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Calculating output from an air displacement pump

  1. Dec 3, 2013 #1
    Can anyone help me with a formula to calculate the output of an airlift pump for a certain head, pipe diameter and pipe length?
    An air lift pump works on the principle of a fill/empty cycle (so many seconds to fill and so many to empty). I can calculate the pressure required for a particular height.
    The higher the water is to be pumped, the longer the pump time has to be, but I can't work out if this is because of the height or the pipe length or a combination of the two.

    If for example, at the pump, with a 1 metre head, The fill time is 5 seconds and the pump time is 10 seconds. At a height of 40 metres, the pump time for the same volume of water increases to around 45 seconds.

    However, at the outlet (open, no restriction) the flow period is reduced and the pause until the next flow event increases. I understand that a part of the pump time will be used to overcome inertia and the weight of the water already in the delivery pipe.

    The pumping design is that a canister has a non-return valve at the entry point. There is a delivery pipe reaching almost to the bottom inside the canister with a non-return valve on top. Compressed air flows into the canister and acts on the surface of the water, forcing it down and up the delivery pipe. The air flow is time controlled to stop before the canister is completely empty. The air is then exhausted from the canister, allowing more water to refill it before the next pump cycle.

    I would like to be able to enter values for:
    Canister volume
    Head (height)
    Delivery pipe diameter
    Delivery pipe length
    Operating air pressure
    To calculate what the pump time would be.
    From this I could then work out an estimated flow rate for a particular sized canister under those operating conditions.

    I cannot find any formulae for normal flow/volume that will give this information.
    Any help would be greatly appreciated, as I have been wrestling with this for some time.
  2. jcsd
  3. Dec 4, 2013 #2
    Here's a link you should check out: http://www.uwex.edu/uwmril/pdf/RuralEnergyIssues/aquaculture/90_Air_Lift_Theory.pdf

    The paper describes an experiment for airlift pumps with small inlet diameters. The theory itself however, still applies. This describes the airlift pump's flow rate in terms of 'Taylor" bubbles which occur in plug flow and slug flow. Don't worry, the author summarizes everything in terms of non-dimensional variables. You'll need the gas flow rate, and then you can rearrange and find the liquid flow rate. Note: everything is dimensionless in his equations, so you'll have to divide the flow rates by the appropriate scaling factor. Also, a flow rate implies a time-value, so you'll be able to calculate the "stroke" time length of the water displacement, probably with some more arranging.

    You'll also need the submergence ratio, which is the length of tube submersion / lift height + length of tube submersion.

    EDIT: I'm assuming that the gas/liquid mixture in the airlift pump is mostly characterized as plug or slug flow. I'm not familiar with airlift pumps, so if the fluid is operating at a different regime, than those equations wouldn't necessarily apply. Also, if you want a really simple estimate, I suppose you could just use Bernoulli's equations {(1/2rhoV^2)+(rho)gH = const.}, of course, it assumes no work/heat loss/friction etc.
    Last edited: Dec 4, 2013
  4. Dec 5, 2013 #3
    My apologies - I performed a typo in the opening line of my post.
    My pump is an air displacement type, not air lift. The operational principle is different to air lift.
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