulse wheel is constrained kinematically: the jet gives up all its kinetic energy only if the bucket runs at exactly half jet speed, because then the water leaves the bucket with zero absolute velocity in the ideal limit. Everything a Pelton designer does — choosing the pitch diameter, choosing the pole number — is an attempt to land close to that ratio while still matching grid frequency. A reaction turbine is constrained thermodynamically: it runs full of water at pressures that fall below atmospheric on the low-pressure side of the runner, and the Thoma parameter is simply the statement that the net positive suction head available, expressed as a fraction of the working head, must exceed a value that experiment shows grows steeply with specific speed. Rearranging that inequality for elevation converts a cavitation criterion into a civil-works dimension: how deep the powerhouse must be dug.

Common pitfall. In Part I, using the gross head for the jet velocity inflates $V_{\text{jet}}$ to 85.6 m/s and drives $\varphi$ down to 0.443, which then looks like a 11 per cent deviation. The net head is the head at the nozzle and is the one that makes the jet. In Part II, dropping the minus sign in the Thoma rearrangement and reporting a setting of $+4.9\ \text{m}$ puts the runner five metres in the air, which would cavitate destructively within hours; a negative answer here is the expected result, not an error.

Source. Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed., §9.2 (Pelton wheel, speed ratio and losses) and §9.6 (cavitation in reaction turbines, Thoma coefficient); White, Fluid Mechanics, 8th ed., §11.7; Fox, McDonald & Pritchard, 10th ed., §10.5 (cavitation and NPSH).

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