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Force Imparted by water gushing out of hole

  1. Apr 19, 2015 #1
    A container is full of water. A hole of cross section area a exists at the wall of the container, x height beneath the surface of water. (So speed would be v = √(2gx). Let's call it v for now). The density of water is ρ.Then why would the force imparted by the water gushing out of the hole on the container be F = av2ρ? I understand it has something to do with change in momentum, per unit time. But when I calculate it that way, I am not able to find change in momentum per unit time but just momentum per unit time.

    So the way I do it is:
    Volume of water water flowing out per second: av (assuming that velocity is per second)
    Mass of water flowing out per second: avρ
    Momentum of water flowing out per second: avρv = av2ρ

    But we haven't computed the (change in momentum/time) but just (momentum/time). So would the answer still be valid?
     
  2. jcsd
  3. Apr 19, 2015 #2
    F=ma
    Dimensionally, Force = Kg*m/s^2
    You have a*v^2*p
    That is m^2*(m^2/s^2)*(Kg/m^3) = Kg*m/s^2
    That is dimensionally correct for force
     
  4. Apr 19, 2015 #3
    Yes but force is dp/dt or Δp/Δt but not simply p/t. So what's happening here? Are we assuming initial momentum to be 0?
     
    Last edited: Apr 19, 2015
  5. Apr 19, 2015 #4
    Anybody there?
     
  6. Apr 19, 2015 #5
    Are you treating velocity as a constant?
    v^2 = 2gx
    v and x are both functions of time.
     
  7. Apr 19, 2015 #6
    I am just asking why the change in momentum is av^2p? Velocity is considered constant in this case. Because we are just concerned about velocity of water when it comes out of the hole. We are not concerned about its velocity after that. But I think its velocity before that would be 0 (since the water would be at rest before flowing out of the hole), and I think that's how we computed the change in momentum. (Since initial momentum would be 0 if velocity is 0). But I just want someone to confirm my reasoning.
     
  8. Apr 20, 2015 #7
    F = pressure x area = xρg*a (for a sufficiently small hole, so that pressure can be considered constant).

    mgx = 1/2*mv2
    x =v2/2g
    F = 1/2*v2ρa

    v is not constant.
     
  9. Apr 20, 2015 #8
    Your reasoning is correct except where you said that you haven’t computed the change in momentum/time, but just the momentum/time.
    In fact, you have calculated the change in momentum with respect to time, and that is why your answer is dimensionally correct for force.
    You can see that is true by reviewing your steps using dimensional analysis:
    1) Volume of water flowing out per second is av.
    Dimensionally: m^2*m/s = m^3/s
    2) Mass of water flowing out per second is avp.
    Dimensionally: m^2*m/s*kg/m^3 = kg/s
    3) Momentum of water flowing out per second is av^2p. Correct that to read the change in momentum of water flowing out per second!
    Dimensionally: m^2*m/s*kg/m^3*m/s = kg*m/s^2
    By doing this you can see in step two you already have kg/s, not just mass but mass per second.
    In step 3 you have kg/s multiplied by m/s = kg*m/s^2 = Force
    I think maybe in step two you forgot you had mass per second, and not just mass, so you thought your final answer was kg*m/s or just momentum/time when in fact what you have is kg*m/s^2 which is force.
    I find that dimensional analysis is very helpful in clarifying calculations of this type.
     
  10. Apr 24, 2015 #9
    To do this problem, you need to find the force that the tank is exerting on the fluid in the horizontal direction. This is equal and opposite to the force that the fluid is exerting on the tank. So, you need to do a momentum balance in the horizontal direction. I assume you are currently learning about momentum balances using the control volume approach. Can you think of a control volume that you can use to help solve this problem? Hint: It can be a pretty big control volume.

    Chet
     
  11. Apr 24, 2015 #10
    I wasn't taught to compute the reaction force using control volume approach. What I was taught was simply: Compute volume per second coming out, then compute mass per second coming out, then compute momentum per second coming out, and voila! We have the reaction force that the fluid coming out of the hole is exerting on the tank. But this method doesn't make any logical sense to me because I always seem to think of force as change in momentum/time elapsed (dp/dt to be more exact). But we just computed momentum coming out per second. How exactly does that equate to change in momentum/time elapsed, or to be more exact: dp/dt?
     
  12. Apr 24, 2015 #11
    The water coming out has zero momentum when in the tank. Its momentum changes from zero to that value you calculate.
    So this is the change in momentum, overall. In each second, some volume of water has the momentum increased by that amount.
     
  13. Apr 24, 2015 #12
    So why do we calculate the momentum that the water would have gained in 1 second? Why not 0.1 sec, 0.0001 sec etc?
     
  14. Apr 24, 2015 #13
    Actually you did. What you calculated is the rate of change in momentum. It does not have any time interval in it. So it applies to any time interval.
    0 For 1 s, the change will be given by that expression For 0.01 s, the change will be 0.01 form that expression. And so on.
    When we say "per second" we just refer to the standard unit of time. Same as velocity is m per second. You don't ask why the velocity is 3m/s and not 0.03m/0.01 s, do you?
     
  15. Apr 24, 2015 #14
    No. We calculated the momentum flowing out per second.
     
  16. Apr 24, 2015 #15
    You assumed an interval of 1s but this is just for "comfort". Take any arbitrary time and do the calculation again. At the end you divide by this time interval to get the rate of change. You see that the rate will be exactly what you got previously. And this time without assuming any specific unit of time.
     
  17. Apr 24, 2015 #16
    Hi andyrk.

    Recently, in a private conversation, you indicated that you were able to relate to my post #12 in the thread Change in Flow of Momentum, and were comfortable with the approach and the answer. The method I am proposing to attack the present problem is almost a carbon copy of what was done in post #12. Are you willing to work with me to solve the present problem in this way, or are you not so sure any more? If you are willing to forego the simplistic explanation that you were taught and to do this thing the right way, I am ready to proceed.

    Chet
     
  18. Apr 25, 2015 #17
    Yes, I am ready to understand it the right way you want to attack this problem with. But I just wanted to confirm that is the approach I was taught to answer this problem, wrong? Or has it been simplified to such an extent that it overlooks all the nitty-gritty details?
     
    Last edited: Apr 25, 2015
  19. Apr 25, 2015 #18
    The result is not incorrect, but it overlooks details that are important to you.

    upload_2015-4-25_11-29-27.png

    I have indicated a control volume in the figure (between the red dashed lines) into which fluid is flowing in, and out of which fluid is flowing out. The rate of flow in is equal to the rate of flow out.

    I have also shown a Before picture in which a small parcel of fluid is about to flow into the control volume at time t. I have also shown an After picture at time t + Δt in which the entering parcel has just fully flowed into the control volume, and another small parcel of equal size has just flowed out.

    If the density of the fluid is ρ, the cross sectional area of the exit hole is A, and the fluid velocity at the exit hole is v, how much mass exits the control volume within the exiting parcel during the time interval Δt? From continuity, how much mass enters the control volume within the entering parcel during the time interval Δt?

    After you answer these questions, we will continue.

    Chet
     

    Attached Files:

  20. May 11, 2016 #19
    Hi Chet ,

    I would like to continue the discussion so as to plug the gap in my basic understanding of fluids . I am interested in understanding how the force exerted by liquid on the container turns out to be ##ρAv^2## .

    ##ρAvΔt##

    Same as before i.e ##ρAvΔt##
     
  21. May 11, 2016 #20
    The horizontal force exerted by the container on the fluid in the tank (control volume) is equal to the rate of change of horizontal momentum of the fluid within the control volume; this is equal to the rate of horizontal momentum exiting the control volume minus the rate of horizontal momentum entering the control volume.
     
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