Concept:
The net head on the turbine is given by,
Hnet = Hg – Hf
Where,
Hg = Gross head = Difference between the water level of the reservoir and the turbine
Hf = Frictional loss
The overall efficiency of the turbine is given by,
\({{\rm{\eta }}_{\rm{o}}} = \frac{{{\rm{Shaft\;power}}}}{{{\rm{Water\;power}}}}\)
The water power is calculated by,
\({\rm{Water\;power}} = \frac{{\rho × g × Q × H}}{{1000}}\)
Q = discharge of water = Area of nozzle × Velocity of water
Calculation:
Given,
d = dia of nozzle = 15 cm
Area of nozzle \(= \frac{\pi }{4}{\left( {0.15} \right)^2} = 0.0177{m^2}\)
Hg = 500 m
Hf = 5 % of velocity head
Hnet = Hg – Hf
Hg = Hnet + friction loss
\(\begin{array}{l} 500 = H_f + \frac{{{V^2}}}{{2g}}\\ 500 = \frac{{0.05{V^2}}}{{2g}} + \frac{{{V^2}}}{{2g}}\\ \frac{{1.05{V^2}}}{{2g}} = 500 \end{array}\)
V = 97.59 m/sec
\({H_f} = \frac{{0.05 × {V^2}}}{{2g}} = \frac{{0.05 × {{\left( {97.59} \right)}^2}}}{{2 × 10}} = 23.809\;m\)
∴ Hnet = Hg – Hf = 500 - 23.809 = 476.2 m
Q = 0.0177 × 97.59 = 1.727 m3/sec
\({\rm{Water\;power}} = \frac{{\rho × g × Q × H}}{{1000}}= \frac{{1000 × 10 × 1.727× 476.2}}{{1000}}=8212.3\ kW\)
W.P = 8212.18 KW
S.P. = W.P × ηo
S.P= 0.8 × 8212.18 = 6569.7 KW