From a waterfall, water is pouring down at the rate of 100kg per second on the blades of turbine. If the height of the fall is 100m, the power delivered to the turbine is approximately equal to
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Answered by
11
mass pouring = 100 kg/s
height = 100m
Maximum energy that can be delivered per second = mgh
⇒ Max. Power = 100kg/s × 9.8 m/s² × 100m
⇒ Max. Power = 98000 J/s
⇒ Max. Power = 98 kJ/s = 98 kW
But all the energy is not transferred from water. The water which fell still has some Kinetic energy. We normally take 50% efficiency. So 50% of maximum Power is delivered.
So power delivered = 98 × (50/100) = 49 kW
height = 100m
Maximum energy that can be delivered per second = mgh
⇒ Max. Power = 100kg/s × 9.8 m/s² × 100m
⇒ Max. Power = 98000 J/s
⇒ Max. Power = 98 kJ/s = 98 kW
But all the energy is not transferred from water. The water which fell still has some Kinetic energy. We normally take 50% efficiency. So 50% of maximum Power is delivered.
So power delivered = 98 × (50/100) = 49 kW
Answered by
7
Hello....
It's your answer......
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Power = work done / time taken
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or P = mgh / t
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P = (m/t) gh = 100 x 10 x 100 = 100 kW
Thank You so much ...
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