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Water tank

  1. Jun 1, 2009 #1
    1. The problem statement, all variables and given/known data

    A tank of height h full of water with it's base 40 meters from the ground, with a hole on the lateral wall at the base. The water comes out of the tank at 6.93 m/s to fill a pool on the ground. The water hits the center of the pool at a distance of 19.5 meters measured horizontally from the tank. What is the tank height h? You can despise the variation of height of the tank while it is being unfilled.


    2. Relevant equations
    Continuity: A₁v₁=A₂v₂
    Bernoulli: P + 1/2ρv² + ρgh = Cte
    free fall of the water (previously calculated): y(t)= 40 -4.9t²
    x(t)= 6.93t

    3. The attempt at a solution

    I used Bernoulli equation, considering that the point at wich the water leaves the tank as a pressure P, is at h=40, and as a speed v=6.93, and the point at wich the water reaches the floor as a pressure P₀ (1 atm = 1.013×10⁵), h=0 and v=28.843(calculated using the law motion of the water in free fall and doing the norm of two velocities: √(V²x + V²y) ).
    With this I would find the pressure P, then I would do: P = P₀ + ρgh <=> h = (P-P₀)/(ρg).

    But this way the pressure P results lower than the atmospheric pressure P₀, wich make things impossible.... what I think I was missing is the fact that as speed increases while the water falls, the pressure decreases.. so the value of the pressure when the water hits the floor must be lower than P₀, but I can't solve the problem this way =\

    I would really apreciate some help thanks ^^
     
  2. jcsd
  3. Jun 1, 2009 #2

    PhanthomJay

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    You appear to be mixing stuff up. The problem asks for the height of the tank (from its base to its top) if the speed of the water exiting at the bottom is 6.93m/s (is this a calculated value or a given value?). Disallowing for extremely small variations in g with height, that exit speed is independent of how high the tank rests above ground; and depends only on the tank height, h, as measured from its base to the top of the water level in the tank. Also, you may assume that the atmospheric pressure is constant over those heights.
     
  4. Jun 1, 2009 #3
    the speed 6.93 is a calculated value but is correct. I know the speed is independent from how high the tank is from the ground, so I tried it two ways. I'll write them down:

    ρ= 1000
    g= 9.8

    Bernoulli at hole = Bernoulli when hitting the ground
    P + 1/2ρv² + ρgh = P + 1/2ρv² + ρgh
    P + 1/2ρ×6.93² + ρg40 = P₀ + 1/2ρ×28.843² + 0
    P = 101260.801

    the pressure at the hole is inferior than atmospheric...

    Now without depending on height under the tank:

    Bernoulli at surface of water = Bernoulli at the hole
    P + 1/2ρv² + ρgh = P + 1/2ρv² + ρgh
    P₀ + 1/2ρ×0² + ρgh = P₀ + 1/2ρ×6.93² + ρg×0
    h = 2.45m wich isnt the right value

    the expected value for h is 12.8m

    I cant see what I am missing =\
     
  5. Jun 2, 2009 #4

    PhanthomJay

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    I guess you may have included air resistance, because I calculate v =6.82 m/s
    ok, so let's now ignore the height under the tank
    assuming v =6.93, it looks like the right value to me
    that a high expectation
    Nor can I.
     
  6. Jun 2, 2009 #5

    Lok

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    2.45 m is correct. And bernoulli is unnecesarry as it can be solved energetically with h=v2/2g.

    For h=12.8 the exit speed must have been around 15.83 m/s.
     
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