Question 2 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: National Examinations — 07-Mec-A6 Fluid Machinery, May 2013. Closed-book, three hours. Section A (Calculative) Q1–Q5, Section B (Descriptive) Q6–Q8; the candidate answers four from A and two from B, six questions of equal value (10 marks each). All eight questions are worked here as a complete study resource.
Reference texts: R. K. Turton, Principles of Turbomachinery; S. L. Dixon & C. A. Hall, Fluid Mechanics and Thermodynamics of Turbomachinery; R. W. Fox, A. T. McDonald, Introduction to Fluid Mechanics; Douglas, Gasiorek & Swaffield, Fluid Mechanics. Constants as printed on the exam attachment pages (g = 9.81 m/s², patm = 100 kPa, pvapour = 2.34 kPa at 20°C, ρwater = 1000 kg/m³).
Question 2: Hydro Turbines — Pelton Wheel and Francis Setting (10 marks)
Given. Net head H = 1118 ft = 340.8 m; pitch diameter D = 95 in = 2.413 m; N = 300 rev/min; output P = 62 000 × 746 = 46.25 MW; gross head 1226 ft = 373.7 m.
Find. Blade-to-jet speed ratio, its deviation from the ideal 0.5, and the volume flow rate.
Approach. Blade speed from πDN/60; ideal jet speed from the net head via Torricelli; ratio and deviation follow; flow from the hydraulic power at net head.
(a) Blade and jet velocities. $U=\dfrac{\pi D N}{60}=\dfrac{\pi(2.413)(300)}{60}=37.9\ \text{m/s}$; ideal jet $V=\sqrt{2gH}=\sqrt{2(9.81)(340.8)}=81.8\ \text{m/s}$. Ratio $\boxed{U/V=0.464}$.
(b) Deviation from the ideal. The ideal ratio for maximum power is 0.5; the deviation is $\dfrac{0.5-0.464}{0.5}=7.3\%$ below ideal. Real wheels run slightly below 0.5 because bucket friction and windage make the true optimum about 0.46–0.47, so a designer deliberately sets U/V a little under 0.5.
(c) Volume flow rate. Taking the output as the hydraulic power developed at the net head (ideal wheel), $Q=\dfrac{P}{\rho g H}=\dfrac{46.25\times10^{6}}{1000\times9.81\times340.8}$, so $\boxed{Q=13.8\ \text{m}^3/\text{s}}$.
Check: no wheel efficiency is quoted, so part (c) assumes the output equals the hydraulic power at the net head (η ≈ 100%). A real Pelton wheel at η ≈ 88% would need about 13.8/0.88 ≈ 15.7 m³/s; the method is unchanged.
Part II — Francis turbine setting
Given. P = 120 MW, N = 125 rev/min (ω = 13.09 rad/s), H = 65 m, ρ = 1000 kg/m³, patm = 100 kPa, pvapour = 2.34 kPa.
Find. Power specific speed, the critical Thoma parameter σc from the chart, and the maximum runner setting above the tailrace.
Approach. Compute the dimensionless power specific speed, read σc off the Francis curve at that value, then apply the cavitation-parameter definition to solve for the runner elevation.
(a) Power specific speed. $\Omega_{sp}=\dfrac{\omega\,P^{1/2}}{\rho^{1/2}(gH)^{5/4}}=\dfrac{13.09\,(120\times10^{6})^{1/2}}{(1000)^{1/2}(9.81\times65)^{5/4}}$, so $\boxed{\Omega_{sp}=1.41\ \text{rad}}$.
(b) Thoma parameter from the chart. Entering the supplied chart at Ωsp = 1.41 on the Francis curve gives a critical cavitation coefficient $\sigma_c\approx0.22$ (boundary of the no-cavitation region).
(c) Runner setting. With $\sigma_c=\dfrac{(p_{atm}-p_{vap})/\rho g-\Delta z}{H}$, solve $\Delta z=\dfrac{p_{atm}-p_{vap}}{\rho g}-\sigma_c H=\dfrac{(100-2.34)\times10^{3}}{1000\times9.81}-0.22(65)=9.96-14.3$, so $\boxed{\Delta z\approx-4.3\ \text{m}}$.
Check: a negative Δz means the runner must sit about 4 m below the tailrace level — normal for a 65 m-head Francis machine. Reading the chart to ±0.02 in σc (0.20–0.24) puts the setting between roughly −3 m and −6 m; the sign and order of magnitude are robust.
