Share this thread: 
#37
May2811, 12:04 PM

Sci Advisor
P: 5,517




#38
May2811, 10:44 PM

P: 1,017

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. 


#39
May2811, 10:47 PM

P: 1,017




#40
May3111, 02:31 PM

P: 688

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 154 & 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. 


#41
May3111, 03:05 PM

Sci Advisor
P: 5,517

"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". 


#42
May3111, 03:33 PM

P: 3,187

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=Lt1...page&q&f=false Note that this effect already works in the laminar flow regime as soon as inertial effects become significant. Cheers, Harald 


#43
May3111, 06:02 PM

P: 1,017

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. 


#44
Jun511, 10:49 PM

P: 26

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. 


#45
Jun611, 12:06 PM

Sci Advisor
P: 5,517




#46
Jun611, 02:29 PM

PF Gold
P: 1,500

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 [itex]\rho g h[/itex] 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 [itex]\rho g h[/itex] 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 Tjunction. 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. 


#47
Jun611, 04:03 PM

P: 26

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 cant 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 


#48
Jun611, 04:22 PM

PF Gold
P: 1,500

Turbulence is nearly a nonfactor. 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. 


#49
Jun611, 06:47 PM

Sci Advisor
P: 5,517

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%E...uille_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 16fold 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 tjunction, and the outlet of the tjunction leads to the needle. All tubes and connectors are of identical diameter (except for the needle), and let's also add a thin guillotinelike 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 counterintuitive, 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 [itex]\rho gh[/itex] the difficulty is apparently in understanding that [itex]\rho gh[/itex] is proportional to the weight of the fluid that presses on the needle. 


#50
Jun611, 07:08 PM

P: 1,017

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. 


#51
Jun611, 08:30 PM

P: 1,017

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). 


#52
Jun611, 09:23 PM

P: 99

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 


#53
Jun611, 10:10 PM

PF Gold
P: 1,500

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. [tex]dp = \rho g \; dy[/tex] If you integrate this from the fluid element of interest (the exit plane in our case) to the very top surface of the fluid in the reservoir (meaning the heighest point of fluid in the control volume), you get: [tex]\Delta p = \rho g \Delta h[/tex] 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 [itex]\rho[/itex] and [itex]g[/itex] and we can't change either of these. Clearly, the only variable is then [itex]\Delta h[/itex] 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: [tex]p_{\textrm{exit}} + \rho g y_{\textrm{exit}} + \frac{1}{2} \rho v_{\textrm{exit}}^2 = p_{\textrm{res}} + \rho g y_{\textrm{res}}[/tex] However, here, [itex]p_{\textrm{exit}}[/itex] and [itex]p_{\textrm{res}}[/itex] 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: [tex]\rho g y_{\textrm{exit}} + \frac{1}{2} \rho v_{\textrm{exit}}^2 = \rho g y_{\textrm{res}}[/tex] We then move the [itex]y[/itex] 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 [itex]y_{\textrm{exit}}[/itex] is the origin and therefore zero and [itex]y_{\textrm{res}}[/itex] just represents the height of the reservoir bag above the exit plane. This leaves us with: [tex]\frac{1}{2} \rho v_{\textrm{exit}}^2 = \rho g y_{\textrm{res}}[/tex] This can obviously be solved for the exit velocity: [tex]v_{\textrm{exit}} = \sqrt{2 g y_{\textrm{res}}}[/tex] The mass flow rate is then trivially: [tex]\dot{m} = \rho v_{\textrm{exit}} A_{\textrm{exit}}[/tex] [tex]\dot{m} = \frac{1}{4} \rho \pi D_{\textrm{exit}}^2 \sqrt{2 g y_{\textrm{res}}} [/tex] Note that the only parameters that we have control over in this scenario are [itex]D_{\textrm{exit}}[/itex] and [itex]y_{\textrm{res}}[/itex]. 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 [itex]\Delta p[/itex] term that is also not dependent on the number of reservoirs or the total amount of fluid in the reservoirs. 


#54
Jun611, 10:52 PM

P: 688




Register to reply 
Related Discussions  
Question About Fluids  Classical Physics  14  
Fluids question!  Introductory Physics Homework  7  
Question regarding ideal fluids vs. nonideal fluids  General Physics  3  
Fluids Question  Introductory Physics Homework  13  
Fluids question  Introductory Physics Homework  3 