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.

 

[SNORGY]

This might be enough...

Regards,

SNORGY.

 

[katmar]

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]

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]

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]

How accurate do you intend your result to be?

Patricia Lougheed

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[kahlilj]

a reasonable estimate is good enough. it does not have to be precise...maybe within 10%? any thoughts?

 

[vpl]

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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[kahlilj]

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]

(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)

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. [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. 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]

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]

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

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

 

[katmar]

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]

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]

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]

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]

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]

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

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

Katmar Software - Engineering & Risk Analysis Software

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[kahlilj]

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]

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.