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Power to raise water from point A to B

  1. Mar 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Water is pumped from a well into a home. The height of the water to be lifted is 20 m, and they want to reach a water flow of up to 3.6 m3/h. The effective efficiency, when one takes into account the energy losses in both the pump in the water flowing in the pipes, is estimated at 40%. How much power should the pump be? (It is assumed that the kinetic energy of the water can be completely neglected in this context.)


    2. Relevant equations

    W=mgh
    Efficiency=W_{applied}/(W_{quarried})
    And I am sure that I am missing an equation(don't know which).

    3. The attempt at a solution

    It is given that the desired flow is up to 3.6 m^3/h = 3600 dm^3/h = 3600 kg/h. And the energy required to transport the water from the well to the home per hour is

    W=mgH/h=3600*9.81*20/h≈ 700 000 W/h. Now the efficiency is 40% which means that the minimum power P form the pump must be:

    0.4P = 700 000 W/h. Hence, P = 1.8*10^6 W/h. Very far from the correct answer(0.5kW). What am I missing?
     
  2. jcsd
  3. Mar 27, 2013 #2
    Check your units. That is the only mistake you have committed.
     
  4. Mar 27, 2013 #3
    Oh. (1.8*10^6)/(3600s)=500 W/second. But in that case I transform from W/hour to W/second. Confusing since it should only be W. Also the answer is in only W
     
  5. Mar 27, 2013 #4
    I also have an other concern that I don't understand from the problem statement:

    It is assumed that the kinetic energy of the water can be completely neglected in this context.

    I don't understand the relevancy of this at all. How is this relevant? Or in what context could it be relevant?
     
  6. Mar 27, 2013 #5
    Are you sue it is W/h. Or is it J/h?

    You can neglect the kinetic energy of the water pumped on the assumption that the water was pumped quite slowly ((i.e. reversibly;like in a reversible thermodynamic process). This assumption is just an ideal scenario.
     
  7. Mar 27, 2013 #6
    I am not sure but I believe so since the answer is 0.5 kW. Does anyone know?
     
  8. Mar 27, 2013 #7
    Yes, 0.5 kW is the correct answer. Remember , Efficiency= Power output/Power input....
    You had to calculate power input in this case which comes out to be 0.5 kW.
     
  9. Mar 27, 2013 #8
    Well, check the dimensions of power. Which one of J/h and W/h have the same dimensions?
     
  10. Mar 27, 2013 #9
    Forgive me for not being smooth here. But I still don't understand. It is something with the units I think.

    The amount of work required from A to B is 700 000 W/h and η=40%=0.4 Therefore

    0.4*Power input = 700 000. Hence power input = 1.8*10^6 W/h. The answer is 500 W = 0.5kW which is given if 1.8*10^6 W/h is divided by 3600 seconds which would yield W/second. So what am I doing wrong?
     
  11. Mar 27, 2013 #10
    Okay, back to the basics. What is the unit of power? (consult your textbook,if you are confused)
    What are its dimensions?
    What are those of W/h? Does it have the same dimensions as those of power?
     
  12. Mar 28, 2013 #11
    Again, kW is the dimension of power. Watt=Joule per second. KiloWatt= Joule *1000 per second.

    Watt per hour as you state is not the dimension of power, watt is.

    The question specifically stated that water "flow up" rate is 3.6 m3 per hour. What will be its rate in m3 per second ? How will you find mass flow up per second then ? How will you get gain in potential energy per second ?

    CHECK YOUR UNITS !!!
     
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