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Thermodynamics problem -- Water pumped into an enclosed tank...

  1. Oct 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Q3.1 (a) A pump forces 1m^3 /min of water horizontally from an open well to a closed tank where the pressure is 0.9 Mpa.compute the work the pump must do upon the water in an hour just to force the water into the tank against the pressure.sketch the system upon which the work iss done before and after the process.



    2. Relevant equations


    3. The attempt at a solution
    here flow rate is 1m^3/min
    p1=0.101325 pa
    p2=0.9 pa
    we don't know v.
    but equation formed will be as:
    w=dpdv
    now i'm where i can get v.
     
  2. jcsd
  3. Oct 13, 2015 #2

    haruspex

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    This is not a thermodynamics question. Titling it so may have deterred some potential responders.
    Where you wrote pa in your attempted solution, I guess you meant MPa.
    You don't need to know the velocity.
    I don't understand your w=dpdv equation. Please explain the terms.
     
  4. Oct 13, 2015 #3
    it's w=(p2-p1)(v2-v1)
     
  5. Oct 13, 2015 #4
    it's w=(p2-p1)(v2-v1)
     
  6. Oct 13, 2015 #5

    haruspex

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    I asked you to explain the terms. What are the v's? If they're velocities, and the p's are pressures, and w is work, then the equation is dimensionally wrong. If the v's are volumes then the dimensions work, but I don't see the relevance to problem. [And anyway, I think you would mean W= d(PV)=PdV+VdP.]
     
  7. Oct 13, 2015 #6
    v=volume
     
  8. Oct 13, 2015 #7

    haruspex

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    Ok.
    You know the flow rate, and you know the time, so...
     
  9. Oct 13, 2015 #8
    so what then?
     
  10. Oct 13, 2015 #9

    haruspex

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    So you can find the volume.
     
  11. Oct 13, 2015 #10
  12. Oct 13, 2015 #11

    haruspex

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    A cubic metre of water per minute, for one hour, gives...?
     
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