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Math Help - power of a water pump

  1. #1
    Senior Member furor celtica's Avatar
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    power of a water pump

    A pump, taking water from a large reservoir, is used to spray a jet of water with speed 20 ms^1 and radius 0.05 metres, from a nozzle level with the surface of the reservoir. Calculate the power of the pump.

    Alright I'll admit I'm pretty lost here and haven't been able to mount a very convincing attempt at a solution. I'm used to force-work-power problems based on the motion of an object so I'm not sure how to model this exercise. Even the wording is unclear; should I take initial speed 20 ms^-1, ending speed 0, distance 0.05? Where is the jet directed, horizontally or vertically?
    Any suggestions would be appreciated.
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  2. #2
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    Re: power of a water pump

    Quote Originally Posted by furor celtica View Post
    A pump, taking water from a large reservoir, is used to spray a jet of water with speed 20 ms^1 and radius 0.05 metres, from a nozzle level with the surface of the reservoir. Calculate the power of the pump.
    water density is 1 kg/L

    1 L = 0.001 m^3

    In a pipe of radius 0.05 m , 1 kg of water occupies a length approximately 0.127 m.

    Traveling at 20 m/s , 1 kg of water takes about 0.006 sec to travel that distance.

    P = \frac{W}{t} = \frac{\Delta E_K}{t} = \frac{\frac{1}{2}mv^2}{t} = \frac{\frac{1}{2}(1)(20^2)}{0.006} \approx 31.4 kW
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  3. #3
    Senior Member furor celtica's Avatar
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    Re: power of a water pump

    thanks bro
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    Re: power of a water pump

    Hello Skeeter,
    Excuse me for using US Eng units

    Velocity of jet = 20mps*3.281=65.6fps
    Jet dia =.05m * 39.37 = 1.97 in (.164 ft)
    Cross-section of jet (,164)^2*pi/4=.0211 ft^2
    Q=V*A =65.6*.0211=1.38cfs
    Q gpm = 1.38*7.5*60 =623 gpm
    Velocity head of jet =(65.6)^2/64.4 =66.8 ft
    Hydraulic Hp = gpm *8.33* 66.8/33000 =10.5Hp 7.84 KW
    No pipe friction loss and no potential head loss and no pump eff loss
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    Re: power of a water pump

    Quote Originally Posted by bjhopper View Post
    Hello Skeeter,
    Excuse me for using US Eng units

    Velocity of jet = 20mps*3.281=65.6fps
    Jet dia =.05m * 39.37 = 1.97 in (.164 ft) ... radius?
    Cross-section of jet (,164)^2*pi/4=.0211 ft^2
    Q=V*A =65.6*.0211=1.38cfs
    Q gpm = 1.38*7.5*60 =623 gpm
    Velocity head of jet =(65.6)^2/64.4 =66.8 ft
    Hydraulic Hp = gpm *8.33* 66.8/33000 =10.5Hp 7.84 KW
    No pipe friction loss and no potential head loss and no pump eff loss
    ...
    Last edited by skeeter; September 7th 2011 at 04:33 AM.
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  6. #6
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    Re: power of a water pump

    Sorry Skeeter I used .05 m as a dia not a radius.Flow increases to 2494 gpm and Hp to 42 same as your answer in KW
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  7. #7
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    Re: power of a water pump

    Thanks for the confirmation from an engineer's perspective.
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