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# Efficiency of a Transformer

Transformers which are connected to the power supplies and loads and are in operation are required to handle load current and power as per the requirements of the load.

Efficiency

Transformers which are connected to the power supplies and loads and are in operation are required to handle load current and power as per the requirements of the load. An unloaded transformer draws only the magnetization current on the primary side, the secondary current being zero. As the load is increased the primary and secondary currents increase as per the load requirements. The volt amperes and wattage handled by the transformer also increases. Due to the presence of no load losses and I2R losses in the windings certain amount of electrical energy gets dissipated as heat inside the transformer.

This gives rise to the concept of efficiency. Efficiency of a power equipment is defined at any load as the ratio of the power output to the power input. Putting in the form of an expression, while the efficiency tells us the fraction of the input power delivered to the load, the deficiency focuses our attention on losses taking place inside transformer. As a matter of fact the losses heat up machine. The temperature rise decides the rating of the equipment. The temperature rise of the machine is a function of heat generated the structural configuration, method of cooling and type of loading (or duty cycle of load). The peak temperature attained directly affects the life of the insulations of the machine for any class of insulation.

These aspects are briefly mentioned under section load test. The losses that take place inside the machine expressed as a fraction of the input is sometimes termed as deficiency. Except in the case of an ideal machine, a certain fraction of the input power gets lost inside the machine while handling the power. Thus the value for the efficiency is always less than one. In the case of a.c. machines the rating is expressed in terms of apparent power. It is nothing but the product of the applied voltage and the current drawn. The actual power delivered is a function of the power factor at which this current is drawn.

As the reactive power shuttles between the source and the load and has a zero average value over a cycle of the supply wave it does not have any direct effect on the efficiency. The reactive power however increases the current handled by the machine and the losses resulting from it. Therefore the losses that take place inside a transformer at any given load play a vital role in determining the efficiency. The losses taking place inside a transformer can be enumerated as below:

1.              Primary copper loss

2.              Secondary copper loss

3.              Iron loss

4.              Dielectric loss

These are explained in sequence below.

Primary and secondary copper losses take place in the respective winding resistances due to the flow of the current in them. The primary and secondary resistances differ from their d.c. values due to skin effect and the temperature rise of the windings. While the average temperature rise can be approximately used, the skin effect is harder to get analytically. The short circuit test gives the value of Re taking into account the skin effect. The iron losses contain two components Hysteresis loss and Eddy current loss. The Hysteresis loss is a function of the material used for the core. Ph = KhB1.6f For constant voltage and constant frequency operation this can be taken to be constant. The eddy current loss in the core arises because of the induced emf in the steel lamination sheets and the eddies of current formed due to it. This again produces a power loss Pe in the lamination. Where t is the thickness of the steel lamination used. As the lamination thickness is much smaller than the depth of penetration of the field, the eddy current loss can be reduced by reducing the thickness of the lamination. Present day laminations are of 0.25 mm thickness and are capable of operation at 2 Tesla.

These reduce the eddy current losses in the core.This loss also remains constant due to constant voltage and frequency of operation. The sum of hysteresis and eddy current losses can be obtained by the open circuit test.The dielectric losses take place in the insulation of the transformer due to the large electric stress. In the case of low voltage transformers this can be neglected. For constant voltage operation this can be assumed to be a constant. The stray load losses arise out of the leakage fluxes of the transformer. These leakage fluxes link the metallic structural parts, tank etc. and produce eddy current losses in them. Thus they take place ’all round’ the transformer instead of a definite place, hence the name ’stray’. Also the leakage flux is directly proportional to the load current unlike the mutual flux which is proportional to the applied voltage.

Hence this loss is called ’stray load’ loss.This can also be estimated experimentally. It can be modeled by another resistance in the series branch in the equivalent circuit. The stray load losses are very low in air-cored transformers due to the absence of the metallic tank. Thus, the different losses fall in to two categories Constant losses (mainly voltage dependant) and Variable losses (current dependant). The expression for the efficiency of the transformer operating at a fractional load x of its rating, at a load power factor of 2 can be written as losses and Pvar the variable losses at full load. For a given power factor an expression for _ in terms of the variable x is thus obtained. By differentiating _ with respect to x and equating the same to zero, the condition for maximum efficiency is obtained. The maximum efficiency it can be easily deduced that this Maximum value increases with increase in power factor and is zero at zero power factor of the load. It may be considered a good practice to select the operating load point to be at the maximum efficiency point.

Thus if a transformer is on full load, for most part of the time then the max can be made to occur at full load by proper selection of constant and variablelosses.However, in the modern transformers the iron losses are so low that it is practically impossible to reduce the full load copper losses to that value. Such a design wastes lot of copper. This point is illustrated with the help of an example below. Two 100 kVA transformers A and B are taken. Both transformers have total full load losses to be 2 kW. The breakup of this loss is chosen to be different for the two transformers. Transformer A: iron loss 1 kW, and copper loss is 1 kW. The maximum efficiency of 98.04%occurs at full load at unity power factor. Transformer B: Iron loss =0.3 kW and full load copper loss =1.7 kW. This also has a full load of 98.04%. Its maximum occurs at a fractional load of q0.31.7 = 0.42. The maximum efficiency at unity power factor being at the corresponding point the transformer A has an efficiency of Transformer A uses iron of more loss per kg at a given flux density, but transformer B uses lesser quantity of copper and works at higher current density.

Power output = Power input  - Total Losses

Power input = Power output + Total Lossess

= Power Output + Pi + Pcu

h = Power Output / Power Input

= Power Output / [Power Output + Pi + Pcu]

Now Power Output  = V2I2cosϕ

h = [ (VA rating) x cosϕ ] / [ (VA rating) x cosϕ + Pi + I22 R2e ] x 100

All day efficiency