# As physics power question?? Help

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#1
Thanks guys Attachment 529311529313
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5 years ago
#2
(Original post by medicapplicant)
Thanks guys Attachment 529311529313
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.

1
5 years ago
#3
(Original post by medicapplicant)
Thanks guys Attachment 529311529313
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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