Tuesday 25 October 2011

efficiency of a transformer(x-mer)

CURRENT – VOLTAGE RELATION IN TRANSFORMER


Consider an ideal transformer.
In an ideal transformer due to no power loss, power input is equal to power output i.e.
Power input = Power out put
Pp = Ps [power = VI]
Vp Ip = Vs Is
Vp/Vs = Is/Ip
This expression indicates that current and voltage have an inverse relation with each other in transformer.

EFFICIENCY OF TRANSFORMER

Efficiency of a device is equal to the ratio of output to input.
Since,

Output = Ps
or
Output = Vs Is
and

Input = Pp
or
Input = Vp Ip

Efficiency = output / input
Efficiency = Ps / Pp
Efficiency = Vs Is / Vp Ip
Efficiency in percent = (Vs Is / Vp Ip) x 100

In actual practice output is not equal to input therefore actual transformers are not 100% efficient. However commercial transformers have very high efficiency in the range of 95% to 99%.
POWER LOSS IN TRANSFORMER
There are various factors that reduces the efficiency of a transformer such as:
EDDY CURRENTS
Due to variation in magnetic flux eddy currents are induced on the surface of iron core which in turn produce heating and therefore reduce the amount of power to the secondary coil.
In order to avoid eddy currents , the core is laminated, made of thin sheets of soft iron. Each sheet is separated from the next by a layer of insulating varnish.
HYSTERESIS LOSS
Each time the direction of magnetization is reversed, some useful energy is wasted in overcoming internal friction. This is known as "hysteresis loss" and it also produces heating in the core.
Hysteresis loss is minimized by using special alloys known as "perm alloy" for core material.
HEATING
Some energy is dissipated as heat in the coil.
This is reduced by using suitablythick wire.
FLUX TRANSPORT FAILURE
Some loss of useful energy occurs because a small amount of the flux associated with the primary coil fails to pass through the secondary.

   

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