How to establish relation between flow rate and height of the water column of the tank? Suppose a water tank has 1" diameter drain at the bottom and is filled with water up to one meter height above the drain. What time it will take the tank to drain out completely. Now say, the tank is filled up to two meter height above the drain then what time it will take the tank to drain out? Will it be double or less than it? Can we establish a relation between flow rate for the given height of water column?  
 A: Use Bernoulli's equation to derive Torricelli's Law (check any website for this) for the velocity out of the hole; $ v = \sqrt{2 g h(t)} $, where g is gravity and $ h(t) $ is the height of the fluid in the tank at any time.
Write a balance on the mass of the fluid in the tank as:
$$ \text{in - out + gen = accumulation} $$
$$ \rho Q_{in} - \rho Q_{out} = \frac{d(\rho V)}{dt} $$
where the generation term is zero, $\rho$ is the fluid density (constant here) and $Q_{in}$ and $Q_{out}$ is the flow rate in and out of the tank, respectively. $Q_{in}$ is zero so we get:
$$ \frac{dV}{dt} = -Q_{out} $$
The flow out is $v A = \sqrt{2 g h(t)} A$, where $A$ is the area of the hole which is calculated by knowing the diameter of the circular hole; given in the problem statement as 1 inch.
The volume of the tank, $V = a_t h(t)$, is the height, $h(t)$ times the area, $a_{t}$.
Putting it all together, we get the separable first order differential equation for the height of the fluid in the tank versus time:
$$ \frac{dh}{dt} = -\frac{A}{a_t}\sqrt{2g}\sqrt{h}$$
Prepare it for Integration
$$ \frac{dh}{\sqrt{h}} = -\frac{A}{a_t}\sqrt{2g}{dt}$$
Integrate the equation. The upper bound for $dh$ is $h(t)$. The lower bound is $h(0)=H$ . For $dt$ we integrate from $t$ to $0$:
$$ 2 (\sqrt{h(t)} - \sqrt{H}) =-\frac{A}{ a_t}\sqrt{2g} t$$
Solve for $h(t)$
$$ h(t)=[\sqrt{H} -\frac{A}{2 a_t}\sqrt{2g} t]^2 $$
To find the time when the tank empties, set $h$ equal to zero and solve for $t$:
$$ T= \sqrt{\frac{H}{2g}} \frac{2 a_t}{A}$$
Plug that back in the previous equation to clarify time dependence on t:
$$ h(t)=H[1 - t/T]^2 $$
The times to empty the same tank for two different starting heights, $H_1$ and $H_2$ is:
$$ \frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}} $$
So, finally, if $H_2 = 2 H_1$ as in the problem statement, then the time to empty the tank is not double, but $\sqrt{2}$ times longer.
Make sense?
Paul Safier
A: The issue is, what is the flow velocity?
It is not simple, and it depends on pressure, which is proportional to height.
For low pressure, viscosity will dominate, and velocity will be proportional to pressure.
For high pressure, velocity will be proportional to square root of pressure.
In any case, the geometry of the opening matters.
The general subject is Orifice Flow.
A: This is a very simple problem. The relation is given by using
$$A \frac{dh}{dt} = \frac{-\pi D^2}{4}v;$$ 
where $D$, $h$, $A$ are the diameter, height and area of the tank and v is velocity.
I will leave the derivation up to you
