Losses in a Transformer

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The following power losses may occur in a practical transformer −

• Iron Loss or Core Loss

• Copper Loss or I²R Loss

• Stray Loss

• Dielectric Loss

In a transformer, these power losses appear in the form of heat and cause two major problems −

• Increases the temperature of the transformer.

• Reduces the efficiency of the transformer.

Iron Loss or Core Loss

Iron loss occurs in the magnetic core of the transformer due to flow of alternating magnetic flux through it. For this reason, the iron loss is also called core loss. We generally use the symbol ($\mathit{P_{i}}$) to represent the iron loss. The iron loss consists of hysteresis loss ($\mathit{P_{h}}$) and eddy current loss ($\mathit{P_{e}}$). Thus, the iron loss is given by the sum of the hysteresis loss and eddy current loss, i.e.

$$\mathrm{\mathrm{Iron\:loss,}\mathit{P_{i}}\:=\:\mathrm{Hysteresies\:loss(\mathit{P_{h}})}\:+\:\mathrm{Eddy\:current\:loss(\mathit{P_{e}})}}$$

The hysteresis loss and eddy current loss (or iron loss) are determined by performing the open-circuit test on the transformer.

The empirical formulae for the hysteresis loss and eddy current loss are given by,

$$\mathrm{\mathit{P_{h}}\:=\:\mathit{k_{h}f\:B_{m}^{x}}\:\cdot \cdot \cdot (1)}$$

$$\mathrm{\mathit{P_{e}}\:=\:\mathit{ke\:B_{m}^{\mathrm{2}}\:f^{\mathrm{2}}t^{\mathrm{2}}}\:\cdot \cdot \cdot (2)}$$

Where,

• The exponent of B_m, i.e. "x" is called the Steinmetz’s constant. Depending on the properties of the core material, its value is ranging from 1.5 to 2.5.

• k_h is a proportionality constant whose value depends upon the volume and quality of the material of core.

• k_e is a proportionality constant which depend on the volume and resistivity of material of the core.

• f is the frequency of the alternating flux in the core.

• B_m is the maximum flux density in the core.

• t is the thickness of each core lamination.

Therefore, the total iron loss or core loss can also be written as,

$$\mathrm{\mathit{P_{i}}\:=\:\mathit{k_{h}f\:B_{m}^{x}}\:+\:\mathit{ke\:B_{m}^{\mathrm{2}}\:f^{\mathrm{2}}t^{\mathrm{2}}}\:\cdot \cdot \cdot (3)}$$

Since the input voltage to the transformer is approximately equal to the induced voltage in the primary winding, i.e.

$$\mathrm{\mathit{V_{\mathrm{1}}}\:=\:\mathit{E_{\mathrm{1}}}\:=\:4.44\:\mathit{f\phi _{m}N_{\mathrm{1}}}}$$

$$\mathrm{\Rightarrow \mathit{V_{\mathrm{1}}}\:=\:4.44\:\mathit{f\:B_{m}AN_{\mathrm{1}}}}$$

Where, A is the cross-sectional area of the transformer core, N₁ is the number of turns in the primary winding and f is the supply frequency.

$$\mathrm{\therefore \mathit{B_{m}}\:=\:\frac{\mathit{V_{\mathrm{1}}}}{4.44\mathit{fAN_{\mathrm{1}}}}\:\cdot \cdot \cdot (4)}$$

Hence, from equations (1) & (4), we get,

$$\mathrm{\mathit{P_{h}}\:=\:\mathit{k_{h}f}\left ( \frac{\mathit{V_{\mathrm{1}}}}{4.44\mathit{fAN_{\mathrm{1}}}} \right )^{x}}$$

$$\mathrm{\Rightarrow \mathit{P_{h}}\:=\:\mathit{k_{h}f}\left ( \frac{\mathrm{1}}{4.44\mathit{AN_{\mathrm{1}}}} \right )^{x}\cdot \left ( \frac{\mathit{V_{\mathrm{1}}}}{\mathit{f}} \right )^{x}}$$

$$\mathrm{\Rightarrow \mathit{P_{h}}\:=\:\mathit{k_{h}}\left ( \frac{\mathrm{1}}{4.44\mathit{AN_{\mathrm{1}}}} \right )^{x}\cdot \mathit{V_{\mathrm{1}}^{x}}\:\mathit{f^{(\mathrm{1}-x)}}\:\cdot \cdot \cdot (5)}$$

Thus, Equation (5) shows that the hysteresis loss depends upon both input voltage and supply frequency.

Again, from equations (2) & (4), we get,

$$\mathrm{\mathit{P_{e}}\:=\:\mathit{k_{e}f^{\mathrm{2}}t^{\mathrm{2}}}\left ( \frac{\mathit{V_{\mathrm{1}}}}{4.44\mathit{fAN_{\mathrm{1}}}} \right )^{\mathrm{2}}}$$

$$\mathrm{\Rightarrow \mathit{P_{e}}\:=\:\mathit{k_{e}\left ( \frac{\mathit{V_{\mathrm{1}}}}{\mathrm{4.44}\mathit{AN_{\mathrm{1}}}} \right )^{\mathrm{2}}\mathit{t^{\mathrm{2}}}\:\cdot \cdot \cdot \mathrm{(6)}}}$$

Hence, from equation (6), we can conclude that the eddy current loss in the transformer is proportional to the square of the input voltage and is independent of the supply frequency.

Therefore, the total core loss can also be written as,

$$\mathrm{\mathit{P_{i}}\:=\:\mathit{k_{h}\left ( \frac{\mathrm{1}}{\mathrm{4.44}\mathit{AN_{\mathrm{1}}}} \right )^{\mathrm{2}}\cdot \mathit{V_{\mathrm{1}}^{\mathit{x}}f^{(\mathrm{1-x})}}\:+\:\mathit{k_{e}}\left ( \frac{V_{\mathrm{1}}}{\mathrm{4.44}\mathit{AN_{\mathrm{1}}}} \right )^{\mathrm{2}}\mathit{t^{\mathrm{2}}}\:\cdot \cdot \cdot \left ( \mathrm{7} \right )}}$$

In practice, transformers are connected to an electric supply of constant frequency and constant voltage, thus, both f and B_m are constant. Therefore, the core or iron loss is practically remains constant at all loads.

We can reduce the hysteresis loss by using steel of high silicon content to construct the core of transformer while the eddy current loss can be minimized by using core of thin laminations instead of solid core. The open-circuit test is performed on a transformer to determine the iron or core loss.

Copper Loss or I²R Loss

Power loss in a transformer that occurs in both the primary and secondary windings due to their Ohmic resistance is called copper loss or I²R loss. We usually represent the copper loss by PC. Therefore, the total copper loss in a transformer is the sum of power loss in the primary winding and power loss in the secondary winding, i.e.,

$$\mathrm{\mathit{P_{c}}\:=\:\mathrm{Copper\:loss\:in\:primary\:+\:Copper\:loss\:in\:secondary}}$$

$$\mathrm{\Rightarrow \mathit{P_{c}}\:=\:\mathit{I_{\mathrm{1}}^{\mathrm{2}}}\mathit{R_{\mathrm{1}}}\:+\:\mathit{I_{\mathrm{2}}^{\mathrm{2}}}\mathit{R_{\mathrm{2}}}\:\cdot \cdot \cdot (8)}$$

Since,

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

$$\mathrm{\Rightarrow \mathit{I_{\mathrm{1}}}\:=\:\left ( \frac{\mathit{N_{\mathrm{2}}}}{\mathit{N_{\mathrm{1}}}} \right )\mathit{I_{\mathrm{2}}}\:\cdot \cdot \cdot (9)}$$

$$\mathrm{\therefore \mathit{P_{c}}\:=\:\left [ \left ( \frac{\mathit{N_{\mathrm{2}}}}{\mathit{N_{\mathrm{1}}}} \right )I_{\mathrm{2}} \right ]^{\mathrm{2}}\:\mathit{R_{\mathrm{1}}}\:+\:\mathit{I_{\mathrm{2}}^{\mathrm{2}}}\mathit{R_{\mathrm{2}}}\:=\:\left [ \left ( \frac{\mathit{N_{\mathrm{2}}}}{\mathit{N_{\mathrm{1}}}} \right )^{\mathrm{2}}\mathit{R_{\mathrm{1}}}\:+\:\mathit{R_{\mathrm{2}}} \right ]\mathit{I_{\mathrm{2}}^{\mathrm{2}}}\:\cdot \cdot \cdot (10)}$$

From Equation (10), it is clear that the copper loss in a transformer varies as the square of the load current. For this reason, the copper loss is also referred as "variable loss" because in practice a transformer is subjected to variable load and hence has variable load current.

We conduct the "short-circuit test" on the transformer to determine the value of its copper loss. In a practical transformer, the copper loss accounts for about 90% of the total power loss in the transformer.

Stray Loss

In practical transformer, a fraction of the total flux follows a path through air and this flux is called leakage flux. This leakage flux produces eddy currents in the conducting or metallic parts like tank of the transformer. These eddy currents cause power loss, which is known as stray loss.

Dielectric Loss

The power loss occurs in insulating materials like oil, solid insulation of the transformer, etc. is known as dielectric loss. The dielectric loss is significant only in transformers working on high voltages.

Although, in practice, the stray loss and dielectric loss are very small, constant and may be neglected.

From the above discussion, we found that a transformer has some losses which are constant and some other are variable. Thus, we may categorize losses in a transformer in two types namely constant losses and variable losses.

Therefore, the total losses in a transformer are the sum of constant losses and variable losses, i.e.,

Total losses in transformer = Constant losses + Variable losses