Transformer losses and maximum efficiency condition

Losses in a transformer

A transformer has no moving parts, so there are only two kinds of loss.

1. Core (iron) losses $P_i$: present whenever the primary is energised at rated voltage and frequency. They are practically constant from no load to full load.
   • Hysteresis loss: $P_h = \eta B_{max}^{1.6} f V$, from the repeated reversal of magnetisation. It is reduced by using silicon steel.
   • Eddy current loss: $P_e \propto B_{max}^2 f^2 t^2$, from currents induced in the core. It is reduced by thin insulated laminations.
2. Copper losses $P_{cu}$: $I_1^2R_1 + I_2^2R_2 = I_2^2R_{02}$ in the windings. They vary as the square of the load current.

Stray (leakage flux) and dielectric losses are small and usually ignored.

Condition for maximum efficiency

Let the load current be $I_2$, secondary voltage $V_2$ (assumed constant), power factor $\cos\phi_2$ and $R_{02}$ the total resistance referred to the secondary.

$$\eta = \frac{V_2I_2\cos\phi_2}{V_2I_2\cos\phi_2 + P_i + I_2^2R_{02}}$$

Divide numerator and denominator by $I_2$:

$$\eta = \frac{V_2\cos\phi_2}{V_2\cos\phi_2 + \dfrac{P_i}{I_2} + I_2R_{02}}$$

For constant $V_2$ and $\cos\phi_2$, $\eta$ is maximum when the denominator is minimum:

$$\frac{d}{dI_2}\left(\frac{P_i}{I_2} + I_2R_{02}\right) = 0 \Rightarrow -\frac{P_i}{I_2^2} + R_{02} = 0$$ $$P_i = I_2^2R_{02}$$

Condition: efficiency is maximum when the variable (copper) loss equals the constant (iron) loss.

Load at maximum efficiency, with $P_{cu,fl}$ the full-load copper loss:

$$x = \frac{I_2}{I_{2,fl}} = \sqrt{\frac{P_i}{P_{cu,fl}}},\qquad \text{kVA at max } \eta = \text{rated kVA}\times\sqrt{\frac{P_i}{P_{cu,fl}}}$$

The second derivative is positive, so this is a minimum of the denominator and a maximum of $\eta$. The condition does not depend on the power factor, but the value of the maximum efficiency is higher at a higher power factor.