A turbine is situated under a huge water fall. The water pours down on the blades of the turbine at the rate of 190 kg per second. The height of the water fall is 30 m.

What is the power delivered to the turbine?

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Solution

$P o w e r = \frac{w o r k d o n e}{t i m e} w o r k d o n e = p o t e n t i a l e n e r g y o f w a t e r a t h e i g h t 30 m p E = m g h p o w e r = \frac{P E}{t i m e} = \frac{m g h}{t} = (\frac{m}{t}) g h = 190 \times 10 \times 30 = 57000 j o u l e$

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A turbine is situated under a huge water fall. The water pours down on the blades of the turbine at the rate of 190 kg per second. The height of the water fall is 30 m. What is the power delivered to the turbine?

Power = work donetimework done = potential energy of water at height 30mpE = mgh power = PEtime = mght = mtgh= 190×10×30=57000 joule

A turbine is situated under a huge water fall. The water pours down on the blades of the turbine at the rate of 190 kg per second. The height of the water fall is 30 m.

What is the power delivered to the turbine?

$P o w e r = \frac{w o r k d o n e}{t i m e} w o r k d o n e = p o t e n t i a l e n e r g y o f w a t e r a t h e i g h t 30 m p E = m g h p o w e r = \frac{P E}{t i m e} = \frac{m g h}{t} = (\frac{m}{t}) g h = 190 \times 10 \times 30 = 57000 j o u l e$