Fluids (Through an IV) Question

[Biebs]

Hi everyone, just some background information--I work in an emergency department as a trauma tech. My co-workers and I got into discussion at work about the fastest method of giving a patient fluid via an IV. We had this really sick patient who needed multiple liters of fluid and I was wondering if you could answer the following question for us.

If you are not familiar with how an IV works, the IV bag is "spiked" with the IV tubing (tubing plugged into the bag), then the distal part of the IV tubing is plugged into the IV catheter (into the patient). The IV tubing also has a needless port/connection maybe about one foot from where it connects into the patient so you can hook another set of IV tubing to run in medications, or give medicine via syringe, etc. So, one of the nurses thought that in order to get the most amount of fluid into the patient the fastest, she would connect the IV tubing into the patient like normal, then also prime another set of IV tubing connected to a second bag of fluid and plug that into the needless port of the first line of IV tubing. She thought that this would be a quicker way to get more fluid into the patient than if you were to run the first bag in (on one set of IV tubing) and then just change the bag immediately after it ran dry.

I didn't believe her and so we did a little experiment. We replicated the same scenario and found that the time to infuse two bags of fluid was definitely faster using a single line of IV tubing and changing the bag once it became dry instead of connecting two lines together and running both bags in together. However I could not come up with an explanation for this. I was thinking that it had something to do with the pressure and Bernoulli's equation. Both bags experience the same potential energy because they are at the same height (but this is doubled when hanging both at the same time?), I am assuming that the pressure is greater when the two lines are connected together (where the two sets of IV tubing meet). Increasing the pressure should decrease the velocity of the fluid right? When running the single bag in with just one line, the potential energy is just 1, because 1 bag instead of two at a time, so lower pressure, higher velocity? Also, maybe fluid turbulence where the two connections meet slows down the rate?

Thanks, this has been bugging me and the ED staff would sure like to know too!

 

[edgepflow]

Bernoulli's equation applies within its ideal limitations (i.e. no friction).

In the case of flow in an IV, friction effects are likely very significant. 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. You would not see much improvement since most of the restriction is in the final needle.

I believe you mentioned you were able to flow even less when you connected two tubes into the needle. What may be happening is that the the second stream was making small vortices (swirling fluid) that may have further restricted the flow.

 

[cjl]

My guess is that having two bags connected to the same line increased the turbulence in the line, which decreased the flow rate. The pressure from two bags was the same as from one bag, since the fluid level was the same (all that matters for the pressure is the height of the fluid surface - how much fluid there is at that height is irrelevant). With one bag, you should get fairly smooth flow down the line, but with two, there would be a fairly substantial disruption of flow at the location where the two flows joined.

 

[Andy Resnick]

Hi everyone, just some background information--I work in an emergency department as a trauma tech. My co-workers and I got into discussion at work about the fastest method of giving a patient fluid via an IV. We had this really sick patient who needed multiple liters of fluid and I was wondering if you could answer the following question for us.

If you are not familiar with how an IV works, the IV bag is "spiked" with the IV tubing (tubing plugged into the bag), then the distal part of the IV tubing is plugged into the IV catheter (into the patient). The IV tubing also has a needless port/connection maybe about one foot from where it connects into the patient so you can hook another set of IV tubing to run in medications, or give medicine via syringe, etc. So, one of the nurses thought that in order to get the most amount of fluid into the patient the fastest, she would connect the IV tubing into the patient like normal, then also prime another set of IV tubing connected to a second bag of fluid and plug that into the needless port of the first line of IV tubing. She thought that this would be a quicker way to get more fluid into the patient than if you were to run the first bag in (on one set of IV tubing) and then just change the bag immediately after it ran dry.

I didn't believe her and so we did a little experiment. We replicated the same scenario and found that the time to infuse two bags of fluid was definitely faster using a single line of IV tubing and changing the bag once it became dry instead of connecting two lines together and running both bags in together. However I could not come up with an explanation for this. I was thinking that it had something to do with the pressure and Bernoulli's equation. Both bags experience the same potential energy because they are at the same height (but this is doubled when hanging both at the same time?), I am assuming that the pressure is greater when the two lines are connected together (where the two sets of IV tubing meet). Increasing the pressure should decrease the velocity of the fluid right? When running the single bag in with just one line, the potential energy is just 1, because 1 bag instead of two at a time, so lower pressure, higher velocity? Also, maybe fluid turbulence where the two connections meet slows down the rate?

Thanks, this has been bugging me and the ED staff would sure like to know too!

To a good approximation, the flow through an IV line is given by Poiseuille flow

http://en.wikipedia.org/wiki/Hagen–Poiseuille_equation

By hooking up multiple bags, you are increasing the pressure drop (using two bags at the same height doubles the driving pressure) and so you increase the flowrate.

To maximize the flowrate, you could do a number of things:

1) increase the needle diameter (if that's the smallest diameter tube- otherwise, increase the diameter of the appropriate tube).
2) Shorten the distance between the entrance port and the exit port
3) Increase the driving pressure (pressurize the bags, for example. Increasing the height of the bag could potentially also be used)
4) Decrease the viscosity of the fluid

Clearly, some of these are more feasible than others. Plus, there is a maximum flowrate that can be delivered without further damaging the patient.

 

[cjl]

To a good approximation, the flow through an IV line is given by Poiseuille flow

http://en.wikipedia.org/wiki/Hagen–Poiseuille_equation

By hooking up multiple bags, you are increasing the pressure drop (using two bags at the same height doubles the driving pressure) and so you increase the flowrate.

No, not at all. Since the problem is a quasi-hydrostatic problem, the pressure from the bags is dependent only on the height of the water, and increasing the reservoir size does not affect the pressure at all.

 

[Andy Resnick]

No, not at all. Since the problem is a quasi-hydrostatic problem, the pressure from the bags is dependent only on the height of the water, and increasing the reservoir size does not affect the pressure at all.

P =F/A = mg/A. Doubling the mass of the fluid doubles the applied pressure. Torricelli showed this quite some time ago, when he observed flow through an orifice.

 

[edgepflow]

P =F/A = mg/A. Doubling the mass of the fluid doubles the applied pressure. Torricelli showed this quite some time ago, when he observed flow through an orifice.

I think cjl was looking at a situation of increasing the length and width of a fixed height swimming pool does not change the pressure at the bottom of the pool.

And I think Andy Resnick was pointing out that if you keep the length and width of the pool fixed and increase the height, the pressure at the bottom of the pool is increased.

 

[cjl]

P =F/A = mg/A. Doubling the mass of the fluid doubles the applied pressure. Torricelli showed this quite some time ago, when he observed flow through an orifice.

Your equation is correct, but you ignore the fact that doubling the quantity of fluid at a given height both doubles the mass and doubles the effective area. Thus, the pressure is unchanged.

For any hydrostatic or quasi-hydrostatic problem, the pressure in a fluid is P = P_external + ρgh in which P_external is the ambient pressure, rho is the density of the fluid, and h is the height of the fluid's surface above the point at which you are trying to find the pressure. The quantity of fluid is irrelevant - only the height matters. Thus, 12 bags at the same height will give the same pressure as one tiny bag. However, one bag suspended ten feet above the patient would provide substantially more pressure (not that I'm recommending this, mind you).

 

[Andy Resnick]

Your equation is correct, but you ignore the fact that doubling the quantity of fluid at a given height both doubles the mass and doubles the effective area. Thus, the pressure is unchanged.

For any hydrostatic or quasi-hydrostatic problem, the pressure in a fluid is P = P_external + ρgh in which P_external is the ambient pressure, rho is the density of the fluid, and h is the height of the fluid's surface above the point at which you are trying to find the pressure. The quantity of fluid is irrelevant - only the height matters. Thus, 12 bags at the same height will give the same pressure as one tiny bag. However, one bag suspended ten feet above the patient would provide substantially more pressure (not that I'm recommending this, mind you).

ρgh=(m/V)gh = mg(h/V) = mg/A.

Doubling the mass and keeping the cross-sectional area constant (i.e. the cross-section of the tube just downstream of the t-junction) means the pressure is doubled.

 

[cjl]

ρgh=(m/V)gh = mg(V/h) = mgA.

Doubling the mass and keeping the cross-sectional area constant (i.e. the cross-section of the tube just downstream of the t-junction) means the pressure is doubled.

No. That isn't how fluids work. The force from the fluid reservoir does not only act on the tube. It also acts on the bottom of the fluid reservoir. Since the projected area of the bottom of the reservoir (the area covered in fluid) plus the area of the tube is equal to the maximum cross sectional area of the fluid, the area of the fluid actually cancels out in the relevant equations. As a result, the cross sectional area of the reservoir is irrelevant, only the height matters.

 

[sams_rhythm]

I have to say I'm with cjl not Andy.

I don't know if this will work, but here goes.

l                 l     l  l
l                 l     l  l
l                 l     l  l
l                 l     l  l
l                 l     l  l
l                 l_____l  l
l__________________________l

Doh. I guess not. Picture a thick and a thin vertical pipe attached at the bottom.


If you fill this container with water, what will the water do? It will stay still because the pressure due to the head of water in either the thick column or the thin column will be the same. It is only the pressure head (that is the vertical height of water) which changes the pressure. The equation for the force should be

Pressure = density* gravitational acceleration * vertical height

By the way, who is Torricelli?

As for the initial question:
Adding a second bag doesn't make any difference, as has been said it is the vertical height between the bag and the needle that sets the pressure force driving the flow rate.

(Flow rate)^2 * (some friction factor for pipe flow) * length of pipe = density* gravitational acceleration * vertical height

You can see this explained better here: http://en.wikipedia.org/wiki/Darcy_friction_factor

The reason that adding the second bag reduced the flow seems strange to me. I suppose you could have found that if the bags weren't at the same height you were actually driving some flow from one bag to the other rather than into the patient. I suppose the pressure force driving fluid into the patient would have been set by the average of the heights of the two bags (althought the average would depend on the length of the two bags).

What do you think?

 

[marcusl]

ρgh=(m/V)gh = mg(h/V) = mg/A.

Doubling the mass and keeping the cross-sectional area constant (i.e. the cross-section of the tube just downstream of the t-junction) means the pressure is doubled.

The equation is correct but not the interpretation. To apply to the area A of the orifice at the bottom of the reservoir, the mass in this equation must be that contained in a virtual cylinder of cross sectional area A directly above the orifice. Mass to the sides (ie. outside of that cylinder) does not contribute any net z axis force.

The equation fragment that you wrote first, whose whole equation reads p=\rho gh, is less prone to misinterpretation. Since there is no mention of mass, or reservoir shape, it is clear that pressure at a certain depth is a constant determined only by liquid density and depth.

 

[Andy Resnick]

The equation is correct but not the interpretation. To apply to the area A of the orifice at the bottom of the reservoir, the mass in this equation must be that contained in a virtual cylinder of cross sectional area A directly above the orifice. Mass to the sides (ie. outside of that cylinder) does not contribute any net z axis force.

The equation fragment that you wrote first, whose whole equation reads p=\rho gh, is less prone to misinterpretation. Since there is no mention of mass, or reservoir shape, it is clear that pressure at a certain depth is a constant determined only by liquid density and depth.

How can you say \rho gh does not refer to mass? What do you think \rho is? The OP was discussing hooking up multiple IV bags and connecting them through various junctions. Considering the pressure at the needle, use of multiple IV bags *must* increase the pressure.

I don't understand the confusion- (hopefully) nobody here would be confused if I claimed that placing one or two beakers of water on a scale results in different readings. What's the difference?

I did the experiment to make sure I was giving the correct answer- I hooked two separatory funnels up to a common outlet and compared the amount of time needed to empty one or both funnels. The time was the same- thus, the flowrate was doubled, thus the driving pressure was doubled.

I'm not asking for anyone to believe me- do the experiment yourself. Physics is an *experimental* science, after all.

 

[marcusl]

Hmm, I'll have to think about your experiment

 

[sams_rhythm]

Andy, no, I disagree. If I fill a sink with 10 cm depth of water or a bath with 10 cm depth of water, the sink empties quicker. This is essentially the same as your experiment. I'd need to know more about whether friction from the walls of your funnels was playing a role to try and understand your experiments.

Pressure only changes in a static fluid due to the vertical distance you move (i.e. the component of the distance which is in the same direction as gravity). If you are diving, the pressure changes when you go down, not if you swim sideways. Also, the pressure is the same if you swim at the same depth in a pool with a large or small horizontal cross-sectional area. This would not be the case if

pressure = mass *g / Area

I agree that the mass of one beaker is half the mass of two beakers. As marcusl says, it is best to stick with the density * height * g equation to avoid confusion. I think the confusion in the way you have manipulated the equations is that you have assumed that V=h * A, but then you say that you double the mass whilst keeping A the same. This could only happen if you doubled h.

Andy, does that make more sense to you?

 

[sams_rhythm]

Biebs, did you find out anything more about this? Do you think using two bags might have led to fluid flowing from one bag to the other?

Did wikipedia make any sense? I think the way that flow is increased in an IV is by using a wider diameter catheter. In fact, I thought they use a wider artery if that is the limiting factor and they really need to pump lots in.

The classic experiment on pipe flow like what is going through your needle and tubes is by Osborne Reynolds. He has a dimensionless number named after him, so he must have been good.

I'm interested to hear what you think seeing as you do this day in, day out.

 

[cjl]

How can you say \rho gh does not refer to mass? What do you think \rho is? The OP was discussing hooking up multiple IV bags and connecting them through various junctions. Considering the pressure at the needle, use of multiple IV bags *must* increase the pressure.

I don't understand the confusion- (hopefully) nobody here would be confused if I claimed that placing one or two beakers of water on a scale results in different readings. What's the difference?

I did the experiment to make sure I was giving the correct answer- I hooked two separatory funnels up to a common outlet and compared the amount of time needed to empty one or both funnels. The time was the same- thus, the flowrate was doubled, thus the driving pressure was doubled.

I'm not asking for anyone to believe me- do the experiment yourself. Physics is an *experimental* science, after all.

Andy, I don't know how else to put it. You are wrong. In a static fluid, the pressure at a given depth is constant. The fluid pushes against everything, including the walls of the container. At any location at which the container is getting narrower, a component of the pressure against the walls will be in a downward direction. Similarly, the walls will push back in an upward direction. The component of this force which is pushing upwards will be proportional to P*A*sin(theta), in which P is the pressure at that location, A is the area of the wall, and theta is the angle it makes to the vertical. Since the projected area that this region would have if projected onto a horizontal surface is A*sin(theta), the effective upwards pressure being exerted on the fluid by the wall is P*A*sin(theta)/(A*sin(theta)) = P. At every location, the wall will be counteracting the force of the fluid above it. As a result, the only force the fluid entering the tube will feel is the force from the column directly above it, thus the quantity of water does not matter. It is irrelevant whether you are feeding it from a funnel or a swimming pool - so long as the water surface is the same height in both, the pressure will be identical.

As for your experiment? I can't explain that without a lot more detail about the apparatus, but suffice it to say that you did not disprove one of the fundamental facts of fluid mechanics with a pair of funnels and a kitchen sink.

Chris
(graduate student in fluid mechanics and propulsion)

 

[Andy Resnick]

Andy, no, I disagree. If I fill a sink with 10 cm depth of water or a bath with 10 cm depth of water, the sink empties quicker. This is essentially the same as your experiment. I'd need to know more about whether friction from the walls of your funnels was playing a role to try and understand your experiments.

<snip>

Andy, does that make more sense to you?

Andy, I don't know how else to put it. You are wrong.

<snip>
As for your experiment? I can't explain that without a lot more detail about the apparatus, but suffice it to say that you did not disprove one of the fundamental facts of fluid mechanics with a pair of funnels and a kitchen sink.

Chris
(graduate student in fluid mechanics and propulsion)

I'm not interested in wasting time arguing with either of you- do the experiment yourself and report the results. Connect two reservoirs to a common nozzle- it's really quite simple.

 

[cjl]

I'm not interested in wasting time arguing with either of you- do the experiment yourself and report the results. Connect two reservoirs to a common nozzle- it's really quite simple.

It's not nearly as simple as you're making it out to be. Depending on the details of your setup, the flow rate could differ even though the fluid level was the same. Flow in a narrow pipe is dominated by friction effects, and depending on the way in which the two were connected together (and by the lengths of pipe on either side of the junction), you could indeed get a different flow rate.

That doesn't change the fact that you're wrong. As I said, much as you may like to think otherwise, you did not just overturn fluid mechanics with a pair of funnels.

 

[marcusl]

Your experimental results are puzzling, but as cjl points out we have insufficient information here to explain them. Your contention that the pressure doubles is not a correct explanation, however. Pressure is an intrinsic quantity in statistical mechanics. Two identical reservoirs, when connected, will produce a system with double the mass as you say but with no change in their pressure. (Imagine removing a partition that separates a single large reservoir.) Mass is extrinsic, pressure intrinsic.

 

[sams_rhythm]

Andy, I did your experiment as I explained with the bath and sink comparison. Why don't you do that one and explain it with your theory.

If you'd like a reference on this try looking up hydrostatics on Wikipedia or a textbook. Fluid mechanics by Kundu is a masters level text, but a high school textbook will also explain it.

With respect, I and the others do work in this professionally.

I'd still be interested to hear more about the original question.

 

[Drakkith]

Andy, I did your experiment as I explained with the bath and sink comparison. Why don't you do that one and explain it with your theory.

If you'd like a reference on this try looking up hydrostatics on Wikipedia or a textbook. Fluid mechanics by Kundu is a masters level text, but a high school textbook will also explain it.

With respect, I and the others do work in this professionally.

I'd still be interested to hear more about the original question.

Can someone explain how a sink emptying quicker than a bathtub means anything? I thought it was obvious that the sink would empty quicker as it has a much less amount of water in it than the tub. Wouldn't you need to make sure both containers were equal in volume and such and have the same sized drains both leading into the same pipe at the same point? And doesn't plumbing have multiple offshoots to different areas in the house which could potentially alter the experiment?

Andy, with your funnel experiment, was the flow from each funnel restricted because of the size of the funnel orifice, or the size of the common tube or outlet or whatever you used? I could see that if the funnel orifices themselves were restricting the flow then adding the combined flow from both funnels might not reach the max flow for the tube for the given pressure. If that makes sense...

 

[sams_rhythm]

We are testing two hypotheses.

Hypothesis 1: Pressure = mass * gravity / Area
In the sink and the bath, the plug holes are exactly the same area. Gravity is the same. The mass of water in a bath is larger. this would suggest that pressure at the plug hole in the bath is greater than the pressure at the plug hole in the sink. Therefore the flow rate of the bath should be quicker. According to Andy's experiment (which is the same set up just in funnel's which I don't have to hand) the bath would empty with a larger flow rate so that it empties in the same time as a sink.

Hypothesis 2: Pressure = fluid density * gravity * vertical height of water
The pressure at the plug hole is the same at the plug hole in either the sink or the bath. Therefore the flow rate is the same. Therefore, as you say Drakkith, the sink will empty faster.

This shows that hypothesis 2 is correct and 1 doesn't make sense.

I think this is the experiment that Andy is talking about:

"Doubling the mass and keeping the cross-sectional area constant (i.e. the cross-section of the tube just downstream of the t-junction) means the pressure is doubled. "

 

[Drakkith]

We are testing two hypotheses.

Hypothesis 1: Pressure = mass * gravity / Area
In the sink and the bath, the plug holes are exactly the same area. Gravity is the same. The mass of water in a bath is larger. this would suggest that pressure at the plug hole in the bath is greater than the pressure at the plug hole in the sink. Therefore the flow rate of the bath should be quicker. According to Andy's experiment (which is the same set up just in funnel's which I don't have to hand) the bath would empty with a larger flow rate so that it empties in the same time as a sink.

I don't follow you. Andy's experiment involved two equally sized funnels flowing into 1 outlet. Your tub and sink experiment has many differences.
EDIT: I DO know what you are saying, I just don't know if the two experiments are similar enough to compare.

I think this is the experiment that Andy is talking about:

"Doubling the mass and keeping the cross-sectional area constant (i.e. the cross-section of the tube just downstream of the t-junction) means the pressure is doubled.

Can I ask about a possible related example? If we take two hydraulic lines, each with 100 pounds being applied by a piston or plunger or whatever on the end of each line, and then these lines merge into a single line of equal diameter to the original lines, we would only have 100 pounds of pressure at the end to move another piston or something? Is this the same concept?

 

[Andy Resnick]

Marcusl,

I spent my morning doing this for you strictly as a professional courtesy- I hope the gesture is not misplaced. Cjl and sams_rhythm, your attitude is deplorable and you should be very thankful that you are not my students.

Here's the setup:

[PLAIN]http://img864.imageshack.us/img864/6328/dsc7101x.jpg


2 500 ml sepratory funnels connected to a common nozzle, as I described. The procedure is simple- fill the funnels and measure the amount of time needed to drain the funnel. Each funnel is at the same height, each funnel was filled to the same level. Each stopcock has a different bore diameter- I have a working lab, and these were the only pieces of glassware available.

Here's the data: individual drainage times

Funnel # 1: 42 +/- 3 s
Funnel # 2: 101 +/- 12 s

And the times when both were allowed to drain simultaneously:

Funnel # 1: 41 +/- 2 s
Funnel # 2: 107 +/- 3 s

The drainage times were unchanged. Thus the flowrate was changed, thus the driving pressure through the nozzle was changed. It doesn't get much simpler than that.

I cannot emphasize enough, this is why *you* need to do actual experiments, not just sit around and think about experiments. I am truly concerned at the attitude expressed by certain individuals in this thread.

 

[sams_rhythm]

Aha. Now I understand you experiment better.

What's wrong with my experiment? I think it is a good demonstration of P = \rho g h
rather than P = m g / A

There is one thing that stops either of our experiments being analogous to an IV. In an IV there will be the largest pressure loss in the needle, downstream of the place where the tubes join. I think this will control the flow.

I think your experiment could easily be modified to show this, if you would be so kind as to humuor me. Try adding a thin tube (or a stopcock which controls the flow rate would do the same) just downstream of the junction. Then I think you will find that the flow rate from one bag will be much the same as from two.

I suppose your description will be along the right lines if the friction downstream of the junction is 0 and if the friction downstream of the junction is dominant then the extra bag will make no difference.

I don't think P = \rho g h explains all of what is going on here anyway. This is not statics and pressure drops due to turbulence and viscosity are dominating the flow.

I would love to hear from the guy with the IV in the first place to see what actually happens there, but I fear our arguing has bored him/scared him off/ or probably he has something better to do like sticking IVs into people in the first place!

(PS I'm afraid I can't resist saying it. I am very thankful I'm not your student, but I am glad to discuss your experiments. I'd love to hear your results and thanks for doing the previous experiment.)

 

[sams_rhythm]

Drakkith, I think it is related very much to what you say. In your example everything is static so it happens just like you say. Once the fluid starts to flow, then there will be pressure loss due to friction at the wall and the pressure downstream will be less than 100 pounds.

I guess for the IV case, the amount of friction in different parts of the line would matter. I suppose if the friction in the bottom part of the line is dominant then a new bag won't change anything, if the friction before the junction is dominant then the new bag will add extra flow. Do you think?

 

[Andy Resnick]

Aha. Now I understand you experiment better.

What's wrong with my experiment? I think it is a good demonstration of P = \rho g h
rather than P = m g / A

There is one thing that stops either of our experiments being analogous to an IV. In an IV there will be the largest pressure loss in the needle, downstream of the place where the tubes join. I think this will control the flow.

I think your experiment could easily be modified to show this, if you would be so kind as to humuor me. Try adding a thin tube (or a stopcock which controls the flow rate would do the same) just downstream of the junction. Then I think you will find that the flow rate from one bag will be much the same as from two.

I suppose your description will be along the right lines if the friction downstream of the junction is 0 and if the friction downstream of the junction is dominant then the extra bag will make no difference.

I don't think P = \rho g h explains all of what is going on here anyway. This is not statics and pressure drops due to turbulence and viscosity are dominating the flow.

I would love to hear from the guy with the IV in the first place to see what actually happens there, but I fear our arguing has bored him/scared him off/ or probably he has something better to do like sticking IVs into people in the first place!

(PS I'm afraid I can't resist saying it. I am very thankful I'm not your student, but I am glad to discuss your experiments. I'd love to hear your results and thanks for doing the previous experiment.)

Are you kidding me? Who do you think you are, expecting me to do experiments for you? After this garbage you wrote:

Andy, I did your experiment as I explained with the bath and sink comparison. Why don't you do that one and explain it with your theory.

If you'd like a reference on this try looking up hydrostatics on Wikipedia or a textbook. Fluid mechanics by Kundu is a masters level text, but a high school textbook will also explain it.

With respect, I and the others do work in this professionally.

And this:

I have to say I'm with cjl not Andy.

<snip>
By the way, who is Torricelli?

As for the initial question:
Adding a second bag doesn't make any difference, as has been said it is the vertical height between the bag and the needle that sets the pressure force driving the flow rate.

(Flow rate)^2 * (some friction factor for pipe flow) * length of pipe = density* gravitational acceleration * vertical height

You can see this explained better here: http://en.wikipedia.org/wiki/Darcy_friction_factor

<snip>

And this:

Andy, no, I disagree. If I fill a sink with 10 cm depth of water or a bath with 10 cm depth of water, the sink empties quicker. This is essentially the same as your experiment. I'd need to know more about whether friction from the walls of your funnels was playing a role to try and understand your experiments.

Pressure only changes in a static fluid due to the vertical distance you move (i.e. the component of the distance which is in the same direction as gravity). If you are diving, the pressure changes when you go down, not if you swim sideways. Also, the pressure is the same if you swim at the same depth in a pool with a large or small horizontal cross-sectional area. This would not be the case if

pressure = mass *g / Area

I agree that the mass of one beaker is half the mass of two beakers. As marcusl says, it is best to stick with the density * height * g equation to avoid confusion. I think the confusion in the way you have manipulated the equations is that you have assumed that V=h * A, but then you say that you double the mass whilst keeping A the same. This could only happen if you doubled h.

Andy, does that make more sense to you?

*Especially* after I wrote this-

P =F/A = mg/A. Doubling the mass of the fluid doubles the applied pressure. Torricelli showed this quite some time ago, when he observed flow through an orifice.

Why on Earth do you expect me to do anything for you? After all , "[you] and the others do work in this professionally."

 

[sams_rhythm]

Sorry. Didn'y mean to annoy you. I thought you were interested. Apologies for any offence caused.

 

[Andy Resnick]

Sorry. Didn'y mean to annoy you. I thought you were interested. Apologies for any offence caused.

Thanks- apology accepted. So let's get on with it:

<snip>
There is one thing that stops either of our experiments being analogous to an IV. In an IV there will be the largest pressure loss in the needle, downstream of the place where the tubes join. I think this will control the flow.

I think your experiment could easily be modified to show this, if you would be so kind as to humuor me. Try adding a thin tube (or a stopcock which controls the flow rate would do the same) just downstream of the junction. Then I think you will find that the flow rate from one bag will be much the same as from two.

That's not relevant, unless (1) the needle is microscopic in diameter, so that the continuum approximation breaks down, (2) the needle is so short that Poiseuille flow does not develop, or (3) the Reynolds number becomes large enough to indicate turbulent flow. None of these conditions are likely to be met in practice.

I suppose your description will be along the right lines if the friction downstream of the junction is 0 and if the friction downstream of the junction is dominant then the extra bag will make no difference.

I don't understand what you mean by friction- the no-slip boundary condition means there is no frictional loss with the wall, and internal friction is accounted for already by the viscosity. Again, unless the fluid is non-Newtonian (viscosity depends on shear rate), this is already accounted for in Poiseuille flow.

I don't think P = \rho g h explains all of what is going on here anyway. This is not statics and pressure drops due to turbulence and viscosity are dominating the flow.

There should not be turbulent flow in normal conditions- especially in an IV line, because turbulent flow introduces cavitation (or at least entrains air bubbles in the flow) which would be disastrous to your body. The expression *does* explain the dynamics, to some degree- the pressure at the nozzle inlet is given by the weight of the fluid. Change the amount of fluid (using multiple bags, etc) and the weight changes.

I would love to hear from the guy with the IV in the first place to see what actually happens there, but I fear our arguing has bored him/scared him off/ or probably he has something better to do like sticking IVs into people in the first place!

He probably got scared off long ago :)

(PS I'm afraid I can't resist saying it. I am very thankful I'm not your student, but I am glad to discuss your experiments. I'd love to hear your results and thanks for doing the previous experiment.)

My students know I love simple questions that have complicated answers. At a AAPT New Faculty Workshop, I had the quite discomforting experience of taking the 'Physics IQ Test" directly from Prof. Berg (U. Maryland):

http://education.jlab.org/scienceseries/physics_iq.html

I won't say what I got (but if I may brag, I did better than most)- I was so crushed, I spent the next few months in correspondence with Dr. Berg trying to understand how I could possibly teach Physics, since there was ample evidence that I understood nothing. He's a very nice individual- his questions are designed to perplex.

 

[sams_rhythm]

Thanks- apology accepted. So let's get on with it:



That's not relevant, unless (1) the needle is microscopic in diameter, so that the continuum approximation breaks down, (2) the needle is so short that Poiseuille flow does not develop, or (3) the Reynolds number becomes large enough to indicate turbulent flow. None of these conditions are likely to be met in practice.

The narrowest part of the line will lead to the biggest pressure loss. If you look at the expression for pressure drop here http://en.m.wikipedia.org/wiki/Hagen–Poiseuille_equation it scales like r^-4 so width often has a very dominant effect. For the reasons you mention, I suppose (guessing here) this throttling position will be upstream from the needle to allow cavitated bubbles to dissolve.

I don't understand what you mean by friction- the no-slip boundary condition means there is no frictional loss with the wall, and internal friction is accounted for already by the viscosity. Again, unless the fluid is non-Newtonian (viscosity depends on shear rate), this is already accounted for in Poiseuille flow.

When I say friction I mean internal friction from shear, transmitted to the wall via the no slip condition. This is the main thing that causes the pressure loss in the pipe.

There should not be turbulent flow in normal conditions- especially in an IV line, because turbulent flow introduces cavitation (or at least entrains air bubbles in the flow) which would be disastrous to your body. The expression *does* explain the dynamics, to some degree- the pressure at the nozzle inlet is given by the weight of the fluid. Change the amount of fluid (using multiple bags, etc) and the weight changes.

I suppose you are right about the laminar flow. I tend to work on larger scales where Reynolds no is big.

He probably got scared off long ago :)



My students know I love simple questions that have complicated answers. At a AAPT New Faculty Workshop, I had the quite discomforting experience of taking the 'Physics IQ Test" directly from Prof. Berg (U. Maryland):

http://education.jlab.org/scienceseries/physics_iq.html

I won't say what I got (but if I may brag, I did better than most)- I was so crushed, I spent the next few months in correspondence with Dr. Berg trying to understand how I could possibly teach Physics, since there was ample evidence that I understood nothing. He's a very nice individual- his questions are designed to perplex.

The weight of fluid drives the flow which is mg, which gives the pressure \rho g h The pressure loss described in posseoulli (?) which is due to wall shear scales like QL/a^2

If you add up the pressure losses along each line, they should match the initial pressure head. Let's say all tubes have the same area and roughness and dominate the pressure losses over any valves etc.. The tube from beaker 1 to the junction is length L_1 and carries flow Q_1. The tube from beaker 2 to the junction is length L_2 and carries flow Q_2. The tube from the junction on is length L_3.

Taking the path from beaker 1.

\rho g h = Q_1 L_1/a^2 + (Q_1+Q_2)L_3/a^2

Taking the path from beaker 1.

\rho g h = Q_2 L_2/a^2 + (Q_1+Q_2)L_3/a^2

Combining these two equations for two unknowns gives
Q_1+Q_2 = a^2 \rho g h (L_1+L_2) /(L_1 L_2+ L_2 L_3+L_3 L_1)

If L_3 is much smaller than L_! or L_2 then the flow is just the combination of the two other flows

Q_1+Q_2 = a^2 \rho g h /L_1 + a^2 \rho g h /L_2

If L_3 is much larger than L1 or L2, then the flow rate depends just on the hydrostatic drop in head and the length of the downstream hose.

Q_1+Q_2 = a^2 \rho g h /L_3

I think you could probably set up an analogous experiment with electric current with bits in series and parallel, but that is probably by the by.

I'll do the experiment once I can get a suitable bit of hose.

Charlie

 

[Skrambles]

You guys are overlooking conservation of momentum at the nozzles. When the velocity of the flow increases due to a decrease in pipe diameter there is a pressure drop that is proportional to the square of the flow rate. That means you would need four times the pressure to double the flow rate if there was a needle or catheter at the end of the pipe. The simplest way to increase the flow would be a larger diameter needle.

 

[cjl]

Here's the setup:

[PLAIN]http://img864.imageshack.us/img864/6328/dsc7101x.jpg 2 500 ml sepratory funnels connected to a common nozzle, as I described. The procedure is simple- fill the funnels and measure the amount of time needed to drain the funnel. Each funnel is at the same height, each funnel was filled to the same level. Each stopcock has a different bore diameter- I have a working lab, and these were the only pieces of glassware available.

Here's the data: individual drainage times

Funnel # 1: 42 +/- 3 s
Funnel # 2: 101 +/- 12 s

And the times when both were allowed to drain simultaneously:

Funnel # 1: 41 +/- 2 s
Funnel # 2: 107 +/- 3 s

The drainage times were unchanged. Thus the flowrate was changed, thus the driving pressure through the nozzle was changed. It doesn't get much simpler than that.

I cannot emphasize enough, this is why *you* need to do actual experiments, not just sit around and think about experiments. I am truly concerned at the attitude expressed by certain individuals in this thread.

I see the problem here. You are misinterpreting the data, since your apparatus is not testing the phenomenon that you believe it is.

The fact that your two funnels drain in dramatically different times indicates that each is being restricted at a separate location. If they were both restricted by the small pipe past the T-joint, they would both drain in approximately the same time. However, the dramatically different drain times indicate a different restriction. I suspect that your stopcocks are the actual flow restricting elements. Since the stopcocks are providing the majority of the flow restriction, they also contain the majority of the pressure drop in the system. As a result, the delta P across the common pipe section shared by both funnels is effectively zero. When both funnels are draining simultaneously, the flow rate through the common pipe section is larger than it was with either funnel draining individually, resulting in a higher pressure loss across the common pipe section than with either funnel individually. However, since the pressure drop was so small to begin with, even the increased pressure drop is relatively minimal compared to the pressure drop across the stopcocks. Because of this, the pressure drop for each side remains basically independent, and dominated by the stopcocks, resulting in the same drain times when both drain simultaneously as when each drains individually. It certainly doesn't demonstrate what you believe it does, since the pressure by the time the fluid gets to the T connection is basically at the ambient pressure, anyways.

In order to properly test this, your apparatus would need the majority of the pressure drop in the system to occur after the two lines have joined. You could achieve this readily with your current setup simply by restricting the flow after the T until the new element in the system became the major restriction (which would be readily apparent, as both funnels would drain in approximately the same time if this were the case, and the time to drain would be substantially longer than either of the individual drain times right now). In that case, the time to drain would be twice as long with both funnels open compared to either one individually.

 

[edgepflow]

I see the problem here. You are misinterpreting the data, since your apparatus is not testing the phenomenon that you believe it is.

The fact that your two funnels drain in dramatically different times indicates that each is being restricted at a separate location. If they were both restricted by the small pipe past the T-joint, they would both drain in approximately the same time. However, the dramatically different drain times indicate a different restriction. I suspect that your stopcocks are the actual flow restricting elements. Since the stopcocks are providing the majority of the flow restriction, they also contain the majority of the pressure drop in the system. As a result, the delta P across the common pipe section shared by both funnels is effectively zero. When both funnels are draining simultaneously, the flow rate through the common pipe section is larger than it was with either funnel draining individually, resulting in a higher pressure loss across the common pipe section than with either funnel individually. However, since the pressure drop was so small to begin with, even the increased pressure drop is relatively minimal compared to the pressure drop across the stopcocks. Because of this, the pressure drop for each side remains basically independent, and dominated by the stopcocks, resulting in the same drain times when both drain simultaneously as when each drains individually. It certainly doesn't demonstrate what you believe it does, since the pressure by the time the fluid gets to the T connection is basically at the ambient pressure, anyways.

In order to properly test this, your apparatus would need the majority of the pressure drop in the system to occur after the two lines have joined. You could achieve this readily with your current setup simply by restricting the flow after the T until the new element in the system became the major restriction (which would be readily apparent, as both funnels would drain in approximately the same time if this were the case, and the time to drain would be substantially longer than either of the individual drain times right now). In that case, the time to drain would be twice as long with both funnels open compared to either one individually.

I posted early to this thread (think the first) but then backed off and calmly tried to consider the different opinions. I think cjl's last post is the correct explanation as I was looking at the relative restrictions in the test and was thinking the same thing.

Someone earlier, I think marcusl, mentioned the idea of removing partitions in a tank will not increase pressure.

I just don't think you can create a pressure multiplying pump by connecting elevated reservoirs in parallel. Take the static case: A pressure gauge in a small reservoir and many separate branches with a valve connected to separate reservoirs at the same height. If I open the valves, one-by-one, I don't think the gauge will read a higher pressure.

 

[sams_rhythm]

cjl, it depends what the experiment is trying to show. It certainly shows that the flow through an IV can in some situations be increased if you add more bags, as long as the pressure drop from each bag is mainly upstream of the junction. It also shows that the pressure drop downstream of the junction doubles when you add another bag which doubles the flow rate.

It does not imo contradict \rho g h
but I don't think anyone is claiming that. It shows that in some sense mg/A holds, but as we both suggested, this requires a specific scenario and fairly careful interpretation of what the terms and assumptions apply to.

As for adding the bag reducing the flow as was originally seen by the nurse, I think it would also be interesting to reproduce this in the lab. If the flow from the junction to the second bag had lower pressure drop than from the junction out, then you could find that the bag reduced the flow in the needle. I would have thought that there would have been one way valves to prevent this happening though. It also suggests that adding a second bag to that line like the guy explained may not be the safest way to increase the flow rate.

 

[Andy Resnick]

I see the problem here.

You most certainly do not. Your comments demonstrate a complete lack of comprehension and understanding. Talking with you further is a waste of time.

 

[Andy Resnick]

If I open the valves, one-by-one, I don't think the gauge will read a higher pressure.

Why don't you do this experiment and report the results?

 

[cjl]

You most certainly do not. Your comments demonstrate a complete lack of comprehension and understanding. Talking with you further is a waste of time.

If that's the case, then would you please explain to me what the problem with my explanation is.

I should note here that I could say the same thing about you. Your comments demonstrate a complete lack of understanding of basic fluid mechanics, including a severe misinterperetation of a simple experiment, and talking with you seems to be a waste of time. I'm mainly continuing to post so that others are not misled by your uncomprehension.

 

[cjl]

cjl, it depends what the experiment is trying to show. It certainly shows that the flow through an IV can in some situations be increased if you add more bags, as long as the pressure drop from each bag is mainly upstream of the junction. It also shows that the pressure drop downstream of the junction doubles when you add another bag which doubles the flow rate.

Well, I wouldn't say that the pressure drop downstream of the junction doubles, since flow rate and pressure are not always linearly related. However, that is a good point - if the flow restriction were prior to the junction, you could indeed increase the flow rate by adding more bags in parallel. I've always been assuming that the needle was the primary restriction.

 

[edgepflow]

Why don't you do this experiment and report the results?

Mr. Resnick,

I would like to do this test, but I am not sure my wife would agree to buying piping and pressure gauges

However, there is a good discussion related to this in "Mechanical Engineering Reference Manual for the PE Exam" 10th Edition, Lindeburg, Section 15-4 & Figure 15.6. The figure shows four unconnected open columns on top of a common reservoir. It states:

"Pressure is independent of the object's area and size and the weight (mass) of water above the object. Figure 15.6 illustrates the hydrostatic paradox. The pressure at the depth h are the same in all four columns because pressure depends on depth not volume."

Please also consider Figure 1 in this link:

http://scubageek.com/articles/wwwparad.html

And this discussion:

http://wishtrain.com/2009/11/194/

Let me know if you feel this applies.

 

[Andy Resnick]

Mr. Resnick,

<snip>
Let me know if you feel this applies.

It very much applies- from your first URL:

"The pressure at a point in a static liquid is due entirely to the weight of liquid (plus the atmosphere) directly above it"

Which is what I have been saying all along. You are neglecting the role of the stopcock/valve in my experiment. When closed, it's a rigid dividing surface- think about what the effect is as fluid drains from the other reservoir, and how that relates to "the weight of liquid directly above it".

 

[harrylin]

[..] your apparatus is not testing the phenomenon that you believe it is.

The fact that your two funnels drain in dramatically different times indicates that each is being restricted at a separate location. If they were both restricted by the small pipe past the T-joint, they would both drain in approximately the same time. However, the dramatically different drain times indicate a different restriction. I suspect that your stopcocks are the actual flow restricting elements. Since the stopcocks are providing the majority of the flow restriction, they also contain the majority of the pressure drop in the system. As a result, the delta P across the common pipe section shared by both funnels is effectively zero. [..]
In order to properly test this, your apparatus would need the majority of the pressure drop in the system to occur after the two lines have joined. [..] In that case, the time to drain would be twice as long with both funnels open compared to either one individually.

Good analysis.

This discussion deviated from the rather surprising (at least for me) observation of the OP. The vortices/turbulence explanation sounds good to me, but I did not expect the effect of a standard T connector to be so pronounced.

Still, this effect was already discovered by Tesla who patented an extreme embodiment of the same, by means of which we can almost block the main flow from by injected flow from the side, slightly in counter sense:

http://www.google.com/patents?id=Lt1PAAAAEBAJ&printsec=drawing&zoom=4#v=onepage&q&f=false

Note that this effect already works in the laminar flow regime as soon as inertial effects become significant.

Cheers,
Harald

 

[cjl]

It very much applies- from your first URL:

"The pressure at a point in a static liquid is due entirely to the weight of liquid (plus the atmosphere) directly above it"

Which is what I have been saying all along. You are neglecting the role of the stopcock/valve in my experiment. When closed, it's a rigid dividing surface- think about what the effect is as fluid drains from the other reservoir, and how that relates to "the weight of liquid directly above it".

It's relatively straightforward. At the opening of each funnel, the pressure (purely from a hydrostatic analysis of the water in the funnel) is equal to ρgh. Note that this is not identical to the weight of the water divided by the cross sectional area of the throat (which would be much greater than ρgh). This is because at every point where the liquid is contacting the walls of the funnel, the walls of the funnel are pushing back on the liquid (an equal and opposite reaction, you could say). However, since this is a hydrostatic analysis, there is no shear between the liquid and wall. As a result, the force must be perpendicular to the wall. In any section of a container in which the diameter is decreasing as you go downwards, the normal to the wall must have an upwards component.

Since the force between the liquid and the wall must be normal to the wall (assuming a hydrostatic problem, again), the force the wall exerts on the liquid has an upwards component. To determine the pressure at the exit, you have to integrate the force all the way around the control volume, then sum the forces and set them equal to zero.

Now, for some surface inclined to the vertical at some angle, the component of the overall pressure force acting upwards is equal to the overall pressure force multiplied by the sin of that angle (since we are defining the angle as relative to the vertical).

To initially simplify the analysis further, I'll first look at the case for which there is a constrained column of water in a container with vertical walls, and a bottom surface inclined to the vertical with some angle (basically, a beveled cylinder or prism). I'll also assume the depth of the water is sufficient such that the difference in depth from one side of the bottom surface to the other is insignificant (basically, consider an infinitesimal sloped element as the base). The weight of the liquid is mg, which is equal to ρghA, in which A is the cross sectional area. Since the bottom section of the tube is inclined however, the area of the inclined region is equal to A/sin(theta), in which theta is the inclination angle to the vertical. The force on this region is equal to P*A/sin(theta). Based on what I've already described above, the vertical component of this is equal to sin(theta)* P*A/sin(theta), which simplifies to P*A. Thus, the vertical component of the pressure force on the inclined surface perfectly balances the weight of the column of water above it by simple fluid statics.

When we consider that the vertical walls in this container could be replaced by a fluid surface and nothing would change about the situation, it can be seen that this situation can be generalized to any situation in which there is a container with fluid and inclined walls. One such case is a funnel.

In a funnel, much of the fluid is directly above an inclined surface. For the same reason as above, any fluid above such a surface has its weight supported by the pressure force on the region of inclined surface directly below it. Thus, the only fluid for which the weight is not fully supported is the fluid directly over the funnel outlet. This column of fluid must necessarily have the same cross sectional area as the funnel outlet. Thus, the pressure at the funnel outlet is equal to the weight of this fluid (ρghA) divided by the area of the outlet (A), giving P = ρgh.

Now, if I take two funnels, each filled to the same level with the same fluid, and attach the outlets of each together with a valve separating them, this clearly shows that the pressure on each side of that valve will be identical, and equal to the density of the fluid multiplied by the gravitational acceleration of the environment in which they reside, multiplied by the height of the free surface of the fluid above the height of the valve. Once again, any time when there is a surface in a static liquid, it can be replaced with a liquid surface at the same pressure and nothing will change about the problem. Therefore, if the valve is removed, it is effectively replacing a solid surface with a liquid surface at the same pressure in this situation, and nothing changes. With two funnels, the pressure is unchanged compared to a single funnel.

Note that all of the analysis to this point assumes a purely static liquid. Thus, there are no inertial effects, nor are there any viscous effects. That is clearly not the case in several of the experiments before. However, for small flow rates and relatively unrestricted piping, it is a reasonable approximation. For example, in your funnels which drain in timescales of minutes, there will not be substantial inertial or viscous effects within the funnel itself.

Now that I've shown from rather basic principles that adding a second funnel does not in fact double the static pressure at the outlet, why does your experiment have a substantially changed flowrate between the various cases?

The answer is in the assumptions. In any real system, especially with flow through a relatively small orifice, there will be substantial viscous losses. These cause a pressure loss through each component, and the magnitude of the loss depends on the specifics of the flow through each element.

In many cases, the loss will primarily happen at one (or a small number of) restricting element. If this is the case, the pressure loss across the restricting element will be very large, while the pressure loss across the other parts of the flow will be minimal. This is what is happening with your setup and with the IV. However, the restrictive component in your setup is prior to the junction (likely the stopcock). As a result, the pressure at the junction is far below the pressure at the bottom of the funnel, and likely very close to ambient (rather than the substantially above ambient pressure that exists at the bottom of the funnel). This means that when both funnels are open, the primary flow restriction (the stopcocks) does not have any more flow through it than it did when the funnels were open separately, since the flow through either individual stopcock is never fed by both funnels.

To illustrate why this is not showing what you believe it is, I'll use a simple numerical example. I'll also assume pressure drop across an element is linearly proportional to flowrate. It isn't, but it will work for the purpose of this example. I'll also assume two identical funnels, rather than the dissimilar ones like you are using.

Suppose that the pressure drop across the stopcock at the bottom of each funnel is 90% of the total pressure loss of the system, and the pressure loss below the junction is 10%. Individually, the funnels drain in 100 seconds. When both funnels are draining simultaneously, the total pressure drop must remain constant, since the outlet pressure and the feed pressure have not changed. However, the flowrate through the area below the junction is now double the flowrate through the stopcocks. Thus, relative to the pressure drop through the stopcocks, the pressure drop through the junction must have doubled. Since it was originally 1/9 the pressure drop through the stopcocks, the new pressure drop through the junction must be 2/9 the pressure drop through either of the stopcocks. Since the total must add up to 100%, this means that with both funnels feeding it, the new distribution of pressure is 18.2 percent through the junction, and 82.8 percent through the stopcocks. This means the flow rate through each stopcock is (82.8/90) of what it was before.

This means that although the feed pressure was unchanged, the flowrate through the junction is up to 1.82 times the flowrate with either funnel individually opened. Similarly, each funnel will now drain in 108.7 seconds, which is only slightly longer than it took individually.

If you bias the numbers even more (less pressure drop through the junction, more through the stopcock), you can make the drain time with both funnels feeding arbitrarily close to the single funnel drain time, even though the pressure drop never changes. However, if you increase the flow restriction after the junction, the drain time substantially rises, and if the majority of the pressure drop occurs after the junction, then the flow rate will be largely unaffected by the number of funnels feeding (and thus, the drain time for two funnels will be basically twice the drain time for one). This clearly shows the flaw in your argument - you assume that the increased total flow rate indicates an increased feed pressure, when in reality, it only indicates where the flow restriction in your setup resides.


(Sorry for the rather extensive essay, but I am trying to explain this as clearly as I can).

I'll try to scrounge up materials to demonstrate this experimentally later tonight. I don't have access to nice glassware like you do, but I should be able to find enough material to make this work.

 

[mixinman7]

Hello, I am new to these forums.

Friction limits velocity. The apparatus has a depreciating return in terminal velocity at a certain pressure input value as a result of the friction (think of a sonic boom). If the fluids are moving too fast, there will be excess turbulence and heat. The input vector of the IV will also become significant when the resulting velocity is high. Based on the original experimental result, I would expect that the IV to be small and the fluid output to be fast. The friction problem could be tested by putting the IV is a denser medium (water) during the test. The flow rate should increase when the vibrations are dampened.

In summary, the increase in pressure is being radiated in energy forms that dampen and resist flow. I could be mistaken.

 

[Andy Resnick]

Hello, I am new to these forums.

Friction limits velocity. The apparatus has a depreciating return in terminal velocity at a certain pressure input value as a result of the friction (think of a sonic boom). If the fluids are moving too fast, there will be excess turbulence and heat. The input vector of the IV will also become significant when the resulting velocity is high. Based on the original experimental result, I would expect that the IV to be small and the fluid output to be fast. The friction problem could be tested by putting the IV is a denser medium (water) during the test. The flow rate should increase when the vibrations are dampened.

My experiment was carried out with a Reynolds number of about 4. How can you possibly justify invoking "excess turbulence and heat"?

I could be mistaken.

Quite.

 

[boneh3ad]

I have to say I agree with cjl here. The key here in the experiment is the use of a stopcock. Stopcocks are designed specifically to restrict flow, even at their fully open position. They are not just simple valves. The difference is that a stopcock is made to have a smaller diameter than the tubing it connects to, so the maximum flow rate that can be sustained by a stopcock is less than that of the larger diameter tube if it was connected without a stopcock.

Your total mass flow coming through those stopcocks will always be less than the total mass flow through just the tubes for a given pressure or fluid arrangement. It is as cjl has said, the stopcocks are governing your flow, not your tubing. The experiment is not actually testing what it was supposed to have tested.

Look at it in terms of a control volume. If you have one bag (or bucket or funnel or whatever reservoir you like) connected to one line, the control volume is the whole system, from the top of the bag to the orifice through which it exits the IV. While it is true that the Bernoulli equation can't handle viscosity, it can be applied to a control volume to get the average velocity - and therefore mass flow rate - through the control volume. In that frame of mind, the pressure at the exit is \rho g h and the average velocity is dependent on that pressure. The mass flow is dependent on that area, A, and the velocity.

Now hook up two bags that T together and come out of the same sized final orifice with all bags and the exit at the same respective heights. The pressure is still \rho g h and the area is still A, so that mass flow rate is still the same.

Were it not for the stopcocks restricting the flow, the exit mass flow rate would be the same in each test for the aforementioned experiment assuming there were no mixing effects at the T-junction. In the case of the OP's question, the mixing effects are certainly a factor and would introduce additional pressure losses, decreasing the total mass flow.

 

[mixinman7]

Hello Andy Resnick.

You have my apologies. I was not referring to your lab test. I was referring only to the test in the original post. IVs are very flexible small diameter tubes with considerable length. I guarantee that turbulence will be a significant factor in a test set up with small diameter tubes with appropriate lengths. With your test, the tubes were of short length, and therefore stiff with a much higher resonant frequency. An interference oscillation will be simple to create in a long flexible, narrow tube, and difficult to create in a short stiff tube of greater diameter. The lab test using only readily available materials is not adequate to test for the interference I suspect produced the flow results in the original post.

After boneh3ad response, I realized that the larger turbulent influence would be at the terminal end, not within the tubing. I would ask whether the original poster had a needle on the end of the IV, and whether it was hanging freely over the edge of the receiving container. Did the needle wobble back and forth while hanging over the bucket edge, and was it a notably more severe wobble with two IVs serving the one needle?

Last, in reference to the original premise, I would suggest that two IVs hooked up to an actual patient might drain faster than two separately. The experiment in the original post might have neglected critical factors such as what the terminal end of the IV was leading to, or the relative heights of the bags or the terminal end. Alternatively, the drain might have been much slower since the capacity resistance of the veins would be factors. I can't predict what will happen. I am only an undergraduate in radiologic science, and am discussing concepts, not math equations (Radiologic technology student).

"How can you possibly justify invoking "excess turbulence and heat"?" by Andy Resnick

Heat would be a negligible factor, but turbulence is notable when considering the length, flexibility, diameter, and terminal end of the IV. I have been reading through much on the forums, and found this topic very interesting. I am intending only to add a concept to the discussion that seems to be absent.

Cheers

 

[boneh3ad]

Turbulence is nearly a non-factor. The Reynolds number is going to be incredibly small given the tiny diameter involved and the relatively low speeds.

Let's say, for grins, the tube ID is 0.5 mm. That would come out to roughly 4 m/s before the flow goes turbulent. That isn't likely to happen considering IV bags are usually only a couple feet higher than the entry point. Additionally, assuming the IV had a drip, the actual pressure is dictated by the fluid height in that drip, not the bag, and that is considerably smaller.

 

[Andy Resnick]

Sigh. Maybe this thread is hopelessly frayed, but I will try one more time to present a correct analysis.

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.

The first conceptual error occurred when posters tried to treat this as a hydrostatic problem. Certainly, a limit can be taken whereby this problem does become static (specifically, use of a IV bag of infinite volume), but this is not useful. This is a problem of *flow*, not static pressures.

The next conceptual error occurred when posters attempted to quantify the role of where the flow is "restricted"- the needle, the valve, the tubing, etc. Disposing of this error requires a more precise statement of the problem, but for now, let's assume the needle has the smallest diameter of any of the other parts of the system. Thus, we can focus only on the role of the needle on regulating the flow of fluid from the IV bag to the patient.

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:

http://en.wikipedia.org/wiki/Hagen–Poiseuille_equation

Note- I am using the flow rate Q as the dependent variable, not the pressure difference.

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.

But let's say that not only are the physical properties of the fluid fixed (the density and viscosity), but the needle diameter is also fixed- we must use a hypodermic needle of a certain size. What can we control? The pressure difference between the needle inlet and needle outlet.

We can't control the pressure at the outlet- it's inside the patient- but we can assume it is a constant. We can also, for the sake of this analysis, set the constant outlet pressure to atmospheric pressure: the IV now drains into open air rather than a vein or subcutaneous tissue. So all we can control is the pressure at the entrance to the needle. What is this pressure?

Let's consider a simple system- two vertical tubes, open to the atmosphere at the top, joined to a single horizontal tube at the base, in the center of which is a t-junction, and the outlet of the t-junction leads to the needle. All tubes and connectors are of identical diameter (except for the needle), and let's also add a thin guillotine-like valve in one of the tubes. This valve is made of unobtanium: infinitely thin, infinitely stiff, and when it is withdrawn no fluid leaks out of the side of the tube. Again, because the needle is the smallest diameter tube, we only need specify the pressure at the needle inlet to fully determine the flow rate.

With the valve open and needle outlet plugged, we fill the two tubes. We should all be able to specify the pressure everywhere. Closing the valve at this point changes nothing- that's clear. But I'll close the valve all the same. Hopefully, you will have noticed I did not specify *where* the valve is. I did not specify the location because it doesn't matter (within reason- putting the valve at the very top would be silly)

Now I unplug the needle and allow fluid to drain freely. What happens? The flow rate through the needle is proportional to the amount of fluid remaining in the open tube- specifically, the *weight* of the fluid remaining in the open tube. 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.

Now let's consider what happens if instead of closing the valve, I leave it open. Initially, there is no difference. But, as fluid starts draining, there *is* a difference- the valve is no longer holding up a mass of fluid, which is now free to press upon the needle. Thus, the weight of fluid pressing on the needle inlet is *greater* than before. In case this seems counter-intuitive, think about what would happen if you open the valve *during* the time fluid is draining.

This is what my experiment showed- that adding a second IV bag increases the weight of fluid pressing in the needle inlet, increasing the pressure, and increasing the flow rate. The roles of turbulence, friction (other than viscosity), location of 'flow restriction' etc. do not matter- the only thing that controls the flow through the needle is the value of the pressure at the needle inlet, and this is indeed given by \rho gh- the difficulty is apparently in understanding that \rho gh is proportional to the weight of the fluid that presses on the needle.

 

[cjl]

Andy,

I've given you the correct explanation, both for the OPs situation and for the results of your experiment about a dozen times now in probably 4 or 5 different ways. The fact that your explanation makes intuitive sense to you doesn't change its factual errors.