Question 2 of 8: Hydro Turbines — Pelton Wheel and Francis Setting

Question 2: Hydro Turbines — Pelton Wheel and Francis Setting (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.

Part I — Pelton wheel

Given. Net head H_net = 1118 ft = 340.8 m; gross head 1226 ft = 373.7 m; P = 62 000 HP = 46.25 MW; N = 300 rev/min; pitch diameter D = 95 in = 2.413 m.

Find. (a) blade-to-jet speed ratio U/V, (b) its deviation from the ideal 0.5, (c) the flow rate.

Approach. The wheel speed follows from the pitch diameter and rpm; the anticipated jet speed follows from the net head via $V=\sqrt{2gH}$; the flow follows from the hydraulic power $P=\rho g Q H_{net}$.

1. (a) Blade and jet velocities. Peripheral (blade) velocity $$U=\frac{\pi D N}{60}=\frac{\pi(2.413)(300)}{60}=37.9\ \text{m/s},$$ anticipated jet velocity from the net head $$V=\sqrt{2gH_{net}}=\sqrt{2(9.81)(340.8)}=81.8\ \text{m/s},\qquad \boxed{\frac{U}{V}=0.464.}$$
2. (b) Deviation from ideal. The frictionless optimum is $U/V=0.5$, so $$\text{deviation}=\frac{0.5-0.464}{0.5}=7.3\%\ \text{below ideal.}$$ The real jet is slower than $\sqrt{2gH_{net}}$ (nozzle velocity coefficient below unity) and bucket friction plus windage move the best-efficiency speed ratio down to about 0.45–0.46, exactly the region observed.
3. (c) Flow rate. Treating the output as ideal hydraulic power on the net head, $$Q=\frac{P}{\rho g H_{net}}=\frac{46.25\times10^{6}}{1000(9.81)(340.8)}=13.8\ \text{m}^3/\text{s}.$$

Part II — Francis turbine setting

Given. P = 120 MW, N = 125 rev/min, H = 65 m, water. (The 217 m³/s, 7 m casing and 5462 mm runner size the machine but are not needed for the dimensionless specific speed.)

Find. (a) power specific speed Ω_sp, (b) Thoma σ from the chart, (c) maximum runner elevation Δz above the tailrace.

[Figure not reproduced: Critical-cavitation chart (page 12), redrawn from the printed log–log attachment. The operating point Ω sp = 1.41 lands on the Francis line at σ c ≈ 0.22. The Francis line runs only from about Ω sp = 0.35 to 2.05 and the Kaplan line begins near 1.49, so the two overlap between roughly 1.5 and 2.0 wi. See the official exam paper.]

Approach. Compute the non-dimensional power specific speed, read the critical σ off the Francis curve, then invert the Thoma definition for the permissible runner height.

1. (a) Power specific speed. With $\omega=2\pi N/60=13.09$ rad/s, $$\Omega_{sp}=\frac{\omega\,P^{1/2}}{\rho^{1/2}(gH)^{5/4}}=\frac{13.09\,(120\times10^{6})^{1/2}}{(1000)^{1/2}(9.81\times65)^{5/4}}=1.41.$$ This value ($\approx$ 1.4 rad) is squarely in the Francis range.
2. (b) Thoma parameter. Entering the chart at $\Omega_{sp}=1.41$ and reading up to the Francis line gives $$\boxed{\sigma_c\approx 0.22.}$$ At this specific speed only the Francis line is drawn (the Kaplan line does not begin until $\Omega_{sp}\approx 1.5$), so there is no ambiguity about which branch to read.
3. (c) Runner setting. The Thoma definition $$\sigma_c=\frac{\dfrac{p_{atm}-p_{vap}}{\rho g}-\Delta z}{H}\quad\Rightarrow\quad \Delta z=\frac{p_{atm}-p_{vap}}{\rho g}-\sigma_c H.$$ The atmospheric–vapour head is $(100-2.34)\times10^{3}/(1000\times9.81)=9.96$ m, so $$\Delta z=9.96-0.22(65)=9.96-14.30=-4.35\ \text{m}.$$ The negative sign is the answer: the runner must be set about 4.3 m below the tailrace water level (submerged) for the pressure at the runner outlet to stay above vapour pressure. Because the required suction head $\sigma_c H=14.3$ m already exceeds the 9.96 m of atmospheric-minus-vapour head available, no positive setting is possible at this head and specific speed.