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!
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.
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.
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.
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.
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.
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.
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.
I have to say I'm with cjl not Andy. I don't know if this will work, but here goes. Code (Text): 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?
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 [tex]p=\rho gh,[/tex] 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 [itex]\rho gh[/itex] does not refer to mass? What do you think [itex]\rho[/itex] 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, 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?
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.
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)
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.
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.