Theoretical vs Experimental flow rate of fluid coming out of a water bottle with a hole in it Say there is a water bottle that is filled with 300 mL of water and has a circular hole with a radius of 2 mm. In this bottle, the water sits 7.8cm above the top of the hole (which has been drilled 1.5cm above the bottom of the bottle).
According to Bernoulli's law the velocity  $v$ of the water flowing out is equal to $\sqrt{2gh}$
Therefore for the setup above, $v=\sqrt{2*9.81\ m/s^2*0.078\ m} = 1.24\ m/s$
Using this, the flow rate can be calculated as $Q\ =\ Av\ =\ π(0.002\ m)^2*1.24\ m/s = 0.000016\ m^3/s = 16\ mL/s$
This doesn't seem accurate considering that the experimental flow rate is equal to 8 mL/s (40 mL over 5 seconds). However I understand that it ignores viscosity (and other things?) 
I'm wondering a few things, firstly, does the theoretical math here apply to the situation I'm describing? The hole in the bottle isn't exactly a pipe and the only examples I've seen with water flow involve pipes.
Secondly, can Poiseuille's Law be used to determine the flow rate instead, with a more accurate result? (From what I understand Q=πPR^4/8nl, however I don't understand what P is, seeing as in Bernoulli's law pressure cancels and as aforementioned this isn't a typical pipe example.) 
Thirdly, I assume the theoretical flow rate will still be different from the experimental flow rate, what factors cause this?
 A: You have two issues at hand. The first is that Bernoulli's law gives the instantaneous flow rate: as water leaves the bottle the height of the water column above the hole also changes. So you compared the flow rate at $t=0$ to the average flow rate over a 5 second interval.  The more accurate comparison would be to calculate the volume of water lost after 5 seconds and compare that to your measured 40mL loss. To do that you need to solve the differential equation:
$$ \frac{dV}{dt} = A_1 \sqrt{\frac{2g(V_0-V)}{A_2}} $$
Where $V$ is the amount of volume the water bottle has lost, $V_0$ is the original amount of volume above the hole, $A_1$ is the area of the hole, and $A_2$ is the cross-sectional area of the bottle. Separating the differentials and integrating both sides:
$$\int_0^V (V_0-V)^{-1/2} dV = \int_0^t A_1 \sqrt{\frac{2g}{A_2}}t dt $$
$$ \sqrt{V_0} - \sqrt{V_0-V} = \left( A_1 \sqrt{\frac{g}{2A_2}}\right) t $$
I had to estimate $A_2$ from your information provided and I assume it is close to 32 cm^2. In that case, $V_0$ = 251.6mL, t = 5s, and solving for $V$ = 72mL, with an average rate of 14 mL/s. Still not much of an improvement in correctly predicting 8 mL/s, which brings me to my second point:
Think of Bernoulli's equation as the best case scenario, analogous to free-fall without air resistance. You get further from this idealization the more:


*

*Viscous your fluid gets.

*compressible your fluid gets.

*unsteady your fluid flow becomes. 


I think item #3 is the largest factor from you realizing the best case scenario. You might try injecting a few drops of food coloring into the water and see if you see turbulence around the hole. Fixing #3 is all about the geometry of your container, so it could be massaged into steady flow by avoiding sharp edges near the fluid flow, etc.
Poiseuille's law applies to the pressure drop for a fluid traveling down a long straight pipe (like fluid flowing in a medical catheter) . I don't believe it will apply to any of your setup here.
A: 
Given $v\ =\ \sqrt{2gh}\ $ and that the change is volume over time $dV/dt$ is given by the surface area of the leaking hole $a$ times the speed of water through the hole $v$ (negative sign to show that it is decreasing):
$$\frac{dV}{dt}\ =\ -av\ =\ -a \sqrt{2gh} $$
since the volume $V$ at any given height $h$ assuming the cross section of the tank is a constant $A$ is given by $V\ =\ Ah$, if we take the derivative of both sides with respect to time we arrive at:
$$\frac{dV}{dt}\ =\ A \frac{dh}{dt}\ $$
Combining this with the previous equation gives:
$$ -a \sqrt{2gh} = A \frac{dh}{dt}$$
$$ \frac{dh}{\sqrt{h}} = -\frac{a \sqrt{2g}}{A} dt $$
Integrating both sides:
$$ \int_{h_0}^{h_f}\frac{dh}{\sqrt{h}} = -\frac{a \sqrt{2g}}{A}\int_{0}^{t} dt $$
$$(2h^{1/2}) \Big|_{h_0}^{h_f}\ =\ -\frac{a \sqrt{2g}}{A}t$$
To make it easier to look at, define $k\ =\ \frac{a \sqrt{2g}}{A}$ then:
$$(2h^{1/2}) \Big|_{h_0}^{h_f}\ =\ -kt$$
Expanding and rearranging to isolate $h_f$:
$$h_f\ =\ \left(-\frac{k}{2}t+\sqrt{h_0} \right)^2$$
We know that the change in volume after time t is:
$\Delta V\ =\ A\Delta h\ =\ A\left[ h_f\ -\ h_0 \right]\ =\ A\left[\left(-\frac{k}{2}t+\sqrt{h_0} \right)^2  - h_0 \right]$
where 
$A$ = $0.003226\ m^2$ 
$k$ = $0.0172\ m^{1/2}s^{-1}$ 
$t$ = $5\ s$ 
$h_0$ = $0.078\ m$ 
Plugging everything in results in $\Delta V  = -71.5\ mL$
