- #1
mirandasatterley
- 62
- 0
"A large storage tank, open at the top and filled with water, develops a small hole in its side at a point 16.0m below the water level. If the rate of flow from the leak is 2.50x10^-3 m^3/min, determine; A) The speed at which the water leaves the hole and (B) The diameter of the hole.
So I used the equation;
P(air) + 1/2 d(density)v1^2 + pgh(total height of water) = P(air) + 1/2dv2^2 + pgh(from the bottom of the container to the leak)
The equation then simplifies to:
gh(total height) = 1/2V2^2 + gh(from bottom to leak)
V2^2 = 2g(h(total) - h(form bottom to leak))
V2^2 = 2g(h(from leak to top of water))
Since this h is known, I can solve for v.
I Just can't figure out the diameter part.
Any help is appreciated.
So I used the equation;
P(air) + 1/2 d(density)v1^2 + pgh(total height of water) = P(air) + 1/2dv2^2 + pgh(from the bottom of the container to the leak)
The equation then simplifies to:
gh(total height) = 1/2V2^2 + gh(from bottom to leak)
V2^2 = 2g(h(total) - h(form bottom to leak))
V2^2 = 2g(h(from leak to top of water))
Since this h is known, I can solve for v.
I Just can't figure out the diameter part.
Any help is appreciated.