Draining a Tank

1. Sep 23, 2009

KillerZ

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

Suppose water is leaking from a tank through a circular hole of area Ah at its bottom. When water leaks through a hole, friction and contraction of the stream near the hole reduce the volume of water leaving the tank per second to $$cA_{h}\sqrt{2gh}$$, where c (0 < c < 1) is a empirical constant. Determine a differential equation for the height h of water at time t for the cubical tank shown. The radius of the hole is 2 in., and g = 32ft/s2

2. Relevant equations

$$cA_{h}\sqrt{2gh}$$

$$\frac{dV}{dt} = -cA_{h}\sqrt{2gh}$$

$$A_{h} = (\pi)r^{2} = (\pi)2^{2} = 4\pi$$

3. The attempt at a solution

I think have to find the volume and take the derivative of it with respect to time.

$$V = h^{3}$$ not sure if this is the right volume

$$\frac{dV}{dt} = 3h^{2}\frac{dh}{dt}$$

$$3h^{2}\frac{dh}{dt} = -cA_{h}\sqrt{2gh}$$

$$\frac{dh}{dt} = \frac{-cA_{h}\sqrt{(2)(32)h}}{3h^{2}}$$

$$\frac{dh}{dt} = \frac{-cA_{h}8\sqrt{h}}{3h^{2}}$$

$$\frac{dh}{dt} = \frac{-c32\pi\sqrt{h}}{3h^{2}}$$

2. Sep 23, 2009

tiny-tim

Hi KillerZ!
No, V is proportional to h, isn't it?

3. Sep 23, 2009

KillerZ

Is it something like V = (a)(b)(h) then? So a and b both are 10 but h is changing.

4. Sep 23, 2009

tiny-tim

That's right!

5. Sep 23, 2009

KillerZ

so like this:

$$A_{h} = (\pi)r^{2} = (\pi)(\frac{2}{12})^{2} = \frac{4}{144}\pi$$ I missed converting in to ft in the first post.

$$(a) = (b) = 10$$

$$V = (a)(b)(h) = 100h$$

$$\frac{dV}{dt} = 100\frac{dh}{dt}$$

$$100\frac{dh}{dt} = -cA_{h}\sqrt{2gh}$$

$$\frac{dh}{dt} = \frac{-cA_{h}\sqrt{(2)(32)h}}{100}$$

$$\frac{dh}{dt} = \frac{-cA_{h}8\sqrt{h}}{100}$$

$$\frac{dh}{dt} = \frac{-c\frac{32}{144}\pi\sqrt{h}}{100}$$

$$\frac{dh}{dt} = \frac{-c\pi\sqrt{h}}{450}$$

Last edited: Sep 23, 2009
6. Sep 23, 2009

Looks good!