Pipe friction loss - Equivalent Length or factor K? 1

[PalomaP]

Trying to understand how to use equivalent length or factor "k", I would like if someone could clarify the following.
Looking at the book: "Practical Centrifugal Pumps" (Paresh Girdhar & Octo Moniz), there is an example for calculating friction losses. However I understand that is mixing up friction calculated by equivalent length and "k" factor. I am attaching the example. I understand this is not right!! Calculating friction loss by "k" factor (hf=k.v^2/2g) is directly losses (NOT EQUIVALENT LENGTH)!!
I would appreciate it, if someone could give an explanation (if there is any) why the book is mixing up these concepts...

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Thanks

 

[katmar]

Unfortunately I cannot see your example, but in principle there is no problem with mixing L/D with K values. As long as each fitting has one or the other and not both you will get the correct answer. The formulas for some fittings are traditionally given as K values (eg reducers and orifices) while others are often given as L/D (eg bends and tees).

Keep in mind that both are estimates of the pressure drops and don't get too hung up on slight differences.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[PalomaP]

In the example, calculate friction loss using h=kv^2/2g and the valued is added to the equivalent length.
I.e:

From chart 140 l/s, 250 mm, sch 40, st
-> 2.30 m per 100m, v^2/2g=0.386

For foot valve with strainer:
k=5.88,k.v^2/2g=5.88x0.386 m -> 2.27 m equivalent lenght of pipe

So according to the example, the friction loss is 2.30x2.27/100=0.05 m.

I believe this is not right. I understand that it should be 2.27m friction!!in stead of 0.05m as in the book...

 

[Montemayor]

Paloma:

Go to

http://www.cheresources.com/eqlength.shtml

download and read Phil's detailed explanations on the L/D method and I believe you will disspell any problems you presently have with basic pipe friction loss.

 

[25362]

The source has apparently mixed up L_e/D with k (used as a multiplier of u²/2g).

 

[katmar]

The example has now become visible, and I am afraid that Mssrs Girdar and Moniz have seriously botched this calculation.

When you multiply the K value by the v²/2g as you have done the result is not the equivalent length of pipe. The result is the head as measured in column of the flowing liquid. Using your example for the foot valve with strainer, which has a K value of 5.88, the result of 5.88 x 0.386 does indeed give 2.27 but this is the pressure drop in meters of water. Girdar and Moniz then add this to the physical pipe length - an excellent example of adding apples and oranges and getting garbage. I hope the rest of the book is not this bad.

It is possible to convert between equivalent length and K values by looking at the Darcy Weisbach formula, i.e.
Head = ([ƒ]_ML/D + K) x v²/2g

where [ƒ]_M is the Moody friction factor.

The L/D value of a fitting can therefore be expressed as K/[ƒ]_M.

In this example [ƒ]_M will be about 0.015 and the L/D of the foot valve and strainer will be 5.88/0.015 = 392, making the equivalent length 392 x 0.25 = 98 meters. This is very different from the 2.27 meters G&M claim. If someone could sell foot valves with this low resistance they would make themselves very rich!

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[PalomaP]

katmar, thanks!! I have lost a few hours trying to find an explanation!! I feel like complaying to the publisher!!