22-Mec-B6 Advanced Fluid Mechanics · December 2019

Question 4 of 8: Hydro Turbine Design

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 4 — Hydro Turbine Design (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. A Francis runner specified by its dimensionless specific speed, net head, rotational speed, hydraulic efficiency, guide-vane flow angle and peripheral velocity factor.

Quantity	Symbol	Value
Power specific speed	$N_s$	$0.9\ \text{rad}$
Effective (net) head	$H$	$160\ \text{m}$
Required rotational speed	$N$	$750\ \text{rev}\,\text{min}^{-1}$
Hydraulic efficiency	$\eta_h$	$0.94$
Flow angle at runner inlet, from the tangent	$\alpha_1$	$18^{\circ}$
Peripheral velocity factor	$U_1/V_{\text{jet}}$	$0.7$
Density of water, gravity (page 19)	$\rho$, $g$	$1000\ \text{kg}\,\text{m}^{-3}$, $9.81\ \text{m}\,\text{s}^{-2}$

Find. The power the machine will produce, the flow it will swallow, the runner diameter, the radial (meridional) velocity at the runner inlet from the velocity diagram, and the runner height needed at the inlet periphery to pass that flow.

Approach. The specific speed is the only equation that contains the power, so invert it first. Everything else then cascades: flow from power and efficiency, diameter from the peripheral velocity factor, the inlet triangle from the Euler equation with zero exit swirl, and the runner height from continuity through the cylindrical inlet periphery.

Angular velocity. $$\omega = \frac{2\pi N}{60} = \frac{2\pi \times 750}{60} = 78.54\ \text{rad}\,\text{s}^{-1}$$
Invert the specific speed to get the power (part a). The page-21 turbine specific speed is $$N_s = \frac{\omega\sqrt{P}}{\rho^{1/2}\,(gH)^{5/4}} \quad\Rightarrow\quad \sqrt{P} = \frac{N_s\,\rho^{1/2}\,(gH)^{5/4}}{\omega}$$ With $gH = 9.81 \times 160 = 1569.6\ \text{m}^2\,\text{s}^{-2}$ and $(gH)^{5/4} = 9879.5$, $$\sqrt{P} = \frac{0.9 \times 31.623 \times 9879.5}{78.54} = 3580.6 \quad\Rightarrow\quad \boxed{P = 12.82\ \text{MW}}$$ The exponent $\tfrac{5}{4}$ is the one to guard: it is written as $1.25$ nowhere in the paper, and using $\tfrac{3}{4}$ (the pump specific speed exponent, also printed on page 21) would give an answer wrong by four orders of magnitude.
Water flow rate (part b). The hydraulic efficiency relates shaft output to the water power available at the net head: $$\boxed{Q = \frac{P}{\eta_h\,\rho g H} = \frac{12.82\times10^{6}}{0.94 \times 1000 \times 9.81 \times 160} = 8.69\ \text{m}^3\,\text{s}^{-1}}$$
Runner diameter (part c). The free-jet velocity corresponding to the head, and hence the runner tip speed, are $$V_{\text{jet}} = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 160} = 56.03\ \text{m}\,\text{s}^{-1},\qquad U_1 = 0.7 \times 56.03 = 39.22\ \text{m}\,\text{s}^{-1}$$ and since $U_1 = \omega D/2$, $$\boxed{D = \frac{2U_1}{\omega} = \frac{2 \times 39.22}{78.54} = 0.999\ \text{m}}$$ A design that lands within a millimetre of one metre is a strong sign that the four given parameters were reverse-engineered from a real machine.
Close the inlet velocity triangle (part d). A well-designed Francis runner discharges with no swirl, so the Euler equation for the runner reduces to a single product and gives the inlet whirl directly: $$\eta_h\,g H = U_1 V_{w1} \quad\Rightarrow\quad V_{w1} = \frac{0.94 \times 9.81 \times 160}{39.22} = 37.62\ \text{m}\,\text{s}^{-1}$$ The flow angle $\alpha_1$ is measured from the tangent (page 15 strikes it between $V_1$ and $u_1$), so the radial component is the whirl times the tangent of that angle: $$\boxed{V_{f1} = V_{w1}\tan\alpha_1 = 37.62 \times \tan 18^{\circ} = 12.22\ \text{m}\,\text{s}^{-1}}$$ and the absolute inlet velocity is $V_1 = V_{w1}/\cos 18^{\circ} = 39.56\ \text{m}\,\text{s}^{-1}$, which the triangle confirms because $V_1\sin 18^{\circ}$ returns the same 12.22 m/s.

Runner inlet velocity triangle drawn to scale. The absolute velocity $V_1$ arrives at 18° to the tangent; subtracting the runner tip speed $U_1$ leaves a short relative velocity that leans slightly forward, because at this duty the tip speed marginally exceeds the whirl the flow carries.

Check — the four given parameters are very slightly over-determined, and the inconsistency is worth stating rather than hiding. The velocity factor and the efficiency both bear on the inlet triangle. Combining $U_1 = 0.7\sqrt{2gH}$ with the zero-exit-swirl Euler condition $\eta_h g H = U_1 V_{w1}$ gives $V_{w1}/U_1 = \eta_h/(2 \times 0.7^2) = 0.959$, so $V_{w1} = 37.62\ \text{m}\,\text{s}^{-1}$ is a little less than $U_1 = 39.22\ \text{m}\,\text{s}^{-1}$ and the true runner blade angle comes out obtuse: $$\beta_1 = \arctan\!\left(\frac{V_{f1}}{V_{w1} - U_1}\right) = \arctan\!\left(\frac{12.22}{-1.60}\right) = 97.5^{\circ}$$ That is a forward-leaning leading edge, which is unusual but not impossible. Exact consistency with a radial-vane inlet ($\beta_1 = 90^{\circ}$, $V_{w1} = U_1$) would require a velocity factor of $\sqrt{\eta_h/2} = 0.686$ rather than the 0.70 stated. The answers above use the paper's own numbers throughout; the 2.0 per cent over-speed is the paper's, not the arithmetic's. Note also that the question calls $\alpha_1$ the "runner blade inlet angle", whereas page 15 defines $\alpha_1$ between $V_1$ and $u_1$, i.e. the absolute flow angle set by the guide vanes. Reading it as the flow angle is the only reading under which parts (d) and (e) are solvable, so that is the reading used.

Runner height at the inlet (part e). The flow enters through the cylindrical periphery of diameter $D$ and height $b$, so continuity gives $$\boxed{b = \frac{Q}{\pi D V_{f1}} = \frac{8.69}{\pi \times 0.999 \times 12.22} = 0.227\ \text{m}}$$ A runner one metre across with a 227 mm inlet height is a squat, high-flow Francis wheel — consistent with a dimensionless specific speed of 0.9, which the page-15 efficiency chart places right at the peak of the Francis band.

Question 4 — final results
Quantity	Symbol	Result
(a) Turbine power output	$P$	$12.82\ \text{MW}$
(b) Water flow rate	$Q$	$8.69\ \text{m}^3\,\text{s}^{-1}$
(c) Free-jet velocity / runner tip speed	$V_{\text{jet}}$, $U_1$	$56.03$ / $39.22\ \text{m}\,\text{s}^{-1}$
(c) Runner diameter	$D$	$0.999\ \text{m}$
(d) Inlet whirl / absolute velocity	$V_{w1}$, $V_1$	$37.62$ / $39.56\ \text{m}\,\text{s}^{-1}$
(d) Radial (meridional) flow velocity at inlet	$V_{f1}$	$12.22\ \text{m}\,\text{s}^{-1}$
(d) Implied runner blade inlet angle	$\beta_1$	$97.5^{\circ}$ (forward-leaning)
(e) Runner height at inlet	$b$	$0.227\ \text{m}$

Topic. Preliminary sizing of a Francis runner from dimensionless specific speed.

Key relations. Power specific speed, $N_s = \omega P^{1/2}/\left[\rho^{1/2}(gH)^{5/4}\right]$; hydraulic efficiency, $P = \eta_h \rho g Q H$; peripheral velocity factor, $U_1 = \varphi\sqrt{2gH}$; tip speed, $U_1 = \omega D/2$; Euler with zero exit swirl, $\eta_h g H = U_1 V_{w1}$; triangle geometry, $V_{f1} = V_{w1}\tan\alpha_1$; continuity, $Q = \pi D b\,V_{f1}$.

Why this works. Specific speed is the one parameter that fixes the shape of a hydraulic machine independently of its size, which is why a designer chooses it first and everything else afterwards. Given the head and the speed the grid demands, the specific speed immediately fixes the power the machine must be built for; efficiency converts that into a flow; the velocity factor — a purely empirical statement that a good runner rim moves at roughly seventy per cent of the free-jet speed — converts the speed into a diameter. From there the Euler equation with zero exit swirl is what makes the problem closed: it says the whole of the useful head is delivered by a single product $U_1 V_{w1}$, so the inlet triangle can be built without knowing anything about the outlet. The height at the periphery is then pure bookkeeping.

Common pitfall. Three catch most candidates. Using the pump specific speed $\omega Q^{1/2}/(gH)^{3/4}$ instead of the power form gives nonsense because the two are not interchangeable. Measuring $\alpha_1$ from the radial direction rather than the tangent turns the tangent into a cotangent and gives $V_{f1} = 116\ \text{m}\,\text{s}^{-1}$, faster than the free jet, which is physically impossible. And using $\tfrac{\pi}{4}D^2$ for the inlet area instead of the cylindrical $\pi D b$ makes the runner height meaningless.

Source. Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed., §9.3–9.5 (Francis turbine, specific speed and preliminary design); White, Fluid Mechanics, 8th ed., §11.6 (reaction turbines and specific speed); Turton, Principles of Turbomachinery, 2nd ed., Ch. 6.

← Question 3 Next: Question 5 →