24"x14" Concentric Reducer 2

[BGD011]

I am trying to find any source/chart about the pressure losses in the above black steel reducer using the cold water as an internal liquid in motion.

Do you know what should be the resistance coefficient (K)? Is it the same for an eccentric reducer?
Can I just use K=1 in the below equation?
(Or the K could be close to 0.05?)

h=k[(v1-v2)^2/2g]

Per this equation it will be the same with the flow in the opposite direction (14"x24")!?

Thank you in advance!

 

[JStephen]

For a gradual expansion (IE, cone shaped expansion), my fluids book shows K=h/(V1^2/(2g)) = 0.6 or so, depending on the inlet condtions and angle. Losses are higher going into an expansion than into a contraction; it's not the same both ways. I would expect that an eccentric reducer would have higher losses than a concentric reducer.

Weldbend has some flow loss data on their fittings, but not on reducers- plus, they don't show a 24x14 anyway.

 

[BGD011]

Thank you JStephen!
You have listed just the velocity v1 in the above equation, is it v2=0!?
I have seen that Weldbend doesn't show a reducer 24x14 in their catalog, they have one 24x16, but we can get that "custom" size.
I was wondering more about the pressure losses through the different size of a reducer in function of high flow.
Thanks one more!

 

[JStephen]

The graph I was looking at shows K as a function of the inlet velocity and angle only. This doesn't mean that V2 is zero, just that the head loss is not dependent on V2. This is also in a part of the graph where K varies quite a bit depending on the flare angle, so you'd need to confirm actual length before using that information.

Weldbend reducers aren't conical sections, they do have a smoothly curved profile. I'm assuming that the conical shape would be a bit higher loss than the reducer.

FYI- we sometimes buy reducers, and sometimes just have a cone formed to the appropriate dimensions.

 

[katmar]

It has always struck me as strange that almost all compilations of K-values include tapered (conical) reducers but very seldom include standard pipe reducers when in actual plants you will see MANY more standard reducers than you will ever see conical ones. Even the very practical and comprehensive Crane 410 manual ignores standard pipe reducers.

The article "Calculate Head Loss Caused by Change in Pipe Size" by William B Hooper, published in Chemical Engineering, Nov 7, 1988, pgs 89-92 does give a correlation specifically for standard reducers. But Hooper includes the caveat "The correlation given in the table looks reasonable, but no published data are available for checking its accuracy".

Hooper is the developer of the "2K" method. Traditionally K-values are assumed to be constant for a given type of fitting, irrespective of size or flowrate. The "2K" method is aimed at correcting the K values of various fitting for varying size and flowrate.

Hooper's correlation for standard pipe reducers in "reducing mode" is

K = (0.1 + 50/Re)((D/d)^4 - 1)

D is upstream diameter (larger diameter)
d is downstream diameter (smaller diameter)
Re is Reynolds number in upstream pipe (diam = D)

When used in "expanding mode" Hooper recommends that you assume a sudden expansion.

This correlation in "reducing mode" will give less friction than a conical reducer, so if you want to be conservative use the conical reducer K-value. It depends on the purpose of your calculation.

My catalog does show a 24"x14" reducer as being a standard, but I would expect them to be hard to come by. If you are going to make up a compound reducer (e.g. 24"x18" plus 18"x14") then you are going to introduce additional turbulence and losses. In this case it might be better to simply assume K=1, but as stated above, it depends on the purpose of your calculation.

Be careful with your h calculation - remember that (V1-V2)^2 is not equal to V1^2 - V2^2.

 

[BGD011]

Thank you very much Katmar on all excellent comments you have made up there!
Best Regards,
BGD011