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Water gun problem, pump stroke -> pressure

  1. May 5, 2016 #1
    Hello all, I have a question about a water gun problem I'm working on.

    The general question is...how can you associate the hand pump stroke length, and the diameter of the pipe the pump is inside, with how much pressure gets built up in the guns pressure chamber?
     
  2. jcsd
  3. May 5, 2016 #2

    billy_joule

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    You can't without knowing the volume of the pressure chamber, obvious a larger pressure tank will tank more pumps to get up to a certain pressure.
    Once you know tank volume then you can find the increase in tank pressure from each stroke from the ideal gas law.
    https://en.wikipedia.org/wiki/Ideal_gas_law
    Each stroke adds the mass of air (at atmospheric pressure & temp) found in the piston volume.
     
  4. May 5, 2016 #3
    Alright, so lets see....lets take this animation for example: http://static.howstuffworks.com/flash/water-blaster-soaker.swf
    ****Only difference will be that originally the only water will be in the reservoir, nothing in the system****

    Lets say the stroke length is 11 [in.] and the cylinder the piston is moving within has a diameter of 2 [in.]. Also we'll assume the volume of the pressure chamber is 250 [in^3]
    *All general numbers*

    The first out-stroke opens the first valve allowing water to fill that volume...the in-stroke would then input a mass of water equal to [itex]\pi \times (2 [in.])^2 \times 11 [in.] = 44 \pi [in.]^3 = 2265.17501 [g][/itex] directly into the pressure chamber? Am I saying that correctly?
     
  5. May 5, 2016 #4
    Ha, obviously I would need to change the volume of the chamber...that's what I get for choosing random numbers!

    Okay lets instead say the Volume of the Pressure Chamber is 5000 in^3 *again just choosing random numbers, as I know this is unrealistic for this application.

    This is where I get a bit confused, the ideal gas law states [itex] P = \frac{mRT}{V} [/itex]

    The original pressure in the chamber would be [itex] P = \frac{mRT}{V} = \frac{(\rho \times V)RT}{V} [/itex] so that would cancel out the Volume? Since I seem to be going in a circle, I'll just wait for clarification
     
  6. May 5, 2016 #5

    billy_joule

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    Ok, I assumed the piston pumped air in, not water. In that case, seeing as water is incompressible the volume of the pressure chamber is effectively reduced by the volume of the piston with each stroke. The corresponding increase in chamber pressure will depend on the ratio of the chamber volume to the piston volume.
     
  7. May 5, 2016 #6
    Got it, so to work with a general example... let's say the volume of the chamber is 5000in^3 originally when filled with just air.

    The *cylindrical* piston has an 11 [in] stroke and we'll say a 2 [in] diameter...so the volume would be that same 2265 in^3

    This volume would effectively be moved into the pressure chamber after each stroke.

    So to find the pressure in the chamber after the first stroke,we are worried about the volume of air remaining in the tank? So (5000-2265) is the new volume of the tank? And we would use that to calculate the pressure?

    For example *forgive me, on my phone* P = (mass of water * temp * R)/volume of air remaining in the tank?

    This is probably incorrect as you stated it would depend on the ratio of volumes...so knowing the ratio is roughly 2.2...where do we go from there?

    I appreciate your time by the way! Never knew a squirt gun could stump me so badly!
     
  8. May 6, 2016 #7

    billy_joule

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    I think we can assume it's an isothermal process as I would expect the thin walls and the presence of the water would allow good heat transfer so the compressed air temp. does not rise above the ambient temp.
    So we have two states, initial and final
    PiVi =niRiTi
    PfVf =nfRfTf
    and the number of moles, gas constant and temperature are the same for both states so:
    niRiTi = nfRfTf
    and so
    PiVi = PfVf
    (which is Boyle's Law)
    And solving for final pressure:
    Pf = Pi (Vi / Vf)
     
  9. May 6, 2016 #8
    Perfect, and that would be the pressure built up in the chamber after corresponding pumps

    Now, if I wanted to find the velocity of water leaving the nozzle when the trigger is pushed...since I know the pressure in the chamber, I could set up a Bernoulli relation correct?

    [itex] \frac{P_{nozzle}}{\rho} + \frac{V_{1}^2}{2} + gz_{1} = \frac{P_{chamber}}{\rho} + \frac{V_{2}^2}{2} + gz_{2} [/itex]

    Where we know:
    > V_2 is 0
    > We can now find P_chamber using Boyles law above
    > P_1 is atmospehric pressure

    Am I correct in this regard?
     
  10. May 6, 2016 #9

    billy_joule

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    Yes, as a first order approximation.
    Any real water gun has losses, the pump is not 100% efficient and there are major & minor losses throughout the system. At a random guess, I'd say only 50%-80% of the work done on the pump handle is converted to kinetic energy of the outgoing water.
     
  11. May 6, 2016 #10
    Absolutely! And I do plan on having this losses accounted for!

    Major head losses = [itex] h_{l} = f \frac{L}{D} \frac{V^2}{2} [/itex]

    Where 'f' is the friction factor found via the Reynolds number and the roughness

    Minor head losses = [itex] h_{ml} = k \frac{V^2}{2} [/itex] where k will depend on the bends in the system and other geometric factors etc

    So I could always modify Bernoulli's and say:

    [itex] (\frac{P_{atm}}{\rho_{air}} + \frac{V_{1}^2}{2} + gz_{1}) - (\frac{P_{chamber}}{\rho_{water}} + \frac{V_{2}^2}{2} + gz_{2}) = f \frac{L}{D} \frac{V^2}{2} + \sum k \frac{V^2}{2} [/itex]

    I can use my textbook to find:
    >the roughness for certain materials
    >friction factor (moody chart)
    >minor loss coefficients for the various bends

    So with all that, I believe I can solve for the true velocity of the water leaving the nozzle correct?
     
  12. May 6, 2016 #11
    Hmm, although, the Reynolds number...

    [itex] Re = \frac{VD}{v} [/itex]

    since I need the velocity FOR the Reynolds number...maybe it isn't quite as easy as I am thinking?
     
  13. May 8, 2016 #12
    Hey billy_joule, Sorry to bother you again, but any more insight on this?

    While before, we could have calculated the velocity leaving the nozzle *ignoring losses*

    Now, taking into account the losses, we have 2 more terms both requiring a velocity [itex]h_{l} = f \frac{L}{D} \frac{V^2 }{2}[/itex] and [itex]h_{ml} = k \frac{V^2}{2}[/itex]

    I'm confused on how to find the friction factor here, as it is a function of Reynolds Number which involves the velocity.

    At this point, should I guess a friction factor *assume the flow is turbulent* and iterate until I get a consistent value for V?
     
  14. May 9, 2016 #13

    billy_joule

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    It looks like you're on the right track.

    Yes, iteration is how I learnt to solve these sorts of problems.
     
  15. May 9, 2016 #14
    Perfect! Also, 1 LAST question that comes from the Boyle's Law relation we came up with earlier

    [itex]P_{f} = P_{i}\frac{V_i}{V_f}[/itex]

    Finding the initial pressure inside of the pressure chamber, using the ideal gas law, leads to [itex]P_{i} = \frac{mRT}{V}[/itex] Which, when replacing [itex]m = \rho V[/itex] leads to [itex]P_{i} = \rho R T[/itex]

    I'm just confused here, as the volume of the tank should DEFINITELY play a factor in this shouldn't it?
     
  16. May 9, 2016 #15
    Oh man....That was just a blonde moment...it took me actually calculating it out to see that it is just atmospheric pressure. That was fun haha!
     
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