## Online Source for Equivalent Lengths of Pipe Fittings

## Online Source for Equivalent Lengths of Pipe Fittings

(OP)

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.

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

## RE: Online Source for Equivalent Lengths of Pipe Fittings

Regards,

SNORGY.

## RE: Online Source for Equivalent Lengths of Pipe Fittings

For generic estimates involving commercial steel pipe I would use L/D = 4 for a full bore ball valve and L/D = 25 for a reduced bore valve.

Ball valves of identical dimensions would have virtually identical pressure drops whether they were made of steel or plastic, but the associated piping could have somewhat different pressure drops per unit length depending on the material. This would of course have an effect on how you express L/D.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

## RE: Online Source for Equivalent Lengths of Pipe Fittings

http://www.pumpworld.com/fitting.htm

& this one too:

http://store.waterpumpsupply.com/friclosthrou.html

One other thing I have a simple pipe system of primarily 6" pipe with some sections of smaller diameter branches (see attached sketch) that are of short length. Should I even bother to convert these short-length sections to the 6" diameter equivalent?

i.e. using the eqn:

Le = L1 * [De^2.6182/(D1^2.6182 + D2^2.6182+...Dn^2.6182)]^1.8539 +...Ln * [De^2.6182/(D1^2.6182 + D2^2.6182+...Dn^2.6182)]^1.8539

## RE: Online Source for Equivalent Lengths of Pipe Fittings

There are 6 branches or approximately equal length - 10'. There are four 4" branches and two 3" branches (which reduce to 2" for half the length). The total length of the 6" straight pipe alone is well over 250' (not including equivalent lengths for 6" fittings).

Is it worth including this small diameter stuff?

## RE: Online Source for Equivalent Lengths of Pipe Fittings

Patricia Lougheed

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## RE: Online Source for Equivalent Lengths of Pipe Fittings

## RE: Online Source for Equivalent Lengths of Pipe Fittings

By the way, Crane Technical Paper 410, "Flow of Fluids," suggests a K value of 3 ft for a full port ball valve.

Patricia Lougheed

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## RE: Online Source for Equivalent Lengths of Pipe Fittings

for example: K = K *(d1/d2)

where d1 is the larger diameter & d2 is the smaller diameter.

## RE: Online Source for Equivalent Lengths of Pipe Fittings

(2) An answer to within 10% for a calculation like this would be as precise as you could ever get. A reasonable estimate would be ±40%

(3) Refering to Patricia's example: the "K = 3 ft" is "Crane-speak" meaning that the K value is obtained by multiplying the Moody friction factor for fully turbulent flow in steel pipe (i.e. ƒ

_{T}) by 3. The value "3" is also the L/D value for commercial steel pipe in turbulent flow. This compares with my slightly more conservative "4" above. Like I said - 40% error is probably reasonable when estimating the pressure drop in fittings. Luckily we can do better for straight pipe.(4) If you want to use a K value for a fitting of a given size in a pipe of a different size then the relationship is not linear - it is to the power 4.

i.e. K

_{1}= K_{2}x (d1/d2)^{4}This is equation 3-24 in Crane TP 410 (but badly formatted in the Metric versions that I have).

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

## RE: Online Source for Equivalent Lengths of Pipe Fittings

I also re-attached the diagram in ".jpg" format.

I don't have access to Crane's, So any info provided here is greatly welcomed & appreciated. In my example above I meant to use equivalent length, Le, & not K. That equation should have been:

Le = L1 * (d1/d2), but is that correct?

So for a 2", 45° ell, the Le is 1.7'. Converting this to 6" equivalent length would give 5.1'. Is this a correct approach?

Thanks all

## RE: Online Source for Equivalent Lengths of Pipe Fittings

It might also be found [albeit illegally] on the web by doing a Google search on "Crane flow of fluids used" (which I did to see if I could find a cheaper priced version). For the record, I didn't place a copy on the web and I did inform Crane of the website

There's also a fair number of free calculators around. You might start by looking at Katmar's site.

Patricia Lougheed

******

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

## RE: Online Source for Equivalent Lengths of Pipe Fittings

Your understanding of equivalent length is correct. This way allows us to use a single number for a particular type of fitting, regardless of its actual size. It is an approximation, but a very useful one. Equivalent lengths remain reasonable estimates even in laminar flow conditions, whereas fixed K values become totally irrelevant.

The "state-of-the-art" at the moment for estimating fitting pressure drops are the 2-K method by Hooper and the 3-K method by Darby. If you search here for them you will find many references.

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

## RE: Online Source for Equivalent Lengths of Pipe Fittings

BTW, I was able to get a copy of the crane tech paper through a coworker in my company that stores these types of documents.

## RE: Online Source for Equivalent Lengths of Pipe Fittings

"Sharing knowledge is the way to immortality"

His Holiness the Dalai Lama.

http://waterhammer.hopout.com.au/

## RE: Online Source for Equivalent Lengths of Pipe Fittings

Equivalent lengths by their very nature get multiplied by the friction factor in the Darcy-Weisbach equation so the calculated pressure drops vary (correctly) with both Reynolds number and pipe size. Using equivalent lengths would be more accurate than either of these two programs if you are using a variety of pipe sizes and you have flows less than fully turbulent in any of the branches.

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

## RE: Online Source for Equivalent Lengths of Pipe Fittings

I calculated a Reynolds number of ~6E4 so my flow is in the transition near turbulent zone. So how do I get an equivalent length for reducer? I have 2" x 3" reducers, 6" x 4" reducing tee & 6" x 3" reducing tees.

## RE: Online Source for Equivalent Lengths of Pipe Fittings

See my comments in thread378-108993: 24"x14" Concentric Reducer

I would model the pressure drop in a reducing tee as a full sized tee followed by a reducer and add the k values.

See thread378-270721: Pressure Loss in Tee

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

## RE: Online Source for Equivalent Lengths of Pipe Fittings

hope this makes sense.

## RE: Online Source for Equivalent Lengths of Pipe Fittings

I'm using the formula: HL (headloss) = (0.00259 * K * Q2)/d4

Q=325 gpm. I used 2" pipe for d. Should I have used 6" instead since I already converted the for 6" pipe when I calculated K?

Thanks all for your help.

## RE: Online Source for Equivalent Lengths of Pipe Fittings

## RE: Online Source for Equivalent Lengths of Pipe Fittings

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

## RE: Online Source for Equivalent Lengths of Pipe Fittings

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.

******

## RE: Online Source for Equivalent Lengths of Pipe Fittings

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

## RE: Online Source for Equivalent Lengths of Pipe Fittings

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.

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

## RE: Online Source for Equivalent Lengths of Pipe Fittings

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

## RE: Online Source for Equivalent Lengths of Pipe Fittings

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?

## RE: Online Source for Equivalent Lengths of Pipe Fittings

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.

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

## RE: Online Source for Equivalent Lengths of Pipe Fittings

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!!