Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Assistance needed with a Torricelli's Law differential equation

  1. Oct 27, 2008 #1
    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:

    [tex] A(y)\frac{dy}{dt} = -a\sqrt{2gy} [/tex]

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

    Length of a chord of a circle: [tex] 2\sqrt{r^2-d^2} [/tex]

    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 [tex] A(y) [/tex] with y going downward is [tex] [5*6 - 5(2\sqrt{3^2-y^2})] [/tex]. The area of the hole in the bottom of the tank is [tex] \pi(1/12)^2 [/tex]. When y = 3 the tank is half full, when y = 0 the tank is empty.

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

    [tex] 30-10\sqrt{9-y^2} \frac{dy}{dt} = -\frac{\pi}{18}[/tex]

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

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

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

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

    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
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted