Equipment pressure drop extrapolation 4

[sheiko]

Hello all,

First let me present myself: i am a french junior process engineer, so please be indulgent with my english.

My questions deals with the better way to extrapolate equipment/instruments pressure drop from known conditions.

Nomenclature:
hf0 and hf are the known and unknown friction heads of equipment,
q0 and q the known and unknown flowrates,
F0 and F the known and unknown Darcy friction factors.
DP100 the known linear pressure drop in feet
dp100 the new linear pressure drop in feet

Reference 1:
The author says that the resistance coefficient K from Crane for fittings and valves is independant of Reynolds number and K=Fturb*(L/D). As a result, using the equivalent length method by summing the straight pipe length and the total equivalent length of fittings with the same friction factor is not rigorous as there is a higher degree of turbulence in the fittings than in the pipe.
However, at the end of the article the author mentions new correlations like the one proposed by Darby that state that the resistance coefficient K varies with the Reynolds number and that is destined to become the new standard.
These two observations seem in disagreement with each other. So, is K a constant as for Crane or a function of the Reynolds number as for Darby?

Reference 2:
Perry 1997 6-16 shows that f=(D/4L)*K, and Perry's 1997 6-17 notes that K for fittings and valves is stable at Re from 2000 to 500 and then increases rapidly as Re decreases below 500.
This observation seems to corroborates the fact that K varies with the Reynolds number.

Now let´s go to the point (equipment/instruments pressure drop extrapolations):

Reference 3:
In this article, it is said that we can safely extrapolate equipment pressure drop from known conditions (for 500<Re<2100 and Re>5000) by: hf=hf0*(q/q0)^2. Obviously this relationship is based on hf=K*(v^2/2g) with K being a constant.
Is this method valid if K depends on the Reynold number as Darby stated?

Reference 4:
In this article, the author uses the equivalent length method: from known hf0 and operating conditions, he calculates a Leq at a selected pipe size D by Leq=100*(hf0/DP100). Then he calculates dp100 with the new operating conditions and determinates hf by hf=dp100*(Leq/100).
If we consider:
DP100=F0*(100/D)*(v0^2/2g)
dp100=F*(100/D)*(v^2/2g)
Then, hf=hf0*(dp100/DP100) <--> hf=hf0*(F/F0)*(q/q0)^2
Do you agree with this method? For me it seems very convenient as we can use it in all flow regime. I would like your point of view as experienced engineers.

Thanking you in advance

References:
1/

http://www.cheresources.com/eqlength.shtml

2/ Perry's Handbook 1997
3/ Anthony, James, Pumping System Head Estimation, Chemical Engineering, February 2005
4/ Yu, Frank, A simple way to estimate equipment pressure drops, Hydrocarbon Processing, August 2005


Kind regards

"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[katmar]

sheiko, thanks for uploading the two scans. I have read them and my opinions remain basically as before.

In the article you termed "Reference 3" James Anthony has done a good job of summarizing the more modern work done in the field. It's not quite up to the standard of the 1968 and 1978 articles by Larry Simpson, but nevertheless a very valuable resource and of course much more relevant than Simpson in the areas where there has been progress.

I'm afraid I cannot say the same of your "Reference 4". I do not see much merit in introducing the hypothetical pipes of arbitrary diameter over simply assuming that the pressure drop varies with the square of the flow for turbulent flow or linearly with flow for laminar flow.

For example he reports that the measured pressure drop on the tube side in his E-1 exchanger is 40 psi. If we predict the pressure drop at 90% flow as 40 x 0.9² we get 32.4 psi. The articles gives results varying from 32.78 to 32.06 psi, depending on the choice of pipe size. Similarly at 50% of flow I would predict 40 x 0.5² and get 10.0 psi. This compares with the article's range of 10.41 to 9.21 psi. Are these numbers really different from each other?

Dr. F Yu and I have a fundamental difference in our approach to engineering calculations. On page 58 he reports the equivalent length for a heat exchanger as 45,848.1 ft and in Point 4 of his Conclusion he states "It is suggested to use at least four significant digits in DP100 to increase the accuracy of equipment pressure drop estimation". Two or three digits are all I need. But then I grew up using a slide rule.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[sheiko]

Again, thank you for your precise answers.

I have appreciated your logical appraoch and will follow your advice for sure!

Regards

"We don't believe things because they are true; Things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[sheiko]

Dear Harvey,

Now taking the risk to deviate from initial discussion (equipment/instruments DP) and to stretch your patience to the limit, i would like to clarify the following (concerning fittings):

In thread378-173164: Crane 410 fittings, it has been written:

"if you want to express the resistance of a fitting in terms of the equivalent length (i.e. L/D) instead of K then you have to calculate

L/D = K/f

and since Crane express their K's as (Constant) x fT you would get

L/D = (Constant) x fT / f

If f is evaluated at the same conditions as fT then of course

L/D = (Constant)"


From my (growing) understanding i would rather say:

L/D = K / f = Kcrane / fT, with L/D being a constant WHATEVER the flow regime (laminar and turbulent).
Then, K = f*(L/D) = f * (Kcrane / fT).

Indeed, if, as you wrote: L/D = (Constant) x fT / f,
then, K = f*(L/D) = (Constant) x fT...a constant! which is indeed not...









"We don't believe things because they are true; Things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[katmar]

Before I address your questions there are two observations I must make.

Firstly, the equivalent length of a fitting is not an absolute constant. For a given class of piping (eg commercial steel) L/D is relatively constant as the flow regime (i.e. Reynolds No) varies and also as the pipe size varies. But it is NOT constant as the pipe roughness varies. A very smooth plastic elbow will have almost exactly the same pressure drop as a rough steel elbow of EXACTLY the same dimensions and subject to the same flow because the pressure drop is not due to the surface roughness. But a metre of smooth plastic pipe will have a lower pressure drop than a metre of rough steel pipe of the same diameter and carrying the same flow. This means that a longer length of plastic pipe would be required to give a pressure drop equivalent to the elbow than would be required of the steel pipe. The result of this is that L/D is higher for a fitting in smooth piping than it is for one in rough piping (but not because of the roughness of the fitting).

Secondly, although I have pointed out some weaknesses in the Crane system, it is still an excellent method for calculating the pressure drop in steel pipe under fully turbulent flow conditions - which just happens to cover about 95% of the world's piping problems. In particular the Crane system is good at predicting how the K value varies with pipe size.

Ok, now let's look at your questions.

sheiko wrote
L/D = K / f = Kcrane / fT, with L/D being a constant WHATEVER the flow regime (laminar and turbulent).
Then, K = f*(L/D) = f * (Kcrane / fT).

Yes, mathematically this relation is correct but we must look at what you are actually trying to do. If we say (Kcrane/fT) is a constant then what you are trying to do is to calculate K as f varies - and this is a reasonable requirement to have. We need to look at why f is varying. If we say f is less because this is a smooth pipe and therefore K will be less then we have made a mistake because the math assumes that L/D is constant and as I said above this is not true for smooth pipe. So you cannot predict a new K for a smooth pipe from this relationship. And of course this applies to any roughness different from steel pipe.

f could also vary because either the flowrate or the physical properties have changed. These changes would vary Re and we know that as Re varies K can vary too. As Re decreases f will generally increase, and from Darby's work we know that decreasing Re also increases K. So we have both f and K increasing together. But we must be careful not to confuse cause and effect. Both f and K are varying because of Re changing - the variation in f is not causing the variation in K. Using your relationship may give you a reasonable answer, but it is not based on good engineering or logic. f and K may not always vary in exact proportion to each other.

sheiko also wrote
Indeed, if, as you wrote: L/D = (Constant) x fT / f,
then, K = f*(L/D) = (Constant) x fT...a constant! which is indeed not...

Actually it is - well, almost. For a given type of fitting what I called (Constant) is the old Crane L/D data and this is close to constant for steel pipe. And fT is constant (for a given pipe size) because it is defined to be at a very specific flow condition. This makes K a constant for a given pipe size and for a given flow condition (i.e. fully turbulent) and this is what has made the Crane method so popular and indeed successful. You can vary the flow quite widely and still have fully turbulent flow - and therefore a constant K.

The Crane method successfully models the change in K with pipe size by linking it to the change in fT (again, with pipe size). But they made it look like the change in fT was the cause of the change in K when in fact it was just a lucky break that they varied at approximately the same rate. And this linking of K to f is what got me so hot under the collar before because it makes people think that they can calculate K for any conceivable condition provided they can calculate f - when in fact f has got nothing to do with the pressure drop in a fitting.

regards
Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[sheiko]

Thank you all,

In an attempt to sum it up,

- Concerning fittings and valves:

L/D = K/f = Kcrane/fT, knowing that:

1/ f varies with Reynolds number, pipe roughness and size
2/ K varies with Reynolds number and pipe size
3/ L/D varies with pipe roughness
4/ fT varies with pipe roughness and size
5/ Kcrane varies with pipe size, as Kcrane=fT*(L/D).
6/ in about of 95% of the cases, f=fT and then, K = Kcrane.

- Concerning equipment and instruments friction head extrapolation:

We should use: hf=hf0*(q/q0)^2, knowing that it is valid if:

1/ We extrapolate within one of the regions: 500<Re<2100 or Re>5000 (as there is little variation of K following Perry's 1997 6-17)
2/ Reference data and operating point are in the same flow regime (500<Re<2100 or Re>5000)
3/ If not possible (to get reference data in the same flow regime than operating point), get an estimate or a reference data at the operating point flow regime (from manufacturer's technical dpt)








"We don't believe things because they are true; Things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[sheiko]

Using mandatorily the fT from the Crane's table on page A-26 of course

Regards

"We don't believe things because they are true, things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[sheiko]

Re-reading the previous posts i would also correct the following:

7/ L/D = K/f is in fact not always true for fittings and valves, because "f and K may not always vary in exact proportion to each other" as the Reynolds number varies and for a given pipe roughness. However, this may give reasonable results for preliminary calculations (even in laminar flow regime).


"We don't believe things because they are true, things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."