Question 5 of 8: Hydro Turbine (10 marks)

Question 5 — 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. A Francis runner of 4.470 m entry diameter turning at 107 rev/min under 65.227 m of head, delivering 54.063 MW at an efficiency of 0.938, with geometry scaled from the photograph on page 14.

Find. The flow rate, the inlet and outlet peripheral and meridional velocities, the inlet and outlet velocity diagrams with $V_1$, $\alpha_1$, $V_2$ and $\alpha_2$, the hydraulic power of the runner, and the blade exit angle implied by the diagram compared with the quoted 165°.

Approach. Get the flow from the power and efficiency, the peripheral speeds from the diameters and rotational speed, the meridional velocities from continuity through the (blocked) inlet annulus and the throat, close the inlet triangle using $\beta_1 = 90^\circ$, then use the Euler turbine equation to find the exit whirl that delivers the runner power and hence the exit triangle.

1. Part (a) — volume flow rate. The efficiency links the mechanical output to the water power supplied, $P = \eta\rho gQH$, so $$Q = \frac{P}{\eta\rho gH} = \frac{54.063 \times 10^{6}}{0.938 \times 1000 \times 9.81 \times 65.227}$$ $$\boxed{Q = 90.07\ \text{m}^{3}\,\text{s}^{-1}}$$ The water power available is $\rho gQH = 57.64\ \text{MW}$, of which 3.57 MW is lost.
2. Part (b) — inlet peripheral and radial velocities. The blade speed at entry is $$U_1 = \frac{\pi D_1 N}{60} = \frac{\pi \times 4.470 \times 107}{60} = 25.04\ \text{m}\,\text{s}^{-1}$$ The inlet flow passes through a cylindrical surface of height $h$, less the 10 % of circumference occupied by blade thickness: $$A_1 = \pi D_1 h\,(1 - 0.10) = \pi \times 4.470 \times 0.894 \times 0.90 = 11.30\ \text{m}^{2}, \qquad V_{r1} = \frac{Q}{A_1}$$ $$\boxed{U_1 = 25.04\ \text{m}\,\text{s}^{-1}, \qquad V_{r1} = 7.972\ \text{m}\,\text{s}^{-1}}$$
3. Part (c) — inlet velocity diagram. A blade inlet angle of $90^\circ$ means the relative velocity is perpendicular to the periphery, that is purely radial, so the whole of the blade speed appears as absolute whirl: $V_{\theta 1} = U_1 = 25.04\ \text{m}\,\text{s}^{-1}$ and $W_1 = V_{r1}$. Combining the two components, $$V_1 = \sqrt{U_1^{2} + V_{r1}^{2}} = \sqrt{25.04^{2} + 7.972^{2}}, \qquad \alpha_1 = \arctan\frac{V_{r1}}{U_1}$$ $$\boxed{V_1 = 26.28\ \text{m}\,\text{s}^{-1}, \qquad \alpha_1 = 17.66^\circ}$$ This $\alpha_1$ is the setting the stationary guide vanes must produce, and Figure 5.1 shows the resulting right-angled triangle.
4. Part (d) — outlet peripheral and meridional velocities. At half the diameter the blade speed halves, $$U_2 = \frac{\pi D_2 N}{60} = 12.52\ \text{m}\,\text{s}^{-1}$$ The question directs that the exit area is the draft-tube throat, a full circle of diameter $D_1$: $$A_2 = \frac{\pi D_{\text{throat}}^{2}}{4} = 15.69\ \text{m}^{2}, \qquad V_{m2} = \frac{Q}{A_2}$$ $$\boxed{U_2 = 12.52\ \text{m}\,\text{s}^{-1}, \qquad V_{m2} = 5.740\ \text{m}\,\text{s}^{-1}}$$ Because the flow has turned into the axial direction by the time it reaches the throat, this "radial" velocity is physically the axial (meridional) component entering the draft tube.
5. Part (e) — hydraulic power developed by the runner. With no separate mechanical or volumetric efficiency given, the single quoted efficiency accounts for every loss between the water and the shaft, so the power the blades take from the water is the stated mechanical output: $$P_{\text{runner}} = \eta\rho gQH = 0.938 \times 1000 \times 9.81 \times 90.07 \times 65.227$$ $$\boxed{P_{\text{runner}} = 54.06\ \text{MW}}$$ This is not merely a restatement of the given data: part (f) needs it as the target for the Euler equation, and it must lie below the zero-exit-swirl ceiling $\rho QU_1V_{\theta 1} = 56.49\ \text{MW}$ for the runner to work at all. It does, by 2.43 MW, which is exactly the whirl energy that must be left in the exit flow.
6. Part (f) — exit swirl angle from the Euler equation. The page-21 hydro-turbine relation is $P = \rho Q\left(U_1V_1\cos\alpha_1 - U_2V_2\cos\alpha_2\right)$. Writing $V\cos\alpha$ as the whirl component and rearranging for the exit whirl, $$V_{\theta 2} = \frac{\rho QU_1V_{\theta 1} - P_{\text{runner}}}{\rho QU_2} = \frac{56.49 \times 10^{6} - 54.06 \times 10^{6}}{1000 \times 90.07 \times 12.52} = 2.153\ \text{m}\,\text{s}^{-1}$$ Combining with the meridional component from part (d), $$\alpha_2 = \arctan\frac{V_{m2}}{V_{\theta 2}} = \arctan\frac{5.740}{2.153}$$ $$\boxed{\alpha_2 = 69.44^\circ}$$ Substituting back, $\rho Q(U_1V_{\theta 1} - U_2V_{\theta 2}) = 54.06\ \text{MW}$ as required. The residual whirl is only 2.15 m/s against a meridional 5.74 m/s, so the flow is indeed close to axial and the question's simplifying assumption $V_2 \approx V_{\text{radial}}$ is self-consistent.
7. Part (g) — exit velocity diagram and the blade angle. The absolute exit velocity closes the triangle, $$V_2 = \sqrt{V_{\theta 2}^{2} + V_{m2}^{2}} = \sqrt{2.153^{2} + 5.740^{2}} = 6.130\ \text{m}\,\text{s}^{-1}$$ and the relative velocity follows by subtracting the blade speed from the whirl, $W_{\theta 2} = V_{\theta 2} - U_2 = 2.153 - 12.52 = -10.37\ \text{m}\,\text{s}^{-1}$, so $$W_2 = \sqrt{10.37^{2} + 5.740^{2}} = 11.85\ \text{m}\,\text{s}^{-1}, \qquad \beta_2 = 180^\circ - \arctan\frac{5.740}{10.37}$$ $$\boxed{V_2 = 6.130\ \text{m}\,\text{s}^{-1}, \qquad \beta_2 = 151.0^\circ}$$ Comparison with the given 165°. The diagram gives 151.0° against the 165° scaled off the photograph, a difference of 14°. The 165° value cannot be right: carrying it through the same construction gives $W_{\theta 2} = -21.42\ \text{m}\,\text{s}^{-1}$, an exit whirl of $-8.90\ \text{m}\,\text{s}^{-1}$ (that is, counter to rotation) and an Euler power of 66.5 MW — more than the 57.6 MW of water power actually supplied, which is impossible. The 165° is therefore an artefact of reading a blade angle off a perspective photograph of a curved, twisted vane; 151.0° is the angle consistent with the stated head, speed, power and efficiency. Real Francis runners are twisted, so a single exit angle is in any case a mean value, and the true blade angle varies from hub to shroud.