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cjl said: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.
cjl said:I see the problem here.
edgepflow said:If I open the valves, one-by-one, I don't think the gauge will read a higher pressure.
Andy Resnick said:You most certainly do not. Your comments demonstrate a complete lack of comprehension and understanding. Talking with you further is a waste of time.
sams_rhythm said: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.
Mr. Resnick,Andy Resnick said:Why don't you do this experiment and report the results?
edgepflow said:Mr. Resnick,
<snip>
Let me know if you feel this applies.
cjl said:[..] 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.
Andy Resnick said: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".
mixinman7 said: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.
mixinman7 said:I could be mistaken.
edgepflow said: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.
Andy_Resnick said: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.
Andy_Resnick said: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:
Andy_Resnick said: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.
Andy_Resnick said:So all we can control is the pressure at the entrance to the needle. What is this pressure?
Andy_Resnick said: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.
I think the reason the drain time of both funnels together is a little less than the sum of the individual times is the extra "inventory" maintaining a slightly higher average elevation "gravity head" during the test. Of course, the addition of the second funnel does not at any time increase the maximum gravity head.cjl said: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).
Good treatment, that is what I would expect. I mentioned several posts ago that adding reservoirs in parallel would not have a pressure multiplying effect, although you will see a small increase in the average flow as cjl measured.boneh3ad said:[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.
edgepflow said:I think the reason the drain time of both funnels together is a little less than the sum of the individual times is the extra "inventory" maintaining a slightly higher average elevation "gravity head" during the test. Of course, the addition of the second funnel does not at any time increase the maximum gravity head.
edgepflow said:If there was a pressure multiplying effect with parallel reservoirs, we could create some pretty cool pressure and flow control loops. Imagine a ring array of reservoirs and actuated valves: if you need more or less flow, just open or close valves. If this were true, there would probably be products like this on the market.
Indeed, and that was already described in the OP.boneh3ad said:Because it is wrong. Adding two bags does not double the flow rate.
That is also wrong except if the Poisseuille friction term is very small compared to the Bernoulli terms.Poiseuille flow, while the correct flow for this situation, is not needed to solve this problem. You don't need to know the velocity profile at a given point in the tubing or needle to know the flow rate, adn that is what Poiseuille flow gives you.
[...] Since it is a control volume, you can still use the Bernoulli equation to get the average exit velocity, despite the fact that there are viscous effects inside). When I say average velocity, I mean the velocity that produces the same mass flow rate out of a given area as integrating the true velocity profile. I promise you, Bernoulli's equation absolutely applies to get that and is used all the time.Right, you only need to take in account that some of the energy goes into friction.
The friction term can be neglected in the case of fast flow through a large opening: Poisseuille flow scales with v while the kinetic energy term scales with v2.boneh3ad said:Except in the case of very fast flow or very small tubes, the friction losses would be very small. if we approach this from the eyes of a mechanical engineer, we could easily introduce a head loss term to fairly accurately account for the effect of viscous losses. You could theoretically do the same for the loss associated with the T-joint.
The important thing is that neither of those additional loss terms will depend on the number of reservoirs or amount of fluid in each reservoir.
harrylin said:Yes of course (that is well understood), except for the issue of the OP which, as I and others suggested, relates to the T connector; and I gave a reference to a practical application of that phenomenon by Tesla.
boneh3ad said:Poiseuille flow, while the correct flow for this situation, is not needed to solve this problem. You don't need to know the velocity profile at a given point in the tubing or needle to know the flow rate, adn that is what Poiseuille flow gives you.
boneh3ad said:If you hold velocity constant (and density is obviously constant), doubling [itex]D[/itex] quadruples the mass flow rate. After all, the formula for area only depends on [itex]D^2[/itex], so doubling [itex]D[/itex] gives you a factor of 4, not 16.
boneh3ad said:This is, in fact, one of the pressures that can be used to answer this problem. It is dependent on the velocity of the fluid at that point plus the static pressure.
boneh3ad said:I promise you, Bernoulli's equation absolutely applies to get that and is used all the time.
boneh3ad said: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).
boneh3ad said:The combined mass flow allowed by the stopcocks was simply lower than the maximum allowed by the tubes,
boneh3ad said:You can show that the force on a differential element is:
[tex]dp = -\rho g \; dy[/tex]
If you integrate this from the fluid element of interest (the exit plane in our case) to the very top surface of the fluid in the reservoir (meaning the heighest point of fluid in the control volume), you get:
[tex]\Delta p = -\rho g \Delta h[/tex]
boneh3ad said: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
Andy Resnick said:I agree Poisueille flow may not be the only way to solve the problem. I also agree that one does need velocity to calculate a flow rate: for example, [itex] Q = \frac{\pi \mu R}{2 \rho}Re[/itex], where Re is the Reynolds number.
Andy Resnick said:I don't see this. The elementary result, [itex] Q = \frac{\pi R^{4}}{8 \mu}\nabla P[/itex] clearly has a 4-th power dependence on R.
Andy Resnick said:I don't follow this either- pressure, which is part of the overall stress tensor, depends on the velocity *gradient*, not the velocity. Clearly, uniform motion of a fluid should not alter the internal pressure of the fluid (in the region where v is constant). The results of measurements in any inertial frame are the same.
Andy Resnick said:Let me make sure I understand- we are talking about the pressure at a specific point- the needle inlet? Because I can change the column height of fluid by adding or removing fluid to/from the bag- that should not be in dispute, either.
Andy Resnick said:I don't understand what you mean by "maximum (flow) allowed by the tubes". Is there some upper limit? What sets this limit? The flow rate through a tube doesn't admit an absolute maximum- if I double the pressure gradient, I double the flow rate. There is some limit when turbulent flow is reached (say, Re = 4000 and higher), but that's not what we are talking about, AFAIK.
Andy Resnick said:You lost me- Why is the velocity zero? Fluid is draining from the reservoir- the top surface is moving down.
boneh3ad said:I may have used some odd terminology, as there certainly isn't really a maximum flow rate for an incompressible, inviscid fluid.
Andy Resnick said:Aha- this is one major problem. I tried to work the problem using Bernoulli's law, and now I am convinced that Bernoulli's law *cannot* be used to correctly analyze this problem, because Bernoulli's law only holds for inviscid flow- Poiseuille flow fails for inviscid flow. I also understand why the introductory textbook picture of Bernoulli's law is presented the way it is, with gravity perpendicular to the flow.
...
The bottom line, Bernoulli's law is not appropriate to model viscous flow. All discussions about "maximum flow capacity", the roles of "flow restrictors" and the like only arise when trying to force inviscid flow equations to handle viscous effects.
boneh3ad said:Aha, but this isn't true. Bernoulli's principle can be used on a viscous flow just fine.
Andy Resnick said:I'm going to ask for a reference on this one- everything I found was considerably more equivocal:
boneh3ad said:Try any basic fluid mechanics textbook geared towards mechanical engineers.