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Thermodynamis - oil pump efficiency?

  1. Aug 5, 2006 #1
    Hi guys, could someone help me on this? I'm stuck :(

    *A motor draws 35kW electricity, it operates at 90% efficiency. The motor is connected to an oil pump which has an inlet diameter of 8cm & an outlet diameter of 12cm. pressure rise in the oil pump is 400kPa. oil enters the pump at 0.1m^3/s. what is the mechanical efficiency of the oil pump?*

    I understand the question, I know the mechanical input power is 31.5kW and that i need to figure out the output power from the info given, then the mech efficiency is that number/31.5kW ... but I've got no clue how to go about finding the output power. sooo.... could I get some pointers pretty please? any help would be greatly appreciated.
  2. jcsd
  3. Aug 9, 2006 #2
    There is some problem with the data. The hydraulic power consumption is 400kPa x 0.1 m^3/s or 400 kN/m^2 x 0.1m^3/s or 40 kN-m/s which is 40kW (then you have pump efficiency). Recheck the problem. The pipe size is redundant data.
  4. Aug 9, 2006 #3
    i left out the oil's density.. dunno if that makes a difference to your query though? word for word the question reads:

    An oil pump is drawing 35kW of electric power while pumping oil with a density of 860kg/m^3 at a rate of 0.1m^3/s. The inlet and outlet diameters of the pipe are 8cm and 12cm, respectively. If the pressure rise of the oil in the pump is 400 kPa and the motor efficiency is 90%, determine the mechanical efficiency of the pump.
  5. Aug 9, 2006 #4
    The density of the oil does matter. Let me try to be the teacher that I am and, rather than giving you the answer, ask you some questions that might get you on the right path. First, I will say that figuring out the mechanical input power is correct as one of the first steps. Now for some questions:

    1) What do you know about mass flow rate? (Are you familiar with the mass flow rate equation?) If so you can solve for an another important input state variable. :wink:
    2) You are also given the total pressure increase from the input to the output. Do you think the level-flow form of the Bernoulli's equation might help you get from input to output once you know the input state variable from (1)?
    3) Wouldn't the output power = output force * output velocity?

    I hope I haven't made it TOO easy! :biggrin:

  6. Aug 10, 2006 #5
    I don't see any point in this particular context, why you require density as the head is expressed in kPa and the flowrate in m^3/s. If the head was given in either ft or meters of water, then you should go for mass flow rate.

    The equation I provided above is dimensionally stable even without density. I still see some problem in the question.
  7. Aug 10, 2006 #6
    it's one my lecturer picked out specifically from a fifth edition textbook... so I'd assume it's OK..

    Thanks rainman, I think i've made some progress with it.. still a little stuck though...

    from the volume rate of 0.1 m^3/s,

    I get mass flow rate = 86 kg/s using m' = pv'

    Also calculated the average input velocity, got 19.89 m/s using V = m'/(pA)

    I think that covers step one of what you gave me?

    then for step two... by level flow do you mean steady flow? just never heard/seen that terminoligy before... but if that is it,

    pressure rise/density - 0.5(Vout^2 - Vin^2) = 0 right?

    which gives you Vout... but why couldn't you calculate that from the same equation you got Vin from? I see you don't get the same value...

    Then what's output force? Pressure rise/outlet cross sectional area?

    I dont get the right answer either way :( can you see where I'm going wrong? thanks again
  8. Aug 10, 2006 #7
    Hello five dots,
    No, I just mean you don't need a potential energy (delta-h) term in Bernoulli's equation. IOW assume h1=h2.
    This is using the ideal version of Bernoulli's equation (no losses). Yet in the real world pump there are frictional and heat losses through the pump, and you need to account for these in your equation, since that term is the fundamental basis for why the pump efficiency will not be 100%. Another issue with using this form of the Bernoulli equation is that it assumes there is no energy added between in and out (only conversion of static pressure to dynamic pressure). Yet this is what a pump does. It adds total pressure energy to the flow (Delta-P-total). So (Ptotal-out)-(Ptotal-in) = 400 kPa.

    You can and you should use the mass flow equation to get Vout, so you can use it in the "lossey" form of the Bernoulli equation (which includes a pressure loss term that you are wishing to solve for).

    Yes. But now that I think about it, you don't even have to go that far, as solving for the pressure loss term will get you to the pump efficiency solution you require.

  9. Aug 11, 2006 #8

    so Vout = 86/(860 x 0.06^2 x pi) = 8.84 m/s


    work (out) per unit mass = pressure rise/density + 0.5(Vout^2 - Vin^2) = 306.31 J/kg

    Output Power = mass flow rate x work per unit mass = 26342.94 W

    efficiency = 26342.94/31500 = 0.84

    Is that right? The answer my lecturer gave was 0.87, but I guess that difference could be round off errors and whatnot.. As long as the method's right I'm happy...

    So, (fingers crossed) is that the right method?

    and thank you very much for your help
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