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Please help calculate flowrate frm blower to push water down

  1. Apr 7, 2015 #1
    How to calculate how much flowrate from blower i can push water pool 2,75m*1,8m down by 1,25m in 1 sec?

    Note : Blower Pressure 0.03 bar - 0.035 bar
    There is an empty room above the water pool with dimension of 2.75m*1.8m*1.7m
    This process happen in a closed chamber.
    For wavepool project

    Thank you very much ! :)
  2. jcsd
  3. Apr 7, 2015 #2

    Simon Bridge

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    Welcome to PF;
    Not enough information.

    What do you mean by "push water down"?
    Generally, blowing air at a pool of water will make it splash about.
  4. Apr 7, 2015 #3
    the blower are not directly facing the surface of water, the chamber is airtight, but there is a hole below the chamber linking it to a bigger pond with same height as the water inside the chamber.

    simulated already, but we need the precise blower specification for the real one.
  5. Apr 7, 2015 #4
    sorry, my point is the water is not splashed
  6. Apr 8, 2015 #5

    Simon Bridge

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    Still not possible to get blower specs from information provided.
  7. Apr 9, 2015 #6
    no need for full specs, i mean just to get the flow rate of blower to get the water down at the rate of 1.25 m/s
    There is already information of :

    surface area of water = 2,75m*1,8m
    designated velocity = 1.25 m/s
    with the temperature of 20 degree Celsius ( room temperature)
    Pressure of the blower = 0.03 bar - 0.035 bar

    Can i use the formula of volumetric flow rate to solve this? or can you please tell me the common formula used for similar problem?
    and are the formulas like Q=AV , or rho=P/R.T, and bernoulli's principle can be used for air or just for liquid?

  8. Apr 9, 2015 #7

    Simon Bridge

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    To do pressure to flow rate you'd normally use something like Bournoulli's equation.

    All we know is you have some pool of some surface area connected somehow to another pool with another surface area and you want to do something (it is unclear what you mean by "get the water down at the rate of 1.25m/s". i.e. down what? Down the adjoining pipe? Down below it's original level?) involving the flow of water from someplace to someplace else. I'm kinda guessing you want to lower the height of one pool ... thus raising the height of the other one?
    So the pressure needed on one end, times the area at that end, has to support the weight of water above the reduced water-level?
    You intend to use a blower to increase the pressure in one chamber?
  9. Apr 9, 2015 #8
    This is a picture i took from internet that describes my problem the most
    It's exactly the same, only without part no.6 ( cover)

    In no.34, there are 2 holes for air to go inside ( from blower ) and go outside

    I use 6 chambers(no.24) using paralleled air flow from one blower but i'm guessing the flow rate are the same for each chamber.

    The surface area of water = 2,75 m*1,8 m ( inside one chamber)

    push the water down means to lower the height of the water inside the chamber ( yes)

    if needed, the hole (no.30) 's dimension is 2,75 m * 0,7 m
  10. Apr 9, 2015 #9

    Simon Bridge

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    OK, you can probably do that just by Newton's laws ... try working it out the pressure needed in the chamber (24) to get a particular water-level difference and you'll start to see what's involved.
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