Water Drainage From a Tank

[corbell777]

A hemispherical tank of radius R is filled with water which is pouring under the influence of gravity out of a circular hole of radius r at the bottom of the tank. Assume that the velocity of fluid flowing from the opening is proportional to sqrt(2gh), where g is gravity and h is the height of the water at time t. Derive a differential equation for the depth of the water at any time.
Change in height of fluid over time = Velocity of fluid leaving tank x area of opening
pi*R*R - pi*(R - h)*(R-h) dh = pi*r*r sqrt(2gh) dt
{(2Rh(to 1/2) - h(to 3/2) )/ r*r*sqrt(2g)} dh = dt
Integration gives
{(20R*h(to 3/2) - 6h(to 5/2) )/(15r*r*sqrt(2g)) = t
This is the answer which I got which matches the answer in the book. But it seems backwards to me.
If h = 0, then the tank is empty. And since it started full, there should be some expression for t. But if you put h = 0 in the above expression, you get t = 0.
Also, if R = h, then the tank is full, which should be at time t = 0, since the tank started full. But if you put R = h into the above expression, you get
t = (14R*h(to 5/2)/(15 r*r*sqrt(2g)).
So I think it should be the other way around. When R = h, I should get t = 0 and when h = 0, I should get the last expression for t.
Can someone explain to me what I'm getting wrong?

 

[galactus]

Let h(t) be the depth of the water at some time t and let x be the radius of the water surface at said level.

Thus, \(\displaystyle dV=-\sqrt{2gh}{\pi}r^{2}dt\)

Where \(\displaystyle dV={\pi}x^{2}dh\)

Thus, \(\displaystyle (R-h)^{2}+x^{2}=R^{2}\)

\(\displaystyle x^{2}=R^{2}-(R^{2}-2hR+h^{2})=2hR-h^{2}\)

\(\displaystyle dV=\pi (2hR-h^{2})dh=-\sqrt{2gh}{\pi}r^{2}dt\)

Thus, we have as the DE:

\(\displaystyle (2hR-h^{2})dh=-\sqrt{2gh}r^{2}dt\)

Now, solving the DE results in:

\(\displaystyle \frac{2\sqrt{gh}h(10R-3h)}{15g}=C-\sqrt{2}r^{2}t\)

Use the initial condition that \(\displaystyle h(0)=R\) to solve for C and we find that

\(\displaystyle C=\frac{14R^{2}\sqrt{gR}}{15g}\)

Sub this back into the DE and we get:

\(\displaystyle \frac{2\sqrt{gh}h(10R-3h)}{15g}=\frac{14R^{2}\sqrt{gR}}{15g}-\sqrt{2}r^{2}t\)

Now, the tank is empty when \(\displaystyle h=0\). Sub this in and we get:

\(\displaystyle t=\frac{7R^{2}\sqrt{2gR}}{15gr^{2}}=\frac{14R^{5/2}}{r^{2}\sqrt{2g}}\)