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

• I

## Main Question or Discussion Point

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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BvU
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Hi,
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.

Chestermiller
Mentor
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.

Chestermiller
Mentor
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.

Chestermiller
Mentor
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.

Chestermiller
Mentor
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.

(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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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.

Gold Member
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".

Chestermiller
Mentor
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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Chestermiller
Mentor 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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BvU
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2019 Award
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.

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?

sophiecentaur
Gold Member
(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.

Chestermiller
Mentor
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.

sophiecentaur
Gold Member
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)

BvU
Homework Helper
2019 Award
(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
sophiecentaur
Gold Member
And to convince yourself the pressure is really radially uniform
Perhaps just to realise that there is no radial flow - so there can't be a radial pressure gradient???

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.

BvU