Question

A Pelton wheel of 1.2 m mean bucket diameter works under a head of 650 m. The jet deflection
is 1650 and its relative velocity is reduced over the buckets by 15% due to friction. If the water is
to leave the bucket without whirl, determine the rotational speed of the wheel. The coefficient of
velocity is C, 50.98
(a) 786.38 rpm
(b) 1425.36 rpm
(c) 320.2 rpm
(d) 480.6 rpm

In the above problem, the ratio of bucket to jet speed can be calculated as
(a) 0.1
(b) 1.5
(c) 0.45
(d) 2.52

In the above problem, for a jet diameter of 100 cm, the impulsive force on a bucket of the Pelton
wheel can be calculated as
(a) 12.26 kN
(b)
62.36 kN
(c) 32.36 kN
(d) 94.26 kN

Answer 1

Explanation:

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Answer 2

Concept:

We first need to determine the velocity of the Jet using the equation Cv√2gh where the Cv is considered to be 0.97

Given:

Head, h = 650 m

Diameter, d = 1.2 m

Jet Deflection = 165°

Loss = 15% = η = 85% = 0.85

Cv = 0.97

Find:

We need to determine the rotational speed and ratio of the bucket to jet speed.

Solution:

Let V1 be the in-let velocity

It is given by the equation = Cv√2gh where h is the head and g is the acceleration due to gravity

Assuming Cv = 0.97

Therefore, V1 = 0.97 √2×9.8×650

V1 = 109.54 m/s

We know, V1 = Vw1 = Velocity of jet = 109.54

Now, u1 is the velocity of bucket = 0.45 × V1

= 0.45 × 109.54

= 49.29 m/s

Now, Vr1 = V1 - u

Vr1 = 109.54 - 49.29

= 60.21 m/s

u1 = u2 = u = 49.29 m/s

We can consider Ф = 15°

Therefore, Vr2 = η × Vr1

= 0.85 × 60.21

=  51.1785

Vw2 = Vr2 × cosФ - u

[51.1785 × cos(15°)] - 49.29

= 0.144 m/s

We know, u = π DN/60

Therefore, 49.29 = π × 1.2 × N/60

N = 49.29 × 60/π × 1.2

N = 784.47 rpm is the rotational speed.

The ratio of bucket velocity to jet velocity = 50/110 - approximate values

Ratio = 0.45

Thus, the rotational speed of the wheel is 784.47 rpm and the ratio of the bucket to jet speed is 0.45.

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