Answer: Hydraulic Efficiency of a Pelton Turbine

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The hydraulic efficiency of a Pelton turbine is defined as the ratio of shaft power to water power, calculated using the formulas: shaft power = (mass flow rate) * (Vw1 + Vw2) * u and water power = density * g * Q * H. The discussion clarifies that while the kinetic energy of water exiting the nozzle is represented by 0.5 * m * V^2, the total energy input to the turbine system is accurately described by density * g * Q * H, which accounts for the gravitational potential energy available from the water source. The distinction between mass and flow rate is crucial for understanding these calculations.

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skpmech
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The hydraulic efficiency of a turbine (let pelton turbine) is given by - (power developed by runner or shaft power)/(water power)

shaft power is - (mass flow rate)*(Vw1+Vw2)*u

water power is - density*g*Q*H

My question is that the pressure head is converted into kinetic energy in nozzle then we should use water power = 0.5*m*V^2
why we use that formula.

my teacher said that energy input to runner= 0.5*m*V^2
and energy input to whole turbine system = density*g*Q*H
please explain.
 
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Not sure I follow your question but..

density*g*Q*H ... That looks like the equation for the theoretical power available from something like a dam. It depends on the density, gravity, height and flow rate. Presumably Q is the flow rate?

The KE of a mass m coming out of the pipe will be 0.5*m*V^2 but power = energy/time. So is m the mass or flow rate?
 

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