Question

In a 25kVA, 2000/200 V transformer the iron and copper losses are 350W and
400W respectively. Calculate the values of iron and copper losses which will give
maximum efficiency and also calculate the value of maximum efficiency.​

Answer 1

Answer:

Dear JD,

This is a good question that pertains to transformers as well as induction motors. Please allow me to add a few comments to the conversation.

As a first approximation, we can assume that the maximum efficiency occurs at about 75% of the rated power. For most electromechanical machines, this is where the copper loss is equal to the core loss. The result is a curve that looks like this.

Continuing on with our first approximation, 75% of 2000 = 1500W.

η=PoutPIn=PoutPOut+PCore+I2R=15001500+100+100≈88%

Know that the core losses are considered a constant, as they are a function of the applied voltage and frequency - both assumed to be constant in this example. The copper losses vary with the electrical load applied to the transformer or the mechanical load applied to a motor. Actually, the copper loses vary as the square of the current - from your electronics studies you will recall the term I2R . We can express this using a power flow diagram:

As other have pointed out, the maximum efficiency occurs at some k percent of the output power:

PCore=PCopper∗k2

For this particular transformer:

100=200∗k2

k=0.7071

POut=0.7071∗2000=1414W

Finally:

η=14141414+100+100=87.6%

Best Wishes,

APDahlen

P.S. You need a bit of calculus to show that PCore=PCopper∗k2