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A Pelton wheel with single jet rotates at 600 rpm. The velocity of the jet from the nozzle is 100 m/s. If the ratio of the bucket velocity to jet velocity is 0.44 and the speed ratio is 0.43, what is the coefficient of velocity of the nozzle?
1. 0.817
2. 0.882
3. 0.975
4. 0.977

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Correct Answer - Option 4 : 0.977

Concept:

The velocity of water jet V through nozzle at inlet in case of a Pelton wheel is given by,

\({\rm{V}} = {\rm{\;}}{{\rm{C}}_{\rm{v}}}\sqrt {2{\rm{g}}{{\rm{H}}_{{\rm{net}}}}}\)

And, the velocity of the wheel U in case of a Pelton wheel is given by,

\({\rm{U}} = {\rm{\;}}\phi \sqrt {2{\rm{g}}{{\rm{H}}_{{\rm{net}}}}} \)

Where, V is the velocity of the water jet at inlet,

U is the velocity of wheel (or bucket),

\({{\rm{C}}_{\rm{v}}}\) is the coefficient velocity (normally varies from 0.97 to 0.99 for Pelton wheel)

φ is the speed ratio,

\({{\rm{H}}_{{\rm{net}}}}\) is the net head available after accounting for all losses and

g is acceleration due to gravity.

Calculation:

Given, the velocity of water jet, V = 100m/s.

Ratio of bucket velocity to the velocity of jet = 0.44. \(\therefore \frac{{\rm{U}}}{{\rm{V}}} = 0.44{\rm{\;\;}}\therefore {\rm{U}} = 0.44 \times 100 = 44{\rm{\;m}}/{\rm{s}}\)

\({\rm{Speed\;ratio}},{\rm{\;}}\phi = 0.43.\)

\(\therefore {\rm{U}} = {\rm{\;}}\phi \sqrt {2{\rm{g}}{{\rm{H}}_{{\rm{net}}}}} {\rm{\;\;}}\therefore \sqrt {2{\rm{g}}{{\rm{H}}_{{\rm{net}}}}} = \frac{{\rm{U}}}{\phi } = \frac{{44}}{{0.43}} = 102.325{\rm{\;m}}/{\rm{s}}\)

\(\therefore {\rm{V}} = {\rm{\;}}{{\rm{C}}_{\rm{v}}}\sqrt {2{\rm{g}}{{\rm{H}}_{{\rm{net}}}}} {\rm{\;}}\therefore {{\rm{C}}_{\rm{v}}} = \frac{{\rm{V}}}{{\sqrt {2{\rm{g}}{{\rm{H}}_{{\rm{net}}}}} }} = \frac{{100}}{{102.325}} = 0.977\)

∴ The coefficient of velocity of the nozzle is 0.977.

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