Types of Losses in a Transformer – Iron and Copper Losses

Digital ElectronicsElectronElectronics & Electrical

The power losses which occur in a transformer are of two types −

  • Core or Iron Losses (Pi)
  • Copper Losses or I2R Losses (Pcu)

The losses in a transformer appears in the form of heat which increases the temperature and reduces the efficiency.

Core or Iron Losses (Pi)

The iron losses occur in the core of the transformer due to the alternating flux. These losses consist of hysteresis loss (Ph) and eddy current loss (Pe), i.e.,

$$\mathrm{P_{i}=P_{h}+P_{e}}$$

The hysteresis and eddy-current losses are given by,

$$\mathrm{Hysteresis\:loss,\: P_{h} \:= \:k_{h}f_{max}^{1.6} \:1.6 V \:Watt}$$

$$\mathrm{Eddy\:current \:loss, \:P_{e} \:= \:K_{e}f^2B_{max}^{2}t^2 V \:Watt}$$

Where,

  • kh = hysteresis coefficient,
  • ke = eddy current coefficient,
  • = supply frequency,
  • Bmax = maximum flux density,
  • t = thickness of each lamination, and
  • V = volume of core.

The exponent 1.6 of the maximum flux density is known as Steinmetz coefficient. Now, the iron losses can also be written as,

$$\mathrm{P_{i}\:=\:k_{h}f_{max}^{1.6} \:1.6 V \:+\:K_{e}f^2B_{max}^{2}t^2 V}$$

As we can see, both the hysteresis and eddy current losses are the function of maximum flux density in core, supply frequency and volume of the core material which is constant. Since, the transformer is connected to a source of constant frequency so that both f and Bmax are constant. Therefore, the iron or core losses are practically constant for a transformer at all loads. The iron losses of a transformer can be determined by open-circuit test.

In order to reduce the hysteresis loss, the core should be made up of high silicon content steel whereas to minimise the eddy current loss, the core is built-up by using thin laminations.

Copper or I2R Losses

The copper losses take place in the primary and secondary windings due to resistances of windings and can be determined by short circuit test.

The total copper loss (Pcu) in a transformer is given by,

$$\mathrm{P_{cu}\:=\: Primary \:winding \:cu \:loss \:+ \:Secondary \:wind\:ng\: cu\: loss}$$

$$\mathrm{⇒\:P_{cu}\:=\:I_{1}^{2}R_{1}\:+\:I_{2}^{2}R_{2}}$$

Since,

$$\mathrm{N_{1}I_{1}\:=\:N_{2}I_{2}}$$

$$\mathrm{⇒\:I_{1}\:=\:\frac{N_{2}}{N_{1}}I_{2}}$$

$$\mathrm{\therefore\:P_{cu}\:=\:(\frac{N_{2}}{N_{1}})^2I_{2}^{2}R_{1}\:+\:I_{2}^{2}R_{2}\:=\:I_2^2((\frac{N_{2}}{N_{1}})^2R_{1}+R_{2})\:=\:I_{2}^{2}R_{02}}$$

Also,

$$\mathrm{P_{cu}\:=\:I_{1}^{2}R_{1}\:+\:I_{2}^{2}R_{2}\:=\:I_{1}^{2}R_{1}+(\frac{N_{1}}{N_{2}})^2I_{1}^{2}R_{1}\:=\:I_{1}^{2}R_{01}}$$

Therefore,

$$\mathrm{P_{cu}\:=\:I_{2}^{2}R_{02}\:=\:I_{1}^{2}R_{01}}$$

Points to Note 

  • Stray Loss – A practical transformer has leakage reactance, which produces eddy currents in the conductors, transformer tank and in other metallic parts which causes the losses in the transformer. These losses are called stray losses.
  • Dielectric Loss – The dielectric losses occur in the insulating material (transformer oil or solid insulation) of the transformer. The dielectric loss is significant only in high voltage transformers.
raja
Published on 18-Aug-2021 06:15:06
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