Cupper losses

Thread Starter

TheSpArK505

Joined Sep 25, 2013
113
Hello guys, hope u fine.

I have encountered a formula for calculating the percentage of cupper losses in a distribution system after employing a capacitor bank. It then jumps to conclude the cupper losses percentage in terms of power factor o_O!! Can anyone explain the jumped formulas please and how can I conclude a cupper losses equation out of power factor!!

SEE ATTACHED FIGURE
 

Attachments

sailorjoe

Joined Jun 4, 2013
361
Spark, the story is really pretty simple.
The first equation relates to the ability to reduce copper losses by reducing the current used to carry the power over a transmission line. Reduce the current by increasing the voltage. But reducing the current has a squared relationship to reduced power, per the first equation. So that solves one problem.

But then, when the loads have low power factors there's more current than normal, so we add capacitor banks to compensate. This raises the power factor, and the improvement in copper losses is again multiplied by the square of the reduced current.

Clear?
 

Thread Starter

TheSpArK505

Joined Sep 25, 2013
113
Thanks to your reply. It seems that u misunderstood me:D. I wanted to know how to derive a cupper losses equation based purely on power factor. So it's a derivation issue.:oops:

@sailorjoe
 

SLK001

Joined Nov 29, 2011
1,543
What is a "cupper" and how does it have losses? Is that a pub thing? Maybe a lossy cupper is a cup with a hole in it. If so, then the losses should be easy to calculate.
 

sailorjoe

Joined Jun 4, 2013
361
Thanks to your reply. It seems that u misunderstood me:D. I wanted to know how to derive a cupper losses equation based purely on power factor. So it's a derivation issue.:oops:

@sailorjoe
OK, understand. The formula you have for %LossReduction is a ratio of comparative power factors, so you don't see current in the equation. Then again, the power factor itself is a ratio of true power to apparent power, so it's still not a direct relationship to current.

True power = P = I^2 R, And Apparent power in a reactive circuit = P = I^2 Z.

In typical power systems, the major reactive loads are inductive, such as large motors for air conditioners and other machinery. So the power factor means that the system needs more current than it would for a purely resistive load. That's why they add capacitor banks, to bring the inductive load back to a purely resistive load. Check out this article. http://www.allaboutcircuits.com/textbook/alternating-current/chpt-11/calculating-power-factor/
 
Last edited:
Top