How does the pressure drop as a fluid flows through a pipe

[Chestermiller]

Have you had a course in introductory physics that includes forces and free body diagrams?

 

[BvU]

It seems to me that you are viewing force as some kind of momentum

Force is the time derivative of momentum. Momentum is transported sideways by viscosity. (Did I say something like that before?)

 

[seratia]

This makes no sense to me either.

I mean, that's what your description was, lol

 

[seratia]

Force is the time derivative of momentum. Momentum is transported sideways by viscosity. (Did I say something like that before?)

To say that momentum is transported sideways is to imply that the force is acting at the center of the fluid, and the radial fluid is carried along by shear stress (thus momentum). When I asked before if the force is acting on the middle, I was told no, it acts against the cross sectional area all at once.

Is this what you mean:

 

[seratia]

Have you had a course in introductory physics that includes forces and free body diagrams?

Yes.

As it applies to this problem, the force pushing the fluid against friction (pushing the fluid from left to right) is getting reduced (I am assuming because of the friction it has incurred to get from the beginning of the journey to the middle of the journey, if for example we look at those two points).

 

[Chestermiller]

I mean, that's what your description was, lol

Please point out where I said anything about momentum.

 

[seratia]

Have you had a course in introductory physics that includes forces and free body diagrams?

This is not a free body diagram. But it gets the point across:

As you can see, force is lessened at point 2.

 

[seratia]

Please point out where I said anything about momentum.

Not about momentum. But about the description of the two plates. Here is what you said:

"To get an idea of viscous friction, think extreme cases like very viscous fluids like molasses, pancake syrup, and corn syrup. Imagine that you have the fluid contained between two horizontal parallel plates, and you are trying to slide the top plate to the right at constant speed. You need to exert a force on the upper plate to the right and a force on the lower plate to the left to hold it in place. OK so far?"

This is how I interpreted it:

It seems you are wanting to oppose the fluid moving with the plates - so "holding it in place" as you said ("it" being the fluid) while you move the plates, the upper plate to the right and the lower plate to the left.

 

[Chestermiller]

Yes.

As it applies to this problem, the force pushing the fluid against friction (pushing the fluid from left to right) is getting reduced (I am assuming because of the friction it has incurred to get from the beginning of the journey to the middle of the journey, if for example we look at those two points).

Since you have had a course in freshman physics, I have a focus problem for us to work on. Consider two parallel plates with viscous fluid situated between them. One plate is at z = 0 and the other plate is at z = h. Both plates are stationary. The y direction is "into the paper. " Pressure is varying in the x direction from high on the left to lower on the right. The fluid is moving from left to right (i.e., in the x direction) with a velocity which depends on z, and with velocity v=0 at z = 0 and z = h.

Focus attention of the layer of fluid in the lower half of the channel, between z and $z + \Delta z$, and draw a free body diagram on the section of this layer situated between x and $x+\Delta x$. In this free body diagram, show the horizontal forces exerted on this fluid section from the layers above and below, and from behind and in-front.

I hope you will be willing to participate in this exercise.

Chet

 

[Chestermiller]

This is not a free body diagram. But it gets the point across:
View attachment 235820

As you can see, force is lessened at point 2.

The frictional force per unit wall area is the same at points 1 and 2.

 

[Chestermiller]

Not about momentum. But about the description of the two plates. Here is what you said:

"To get an idea of viscous friction, think extreme cases like very viscous fluids like molasses, pancake syrup, and corn syrup. Imagine that you have the fluid contained between two horizontal parallel plates, and you are trying to slide the top plate to the right at constant speed. You need to exert a force on the upper plate to the right and a force on the lower plate to the left to hold it in place. OK so far?"

This is how I interpreted it:
View attachment 235821

It seems you are wanting to oppose the fluid moving with the plates - so "holding it in place" as you said ("it" being the fluid) while you move the plates, the upper plate to the right and the lower plate to the left.

No. The lower plate is not moving.

 

[BvU]

is to imply that the force is acting at the center of the fluid [edit, BvU:] only

No, it is not implied. Wherer do you get that idea ?

 

[boneh3ad]

it applies to this problem, the force pushing the fluid against friction (pushing the fluid from left to right) is getting reduced.

Why do you think the force is getting reduced? Across a small fluid element, the net forforce due to pressure is bases on the pressure difference from one side to the other, not simply the pressure on one side. This, taken to it's infinitesimal limit, implies that force is proportional to thrthe press gradient, not thethe press itself. So, a constant force requires qnconstsnt pressure gradient, not a constant pressure.

 

[Chestermiller]

To say that momentum is transported sideways is to imply that the force is acting at the center of the fluid, and the radial fluid is carried along by shear stress (thus momentum). When I asked before if the force is acting on the middle, I was told no, it acts against the cross sectional area all at once.

Is this what you mean:

View attachment 235817

This is close to what I'm saying, except that there should be an arrow on the left for each of the shells.