1. Torricelli's Law

I am working through Penney and Edwards Differential Equations and Boundary Value Problems 4th edition.

On section 1.4 Problem 60, I am not getting what they have for an answer, and I am not completely certain that it is my fault as they make their fair share of errors.

A cylindrical tank with length 5 ft and radius 3 ft is situated with its axis horizontal. If a circular bottom hole with a radius of 1 in. is opened and the tank is initially half full of xylene, how long will it take for the liquid to drain completely?

The answer they get is 6 min 3 sec.

I get 8 min 32 sec.

You are supposed to use Toricell's Law which states: A(y)dy/dt = -a * sqrt(2gy) where g is 32ft/sec^2.

To do this, I observed that the area of a cross section A(y) is a constant 9 * pi. a = (1/12) ^ 2 * pi.
After this, it is a routine differential equation. And solving for t in the equation y(t) = 0 yields 162 * sqrt(10) sec or 8 min 32 sec.

2. Re: Torricelli's Law

Originally Posted by grandunification
I am working through Penney and Edwards Differential Equations and Boundary Value Problems 4th edition.

On section 1.4 Problem 60, I am not getting what they have for an answer, and I am not completely certain that it is my fault as they make their fair share of errors.

A cylindrical tank with length 5 ft and radius 3 ft is situated with its axis horizontal. If a circular bottom hole with a radius of 1 in. is opened and the tank is initially half full of xylene, how long will it take for the liquid to drain completely?

The answer they get is 6 min 3 sec.

I get 8 min 32 sec.

You are supposed to use Toricell's Law which states: A(y)dy/dt = -a * sqrt(2gy) where g is 32ft/sec^2.

To do this, I observed that the area of a cross section A(y) is a constant 9 * pi. a = (1/12) ^ 2 * pi.
After this, it is a routine differential equation. And solving for t in the equation y(t) = 0 yields 162 * sqrt(10) sec or 8 min 32 sec.

If the cylinder is on it's side, which I think is what is meant by it's axis is horizontal, then the cross sectional is going to vary as the tank drains.

3. Re: Torricelli's Law

I think that is certainly what is going on. However, I am not sure what the cross sectional area should be.

4. Re: Toricelli's Law

When $\displaystyle y=0$, we can say the cylinder is empty. When $\displaystyle y = 3$, the cylinder is half full. So, drop a line from the center of one of the end circles so that it makes a right angle with the horizontal line that is at the height of the liquid, then to the edge of the circle where the liquid intersects the circle, then back to the center of the circle. This makes a triangle with one side $\displaystyle 3-y$ and hypotenuse 3. The other side is then $\displaystyle \sqrt{3^2-(3-y)^2} = \sqrt{6y-y^2}$. The whole length of the liquid along the circle is twice that length, or $\displaystyle 2\sqrt{6y-y^2}$. The cross-sectional area is that length times the length of the cylinder: $\displaystyle 10\sqrt{6y-y^2}$. So, $\displaystyle A(y) = 10\sqrt{6y-y^2}$.