8/6/2019 Efficiency of Transformers

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Efficiency of transformers

In practice, real transformers are less than 100% efficient.

• First, there are resistive losses in the coils (losing power I 2.r). For a given material, the resistance of the coils can bereduced by making their cross section large. The resistivity can also be made low by using high purity copper.(See Drift velocity and Ohm's law .)

• Second, there are some eddy current losses in the core. These can be reduced by laminating the core. Laminations

reduce the area of circuits in the core, and so reduce the Faraday emf, and so the current flowing in the core, and sothe energy thus lost.

• Third, there are hysteresis losses in the core. The magentisation and demagnetisation curves for magnetic materialsare often a little different (hysteresis or history depedence) and this means that the energy required to magnetise thecore (while the current is increasing) is not entirely recovered during demagnetisation. The difference in energy islost as heat in the core.

• Finally, the geometric design as well as the material of the core may be optimised to ensure that the magnetic fluxin each coil of the secondary is nearly the same as that in each coil of the primary.

General

The efficiency of a transformer is very important in terms of energy lost. Efficiency deals with the power losses of a transformer. The less efficient a transformer is, the more heat it dissipates. Efficiency, inmathematical terms, is the ratio of the power out to the power in of a transformer, where the power in isequal to the power out plus the losses. The symbol of efficiency is η. For example:

η = ( Power out/ Power in ) x 100 and Power in = Power out + losses

Why is it Important to Consider Efficiency?During operation, a transformer’s main purpose is to transfer electrical energy from the primary coil to thesecondary coil. The iron core and copper coils of the transformer will convert some of the electrical energyinto heat energy. Because of this conversion, the transformer dissipates heat during operation. Heat

produced by the transformer represents steel excitation losses and copper losses, and makes the transformer less efficient.

A ll Day Efficiency of a Distribution Transformer The commercial Efficiency of a transformer is given by the ratio of output power to input power

Efficiency = output power in watts/input power in watts

The losses in the transformer can be classified into copper losses and iron losses. The copper losses (hysteresis and Eddy CurrentLosses) are independent of the load. The iron losses though are dependent on the load.

In the case of the distribution transformers, the load is continually varying. It is low in the day time and high in the evenings andnight. Therefore, efficiency measured at any one point of the day would not be an accurate reflection of the transformer’scapability.

Hence, We have the all day efficiency measurement of the distribution transformers

The formula for the all day efficiency of the distribution transformers is

Efficiency = Output in kWh for 24 hours/ 1nput in kWh for 24 hours

The All day Efficiency is always lesser than the commercial efficiency of the transformer.