1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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