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    Can some one please help me to understand how to do questions 2.2 and 2.3
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    (Original post by medicapplicant)
    Can some one please help me to understand how to do questions 2.2 and 2.3
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    The pipe is circular and so forms a cylinder.

    The volume of a cylinder is Area x length = \pi r^2 l

    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.

    V = \pi r^{2} l

    r = \sqrt{(\frac{V}{\pi l}})

    Radius of pipe is 16.3cm.
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    (Original post by medicapplicant)
    Can some one please help me to understand how to do questions 2.2 and 2.3
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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) = \rm mgh = 3500 x 9.81 x 90 = 3.087x10^{6} Joules

    K.E. of water exiting turbine = \rm \frac{1}{2}mv^{2} = \frac{1}{2}(3500 x 12^{2}) = 0.252x10^{6} Joules


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

    \rm GPE - KE_{exit} = 3.087x10^{6} - 0.252 x 10^{6} = 2.835x10^{6} Joules



    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

    \rm 2.835x10^{6} \ Joules \ per \ second = 2.835x10^{6} \ Watts = 2.835 \ MW


    Rounding your answer to 2.8MW would be OK.
 
 
 
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