I am trying to deduce a differencial equation that is satisfied by height h of water above the bottom of a spherical tank with radius $R = 50cms$, and a hole in the bottom of radius $r = 5cms$.
I need to find out how long it will take the tank to empty. The information I have is:
$ dV = \pi(2Rh - h^2)dh$
$dV = 0.6\pi r^2\sqrt{2gh}dt$
$v = 0.6 \sqrt{2gh}$
The problem i am trying to solve is: using the first two equations above, deduce a differential equation satisfied by the height h of water above the bottom of the tank. Assuming the outlet is opened at t = 0 when h = R, how long will it take for the tank to empty, given that R = 50cms and r0 = 5cms?
I know I need to combine those two equations, but I have absolutly no idea what to do next... If someone could help it would be much appreciated.
Thanks :)
First of all notice that $ v =- 0.6 \sqrt{2gh}$
Upon equating the two version of dV, we get
$$ \pi(2Rh - h^2)dh=0.6\pi r^2\sqrt{2gh}dt$$
$$ \frac {\pi(2Rh - h^2)dh}{\pi r^2\sqrt{2gh}}= -0.6dt$$
This is a separable differential equation subject to $ h(0) =R$
Integrate and solve for h as a function of time.