Equipment pressure drop extrapolation
Equipment pressure drop extrapolation
(OP)
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
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."





RE: Equipment pressure drop extrapolation
My quick summary : K values are dependent on Reynolds Number. This especially true if you are working at Reynolds Numbers below 4000.
The variation in K is usually not a problem because the overwhelming majority of flow calculations are for fully developed turbulent flow, and under these circumstances it is a reasonable assumption that K values are independent of the Reynolds Number. I believe that a weakness of the (otherwise excellent) Crane 410 manual is that it does not make this distinction clear.
I have not seen the two articles in your References 3 and 4, but my feeling towards this type of calculation is that it is irrelevant in this day and age. There was a time (viz. the era of slide rules) when it was a useful time saver to be able to make these quick ratio calculations. But now we have software (even if it is only a spreadsheet) and it is just as easy to do the calculation properly as it is to take the shortcut. If these articles are aimed at equipment rather than at piping then taking a ratio to the square of the flowrate is probably OK in the absence of better information, but it would be better to get the right information from the manufacturers.
Harvey
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Equipment pressure drop extrapolation
My aim is indeed to develop a general spreadsheet allowing the estimation of equipment/instruments pressure drop at new operating conditions given a given known condition.
The advantage i see with the equivalent length method (reference 4) is that we could extrapolate pressure drop for different fluids (for example known pressure drop with water and new conditions with gasoline) and different flow regimes (for example known pressure drop in turbulent regime whereas new flow regime is laminar) thanks to the ratio of the friction factors (which takes into acount fluid properties and flow regime).
Could you tell me if this method is theoreticaly valid?
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."
RE: Equipment pressure drop extrapolation
Have a look in the FAQs section at the top of this forum. You will find some excellent work done by Quark on how to calculate the friction factor. If you use the Churchill equation there is no need to bother with ratios.
Harvey
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Equipment pressure drop extrapolation
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
RE: Equipment pressure drop extrapolation
But, if i only calculate the new friction factor, how can i determine the equipment/instruments pressure drop at new conditions without equivalent length? As i mentionned in my first post, the equivalent length is calculated thanks to the known (old) friction factor.
Indeed, the problem is the following:
Known: hf0, rho0, mu0, q0, rho, mu, q, arbitrary D
Unknown: hf
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."
RE: Equipment pressure drop extrapolation
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
RE: Equipment pressure drop extrapolation
I mean by new: at new operating conditions
Please read the entire discussion since the beginning.
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."
RE: Equipment pressure drop extrapolation
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
RE: Equipment pressure drop extrapolation
I thought i had understood his answers. But maybe i am wrong (as i mentionned in the first post my english is very poor...)
But, what do you mean? that we can estimate hf without calculating f0 and f (using the method in refernce 4 of course)?
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."
RE: Equipment pressure drop extrapolation
If you did it this way (i.e. two separate steps) you would first get the equivalent length as above and then (in the second step) you could calculate the friction factor for the new fluid using Churchill, and then you could calculate either the flowrate or the pressure drop (with the other one known).
This procedure has the implicit assumption that the equivalent length remains the same for the two different fluids. Indeed, the equivalent length of fittings like elbows and tees does remain remarkably constant for laminar and turbulent flows, but I would be reluctant to make this assumption for valves, reducers and orifices.
In short, I think that you are putting a lot of effort into solving a problem that doesn't really exist. Or maybe I have misunderstood what you are trying to achieve.
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Equipment pressure drop extrapolation
As mentionned in the first post, this discussion deals with equipment/instruments pressure drop not fittings nor valves. So items for which we don't know equivalent lengths nor resistance coefficients. For example: heat exchangers, ...without asking manufacturers.
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."
RE: Equipment pressure drop extrapolation
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Equipment pressure drop extrapolation
I made reference to 1/ and 2/ just to introduce the subject: The equipment/instruments extrapolation method is often based on the fittings/valves pressure drop calculation method (eg: hf=K*v^2/2g) and also because the common extrapolation method for equipement/instruments pressure drop: hf=hf0*(q/q0)^2, assumes that K is always constant, which is false as you previously and correctly pointed out.
I am waiting for your conclusion on this point (equipement and instruments and not valves and fittings for which i use Darby and Hooper). Thanks again.
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."
RE: Equipment pressure drop extrapolation
If the flow is fully turbulent, the head loss is proportional to the square of the velocity. i.e h=k(v^2)/(2g)
If the flow is transitional, something else happens and this is represented by saying h=k(v^2)/(2g) where k varies with Reynolds number.
If you know whether you are fully laminar or fully turbulent, then it is relatively simple and reliable to extrapolate your values for k.
A word of caution when discussing your Reynolds number. The Reynolds number includes a characteristic dimension. This could be any dimension, the radius, diameter, circumference of the pipe, the length of the pipe, etc, etc. When discussing Reynolds number you need to be clear what the characteristic length is.
You say you are extrapolating data for an instrument. Within one unit, laminar, turbulent and transitional conditions could all exist. You will need to be clear how the fluid passes through the instrument and the flow conditions within the instrument.
RE: Equipment pressure drop extrapolation
The selected characteristic length is the internal diameter.
I also understand that you agree with the following formula:
hf=hf0*(f/f0)*(q/q0)^2 as an extrapolation for determining equipment/instruments pressure drop. Indeed, if:
- Flow is laminarw both in known and new conditions and in the pipe upstream and downstream the piece of equipment:
f=16/Re anf f0=16/Re0 => hf=hf0*(q/q0)
- Flow is fully turbulent both in known and new conditions and in the pipe upstream and downstream the piece of equipment:
f and f0 independant of Re => hf=hf0*(q/q0)^2
But what if the flow regimes are different in known and new conditions, or the flow regimes are in critical and transition zones both in known and new conditions?
My suggestion is: hf=hf0*(f/f0)*(q/q0)^2 which appears (to my humble eyes) to be a generalization on what you've just said.
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."
RE: Equipment pressure drop extrapolation
If the flow regimes are different in known and new conditions, then extrapolation is unlikely to predict the pressure loss across the instrument - you will need experimental data for the flow regime you are interested in.
If the flow is in the critical or transitional zone, similarly I would rely on experimental data rather than extrapolation. Remember, extrapolation is a dangerous tool which is easily abused.
You say you want to develop a spread sheet to predict head loss across your instrument at different flowr rates. I would get as many test results as available, put these in a chart and set up the spread sheet to linearly interpolate between these, stopping the spread sheet from providing a result outside the range of test data.
You also haven't told us the nature of the instrument. Is it a simple pipe, does it include weirs, are their obstructions, multiple routes for the fluid to flow throught etc. etc.
RE: Equipment pressure drop extrapolation
Reminder (Nomenclature):
Subscript 0: refers to the given old flow condition or/and fluid.
Without subscript: refers to the new flow condition or/and fluid (also known).
hf0 and hf are the known and unknown friction heads of a given peice of equipemnt or instrument (heat exchanger for example),
q0 and q the old and new (both known) flowrates,
F0 and F the old and new (both known) Darcy friction factors
Re0 and Re the old and new (both known) Reynolds numbers
Known: hf0, rho0, mu0, q0, rho, mu, q, arbitrary D
Unknown: hf
I agree that the use of experimental data prevails on the use of correlations. However, as a process engineer working in an engineering company, experimental data are not always available, and without asking manufacturers, an accurate extrapolation method has to be found. Moreover you have not provided arguments that infirm the validity of the proposed extrapolation into all flow regimes.
Thus, are you sure that the formula hf=hf0*(f/f0)*(q/q0)^2 is theoretically incorrect when the old and new flow regimes or/and fluids differ (Re0>4000 and Re<2000 for example)?
I insist on this because in the case of same flow regimes and/or fluid for old and new conditions, it gives:
- laminar (Re0 and Re <2000) => hf=hf0*(q/q0)
- turbulent (Re0 and Re >4000) => hf=hf0*(q/q0)^2
Which you will probably agree on
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."
RE: Equipment pressure drop extrapolation
I strongly recommend you ask the manufacturers. Not their sales department, their technical department.
RE: Equipment pressure drop extrapolation
In addition to this thread (on which i wait for your conclusion), i have also read the thread378-173164: Crane 410 fittings: Crane 410 fittings.
You have concluded that the K values (resistance coeffs) are not constants as they vary with Reynolds number, and i agree (experimentally proven).
My feeling is also that the only constant in all that is L/D.
Moreover, Pleckner seems also right when he says that no one should try to convert the Crane K value into an equivalent length, as there is no physical link between the two. Indeed, L/D and fT, AS DEFINED IN THE CRANE PAPER, seem rather mathematical parameters than real physical parameters of a physical model.
That's why i also choose to let the past behind (symbolized by the Crane method, only valid for fully turbulent) and go ahead with Darby's 3-K method, without forgetting the good old equivalent length method for preliminary calcs. Why bother anymore with Crane K values? as they are not always applicable nor convenient to use...
Now i would like to know the result of your reflexion on the earlier proposed EQUIPMENT pressure drop "extrapolation" method as it is a close subject to me.
P.S: I hope my english was good enough this time.
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."
RE: Equipment pressure drop extrapolation
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
RE: Equipment pressure drop extrapolation
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."
RE: Equipment pressure drop extrapolation
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
RE: Equipment pressure drop extrapolation
Thanks for all!
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."
RE: Equipment pressure drop extrapolation
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."
RE: Equipment pressure drop extrapolation
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
RE: Equipment pressure drop extrapolation
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.
RE: Equipment pressure drop extrapolation
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
RE: Equipment pressure drop extrapolation
"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."
RE: Equipment pressure drop extrapolation
Herewith the reference 4
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."
RE: Equipment pressure drop extrapolation
From Plant Engineer's Reference Book (2nd Edition), edited by Snow, Denis A, 2002, Page 24/6
RE: Equipment pressure drop extrapolation
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."
RE: Equipment pressure drop extrapolation
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.
RE: Equipment pressure drop extrapolation
"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."
RE: Equipment pressure drop extrapolation
RE: Equipment pressure drop extrapolation
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."
RE: Equipment pressure drop extrapolation
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
RE: Equipment pressure drop extrapolation
Harvey
Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com
RE: Equipment pressure drop extrapolation
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."
RE: Equipment pressure drop extrapolation
RE: Equipment pressure drop extrapolation
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
RE: Equipment pressure drop extrapolation
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."
RE: Equipment pressure drop extrapolation
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: 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."
RE: Equipment pressure drop extrapolation
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
RE: Equipment pressure drop extrapolation
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."
RE: Equipment pressure drop extrapolation
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."
RE: Equipment pressure drop extrapolation
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."