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Turbine Question :|

  1. Mar 22, 2010 #1
    A turbine is installed in a vertical pipe line, 150mm diameter. A gauge is connected to the pipe at a point A above the turbine and another at point B below the turbine, the vertical distance between A and B being 1.2m. When the discharge is 0.06m^3/s, the gauages read 400kN/m^3 and 35kN/m^3 respectively.

    Calculate the output power from the turbine, assuming it to be 85% efficient.



    I've found the Cross sectional area = 0.01767m^2

    Vol per sec = 0.06
    therefore 0.06 = m/roh
    0.06 = m/10^3
    therefore mass flow rate = 60


    Q = AV
    0.06 = 0.01767V
    therefore V = 3.395m/s

    Then gone on to use Bernoulli's ... unsucessfully!

    Any help? Am I on the right tracks or?
     
  2. jcsd
  3. Mar 22, 2010 #2

    Filip Larsen

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    You may want to think about how pressure and flow rates relate to fluid power.
     
  4. Mar 22, 2010 #3
    I know I need the pressure values to put into the bernoullis equation but that is what I am stuck on finding/working out
     
  5. Mar 22, 2010 #4

    Filip Larsen

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    What quantity do the gauges measure? I assume its pressure (kN/m^2) and that you just wrote the unit as kN/m^3 by mistake.

    If it is pressure they measure, you have the pressure and flow rate at two points in the flow so you can very easily calculate the fluid power those points without need of Bernoulli's equation. You also know how much pressure (and hence fluid power) increases over a 1.2 m drop so you can calculate how much power the fluid should have at point B if the turbine wasn't there. The difference between that value and the actual measured power at B is equal to the fluid power absorbed by the turbine (well, and pipe resistance, but that seems to be ignored in the problem text).
     
  6. Mar 22, 2010 #5
    So I presume that the flow rate that I have is correct?
     
  7. Mar 22, 2010 #6

    Filip Larsen

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    Yes, if the volume flow rate is 0.060 m^3/s and it is water we are talking about (density 1000 kg/m^3) then a mass flow rate of 60 kg/s is correct.
     
  8. Mar 22, 2010 #7
    Thanks a lot!
     
  9. Mar 22, 2010 #8
    I know how much the fluid power/pressure increases by 1.2m drop but how do I work it out if the turbine isn't there?
     
  10. Mar 22, 2010 #9

    Filip Larsen

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    If there were no turbine (i.e. no pressure resistance between A and B), the pressure at B should be the pressure PA at A plus the pressure increase Ph from the drop. If we now include the turbine and the measured pressure PB at B, you now also have to include the pressure drop across the turbine PT so that you get the relation PA - PT + Ph = PB from which you can calculate PT and the fluid power that it corresponds to (using the mass flow rate).
     
  11. Mar 22, 2010 #10
    That makes more sense to me now, thank you :)
     
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