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

• seratia
In summary, assuming laminar viscous flow in an incompressible fluid, a pressure difference between the two ends of a pipe causes a net force to act on the fluid, resulting in it flowing from left to right. As the fluid travels along the pipe, it experiences a net force of 0 due to opposing forces from friction. The pressure gradually drops along the pipe due to viscous frictional forces and momentum transport in the transverse direction. This can be explained by a viscous shear stress that varies radially and a velocity profile that transitions from uniform to parabolic. The main pressure gradient is along the axis of the pipe, and the force acting on the fluid decreases as it flows from the higher pressure side to the lower pressure side
seratia
Assuming laminar viscous (meaning not frictionless) flow.

Here is what I know about fluids flowing: You have a pressure difference between the two ends of the pipe. This causes a net force acting on the left side of the fluid in the pipe. Therefore, this incompressible fluid flows from left to right.

Where it becomes abstract for me is following a section of this fluid as it travel along the pipe and what happens to it
(A) I think it experiences a net force of 0 at all points because it is not accelerating. I am assuming the opposing force due to friction is keeping the net force at 0.
(B) I am having trouble understanding why pressure gradually drops along the pipe or hose. The rightward pointing force acting on it is getting reduced. How exactly does this happen? If the fluid is incompressible, why isn't this section still feeling the same force as it did as the initial point in the hose?

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Hi,
seratia said:
You have a pressure difference between the two ends of the pipe. This causes a net force acting on the left side of the fluid in the pipe
You sure this is cause and effect? Or is it in reverse ?

(A) Right. Newton. Correct. Friction: Viscosity.
(B) Basically: momentum transport in transverse direction. I like the explanation in Bird, Stewart & Lightfoot. But the lecture by John Biddle is very instructive as well.

At the wall of the pipe, there is a viscous frictional force acting in the direction opposite to that imposed by the pressure forces. Internally, the viscous friction is transmitted to each of the shells of fluid from the pipe wall to the center of the pipe. This is caused by a viscous shear stress that varies radially. If you do a force balance on each shell of fluid flowing through the pipe, you will find that each shell has the same pressures at its ends, and has viscous shear stresses on its surfaces which sum to balance the pressure forces.

The velocity profile across the pipe transitions from uniform at the entrance to about parabolic further downstream. To determine the total pressure, you need to integrate across the velocity profile.

Dr Dr news said:
The velocity profile across the pipe transitions from uniform at the entrance to about parabolic further downstream. To determine the total pressure, you need to integrate across the velocity profile.
The OP seemed to be asking about the pressure drop with fully developed flow.

Just a guess: a closed pipe where the liquid cannot escape exerts the same force in all directions(ignoring gravitational effects.) when the pipe is opened, some of the force on the walls transfers toward the open ends of the pipe, and the pressure on the sides is reduced. Think of an air compressor tank, and it's relief valve.

KIRBY said:
Just a guess: a closed pipe where the liquid cannot escape exerts the same force in all directions(ignoring gravitational effects.) when the pipe is opened, some of the force on the walls transfers toward the open ends of the pipe, and the pressure on the sides is reduced. Think of an air compressor tank, and it's relief valve.
If it were not for viscous stresses, the fluid would be accelerating in this scenario.

I thought the pressure was measured at the side of the pipe.

KIRBY said:
I thought the pressure was measured at the side of the pipe.
This has nothing to do with how the pressure is measured. The main pressure gradient is along the axis of the pipe. Please, if you are not sure about something, avoid making guesses. If they are wrong, they just go to confusing the member who originally posted.

Are we not all working toward understanding? I will refrain.

BvU said:
(B) Basically: momentum transport in transverse direction. I like the explanation in Bird, Stewart & Lightfoot. But the lecture by John Biddle is very instructive as well.

There seemed to be a lot derivation and equation in those sources, without much "interpretation". I guess I touched an area outside of my scope with this one. It just doesn't logically make sense for the force acting on the fluid to be decreasing as the fluid flows from the higher pressure side to the lower pressure side.

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BvU said:
You sure this is cause and effect? Or is it in reverse ?

Pressure and force are essentially the same thing, if you simplify it by assuming the area is constant along the pipe or hose or artery. So yes, a force difference causes you to have a net force.

KIRBY said:
Are we not all working toward understanding? I will refrain.

Some of us here are not "working" toward understanding because we know the answer and are trying to help the rest of the participants work toward an understanding. Wild guesses posed as answers aren't helpful in that regard.

Chestermiller
As I said, "I will refrain".

seratia said:
Pressure and force are essentially the same thing, if you simplify it by assuming the area is constant along the pipe or hose or artery. So yes, a force difference causes you to have a net force.
Since you are so focused with regard to forces, have you drawn a free body diagram showing the forces acting on the plug of fluid between r=0 and arbitrary radius r. Or do you feel that you have advanced to the point where you no longer need to use free body diagrams?

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Force Balance on Free Body:

$$\pi r^2p_1-\pi r^2p_2-2\pi rL\tau_{rz}=0$$or
$$\tau_{rz}=\frac{(p_1-p_2)}{L}\frac{r}{2}$$

where ##\tau_{rz}## is the viscous shear stress at radial location r

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seratia said:
How exactly does this happen
Viscosity: A ##\ \overrightarrow {\nabla v} \ ## in e.g the ##r##-direction causes a shear stress ##\vec \tau_{rz}## in the ##z##-direction.

Chestermiller said:
View attachment 235454

Force Balance on Free Body:

$$\pi r^2p_1-\pi r^2p_2-2\pi rL\tau_{rz}=0$$or
$$\tau_{rz}=\frac{(p_1-p_2)}{L}\frac{r}{2}$$

where ##\tau_{rz}## is the viscous shear stress at radial location r
So your point is that net force = 0. Got it.

Now to relate this all back to the original question. BvU cited "momentum transport in the transverse direction" as the reason for pressure loss along the cylinder. I take this to mean the parabolic flow, where the center fluid slides against the radial fluids and cause them to flow, although with less velocity as the distance from the center increases. Basically, like a deck of cards being pushed at the top, causing the deck to go from vertical to sideways stacked. So I have two questions:

(A) The above model assumes only the center of the fluid is being pushed, and that the radial parts are all just reactive and move due to friction. Is that really true or just an assumption?
(B) Momentum transport - so basically the center fluid loses momentum due to sliding past radial fluid as it makes its journey through the cylinder. (a) Momentum implies velocity. But as we know, flow rate is continuous along the cylinder, meaning velocity is the same (assuming the cylinder diameter is constant). So I'm confused. The center flowing fluid doesn't slow down. So how is it losing momentum?

seratia said:
(A) The above model assumes only the center of the fluid is being pushed, and that the radial parts are all just reactive and move due to friction. Is that really true or just an assumption?
I think the model assumes everything has settled down to a steady state and no one is 'pushing' any particular part of the fluid. The pressures and flows are just the result of what is happening higher upstream. I see many parallels here like "How does a resistor in a circuit know how much current to let through?" and (more complex) "How can you work out how an amplifier with feedback makes the gain what it is?". It is not intuitive to take the steady state as the starter for working things out - but it works.

seratia said:
So your point is that net force = 0. Got it.

Now to relate this all back to the original question. BvU cited "momentum transport in the transverse direction" as the reason for pressure loss along the cylinder. I take this to mean the parabolic flow, where the center fluid slides against the radial fluids and cause them to flow, although with less velocity as the distance from the center increases. Basically, like a deck of cards being pushed at the top, causing the deck to go from vertical to sideways stacked.
It's more like the outer fluid is slowing down the fluid in the inner shells. Don't forget, the axial velocity at the wall is zero.
So I have two questions:
(A) The above model assumes only the center of the fluid is being pushed, and that the radial parts are all just reactive and move due to friction. Is that really true or just an assumption?
Who says only the center fluid is being pushed? The pressure is distributed uniformly over the entire cross section, so all the fluid is being pushed by the pressure gradient.

(B) Momentum transport - so basically the center fluid loses momentum due to sliding past radial fluid as it makes its journey through the cylinder.
The fluid in any shell is dragged backward by the fluid in outer shells and is dragged forward by fluid in inner shells.
(a) Momentum implies velocity. But as we know, flow rate is continuous along the cylinder, meaning velocity is the same (assuming the cylinder diameter is constant). So I'm confused. The center flowing fluid doesn't slow down. So how is it losing momentum?
The center flowing fluid is dragged backward by fluid in the shell right next door.

Sometimes the term radial transport of axial momentum is used to describe the frictional shear stress. On a molecular level, this makes sense because momentum is exchanged by collisions of molecules (and exchange of molecules) between shells. At the continuum level, this translates into a shear stress.

Chestermiller said:
so all the fluid is being pushed by the pressure gradient.
That's a longitudinal pressure gradient, of course. (Implied ok but it's worth while making it explicit)

seratia said:
(A) The above model assumes only the center of the fluid is being pushed
I can understand you make this assumption from the picture, but I challenge you to point out where it is being stated. And to convince yourself the pressure is really radially uniform at a given point along the pipe.

sophiecentaur
BvU said:
And to convince yourself the pressure is really radially uniform
Perhaps just to realize that there is no radial flow - so there can't be a radial pressure gradient?

Chestermiller said:
Who says only the center fluid is being pushed? The pressure is distributed uniformly over the entire cross section, so all the fluid is being pushed by the pressure gradient.
But if the "the outer fluid is slowing down the fluid in the inner shells" it would seem that the flow of the inner shell is the causal event.

The outer fluid needs pushing too

seratia said:
But if the "the outer fluid is slowing down the fluid in the inner shells" it would seem that the flow of the inner shell is the causal event.
I have no idea what this means. I still don't understand what your issue is. I have asked you to furnish a free body diagram which illustrates your issue. I have provided a free body diagram of my own showing the forces involved. Maybe you can comply and provide a free body diagram, perhaps involving a shell of fluid, to explain yourself more fully. It will be difficult to help you otherwise. If you don't provide a diagram by the end of today, I am going to close this thread.

Chestermiller said:
I have no idea what this means. I still don't understand what your issue is. I have asked you to furnish a free body diagram which illustrates your issue. I have provided a free body diagram of my own showing the forces involved. Maybe you can comply and provide a free body diagram, perhaps involving a shell of fluid, to explain yourself more fully. It will be difficult to help you otherwise. If you don't provide a diagram by the end of today, I am going to close this thread.

I was just quoting you there. You drew a diagram of a inner shell of fluid being acted on by forces, and explained that the outer shells are trying to slow it down. So I thought "okay maybe the force is acting on the inner shell (the causal event), and the outer shells are getting dragged along by friction, rather than the force acting equally on all sides, pushing against everything all at once.

I'm not sure where any of that fits in the grand scheme of things or how we got here.

My main issue/question was why does a viscous laminar flow experience a pressure drop during its journey down the cylinder. Is there a conceptual answer, rather than mathematics and derivations. I got the textbook that was mentioned, but it kept going on and on without any foreseeable conclusion in sight. Derivation after derivation. It wasn't leading to any conclusions or concluding statements such as "...and that's why, ladies and gentlemen, pressure drops along the path of flow"

I'm actually more of a medical student. Just wanted to research for why pressure drops along the CV system, rather than just taking things at face value.

seratia said:
I got the textbook that was mentioned, but it kept going on and on without any foreseeable conclusion in sight
Pity; good book for later, then. What about the youtube lecture ? He does have a point where the only thing that changes (in ##dm/dt, Q, \vec v, P##) can be ##P##.

Physics: friction means you need to do work to keep things moving steadily. Energy is dissipated. In fluid flow, the rate of mechanical work (power) is ##{d\over dt} (pV) = p\dot V + V{dp\over dt}##. The only non-constant (i.e. wrt postion) along the flow line in steady state is p. But again, unfortunate for you, that's formulas.

seratia said:
I was just quoting you there. You drew a diagram of a inner shell of fluid being acted on by forces, and explained that the outer shells are trying to slow it down. So I thought "okay maybe the force is acting on the inner shell (the causal event), and the outer shells are getting dragged along by friction, rather than the force acting equally on all sides, pushing against everything all at once.

I'm not sure where any of that fits in the grand scheme of things or how we got here.

My main issue/question was why does a viscous laminar flow experience a pressure drop during its journey down the cylinder. Is there a conceptual answer, rather than mathematics and derivations. I got the textbook that was mentioned, but it kept going on and on without any foreseeable conclusion in sight. Derivation after derivation. It wasn't leading to any conclusions or concluding statements such as "...and that's why, ladies and gentlemen, pressure drops along the path of flow"

I'm actually more of a medical student. Just wanted to research for why pressure drops along the CV system, rather than just taking things at face value.
Actually what I drew in the diagram was a plug of fluid of any arbitrary radius (not a shell). And the diagram shows that the pressure forces at the end of the plug are exactly balanced by the viscous shear force on the surface of the plug. So there is no net force on the plug. However, the viscous shear stress force has to increase with radius of the plug because the plug is larger, and thus the pressure forces are larger.

This all happens because the fluid "sticks" at the wall, and the axial fluid velocity is zero at the wall. So the outer shells of fluid act to try to slow down the inner shells of fluid, and the inner shells of fluid act to try to speed up the outer shells of fluid. This is like your deck of cards in which you push on the edge of the deck, but the cards on the very top and very bottom of the deck are not allowed to move.

seratia said:
I'm actually more of a medical student. Just wanted to research for why pressure drops along the CV system, rather than just taking things at face value.
Kudos for that !

seratia
BvU said:
Kudos for that !

Thanks to both of you for the help. I don't want to drag this on further. For now, I'll just re watch and reread the sources you suggested and reread the comments and hopefully make sense of it all after a few more times.

seratia said:
My main issue/question was why does a viscous laminar flow experience a pressure drop during its journey down the cylinder. Is there a conceptual answer, rather than mathematics and derivations.

Let's assume the flow is moving in one direction, say, to the right. The walls of the tube are exerting a force on the fluid opposing the flow, i.e. to the left. If there was no pressure drop, then there would be no force pushing the fluid to the right against viscosity, and the flow would stop. Since the flow is actually a constant velocity, the pressure drop exists to exactly counteract the viscous force due to the wall.

That force due to the pressure gradient acts everywhere through the cross-section. The force opposing the motion due to the wall acts at the wall and is transmitted through the fluid by viscosity.

Let's assume the flow is moving in one direction, say, to the right. The walls of the tube are exerting a force on the fluid opposing the flow, i.e. to the left. If there was no pressure drop, then there would be no force pushing the fluid to the right against viscosity, and the flow would stop. Since the flow is actually a constant velocity, the pressure drop exists to exactly counteract the viscous force due to the wall.

That force due to the pressure gradient acts everywhere through the cross-section. The force opposing the motion due to the wall acts at the wall and is transmitted through the fluid by viscosity.

Let me digest this for a minute, please. Thank you for the explanation.

and the flow would stop.
The way I read it, Newton 3 would seem to be violated. With no viscosity and constant pressure would not the liquid just go faster until rate of momentum transfer reaches an equilibrium condition?

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seratia said:
Let me digest this for a minute, please. Thank you for the explanation.
Consider the rigid body analogy:

If you push a block along the floor at constant speed against friction, you will also have a horizontal gradient of compressive stresses within that block. Each slice of the block has to transmit the force to counter the friction of the block part in front of the slice. The further forward you go, the less friction in front you have left.

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