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Yet another one

  1. Jul 8, 2004 #1
    Kerosene (sp. gr. .82) leaks out of a hole 0.5 in. in diameter at the bottom of a tank at a rate of 10 gal/min. How high is the kerosene in the tank??
    (hint, 1gal = 0.1337 ft^3)

    So I used this formula:

    R = sqrt(2*g*h) * A

    and came up with 15004ft. which is completely off mark.

    any ideas? :surprise:
  2. jcsd
  3. Jul 8, 2004 #2


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    What is that formula you used? What do the variables represent? I would think you'd use the conservation of energy. Imagine that at the bottom, the tank was attached to a horizontal hose of 0.5 in diameter. The kerosene in this tube, after it's all emptied, will of course be going at the same speed as it did as it left the tank, so it has a constant kinetic energy (ignore friction and other resistive forces). The closed tank has just a bunch of stored potential energy. So, one form of energy turns entirely into another.

    Let h be the height of the tank
    Let p be the density of the liquid
    Let A be the constant cross-sectional area of the cylindrical tank (I assumed a bunch of stuff here, but I think I'm free to make these assumptions)
    Let E represent the total initial energy of the liquid (gravitational potential energy)
    Let E' represent the total final energy of the liquid (translational kinetic energy)

    [tex]E = E'[/tex]

    Now, the potential energy of a small slice of the liquid is:

    dE = dm * g * y, where y is the height of the slice
    dE = pAdy * g * y


    [tex]\int _0 ^h (gpA)(y)dy = \frac{1}{2}(pAh)v^2[/tex]

    [tex]\frac{gh^2}{2} = \frac{hv^2}{2}[/tex]

    [tex]gh = v^2[/tex]

    Now, you can calculate v. If it flows out with a diameter of 0.5 in. at a rate of 10 gal/min, I think you can calculate the speed with which it flows out. After 1 minute, 1 gallon flows out. In that time, [itex]0.0625\pi in.^2 \times D[/itex] flows out. You can solve for D. The speed, obviously, will be v = D/min. So

    [tex]gh = D^2[/tex]

    [tex]h = \frac{D^2}{g} = \frac{\left (\frac{1\mbox{ gallon}}{0.0625\pi \mbox{ in.}^2} \right )^2}{g}[/tex]
    Last edited: Jul 9, 2004
  4. Jul 8, 2004 #3
    Thanks so much for your help, but it appears as if you are trying to solve for the velocity, and we are looking for the height of the liquid.

    The formula I used was essentially the same as the one you just described.

    This one is driving me nuts!

  5. Jul 8, 2004 #4


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    You find velocity,v from Q = vA, where Q = 10 gal/min coverted into the right units, and A is the area of the opening.

    Having calculated v, you can then find h from
    [tex]~h =v^2/g [/tex]
  6. Jul 9, 2004 #5


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    So I guess you just didn't bother to read what I wrote?
  7. Jul 9, 2004 #6
    I'm sorry that I obviously misunderstood what you were trying to say. However, with your help, I did figure it out. The main problem was that I was working with mismatched units.

    I wasn't trying to offend you, I was just confused, that's all. But thanks for both of your help, that one took me quite a bit of time to figure out. Thanks! :biggrin:
  8. Jul 9, 2004 #7


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    Don't worry, I wasn't offended at all, just shocked that you blatantly missed the second half of my post, the half that just about gives you a final numerical answer. If you decided to use my work, you would have been about 5 seconds away from having a final answer, so I'm surprised you blatantly missed that, but on the other hand, I suppose it's good as it may have forced you to work more on the problem yourself. Again, I'm not offended, and was glad to help. :)
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