# Assistance needed with a Torricelli's Law differential equation

1. Oct 27, 2008

### bitrex

I would be grateful for any assistance with the following Torricelli's law problem :yuck:

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

I have a cylindrical tank that is oriented horizontally, with an end radius of 3 feet and a length of 5 feet. (It's a math text, so no SI units for me.) The tank is half-full of liquid. A circular hole with a radius of 1 inch is opened in the bottom of the tank. How long will it take for the tank to drain?

2. Relevant equations

Torricelli's Law:

$$A(y)\frac{dy}{dt} = -a\sqrt{2gy}$$

Where $$A(y)$$ is the surface area of the liquid in the container as a function of height y, $$a$$ is the area of the hole at the bottom of the tank, and g = $$32 ft/sec^2$$

Length of a chord of a circle: $$2\sqrt{r^2-d^2}$$

3. The attempt at a solution

The surface area of the liquid in the tank is a series of rectangles going downwards following the contour of the circular ends, so $$A(y)$$ with y going downward is $$[5*6 - 5(2\sqrt{3^2-y^2})]$$. The area of the hole in the bottom of the tank is $$\pi(1/12)^2$$. When y = 3 the tank is half full, when y = 0 the tank is empty.

So the differential equation is $$[30-5(2\sqrt{9-y^2})]\frac{dy}/{dt} = -\pi(1/2)^2\sqrt{64y}$$

$$30-10\sqrt{9-y^2} \frac{dy}{dt} = -\frac{\pi}{18}$$

$$30-(90-10y^2)^{1/2} \frac{dy}{dt} = -\frac{\pi}{18}$$

$$\frac{30}{y^{1/2}} - \frac{(90-10y^2)^{1/2}}{y^{1/2}} \frac{dy}{dt} = -\frac{\pi}{18}$$

$$30y^{-1/2} - (90-10y^2)y^{-1} \frac{dy}{dt} = -\frac{\pi}{18}$$

$$30y^{-1/2} - 90y^{-1} - 10y dy = -\frac{\pi}{18} dt$$

I can do this integration, but unfortunately after finding the constant of integration I need to set y = 0 to find the time t for the tank to drain. This is a problem because after integrating the left hand side I have a term that involves ln(y). I've gone wrong somewhere either in the setup or execution of this problem but I can't see it. Thanks in advance for any help!

Edit: I see I made a mistake on the last step where I didn't multiply the first term by dt...perhaps I should have gotten rid of that term earlier? Still doesn't help the Ln(0) issue though...

Last edited: Oct 27, 2008