Non-newtonian Pressure Drop 1

[777gne]

Now that I have my pressure drop spreadsheet all polished up I get thrown a curve ball; I got handed some rheology data for a several shear thinning fluids. I'm a little rusty with non-newtonian fluids. Here's what I think I know, I'm hoping someone can give me a nudge...

The data that I have is:
Shear Rate in Sec^-1
Apparent Viscosity in centipoise
% Soilds
Specific Gravity
Specific Heat
Thermal Conductivity

...and a legend that reads n = 30,210*y^-.6310

This is a good start - at least I can determine the apparent viscosity at a given shear rate. I don't know if I have enough information to calculate pressure drops in piping though(?)

I believe that I need the constant of proportionality, usually shown as K, or the shear stress so I can calculate K. Can anyone verify this? Is this something that the lab who did the rheology would be expected to have access to?

The pressure drop equation that I have is:

delta P = (3 * n + 1 / n)^n * (Q / pi * r^3)^n * (2 * L * K / r)


- Is there a way to determine K from the info that I do have?
- Is there another means of determining pressure drop using the info that I do have?

 

[FACS]

Maybe look in here for a hint.

http://www.glossary.oilfield.slb.com/Display.cfm?Term=power-law fluid

Research the "Power Law Model" and "Bingham-Plastic Model". This is the only thing I remember from my Rheology days and friction loss in Non-Newtonian fluids

Charlie

http://www.facsco.com

 

[777gne]

Well, upon closer inspection I made a mess out of this one (I really wish we could edit our posts in this forum!).

The apparent viscosity is represented by the greek letter Eta (I'll use N). Lowercase "n" is what we are using in the pressure drop equation and is called the "flow index". Yet another variable I'm missing. Ah, such fun with non-newtonian fluids. I guess I need to give the lab a call and admit my ignorance.


Still open to comments, or advice from those with experience in this area, I reposted my known data and equations below with the corrected info.

The data that I have is:
Shear Rate in Sec^-1
Apparent Viscosity (N) in centipoise
% Soilds
Specific Gravity
Specific Heat
Thermal Conductivity

legend states that N = 30,210*y^-.6310


The pressure drop equation that I have is:
delta P = (3 * n + 1 / n)^n * (Q / pi * r^3)^n * (2 * L * K / r)

where:
n = flow index (hopefully the lab can tell me this)
Q = flow rate m^3/s
r = radius of pipe m
L = length of pipe m
K = proportionality coefficient units of Pa s^n (hoping the lab can tell me this)


I also have the following equation which I believe is related to the Power Law Expression:
sigma = K * y^n

where:
sigma = shear stress

 

[777gne]

Thanks Charlie. I just saw your post. I'm in deeper than I thought here, so getting ready to make a phone call.

Side Note:
For interested parties who are reading I found what appears to be an in-depth book on rheology here:

http://www.egr.msu.edu/~steffe/freebook/STEFFE.pdf

 

[Latexman]

The problem is n is being used for viscosity AND the flow index, which is traditional. It's confusing.



Good luck,
Latexman

 

[Latexman]

[ν] = viscosity
[τ] = shear stress
[γ] = shear rate
k = proportionality coefficient
n = flow index

Definition of viscosity
[ν] = [τ]/ [γ]

Definition of a power law fluid
[τ] = k[γ]ⁿ

Substitute bottom equation into top equation
[ν] = k[γ]ⁿ/ [γ]

Simplify
[ν] = k[γ]^n-1

So you had k and n all along.

[ν] = 30,210[γ]^-0.631

k = 30,210

n-1 = -0.631 and n = 0.369


Good luck,
Latexman

 

[Latexman]

A word of caution is needed also. Be extremely diligent that the correct units are used in the equations and they lead to the dimensions you are looking for. The power in the power law and being unfamiliar with the non-Newtonian pressure drop equation makes this a little tricky.

Good luck,
Latexman

 

[777gne]

Latexman -

Thank you!


...need some time to absorb and run a few numbers. Will post results/questions later...

 

[777gne]

Latexman,

First, thank you for shinning some light on this subject. Your post made sense and I was able to find and cross reference everything you stated in my food engineering book.

I'm a little unsure about the units of the proportionality coefficient. My book states units of Pasⁿ, for example tomato paste K = 70 Pasⁿ. Am I to interpret this as K = 70ⁿ Pas? If not, what then??

One other question - have you ever see the pressure drop equation that I posted? It also came from the same book and states that it is "based on the power law expression to find pressure drops for laminar pipe flow". I'm getting some strange results.

Given:
v = 48,158 * ?^-.6331 Cp
v = 4,816 * ?^-.6331 Pa
k = 4,816 Pasⁿ (huh?)
n = .3669
Q = .0019 m³/s (30 GPM)
r = .03 m (radius of pipe)
L = 6.1 m (length of pipe, 20ft)

I end up with a pressure drop of 1.959 MPa! No way that can be correct so I'm doing something wrong (maybe as you cautioned I have issues with my units), or this equation is no good. I find it interesting that the shear rate or apparent viscosity doesn't factor into the equation for pressure drop (?).

It's been a really long week, maybe I'm missing the obvious.

 

[Latexman]

I'll respond tomorrow from the office where I have my handy-dandy, power law, pressure drop spreadsheet.

Good luck,
Latexman

 

[BigInch]

Latexman,

If you would like to post your spreadsheet on

http://www.rapidshare.com

(free), you could make a few people very happy.

Just a thought.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[Latexman]

777gne,

250 lb/min (30 gpm and s.g. = 1)
62.4 lb/ft^3
48158 cP.sec^(n-1)
0.3669
2.362 in. diameter
20 ft. length

dP = 16.8 psi = 116 KPa = 0.116 MPa

Good luck,
Latexman

 

[Latexman]

dP = ((3.n+1)/n)^n * (Q/pi/r^3)^n * (2.L.K/r)


Good luck,
Latexman

 

[777gne]

Took another stab at it today... still scratching head. I'm unable to get anything close to your result above.

I would love to take a look at the spread sheet you're using to see how it is calculating the result you posted above and especially what units and conversions it's performing.

 

[777gne]

Also a few notes, and I'm not trying to point any fingers here (did I mention how much I appreciate the help?)... but.. I think you have possibly posted a couple of errors above.

1) The flow rate needs to be volumetric for the units to work out.

2) By definition 1000 cp = 1 Pa.s, so I don't think we want to use cp.s for our K value, just cp.

3) You state the units of K are viscosity^(n-1). The material I have states the units to be viscosity^n. This is a point I'm still not clear on. Let's say K = 1000 cp^.3, is this to say that K = 1000^.3 cp? It seems unusual the way it is written. (or by your method above 1000^(.3-1))

I have a good feeling that #3 is causing me all the grief here.

 

[Latexman]

Did you notice the difference between your equation and mine?

Mine: dP = [red]((3.n+1)[/red]/n)^n * (Q/pi[red]/r^3[/red])^n * (2.L.K/r)

Yours: dP = [red](3.n+1/n)[/red]^n * (Q/pi[red].r^3[/red])^n * (2.L.K/r)

I don't know how you are actually evaluating your equation, but it's wrong the way it's written using the traditional math operator hierarchy (sp?).


Good luck,
Latexman

 

[Latexman]

1) It's just a conversion difference when density is known. 30 gpm = 250 lb/min = 15,000 lb/hr @ s.g = 1. I assumed s.g. = 1 since I didn't see it specified.

2) I didn't use cP.s as the units for k. I used cP.sec^(n-1). These units are convenient to me since the apparant viscosity I get is in cP. My program takes these units and does the proper conversion.

3) The required units are viscosity x time^(n-1). Shear rate is almost always in units of 1/second. You left out the time unit.

Good luck,
Latexman

 

[Latexman]

Are you working in metric or King George (English/U.S.) units?

Good luck,
Latexman

 

[Latexman]

1 cP = 0.01 g/cm/sec or dyne.sec/cm²

1 cP = 0.000672 lb_m/ft/sec or 0.0000209 lb_f.sec/ft²

Good luck,
Latexman

 

[777gne]

I didn't write it properly in the forum but order of ops I'm using is standard. I went back and solved it again using SI units and was able to get the same result that you did. I made an decimal point error converting Cp to Pa.s (see my post far above). Which was part of the problem. The other part was not knowing what to do with units of Pa.s^n. I ignored ^n. If you attempt to solve this by hand and write out your units there is no way to eliminate it.

I still do not understand why the units of K are written this way. Do you know why the units are expressed in this way?


Thank you for your help with this problem.



p.s. Did you ever find a solution to the question that you asked in this thread:

http://eng-tips.com/viewthread.cfm?qid=140161&page=1