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# As physics power question?? Help watch

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2. (Original post by medicapplicant)
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The pipe is circular and so forms a cylinder.

The volume of a cylinder is Area x length =

You calculated the maximum speed of water exiting the pipe (42m/s) in the first part.

i.e. a 42 metre long column of water will pass the same point every second.

We are given the volume of that cylinder (3.5m3) so now it's a case of rearranging the volume formula to make the radius the subject and then plug in the values.

3. (Original post by medicapplicant)
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Q2.3

Maximum possible power output will be when the G.P.E. of the water in the lake entering the pipe intake at 3.5m3 per second is all converted to kinetic energy after a drop of 90m and then converted to electrical energy with no losses.

We are told that the exit speed of the water after the turbine is 12ms-1

Which means the residual K.E. of 3.5m3 of water exiting the turbine at 12ms-1 must be subtracted from the G.P.E. of the water at the entrance to the pipe with the difference assumed to have all been converted to electrical energy.

Water has a mass of 1kg per litre at standard temperature and pressure.

3.5m3 = 3500 litres = 3500kg. (1m3 = 1000 litres = 1000 kg = 1 metric tonne)

G.P.E. (per 3.5m3 of water) =

K.E. of water exiting turbine =

Which means the conversion of energy for each 3.5m3 of water is:

We are also told that water flows through the pipe at a rate of 3.5m3/s i.e. 3500kg/s

Power = Energy/Time (work done per second)

Hence the maximum power output is

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