# what really happens at microscopic level in Bernoulli's principle

Discussion in 'Homework Help' started by PG1995, Aug 25, 2012.

1. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Hi

Probably there are errors in my description below but I believe you can see what I'm driving at. In simple words, I want to know what happens at microscopic levels which reduces the pressure in region B of the tube when speed of flow gets doubled. Please help me with it. Thank you very much for your time and help.

Suppose the tube has rectangular cross section like this.

Let's assume that the central region, B, can hold five liters of water and both regions, A and C, can hold 10 liters of water. Suppose that water is entering the tube at the rate of 10 liters per second which means every second 10 liters of water enters A region and at the same time 10 liters leave the C region. This would simply mean that water in the region B has to have double the speed compared to other two regions. Let's say in regions A and C speed of water is 50 m/s and therefore speed in the region B is 100 m/s.

At microscopic level liquid pressure is result of collisions with the walls of tube. The more forceful collisions are, the more pressure will be exerted. I believe the liquid pressure has also to do something with the frequency of collisions. For instance, if the number of collisions of individual molecules with a certain wall of the tube becomes halved, then I believe the pressure would also get halved. Please correct me if I'm wrong.

Although the speed of the water in B region is twice (which was assumed to be 100 m/s), I don't think it would affect the individual velocity of molecules (let's say average velocity of water molecules is 350 m/s). Only the "y" component of molecules' velocity contribute to the pressure.

Suppose that if the tube were of uniform cross section as that of region B and water were flowing at rate of 50 m/s and volume of water flowing per second through region B was 5 liters (please note that region B was assumed to hold only 5 liters of water), then the number of collisions of water molecules every second per wall of region B was 500. But in our case the tube is not of uniform cross section and speed of water flow in region B is 100 m/s which results in half the number of collisions per second, i.e. 250. But twice the speed in region B means that twice the volume of water flows per second, i.e. 10 liters. This twice the volume would keep the number of collisions per second per wall of region B constant, i.e. 500. Right?

If the number of collisions per wall of region B per second does not change when the speed of the flow through region B gets doubled, then I think pressure should not change either; it should remain constant.

Regards
PG

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2. ### steveb Senior Member

Jul 3, 2008
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One way to understand this is to consider conservation of energy, which is discussed in most text books and the wiki article. The basic idea is to consider where the increase in kinetic energy of the fluid is coming from. However, you seem to be looked for a more intuitive explanation from the point of view of molecules and their motion relative to collisions with the container wall.

Perhaps, a simple way to grasp it is to consider that fluids exert pressure by collisions with the container, and those collisions do not necessarily happen perpendicular to the wall since this is random motion we are talking about. Some molecules hit perpendicular to the wall, but most are oblique collisions. A glancing collision exerts less pressure, but the net pressure is the average of all collisions. However, if the fluid is moving with a net faster velocity, then this velocity increases one component of the velocity vector of each molecule. This means that (on average) more molecules are hitting more obliquely, and the static pressure is decreased, as a result.

Perhaps that is not quite the full answer though. It likely also relates to the number of collisions with the container that happen which would be less when the speed is faster.

Last edited: Aug 25, 2012
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3. ### WBahn Moderator

Mar 31, 2012
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4,917
Coming at it from a different angle, consider a small packet of fluid as it moves through the system. If it moves from region of low velocity to higher velocity, that means that that small packet of fluid had to accelerate, right? But in order to accelerate, that means that there needed to be an unbalanced force acting on it; specifically, the pressure behind it had to be greater than the pressure in front of it. In other words, for the net velocity to change there has to be a pressure differential.

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4. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Thank you, Steve, WBahn.

It seems that I have already tried to address the main points of your reply in my first post above. Anyway, I'm not sure if I was right about what conclusions as I pointed out at the very beginning in my earlier post.

1: First of all, as I also said previously, in my humble opinion (which might be completely incorrect) that the glancing collision exert the same pressure as it would if there were no x-component of velocity. Please see this diagram.

2: Yes, when the flow speed is doubled the molecules will hit more obliquely but it won't affect y-component that is responsible for the pressure (please see this diagram again).

3: When the flow speed is doubled the number of molecules colliding with the wall per second won't change. Here I will quote, with minor edits, what I said in my earlier post:
Now you know how I'm visualizing the situation and so I believe you can help me in a better way. Thank you very much.

Best wishes
PG

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5. ### steveb Senior Member

Jul 3, 2008
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469
Perhaps my explanation is not perfectly correct, but certainly the pressure is different, and it seems an argument along the lines I made should reveal the answer. But, I think you are correct that a more detailed answer is needed to be convincing. I'll give it some more thought and see if I can determine which of the two reasons is more significant, and what the underlying assumptions need to be, for each one to make sense.

I understand your point about the y-component of velocity being unchanged based on the simple argument, and I agree with that. What is less clear, is what this does to the average force. Generally, the oblique collision would need to be analyzed considering conservation of energy and conservation of momentum and a key question is whether the collision can be considered elastic. Still, you may be correct, especially if the collisions are elastic. I can't visualize the answer in my head, but a calculation should be easy enough.

I dont' think I agree with your point about the number of collisions being the same, although, I agree that a factor of 2 change in speed does not imply a factor of 2 change in the number of collisions. I think rather, the difference is much smaller than that and would depend on the average thermal speed of the molecules as compared to the overall bulk water velocity.

Anyway, good thinking on your part, and let's give it more thought.

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6. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Thank you very much, Steve.

I'm happy that some of the points I outlined are correct. Yes, it needs more thinking. These days I'm really stuck in many things but on the other hand there is no hurry. Thanks a lot for your help and time.

Best wishes
PG

7. ### steveb Senior Member

Jul 3, 2008
2,433
469
As i think about this more, I feel that one needs to carefully specify the assumptions that are being made about the fluid under consideration. First, is it a liquid or a gas, and second, if liquid, is it incompressible, and if gaseous, is it an ideal gas. There are also assumption about whether there is friction and turbulent flow and various other things along these line.

Mentally, I was visualizing a gaseous fluid and this threw me off, but looking at your example of water, one could make the assumption of an incompressible and frictionless liquid, and see what this implies. If we were to allow nonideal effects, then the number of collisions and the kinetic velocity might come into play, but if the fluid is incompressible, then we need another explanation.

Let's first ask about the water pressure in a simple container, and forget about flow and changing cross sections. We know that the pressure on the surface is atmospheric pressure, and then pressure increases with depth. However, in some sense the water's pressure can be viewed as a simple response to applied forces. Unlike a gas that will fill a container, the ideal liquid simply applies force to oppose whatever force is applied to it. The top surface has the air pushing with atmospheric pressure. The bottom surface has the bottom resisting the weight of the atmosphere and the weight of the water, so the water at the bottom gives pressure that is higher than at the top. The side walls have varying pressure dependent on depth.

Hence, rather than thinking in terms of collisions, think in terms of springs. The molecules are like very stiff springs. If you push on them hard, they push back hard but do not compress very much. So, you may have the same number of thermal collisions happening, but as you apply more pressure to the liquid, those collisions have more force. I think this is the best way to visualize the situation from the point of view of collisions on a side wall. So there are approximately the same number of collisions, but if the pressure goes down when velocity is higher, then the collisions are not as forceful. (that is, the "springs" are not compressed as much, on average)

The explanation would be augmented for gasses and nonideal liquids, as various other effects come into play, but it seems that this effect is the dominant one that answers your question. This is clear because Bernoulli's equation for incompressible fluids predicts very large changes in pressure, when the speeds are largely different. So, unless that pressure could be directly transmitted to affect the average impulsive collision force, this large change would not be possible.

Last edited: Aug 27, 2012
8. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Thank you, Steve, for the explanation given in last reply but I'm afraid it didn't resolve the question that what really happens at the microscopic level which affects the pressure in Bernoulli principle.

When I first read about the principle I was quite confident that in a narrower region reduction in pressure happens due to two reasons: (1) the collisions become less forceful and/or (2) there are less collisions per unit time. But now I'm utterly at loss to understand the exact reasons because my previous reasoning has to totally fallen apart into pieces. Now this is about time when I desperately need your help!

On a side note, I was wondering that why most textbooks keep on recycling the same material using different words and colorful diagrams. I'm not particularly talking about the Bernoulli principle here.

Best wishes
PG

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9. ### steveb Senior Member

Jul 3, 2008
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OK, so my above post is trying to address exactly your confusion. What I'm claiming above is that (with your assumptions) the rate of colisions is the same, but the force of the collisions depends on the pressure being applied to the water.

This is why I mentioned thinking about springs rather than collisions. The nearly incomressable water has molecule interactions that have spring like effects. The very stiff spring like force is what keeps you from compressing the matter.

So, in your example, the lower pressure in the central region with higher speed, will allow the molecules to get just a little farther appart, because it is not truly incompressible. The farther spacing allows the spring force to be reduced (on average) and hence the collisions are less forceful (on average).

It is very hard to visualize because the assumption of incompressibility is not truly valid. So, if we assume density is constant and there is no compression, then the number of collisions would be the same, but the force from each of those collision is smaller when pressure is lower. This is the dominant effect. To say it another way, not all of the energy is due to molecular vibrations/collisions (this thermal energy is kinetic energy, by the way). Some of the energy is potential energy due to compression and the energy balance inherent in Bernoulli's equation should have a potential energy of the compression of the fluid, but we ignore that with the assumption of incompressible fluid. We may approximate the fluid as incompressible, but it is not truely incompressible. Squeezing a fluid (or a solid for that matter), stores potential energy.

I can't answer the question in your side note for the general case. Through the years I've also not always been impressed with many text books. Occationally, you do find a good one, but more often than not, you have to dig for the answers.

Last edited: Aug 31, 2012
10. ### steveb Senior Member

Jul 3, 2008
2,433
469
I thought of another way to try and explain this.

Let's pretend that we could have a liquid with no thermal energy. In other words, imagine cooling a liquid down to absolute zero, so that no molecular vibrations happen. Let's forget the fact that it would freeze to a solid, and just pretend that the liquid properties are retained.

Now, what affect would this have on Bernoulli's equation? I submit that it has no effect at all because Bernoulli's equation is a simple energy balance equation that does not care about the physics at the molecular level. Since energy conservation always holds, the details don't matter.

Now, if we think about this, your attempt to describe pressure as being due only to molecular collisions would fail to make any sense. What is left then? I submit that what is left is the spring like interactions of the molecules. The molecules do not like to be forced together, due to electrostatic repulsion. THis is what makes the liquid nearly incompressible. But, it still can be compressed a little, and this puts a pressure force into the medium.

Now, carry this idea out to the real world where there is thermal motion and molecular collisions. Hopefully, you can see that there can be additional pressure that goes beyond the thermal collisions, and this additional pressure is the spring like action in trying to force the molecules to force together, when they don't like to do that.

If we compare to a gas, the gas gives pressure mostly due to thermal collisions, since the molecules are far apart and not bound by electrostatic forces. This is why gasses are highly compressible. When gas is compressed, then the atoms are closer together and there are more collisions, hence more pressure. The liquid is different though. Since it does not compress (very much), the number of collisions is about the same and it is the electrostatic interactions that give spring effect, which prevents compression and increases pressure.

11. ### mlog Member

Feb 11, 2012
276
36
I know this thread has been here for awhile, but it seems more appropriate for the Physics forum. Just my opinion.

12. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Hi Steve

I have also thought about this several times today. I have a rough explanation which needs little refinement. I have also read your explanations many a time and they have provided me with clues as to what is really happening... let's see. I will put into words soon. Once again, thanks for always being there to help me.

Best wishes
PG

13. ### obituary New Member

Sep 20, 2012
1
1
I have just met this forum and found this question quite interesting. I realized I missed a factor 2 at some point, but I decided to post this anyway as I think that the philosophy of the resolution is correct.

In short, the molecules impact the wall because they move towards it. If velocity is seen as having axial (along the flow displacement) and lateral components, then it is easy to understand that part of the corresponding lateral kinetic energy (with origin in thermal vibrations) is converted to kinetic energy when the flow accelerates due to its incompressibility. As a consequence, less particles hit the wall per unit time (a factor lineal with speed) and with less velocity (another lineal factor, giving the square law of Bernouillis equation).

This is valid for liquids and gases at low speeds (with respect to their Mach number).

See the photo here for the resolution:

http://imageshack.us/a/img20/6518/bernouilli2.jpg

Hope this is what you asked for.

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14. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Welcome to the forums, obituary! And thank you very much for making an attempt at the solution. I don't know if this is exactly what I was after. Let Steve have a look on it.

I had been thinking about it but couldn't come up with plausible reasoning. I have arranged the pages of your solution in vertical order for easy reading. I have to go through it several times in order to understand it (it's just that I'm little dumb!). Once again, thanks.

I think I would just mention a few details about what I had in mind. Pressure is energy per unit volume so if there is a decrease in pressure then there is less energy per unit volume. Suppose before there is any decrease in pressure, the energy per unit volume is 10 J (4 J of K.E. + 6 J of P.E.). As the gas or liquid passes through a narrower region, there is a decrease in pressure and consequently a decrease in energy per unit volume; say the energy per unit decrease from 10 J to 8 J. But while passing though the narrower region, there must be an increase in the kinetic energy therefore the kinetic energy per unit volume should be more than 4 J now, say it's 6 J. If K.E. is now 6 J then the P.E. per unit volume would be 2 J because 6J+2J=8J. This is where I was stuck. I was unable to visualize how P.E. can decrease and how it will affect the pressure. By the way, I also had Joule-Thomson effect in my mind when I was thinking about this problem. I thought perhaps Joule-Thomson effect had some role to play. Moreover, I also had in mind that the molecules also have rotational kinetic energy.

Best wishes
PG

Last edited: Sep 23, 2012
15. ### BobLeichhardt New Member

Dec 1, 2013
1
0
Hi, this thread is getting a bit old but I have struggled over this same question and I think I have got some way to figuring it out.

The basic question is, what happens at a molecular level to explain Benoulli's Principle? In other words, how do you explain (at the molecular collision level) the drop in pressure on the side walls of a conduit when the flow in the conduit speeds up? If the gas or liquid is not compressed or rarified, why should the pressure perpendicular to the flow decrease?

Conceptually, the pressure exerted by the fluid is due to the molecules in the fluid bouncing off the side of the vessel. The force exerted is proportional to that component of the molecule's motion that is perpendicular to the wall of the vessel. Now regardless of the speed that the fluid is moving parallel to the wall of the vessel, this component of the molecule's motion perpendicular to the flow will remain unchanged. So why does the pressure perpendicular to the direction of flow decrease as the fluid speeds up?

I think the answer is explained like this:
Consider one molecule in the fluid. Lets say that its random motion causes it to bounce off the wall of the conduit every millisecond, on average. If the fluid is moving at a particular velocity the molecule will strike the wall of the container repeatedly every millisecond on average, with some average spacing distance from bounce to bounce along the conduit. If we now increase the speed that the fluid is moving down the conduit, without changing the density or temperature of the fluid, the molecule will still strike the wall of the conduit every millisecond, but the average distance between each strike of the wall in the direction of flow will now be greater because the fluid is moving faster past the wall. This will apply to all the molecules in the fluid. So the molecules are still bouncing off the wall of the conduit in the same 'molecule bouncing off the wall' like way that they were before, but the distances between bounces in the direction of fluid flow is, on average, greater than it was when the fluid was moving slower. Because the pressure on the wall of the vessel is proportional to the amount of bounces that occur over a given area of the wall's surface, the increase in the distance between bounces as the speed of the fluid increases will mean that the net pressure on the wall will be reduce.

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17. ### Morwen New Member

Jan 4, 2016
2
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I have puzzled over this for decades.
Thank you all for a useful discussion. It helps to separate out a few different approaches.
Unfortunarely, none yet seem to work for me.

Replying to last entry in 2013, I find the idea of longer between hits useful, but at higher speed would we not just get proportionately more molecules hitting, thus cancelling out the longer distance?

I think that there must he another explanation.
Will try to think from point of view of starting the flow, rather than currently existing steady flow.
Think of how the molecules get squeezed down into narrower tube.