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A penstock of 1 m diameter and 5 km length is used to supply water from a reservoir to an impulse turbine. A nozzle of 15 cm diameter is fixed at the end of the penstock. The elevation difference between the turbine and water level in the reservoir is 500 m. Consider the head loss due to friction as 5% of the velocity head available at the jet. Assume unit weight of water = 10 kN/m3 and acceleration due to gravity (g) = 10 m/s2. If the overall efficiency is 80%, power generated (expressed in kW and rounded to nearest integer) is _______________

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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

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