Online Source for Equivalent Lengths of Pipe Fittings 1

[kahlilj]

Does anyone know a good source on the web for accurate equivalent lengths of pipe fittings. I have searched & located several sites, but they all seem to vary significantly & especially compared to sample values in my mechanical engineering reference manual. I especially want to know the equivalent length for 4" & 6" ball valves.

p.s. I do not have name plate data to contact the manufacturer for the Cv values either.

 

[kahlilj]

can anyone help please?

 

[katmar]

325 gpm is a very high flow for a 2" pipe. It will give you a velocity of over 30 ft/s and will result in high pressure drops. Now that you have access to Crane 410 I suggest you work through all the examples in Chapter 4 to get some experience in using these formulas.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[vpl]

kahlilj

You need to work in consistent diameters. If you got the equivalent length of the 2" pipe in terms of the 6" pipe, then your diameter should be 6". However, it'd really help if you'd post your work so we could see it, otherwise we're just guessing what you're doing.

Patricia Lougheed

******

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[kahlilj]

Thanks for your suggestions Harvey & Patricia. i sometimes get ahead of myself without thinking about some fundamental things like Harvey mentioned the flow rates through the 2" pipe.

The flow should be split. the main header is 6" and there are three branches that it feeds" two 4" branches & one 3" branch. The total length for each branch is only about 10 feet. That 3" branch (pipe) reduces to 2" pipe. The other branches remain the same.

So for my corrected flows to each branch I did the following:
total flow * (branch diameter/sum of branch diameters)
so for the 4" branches that gave me:
325 *(4/11) = 118 gpm & fluid velocity of ~3 ft/s
for 3" branch the flow was 89 gpm & fluid velocity of ~4ft/s.

The K value for the 3" ball valve using Crane formula A-28 (K= 3*f) and using example 4-19 on page 4-12 was:
K = (3*f)*(1/beta)^4
=(3*.018)*(1/0.51)^4
which gives me K = 0.8

I then calculated frictional (or velocity) head loss using:
Hf = (K*v^2)/(2*g)
= (0.8 *4^2)/(2*32.2)
=0.19

& Total Head loss, Hl from Crane formula 3-14 on page 3-4 for liquid flow thru valve/fittings:
Hl = (0.00259 * K * Q^2)/d^4
= 0.2 ft oh total head loss through the 3" valve

Does this seem the right approach/setup?

Thanks all for your help!
Kahlil

 

[katmar]

If you do not know the actual flows through the branches then splitting the total flow in proportion to the branch cross sectional areas (i.e. diameter squared) might be a better assumption than in proportion to the diameters. This would give you equal velocities in all branches.

Why have you multiplied the K value for the ball valve by (1/beta)^4? In example 4-19 Crane uses this ratio to convert the K value for the 2" line to a K value that can be used with the flow (or velocity) in the 3" line.

The calculations you have done for Hf and Hl are in fact the same calculation. It is just that one is in terms of velocity and the other in terms of flow. Both would give you the pressure drop through the valve, except that you have the wrong K value for a full bore ball valve.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[sheiko]

Katmar is the boss. Dot.

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

 

[kahlilj]

katmar (harvey?),
I actually tried multiplying the flow by ratio of cross sectional area for each branch, but the sum of the branch flows didn't equal the (total 325gpm) flow. for the 4" branches the flow was ~43 gpm & the 3" branch had 24 gpm flow. Unless I did something wrong?

I multiplied the k value for the 3" valve to convert it into flow for the 6" main line (header). I thought I needed to convert all losses for all lines to the same size? i.e. the head loss for 4" branch needed to be converted into an equivalent for 6" line & the same for the 3" branch.

Have I misunderstood when to apply this conversion?

 

[katmar]

If you split the flow in proportion to the areas then by definition the sum of the branch flows has to be equal to the total flow. Check your calculation.

Crane example 4-19 is for the case where you have the same mass flowrate all along the pipeline, but the diameter of the pipe changes somewhere. This method allows you to treat the compound pipeline (i.e. more than one diameter) as a single entity by bringing all the K values to a common basis, and applying them to a single velocity. In your case the flowrate is not constant because it splits at the header. You have to calculate the pressure drop for each segment having a constant flowrate, and then add the pressure drops for the pipes that are in series.

You need to calculate a pressure drop in the main, where you have the flow of 325 gpm, from its source to the header. Call this pressure drop "M". Then calculate the pressure drop for each branch from the header to the piece of equipment it serves - based on the flowrate in that branch. Call these pressure drops "A", "B" and "C". The pressure drop from the source to the first piece of equipment will be M+A, to the second M+B and so on.

You could use the Crane example 4-19 method on the branch that has a 3" and a 2" section to allow it to be treated as a single pipe - assuming the flowrate is constant in mass terms for both sections.

Over and out.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[kahlilj]

thanks for explaining about the pressure drop through the flowrates. i will fix this calculation.

i see the error in my calculation to get the branch flowrate. i was using a total area based on the 6" header plus that individual branch instead of the total area based on the branches (3" & 4"). In short I multiplied my flowrate by [pi*4^2/4]/[(pi*6^2/4)+ (pi*4^2/4)]. That is wrong. Instead it should be:
[pi*4^2/4]/[(pi*3^2/4)+ (pi*4^2/4)].

Using the corrected formula I got mass flows of ~127gpm for the 4" branches & 71gpm for the 3" branch. totals: ~325gpm

thanks katmar!!