Fluid dynamics, water emptying out of cylinder

[Funktimus]

Homework Statement

Tank is filled with water to a height h_0 = 1.00 meters
Cross section area of tank (A_1 = .785 m^2)
Hole at bottom of tank with an area, A_2 = .002 m^2

Homework Equations

How much time does it tank to half empty the tank? find t_(1/2).

Question also provides "a useful antiderivative."
/ = integral sign
/ x^(-1/2) dx = 2x^(1/2)

The Attempt at a Solution

1.
V = A_2 * v_hole
dV/dt = A_2 * v_hole

2. getting an equation for velocity coming out of hole
v_hole = [2gh + (v_surf)^2] ; got that from using Bernoulli's principle.

3. equation of continuity to get v_surf
v_surf = v_hole * A_hole / A_surf

thus,

v_hole = (2gh)^(1/2) , since [1 - (A_2/A_1)^2] = ~1

4. rate of change of the level of water in the tank
dh/dt = (-v_hole * A_2)/(A_1)
dh/dt = -.0113

5.
h_2 - h_0 = -.0113 * t_(1/2)
a half empty tank means, h goes down 1/2, 1/2 of h = 1, h=-.5
t_(1/2) = -.5m / -.0013 m/s = 44.3 s

44.3 s is wrong though. I don't get how come. I went through the hints for this problem. It mentions

dy/dx = f(x) * g(y) ... f(x) and g(y) are known functions

/ dy/g(y) = / f(x)dx

I don't get which known functions they are referring to. And it's also been probably 2 years since I ever used calculus, so those equations don't make much sense to me.

Any help would be appreciated.

 

[MexChemE]

First, we establish an unsteady state mass balance on the tank
$$\frac{dm}{dt} = -w$$
We can express mass inside the tank as $m = \rho V = \rho A_1 h$, where density and cross section area are constant, so the only variable changing in time is the height of liquid inside the tank. We can also express the mass flow rate out of the tank as $w = \rho A_2 v$. Now we need to relate the exiting velocity with the height of the liquid. We can go ahead and use Torricelli's law, $v = v = \sqrt{2gh}$, turning our unsteady state model into a quasi-steady state one.
$$A_1 \frac{dh}{dt} = - A_2 \sqrt{2gh}$$
We want to know the time it takes to empty half of the tank, so we can separate variables and integrate the differential equation from 0 to $t_{\frac{1}{2}}$ and $h_0$ to $\frac{h_0}{2}$
$$\int_0^{t_{\frac{1}{2}}} dt = - \frac{1}{\sqrt{2g}} \frac{A_1}{A_2} \int_{h_0}^{\frac{h_0}{2}} \frac{dh}{\sqrt{h}}$$
$$t_{\frac{1}{2}} = \frac{2}{\sqrt{2g}} \frac{A_1}{A_2} \left(\sqrt{h_0} - \sqrt{\frac{h_0}{2}} \right)$$
Finally, we just plug in the values and solve
$$t_{\frac{1}{2}} = \frac{2}{\sqrt{2 \left(9.8 \ \frac{m}{s^2} \right)}} \frac{0.785 \ m^2}{0.002 \ m^2} \left(\sqrt{1 \ m} - \sqrt{\frac{1 \ m}{2}} \right) = 51.93 \ s$$