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#1 2006-01-27 15:34:05

Registered: 2006-01-27
Posts: 8

Just a simple problem?

Can anyone help with this one.

You want to drill a hole in the tank that will take the most time to drain, and then drill a hole in the tank that will take the second most time to drain.

Tank Height is 20 feet, tank diameter is 8 feet, hole diameter of 2 inches. It takes 42 minutes and 45 seconds to drain the tank.

If someone could just help me set up an equasion that would be wonderful
Thanx for any help
- stormswimmer


#2 2006-01-27 16:07:02

Registered: 2005-08-22
Posts: 1,504

Re: Just a simple problem?

What exactly are we trying to find here? All I see is desription, no question.

A logarithm is just a misspelled algorithm.


#3 2006-01-27 21:07:19

Registered: 2005-01-21
Posts: 7,608

Re: Just a simple problem?

To completely drain the tank, the hole has to be at the lowest point, so how can there be two holes of the same size that take different times?

Anyway, to work out an equation of "draining time" we will first need an equation of how fast the water flows out of the hole vs water pressure (or height of water in the tank)

Is this the kind of thing you are after?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman


#4 2006-01-28 02:44:44

Registered: 2006-01-27
Posts: 8

Re: Just a simple problem?

I am looking for how quickly can the tank be completely drained.

Maybe this would help
The velocity = 8.02 times the square root of the fluid level (in feet) remaining in the tank

The amount of fluid drained through the hole (in one second) is velocity multiplied by the area of hole.


#5 2006-01-28 03:02:12

Registered: 2005-11-24
Posts: 457

Re: Just a simple problem?

I couldn't quite grasp all of that either, but here are some relationships that might help you along your way.

  The velocity of water leaving either hole is;

  v = √2gh,  where h is Δh from the top of the tank to the center of the hole.

  So the volume leaving either hole would be;

  dV = -√(2gΔhA²)dt,  where A is the area of the hole.

  Perhaps someone here is that is very good with integration and related rates can get a hold of this one.  I would start with the assumption that one hole is indeed centered one inch above the base and then try to solve for the location of the second one.


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