## Fluids (Through an IV) Question

 Quote by edgepflow I have a feeling that most of the flow restriction is in the final needle that dispenses the IV into the patient. So you could imagine putting even 10 hoses connecting together into the one needle.
Agreed. No matter how many bags you hook up, it will ultimately come down to how much liquid you can get through the tubes and the needle. Larger tube and needle = more flow. But in the case of IV tubes, i'd say there are designed to be the correct size for one bag and produce the correct flow rate into a patient.

Adding an extra bag, it would be more than likely that one bag would flow at a greater rate than the other as the line is designed for a single bag?

You could pimp out the IV bag with some small garden hose and a larger needle, but i wouldn't like to be that patient! :P

 Quote by Andy_Resnick The OP asked a simple question- how can s/he maximize the delivery of fluid through an IV line to a patient? My answer is "use of two (identical) IV bags will deliver fluid at twice the rate of a single bag", and this answer seems to have caused a lot of confusion.
Because it is wrong. Adding two bags does not double the flow rate.

 Quote by Andy_Resnick What controls the flow through the needle? Lots of parameters: the fluid density, the fluid viscosity, the radius of the needle, and the difference in pressure between the ends of the needle. This is the principal result of Poiseuille flow:
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.

 Quote by Andy_Resnick As a practical matter, the flow rate can most efficiently be increased by increasing the needle diameter- doubling the diameter results in a 16-fold increase in flow rate, keeping everything else constant.
If you hold velocity constant (and density is obviously constant), doubling $D$ quadruples the mass flow rate. After all, the formula for area only depends on $D^2$, so doubling $D$ gives you a factor of 4, not 16.

 Quote by Andy_Resnick So all we can control is the pressure at the entrance to the needle. What is this pressure?
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. Of course, that can be easily deduced by simplifying the whole problem and looking at it as a control volume containing everything from the bag (or bags) to the needle. Since it is a control volume, you can still use the Bernoulli equation to get the average exit velocity, despite the fact that there are viscous effects inside). When I say average velocity, I mean the velocity that produces the same mass flow rate out of a given area as integrating the true velocity profile. I promise you, Bernoulli's equation absolutely applies to get that and is used all the time.

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).

Again, your original experiment was flawed based on the fact that the stopcocks were governing the flow rate, not the tube size. The combined mass flow allowed by the stopcocks was simply lower than the maximum allowed by the tubes, thereby allowing the bags to drain at the same rate together as a single bag. You would need a valve capable of passing at least the same mass flow as the tube in order to truly test the variable you think you are. These valves do exist aplenty, but stopcocks are not an example of them.

 Quote by Andy_Resnick This should be obvious as well, since the IV won't drain at all if we were on the International Space Station and tried to do this- there, we would have to apply a pressure to the IV bag to force fluid through the needle.
You have identified the proper force that is driving the flow. You just have a fundamental misunderstanding about how it is applied. It all comes down to a force balance. You can show that the force on a differential element is:
$$dp = -\rho g \; dy$$
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:
$$\Delta p = -\rho g \Delta h$$
Note that while this is still fundamentally a gravity force on the fluid, it has nothing to do with mass directly, but only with density, which is invariant with respect to the size of the reservoir or the number of reservoirs. In other words, we know $\rho$ and $g$ and we can't change either of these. Clearly, the only variable is then $\Delta h$ and the pressure (which in your case is simply the pressure drop since there is atmospheric pressure on both ends) is simply the resulting hydrostatic pressure.

What does this mean? It means the size of the reservoir and amount of fluid plays no role in the resulting pressure and flow rate other than how full the container is, since the fluid in a fuller container will reach higher.

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 can write Bernoulli as:
$$p_{\textrm{exit}} + \rho g y_{\textrm{exit}} + \frac{1}{2} \rho v_{\textrm{exit}}^2 = p_{\textrm{res}} + \rho g y_{\textrm{res}}$$
However, here, $p_{\textrm{exit}}$ and $p_{\textrm{res}}$ are both atmospheric pressure (in the case of a real IV, you would have a term representing the difference between blood pressure and atmospheric pressure). These terms cancel out, leaving:
$$\rho g y_{\textrm{exit}} + \frac{1}{2} \rho v_{\textrm{exit}}^2 = \rho g y_{\textrm{res}}$$
We then move the $y$ terms to the same side. Note that these two terms are arbitrary, and as long as they are taken with respect to the same origin, they can be anything since only their difference matters. For simplicity, we will assume that $y_{\textrm{exit}}$ is the origin and therefore zero and $y_{\textrm{res}}$ just represents the height of the reservoir bag above the exit plane. This leaves us with:
$$\frac{1}{2} \rho v_{\textrm{exit}}^2 = \rho g y_{\textrm{res}}$$
This can obviously be solved for the exit velocity:
$$v_{\textrm{exit}} = \sqrt{2 g y_{\textrm{res}}}$$
The mass flow rate is then trivially:
$$\dot{m} = \rho v_{\textrm{exit}} A_{\textrm{exit}}$$
$$\dot{m} = \frac{1}{4} \rho \pi D_{\textrm{exit}}^2 \sqrt{2 g y_{\textrm{res}}}$$

Note that the only parameters that we have control over in this scenario are $D_{\textrm{exit}}$ and $y_{\textrm{res}}$. It does not depend on the number of reservoirs, the total amount of fluid in the reservoirs or anything other than these parameters.

Also note that while your assumptions made the pressure terms drop out, leaving them in by either actually assuming blood pressure or applying a pressure to the reservoir surface(s) would be easy and would just result in a $\Delta p$ term that is also not dependent on the number of reservoirs or the total amount of fluid in the reservoirs.

 Quote by cjl Oh, and just to demonstrate it, I just performed the experiment with the flow restriction primarily (but not completely) at the common exit of the two funnels. I couldn't find two identically sized funnels, so the drain times individually were quite different, but I think the results speak for themselves: Funnel 1 drain time: 1:38.23 1:35.97 1:39.60 1:39.75 Funnel 2 drain time: 27.54 27.74 28.11 27.54 Both funnels together: 1:57.55 1:58.65 2:02.85 1:55.63 As you can see, the flow rate stays roughly constant, while the drain time for both funnels together is only slightly less than the sum of the drain times for each funnel individually. It is also substantially longer than the individual drain time of either funnel (as I predicted several pages ago).
I think the reason the drain time of both funnels together is a little less than the sum of the individual times is the extra "inventory" maintaining a slightly higher average elevation "gravity head" during the test. Of course, the addition of the second funnel does not at any time increase the maximum gravity head.

 Quote by boneh3ad $$\dot{m} = \rho v_{\textrm{exit}} A_{\textrm{exit}}$$ $$\dot{m} = \frac{1}{4} \rho \pi D_{\textrm{exit}}^2 \sqrt{2 g y_{\textrm{res}}}$$ Note that the only parameters that we have control over in this scenario are $D_{\textrm{exit}}$ and $y_{\textrm{res}}$. It does not depend on the number of reservoirs, the total amount of fluid in the reservoirs or anything other than these parameters.
Good treatment, that is what I would expect. I mentioned several posts ago that adding reservoirs in parallel would not have a pressure multiplying effect, although you will see a small increase in the average flow as cjl measured.

If there was a pressure multiplying effect with parallel reservoirs, we could create some pretty cool pressure and flow control loops. Imagine a ring array of reservoirs and actuated valves: if you need more or less flow, just open or close valves. If this were true, there would probably be products like this on the market.

 Quote by edgepflow I think the reason the drain time of both funnels together is a little less than the sum of the individual times is the extra "inventory" maintaining a slightly higher average elevation "gravity head" during the test. Of course, the addition of the second funnel does not at any time increase the maximum gravity head.
I agree with this as the most likely explanation - the way the experiment was set up, the last portion of the drainage was largely independent of reservoir volume, so it always took about the same time. Since the reservoir drainage took longer in the case where both funnels were open, the average head was slightly higher for the two-funnel case than for the one-funnel case. It's a relatively small effect, but enough to be visible in the results as a slight disparity between the drain time of both funnels together and the sum of the individual drain times.

 Quote by edgepflow If there was a pressure multiplying effect with parallel reservoirs, we could create some pretty cool pressure and flow control loops. Imagine a ring array of reservoirs and actuated valves: if you need more or less flow, just open or close valves. If this were true, there would probably be products like this on the market.
Interestingly enough, it would also allow for free energy and perpetual motion.

 Quote by boneh3ad Because it is wrong. Adding two bags does not double the flow rate.
Indeed, and that was already described in the OP.
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.
 Right, you only need to take in account that some of the energy goes into friction.
[...] Since it is a control volume, you can still use the Bernoulli equation to get the average exit velocity, despite the fact that there are viscous effects inside). When I say average velocity, I mean the velocity that produces the same mass flow rate out of a given area as integrating the true velocity profile. I promise you, Bernoulli's equation absolutely applies to get that and is used all the time.
That is also wrong except if the Poisseuille friction term is very small compared to the Bernoulli terms.

Regards,
Harald
 harrylin, Except in the case of very fast flow or very small tubes, the friction losses would be very small. Regardless, if we approach this from the eyes of a mechanical engineer, we could easily introduce a head loss term to fairly accurately account for the effect of viscous losses. You could theoretically do the same for the loss associated with the T-joint. The important thing is that neither of those additional loss terms will depend on the number of reservoirs or amount of fluid in each reservoir.

 Quote by boneh3ad Except in the case of very fast flow or very small tubes, the friction losses would be very small. if we approach this from the eyes of a mechanical engineer, we could easily introduce a head loss term to fairly accurately account for the effect of viscous losses. You could theoretically do the same for the loss associated with the T-joint.
The friction term can be neglected in the case of fast flow through a large opening: Poisseuille flow scales with v while the kinetic energy term scales with v2.
 The important thing is that neither of those additional loss terms will depend on the number of reservoirs or amount of fluid in each reservoir.
Yes of course (that is well understood), except for the issue of the OP which, as I and others suggested, relates to the T connector; and I gave a reference to a practical application of that phenomenon by Tesla.

 Quote by harrylin Yes of course (that is well understood), except for the issue of the OP which, as I and others suggested, relates to the T connector; and I gave a reference to a practical application of that phenomenon by Tesla.
And I fully agree with that assessment. That is very likely the cause of a decrease in flow rate in the OP's case rather than seeing no change.

Recognitions:
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, $Q = \frac{\pi \mu R}{2 \rho}Re$, where Re is the Reynolds number.

 Quote by boneh3ad If you hold velocity constant (and density is obviously constant), doubling $D$ quadruples the mass flow rate. After all, the formula for area only depends on $D^2$, so doubling $D$ gives you a factor of 4, not 16.
I don't see this. The elementary result, $Q = \frac{\pi R^{4}}{8 \mu}\nabla P$ clearly has a 4-th 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.

 Quote by boneh3ad 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: $$dp = -\rho g \; dy$$ 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: $$\Delta p = -\rho g \Delta h$$
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 $\Delta p = -\rho g \Delta h$, 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 $\Delta p = \rho g \Delta h/l * x$, 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.

 Quote by Andy Resnick 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, $Q = \frac{\pi \mu R}{2 \rho}Re$, where Re is the Reynolds number.
Well, you can use a Reynolds number but you end up using it for it's velocity component so to speak. You need some form of velocity to get the flow rate. The difference is that you don't need a velocity profile. You can simply use the average velocity over the entire exit area to get mass flow rate. After all, the average velocity, by definition, could be obtained from integrating that velocity profile given by Poiseuille flow. In fact, in integrating the Poiseuille solution across the exit area, you would end up with an average velocity term if you so chose to isolate it.

The important takeaway here is that you use a Poiseuille flow solution to get the velocity profile, which you don't need. The volumetric flow rate that is quoted on Wikipedia is just one of the takeaways from integrating the velocity profile across a plane of interest.

 Quote by Andy Resnick I don't see this. The elementary result, $Q = \frac{\pi R^{4}}{8 \mu}\nabla P$ clearly has a 4-th power dependence on R.
The problem is that $Q$ has an $R^2$ term in it as well. At its most basic for incompressible flow, the mass flow rate can be written as $\dot{m} = \rho v A$. In this case, $v \propto \Delta p$ and $A \propto D^2$. That means that for a given $\Delta p$, you are left with $\dot{m} \propto D^2$.

 Quote by Andy Resnick 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.
The stress tensor and pressure are not synonymous. The stress tensor does have a dependence on the velocity gradient. The pressure, while a component of the stress tensor, does not. The stress tensor technically speaking is defined as:

$$\boldsymbol{\tau} = -\nabla p \;\mathbb{I} + \mu \nabla^2 \vec{v}$$

The pressure term shows up as a gradient, and has nothing to do with the velocity gradient and everything to do with externally enforced pressure gradients or hydrostatic pressure. It has no explicit dependence on velocity.

 Quote by Andy Resnick 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.
You absolutely can add fluid to the bag to change the height. I was taking the point to be the needle outlet, but if you were trying to figure out the inlet properties, you could easily do it that way as well and then use your result to do a similar analysis across the needle. I just skipped a step. The difference would be if there was a height difference between the entrance and exit of the needle (unlikely) or an area difference (possible).

 Quote by Andy Resnick 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.
I may have used some odd terminology, as there certainly isn't really a maximum flow rate for an incompressible, inviscid fluid. Of course, for very high velocities, viscosity could counteract any additional pressure and compressibility effects can choke the flow to prevent additional mass flow, but neither of those situations apply here.

What I mean is this: imagine you have a garden hose and you are dripping water through it with a turkey baster. That is essentially what you are doing with the stopcock. Your mass flow rate is dictated by the stopcock, and for the same conditions, the stopcock admits less flow than the tubes would. Your stopcock is throttling your flow, in essence. Based on your results, I infer that even the combined mass flow of both stopcocks is likely not as much as the tube could admit under the given flow conditions. In all likelihood, that means there is a bit of unused space in the tubes when the water is draining from your funnels, or in other words, the entire cross section is not filled with water. You are only getting as much water through that system as the stopcocks will allow, which is apparently less than what the tubes would allow.

 Quote by Andy Resnick You lost me- Why is the velocity zero? Fluid is draining from the reservoir- the top surface is moving down.
Looking at an instant in time, the velocity is zero in the reservoir but nonzero at the exit. Even in general, you take the reservoir velocity to be zero in these cases because it is very tiny compared with the rest of the system. You can look at a handful of books and find examples of this, but it is a reasonable approximation except perhaps when the bag is nearly empty. At that point crazy things happen anyway. The top surface is of course moving down, so when looking over a period of time, you would have to take the $\Delta h$ term as a time varying quantity and you would end up with a differential equation.

If you did take $\Delta h$ to be time varying, you would notice that with two bags, the fluid level falls more slowly, so the pressure would stay marginally higher, but since we are only talking a matter of inches here, it still won't speed the bags up enough to pass the single bag, especially because even if the two bags did drain faster, they would catch up to the height of the single bag and end up draining at half the rate again.

In other words, you would expect that in practice, neglecting the effect of the junction downstream, the two bags would probably drain in something like 1.75 times as long as the single bag rather than a perfect 2. Finding the actual value is nontrivial and would need to be numerically integrated, so I don't really plan to do that since I ought to be finishing this paper that needs to be submitted to AIAA by next week.

Recognitions:
 Quote by boneh3ad I may have used some odd terminology, as there certainly isn't really a maximum flow rate for an incompressible, inviscid fluid.
Aha- this is one major problem. I tried to work the problem using Bernoulli's law, and now I am convinced that Bernoulli's law *cannot* be used to correctly analyze this problem, because Bernoulli's law only holds for inviscid flow- Poiseuille flow fails for inviscid flow. I also understand why the introductory textbook picture of Bernoulli's law is presented the way it is, with gravity perpendicular to the flow.

Bernoulli's law is simply conservation of mechanical energy. Let's compare two control volumes, one at the top of the reservoir (say, a large diameter thin cylinder) and one within the needle, terminating at the outlet (a long slender cylinder). In some time dt the volume dV exits the needle. Each control volume dV is the same (since conservation of mass holds and we assume an incompressible fluid) and has a mass dm. Conservation of mechanical energy gives:

P + 1/2 dm v^2 + dm gy = constant.

replacing dm = $\rho \Delta V$ and v = dl/dt = 1/A dV/dt = Q/A (since dV = A dl, and Q = dV/dt), we get for the upper control volume:

P$_{atm}$ + $\frac{1}{2} \rho\Delta V\frac{Q^{2}}{A^{2}_{1}}+\rho\Delta Vgh$, where h is the height above the needle outlet, and we assumed that l_1 was so thin that the entire control volume is held at atmospheric pressure.

Writing the expression for the lower volume is not so simple, as we will see- since we want A_1 >> A_2 (the diameter of the reserviour is much larger than the needle), the corresponding height l_2 >> l_1, and so we must be careful about assigning a pressure and potential energy. For now, let's do the obvious linearization:

$P_{atm} +\delta P + \frac{1}{2} \rho\Delta V\frac{Q^{2}}{A^{2}_{2}}+\rho\Delta Vg \frac{l_{2}}{2}$

and thus Bernoulli's law gives:

$\delta P = \rho\Delta V [ \frac{Q^{2}}{2}(\frac{1}{A^{2}_{1}}-\frac{1}{A^{2}_{2}}) + g(h-\frac{l_{2}}{2})]$

Now we take some limits: A_1 >> A_2, h >> l_2:

$\delta P = \rho\Delta V [ -\frac{Q^{2}}{2}\frac{1}{A^{2}_{2}} + gh)]$.

From this, we can generate Torricelli's result v = $\sqrt{2gh}$ by setting dP = 0. This corresponds to putting a hole in the side of the container, so there is no pressure difference across the ends of the lower control volume. But, our control volume in the needle is vertical, and so there is a pressure difference between the top and bottom face. Since we can't use the result from Poiseuille flow, it's not entirely obvious what to put for this pressure difference.

The bottom line, Bernoulli's law is not appropriate to model viscous flow. All discussions about "maximum flow capacity", the roles of "flow restrictors" and the like only arise when trying to force inviscid flow equations to handle viscous effects.

 Quote by Andy Resnick Aha- this is one major problem. I tried to work the problem using Bernoulli's law, and now I am convinced that Bernoulli's law *cannot* be used to correctly analyze this problem, because Bernoulli's law only holds for inviscid flow- Poiseuille flow fails for inviscid flow. I also understand why the introductory textbook picture of Bernoulli's law is presented the way it is, with gravity perpendicular to the flow. ... The bottom line, Bernoulli's law is not appropriate to model viscous flow. All discussions about "maximum flow capacity", the roles of "flow restrictors" and the like only arise when trying to force inviscid flow equations to handle viscous effects.
Aha, but this isn't true. Bernoulli's principle can be used on a viscous flow just fine. The kicker is that it doesn't give you the exact velocity profile and needs correction factors (typically called head loss) to correct for viscous losses. As long as you know the limitations, you can successfully use it to analyze a viscous problem. If you would like, I could go look up the correction factor for friction for our system. Otherwise, you can do it. It is called Darcy's friction factor or Colebrook's friction factor, depending on how accurate of a result you want. Colebrook's value is more accurate but is an implicit equation.

Anyway, all that friction loss will do is change the total pressure drop between the bag and the needle, meaning a lower exit velocity. It doesn't change the fundamental dependencies of the solution.

Also, the flow restriction talk has nothing to do with viscous or inviscid flow. It has to do with the stopcock not physically being able to physically pass the same amount of mass through it for a given flow condition. That has nothing to do with viscosity and everything to do with simply the size of the hole. Stopcocks are designed to restrict flow rate. YOur stopcocks are doing their jobs.

Recognitions:
 Quote by boneh3ad Aha, but this isn't true. Bernoulli's principle can be used on a viscous flow just fine.
I'm going to ask for a reference on this one- everything I found was considerably more equivocal:

http://arxiv.org/pdf/0809.1309

http://www.engineersedge.com/fluid_f..._bernoulli.htm

The second reference has the fluid exiting out of a hole in the side, which makes direct comparison to our situation questionable. The third reference requires empirical input to solve- which begs the question why it is preferred in our simple system.

 Quote by Andy Resnick I'm going to ask for a reference on this one- everything I found was considerably more equivocal:
Try any basic fluid mechanics textbook geared towards mechanical engineers. You can change the viscous term into a head loss, which shows up in the form of a length so that it can be directly compared to a change in height. This is done commonly with the Darcy–Weisbach equation for the case of the head loss due to friction in the pipe (or otherwise known as viscous dissipation).

A book I happen to have on hand here that I can cite is "Fundamentals of Fluid Mechanics" by Munson, Young and Okiishi (http://www.amazon.com/Fundamentals-F.../dp/0471675822). Particularly check the section about pipe flow. You essentially get a corrected Bernoulli equation.

Other assumptions necessary, such as fully-developed flow, can also be accounted for in terms of an energy loss term that shows up as a head loss. The same can be said for the T where the IV bags come together, and that is where the slowup comes from in the OP's example, since friction is present in both cases.

Still, none of them would support your original claim that two bags would create twice the mass flow rate. The only way that you can get more flow rate is by raising the bags higher, putting pressure on the bags or increasing the exit diameter of the needle. That goes for the viscous case, the inviscid case, and any other incompressible case you can think of that involves obeying the laws of physics.
 In practice, the Bernoulli Equation is corrected for viscous flow with a "discharge coefficent." Similar to the equation boneh3ad derived earlier, we can express the volumetric flow rate as: Vdot = Cd * Aorf * sqrt (2gh) where, Vdot = volumetric flow rate Cd = discharge coefficient of opening = actual discharge flow / theoretical discharge flow Aorf = area of discharge opening h = fluid level. As was derived earlier by boneh3ad and Andy Resnick, the surface area of the reservoir does not appear in this expression.

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