Question 3 of 8: Hydro Turbines — Pelton Wheel and Francis Setting
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examination 07-Mec-A6-1 Fluid Machinery (May 2016) — closed book, three hours, 60 marks. Section A is calculative (Q1–Q5) and Section B is descriptive (Q6–Q8); the rubric asks for four of Section A plus two of Section B (six questions, each of equal value, 10 marks). All eight questions are solved in full as a study resource. General constants supplied with the paper: g = 9.81 m/s², patm = 100 kPa, pvapour = 2.34 kPa (20 °C), ρwater = 1000 kg/m³, ρair = 1.21 kg/m³ (15 °C), cp,air = 1.005, cv,air = 0.718 kJ/kg·K.
Reference texts. S. L. Dixon & C. A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed.); R. K. Turton, Principles of Turbomachinery; R. H. Sabersky, A. J. Acosta, E. G. Hauptmann & E. M. Gates, Fluid Flow: A First Course in Fluid Mechanics (4th ed.) — pump-characteristic and cavitation-parameter charts; R. W. Fox, A. T. McDonald & P. J. Pritchard, Introduction to Fluid Mechanics (pump energy equation and affinity laws); Y. A. Çengel & J. M. Cimbala, Fluid Mechanics: Fundamentals and Applications.
Question 3: Hydro Turbines — Pelton Wheel and Francis Setting (10 marks)
Approach. Convert to SI, get the wheel peripheral speed from the pitch diameter and the ideal jet speed from the net head, compare their ratio to the ideal 0.5, then obtain flow from the hydraulic power on the net head.
SI conversions. $H_{net}=1118(0.3048)=340.8\ \text{m}$; $D=95(0.0254)=2.413\ \text{m}$; $P=62000(746)=46.3\ \text{MW}$.
Blade and jet velocities. $U=\dfrac{\pi D N}{60}=\dfrac{\pi(2.413)(300)}{60}=37.90\ \text{m/s}$; ideal jet $V=\sqrt{2gH_{net}}=\sqrt{2(9.81)(340.8)}=81.77\ \text{m/s}$.
(b) Deviation from ideal. The ideal ratio is 0.5, so $\dfrac{0.5-0.464}{0.5}=\boxed{7.3\%\ \text{below ideal}}$. This is expected: nozzle friction makes the actual jet speed a little below $\sqrt{2gH}$ (a velocity coefficient $C_v\approx0.97$), and running slightly below U/V = 0.5 keeps the wheel near peak efficiency while allowing the water to leave the buckets with a small residual velocity.
(c) Flow rate. Treating the net head as fully available to the jet (ideal), $Q=\dfrac{P}{\rho g H_{net}}=\dfrac{46.3\times10^{6}}{1000(9.81)(340.8)}=\boxed{13.84\ \text{m}^3/\text{s}}$.
Part II — Francis turbine setting
Given. P = 120 MW, N = 125 rev/min, H = 65 m, Qmax = 217 m³/s, runner D = 5.462 m.
Find. (a) power specific speed, (b) Thoma σ from the chart, (c) runner setting relative to tailrace.
Because σH (about 15 m) exceeds the 9.96 m of atmospheric-minus-vapour head, Δz is negative: the runner must be set about 5 m below the tailrace water level.
Approach. Form the non-dimensional power specific speed, read the Thoma parameter for a Francis machine from the Page 14 chart, and apply the cavitation-setting relation.
(a) Power specific speed. $\omega=\dfrac{2\pi(125)}{60}=13.09\ \text{rad/s}$. $$\Omega_{sp}=\frac{\omega\,P^{1/2}}{\rho^{1/2}(gH)^{5/4}}=\frac{13.09\,(120\times10^{6})^{1/2}}{1000^{1/2}\,(9.81\times65)^{5/4}}=\boxed{1.42}.$$
(b) Thoma parameter. At $\Omega_{sp}\approx1.42$ the Page 14 chart has only one line (the Kaplan line starts near $\Omega_{sp}\approx1.5$), and the Francis line there reads $\boxed{\sigma\approx0.23}$ (for reference it passes about 0.13 at $\Omega_{sp}=1.0$ and 0.40 at 2.0).
(c) Runner setting. The available suction head is $\dfrac{p_{atm}-p_{vap}}{\rho g}=\dfrac{100-2.34\ \text{kPa}}{9.81}=9.96\ \text{m}$, so $$\Delta z=\frac{p_{atm}-p_{vap}}{\rho g}-\sigma H=9.96-0.23(65)=9.96-14.95=\boxed{-5.0\ \text{m}}.$$ The negative result means the runner must be set at least about 5.0 m below the tailrace water level (submerged) to avoid cavitation, which is normal for a large Francis unit at this specific speed. The reading is sensitive: each 0.01 change in σ moves the setting by 0.65 m, so a designer would add a margin of submergence.
