22-Mec-B6 Advanced Fluid Mechanics · December 2018

Question 4 of 8: Francis Hydro Turbine

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

Notes on this paper

Paper format: National Examinations, December 2018 — 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); candidates answer 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 on pages 11–17, and the nomenclature, general constants and reference equations on pages 18–22. All eight questions are solved below.

Check — angle conventions are taken from the paper's own attachments, and they are not the same across the paper. The compressor velocity diagram on page 11 strikes $\alpha_1$ and $\beta_1$ off the axial component $C_{X1}$, so every blade and vane angle in Questions 1 and 2 is measured from the axial direction. The steam-turbine diagram on page 16 strikes $\theta$, $\phi$, $\gamma$ and $\delta$ off the tangential (blade-motion) direction, and the Francis diagram on page 15 defines $\alpha_1$ between $V_1$ and $u_1$ — also tangential. Questions 4 and 5 therefore use the tangential reference. Every constant below is the paper's own page-19 value: $g = 9.81\ \text{m}\,\text{s}^{-2}$, $\rho_{\text{water}} = 1000\ \text{kg}\,\text{m}^{-3}$, $c_p = 1005\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$ and $c_v = 718\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$, from which $R = c_p - c_v = 287\ \text{J}\,\text{kg}^{-1}\text{K}^{-1}$ and $k = c_p/c_v = 1.3997$ — not a textbook 1.40.

Reference texts

Dixon, S. L. and Hall, C. A., Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed. — Ch. 5 (axial compressors), Ch. 4 (axial turbines), Ch. 9 (hydraulic turbines).
Cohen, H., Rogers, G. F. C. and Saravanamuttoo, H. I. H., Gas Turbine Theory, 7th ed. — Ch. 5 (axial flow compressors), Ch. 6 (compressor and turbine matching, surge and stall), Ch. 3 (jet propulsion cycles).
Turton, R. K., Principles of Turbomachinery, 2nd ed. — Ch. 2 (Euler equation), Ch. 8 (fans, blade form and control).
White, F. M., Fluid Mechanics, 8th ed. — Ch. 11 (turbomachinery: specific speed, Francis and Kaplan turbines, cavitation).
Fox, R. W. and McDonald, A. T., Introduction to Fluid Mechanics, 10th ed. — Ch. 10 (fluid machinery, similarity rules).
Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, 9th ed. — Ch. 9 (gas power cycles, Brayton and turbojet).

Question 4 — Francis Hydro Turbine (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.

Quantity	Symbol	Value
Head available	$H$	$100\ \text{m}$
Shaft power required	$P$	$4\ \text{MW}$
Specific speed (page-21 definition)	$N_s$	$0.80$
Hydraulic efficiency	$\eta_h$	$0.90$
Peripheral velocity factor	$U_1/V_{JET}$	$0.70$
Outlet-to-inlet vane diameter ratio	$D_2/D_1$	$0.75$
Guide vane angle (from tangential)	$\alpha_1$	$20^{\circ}$
Inlet vane angle	$\beta_1$	$90^{\circ}$
Exit flow angle	$\alpha_2$	$90^{\circ}$ (no exit whirl)
Exit radial velocity	$V_{f2}$	$1.2\,V_{f1}$

Find. Rotational speed, spouting velocity, runner peripheral speeds and diameters at inlet and outlet, volume flow, and the vane heights at inlet and outlet together with the outlet vane angle.

Velocity diagrams for the preliminary Francis design, drawn with the page-15 convention in which $\alpha$ and $\beta$ are struck off the tangential direction $u$. At inlet $\beta_1 = 90^{\circ}$ makes the relative velocity purely radial, so the whirl equals the blade speed; at outlet $\alpha_2 = 90^{\circ}$ makes the absolute velocity purely radial, so there is no exit whirl.

Approach. Invert the specific-speed definition to get the running speed, use the spouting velocity and the stated velocity factor to fix the inlet peripheral speed and hence both diameters, get the flow from power and efficiency, and close the two velocity triangles — each of which is fully determined by one stated angle — to obtain the vane heights and the outlet vane angle.

Rotational speed from the specific speed (part a). The page-21 turbine specific speed is $N_s = \omega P^{1/2}/[\rho^{1/2}(gH)^{5/4}]$, which rearranges to $$\omega = \frac{N_s\,\rho^{1/2}\,(gH)^{5/4}}{P^{1/2}} = \frac{0.80 \times 31.623 \times (981)^{1.25}}{2000} = 69.45\ \text{rad/s}$$ $$\boxed{N = \frac{60\,\omega}{2\pi} = 663\ \text{rev/min}}$$ Six hundred and sixty revolutions a minute is a direct-drive speed a small generator can accept, which is part of why $N_s = 0.80$ was specified.
Spouting velocity (part b). The maximum velocity the available head can produce is the free-jet value obtained by converting the whole head to velocity head: $$\boxed{V_{JET} = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 100} = 44.29\ \text{m/s}}$$
Peripheral speeds at inlet and outlet (part c). The stated velocity factor gives the runner tip speed, and the diameter ratio scales it to the outlet, since both radii turn at the same $\omega$: $$U_1 = 0.70\,V_{JET} = 0.70 \times 44.294 = 31.01\ \text{m/s},\qquad U_2 = 0.75\,U_1 = 23.25\ \text{m/s}$$ $$\boxed{U_1 = 31.0\ \text{m/s}\qquad U_2 = 23.3\ \text{m/s}}$$
Runner diameters (part d). With $U = \omega r$ and $D = 2r$, $$D_1 = \frac{2U_1}{\omega} = \frac{2 \times 31.006}{69.446} = 0.893\ \text{m},\qquad D_2 = 0.75\,D_1 = 0.670\ \text{m}$$ $$\boxed{D_1 = 0.893\ \text{m (inlet)}\qquad D_2 = 0.670\ \text{m (outlet)}}$$ A runner under a metre across for four megawatts is exactly what a 100 m head buys; the same power at the 15.67 m head of Question 3 needed a Kaplan several times the size.
Volume flow rate (part e). The hydraulic efficiency relates the shaft power to the water power, so $$Q = \frac{P}{\eta_h\,\rho\,g\,H} = \frac{4 \times 10^{6}}{0.90 \times 1000 \times 9.81 \times 100}$$ $$\boxed{Q = 4.53\ \text{m}^3/\text{s}}$$
Close the inlet triangle. The condition $\beta_1 = 90^{\circ}$ means the relative velocity enters radially, so it has no tangential component and the whirl must equal the blade speed: $V_{w1} = U_1 = 31.01\ \text{m/s}$. With $\alpha_1$ struck off the tangential direction (page 15), the radial component follows: $$\begin{aligned}V_1 &= \frac{V_{w1}}{\cos\alpha_1} = \frac{31.006}{\cos 20^{\circ}} = 33.00\ \text{m/s} \\ V_{f1} &= V_1\sin\alpha_1 = 33.00 \times 0.3420 = 11.29\ \text{m/s}\end{aligned}$$ Equivalently $V_{f1} = U_1\tan\alpha_1$, which is the quicker route and the one to use under exam pressure.
Vane height at inlet (part f). The whole flow crosses the cylindrical surface at the runner periphery, whose area is the circumference times the vane height — never a disc area: $$b_1 = \frac{Q}{\pi D_1 V_{f1}} = \frac{4.5305}{\pi \times 0.89296 \times 11.285}$$ $$\boxed{b_1 = 0.143\ \text{m} = 143\ \text{mm}}$$
Outlet vane angle and vane height (part g). The exit radial velocity is stated as 1.2 times the inlet value, and $\alpha_2 = 90^{\circ}$ means the absolute exit velocity is entirely radial, so $$V_{f2} = 1.2 \times 11.285 = 13.54\ \text{m/s} = V_2$$ The relative velocity at exit must close the triangle against the blade speed $U_2$, so the vane angle measured from the tangential direction is $$\beta_2 = \arctan\!\left(\frac{V_{f2}}{U_2}\right) = \arctan\!\left(\frac{13.542}{23.255}\right) = 30.2^{\circ}$$ and the outlet vane height follows from continuity through the outlet periphery: $$b_2 = \frac{Q}{\pi D_2 V_{f2}} = \frac{4.5305}{\pi \times 0.66972 \times 13.542}$$ $$\boxed{\beta_2 = 30.2^{\circ}\qquad b_2 = 0.159\ \text{m} = 159\ \text{mm}}$$ The outlet passage is taller than the inlet passage even though the diameter is smaller, which is the characteristic flare of a Francis runner towards the draft tube.

Check — the paper's assumed parameters over-determine the machine, and the small inconsistency is worth stating. With $\beta_1 = 90^{\circ}$ the whirl equals the blade speed, so Euler's equation with no exit whirl gives the specific work as $U_1^2$. Since $U_1 = 0.7\sqrt{2gH}$, that work is $0.49 \times 2gH = 0.98\,gH$, i.e. the velocity-factor and vane-angle assumptions between them imply a hydraulic efficiency of exactly 98 per cent, not the 90 per cent stated. The two are used where each belongs, as the question's own ordering intends: the stated efficiency of 0.90 fixes the flow in part (e), and the triangle geometry fixes the velocities and vane heights. A designer settling this would reduce the velocity factor to $U_1/V_{JET} = \sqrt{\eta_h/2} = 0.671$, which makes the two consistent and changes $D_1$ to 0.856 m. Note also that the question's line "$d_1 = \frac{3}{4}d_2$ where $d_2$ is inlet vane diameter" inverts the subscripts used on its own page-15 attachment (where 1 is the inlet); it is read here in the only physically sensible way, namely that the outlet diameter is three quarters of the inlet diameter.

Quantity	Symbol	Result
Rotational speed	$N$ ($\omega$)	$663\ \text{rev/min}$ ($69.45\ \text{rad/s}$)
Spouting (jet) velocity	$V_{JET}$	$44.29\ \text{m/s}$
Runner tip speed / outlet speed	$U_1$, $U_2$	$31.01$ / $23.25\ \text{m/s}$
Runner inlet / outlet diameter	$D_1$, $D_2$	$0.893$ / $0.670\ \text{m}$
Volume flow rate	$Q$	$4.53\ \text{m}^3/\text{s}$
Inlet absolute / radial velocity	$V_1$, $V_{f1}$	$33.00$ / $11.29\ \text{m/s}$
Vane height at inlet	$b_1$	$0.143\ \text{m}$ ($143\ \text{mm}$)
Exit radial velocity	$V_{f2}$	$13.54\ \text{m/s}$
Outlet vane angle	$\beta_2$	$30.2^{\circ}$
Vane height at outlet	$b_2$	$0.159\ \text{m}$ ($159\ \text{mm}$)

Topic. Preliminary sizing of a radial inward-flow (Francis) runner from specific speed, head and power.

Key relations. Turbine specific speed $N_s = \omega P^{1/2}/[\rho^{1/2}(gH)^{5/4}]$; spouting velocity $\sqrt{2gH}$; $U = \omega r$; water power $P = \eta_h\rho g Q H$; Euler turbine work $w = U_1V_{w1} - U_2V_{w2}$; continuity through a cylindrical periphery $Q = \pi D\,b\,V_f$.

Why this works. Specific speed is the one dimensionless group that survives when head, power and speed are the only things known, and because it is dimensionless it selects a shape of machine rather than a size. Fixing $N_s$ therefore fixes $\omega$, and $\omega$ plus the peripheral velocity factor fixes the diameter — the machine is sized before any blade has been drawn. The two stated vane angles then each collapse one triangle to a right angle: $\beta_1 = 90^{\circ}$ removes the relative whirl at inlet, so $V_{w1} = U_1$; $\alpha_2 = 90^{\circ}$ removes the absolute whirl at outlet, so all the angular momentum given up by the water is extracted, which is the condition for maximum work and for a draft tube free of swirl. The vane heights come from continuity applied to the surface the water actually crosses, which for a radial machine is a cylinder of diameter $D$ and height $b$, not a circle of diameter $D$.

Common pitfall. Measuring $\alpha_1$ from the radial direction instead of the tangential: the page-15 attachment places $\alpha_1$ between $V_1$ and $u_1$, so the radial component is $U_1\tan\alpha_1 = 11.3\ \text{m/s}$; taking the complement gives $U_1\cot\alpha_1 = 85\ \text{m/s}$ and a vane height an order of magnitude too small. The other classic error is using $\frac{\pi}{4}D^2$ for the flow area at the runner periphery instead of $\pi D b$.

Source. Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery, 7th ed., §9.3–9.5 (Francis turbines, preliminary design); White, Fluid Mechanics, 8th ed., §11.4 and §11.6; Turton, Principles of Turbomachinery, 2nd ed., Ch. 2.

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