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Pelton turbine design (1.8 m³/s, 605 m)

Short answer

Single-jet Pelton wheel: power 9.29 MW at 500 rpm (12 poles). Jet diameter 147 mm, pitch circle diameter 1.91 m, jet ratio 13.07. Specific speed 16.1. 22 buckets; bucket width 0.73 m, depth 0.18 m.

Design a pelton turbine with following data: Overall efficiency = 87%, Discharge (Qd) = 1.8 m³/s, Net head (H) = 605 m.

A Pelton wheel is designed by finding the jet velocity and jet diameter, choosing a synchronous speed that gives a suitable jet ratio m=D/dm = D/d (10 to 14 for good efficiency), and then sizing the wheel and buckets. Of the two nearest synchronous speeds (500 and 600 rpm), 500 rpm is chosen because it gives a jet ratio nearer 12.

Data: Q=1.8Q = 1.8 m³/s, H=605H = 605 m, ηo=0.87\eta_o = 0.87, f=50f = 50 Hz. Assumed from the standard ranges: Cv=0.98C_v = 0.98 (0.97 to 0.99) and speed ratio ϕ=0.46\phi = 0.46 (0.43 to 0.47).

1. Power developed

P=ηoρgQH=0.87×9.81×1.8×605=9294.3 kWP = \eta_o \rho g Q H = 0.87 \times 9.81 \times 1.8 \times 605 = 9294.3\ \text{kW}

2. Jet velocity and bucket (peripheral) velocity

2gH=2×9.81×605=108.95 m/sV1=Cv2gH=0.98×108.95=106.77 m/su=ϕ2gH=0.46×108.95=50.12 m/s\begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 605} = 108.95\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.98 \times 108.95 = 106.77\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 108.95 = 50.12\ \text{m/s} \end{aligned}

3. Jet diameter (single jet)

Q=π4d2V1d=4×1.8π×106.77=0.1465 m≈147 mm\begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 1.8}{\pi \times 106.77}} = 0.1465\ \text{m} \approx 147\ \text{mm} \end{aligned}

4. Speed of the wheel

Trial with jet ratio m=12m = 12: D=12d=1.758D = 12d = 1.758 m, so

N=60uπD=60×50.12π×1.758=544.4 rpmN = \frac{60u}{\pi D} = \frac{60 \times 50.12}{\pi \times 1.758} = 544.4\ \text{rpm}

Nearest synchronous speed for 50 Hz: N=120f/p=6000/12=500N = 120f/p = 6000/12 = 500 rpm (12 poles).

5. Wheel (pitch circle) diameter and jet ratio

D=60uπN=60×50.12π×500=1.914 mm=Dd=1.9140.1465=13.07\begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 50.12}{\pi \times 500} = 1.914\ \text{m} \\ m &= \frac{D}{d} = \frac{1.914}{0.1465} = 13.07 \end{aligned}

mm lies within 10 to 14, so a single-jet wheel is satisfactory.

6. Specific speed check

Ns=NPH5/4=500×96.413000.5=16.1N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{500 \times 96.41}{3000.5} = 16.1

This is within the single-jet Pelton range (about 8.5 to 30), so the Pelton wheel is the correct choice.

7. Number of buckets (Tygun's formula)

z=15+D2d=15+13.072=21.53≈22z = 15 + \frac{D}{2d} = 15 + \frac{13.07}{2} = 21.53 \approx 22

8. Bucket size (usual proportions)

  • Bucket width B≈5d=0.733B \approx 5d = 0.733 m
  • Bucket depth T≈1.2d=0.176T \approx 1.2d = 0.176 m
  • Bucket radial length L≈2.5d=0.366L \approx 2.5d = 0.366 m

Summary of design

ItemValue
Power9.29 MW
Speed500 rpm
Number of jets1
Jet diameter dd147 mm
Wheel diameter DD1.91 m
Jet ratio mm13.07
Specific speed NsN_s16.1
Number of buckets22
Bucket width × depth0.73 m × 0.18 m

✓ Answer checked by Claude Opus 5.5 (high effort)

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