Water flow question

uhuge

Neglecting evaporation, if you have a cubic mile of water (1 mile by 1 mile by 1 mile) how long would it take to drain if at the bottom there is a one square inch opening?

I am verifying my answer with some really bright people. You

Tide

Why don't you at least tell us how you arrived at your answer? :)

uhuge

Well barring evaporation, I used Toricelli's theorem. Velocity = the sqr root of 2gh. THis gave me the exit velocity at the opening.
Realizing that flow rate is equal to velocity time area of the opening, we can solve for time to drain a certain amount of the tank. The problem is h (height of the fluid) is constantly changing so the formula must be continueously recalculated.
I have a number, I just want to see how close I actually am with my estimation.

Tide

I think you should get something like:

[tex]t = \frac {2A}{a} \sqrt {\frac {H}{2g}}[/tex]

where A is the cross sectional area of the water, a is the area of the opening and H is the starting height of the water. Is that consistent with your analysis?

uhuge

this is fairly close to what I estimated in terms of ballparks. I came up with 2312 years and some change, in running your formula I got right at 2100 years.

uhuge

I think i made a mistake in that when I do your calculations I get 21005.32 years. Is this what you get?

Tide

uhuge,

I got about 2300 years when I used my formula. Check your units carefully.

Dr.Brain

Instead of remembering the formula , just apply the Bernoulli's Theorem at the top of the water surface and at the bottom ( bottom means the square peg) . And then calculate your answer and remember converting given units in standard form.

BJ

quark

2300 is about right(the average velocity can be presumed as half the velocity calculated from initial gravity head). If the ridiculously high velocity won't damage the system, it may take another 255 years due to a 0.9 coefficient of discharge.