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]

If you are feeling brave, and you have a few hours to spare, this topic was debated in great depth in thread378-173164

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

 

[sheiko]

Thanks Harvey,

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."

 

[katmar]

Yes, of course you can get the ratio correct if it is based on the correct underlying behavior and properties that impact on the friction factors. But my contention remains as before - if you are going to go to the trouble of getting the ratio behavior correct, why not just use the same relationships to do the calculation properly? If you want the friction factor for gasoline in the laminar regime just calculate it. Don't make it unnecessarily complicated by first calculating the friction factor for water in turbulent flow and then trying to predict from that. It might be intellectually stimulating, but it isn't necessary.

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

 

[BigInch]

If? Always Churchill. Its SOOOOooooo..... much better.

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

 

[sheiko]

Again thanks a lot Harvey for your insight.

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."

 

[BigInch]

The friction factor is not new. Its just a non-iterative procedure to calculate the same old ff.

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

 

[sheiko]

BigInch,

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."

 

[BigInch]

OK, then try restating your remaining questions in light of what Harvy has said already.

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

 

[sheiko]

Thank you,

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."

 

[katmar]

I would rather split the calculation into two separate steps. If you had a situation where you knew the flowrate, pressure drop and pipe diameter for the first fluid you would be able to calculate the friction factor and the equivalent length of the piping system. But this is a very unusual way of getting to the equivalent length. In the majority of calculations you would measure the straight pipe and count the fittings.

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

 

[sheiko]

Harvey,

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."

 

[katmar]

That's clearer then. You put me on the wrong track with your discussions in References 1 and 2 to the K values for fittings and valves. I will have to re-read and re-think a bit, and I am unfortunately rather busy at the moment.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[sheiko]

Sorry,

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."

 

[rcooper]

If the flow is laminar, the head loss is proportional to the fluid velocity. i.e. h=Kv (or h=K(v^2)/(2g) where K is proportional to the inverse of the Reynolds number)

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.

 

[sheiko]

Thanks rcooper,

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."

 

[rcooper]

"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?"

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.

 

[sheiko]

Thank your for your reply,

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."

 

[rcooper]

"... and without asking manufacturers ..."

I strongly recommend you ask the manufacturers. Not their sales department, their technical department.

 

[sheiko]

Harvey,

In addition to this thread (on which i wait for your conclusion), i have also read the thread378-173164: 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."