11 replies to this topic

[Nate River]

Hello Everyone,

I'm currently modeling a flare system and I want to calculate a pressure drop given an equivalent length of pipe. Right now I'm using Crane Technical Paper No. 410 for equations and general information regarding flow theory. I've also read an article about equivalent length pressure drop calculations on this page http://www.cheresour.../eqlength.shtml. The problem is that neither give an example about how to convert equivalent length into a pressure drop given a compressible fluid. Anyway, here are the details:

f (friction factor) = 0.009
L (equivalent length) = 1416.7 feet
W (flow rate) = 1.25E6 lb/hr
d (diameter) = 42 inches
T (isothermal piping) = 515 deg F
MW (molecular weight) = 65.0 g/mol
Z (compressibility factor) = 0.98

The equation I'm currently looking at is on page 3-2 of Crane, equation 3-5, and is a form of Darcy's formula. It is as follows:

deltaP = 0.00000336*f*L*W^2*V'/d^5

where V' is specific volume in cubic feet per pound. (All other variables should use the values and units that I've listed above)

Any help would be greatly appreciated.

-Nate

[Art Montemayor]

Nate:

Please help me out. I’m trying to figure out what it is that you need from us. Can you specifically ask us a question regarding what it is that is troubling you?

You cite the correct equations for calculating the pressure drop of a compressible fluid – within the guidelines and limitations imposed by Darcy’s equation. Phil Leckner’s equation and the one you cite for Crane TP #410 are one and the same. Read the limitations of Darcy’s Formula for compressible flow as found in Crane TP #410, page 3-3. If you can live with these limitations, then you have a very easy calculation – if you have the correct data input.

Where did you get the value of 0.009 for the “friction factor”? I usually get values in the range of 0.0015 for the Darcy Friction Factor for a gas viscosity of 0.015 cP. Are you sure you are calculating the Darcy – and not the Fanning – Friction Factor? I get very suspicious when engineers don’t label their data appropriately and I want to make sure we have the right input data.

Phil goes to a lot of trouble to fully explain and illustrate how to use equivalent lengths correctly in the Darcy equation. I don’t understand why you can’t directly use what he shows you. You don’t need the molecular weight or the compressibility factor, so I don’t know why you give them. What you are going to need to identify is the density (or the inverse value which is the specific volume) at the flowing temperature.

You are wrong when you state “The problem is that neither give an example about how to convert equivalent length into a pressure drop given a compressible fluid”. Phil clearly gives you the correct equation to use (with limitations). Crane certainly does give you examples. What is it that you need from us? Do you want the answer? Or do you want us to guide you in what you have calculated? If the latter, show us your calculations.

Await your reply.

[Nate River]

Thank you for the quick response.
I will outline the problem I'm working on more fully. We are installing a new knock-out drum in our flare system and I was given an equivalent length of 42" Schedule 40 pipe that was calculated from an outside engineering firm to account for pressure drop. I was also given a pressure drop, for the given conditions, from a different engineering firm. So what I'm trying to do here is match up the equivalent length that I was given with the pressure drop I was given. (In the program I'm using, called Visual Flow, I account for the pressure drops of knock out pots as equivalent lengths.)
The problem I was encountering was solving for the density (or specific volume). The problem that persists is that I do not know the inlet or outlet pressures. The reason I included the compressibility and molecular weight is because I was given that information and I was wondering if there's a way to do a PV=znRT calculation to get my densities. (density = n*MW/V = P*MW/(Z*R*T), but then I don't know n!) The reason why I have such a small Darcy friction factor is because of the 42" pipe (which isn't commonly encountered). I've taken some liberty and have extrapolated the data on page A-26 of Crane to get my friction factor of 0.009. I was feeling pretty stuck on this problem this morning, and was hoping someone could at least point me in the right direction.
Thanks!

[Art Montemayor]

Nate:

This is very confusing. You are posting in a student Forum and yet you say you are “installing a new knock-out drum in our flare system”. Are you installing it for your university?

You also state: “I was given an equivalent length of 42" Schedule 40 pipe that was calculated from an outside engineering firm to account for pressure drop”. Is the equivalent length of pipe meant to represent the Knock Out drum? Why were you given an equivalent length? What were you asked to do?

You say you were “given a pressure drop, for the given conditions, from a different engineering firm”. OK, but what is the meaning of that? Are you a student or out in industry? What is the meaning of the pressure drop – is it for the system or the Knock Out drum? Again, what were you told to do?

If you want to calculate the pressure drop you must have the identification of the vapor or gas you are designing for in the flare system. From that you identify the density and the viscosity. From these values and the pipe surface characteristics you can obtain the friction factor using Chen’s, Churchill’s, or Serghides’ equation for the friction factor.

Please be specific about what it is you are doing – or trying to do. We’ll try to help when we can understand what it is we are being asked.

[Nate River]

My apologies. I am an undergrad university Chem E major working as a summer intern at a refinery. My task was to compare a given equivalent length (from one firm), meant to symbolize the pressure drop across a knock out drum, with a pressure drop given to me by another firm. The reason why one of the firms gave me the pressure drop in terms of equivalent length is because the software program we are using can only account for knock out drum pressure drops in terms of equivalent length. We specifically asked for the pressure drop in terms of equivalent length. However, before I proceed to update the model I want to make sure the given equivalent length and given pressure drop correlate and give me the same answer, within reason. So what I have been trying to do is convert my equivalent length into a pressure drop. At the moment the only data I have on the flow composition is listed in my first post.

[latexman]

Nate,

#1
Master the basics before you run the computer. Garbage in = garbage out, and if you don't know the basics you will not know if you have garbage or sound answers. My advice for you based on my perception of your level of education is . . . do it by hand!

#2
V' = (V/n)/MW
PV=znRT rearranges to V/n = zRT/P
Combining the two gives V' = (zRT/P)/MW = zRT/(P.MW)
To calculate V' at a certain point in the process, you need an absolute pressure at the same point in the process, and you have not mentioned that anywhere. What is P?

#3
With P, you can use the equation you cited. Use Contractor A's L (equivalent length) in the equation to calculate Contractor A's deltaP. Then compare Contractor A's deltaP to Contractor B's deltaP. Or, rearrange the equation and compare Contractor A's equivalent length to Contractor B's equivalent length.

[Nate River]

Thanks for the advice. Someone else who works here told me the same thing about models; "garbage in = garbage out". Haha.
It seems that the bottom line is that I don't have any values for P (which I mentioned in my previous post). Unfortunately no P values were given with the equivalent length from Contractor A. I see that with P values I could have gotten my specific volume and used the Darcy formula, as previously stated.
The good news is that I received a print out of a spreadsheet used by Contractor B that has pressure values at the inlet and outlet positions of the knock out drum. (I'm not sure how they got them) I'll use those to get specific volumes at these points and average them as suggested by Crane page 3-3 (since the pressure drop falls in that 10 to 40% range). Then I'll calculate an equivalent length (as suggested in suggestion #3) and compare equivalent lengths. It's all about getting the info/resources...

Thanks for all the help guys!

-Nate

[latexman]

It sounds like you are on the right track now. Please let us know how Contractor A compared to Contractor B. I'm curious. Don't be surprised if they are somewhat different.

[JBradley]

Sorry if i'm about to stir things up here!! A while back I had lots of difficulties trying to get a simple way to deal with compressible flow and so am really intrested in this posting.

The best I got (from Wikipedia!) was that the flow can be considered as incompressible if Ma<0.33. Luckily, this is the case we have here.

So guys - what formula do you use if Ma>0.33. Looking in textbooks all I get is lots of calculus (and you definately need the compressibility factor!)

[pleckner]

For gas/vapors you should be using (with an exception I will give below) either the Isotheraml gas equations or the adiabatic gas equations. The Isothermal gas equations are significantly simplier to use than the adiabatic gas equations and are more than adequate for most applications. Saying this, CRANE TP410 actually gives examples for vapor/gas using the adiabatic equation, which they simplify with the use of the adiabatic expansion factor. There is nothing wrong with using this equation within the limits given in TP410.

Now for the exception. IF the pressure drop in the line or line segment will be no less than 10% of the inlet pressure, then you can use the incompressible fluid (liqiuid) equation, better known as the Darcy equation that I use in my article @Nate River references in his post above. You can still get away with the Darcy equation if the pressure drop in the line or line segment will be no less than 40% of the inlet pressure AND you use the average density of the vapor/gas.

The reason you want to use the gas/vapor equations is that gas/vapor expands as it flows down the pipe and pressure is lost, thus there is a constant change in fluid density. The gas/vapor equations take this expansion into account whereas the imcompressible fluid flow equation does not.

[JBradley]

Thanks for that - as ever Crane is the last place I look but has given me an answer!!

[latexman]

In the 28 years of my CHE experience, I opened Crane TP410 much more frequently than any other reference. If it's not two-phase flow, that's what I open first!