Quote by boneh3ad

There's a lot in your post I don't understand maybe you can help walk me through it:
Quote by boneh3ad
Poiseuille flow, while the correct flow for this situation, is not needed to solve this problem. You don't need to know the velocity profile at a given point in the tubing or needle to know the flow rate, adn that is what Poiseuille flow gives you.

I agree Poisueille flow may not be the only way to solve the problem. I also agree that one does need velocity to calculate a flow rate: for example, [itex] Q = \frac{\pi \mu R}{2 \rho}Re[/itex], where Re is the Reynolds number.
Quote by boneh3ad
If you hold velocity constant (and density is obviously constant), doubling [itex]D[/itex] quadruples the mass flow rate. After all, the formula for area only depends on [itex]D^2[/itex], so doubling [itex]D[/itex] gives you a factor of 4, not 16.

I don't see this. The elementary result, [itex] Q = \frac{\pi R^{4}}{8 \mu}\nabla P[/itex] clearly has a 4th power dependence on R.
Quote by boneh3ad
This is, in fact, one of the pressures that can be used to answer this problem. It is dependent on the velocity of the fluid at that point plus the static pressure.

I don't follow this either pressure, which is part of the overall stress tensor, depends on the velocity *gradient*, not the velocity. Clearly, uniform motion of a fluid should not alter the internal pressure of the fluid (in the region where v is constant). The results of measurements in any inertial frame are the same.
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I promise you, Bernoulli's equation absolutely applies to get that and is used all the time.

I think Bernoulli's equation could probably be used as well.
Quote by boneh3ad
So, what does this simplified system mean? It means you can ignore the velocity going through the system. And only worry about the pressure from the reservoir and the exit pressure. As you stated, the exit pressure is fixed. The pressure of the reservoir, however, is entirely determined by two things: height of the reservoir and atmospheric and other external pressures. It has nothing to do with the amount of fluid in the bag or the number of bags hooked up. It only depends on the highest point the fluid reaches above the exit. In other words, the only way to increase the pressure without changing the diameter of the tubes or the needle is to either lift the bag higher or squeeze it (either with your hands or an external pressure of some kind, it doesn't matter).

Let me make sure I understand we are talking about the pressure at a specific point the needle inlet? Because I can change the column height of fluid by adding or removing fluid to/from the bag that should not be in dispute, either.
Quote by boneh3ad
The combined mass flow allowed by the stopcocks was simply lower than the maximum allowed by the tubes,

I don't understand what you mean by "maximum (flow) allowed by the tubes". Is there some upper limit? What sets this limit? The flow rate through a tube doesn't admit an absolute maximum if I double the pressure gradient, I double the flow rate. There is some limit when turbulent flow is reached (say, Re = 4000 and higher), but that's not what we are talking about, AFAIK.
Quote by boneh3ad
You can show that the force on a differential element is:
[tex]dp = \rho g \; dy[/tex]
If you integrate this from the fluid element of interest (the exit plane in our case) to the very top surface of the fluid in the reservoir (meaning the heighest point of fluid in the control volume), you get:
[tex]\Delta p = \rho g \Delta h[/tex]

Ah I think this is the problem. The pressure at both the top surface and the bottom surface is atmospheric pressure. This causes a lot of confusion the pressure at the exit plane of the needle is atmospheric pressure. The pressure at the top of the reservoir is also atmospheric pressure. So what we really have is the pressure from the top of the reservoir to the needle inlet is [itex]\Delta p = \rho g \Delta h[/itex], which 'h' taken as negative (since the pressure at the needle is *higher* than the pressure at the top free surface), and the pressure (over atmospheric) along the needle is given by [itex]\Delta p = \rho g \Delta h/l * x[/itex], where 'x' is the distance along the needle (and varies from 0 to l, the length of the needle) I may have reverse the sense of direction within the needle.
Quote by boneh3ad
Of course, this is a hydrostatics example. For this problem, we need a moving fluid. Enter Bernoulli's equation, which has the same form as the hydrostatic equation, only including motion terms. If we look at the same two points at the top of the reservoir where the velocity is zero and at the exit of the needle

You lost me Why is the velocity zero? Fluid is draining from the reservoir the top surface is moving down.