potentialenergy= mgh=1kg*9.8N/kg*26m
massflow*g*h/time * .80=54*10^6
Flowrate=54^10^6/gh * 1/.80
In a hydroelectric dam, water falls 26.0 m and then spins a turbine to generate electricity. What is ∆U of 1.0 kg of water? Suppose the dam is 80% efficient at converting the water's potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 54.0 MW of electricity? This is a typical value for a small hydroelectric dam.
3 answers
If what you are calling ∆U includes the gravitational potential energy*, then it is -(density*g*(delta H)
= -(10^3 kg/m^3)*26 m*9.81 m/s^2
The product will be in Joules.
For P = 54*10^6 Watts, you need a mass flow rate such that
P = (mass flow rate)*(∆U)
*Usually, in thermodynamics, U represents internal energy only, which has nothing to do with elevation.
= -(10^3 kg/m^3)*26 m*9.81 m/s^2
The product will be in Joules.
For P = 54*10^6 Watts, you need a mass flow rate such that
P = (mass flow rate)*(∆U)
*Usually, in thermodynamics, U represents internal energy only, which has nothing to do with elevation.
I forgot to include the 80% efficiency factor, but BobPursley did it right
1 answer