- #1

axim

- 17

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- TL;DR Summary
- Question about how the flow will establish in a parallel pipe branch when static pressure after the branch is higher than before

Hello!

I have a question regarding the application of the bernoulli equation and calculation of the flow through a parallel pipe branch. It's more the basic understanding how the flow will establish.

You can find a sketch attached to follow my explanation.

Let's assume I have a pipe with diameter A, which gets larger to diameter B - let's assume in width so the height is negligible - and then smaller to diameter A again. There is a liquid flowing through the pipe, e.g. water.

There is a small head loss in the pipe due to friction, so pOut will be smaller than pIn obviously.

Now at the part where the diameter expands from A to B, the static pressure has to increase because the kinetic energy is reduced due to the larger diameter according to the Bernoulli equation. If the friction is low, this will "overcompensate" the decrease due to friction and I can see a higher static pressure at point 2 than in point 1. Note: If the diameter had stayed the same, static pressure at point 2 would have been lower than in point 1 because of the friction.

The total pressure that consists of the static pressure and dynamic pressure has decreased nevertheless due to the friction. I hope it's correct until here.

Now I wonder about two scenarios, if a branch is added between point 1 and point 2.

Scenario 1: The branch is flat on the ground.

According to pipe network calculation theory, the head losses in both branches should be equal and the flow will align to match that. So this means there must be a flow in the green painted branch from point 1 to 2. But, intuitively, I'd say that if I open the branch when it was in steady flow before, the fluid would rather try move from point 2 to back to point 1 through the branch as the static pressure is higher at point 2? Or is this some transient behaviour and in the end there will be a flow from left to right in both branches? In hardy cross it would not be possible to have some kind of reverse loop to my understanding. Also QIn must still equal QOut and the total pressure pIn is higher than pOut, so there could only be some internal loop.

I know if I apply Bernoulli between 1 and the upper branch and between 1 and the lower branch it must yield the same energy for both branches - but I don't know if the flow will establish at all like I expect or not to be ablw to draw the streamlines?

Scenario 2:

I modified the example 1. I have basically a similar setup, but this time the branch is added vertical and not flat.

Let's imagine the branch is not opened yet. At point 1 if I insert a vertical static pressure measuring tube (at the wall, parallel to the flow, I just imagine a thin long glass pipe), the water level would rise as high as the static pressure allows - it is not high enough to flow through the branch to point 2 (or out of the measuring tube). If I measure at point 2, the water would rise high enough to be able to flow through the branch and then down to point 1.

Here, I have the same confusion. Seen as a pipe network, I would expect the flow to be from left to right in the main pipe (lower branch) and also in the upper branch (with different velocities depending on friction and branch geometry). But, it seems it will rather flow back from 2 to 1. I could only image it flowing from 1 to 2 if I regard the total pressure which is higher in 1 than in 2.

So here I'm really confused how the flow will establish:

A) Flow from 1 to 2 - but the static pressure when opening the branch isn't high enough in 1 to pass the elevation?

B) No Flow - I can't imagine that as the static pressure in 2 is high enough to lift water as high as the horizontal part of the branch

C) Flow from 2 to 1 - Doesn't match my expectation about pipe analysis and would violate that there should be the same head loss through two parallel branches

I hope you could understand my problem and I hope to get some understanding how the flows will establish and why!

Thanks a lot in advance

PS: I know usually the kinetic pressure part is small and can often be neglected, but I want to understand it completely first.

I have a question regarding the application of the bernoulli equation and calculation of the flow through a parallel pipe branch. It's more the basic understanding how the flow will establish.

You can find a sketch attached to follow my explanation.

Let's assume I have a pipe with diameter A, which gets larger to diameter B - let's assume in width so the height is negligible - and then smaller to diameter A again. There is a liquid flowing through the pipe, e.g. water.

There is a small head loss in the pipe due to friction, so pOut will be smaller than pIn obviously.

Now at the part where the diameter expands from A to B, the static pressure has to increase because the kinetic energy is reduced due to the larger diameter according to the Bernoulli equation. If the friction is low, this will "overcompensate" the decrease due to friction and I can see a higher static pressure at point 2 than in point 1. Note: If the diameter had stayed the same, static pressure at point 2 would have been lower than in point 1 because of the friction.

The total pressure that consists of the static pressure and dynamic pressure has decreased nevertheless due to the friction. I hope it's correct until here.

Now I wonder about two scenarios, if a branch is added between point 1 and point 2.

Scenario 1: The branch is flat on the ground.

According to pipe network calculation theory, the head losses in both branches should be equal and the flow will align to match that. So this means there must be a flow in the green painted branch from point 1 to 2. But, intuitively, I'd say that if I open the branch when it was in steady flow before, the fluid would rather try move from point 2 to back to point 1 through the branch as the static pressure is higher at point 2? Or is this some transient behaviour and in the end there will be a flow from left to right in both branches? In hardy cross it would not be possible to have some kind of reverse loop to my understanding. Also QIn must still equal QOut and the total pressure pIn is higher than pOut, so there could only be some internal loop.

I know if I apply Bernoulli between 1 and the upper branch and between 1 and the lower branch it must yield the same energy for both branches - but I don't know if the flow will establish at all like I expect or not to be ablw to draw the streamlines?

Scenario 2:

I modified the example 1. I have basically a similar setup, but this time the branch is added vertical and not flat.

Let's imagine the branch is not opened yet. At point 1 if I insert a vertical static pressure measuring tube (at the wall, parallel to the flow, I just imagine a thin long glass pipe), the water level would rise as high as the static pressure allows - it is not high enough to flow through the branch to point 2 (or out of the measuring tube). If I measure at point 2, the water would rise high enough to be able to flow through the branch and then down to point 1.

Here, I have the same confusion. Seen as a pipe network, I would expect the flow to be from left to right in the main pipe (lower branch) and also in the upper branch (with different velocities depending on friction and branch geometry). But, it seems it will rather flow back from 2 to 1. I could only image it flowing from 1 to 2 if I regard the total pressure which is higher in 1 than in 2.

So here I'm really confused how the flow will establish:

A) Flow from 1 to 2 - but the static pressure when opening the branch isn't high enough in 1 to pass the elevation?

B) No Flow - I can't imagine that as the static pressure in 2 is high enough to lift water as high as the horizontal part of the branch

C) Flow from 2 to 1 - Doesn't match my expectation about pipe analysis and would violate that there should be the same head loss through two parallel branches

I hope you could understand my problem and I hope to get some understanding how the flows will establish and why!

Thanks a lot in advance

PS: I know usually the kinetic pressure part is small and can often be neglected, but I want to understand it completely first.