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Water in a tank

  1. Apr 5, 2007 #1
    1. The problem statement, all variables and given/known data
    The water tank is open to the atmosphere and has two holes in it, one at 0.80m and the other at 3.6m above the floor on which the tank rests. if the two streams of water strike the floor in the same place, what i the depth of water in the tank.
    tank height is 5 m.

    2. Relevant equations

    3. The attempt at a solution
    i know if you find the velocities out from both holes, you can use those to determine the hieght of the water, however, when i did that i got 2.024m which in impossilbe. any help would be great. thanks
  2. jcsd
  3. Apr 5, 2007 #2
    these are great problems--whats the source?

    You have made the connection between height of water and velocity as it exits hole via Bernoullis eqn. The difference in height between the two holes can be used in kinematics problem, treat water drop as a projectile with arbitrary mass, I think. That help? Show some work and we can likely get this sorted out.
  4. Apr 5, 2007 #3
    these problems suck balls man! i sat here for a good 3 hours trying to figure it out. but no good. man i donno how to type out the work using regular stuff. i which it had symbols on the forum. the source of these problems is my physics book. by james walker.
  5. Apr 5, 2007 #4


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    Okay, so where are you upto in this problem, what have you done thus far?
  6. Apr 5, 2007 #5
    you know that from Bernollis: here d=density(rho)


    from kinematics we can obtain relation between exit velocities.

    (the stream from the bottom must be going considerably faster to hit same spot). Then just slogging thru algebra in Bernoullis eqn for height,x, where
    x+3.6 and x+.8 pressure difference must equal velocity squared difference.
  7. Apr 5, 2007 #6
    .5[itex]\rho g h[/itex] +P = .5[itex]\rho v1^2[/itex] + [itex]\rho g h1[/itex] + P + 5[itex]\rho v2^2[/itex] + [itex]\rho g h2[/itex] + P

    this is the main equation i used to derive the velocities related to the height. but i donno if its in the right set up
  8. Apr 5, 2007 #7
    looks good except there should be an equal sign after the middle P.

    Substitute for h1=h-0.8
    and for h2=h-3.6 now from the kinematics part, you can get v1=kv2
    which will allow you to solve for the velocities, and then go back to eqn to compute height, h
  9. Apr 5, 2007 #8
    ahhh.... i knew the hieghts should have been h-h1 and h-h2. i tried that but i donno how u got v1 to equal kv2, i keep on getting v1^2 + g(h-.8)=v2^2 + g(h-3.6)
    what am i doing wrong?
  10. Apr 5, 2007 #9
    nothing, thats where the kinematics comes in

    pretend water is arbitrary mass

    x the distance it lands from tank=v0*t0
    water from second hole travels same distance v1*t1

    from acceleration we know 1/2g*t0^2=0.8 and 1/2 g*t1^2=3.6

    and so forth.

    edited to correct omission of t squared
    Last edited: Apr 5, 2007
  11. Apr 5, 2007 #10
    I had time to sit down and look at problem more in depth :uhh: , I think you need to set up problem using following lines:

    a) v0^2=2gh
    b) v1^2=2g(h+2.8)

    now we need eqn relating v0 and v1, again considering kinematics

    x=v1*t1=v0*t0 so (v0/v1)^2=(t1/t0)^2

    from acceleration outside the tank, we know 1/2g(t0^2)=3.6

    and similarly, 1/2g(t1^2)=0.8 this gives (t1/t0)^2=0.8/3.6

    Divide eqn a by b and equate to .8/3.6 you should be able to solve for h, which is height of column above top hole. There may be much better method, but this seems at least reasonable, if not quickest way to get solution.
  12. Apr 5, 2007 #11
    The answer is 4.4 and the key is to set the d1=vt and d2=vt to each other so v1t1=v2d2 start there and see if you get that answer
  13. Apr 5, 2007 #12
    to OP,

    Maybe it was simultaeous post, this is the approach i suggested as well 3 posts ago. Hope it wasn't too unclear.
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