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I have been going through the Handbook of Electricity Metering (11th edition) for my new job. They give the following equation for transformer watt loss due to core/line losses. I am not an EE so this is all new to me. I am trying to understand where these equations come from and intuitively understand what they are doing.

For instance on the iron-losses equation, I can see it is multiplying the no-load watts by some sort of efficiency ratio. Why those two voltage values? Why are they squared?

On the copper loss equation, why does it use 50% of meter class amperes?

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ocrdu
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compto2017
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  • Please describe (maybe using a picture) of what **meter rated voltage** and **rated transformer volts** refer to. I think generally, they may be referring to a 3 phase star would primary because they appear to be converting line voltage to phase voltage with the sqrt(3). But, without access to the book you refer to many here will be in the dark about the formulas. – Andy aka Apr 11 '23 at 13:54

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These equations are of the form

losses = constant * parameter^power

I can't help you with the constant term, that's buried somewhere in the assumptions around what 'meter rated voltage' and suchlike mean.

I can help with which parameter, and what power it's raised to.

Core losses have two elements, magnetic hysteresis losses, and eddy current losses.

The eddy current losses depend on the voltage induced by the changing flux driving a current round little conductive loops in the core. As such, they depend on the core flux, which only depends on the voltage. As the losses depend on the square of the current that those induced voltages cause to flow, the loss is proportional to voltage^2.

The hysteresis losses depend on taking the magnetic core round a BH loop. The amount of energy required to perform each loop is proportional to the area of the loop, or to B*H. If the transformer is operated below saturation (as all good transformers are), then B is proportional to H, and the power goes as the flux squared. As the flux depends only on the terminal voltage, this loss is also proportional to voltage^2.

The copper losses are proportional to I2R, where R is the (roughly constant) winding resistance. The copper loss is therefore proportional to load current^2.

Neil_UK
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