## Featured Articles

Check out the latest featured articles.

## New Article

Product Viscosity vs. Shear

## Featured File

Vertical Tank Selection

## New Blog Entry

Low Flow in Pipes- posted in Ankur's blog

0

# Clarification On Article About Minor Loss Calculation Method From Cher

fluid flow k values minor losses

5 replies to this topic
|

### #1 jayari

jayari

Veteran Member

• Members
• 44 posts

Posted 11 March 2020 - 06:57 PM

“Using Equivalent Lenghts of Valves and Fittings” by P. Leckner, http://www.cheresour...d-fittings?pg=2

From page 2, section titled Relationship Between K, Friction Factor, and Equivalent Length:

“Pipe friction in the inlet and outlet straight portions of the valve or fitting is very small when compared to the other three. Since friction factor and Reynolds Number are mainly related to pipe friction, K can be considered to be independent of both friction factor and Reynolds Number. Therefore, K is treated as a constant for any given valve or fitting under all flow conditions, including laminar flow” (emphasis is mine).

K = f * (L/D) {Eqn. 4}

“In Equation 4, f therefore varies only with valve and fitting size and is independent of Reynolds Number. This only occurs if the fluid flow is in the zone of complete turbulence…” (emphasis is mine).

The bolded statements seem contradictory to me. The first quoted section seems to make an unconditional statement on when K values can be used (laminar, transition(?), and turbulent flow regimes). The second quoted section reads as if the K value is only applicable if the flow regime is turbulent because when the flow is turbulent the K value of for a 2" vs 6" version of a particular piping component will be different because the component's friction factor (f) values are different for the two different sizes.

I'd appreciate if someone could point out what I'm missing.

Edited by jayari, 11 March 2020 - 06:57 PM.

### #2 breizh

breizh

Gold Member

• 5,628 posts

Posted 11 March 2020 - 10:22 PM

Hi,

Let you find Katmar's  explaination about pressure drop calculation methods for fittings .

Using the search engine in this forum is also an option .

more :underneath .

https://neutrium.net.../pressure-drop/

Good luck

Breizh

Edited by breizh, 15 March 2020 - 12:31 AM.

### #3 jayari

jayari

Veteran Member

• Members
• 44 posts

Posted 12 March 2020 - 08:08 AM

The neutrium.net domain as a whole hasn't been working for me this last week when I was searching my old bookmarks. Does the website fully load for you?

Thanks for linking me to Katmar's website - I read those articles and a few other threads on this site and Eng-tips.com about minor loss calculation. I am focused on understanding this article from cheresources at this time because either there is an inconsistency that the author should be alerted to or I need to fix my understanding of what is written (this is the more likely situation lol).

### #4 Pilesar

Pilesar

Gold Member

• Members
• 1,041 posts

Posted 12 March 2020 - 09:25 AM

jayari,

You are making the same point the author of the article makes. There is a contradiction in the definitions so they are NOT the same. Read the article closely:

"f in Equation 4 is not the same f as in the Darcy equation for straight pipe, which is a function of Reynolds Number."

### #5 jayari

jayari

Veteran Member

• Members
• 44 posts

Posted 12 March 2020 - 10:55 AM

@Pilesar

My question, stated another way, is if K = ft * Leq/D and ft is defined as the friction factor in a zone of complete turbulence, then how can the K value calculated in this way be applied to any situation other than turbulent flow?

If I were to perform a calculation with my understanding of the article now for a laminar flow situation, it would look something like this:

hL = (fL/D + SK) * (v2/2g)

Where f, L, D are for the pipe and the K values are for the fittings in the pipe layout. In my equation above, I would have an f value for laminar flow applied to the pipe component and K values for the fittings that are defined with a variable that necessitates turbulence (ft). This is where I think my confusion with the highlighted portions of the text are applicable - the first portion says the K value may be used regardless of the flow regime while the second portion says turbulence is a requirement. If you read the whole section on page 2, you will see multiple instances of the author emphasizing the turbulent regime as a necessary condition for the applicability of K = f* Leq/D

I understood the author's intent to separate f from ft because in the author's view, the head loss through fittings and pipe are different and should be calculated separately, as they detail in the following section and into page 3. Even if you follow the author's prescribed method of performing the calculations, if the flow is in the laminar regime, you end up adding a head loss calculated with a laminar pipe friction factor and a head loss that uses a turbulent flow regime K value for the fittings.

Edited by jayari, 12 March 2020 - 10:56 AM.

### #6 breizh

breizh

Gold Member

• 5,628 posts

Posted 13 March 2020 - 01:40 AM

Hi ,

Yes time to time I can get access to Neutrium , seems more difficult than before (time delay).

Note: I've attached an extract from Ron Darby's Book :Chemical engineering fluid mechanics 2nd edition where 4 methods are compared .

Good luck.

Breizh

Edited by breizh, 13 March 2020 - 06:34 AM.