Computing Copper Loss: Examples & Reduction Strategies

[energysaver]

Good day. just want to ask anyone about copper loss. How is it computed for example in a manufacturing plant or in a dept mall.Could anyone give an illustration/example?

Also , is it true that copper loss can be reduced by using big wires?What if the wire is big enough?

thank you. hope you can help me...

 

[Jiggy-Ninja]

Good day. just want to ask anyone about copper loss. How is it computed for example in a manufacturing plant or in a dept mall.Could anyone give an illustration/example?

Also , is it true that copper loss can be reduced by using big wires?What if the wire is big enough?

thank you. hope you can help me...

Copper loss is the power lost because of the resistance in the wires. P = I²R.

Big wires have less resistance, but with AC current there is such a thing as the "skin effect", which means that most of the current is concentrated around the surface of the conductor, lowering the effective cross sectional area of the conductor and increasing its resistance.

 

[energysaver]

what estimated percentage of a copper loss in a certain manufacturing plant? if i install capacitors in loads , would the demand at the meter be reduced by 10% by eliminating copper loss?...

thank you...

 

[Bob S]

You are correct. If you have an inductive load (induction motor, for example) with a poor power factor, a capacitor could reduce the reactive current and the resultant current demand at the meter. Power meters usually measure only real power, and not reactive power, however, so as you point out, only the extra I²R losses in the copper are reduced.

Bob S

 

[adool_617]

Hello

as our colleagues says in the above replies , copper losses is equivalent to I^2 X R

but what type of application are you going to calculate copper losses for ??

for example , if you are going to calculate copper losses for a flat busbar just calculate the resistance by knowing the cross sectional area of the bar and its length and the grade of copper you are using and then calculate it by resistivity x length divided by area then multiply the result by the current squared

but if you need to calculate this losses for temperature rise or when considering the heat effect for a transformer as an example then take into your consideration the current density , and the heat dissipation to the surrounding , and there are specific tables which can guide you to select the value of the current density in the first place as if you only looks at the value of the losses , it will not give you good indication for temperature rise

hope information was useful for you