Transmission Factor ? 5

[BCG1]

I am trying to calculated the Darcy friction factor and got stumped.

The specification that I am following uses the equation below to solve for the transmission factor:

F(oil) = 3.6 Log (Re/8)

Where is this equation coming from?

Then it uses the equation: F = 2 / f(d)^1/2 to solve for the friction factor f(d). That part I can verify.

A little background, I am trying to cacluate friction losses on diesel flow through a steel pipeline. Reynolds number is in the neighborhood of 45,000. I think the specification has assumed smooth pipe, which may/may not be correct.

Any insight would be much appreciated....can't seem to find anything in my hydraulics books.

BCG1

 

[vzeos]

BCG1,

The term Transmission Factor was coined in 1956 in the U. S. Bureau of Mines Monograph No 9, and it is defined as 1/[√]f. The reason for establishing this usage was due to the tendency of authors at the time to refer to f and 1/[√]f both as the friction factor, leading to a great deal of confusion.

I am aware that there are some authors out there who think that the Transmission Factor is defined as 2/[√]f. This misconception apparently comes from the lack of understanding the difference between the Darcy transmission factor, 1/[√]f[sub]D[/sub], and the Fanning friction factor, 1/[√]f[sub]F[/sub]. The Fanning transmission factor is numerically twice the Darcy transmission factor so that if 1/[√]f[sub]D[/sub]= 8.0 at a particular Reynolds number then 1/[√]f[sub]F[/sub]=16.0 at the same Reynolds number. This is due to the fact that the Darcy defining equation was developed using the pipe diameter and the Fanning defining equation was developed using the the wetted perimeter. In practice it really doesn’t matter which transmission factor convention you use as long as you use the consistent flow equation constant for that convention. For example:

Q= C x [√]((dP / (Ro x L)) x (D [sup] 2.5 [/sup]) x 1/[√]f

C = 68.04138 for Darcy
C = 34.02069 for Fanning
Q = gpm
L = feet
D = inches
dP = psid
Ro = lb/cu ft,1

The function, T[sub]oil[/sub] = 1/[√]f[sub]oil[/sub]=3.6 Log (Re/8), is an empirical representation of Prandlt’s theoretical smooth pipe law, 1/[√]f = 4Log(Re/1/[√]f) -0.6, both of which are Fanning transmission factor equations. T[sub]oil[/sub] provides good transmission factor values for the range 3,000>Re>3,000,000.

The following links may be useful:

http://emka.xs4all.nl/dP/index.htm

http://kirkmansoftware.com/

 

[zdas04]

vzeos,
Thank you for a very clear and authoritative explanation.

David

 

[katmar]

I would like to second David's vote of thanks. This excellent post has been carefully stored in the friction factor section of my filing system.

Harvey

 

[BCG1]

Ditto

Thank you very much!!

 

[BigInch]

Excellent. My thanks as well.

For basic system curve calculations for diesel pipelines, I have used this spreadsheet based on the Colebook Friction factors for multiple product pipelines gas, diesel, jet fuels, etc. for the last 10 years. Its accurate as long as you get the viscosities right for your products. Using a pipe roughness of 0.0018 will be a bit conservative, but pipelines often roughen up over time (but sometimes they can actually get even smoother!). For new pipe, 0.0008 to 0.0012 may match your initial flow curves better.

It's nothing special, but will give you something to to check your transmission factor equation against.

I have posted it here for all that wish to download,

http://rapidshare.com/files/16596703/Pipeline_System_Curve_Colebrook.XLS.html

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com