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Let us suppose that a fluid is flowing through a pipe (fully filled) of uniform cross section. The fluid is ideal and hence must flow in a streamlined path and must be in a steady state. This means that the path of particles of fluid must never intersect (streamlined flow) and hence the velocity of all particles must be parallel to the walls of the pipe and equal at point (steady flow).

Note: The figure below is a horizontal cross section of the pipe

Since the path is streamlined, the velocity of particles are parallel to each other and to the wall, therefore the particle A will also have velocity parallel to the walls and hence no component of velocity of A is towards the wall.

  1. So how will A exert any pressure on the wall since there is no component of velocity in the direction of wall (it will not strike the wall and hence not exert pressure on it.)

Moreover since the velocities of particle A and particle B are parallel (both particles are in the same horizontal plane), they will not exert pressure on each other? Is it wrong.

What am I getting wrong about streamlined flow?

  1. If the pressure is due to the vibration of a particle which will result in collision with the wall, then my further question is that according to Bernoulli equation, the pressure is different at different velocity but since the the pressure is caused by the vibration of fluid particle and not due to its velocity perpendicular component (which is a contradiction to my reasoning above that perpendicular component of velocity will be zero) then why will the pressure change at all when it is flowing with different velocity (due to increase/decrease in cross sectional area)?

Edit: I got this question because I was watching a video on Bernoulli Equation on an molecular scale.(https://youtu.be/TcMgkU3pFBY) Here, they explain how there is low pressure in smaller cross-sectional (and higher velocity) area due to less perpendicular velocity and hence collisions with the wall are less. But in the case of an ideal fluid the flow should be streamlined hence there should not be any perpendicular velocity(?), which leaves the explanation in the video incomplete for ideal fluids. Am I messing up pressure on the wall with the pressure inside the fluid. If not and considering there will be no perpendicular component , then how will we explain the pressure change on walls due to Bernoulli Principle.

enter image description here

Because if the inter-molecular repulsion at X is some quantity, then the inter-molecular repulsion at Y must be less that quantity(as pressure is less due to Bernoulli Principle at Y), which seems contradictory to me.

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  • $\begingroup$ You may find an answer here physics.stackexchange.com/q/519965/275176 $\endgroup$
    – Linkin
    Dec 5, 2020 at 4:49
  • $\begingroup$ you cannot constraint the particles in one single direction. That will violate the second law of thermodynamics. $\endgroup$
    – Anonymous
    Dec 5, 2020 at 4:51
  • $\begingroup$ They will have different velocities but you are watching the horizontal component effectively(somewhat like superposition principle i suppose).. This is because all the particles are identical so if a particle is colliding you cannot observe it macroscopic level $\endgroup$
    – Anonymous
    Dec 5, 2020 at 4:54
  • $\begingroup$ @JustJohan , yes I have been through it but it doesn't answer my queries to my satisfaction $\endgroup$
    – Satwik
    Dec 5, 2020 at 5:07

6 Answers 6

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When we talk about the velocity of fluid particles (or parcels), we are not referring to individual molecules. The individual molecules have velocities in all directions, and thus exert pressure on the wall. For fluid particles, we are talking about the organized velocity of the molecules, or more precisely their vectorial average which, in flow, has a bias in the direction of flow. The velocities of the molecules are the random motion superimposed on this mean.

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  • $\begingroup$ I'm a bit confused now, an answer states that there will be no force but yours and including 2 others state that the molecules have velocity in random direction but at an macroscopic level tends to flow in a certain direction. With no intension of disrespecting, are you sure about your answer so that I may be confirmed. $\endgroup$
    – Satwik
    Dec 5, 2020 at 15:40
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    $\begingroup$ I confirm what I said. It is silly to say there will be no pressure on the walls of the tube. $\endgroup$
    – Chet Miller
    Dec 5, 2020 at 15:55
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    $\begingroup$ @Satwik This is the correct answer. Nothing else needs to be said about it. $\endgroup$
    – D. Halsey
    Dec 5, 2020 at 23:21
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    $\begingroup$ I have a question ,how about we take the most ideal case possible ,like the op is asking .so each and every particle moves in a straight line ? Is such an ideal case possible .I'm no expert ,thank you . $\endgroup$
    – Glowingbluejuicebox
    Dec 6, 2020 at 0:37
  • $\begingroup$ @ChetMiller , I think I am satisfied by your answer because if there is velocity of a single particle in all direction then there is some perpendicular component also, hence the question in my edit part of the question need not arise. Your explanation also answers the edit part, doesn't it? $\endgroup$
    – Satwik
    Dec 6, 2020 at 2:50
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We can imagine a liquid that for a remarkable coincidence all vibrations are in the same direction of the flow, or a gas that all molecules have the same velocity of the flow. In this case, the pressure on the walls would be zero. What means that pressure (as temperature) are macroscopic concepts, that relies in statistical mechanics, and the probabilities of events have to be considered.

Events like that doesn't happen because their probability is vanishing small.

About the second question, it is better to think in a greater pressure in the pipe with large diameter. Suppose we are in the frame with the same velocity of the fluid in the larger pipe. For us, the pipe is moving, and the region of diameter reduction is coming to us. The effect is similar of a piston that compresses a fluid, increasing the pressure.

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Pressure on the wall is due to two reasons :

  1. Almost contact forces
  2. Hits from each molecule

Now if we assume the flow to be perfectly streamline, the hits are almost zero and thus the walls experience pressure due to almost contact forces (shown diagrammatically in Just Johan's answer)

So, even in a streamlined flow , pressure is not zero on the walls.

You asked about pressure at A and B on the basis of Bernoulli's Principle in chat, so users who are interested in this may see this chat.

Note : I linked the discussion (in chat) and gave a short answer because I can't paste all those discussions in my answer. The image is also taken from the chat between me and @Satwik.

Hope it helps 🙂.

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  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Ankit
    Dec 5, 2020 at 11:08
  • $\begingroup$ Glad to know that someone downvoted :) but please make me inform of my mistake so that I may get to know something 🙂 $\endgroup$
    – Ankit
    Dec 5, 2020 at 14:59
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Let's take a case where the fluid is not accelerating and thus there is no pressure difference across the ends of the tubes and also the net force on each particle of water is zero (will be needing this).

enter image description here

The forces acting on any particle near the boundary are the intermolecular repulsion forces between it and the particles in the middle and other boundary particles.

enter image description here

As the fluid is not accelerating we need a balance of forces for constant velocity, this force is provided by the tube thus we get pressure.

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  • $\begingroup$ Yes, thanks but it does not answer my 2nd part and the edited part to the question, does it? $\endgroup$
    – Satwik
    Dec 6, 2020 at 2:47
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No, your concept about streamline flow is OK. But, you missed a bit confusing concept about the pressure microscopically. As the definition of Pressure is

The magnitude of the normal force per unit surface area is called Pressure. And Pressure is a scaler quantity.

Microscopically, the pressure exerted by a fluid on a surface in contact with it is caused by the collision of molecules of the fluid with the surface. As a result of collision, the component of a molecule's momentum perpendicular to the surface is reversed. The surface must exert an impulsive force on the molecule, and by Newton's Third law the molecules exert an equal force perpendicular to the surface. The net result of the reaction force exerted by many molecules on the surface give rise to the pressure on the surface.

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    $\begingroup$ thank you for your answer. My question is that if the velocity of particle is parallel to the wall ,why will it collide at all? If you are talking about the vibration of a particle which will result in collision with the wall, then my further question is that according to Bernoulli equation, the pressure is different at different velocity but since the the pressure is caused by the vibration of fluid particle(and not due to its velocity) then why will the pressure change at all? I will add this to my question $\endgroup$
    – Satwik
    Dec 5, 2020 at 4:57
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You cannot constraint the particles of water to flow in a single direction that will violate the laws of thermodynamics(second law)

They will have different velocities but you are watching the horizontal component effectively(somewhat like superposition principle i suppose). This is because all the particles are identical so if a particle is colliding with walls you cannot observe it at macroscopic level. So you just see their flow and that is enough for you to explain other phenomenon.

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  • $\begingroup$ This might be the explanation but then, doesn't it change the definition of streamline flow $\endgroup$
    – Satwik
    Dec 5, 2020 at 5:05
  • $\begingroup$ @Satwik sorry i don't know much but isn't the streamlined flow a macroscopic definition of movement of layers of fluid. $\endgroup$
    – Anonymous
    Dec 5, 2020 at 6:14
  • $\begingroup$ so according to me streamlined flow is also a superposition of microscopic observations to bring it into macroscopic world :) $\endgroup$
    – Anonymous
    Dec 5, 2020 at 6:17
  • $\begingroup$ @Anonymous Could you please add in to your answer how does the second law get violated? $\endgroup$
    – Mitchell
    Dec 5, 2020 at 13:03
  • $\begingroup$ @Mitchell Sure. $\endgroup$
    – Anonymous
    Dec 5, 2020 at 13:29

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