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Math Help - work done by a pump

  1. #1
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    work done by a pump

    an engine is pumping water from a large tank delivering it through a pipe of diameter 0.04m at a rate of 100 litres per second. Find the work done by the pump (mass of 1 litre is 1 kg)

    at a loss with this question.

    im assuming pump is horizontal so no PE (?)

    then 100 litres= 0.1m^3
    and volume is L(pi(0.02)^2=0.001256L (L = length of pipe)


    so 0.001256L=.1m^3
    L=79.62

    so in 1 second 100 kg of water moves 79.62 so

    KE=(0.5)(100)(79.62)^2=3.17x10^5 which is correct answer, but have i just stumbled across this by accident or is my reasoning ok?
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  2. #2
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    Re: work done by a pump

    net work = change in kinetic energy

    you used the work/kinetic energy theorem
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    Re: work done by a pump

    Hi jiboom,
    hydraulic Hp= gpm*H*s*8.33/33000
    where H =total discharge head-total suction head assuming the level in tank is above pump.If you do not have tnese heads you have to calculate head losses in suction and discharge piping systems.
    s is specific gravity of fluid .If you need metric units make the conversions or get it on line
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    Re: work done by a pump

    thanks for replies. I have another one,but this time i have no idea what to do:

    water is pumped at the rate of 1.2cubic metres per minute from a large tank on the ground, up to a point 8 metres above the level of the water in the tank. It emerges as a horizontal jet from a pipe with a cross section of 5x10^(-3) square metres. if the efficiency of the apparatus is 60%, find the energy supplied to the pump per second.

    anyhelp would be gratefully received
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  5. #5
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    Re: work done by a pump

    Quote Originally Posted by jiboom View Post
    water is pumped at the rate of 1.2cubic metres per minute from a large tank on the ground, up to a point 8 metres above the level of the water in the tank. It emerges as a horizontal jet from a pipe with a cross section of 5x10^(-3) square metres. if the efficiency of the apparatus is 60%, find the energy supplied to the pump per second.
    efficiency is the ratio, E = (work done)/(work input)

    work done in one second will be the gain in gravitational potential energy of (1.2/60) cubic meters of water as it is raised 8 meters by the pump.

    you were given the efficiency and you can calculate the gain in potential energy ... determine the work input.
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  6. #6
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    Re: work done by a pump

    Hi jiboom,
    My reply to your first question was an engineering answer.In your new question in wnich no pipe information is given except a discharge cross-section and want a simple answer you can assume that pipe losses are negligible but you cannot neglect the loss in head in the discharge jet.The latter would depend on type of opening (orifice or nozzle)


    bjh
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