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 :)

  • $\begingroup$ Use the first 2 equation to remove $dV$ so you have and equation relating $dh$ to $dt$. Get all $h$'s over to one side and integrate. $\endgroup$ – user121049 Mar 8 '18 at 10:02

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.

  • $\begingroup$ Hey, Thanks heaps for that. Its much appreciated. The problem I am facing now is I have no idea how to integrate this so I have it with h as a function of t. How would i go about integrating the left hand side? Thanks heaps again :) $\endgroup$ – George White Mar 8 '18 at 23:51

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