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.