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Water gushing out the barrel

  1. Nov 6, 2007 #1
    [SOLVED] Water gushing out the barrel

    Here is the problem:

    We have two barrels with holes in it, only one of the following figures is the right one. Which one?


    I think the following:

    We know that both holes have the same energy namely:

    rho.g.H (pressure energy) and m.g.h (potential energy)

    Now for the upper hole the potential energy of the fluid is higher and the pressure energy lower.

    For the lower hole the potential energy is half of the potential energy of the upper hole and the pressure energy is double the pressure energy of the upper hole.

    This means both holes have the same energy no?

    This means they should both use this energy to fullfill the same work. The same length of water that is gushing out the barrel.

    So I would think the bottom one is correct.

    But I would like people to confirm on this, I need a real professional who knows the correct answer.

    I think the top one is really plausible too...
  2. jcsd
  3. Nov 7, 2007 #2
    The lower barrel is the correct one. What we need here is the velocity on the horizontal direction.
  4. Nov 7, 2007 #3


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    Think of it this way...

    The bottom hole will have the greater pressure head ([tex]\rho g h[/tex]). Since all three holes exhaust to atmosphere, that means the bottom hole has the greater pressure differential across it. That means the bottom hole will have the highest flow. Also, since all of the holes are assumed to be the same size, that means the greater pressure differential and thus greater flow rate, will have a higher fluid velocity.

    That means the bottom picture would be the correct one.
  5. Nov 7, 2007 #4


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    The bottom one is the correct choice for the reasons given by Fred.
  6. Nov 7, 2007 #5
    Well, the top one has also a higher pressure in the lower hole.

    You can easily see that the shape of the lower hole is more straight than the shape of the upper hole. (in figure 1)

    So why isn't it the first one?

    Well I have the solution from a collegue of mine:

    This is the base formular:

    rho*g*h = cte 1/2*rho*v^2

    Now you can just calculate the intersection points of the gushes of water.
    They are not going through one point.

    THOUGH I think that formular isn't correct, because it doesn't take in account the intersectional area of the holes???

    But for this problem it is correct because the holes are of equal size...

    Anyone who knows the right formular with intersectional area of the holes?
    Last edited: Nov 7, 2007
  7. Nov 8, 2007 #6


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    Because the velocity at the outlet is greater as you move down the barrel due to the increase in hydrostatic pressure.

    The formula for determining the velocity is given by,

    v = SQRT(2gh)


    h = height from top of fluid level to the orifice discharge level
    g = gravitational acceleration

    "The velocity out from the tank is equal to speed of a freely body falling the distance h", also known as Torricelli's theorem.

  8. Nov 22, 2007 #7
    the lower barrel is the right one. when u calculate the pressure of the fluid coming out of the nozzle, u will also have to add the pressure head of the liquid above the nozzle, which is equal to density of the fluid X Height of the fluid level above the nozzle. So the total pressure will be high for the lower most nozzle, n some what less for the 2nd nozzle and the 3 rd one. hence the exit velocity will be more for the bottom most nozzle..
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