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

[seratia]

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.

 

[boneh3ad]

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.

 

[seratia]

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.

 

[sophiecentaur]

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?

 

[A.T.]

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.

 

[sophiecentaur]

Consider the rigid body analogy:

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

You could take this further and use a Rubber / Sponge block (highly deformable). The end of the block in the direction of motion will be less compressed than the end where the applied force is. At any point along the block there will be the equilibrium but the forward pressure will be higher at the end that's pushed. You could look at it in terms of the force being 'used up' against friction as you look further along the block. At the end there will be no horizontal force. (Constant velocity situation)

 

[A.T.]

You could take this further and use a Rubber / Sponge block (highly deformable).

I wouldn't go that route because the water in the pipe is not compressible. What is different between water and rigid block, is the nature of the internal shear forces. But the mechanism of the pressure gradient along the movement direction is similar.

 

[sophiecentaur]

I wouldn't go that route because the water in the pipe is not compressible.

"One analogy too far?"
But the velocity profile across the bore is analogous to the distortion along the block, althsoough a different axis could be involved. The water is not compressible but it does distort due to the velocity gradient.

 

[jbriggs444]

"One analogy too far?"
But the velocity profile across the bore is analogous to the distortion along the block, althsoough a different axis could be involved. The water is not compressible but it does distort due to the velocity gradient.

Yes, that seems to be one analogy too far.

The spongy block experiences a longitudinal pressure gradient. So it compresses more in back where the pressure is high and less in front where the pressure is low. Similarly, the water is compressed slightly near the tube ingress and less so near the tube egress.

The spongy block experiences an axial shear stress. You are pushing in the middle, but friction acts on the bottom face. This stress is not uniform across the block. The block deflects under this stress. The deflection is not uniform across the axis of the spongy block.

A better analogy would compare the non-uniform shear deflection in the block to the non-uniform velocity gradient in a cross-section of the flow. Solids deflect under shear stress. Fluids have differential flow rates.

 

[seratia]

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.

You mean like this?

 

[seratia]

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.

I'm talking about within the pipe, not between the inside of the pipe and the outside of the pipe at the opening.I can understand the pressure being lower at point D. But why is it getting lower from A to B to C?

 

[jbriggs444]

I'm talking about within the pipe, not between the inside of the pipe and the outside of the pipe at the opening.I can understand the pressure being lower at point D. But why is it getting lower from A to B to C?

Some force must keep the fluid in section A-B moving despite the viscous friction with the walls. That's the pressure drop from A to B.

Some force must keep the fluid in section B-C moving despite the viscous friction with the walls. That's the pressure drop from B to C.

Some force must keep the fluid in section C-D moving despite the viscous friction with the walls. That's the pressure drop from C to D.

In the absence of a free body diagram for the fluid in section B-C [for instance], can you at least enumerate the external forces (and momentum flows) acting on the fluid in that section?

 

[seratia]

Some force must keep the fluid in section A-B moving despite the viscous friction with the walls. That's the pressure drop from A to B.

Some force must keep the fluid in section B-C moving despite the viscous friction with the walls. That's the pressure drop from B to C.

Some force must keep the fluid in section C-D moving despite the viscous friction with the walls. That's the pressure drop from C to D.

In the absence of a free body diagram for the fluid in section B-C [for instance], can you at least enumerate the external forces (and momentum flows) acting on the fluid in that section?

Makes sense. How it comes about/gets established doesn't.

 

[jbriggs444]

Makes sense. How it comes about/gets established doesn't.

It is all about establishing an equilibrium. If the pressures are not right so that the flows are even through each cross-section then fluid will start building up or emptying out somewhere. If it builds up, pressure increases, ingress flow decreases, egress flow increases and the build up stops when an equilibrium is reached.

The normal assumption is that the system will settle down to a stable equilibrium. We should probably not worry about instabilities (such as turbulent flow).

Rather than worry about the way that the system settles down or "relaxes" toward the equilibrium state, we can just solve for the conditions that must hold if an equilibrium is reached. Doing things this way, concerns for cause and effect become irrelevant. We can consider our equations to be about correlations instead.

 

[sophiecentaur]

Rather than worry about the way that the system settles down or "relaxes" toward the equilibrium state, we can just solve for the conditions that must hold if an equilibrium is reached. Doing things this way, concerns for cause and effect become irrelevant. We can consider our equations to be about correlations instead.

+1
This approach works well throughout science and it gives the right results so many times. It 'offends' intuition because it seems to get away without having identified an agency.

 

[seratia]

Rather than worry about the way that the system settles down or "relaxes" toward the equilibrium state, we can just solve for the conditions that must hold if an equilibrium is reached.

I meant more in the sense of if a scientist was investigating and wanted to know why the pressure at point C is less than point A, when discovering it or the first time in human history. Saying that "because otherwise flow wouldn't occur" doesn't explain why there is a difference, just that it causes flow.

Although, the explanation by A.T. makes sense, if I understood it correctly as in the drawing I made.

 

[jbriggs444]

I meant more in the sense of if a scientist was investigating and wanted to know why the pressure at point C is less than point A, when discovering it or the first time in human history. Saying that "because otherwise flow wouldn't occur" doesn't explain why there is a difference, just that it causes flow.

Score +1 for @sophiecentaur. Your intuition has been offended. You need to discard cause and effect reasoning and accept "because otherwise flow would not be occurring".

 

[gmax137]

i must be missing something here. the pressure drops along the pipe because energy is lost to friction. pressure is energy per unit volume. see boneh3ad's insight on bernoulli.

 

[A.T.]

You mean like this?

Yes, If the bottom arrows represent the total friction of the block part they cover (the lowest arrow is the total friction of the block).

 

[James Demers]

I think the OP is trying to apply rules appropriate to the static situation (no flow, pressure is equal throughout the pipe) to the dynamic situation, where the fluid is in motion. There's nothing "common sense" about the latter, at least not when it comes to the math.
Imagine 20 psi at one end of a mile-long 1-inch pipe, and explaining "what happened" to the pressure at the far end, where the water is barely dripping out. Close off the far end, though, and the pressure jumps instantly to 20 psi. Chestermiller can explain it, but it's baffling to the amatuers in the room.

 

[jbriggs444]

I think the OP is trying to apply rules appropriate to the static situation (no flow, pressure is equal throughout the pipe) to the dynamic situation, where the fluid is in motion. There's nothing "common sense" about the latter, at least not when it comes to the math.
Imagine 20 psi at one end of a mile-long 1-inch pipe, and explaining "what happened" to the pressure at the far end, where the water is barely dripping out. Close off the far end, though, and the pressure jumps instantly to 20 psi. Chestermiller can explain it, but it's baffling to the amatuers in the room.

If you close off the far end, the pressure at the near end does not jump to 20 psi. It was already at 20 psi. If you close off the far end, the pressure at the far end does not jump instantly to 20 psi. It jumps instantly to an extreme value and the pipe may break.

It is only after the system has settled down into a new equilibrium after some seconds (speed of sound in water is in the neighborhood of one mile per second) that it attains a condition of 20 psi throughout.

Which, I suppose, makes your point about the details not being intuitively obvious.

 

[zul8tr]

From original post:
"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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(A) for steady state in equilibrium the net force is zero but a net hydraulic gradient per unit length is developed and that results in the unit driving force
(B) The hydraulic gradient drops in the flow direction to maintain the driving force to move the fluid along, if the flow is reduced to zero then the pressure driving force = 0.

A bit more detail:
In steady state (dv/dt partial = 0) incompressible flow all forces are in balance and if the pipe diameter is assumed constant then no convective acceleration (v dv/dx partial = 0) exists since mean velocity is constant and a hydrostatic pressure distribution (straight and parallel streamlines) exists at any cross section. Then in this case a linear hydraulic gradient is established that always drops in the flow direction. This hydraulic gradient has a constant unit pressure drop along the pipe or a unit pressure difference and that here is the constant unit driving force developed on each cross section area all along the pipe, that is exactly balanced by the unit fluid friction force (no gravity force if the pipe is horizontal). The hydraulic gradient is a function of the total system geometry, fluid type, etc and the beginning and end points (boundary conditions) that establishes the total system mass flow achieved.

There is no energy lost it is just converted to other forms and not recoverable in this irreversible process resulting from friction.

Above discussion a small diameter pipe was considered with pressure measured at the center of the pipe, but if the pipe diameter is large and with a hydrostatic pressure distribution at any cross section then there is a pressure difference from the top to the bottom of the pipe (say horizontal pipe for this discussion) and that internal transverse force is transferred into the pipe wall as stresses and strains.This does not generally change the unit pressure driving force in the pipe length direction. In any pipe flowing with say a Newtonian fluid the velocity varies from zero at the pipe wall (stick condition) to generally max velocity at the pipe center with a transverse velocity distribution depending on the state of flow: with laminar flow it is parabolic and logarithmic for full turbulent flow. Other type very different fluids can have very different velocity distributions but the stick condition at the wall is always present.

In the CV system with flexible walled arteries down to super small capillaries then back to veins and to the heart pumping away the hydraulic process is the same but much more complected with variable hydraulic gradients in variable pulsing flow conduit diameters responding to the pulsing pressure changes by the heart.

 

[seratia]

What exactly does pressure loss due to friction mean? How do you even visualize that?

 

[jbriggs444]

What exactly does pressure loss due to friction mean? How do you even visualize that?

Can you visualize the tension of a fuzzy rope drawn through a slightly tight tube?

 

[seratia]

Can you visualize the tension of a fuzzy rope drawn through a slightly tight tube?

I can. But that doesn't mean I can relate it to pressure drop.

And also: Why can't a constant pressure drive the fluid through the pipe? Why does it have to decrease for there to exist a flow. When I drag a book across the floor at constant speed, I do so with constant pressure. Why must there be a pressure loss in order for the fluid to flow?

 

[Chestermiller]

I can. But that doesn't mean I can relate it to pressure drop.

And also: Why can't a constant pressure drive the fluid through the pipe? Why does it have to decrease for there to exist a flow. When I drag a book across the floor at constant speed, I do so with constant pressure. Why must there be a pressure loss in order for the fluid to flow?

When you push a book across the floor at constant speed, the force you are exerting on the back end of the book is higher than the (no) force you are exerting on the leading end of the book. So there actually is a pressure loss.

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?

 

[seratia]

When you push a book across the floor at constant speed, the force you are exerting on the back end of the book is higher than the (no) force you are exerting on the leading end of the book. So there actually is a pressure loss.

That makes sense. Someone else in this thread also told me something like this actually, which I forgot about. This is the best example so far. So the middle of the book has half the pressure?

I think I understand - so the fluids pressure stays proportional to how much fluid it has in front of it?! If so, that makes total sense! But I wouldn't call that "loss due to friction".

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?

When you say "hold it in place", what do you mean?

 

[Chestermiller]

That makes sense. Someone else in this thread also told me something like this actually, which I forgot about. This is the best example so far. So the middle of the book has half the pressure?

Yes.

I think I understand - so the fluids pressure stays proportional to how much fluid it has in front of it?! If so, that makes total sense!

Only if the exit is at zero pressure. What you really want to look at the the difference in pressure between two locations along the pipe axis; like your example of full pressure at the back, and half the pressure at the middle. This is proportional to the "frictional" force at the pipe wall acting on the fluid between the two locations.

But I wouldn't call that "loss due to friction".

But that is exactly what it is: loss in pressure between two locations due to viscous friction at the wall.

When you say "hold it in place", what do you mean?

Don't allow it to move.

 

[seratia]

But that is exactly what it is: loss in pressure between two locations due to viscous friction at the wall.

It seems to me that you are viewing force as some kind of momentum, reducing as it encounters friction along its journey.

Don't allow it to move.

Like this?

 

[Chestermiller]

It seems to me that you are viewing force as some kind of momentum, reducing as it encounters friction along its journey.

This makes no sense to me. A force is just a force.

Like this?

View attachment 235814

This makes no sense to me either.