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For a fluid moving through a cylindrical tube in a streamlined motion, a fluid layer in immediate contact with the wall of the tube remains effectively at rest. The layers away from the wall has gradually increasing velocities attaining the maximum at the axis of the cylinder.

The molecules of the fluid layer $L_1$ in immediate contact with the wall are at rest because fluid molecules of this layer collide with those of the wall and make them jiggle. In this process, they transfer energy and momentum to the wall. But instead of coming to rest, the fluid molecules of $L_1$ must gain a backward momentum due to recoil. But that is being cancelled by the collisions with the molecules of the forward moving layer $L_2$ just below it (which might be causing it to be at rest). But why doesn't the layer $L_2$ also come to rest?

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  • $\begingroup$ Because the fluid molecules in layer $L_2$ on average have more axial momentum than those in layer $L_1$. So, on average, the $L_2$ molecules will still have more axial momentum than the $L_1$ molecules after the collisions have taken place. $\endgroup$ – Time4Tea Nov 21 '18 at 13:50
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    $\begingroup$ Because there is a pressure gradient in the tube to drive the flow. If you removed the pressure gradient, the flow would gradually come to a stop (as you correctly described). $\endgroup$ – Chet Miller Nov 21 '18 at 15:43
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But that is being cancelled by the collisions with the molecules of the forward moving layer L2 just below it (which might be causing it to be at rest). But why doesn't the layer L2 also come to rest?

Think of it transactionally. Is the rate that an air molecule can transfer momentum to other air the same as the rate that air can momentum energy to a solid object? No, the solid object acts effectively as an infinite mass, while another air molecule will have similar mass and thus we would expect the scattering to be much more uniform.

Thus while air at L1 is continually radiating momentum into the wall at R1, L2 is radiating momentum into L1 at R2. As long as R2 is smaller than R1, you would expect there to be a difference in speed between the layers.

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Another way to think about it is to imagine the fluid in the pipe as consisting of a very large number of extremely thin cylindrically symmetric sheaths of fluid nestled within one another. The outermost sheath is in direct contact with the pipe walls and the innermost one follows the central axis of the pipe.

Now imagine that these sheaths are capable of slipping past one another in the direction of the pipe's length, but there is a little friction at the interface between them. If we push on the innermost sheath, it starts to move in response but it also pulls on the next sheath out because of the friction that couples them together.

That sheath starts moving in response to the force but it too is in contact via friction with the next sheath surrounding it and eventually it too is set into motion.

This process continues until we try to set the outermost sheath- the one in contact with the pipe wall- into motion. Because the outermost sheath is essentially "glued" to the pipe wall- which you can think of as a sheath which cannot move- it resists being slid down the pipe and pulls back on the moving sheath immediately inside it.

In this way, fluid friction- viscosity- couples together the motion of those concentric sheaths of liquid in just such a way as to yield a velocity maximum in the center of the pipe and zero velocity at the inside surface of the pipe.

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