# rate of flow out of a reservoir

A reservoir contains a fluid, and there is an aperture at the bottom through which the fluid flows out as shown in the illustration:

I understand that the rate of flow is based on at least these parameters: the amount of fluid in the reservoir due to the weight of the fluid, the size of the apperture, and the density of the fluid. But how exactly do these relate to the flow rate? That is:

$$Q=f(V,A,\rho)$$

What is $f$?

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In addition to the density of the fluid, you have to account for the viscocity of the fluid, the length of the pipe connected to the drain (unless it's just a hole), the area of the horizontal cross-section of the tub (the pressure at the drain is determined by the height of the water, not the volume), and possibly some other things. Fluid dynamics usually ends up being messier than you think it should. – krs013 May 21 '13 at 5:10

The basic equation for inviscid flow is $Q = AVK$ where $Q$ is flow rate in mass/time, $A$ is orifice area, and $V$ is flow velocity. $K$ is a factor that depends on the shape of the orifice.
Velocity $V$ generally follows $V = \sqrt{2gh}$ where $g$ is acceleration due to gravity, and $h$ is the "head", or height of fluid above the orifice.
This says inviscid flow rate varies as $h^{1/2}$. For viscous flow, you would expect it to vary more as $h^1$.