K factor in calculating the Load Loss Factor

[samy81]

Hello Everyone,
I hope my message finds you well.

I am trying to estimate the annual average losses in a feeder by using the Load loss factor as follow:

Annual Average losses = Annual peak losses(kwh)× Load Loss Factor

where:

Load Loss Factor= [(K)×(Load Factor) +(1-k)× (Load Factor)^2 ]

where K is defined as the constant K (from 0 to 1).

As per what I was told, the factor k depends on the country and region. is there any documentation that lists the constant K? is there a way to estimate this value?

Thank you in advance and best regards,
Samy

 

[Skogsgurra]

I know k-factor from sizing transformers with regard to harmonics contents. But haven't seen it used the way you describe.
There may be different k-factors but this ABB document says a lot about it when dealing with transformers.

https://library.e.abb.com/public/be...mers_with_non_sinusoidal_currents_KFactor.pdf

I guess that it can be used also for transmission in general. But never seen it used that way. The Fluke 41 analyzer does k-faktor automatically. Yes, it is old. But that's the one I am used to.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[aussiejohn2]

G'day Samy81,

Typically k is considered to lie in the range 0.15 to 0.3. It is usually lower further down in the system (less diversity).

John.

 

[jghrist]

I second aussiejohn2. The range of 0.15 to 0.3 comes from the Westinghouse Distribution Reference Book, 1959. It depends on load characteristics. If you have hourly loads for your system, you can calculate the load factor as average load divided by peak load and loss factor as average load squared divided by peak load.

 

[7anoter4]

In The Calculation of the Temperature Rise and Load Capability of Cable Systems by J.H.Neher and M.H.McGrath

http://www.mosaic-industries.com/em...uct-development/cables/neher-mcgrath-1957.pdf

this formula appears as equation no.3:
LF=0.3*lf+0.7*sqrt(1-0.3)*lf^2
where LF = loss factor per unit
lf=load factor per unit

 

[7anoter4]

Sorry:
LF=0.3*lf+0.7*lf^2