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Online Source for Equivalent Lengths of Pipe FittingsHelpful Member! 

kahlilj (Mechanical)
25 May 11 17:55
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.
SNORGY (Mechanical)
25 May 11 18:58
Helpful Member!  katmar (Chemical)
26 May 11 6:23
If you do not have the valve nameplate data then any estimate is as good as any other, particularly if you have reduced bore valves. There is no guaranteed consistency in reduced bore ball valve internal dimensions from one manufacturer to another.

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"

kahlilj (Mechanical)
26 May 11 10:12
Thanks Snorgy & Katmar.  The engineering toolbox was actually one of the first sites I checked, but when I didn't see a factor for all valves began to check around & noticed differences in factor values for the same fittings such as this site:
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
 
kahlilj (Mechanical)
31 May 11 15:39
Anyone have any thought to my previous question: "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?"

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?
vpl (Nuclear)
1 Jun 11 16:05
How accurate do you intend your result to be?

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.

kahlilj (Mechanical)
1 Jun 11 16:10
a reasonable estimate is good enough. it does not have to be precise...maybe within 10%?  any thoughts?
vpl (Nuclear)
1 Jun 11 16:17
I'd recommend taking the time to convert them.

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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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.

kahlilj (Mechanical)
1 Jun 11 17:18
thanks....  for the ball valve k value if i wanted to convert from one size to another do i multiply it by the ratio of the diameters?

for example: K = K *(d1/d2)   
where d1 is the larger diameter & d2 is the smaller diameter.
katmar (Chemical)
2 Jun 11 5:49
(1) You will get more response if you use a more commonly used file format. Not everybody cam handle a .vsd file. Rather use .jpg, .gif or .png.

(2)

Quote:

a reasonable estimate is good enough. it does not have to be precise...maybe within 10%?  any thoughts?
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. ƒT) 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.  K1 = K2 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"

kahlilj (Mechanical)
2 Jun 11 13:44
Thanks Katmar, points well taken.  Whatever the figure may be for a reasonable guess (40%?) is ok, so forget the 10% reference.  :)
 
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
vpl (Nuclear)
2 Jun 11 15:12
The Crane technical paper is available from them for a measly $60 US.  You might ask your company to buy it for you. If it is within your financial capability, it is well worth purchasing as a general reference.

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.

katmar (Chemical)
3 Jun 11 3:18
The need to calculate each branch individually depends on the flow to each branch. As a generalization I would recommend that you do calculate each branch, with its particular flow.

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"

kahlilj (Mechanical)
3 Jun 11 11:00
Thanks again Katmar.  

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.
stanier (Mechanical)
3 Jun 11 19:47
Suggest you download freeware Epanet or PSIM and model it using more modern & accurate techniques. the use of equivalent lengths is not very accurate.

"Sharing knowledge is the way to immortality"
His Holiness the Dalai Lama.

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

katmar (Chemical)
6 Jun 11 7:10
The resistance factors for fittings vary with Reynolds number and with pipe size. PSIM uses the Crane method for most fittings.  This takes the pipe size into account, but is applicable to fully turbulent flow only. Epanet uses fixed K values which take neither Reynolds number nor pipe size into account.

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.

Katmar Software - Engineering & Risk Analysis Software
http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

kahlilj (Mechanical)
15 Jun 11 18:45
I've been away on vacation so sorry for the delayed response, but thanks Katmar & others as well.

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.   
katmar (Chemical)
16 Jun 11 9:43
Just when I am extolling the virtues of the equivalent length method you catch me out by asking about a fitting that does not really work well as an L/D - mainly because they have such different ratios for (inlet diam)/(outlet diam). For pipe reducers it is better to work with k values.

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

Katmar Software - Engineering & Risk Analysis Software
http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

kahlilj (Mechanical)
16 Jun 11 16:22
thanks. I'm trying to put together a spreadsheet that will summarize my calcs, but I'm a little confused on when to apply the resistance coefficient eqn (Crane TP410 eqn 3-24 you referenced earlier) for different pipe sizes when I'm ultimately trying to determine the head loss (Hl). should i apply this ratio to the equivalent length formula & then determine K from f*L/D?  Or should I determine K from the Crane paper appendix  & then apply this beta (ratio coefficient)?  

hope this makes sense.
kahlilj (Mechanical)
20 Jun 11 17:19
Does this seem a reasonable K & HL values? for a 6 ft length of 2" pipe I'm modeling in a 6" pipe system I get a K value of 4.1 & my Head loss is ~61 feet.  That seems a bit high to me.

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.
kahlilj (Mechanical)
21 Jun 11 15:41
can anyone help please?
katmar (Chemical)
22 Jun 11 2:57
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 (Nuclear)
23 Jun 11 18:27
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: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

kahlilj (Mechanical)
24 Jun 11 13:43
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 (Chemical)
25 Jun 11 14:46
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 (Chemical)
26 Jun 11 9:27
Katmar is the boss. Dot.

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

kahlilj (Mechanical)
27 Jun 11 10:11
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 (Chemical)
27 Jun 11 16:04
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 (Mechanical)
30 Jun 11 17:03
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!!

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