How is the power developed by the Pelton turbine calculated?

P = overall efficiency x rho g Q H = 0.85 x 9.81 x 5 x 312.5 = 13028.9 kW

P = rho g Q H with no efficiency

P = eta x 0.5 rho Q V1^2 with V1 = sqrt(2gH) only, ignoring head

How are the jet velocity V1 and bucket velocity u found?

V1 = Cv sqrt(2gH) = 76.74 m/s and u = phi sqrt(2gH) = 36.02 m/s

V1 = sqrt(2gH) with no coefficient, and u = V1

V1 = phi sqrt(2gH) and u = Cv sqrt(2gH)

How is the jet diameter d found for a single jet?

From Q = (pi/4) d^2 V1, so d = sqrt(4Q/(pi V1)) = 0.2880 m (288 mm)

From Q = pi d^2 u, using the bucket velocity

From P = pi d^2 V1, using the power

How are the wheel speed and diameter fixed?

Pick a synchronous speed N = 120f/p (200 rpm), then D = 60u/(pi N) = 3.440 m, giving m = D/d = 11.94, within 10 to 14

Use any speed and accept any m, because only d matters

Take D = 12d always and N follows with no synchronous restriction

Choose N from Ns = 100 and ignore m

What do the specific speed check and bucket count show?

Ns = N sqrt(P)/H^(5/4) = 17.4, inside the single-jet range 8.5 to 30; z = 15 + m/2 = 20.97, so 21 buckets

Ns = 17.4 is above 30, so several jets are needed; z = 15 + m = 25 buckets

Ns must equal 100 exactly; z = D/d buckets

Ns is irrelevant to Pelton choice; z = 2m = 23 buckets

2080 Chaitra · 8 marks

Pelton turbine design (5 m³/s, 312.5 m)

Short answer

Single-jet Pelton wheel: power 13.03 MW at 200 rpm (30 poles). Jet diameter 288 mm, wheel diameter 3.44 m, jet ratio 11.94. Specific speed 17.4. 21 buckets; bucket width 1.44 m, depth 0.35 m.

Design of a Pelton turbine for a hydropower plant having net head of 312.5 m and discharge 5 cumecs. Take efficiency of turbine 85%, frequency 50 Hz and velocity coefficient 0.98.

A Pelton wheel is designed by finding the jet velocity and jet diameter, choosing a synchronous speed that gives a suitable jet ratio $m = D/d$ (10 to 14 for good efficiency), and then sizing the wheel and buckets.

Data: $Q = 5$ m³/s, $H = 312.5$ m, $\eta_o = 0.85$ , $f = 50$ Hz, coefficient of velocity of the jet $C_v = 0.98$ (given). Assumed from the standard range: speed ratio $\phi = 0.46$ (0.43 to 0.47).

1. Power developed

P = \eta_o \rho g Q H = 0.85 \times 9.81 \times 5 \times 312.5 = 13028.9\ \text{kW}

2. Jet velocity and bucket (peripheral) velocity

\begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 312.5} = 78.30\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.98 \times 78.30 = 76.74\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 78.30 = 36.02\ \text{m/s} \end{aligned}

3. Jet diameter (single jet)

\begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 5}{\pi \times 76.74}} = 0.2880\ \text{m} \approx 288\ \text{mm} \end{aligned}

4. Speed of the wheel

Trial with jet ratio $m = 12$ : $D = 12d = 3.456$ m, so

N = \frac{60u}{\pi D} = \frac{60 \times 36.02}{\pi \times 3.456} = 199.0\ \text{rpm}

Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/30 = 200$ rpm (30 poles).

5. Wheel (pitch circle) diameter and jet ratio

\begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 36.02}{\pi \times 200} = 3.440\ \text{m} \\ m &= \frac{D}{d} = \frac{3.440}{0.2880} = 11.94 \end{aligned}

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

6. Specific speed check

N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{200 \times 114.14}{1313.9} = 17.4

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 + \frac{D}{2d} = 15 + \frac{11.94}{2} = 20.97 \approx 21

8. Bucket size (usual proportions)

Bucket width $B \approx 5d = 1.440$ m
Bucket depth $T \approx 1.2d = 0.346$ m
Bucket radial length $L \approx 2.5d = 0.720$ m

Summary of design

Item	Value
Power	13.03 MW
Speed	200 rpm
Number of jets	1
Jet diameter $d$	288 mm
Wheel diameter $D$	3.44 m
Jet ratio $m$	11.94
Specific speed $N_s$	17.4
Number of buckets	21
Bucket width × depth	1.44 m × 0.35 m

✓ Answer checked by Claude Opus 5.5 (high effort)

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