Question 4 of 8: Multi-Jet Pelton Turbine

Question 4: Multi-Jet Pelton 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 multi-jet Pelton wheel:

Find. The wheel speed in rev/min, the power output in MW, the number of nozzles that the stated jet-diameter ratio implies, and the dimensionless specific speed of the machine.

Approach. The nozzle coefficient converts the head into a real jet velocity; the desirable speed ratio then gives the bucket speed and hence the wheel speed. Power comes from the head, flow and efficiency. Dividing the total flow by what one jet of the stated diameter can pass gives the nozzle count, and the specific speed follows from the page-15 definition.

1. Part (a) — jet velocity. The nozzle converts the head into velocity with the loss folded into the coefficient K: $$V=K\sqrt{2gH}=0.98\times\sqrt{2\times 9.81\times 200}=0.98\times 62.64=61.39\ \text{m/s}$$
2. Bucket speed and wheel speed. The desirable ratio fixes the peripheral speed, $$U=0.47\,V=0.47\times 61.39=28.85\ \text{m/s}$$ and the wheel turns at $$N=\frac{60U}{\pi D}=\frac{60\times 28.85}{\pi\times 1.47}$$ $$\boxed{N=375\ \text{rev/min}}$$ The ratio 0.47 is deliberately a little below the frictionless optimum of 0.5, which is where a real bucket — whose exit angle cannot be a full 180° and whose surfaces are not smooth — actually peaks.
3. Part (b) — power output. Taking the quoted 88 % as the overall water-to-shaft efficiency, $$P=\eta\,\rho g Q H=0.88\times 1000\times 9.81\times 4\times 200$$ $$\boxed{P=6.91\ \text{MW}}$$
4. Part (c) — jet diameter and per-jet flow. The desirable jet-to-wheel diameter ratio gives $$d=0.113\,D=0.113\times 1.47=0.1661\ \text{m}\qquad\Rightarrow\qquad A_j=\frac{\pi}{4}d^2=0.02167\ \text{m}^2$$ so one nozzle passes $$q=A_jV=0.02167\times 61.39=1.330\ \text{m}^3/\text{s}$$
5. Number of nozzles. The total flow divided by the flow one jet can carry is $$n=\frac{Q}{q}=\frac{4.0}{1.330}=3.01$$ $$\boxed{n=3\ \text{nozzles}}$$ That the result lands within one per cent of a whole number is the internal check that the two “desirable” ratios and the given flow were chosen consistently — a designer reading 2.6 or 3.4 would have to revisit d/D.
6. Part (d) — specific speed of the machine. Using the page-15 turbine definition with the whole-machine power, $$\omega=\frac{2\pi\times 374.9}{60}=39.26\ \text{rad/s},\qquad N_s=\frac{\omega P^{1/2}}{\rho^{1/2}(gH)^{5/4}} =\frac{39.26\times\sqrt{6.906\times 10^{6}}}{31.62\times(1962)^{1.25}}$$ $$\boxed{N_s=0.250}$$
7. Interpret the value. A specific speed of 0.25 sits squarely in the Pelton band of the page-10 efficiency chart, which peaks near 0.1–0.2 and is still healthy at 0.25 — so the impulse choice is right for this head. Evaluated per jet, as some texts prefer, the figure is $N_s/\sqrt{3}=0.144$: multi-jetting is precisely the device that lets a higher-specific-speed shaft be driven by a machine that is, jet for jet, a low-specific-speed impulse wheel.