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Pool Circulation System Pumps

  1. Apr 25, 2010 #1

    danago

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    Hey.
    At the moment, for my fluid mechanics course, i am working on an assignment where i have been instructed to design a basic fluid system which involves the use of an operating centrifugal pump. The system i have chosen is a pool circulation and filtration system.

    I have finished with most of the system design work and calculations involving head requirements, NPSH available etc., however i now need to select a pump which would be suitable for the job.

    I have searched through the websites of quite a few pool pump manufacturers, and what i have found is that most of them have published a "head supplied vs. flow rate" curve, but none of them provide NPSH or efficiency curves.

    Does anybody know why NPSH curves are not published for smaller scale pumps like these? If i can't get hold of any, are there any rough guidelines that i should follow when sizing a pump to ensure that cavitation does not occur, and that the pump isnt operating too inefficiently?

    Clearly i am a beginner to this, so any help is very much appreciated :smile:
    Thanks,
    Dan.
     
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  3. Apr 25, 2010 #2

    Q_Goest

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    Hi danago,
    Consider that the fluid is water at roughly atmospheric pressure at the surface of the pool. Water boils at roughly 0.2 psia (give or take a few tenths of a psi) at atmospheric temperature. Let's say the pump is another 3+ feet below the surface of the pool, adding another 1.5 psi of head. So in order for the pump to cavitate, the pressure at the pump inlet has to drop from 14.7 + 1.5 = 16.2 psia down to about 0.2 psi right? That's a difference of 16 psi. Granted, there is an NPSH of 1 or so psi at the pump suction, but that only drops you down to 15 psi, give or take. In other words, the frictional pressure loss upstream of the pump has to be 15 psi in order for cavitation to occur. A typical inlet line on such a pump is around 1 or 2 inches. If that were to drop down to about 1/4", you might have a 15 psi pressure drop to the pump suction and cavitation would occur. But that isn't realistic. I suspect they don't provide NPSH simply because there's no reason to suspect that a system could possibly be designed that would create a 15 psi pressure drop through the suction line. It just isn't worth considering.

    That said, there are many pump situations in industry where NPSH must be considered. We always look at NPSH for any pump being installed. It's just that in this case, there is a very standard way of installing these pumps and the available NPSH is relatively large compared to what is typically available. Your typical swimming pool owner not only doesn't understand NPSH, but doesn't need to in order to install a pump. It would just confuse the swimming pool owner and create problems where none exist.

    Actually, I'm a bit surprised you found pump curves. Typical swimming pool owners don't generally understand anything about pump curves, let alone want to look for them and understand what they would mean.
     
  4. Apr 26, 2010 #3

    danago

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    Thanks very much for the reply! I suspected that the reason might have been something along those lines.

    Ive actually found that most of the pool pump manufacturers have supplied some form of pump curve, as minimal as most of them are. The most detailed one i have found was a head vs. flow curve for a few pump models, with only one "best efficiency" curve drawn over it.

    Do you have any idea what the typical operating efficiencies for small pumps like pool pumps are?
     
  5. Apr 26, 2010 #4

    Q_Goest

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    I don't, no. I suspect that because they're mass produced and have relatively large clearances between the impeller and housing that the efficiencies are not that great, maybe 50% to 70% range, but I don't know. If you have a curve though, you can easily determine efficiency from them, assuming you have a water properties data base such as a steam table. If you have a curve you can post, I'll see what the efficiency looks like. Only caveat is that the motor power is assumed to be 100% of the actual power input which may or may not be correct.
     
  6. Apr 26, 2010 #5

    danago

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    Well here are the head curves and electrical data:
    [PLAIN]http://img706.imageshack.us/img706/8043/headcurve.jpg [Broken]
    [PLAIN]http://img684.imageshack.us/img684/9941/elecdata.jpg [Broken]

    The one i am interested in is the SLS100, with a desired flow rate of about 125 L/min.

    So to calculate the efficiency with respect to motor power input, would it simply be [tex]\eta=\frac{QHg\rho}{P_{motor}}[/tex]? Where H is the head supplied and Q is the flow rate, rho is the density and g is the gravitational acceleration constant.
     
    Last edited by a moderator: May 4, 2017
  7. Apr 26, 2010 #6

    Q_Goest

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    For 125 l/min and a head of 11.5 meters (about 16.1 psig) I get a power needed of 0.44 hp. Efficiency then is .44 divided by actual power, in this case either 0.7 or 1 depending on what they mean by "input power" and "output power".
     
  8. Apr 27, 2010 #7

    danago

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    Sorry, but where did the 0.44 hp come from?
     
  9. Apr 27, 2010 #8

    Q_Goest

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    Sorry... I design pumps for a living and just assumed you wanted a check of your equation. Is 0.44 hp not what you get using your equation?

    [tex]P_{theor}={QHg\rho}[/tex]


    Once you find theoretical hp, divide by actual hp required and you get isentropic efficiency, which is what you've done here:

    [tex]\eta=\frac{QHg\rho}{P_{motor}}[/tex]
     
  10. Apr 27, 2010 #9

    danago

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    Im getting:

    P = (0.002083)(11.5)(9.81)(1000) = 235W

    After converting flow rate to m^3/h. Now im not really familiar with using hp (im from Australia, we seem to use watts much more than hp at university), but a quick search led me to believe that 1hp = 750W, which, after doing the conversion, does not give me 0.44 hp.

    Have i overlooked something really simple?
     
  11. Apr 27, 2010 #10

    Q_Goest

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    Sorry, my bad. I have a program that I forgot to change isentropic efficiency in. I had 70% listed. Changing that to 100% gives me .235 kW or .32 hp. Hope that clears it up.
     
  12. Apr 27, 2010 #11

    danago

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    Yep, that makes much more sense :smile:

    Thanks again for your help, very much appreciated!
     
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