Can some one please help me to understand how to do questions 2.2 and 2.3
Thanks guys Attachment 529311529313
As physics power question?? Help watch
- Thread Starter
- 08-05-2016 17:38
- Study Helper
- 08-05-2016 18:36
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.
Radius of pipe is 16.3cm.Last edited by uberteknik; 08-05-2016 at 20:07.
- Study Helper
- 09-05-2016 10:12
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
Rounding your answer to 2.8MW would be OK.Last edited by uberteknik; 09-05-2016 at 10:50.