Flow through a cylinder with pores I'm building a model to study the flow of fluid through a cylind

Physical meaning of $\frac{\pi }{8}$ in Poiseuille's equation
Recently I read about the Poiseuille's equation which relates the flow rate of a viscous fluid to coefficient of viscosity ($\nu$), pressure per unit length($\frac{P}{l}$) and radius of the tube ($r$) in which the fluid is flowing. The equation is
$\frac{V}{t}=\frac{\pi P{r}^4}{8\nu l},$
where $V$ denotes volume of the fluid.
One thing which I don't understand in this equation is that why do we have the constant term as $\frac{\pi }{8}$ . Is there some physical significance of this specific number here or is it just a mathematical convention?

Proving that the water leaving a vertical pipe is exponential (decay)
How can I prove that the rate of which water leaves a vertical cylindrical container (through a hole at the bottom) is exponential of the form :
$A{e}^{kx}$
I know that Torricelli's law is:
$\sqrt{2gh}$
But this only proves a square root relationship. I have data points every 10 seconds and graphed it suggests a decay function. I know the distance between the pipe is 1.5M and the internal diameter is 5cm. The hole diameter is 0.25cm, if this helps. I need to prove that the water leaving the pipe is exponentially decaying.

What happens to pipe length when the pipe diameter changes?
Intuitively, when the diameter of pipe is decreased, there will be more friction loss, more water pressure, and a higher flow rate. Is there a direct relationship/equation derived to see the affect of the pipe lengths?
Deriving from Hagen-Poiseuille's equation, we get:
$Q=\frac{\mathrm{\Delta }P\pi r^4}{8\mu L}$
where :
$Q=$ flow rate
$\mathrm{\Delta }P=$ change in fluid pressure
$r=$ radius of pipe,
$\mu =$ dynamic viscosity of fluid,
$L=$ length of pipe
To keep similar flow rates, are we able to use Poiseuille's derived formula to find the new lengths of pipe with a change in $r$ (pipe radius)?

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 $\sqrt{2gh}$
Using this, the flow rate can be calculated as $Q=Av=\pi (0.002m{)}^2\ast 1.24m/s=0.000016m^3/s=16mL/s$
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?

What determines the dominant pressure-flow relationship for a gas across a flow restriction?
If one measures the pressure drop across any gas flow restriction you can generally fit the relationship to
$\mathrm{\Delta }P=K_2{Q}^2+K_{1}Q$
where $\mathrm{\Delta }P$ is the pressure drop and $Q$ is the volumetric flow
and what I've observed is that if the restriction is orifice-like, $K_2>>K_1$ and if the restriction is somewhat more of a complex, tortuous path, $K_1>>K_2$ and $K_2$ tends towards zero.
I get that the Bernoulli equation will dominate when velocities are large and so the square relationship component. But what's determining the $K_1$ component behavior? Is this due to viscosity effrects becoming dominant? Does the Pouiselle relationship become dominant?

How to convert this derivation of Poiseuille's law into the standard one?
I am trying to derive Poiseuille's law. I have reached a point in the derivation where I have:
$V=\frac{(p1-p2)(R^2)}{4lu}$
Where $l$ is length, u is viscosity, p is pressure, v is flow velocity and R is radius. What I am stuck on is shifting this to the volumetric flow rate:
$V\pi R^2=Q=\frac{(p1-p2)(R^4)}{4lu}.$
$V\pi R^2=Q=\frac{(p1-p2)(R^4)}{4lu}.$
However this is incorrect as Poiseuille's law is divided by 8. I know that I am probably missing something obvious, but I can't think of a reason as to why the whole equation needs to be halved. Any help understanding why (or whether my original derivation for V was inaccurate) would be appreciated.

Is there a contradiction between the continuity equation and Poiseuilles Law?
The continuity equation states that flow rate should be conserved in different areas of a pipe:
$Q=v_1{A}_1=v_2{A}_2=v\pi r^2$
We can see from this equation that velocity and pipe radius are inversely proportional. If radius is doubled, velocity of flow is quartered.
Another way I was taught to describe flow rate is through Poiseuilles Law:
$Q=\frac{\pi r^4\mathrm{\Delta }P}{8\eta L}$
So if I were to plug in the continuity equations definition of flow rate into Poiseuilles Law:
$vA=v\pi r^2=\frac{\pi r^4\mathrm{\Delta }P}{8\eta L}$
Therefore:
$v=\frac{r^2\mathrm{\Delta }P}{8\eta L}$
Now in this case, the velocity is proportional to the radius of the pipe. If the radius is doubled, then velocity is qaudrupled.
What am I misunderstanding here? I would prefer a conceptual explanation because I feel that these equations are probably used with different assumptions/in different contexts.