Non-Newtonian fluid ratholing 1

Mechanical Farmer · Mar 12, 2025

Hello,

Long time listener, first time caller.

I have been diving into the behavior of non-Newtonian (shear thinning, specifically) and believe that I am experiencing rat-holing in a Pseudoplastic flow.

My question is this: Provided this is a correct assumption, is there a way to determine the thickness of the "stagnant" layer on the pipe walls. Is there a way to determine a radial pressure gradient from the centerline of the pipe to the pipe walls?

Mechanical Farmer · Wednesday at 7:59 PM

TiCl4 said:
Also, if you do the steps above, make sure to graph velocity, v, versus r (distance from center). This will give you a velocity profile in your pipe. Give the extreme shear-thinning nature here, I would expect to see velocity drop off from the center and very quickly approach 0.

You won't ever have "stagnant" material in the pipe, but you'll get an estimate of how long each "layer" at differential radius "r1-r2" will stay in the pipe. If flow is down in the mm/min range near the edge, you could be looking at a lifespan of weeks or months in the pipe!

let me get to work.

Mechanical Farmer · Wednesday at 8:04 PM

TiCl4 said:
OP, since you have a rheology profile for your liquid, follow the below steps to get an equation that describes your fluid. Based on your graph, I would say your material is simply shear thinning, not a Bingham plastic, though starting out at a really high 0-shear of >1,000,000 cP.

Ostwald-de Waele equation:

τv = K [du/dy]N Eqn 1

Tv = Shear stress, N/m2

K = Flow consistency index

[du/dy] = shear rate, 1/s

N = Flow Behavior Index

Apparent Viscosity

Ƞ = τv / [du/dy] Eqn 2

Ƞ = apparent viscosity, Pa*s

Ƞ = K [du/dy]N / [du/dy] Eqn 3. Combines Eq 1 and 2.

Ƞ = K [du/dy]N-1 Eqn 4.

Linearize by log function.

Log[Ƞ] = (N-1)*Log( [du/dy] ) + Log(K) Eqn 5.

y = a * x + b,

Used excel to graph Log[Ƞ] vs Log( [du/dy] ), added trendline to get parameters, where a = (N-1) and b = Log(K)

Note that shear rate is described as the differential in fluid velocity along an axis (for a pipe, from the center to the wall). Your fluid will be experiencing the same "shear" from inside to outside. However, due to the extremely high viscosity, your will have an extreme drop in velocity as you move towards the wall. A Bingham fluid will develop a "true" zero-velocity plug that will "rathole", but this high viscosity fluid may effectively function the same way, with flow near the wall of the pipe being down in the mm/min range.

To estimate flow profile, you'll need K and n, which you found above. Going off of the developed shell balance from (https://scholarworks.utrgv.edu/cgi/viewcontent.cgi?article=1410&context=mss_fac), you can see your velocity profile using the equation below and your K/n parameters.

Do the math and report back!
View attachment 6774

also, the graph shared was for a "worst case" material composition the "best case" is a bit different.

pierreick · Wednesday at 9:00 PM

hi,
You may want to read this paper and watch this video.
Book to support the video: Introduction to chemical engineering Fluid mechanics by William M. Deen

Pierre

TiCl4 · Wednesday at 9:39 PM

Notation correction: In the last part of my prior post, the picture for velocity profile includes mu and K. The author of that paper denoted K as the dP/dz, and used mu for what I called "K". Just an FYI, as the math won't math out otherwise. If you know what pressure drop you are getting through your pipeline, you can use that as your K value.

Regardless, your velocity profile will likely look fairly flat if you are starting a 1 million cP. You will have a very slow moving fluid near the wall, and in the center you will have a relatively uniform plug flow of liquid. Look at the profile in the reference I suggested.

Again, from the same reference I linked above. As they increase mu (my K in the calculation Ƞ = K [du/dy]^(N-1) above, meaning as they increase viscosity), the profile becomes flatter and flatter. Notice their mu is 0.35 - 0.5, while I expect your mu (my K) to be quite a bit higher.

Functionally, this means that as you increase pressure drop by pumping more volume/minute, the higher the dispersity between center velocity and wall velocity. The slower you pump, the more evenly matched the outer and inner portions of liquid flow will be, and the residence time of the material "layers" in the pipe will begin to converge.

Regardless, you have a yield stress of between 0.5 (best case) and 15-ish Pa, worst case. I don't think your layer of "solid" material is thick enough to worry about. This will functionally behave more like a typical shear-thinning liquid and not like a Bingham fluid.

Mechanical Farmer · Thursday at 1:53 PM

TiCl4 said:
OP, since you have a rheology profile for your liquid, follow the below steps to get an equation that describes your fluid. Based on your graph, I would say your material is simply shear thinning, not a Bingham plastic, though starting out at a really high 0-shear of >1,000,000 cP.

Ostwald-de Waele equation:

τv = K [du/dy]N Eqn 1

Tv = Shear stress, N/m2

K = Flow consistency index

[du/dy] = shear rate, 1/s

N = Flow Behavior Index

Apparent Viscosity

Ƞ = τv / [du/dy] Eqn 2

Ƞ = apparent viscosity, Pa*s

Ƞ = K [du/dy]N / [du/dy] Eqn 3. Combines Eq 1 and 2.

Ƞ = K [du/dy]N-1 Eqn 4.

Linearize by log function.

Log[Ƞ] = (N-1)*Log( [du/dy] ) + Log(K) Eqn 5.

y = a * x + b,

Used excel to graph Log[Ƞ] vs Log( [du/dy] ), added trendline to get parameters, where a = (N-1) and b = Log(K)

Note that shear rate is described as the differential in fluid velocity along an axis (for a pipe, from the center to the wall). Your fluid will be experiencing the same "shear" from inside to outside. However, due to the extremely high viscosity, your will have an extreme drop in velocity as you move towards the wall. A Bingham fluid will develop a "true" zero-velocity plug that will "rathole", but this high viscosity fluid may effectively function the same way, with flow near the wall of the pipe being down in the mm/min range.

To estimate flow profile, you'll need K and n, which you found above. Going off of the developed shell balance from (https://scholarworks.utrgv.edu/cgi/viewcontent.cgi?article=1410&context=mss_fac), you can see your velocity profile using the equation below and your K/n parameters. Note the author calls my "K" "mu" and uses K for dP/dz (i.e. pressure drop in Pa/meter)

Do the math and report back!
View attachment 6774

Not sure if I did this correctly.
Pipe size seems to have no effect on the shape of the graph, even when i mixed inches and meters the graph plotted the same, geometrically, just shifting magnitudes along the x.
Additionally, at smaller pipe sizes I am getting negative velocities. .

vo

v(r)

K,n,u, and R are taken from a separate worksheet.

Mechanical Farmer · Thursday at 2:03 PM

pierreick said:
hi,
You may want to read this paper and watch this video.
Book to support the video: Introduction to chemical engineering Fluid mechanics by William M. Deen

Pierre

Small World, I have actually been consulting with the man presenting this lecture in regard to some of these issues.

TiCl4 · Thursday at 3:32 PM

Farmer,

Look to see my note above (Notation correction: In the last part of my prior post, the picture for velocity profile includes mu and K. The author of that paper denoted K as the dP/dz, and used mu for what I called "K". Just an FYI, as the math won't math out otherwise. If you know what pressure drop you are getting through your pipeline, you can use that as your K value.)

What I call "K" the author calls mu, and they use pressure drop as K. You are using K as Pa*s, when it needs to be pressure drop along the pipe length (dP/dz). This may be the issue with the lack of "flattening" as your pipe size changes, and is likely the issue with the overall v0 being really small (.0005 m/s). For standard units, you'd take your pressure drop per meter in Pa/m, and your K value is 9.17. From the formula's perspective, it only has 9 Pascals of pressure drop per meter available to flow the material, meaning it goes very, very slowly. Try fixing K and report back. You can look at your pump discharge pressure and at your terminal pressure at the pipe exit to determine dP/meter.

Doing an evaluation on the velocity formulae, v0 can never be negative, as none of the parameters can be negative values.. Likewise, v can never be negative unless r/R is larger than 1. Check to make sure you are adjusting r to be smaller than R when lowering your pipe radius (R).

However, that profile looks essentially correct in shape when compared to the profiles suggested by the literature. You have a central plug flow of thinned material and an outer shell of very slow, viscous goop that stays behind.

Mechanical Farmer · Thursday at 6:51 PM

TiCl4 said:
Farmer,

Look to see my note above (Notation correction: In the last part of my prior post, the picture for velocity profile includes mu and K. The author of that paper denoted K as the dP/dz, and used mu for what I called "K". Just an FYI, as the math won't math out otherwise. If you know what pressure drop you are getting through your pipeline, you can use that as your K value.)

What I call "K" the author calls mu, and they use pressure drop as K. You are using K as Pa*s, when it needs to be pressure drop along the pipe length (dP/dz). This may be the issue with the lack of "flattening" as your pipe size changes, and is likely the issue with the overall v0 being really small (.0005 m/s). For standard units, you'd take your pressure drop per meter in Pa/m, and your K value is 9.17. From the formula's perspective, it only has 9 Pascals of pressure drop per meter available to flow the material, meaning it goes very, very slowly. Try fixing K and report back. You can look at your pump discharge pressure and at your terminal pressure at the pipe exit to determine dP/meter.

Doing an evaluation on the velocity formulae, v0 can never be negative, as none of the parameters can be negative values.. Likewise, v can never be negative unless r/R is larger than 1. Check to make sure you are adjusting r to be smaller than R when lowering your pipe radius (R).

However, that profile looks essentially correct in shape when compared to the profiles suggested by the literature. You have a central plug flow of thinned material and an outer shell of very slow, viscous goop that stays behind.

The profile remains the same. I think the issue is that my flow consistency should be changing as the shear rate changes. (the difference between 4" and 1" pipe is 2 orders of magnitude in reciprocal seconds.) Unfortunately when I run the numbers in the appropriate range or shear rate, i am getting a value of n<0 so i need to go back and confirm those calcs.

TiCl4 · Thursday at 10:42 PM

You have an actual pressure drop of 1.5 bar per meter of pipe?

dvd · 2025-03-21T01:34:41+0000

I have just skimmed through this, but have you ruled out thermal effects and phase transition?

pierreick · 2025-03-21T05:06:32+0000

I don't know why but your product looks like cellulose!
Pierre

Mechanical Farmer · 2025-03-21T14:14:47+0000

pierreick said:
I don't know why but your product looks like cellulose!
Pierre

can you elaborate in what way it looks like cellulose?
Are there any concerning implications here?

Mechanical Farmer · 2025-03-21T14:19:25+0000

dvd said:
I have just skimmed through this, but have you ruled out thermal effects and phase transition?

No, I haven't ruled that out. Stagnant wall conditions would contribute to hot spots. I have explored the idea that this may have a non-monotonic radial temperature gradient as well. There just so much I don't know/understand right

Mechanical Farmer · 2025-03-21T14:27:00+0000

TiCl4 said:
You have an actual pressure drop of 1.5 bar per meter of pipe?

in process, not likely. but pressure doesn't really have an effect on the shape of the profile.

Mechanical Farmer · 2025-03-21T14:51:26+0000

TiCl4 said:
You have an actual pressure drop of 1.5 bar per meter of pipe?

Should velocity be in terms of reciprocal seconds?
looking at v(r); n is dimensionless and r and R units are both meters and therefore cancel out, so v(r) is in the same units of vo
looking at vo; R is in terms of m outside of the brackets, inside of the brackets K/mu is (Pa/m)/(Pa-s) which reduced to 1/ms which reduces further to 1/s with R.

TiCl4 · 2025-03-21T15:25:07+0000

Farmer,

There were two points to the exercise. Firstly, to get an idea of your flow profile shape. I think it's pretty clear that you have, as you initially suspected, a "pluglike" flow of faster core material with a shell of slow/sluggish material.

Secondly, for your worst case material, it is entirely conceivable that you can open up a small sample valve on a tee in the pipe and get no flow due to the extreme viscosity. Even though there is a pressure differential between the open valve (atmosphere) and the liquid inside of the pipe, the pressure at that point might not be enough to overcome the cohesive forces of the liquid, which are quite large in high viscosity materials.

Your worst-case material does have a small amount of "Binghamness" with an initial shear stress required of 14 Pa. It's not much, but that, combined with the high viscosity, makes your observations of no flow out of sample ports plausible.

I want to bring back your earlier observation:

Mechanical Farmer said:
Shear would be highest at the walls. There would still be a zero-slip condition at the wall, but from there we would have a very rapid increase in fluid velocity with decreasing shear as we approach the centerline of the pipe. At some point between the pipe wall and centerline shear stress would drop below the yield stress and we would essentially be moving a solid plug along the center of the pipe.

Not quite, but close. All materials assume a zero-slip at the wall. However, most materials will have a steady increase of fluid velocity towards the center. A true Bingham fluid will have a shell of immobile material, with velocity 0 m/s, and with a layer thickness determined by the required shear to begin flow. After that is achieved, you would have Your "worst case" fluid looks to be of this type, though likely with a small immobile layer because 14 Pa is not a lot of stress to overcome. Your "best case" fluid does not appear to be a Bingham fluid, so the profile should match quite closely in shape with what you have calculated. So your two fluids should look similar to below, with the "worst-case" fluid being the orange profile. The area of zero velocity is likely much smaller due than pictured, but I just made this for a visual explanation of a difference between a thick, shear-thinning liquid and a similarly thick, shear-thinning Bingham liquid.

Mechanical Farmer · 2025-03-21T20:23:47+0000

TiCl4 said:
Farmer,

There were two points to the exercise. Firstly, to get an idea of your flow profile shape. I think it's pretty clear that you have, as you initially suspected, a "pluglike" flow of faster core material with a shell of slow/sluggish material.

Secondly, for your worst case material, it is entirely conceivable that you can open up a small sample valve on a tee in the pipe and get no flow due to the extreme viscosity. Even though there is a pressure differential between the open valve (atmosphere) and the liquid inside of the pipe, the pressure at that point might not be enough to overcome the cohesive forces of the liquid, which are quite large in high viscosity materials.

Your worst-case material does have a small amount of "Binghamness" with an initial shear stress required of 14 Pa. It's not much, but that, combined with the high viscosity, makes your observations of no flow out of sample ports plausible.

I want to bring back your earlier observation:

Not quite, but close. All materials assume a zero-slip at the wall. However, most materials will have a steady increase of fluid velocity towards the center. A true Bingham fluid will have a shell of immobile material, with velocity 0 m/s, and with a layer thickness determined by the required shear to begin flow. After that is achieved, you would have Your "worst case" fluid looks to be of this type, though likely with a small immobile layer because 14 Pa is not a lot of stress to overcome. Your "best case" fluid does not appear to be a Bingham fluid, so the profile should match quite closely in shape with what you have calculated. So your two fluids should look similar to below, with the "worst-case" fluid being the orange profile. The area of zero velocity is likely much smaller due than pictured, but I just made this for a visual explanation of a difference between a thick, shear-thinning liquid and a similarly thick, shear-thinning Bingham liquid.

View attachment 6903

A couple more questions here:

1: I am struggling to understand how I haven't overcome the yield stress at the pipe wall when shear is highest at the walls.

2: I assume you are making judgements about the fluid types the best and worst case based on the rheology curves. What are you seeing to make those determinations? I would have guessed the opposite if anything with the best case having to overcome a shear to induce thinning.

TiCl4 · 2025-03-21T21:06:54+0000

Best case has a near 0-0 intercept on the green line (stress-shear curve). That tells me it's a standard shear thinning fluid (shear thinning due to the clear viscosity drop with shear rate. The worst case has a non-zero, but still small stress intercept near 15 ish Pa (this is called the yield stress, or the minimum required stress before the fluid begins to flow. This indicates some level of Binghamness in the fluid, but not much. For comparison, Google suggests yield stress for toothpaste is in the 55-200 Pa range.

A few explanations for why you don't see flow when opening your valve with material in the pipe.
1. Is there pressure on the line? If no pressure, you certainly won't see any flow.
2. "Old" or "aged" material. Due to the very high viscosity, material at the walls will be in the pipes for quite some time. If you have heat tracing on the line or if your material naturally degrades at ambient temp, could it be "aging" and hardening over time?

pierreick · 2025-03-22T00:06:21+0000

Mechanical Farmer said:
can you elaborate in what way it looks like cellulose?
Are there any concerning implications here?

Rheology of Microfibrillated Cellulose Suspensions in Pressure-Driven Flow

Rheology of Microfibrillated Cellulose (MFC) suspensions is useful for designing equipment to transport, mix, or process them. Pressure-driven flow behavior is particularly important for MFC suspensions if they are to be pumped, extruded or coated. Herein, we report use of slot and pipe...

www.degruyter.com

pierreick · 2025-03-22T11:45:23+0000

Why don't you tell us what your product is?
Pierre

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Non-Newtonian fluid ratholing 1

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