How Much Water Flow Is Needed to Generate 780 kW from a 27m Dam?

AI Thread Summary
To generate 780 kW of electrical energy from a 27m dam with a 65% efficiency, the required water flow rate must be calculated. The mass of water falling per second and the kinetic energy gained from the 27m drop are critical factors in this calculation. The energy conversion efficiency means that only 65% of the potential energy is converted into electrical energy, necessitating an adjustment for losses. The discussion emphasizes the need to determine the flow rate in liters per second to meet the energy output requirement. Understanding these relationships is essential for solving the problem effectively.
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Homework Statement



A small dam produces electrical power. The water falls a distance of 27m to turn a turbine. If the efficiency to produce electrical energy is 65%, at what rate must water flow over the dam to produce 780 kW of electrical energy?


Homework Equations



v2 = u2 + 2as

P = \tau x \omega

The Attempt at a Solution



I've basically been at this one on and off for a few hours, and have tied myself so completely in knots that I have no idea what I'm doing anymore. Please help someone!
 
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Consider a litre of water going over that dam in one second.
What is its mass?
How much kinetic energy will it pick up by falling 27 metres?
The energy conversion is not totally efficent so reduce that by 35%.
Now how many litres per second do I need?
 
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