Answer: Hydraulic Efficiency of a Pelton Turbine

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The hydraulic efficiency of a Pelton turbine is calculated using the ratio of shaft power to water power, with shaft power defined as the product of mass flow rate, velocity components, and wheel speed. Water power is determined by the density, gravitational acceleration, flow rate, and height of the water source. The discussion highlights confusion regarding the relationship between kinetic energy and water power, specifically questioning why water power is expressed as density times gravity times flow rate times height instead of kinetic energy. It clarifies that while kinetic energy is relevant, power is defined as energy per time, necessitating the use of flow rate in calculations. Overall, the conversation emphasizes the distinction between the energy input to the runner and the total energy input to the turbine system.
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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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