I Maximizing Flow: Understanding the Impact of Fittings on Water Velocity

[Andy Resnick]

The difference between these two scenarios is often misapplied, but I don't think we're talking about two different scenarios...though frankly it sounds like you switched back and forth between them in your answer.

That could be. My understanding is that when unsteady flow is created (for example, just after a discontinuous expansion or flow through an orifice in a pipe, either one resulting in a vena contracta), the flow field that exists prior to resumption of steady flow is associated with viscous energy losses. My (ancient) book spends a chapter or so on hydraulic grade lines and energy grade lines. It describes "minor losses" and 'head losses', but isn't very specific on *what* is being lost.

 

[russ_watters]

That could be. My understanding is that when unsteady flow is created (for example, just after a discontinuous expansion or flow through an orifice in a pipe, either one resulting in a vena contracta), the flow field that exists prior to resumption of steady flow is associated with viscous energy losses. My (ancient) book spends a chapter or so on hydraulic grade lines and energy grade lines. It describes "minor losses" and 'head losses', but isn't very specific on *what* is being lost.

Unsteady flow is what you have when the flow field is in the process of changing, caused by the system setup changing. E.G., for the second it takes to bend a hose into a kink or few seconds it takes to actuate a valve or variable orifice. Durign this time, the volumetric flow rate is changing and is indeed also not constant along the pipe.

The OP asks about flow through fittings, which are welded/soldered in place before the system is even turned on. So I don't see a way to apply an unsteady flow to it. And I don't think it is really what you mean:

It sounds like you are describing undeveloped or developing flow. This is where the velocity profile across the pipe is changing along the pipe due to a discontinuity (in the image in the wiki, an entrance). After the discontinuity, where the pipe becomes uniform again, the velocity profile is still changing, but the average velocity is not.
https://en.wikipedia.org/wiki/Entrance_length#Hydrodynamic_entrance_length

 

[Asymptotic]

What happens with rotameters in turbulent flow? Do they over-estimate velocity? Is there any data on this?

Do a search on "flow meter installation guidelines".

How much turbulence affects flow meters depends on how they operate, and ranges from 'practically no effect' to 'significant'. Turbine (paddlewheel) sensors are ubiquitous, but sensitive to turbulent flow, and require (sometimes significant; up to 50, although usually no less than 10 upstream/5 downstream pipe diameters) of straight run to operate well.

Read sensor manufacturer literature (and post #21 and #32) for clues to the 'whys', but changes in average flow rate isn't one of them.

 

[boneh3ad]

That could be. My understanding is that when unsteady flow is created (for example, just after a discontinuous expansion or flow through an orifice in a pipe, either one resulting in a vena contracta), the flow field that exists prior to resumption of steady flow is associated with viscous energy losses. My (ancient) book spends a chapter or so on hydraulic grade lines and energy grade lines. It describes "minor losses" and 'head losses', but isn't very specific on *what* is being lost.

What is being lost is "total head", which is a code word for total pressure. If you think about it in terms of the modified Bernoulli equation, the total head would be
h_{total}=\dfrac{p}{\rho g} + \dfrac{V^2}{2g} + z,
where $z$ is a height. The relationship between two points in the flow is then
\dfrac{p_1}{\rho g} + \dfrac{V_1^2}{2g} + z_1 = \dfrac{p_2}{\rho g} + \dfrac{V_2^2}{2g} + z_2 + h_{loss}.
Generally, $V$ is set by conservation of mass through the system, so the losses are manifested as a loss in either pressure or elevation.

 

[Andy Resnick]

What is being lost is "total head", which is a code word for total pressure.

It sounds like you are describing undeveloped or developing flow. This is where the velocity profile across the pipe is changing along the pipe due to a discontinuity (in the image in the wiki, an entrance). After the discontinuity, where the pipe becomes uniform again, the velocity profile is still changing, but the average velocity is not.

I guess this is the source of my confusion: if there is a pressure loss associated with, for example, a sudden constriction, how is this manifested in the flow?

http://images.slideplayer.com/24/7380603/slides/slide_94.jpg

The figure above mentions that velocity increases but the kinetic energy decreases...??

 

[boneh3ad]

I guess this is the source of my confusion: if there is a pressure loss associated with, for example, a sudden constriction, how is this manifested in the flow?

http://images.slideplayer.com/24/7380603/slides/slide_94.jpg

The figure above mentions that velocity increases but the kinetic energy decreases...??

A sudden constriction generally causes separation bubbles to form in the corners, and these recirculating regions lead to very high viscous dissipation. Physically, that is what is causing the extra head loss in a sudden contraction (or expansion, for that matter). The slide here is fairly vague in what they are describing as energy. They don't mean kinetic energy, as that is clearly increasing with the velocity increase. They are referencing the total available pool of energy available to the flow, e.g. the sum of kinetic energy, gravitational potential energy, and "pressure energy".

In the picture they drew, the height isn't changing so we can ignore gravity. If you ignore losses, the velocity increase predicted by conservation of mass would result in a pressure drop predicted by the Bernoulli equation. If you then include losses due to, for example, the contraction, the velocity change would be the same (conservation of mass still applies), but the pressure after the contraction would be lower than that predicted by Bernoulli alone. That is the nature of the equation I provided earlier. It basically says the sum of all of the forms of energy at point 1 equals the sum of all of those terms plus terms representing losses at point 2.

 

[russ_watters]

I guess this is the source of my confusion: if there is a pressure loss associated with, for example, a sudden constriction, how is this manifested in the flow?

http://images.slideplayer.com/24/7380603/slides/slide_94.jpg

The figure above mentions that velocity increases but the kinetic energy decreases...??

That slide is very poorly worded: there are two sources of pressure drop, but only one source of pressure loss.

One drop is due to conservation of energy, e.g., the Venturi Effect. It is recoverable if the pipe expands later.

The other drop is due to drag, which is complicated for an orifice as described above. This is a true loss.

But you asked "how is this manifested", and I'm not sure what you mean. How is the energy loss when you push a box across the floor "manifested"?

 

[boneh3ad]

But you asked "how is this manifested", and I'm not sure what you mean. How is the energy loss when you push a box across the floor "manifested"?

When you push a box along the floor, the friction dissipates some of that energy as heat. Loosely speaking, the same general idea applies in fluids, and any process that results in higher viscous losses, such as separation, is going to result in greater energy dissipation.

 

[Andy Resnick]

The other drop is due to drag, which is complicated for an orifice as described above. This is a true loss.

But you asked "how is this manifested", and I'm not sure what you mean. How is the energy loss when you push a box across the floor "manifested"?

A sudden constriction generally causes separation bubbles to form in the corners, and these recirculating regions lead to very high viscous dissipation. Physically, that is what is causing the extra head loss in a sudden contraction (or expansion, for that matter). The slide here is fairly vague in what they are describing as energy. They don't mean kinetic energy, as that is clearly increasing with the velocity increase. They are referencing the total available pool of energy available to the flow, e.g. the sum of kinetic energy, gravitational potential energy, and "pressure energy".

Hopefully you have anticipated my next question- let me thread a horizontal pipe through a black box, and set the inlet pressure to be constant. At the outlet where the pipe opens to the outside air, fluid comes out at a constant rate. We may be tempted to measure that flow rate and compare it with, for example, a prediction assuming Poiseuille flow.

However, unbeknownst to you, inside the black box I welded two fittings; one is a constrictor that decreases the diameter and then a second fitting that increases the diameter back to the original diameter. I guess I could be even more simple and insert an orifice rather than weld two fittings. Maybe I left the pipe diameter constant but roughened the interior pipe walls- you have no idea.

But regardless of what I did, the inlet pressure is unchanged and the outlet pressure is unchanged. I suppose the volume flow (say, m^3/s) changes, so our predicted flow rate won't match. We would ascribe that to some sort of 'pressure loss' ('minor loss') within the hidden section of pipe, right?

The OP asked if there is a loss of kinetic energy or velocity through a fitting. Here, volume flow is reduced in comparison to Poiseuille flow due to a pressure loss. It would seem that would relate to loss of kinetic energy and/or velocity, right?

 

[physea]

So how can I explain that the difference between Darcy losses and experimentally calculated, is consistently increasing in a exponential-like way?
The set up is just a pump that pumps water inside a pipe with bends and the end of the pipe let's the water flow out freely.

The only factor I found that causes such systematic error is velocity!

 

[boneh3ad]

In the black box example, you are correct. The losses inside would mean that, for a constant inlet and outlet pressure, the flow rate would drop relative to the case of Poiseuille flow. That's not how I interpreted the OP's question, though, nor did anyone else.

If that's what he meant then it illustrates the importance of being as exact as possible when describing you problem. We all believed he was asking whether a fitting or other loss will slow down the downstream flow relative to the upstream flow without area change

 

[russ_watters]

Hopefully you have anticipated my next question- let me thread a horizontal pipe through a black box, and set the inlet pressure to be constant. At the outlet where the pipe opens to the outside air, fluid comes out at a constant rate. We may be tempted to measure that flow rate and compare it with, for example, a prediction assuming Poiseuille flow.

However, unbeknownst to you, inside the black box I welded two fittings; one is a constrictor that decreases the diameter and then a second fitting that increases the diameter back to the original diameter. I guess I could be even more simple and insert an orifice rather than weld two fittings. Maybe I left the pipe diameter constant but roughened the interior pipe walls- you have no idea.

So it is a box with a balancing valve in it, of unknown setting. Fair enough.

But regardless of what I did, the inlet pressure is unchanged and the outlet pressure is unchanged. I suppose the volume flow (say, m^3/s) changes, so our predicted flow rate won't match. We would ascribe that to some sort of 'pressure loss' ('minor loss') within the hidden section of pipe, right?

The OP asked if there is a loss of kinetic energy or velocity through a fitting. Here, volume flow is reduced in comparison to Poiseuille flow due to a pressure loss. It would seem that would relate to loss of kinetic energy and/or velocity, right?

You're making the same mistake the OP did, and maybe it means that your previous question was asked backwards. But I'll come back to that.

First off, if the flow rate into and out of the box are the same and the pressure on the inlet and outlet of the box are the same, then you haven't lost anything! To "lose" means you don't have something anymore. To "lose" it in the box means it's still in the box! So what did we lose in the box? We lost *pressure*. The *pressure* at the inlet of the box is higher than the pressure at the outlet of the box.

Now, depending on what is causing the flow to happen in the first place, odds are good you also will have a lower flow with the second scenario than the first, but that isn't necessarily true. We don't know enough about the scenarios (namely, what is causing the flow to happen -- a pump? under control? What kind) to say whether the flow rates will be different.

So you are indeed mixing and matching what happens in one scenario vs what happens between two different scenarios. "Loss" is not referring to how one scenario differs from another, it is about what happens in a particular scenario.

And back to "manifest". You asked how pressure drop manifests, but I think that's the answer to the question you really needed to ask: how does energy loss manifest? Via permanent pressure drop (loss) across a fitting.

 

[Chestermiller]

So how can I explain that the difference between Darcy losses and experimentally calculated, is consistently increasing in a exponential-like way?
The set up is just a pump that pumps water inside a pipe with bends and the end of the pipe let's the water flow out freely.

The only factor I found that causes such systematic error is velocity!

In my judgment, based on your prior responses, the reason the calculation does not match the observations adequately is that you did the calculation incorrectly. Since you have not seen fit to share the details of the system design and operating parameters with us, we can't help you find your error. It seems to me you are just continuing to waste everyone's time.

 

[russ_watters]

Or, like I said in your other thread, maybe it's a feature, not a bug; two similar quadratic functions with different constants will diverge.

But as said, unless we see the data and calculations, we're only guessing at what the problem might be.

 

[Andy Resnick]

So it is a box with a balancing valve in it, of unknown setting. Fair enough.

You're making the same mistake the OP did, and maybe it means that your previous question was asked backwards. But I'll come back to that.

First off, if the flow rate into and out of the box are the same and the pressure on the inlet and outlet of the box are the same, then you haven't lost anything! To "lose" means you don't have something anymore. To "lose" it in the box means it's still in the box! So what did we lose in the box? We lost *pressure*. The *pressure* at the inlet of the box is higher than the pressure at the outlet of the box.

Now, depending on what is causing the flow to happen in the first place, odds are good you also will have a lower flow with the second scenario than the first, but that isn't necessarily true. We don't know enough about the scenarios (namely, what is causing the flow to happen -- a pump? under control? What kind) to say whether the flow rates will be different.

So you are indeed mixing and matching what happens in one scenario vs what happens between two different scenarios. "Loss" is not referring to how one scenario differs from another, it is about what happens in a particular scenario.

And back to "manifest". You asked how pressure drop manifests, but I think that's the answer to the question you really needed to ask: how does energy loss manifest? Via permanent pressure drop (loss) across a fitting.

I wonder if we are speaking past each other.

As I set up the problem, it should have been clear that I am considering pressure-driven flow: Poiseuille flow applies for axisymmetric laminar viscous flow through a straight circular impermable tube of radius R, resulting in a flow rate Q= π/8μ dP/dz R⁴. I can hold the inlet and outlet pressures constant.

As for: "So you are indeed mixing and matching what happens in one scenario vs what happens between two different scenarios. "Loss" is not referring to how one scenario differs from another, it is about what happens in a particular scenario.", I guess that makes sense. For example, in Poiseuille flow there is a pressure loss between inlet and outlet. But isn't the point of "head loss coefficients" to compare to some ideal configuration?

In the end, we seem to be agreeing with each other: fittings can introduce a permanent pressure drop. There should also be a loss of momentum in addition to energy, because there is an interaction between the fluid and fitting, resulting in a transfer of energy and momentum from the fluid to the fitting.

 

[Andy Resnick]

If that's what he meant then it illustrates the importance of being as exact as possible when describing you problem. We all believed he was asking whether a fitting or other loss will slow down the downstream flow relative to the upstream flow without area change

I don't know about 'we ALL believed', but I agree the OP was ambiguous.

 

[Chestermiller]

The OP has last been on site at 7am today, and still has not provided us either a diagram of the system or his calculation. How do responders feel about closing this thread?

 

[physea]

The OP has last been on site at 7am today, and still has not provided us either a diagram of the system or his calculation. How do responders feel about closing this thread?

wow

 

[russ_watters]

I wonder if we are speaking past each other.

I don't think so. I'll try to focus:

In the end, we seem to be agreeing with each other: fittings can introduce a permanent pressure drop. There should also be a loss of momentum in addition to energy, because there is an interaction between the fluid and fitting, resulting in a transfer of energy and momentum from the fluid to the fitting.

This is wrong: there is no [permanent] loss of momentum, only a loss of energy across the fitting.

 

[russ_watters]

The OP has last been on site at 7am today, and still has not provided us either a diagram of the system or his calculation. How do responders feel about closing this thread?

I'd agree. And he can pm you (Or me) the requested info to reopen the thread with.