Can you please explain Bernoulli's equation?

[Callmelucky]

The fluid element is being compressed by its container and the fluid surrounding the element. It's under pressure. Thus, when we isolate the element from its surroundings to analyze it, we need to replace the effect of the surroundings as external forces acting on the element. The net effect of all those forces (and others not shown-like weight) dictates whether or not the element is accelerating in a particular direction.

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When you first started this, I was thinking you were seeing some seeming contradictions that can arise in certain systems where Bernoulli's gives spectacularly unrealistic results because of the inviscid assumption applied to an entire system. But, it just dawned on me (when you kept asking about $F_2$ and mentioned about calculating it in high school) that you probably just don't understand how we apply Newtons Second Law to analyze fluid systems. The average Mech. Engineers don't get into fluid mechanics until sophomore\junior year in college (I think). Sorry, If I caused any confusion in that regard. So lets back up, and I think we can get this resolved.

Haha, I mean I don't know what I don't know. I read the link someone posted in comments and then I realised that I was thinking completely in the wrong direction. I figured that it's not pressure that fluid is causing on the pipes but the pressure in fluid, that was like heureka moment.

But last comment you posted

Thus, when we isolate the element from its surroundings to analyze it, we need to replace the effect of the surroundings as external forces acting on the element. The net effect of all those forces (and others not shown-like weight) dictates whether or not the element is accelerating in a particular direction.

lost me completly.

Btw thank you for being patient and for explaining me stuff.

Can you tell me what you think I don't know?

 

[erobz]

lost me completly. Mostly because I thought that pressure is acting towards inside and not towards outside.

Imagine that parcel of fluid were instead a spring inside a non-rigid box. The fluid surrounding that package is under pressure. Its pushing on the boundaries of the box which contains the spring. The spring is being compressed. If we take the spring out of the fluid would decompress (expand). So...if we are going to analyze it as it were inside the fluid, then outside the surrounding fluid we need to apply forces to the spring to keep it in its compressed state. When we isolate a system, this is what we are doing. We are removing everything around that system, and replacing what everything does to that system with forces.

 

[Callmelucky]

Imagine that parcel of fluid were instead a spring inside a non-rigid box. The fluid surrounding that package is under pressure. Its pushing on the boundaries of the box which contains the spring. The spring is being compressed. If we take the spring out of the fluid would decompress (expand). So...if we are going to analyze it as it were inside the fluid, then outside the surrounding fluid we need to apply forces to the spring to keep it in its compressed state.

I got that, that is why I deleted that part

 

[erobz]

I got that, that is why I deleted that part

What part did you delete?

 

[Callmelucky]

What part did you delete?

the one where I wrote that I thought that pressure is acting inwards not outwards

 

[erobz]

the one where I wrote that I thought that pressure is acting inwards not outwards

I'm referring to the part where you said "lost me completely"... I'm trying to re-explain that part.

 

[erobz]

Can you tell me what you think I don't know?

What do you think I'm trying to do?

 

[Callmelucky]

What do you think I'm trying to do?

lol, I posted this earlier

 

[erobz]

lol, I posted this earlier

Ok, well I'm trying to establish whether any of this explanation is landing before we keep moving. Does the diagram in #35 make sense to you?

 

[Lnewqban]

So could it be that static pressure drops because the tube becomes narrower(static pressure drops with decrease of diameter of tube)?

Mechanical energy within the fluid remains the same, if we consider the Bernoulli’s idealization of an isolate system (no energy supplied in or sustracted out the fluid).
What and how we measure the three ways in which that energy manifests itself is not the same as that energy.

What you have described in your explanation is what we can see when fluid comes to a stop inside the vertical pipe or the L shaped pipe that you have mentioned.
The same applies to the picture that you have posted showing the transparent tubes arrangement.

If the height in the left side vertical tube is significantly less than in the right hand one, some residual velocity must remain at the point where vertical and horizontal tubes meet, which is stealing some potential energy from the vertical column.

Regarding discussed F1 and F2, independent from the source that induces that imbalance, they must exist if the fluid is moving in an accelerated manner.
If the fluid is static or moving at constant speed, F1 - F2 = 0.

 

[erobz]

If the fluid is static or moving at constant speed, F1=F2=0.

Typo?

$F_1 - F_2 = 0$

 

[Lnewqban]

Typo?

$F_1 - F_2 = 0$

Yes, thank you!
Fixed.

 

[Chestermiller]

So, the Euler differential force causes $F_2$?

edit: I mean, the bernoully equation is consequence of Euler diff. force balance eq...?

That's what I said. Have you Googled Euler's fluid dynamics equation?

 

[Callmelucky]

yes, and didn't understand much of what wikipedia said

But I feel better now that I know it's something I am not supposed to know. Thank you very much for explaining this to me

As I said earlier, I just wanted to know where it comes from(I don't need to completly understand it) becauee it's not said in textbook but F2 is there so that bugged me a lot.
If it's too complicated and way beyond high school curriculum then I accept that $F_2$ is there and I don't need to completly understand it , but I asked this question because I thought it's something simple that is not explained in tectbook

 

[Callmelucky]

Thanks everyone for trying to help.
Maybe my question about $F_2$ and @Chestermiller explanation could be moved to the top, so that everyone can see it without reading 50 comments before, because I believe others have same question.

 

[Callmelucky]

Ok, well I'm trying to establish whether any of this explanation is landing before we keep moving. Does the diagram in #35 make sense to you?

I do understand that

 

[erobz]

I do understand that

Then that should answer your question about the existence of $F_2$?

 

[erobz]

If it's too complicated and way beyond high school curriculum then I accept that $F_2$ is there and I don't need to completly understand it , but I asked this question because I thought it's something simple that is not explained in tectbook

The existence of $F_2$ is not beyond high school curriculum. A mountain is being made of a mole hill. Why do you have no trouble believing $F_1$ exists?

 

[Callmelucky]

The existence of $F_2$ is not beyond high school curriculum. A mountain is being made of a mole hill. Why do you have no trouble believing $F_1$ exists?

Because that is force we are creating(actually pump) to move fluid around(then again you showed me that there is no work done which makes sense because there is no friction so all we need is initial push to keep fluid moving forever, but that is in pipes that have same diameter all around) but then again if energy is conserved, for pipes that do become narrow if some energy is taken there it will be given back when pipe become wide again, so I don't get where $F_2$ comes from, because if there i $F_2$ then we do need pump

 

[Callmelucky]

Then $F_1$ is used only to overcome air, but then again, air is fluid, and there is no friction, so taken mass of air into consideration the only thing that can happen is smaller acceleration of fluids(water and air in this case).

But that is if pipe has same diameter all the way around, there won't be any changes in kinetic, potential(in no difference in height) and pressure.

If pipe does have different diameters in different areas then pressure and kinetic energy will change but ultimately they will be conserved.

 

[Callmelucky]

So I don't get why would anyone want to apply force on both ends of pipe, the only thing that can happen is for pipe to explode eventually.

But yet again, what is the point of puttin that in textbook at the beginning where student is learning the concept.

 

[Callmelucky]

only thing $F_1$ can be used for is to overcome gravity.

Or maybe they didn't think that air outside the pipe is ideal fluid. Maybe that air that enters the pipe does have some friction with pipes so $F_1$ is used to overcome that.

But then again that should be mentioned.

 

[Lnewqban]

So I don't get why would anyone want to apply force on both ends of pipe, the only thing that can happen is for pipe to explode eventually.

But yet again, what is the point of puttin that in textbook at the beginning where student is learning the concept.

You are correct, excessive internal static pressure can make a pipe fail.

You can have a U-shaped vertical tube with a closed valve that keeps a height differential between left and right columns.
For that set up, fluid is not moving, therefore it has no dynamic energy, only potential and pressure.

After we suddenly open that valve, the volume contained between the two columns (the U of the arrangement) will be feeling a different amount of pressure between left and right ends or cross-sections.
That pressure differential will tend to naturally get balanced (due to energy loss through friction and turbulence), but it will give impulse to our volume of fluid initially inside the bottom of the U tube.

These additional links may help you:
http://hyperphysics.phy-astr.gsu.edu/hbase/press.html#ed

http://hyperphysics.phy-astr.gsu.edu/hbase/bercon.html

 

[Callmelucky]

You are correct, excessive internal static pressure can make a pipe fail.

You can have a U-shaped vertical tube with a closed valve that keeps a height differential between left and right columns.
For that set up, fluid is not moving, therefore it has no dynamic energy, only potential and pressure.

After we suddenly open that valve, the volume contained between the two columns (the U of the arrangement) will be feeling a different amount of pressure between left and right ends or cross-sections.
That pressure differential will tend to naturally get balanced (due to energy loss through friction and turbulence), but it will give impulse to our volume of fluid initially inside the bottom of the U tube.

These additional links may help you:
http://hyperphysics.phy-astr.gsu.edu/hbase/press.html#ed

http://hyperphysics.phy-astr.gsu.edu/hbase/bercon.html

can I access other topics as well from their site, as far I could see that is from Alabama University

edit: I was wrong it's hyper Physics. Mental maps got me confused, didn't know they were interactive.

Thanks for sharing.

 

[Callmelucky]

Pease, see:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/14-6-bernoullis-equation/

This is awesome, thank you.

 

[Lnewqban]

This diagram shows that you can have pressure on both sides of the volume of fluid you are analyzing.
Let's assume that your book represents only the volume of oil contained in the tube and transition.

As you can see, nothing is moving as represented; therefore, the value of static pressure in cross-sections S1 and S2 must be equal to the height pressure created by the column of water on the left leg.

Once the valve is opened, the pressure on S2 is the atmospheric pressure, which is transferred through the volume of ethanol.
The higher value of the static pressure acting on S1 will move oil and ethanol up through the horizontal and right vertical tube.

As the water level descends, the pressure on S1 gets smaller, while the pressure on S2 increases.
Naturally, the levels of the open surfaces inside the left and right vertical tubes, will tend to equalize.

When that state of balance is eventually reached (after some time of back-and-forth oscillations of the fluid inside the U-shaped tube), pressure on S1 will have a greater value than the pressure on S2 due to the height difference between both sections.

 

[erobz]

Here is another example: The loop is initially charged to some desired pressure $P$ by the fill line. It is part of very commonly used hydronic heating systems in homes across the globe. They are also used heavily in industry as part of process cooling systems.

Any elemental slice you pick in the loop, there is a pressure acting on either side of it.

And furthermore, you don't need these special systems. In any piping system with fluid flowing there is a differential pressure across virtually any section you wish to examine, and that's because they all have what Bernoulli's doesn't have...friction.

So, if you are trying to determine head loss for a particular component, Bernoulli's still can be useful. If you are measuring pressures across a fitting (and flow rate), Bernoulli's says $X$ for the pressure differential, but we are measuring $Y$, the difference is...the head loss from friction.