Non-Newtonian Pressure Drop (revisited) 1

[777gne]

After quite a bit of help from Latexman last year (this thread:

http://eng-tips.com/viewthread.cfm?qid=177111&page=1)

I have been calculating frictional losses for power law fluids rather successfully. Most of the calculations I have double checked by a fluids lab that I work with and my calculated pressure drop is typically 10% higher than theirs. I don't know the exact method they're using but have to assume the approach is different otherwise I've been getting lucky!

I've been using the following formula:
delta P = [(3 * n + 1 / n)^n * (Q / pi * r^3)^n * (2 * L * K / r)] / 144

where:
n = flow index (unitless) - obtained from rheology test
Q = flow rate ft^3/s
r = radius of pipe ft
L = length of pipe ft
K = proportionality coefficient (unitless) - obtained from rheology test
delta P = frictional loss psi


As I mentioned before it's been smooth sailing the handful of times I've used this forumula, however I ran into a bump in the road recently. The rheology data usually gives the apparent viscosity in a format such as: v = 100,000*y^-.0500 (just an example) which is common to express v in terms of the expected shear rate (probably not news to anyone reading this).

As Latexman pointed out, by definition the exponent above reflects n-1, so in the example above n = .95. The problem I recently encountered involves a negative value of n. From the rheology test v = 394,623*y^-1.0062. So I'm dealing with a Dilitant fluid now and not sure how to approach frictional loss calcs.

Any insight/advice is most appreciated.


I do have reason to believe that the lab may have possibly made an error on the analysis, but I'm going to assume for now that they have it right.

 

[Latexman]

The frictional loss calcs of a dilatant fluid are handled mathmatically the same as a shear thinning fluid. The behavior is different, but the equations are the same.

Is the dispersed phase to continuous phase ratio higher for this sample, or are the phases reversed (i.e. water in oil vs. oil in water)? Its common for these things to lead to dilatant behavior.

Dilatant fluids resist flow more the faster you push them. I've seen some dilatant fluids get as thick as a brick when pushed too hard.

Good luck,
Latexman

 

[IRstuff]

So, like corn starch and water. That's always a trip. Push slowly, and you sink. Push hard, and it's a solid.

TTFN

FAQ731-376

 

[btrueblood]

...or put a pan of it on a speaker cone and watch it stand up. No kidding:

http://www.youtube.com/watch?v=vCHPo3EA7oE

 

[BigInch]

Nature's alternative to quicksand.

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[harvey13]

777gne

I have tried to follow your conversation with Latexman, but there are too many gaps for me in the development of this equation and the thread from last year. Latexman had a different variation of your equation that I think was due to the units of the proportionality coefficient.

I haven’t been able to work last years calculations to get the units right. Could you use this equation with the values from last year to calculate the 16.8 psig dp? This would help a lot. I think you had these values:
Q = 250 lb/min = 0.0668 cu ft/min
n = 0.3669
r = 0.0984 ft
L = 20 ft
K = 30,210, I could not follow how the discussion of the units for this was resolved.

Thanks
harvey13

 

[gne77]

I apologize for the non-response. For whatever reason I have not been able to log in on the site. I contacted the "admin" several times with no response what so ever! Ultimately I had to create a new user name (formerly 777gne). Hopefully this thread can come back to life and earn more than 2 stars I appreciate the responses and I need to revisit my dilitant fluid application to see if the result makes sense.


Harvey: I can see why you're confused.
1. Reading that old thread gives me a headache
2. I have made a typo above - K is not unitless, sorry copy & paste error.

To first address K it WILL be given with units of Pa.s^[n-1] or cp.s^[n-1] (possibly other units). In the problem you mentioned I had chose to convert to lb.s/ft^2. Now onto the example problem. The following are the known's (you have a couple of mistakes there)...

Equation (pressure drop in psi):
[(3 * n + 1 / n)^n * (Q / pi * r^3)^n * (2 * L * K / r)] / 144

Known:
sg = 1
Q = 30 gpm = 250 lb/min = 0.0668 cu ft/sec
n = 0.3669
r = 0.0984 ft
L = 20 ft
K = 48,158 cp.s^[n-1] = .996 lb.s/ft^2

First Part:
(3 * .3669 + 1 / .3669)^.3669 = 1.897

Second Part:
(.0668 / pi * .0984^3)^.3669 = 3.125

Third Part:
(2 * 20 * .996 / .0984) = 405

Put it all together:
[1.897 * 3.125 * 409] / 144 = 16.7 psi


If you put the units in it does work out; give it a try. The only thing I never figured out is the difference between Latexman's second part and my second part... some mysteries remain unsolved


Like I mentioned this has been working for me well enough, I have always been on the conservative side by 5-10%. I have since come across some more literature on the subject which discuss an "Generalized Reynolds Number" and how this comes into play. One document I purchased goes even further to illustrate issues with calculations of non-newtonian pressure losses through that tend to be overly conservative. I need to dig deeper there.

I am interested if Latexman (or others) have any input on the Generalized Reynolds number.

 

[Latexman]

gne77,

Back in the first post you said "I didn't write it properly in the forum but order of ops I'm using is standard", and since we got the same numbers, I figured the mystery was solved.

I wrote the second term as (Q/pi/r^3)^n . Adding paraentheses for clarity, this is the same as (Q/(pi*r^3))^n.

You wrote the second term as (Q / pi * r^3)^n. Is this the same as (Q/(pi*r^3))^n?

Good luck,
Latexman

 

[gne77]

Latexman-

Yep I did it again. My excel formula is in the format: (Q/(pi*r^3))^n, which as you pointed out is not the same as (Q / pi * r^3)^n. Its also pretty clear now why yours is written slightly different. It would have helped if I wrote it correctly to begin with.

In case anyone missed it, the correct format for the second part of the equation is:
(Q/(pi*r^3))^n

or

(Q/pi/r^3)^n

 

[Latexman]

Agreed!

Good luck,
Latexman

 

[gne77]

Thank you for all of your input on this subject; it's been very valuable to me.

 

[harvey13]

Gne77

I’ve been away from “my” computer for over a month, and since I’m old, I couldn’t remember by password and I wasn’t able to respond until today, so a belated thanks for the clarification.

In the example from the 7JUL post I have a question. You have the “First Part” written as:

(3 * .3669 + 1/.3669)^.3669. I get this to equal 1.636.

To get the value of 1.897 for the “First Part” I have to add two parentheses to this part of the equation:

((3 * .3669 + 1)/ .3669)^.3669 = 1.897. Is this correct?

With Latexman’s comments from 8JUL about the “Second Part” I now have the complete equation as:

[{((3 * n + 1)/n)^n} * {(Q/(pi * r^3))^n} * {(2 * L * K/r)}]/144

Thanks again,
Harvey

 

[Ninja1980]

I’ve been following this thread with some interest.

The equations for pressure drop are all for the laminar flow regime. How is pressure drop calculated for turbulent flow??

 

[Latexman]

That question is off topic, I'd recommend starting a new thread, but first you should use the search functionallity to see if your question has already been answered in the archives.

Good luck,
Latexman