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]

First of all let me apologize if it seemed that I was questioning your command of the English language. Your English is excellent, and better than many supposedly native English speakers who have posted here before. Indeed, I could be accused of speaking the Durban sub-dialect of Sarf Efrikan myself. But it passes for English and I don't think language is a problem between us. And on re-reading your posts it is pretty clear that you were concerned with equipment pressure drops and not with pipe fittings. Anyway, on to the engineering.

As I said before, I have not read the original articles for References 3 and 4. If the conclusion in Reference 3 is that you can use the proposed relationship to predict within the range 500<Re<2100 or in the range Re>5000 separately I could probably go along with that as an estimating method. Note that I say separately because I suspect the author is not suggesting that you can use a data point from the first range to predict a pressure drop in the second range (or vice versa). But as long as your measured data point and the estimated data point are in the same range it might be OK.

As far as I can tell from your summary of Reference 3, there is no mention of friction factors in that method. I like that if it is true.

The posting of email addresses in these forums is not allowed, but if you dig around in my signature you will find how to contact me. If you want to email a scan of the article I could make more informed comments.

Now to Reference 4. My objection to this method would be the very same argument I used in the Crane 410 thread that I referenced previously. The friction factor has nothing to do with the pressure drop through an elbow or tee, and much less does it have anything to do with the pressure drop through an instrument.

To understand how Crane 410 got to where it is now, it is important to know that in the early versions of the manual the resistances of fittings were reported as Equivalent Lengths and not as K values. The experimental data was originally obtained in commercial steel pipe, so when Crane decided to switch to using K values they converted their historical data from Equiv Length to K using the relationship in your first post and using the friction factor for fully developed turbulent flow in steel pipe. It is therefore perfectly acceptable to reconvert a Crane 410 K value back to an equivalent length provided we realize that the length obtained is of steel pipe in turbulent flow.

In the early days Crane noted that the equivalent length data they had accumulated worked fairly well in all flow regimes (for steel pipe), but when they used the very specific friction factor for fully developed turbulent flow to convert the data they unfortunately (and inadvertently) narrowed the applicability of their data down to that same very specific instance. You can no longer use Crane data for laminar flow the way you could use their old Le/D data.

This use of the friction factor for fully developed turbulent flow in steel pipe to calculate K values was the basis of my ranting in the earlier thread. It has resulted in many inexperienced (and some not so inexperienced) engineers believing that the K value is somehow linked to the friction factor and therefore to the pipe roughness, when in fact this is not so.

It seems to me that the Author of Reference 4 has fallen into the same trap that caught Crane. What is the friction factor of a metering orifice? It has no meaning. Even Churchill could not calculate it. Therefore I believe that using the friction factor as the basis of the prediction method invalidates the method completely. But you should remember that millions of people use the position of the stars at the moment of their birth as the ultimate predictor of everything that happens in their life, so maybe I have just failed to see the connection. I am a bit narrow minded this way. We do not believe things because they are true. Things are true because we believe them. Enough philosophy - I'm off to have a beer.

regards
Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[sheiko]

Thanks for your availability and the level of detail you provide each time,

Yes indeed, the range of validity of the method in reference 3 is 500<Re<2100 (upper region of laminar flow) or Re>5000. So we consider them separately.

To put it in a nutshell, the preferred method to extrapolate the pressure drop of a piece of equipment or instrument is based on reference 3: hf=K*(v^2/2g) with K constant and not on reference 4, based on hf=F*(L/D)*(v^2/2g), WHICHEVER THE FLOW REGIME in both old and new conditions.

However,

1/ The method in reference 3 assumes that K is a constant which you disagreed on for fittings

2/ Most of all, if "The friction factor has nothing to do with the pressure drop through an elbow or tee, and much less does it have anything to do with the pressure drop through an instrument", then how people (icluding Crane engineers for the earlier version of the manual) could estimate and tabulate equivalent lengths without considering that an item causing pressure drop can be "modelized" by a piece of pipe? For me, we can conceptualize the equipment problem the same way...and thus estimate equivalent lengths for equipement as well...no?

I like to remind the principle of the equivalent length method for our problem:

Known: hf0, rho0, mu0, q0, rho, mu, q, selected D
Unknown: hf

1/ Calculate the old DP100 at the old conditions (average flowrate and fluid properties)
2/ Calculate a Leq at any selected pipe size D by Leq=100*(hf0/DP100)
3/ Calculate new dp100 at the new conditions (average flowrate and fluid properties) and the same selected pipe size D
4/ Determine hf by hf=dp100*(Leq/100)
Considering that:
DP100=F0*(100/D)*(v0^2/2g)
dp100=F*(100/D)*(v^2/2g)
Then, in one step: hf=hf0*(dp100/DP100) <=> hf=hf0*(F/F0)*(q/q0)^2

And i also remind the main advantage: Be abble to predict pressure drop with different fluids and different flow regimes between the old and the new operating conditions, which is not allowed by the method based on reference 3: hf=hf0*(q/q0)^2...



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

 

[katmar]

Reference 3.
We have agreed that K varies with Re, but if you restrict the ranges then K can be regarded as "constant" within a particular range. That was why I said you should not use data in one range to make predictions in a different range - the K values would not be equal for the 2 different ranges.

Reference 4.
There is no "friction factor" that causes the pressure drop through a fitting like an elbow. It is convenient to ratio the pressure drop through a fitting to the pressure drop along a pipe of equal diameter (and the pipe would have a friction factor) and to call this ratio the equivalent length. But it is just a convenient calculation method. It does not imply that there is a friction factor associated with the elbow.

You could calculate the equivalent length of pipe that would give the same pressure drop as an instrument or piece of equipment under given flow conditions. You would have to apply some friction factor to the pipe calculation to get this equivalent length. But you are making the same mistake as the others by wanting to calculate a new friction factor for this imaginary piece of pipe for the new conditions and then believing that this new friction factor is representative of the pressure drop through the instrument or piece of equipment.

There is no reason to believe that the pressure drop through the instrument will vary in the same way as the friction factor in the imaginary pipe because the pressure drop through the instrument is not caused by the same mechanism as the pressure drop in the pipe. The pressure drop in a pipe is caused by friction with the pipe wall, but the pressure drop through an instrument or piece of equipment is (mostly) caused by form factors i.e. by changes in flow direction and velocity.

This is not a simple concept and it has tripped up MANY engineers. Unfortunately doing it this way does sometimes give reasonable answers and it seduces people into believing it. Just like your horoscope will be correct on some days. But that does not make it a correct principle.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[sheiko]

Very clear

Thanks for all!

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

 

[sheiko]

Thanks Harvey,

Before closing the thread i have the following last questions:

Then, does a general method exist in order to extrapolate equipement/instruments DP if:

1/ old and new conditions are in laminar flow?

2/ old and new conditions are in different flow regimes?

3/ old and new situations involve different fluids?

4/ 1/ and 3/ together?

5/ 2/ and 3/ together?

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

 

[katmar]

If I was faced with doing this type of calculation myself this is what I would do :-

1/ All laminar flow - resistance coefficients can change very rapidly in laminar flow and I would go back to first principles and not try to extrapolate. If I had to make a quick estimate and the flow did not change by more than 30% I would ratio the pressure drop by the square of the flow ratio, but then I would redo the calc properly as soon as I had a chance because I know the answer would not be accurate.

2/ I would not try to extrapolate from one regime to another.

3/ Different fluids - In laminar flow changing the fluid can have a dramatic effect on the Reynolds number and the K value, so I would not try to extrapolate.

In turbulent flow fairly large changes (eg a factor of 2) to the viscosity makes virtually no difference to the pressure drop. Density changes also make almost no change to the pressure drop if the volumetric flow is unchanged, provided the pressure drop is measured in height of the flowing fluid. In kPa or psi of course the pressure drop does change.

4/ Would not do this.

5/ Would not do this.


Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[rcooper]

katmar,

In 1/ if it is laminar flow, the pressure drop is increased by the ratio of the flow rates, not the square of the flow rates.

 

[katmar]

rcooper, what you have said is true for pipe but not for instruments and equipment.

For example the permanent pressure drop through a metering orifice is quite close to being proportional to the square of the flow in the laminar regime. For fittings like valves the proportionality varies from the square of the flow at the high flowrate end of the laminar regime down to being proportional to the flow at very low flows.

That is why I said that I would use this method for quick estimates only and then do the calculation properly as soon as I got a chance. I know that my estimate would be wrong.

You are correct for pipe because in laminar flow the friction factor is proportional to 1/Re, but for the type of equipment that sheiko was interested in it is not (necessarily) true that K is proportional to 1/Re.

Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[sheiko]

I have submitted the reference 4 to a senior engineer who said:
"I looked over the article today in Reference 4 and the suggestions given by Frank Yu are simple, sound and useful. Go ahead and use them"

Please find attached this reference


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

 

[sheiko]

I mistake that was the reference 3

Herewith the reference 4

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

 

[rcooper]

"The simplest method of measuring flow rate is to use an orifice plate, this is a plate fitted into a pipe which reduces the diameter along a short section of pipe. For a liquid whose flow rate is slow enough to be laminar the pressure difference across the orifice plate is linearly proportional to the rate of flow. For flows which are not laminar but are turbulent the pressure difference across the orifice plate is proportional to the square of the flow rate."

From Plant Engineer's Reference Book (2nd Edition), edited by Snow, Denis A, 2002, Page 24/6

 

[sheiko]

rcooper,

I may be wrong, but maybe the flow through an orifice plate is closer to that in a pipe that in a piece of equipement like a heat exchanger for example (no changes in direction, ...)

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

"We don't believe things because they are true; things are true because we beleive them."

 

[rcooper]

What are the posters' opinions on the 2-K (Hooper) method for finding K throughout the laminar and turbulent ranges?

K=Kl/Re+Kinf(1+1/D)

where Kl is the laminar K and dominates at low Reynolds numbers

Kinf is the turbulent K and dominates at high Reynolds numbers

when Kl dominates, K is proportional to 1/Re, so the head loss is proportional to the flow rate.

when Kinf dominates, K is nearly independant of Re so the head loss is proportional to the square of the flow rate.

 

[sheiko]

This topic was partly debated in thread378-173164: Crane 410 fittings: This 2-K method is good and the Darby's 3-K method is even better.


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

 

[rcooper]

So we have agreement that at low Reynolds numbers (laminar flow) the pressure loss through a fitting is proportional to the flow rate, and at high Reynolds numbers (turbulent flow) the pressure loss is proportional to the square of the flow rate.

 

[sheiko]

Yes, as far as i am concerned.

Let's see katmar opinion on your post: 18 Apr 08 5:29 and on mine: 18 Apr 08 4:07 .

"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]

rcooper, your quotation from Snow is correct but not complete. But I believe your later post referencing Hooper is the real answer.

Before I made my post where I claimed that the permanent pressure drop through an orifice is proportional to the square of the flow I tested it in a program that is based on Hooper. I only tested down to Re = 100 and for the Re range from 100 to 2100 the pressure drop is very close to proportional to the square of the flow.

When you posted the Snow reference I checked again and it is true that for VERY low Re the PD is proportional to the flow and not the square. For example when Re (based on the pipe) increases from 1 to 2 the Hooper K value decreases from 196 to 103 (assumed d/D = 0.6). i.e. it is virtually halved. If you plug these numbers into Bernoulli or Darcy where the velocity is squared, the fact that K is basically halved means that the PD will be proportional to the flow and not the square of the flow.

I also dug deep into my aging memory to recall that many years ago I used some micro orifices where the diameter was deliberately made so small that the flow was laminar and the pressure drop was linear with the flow. There is a flow metering device (which I have not used personally) which takes advantage of this by placing many micro pipes in parallel to take advantage of this linear behavior. I'm sure someone here must have used them.

So I believe you and I were both correct, but also both incorrect (or at least incomplete) with our answers.

The Hooper method I used was published in Chemical Engineering, Nov 7, 1988. pgs 89-92.

Thanks for a very interesting exchange.

Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[katmar]

rcooper and sheiko - you both posted while I was slowly typing my response to the earlier posts so my answer probably reads a bit out of context with your latest posts. Hopefully it makes sense.

Harvey

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[sheiko]

Harvey,

I have also posted a comment and an attachment for info. in my posts dated 18 Apr 08 4:07 and 18 Apr 08 4:14.

"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."

 

[rcooper]

I guess that's what happens when two industries collide. Some of my work involves ultrafiltration where the orifices are very small. Definitely laminar flow. It's just our idea of what is small that differed.