Transformer losses and condition for maximum efficiency

A transformer has no moving parts, so it has no friction or windage loss. Its losses are core (iron) losses and copper losses.

Losses in a transformer

1. Core or iron loss ($P_i$) – occurs in the core due to the alternating flux. It depends on $B_m$ and $f$, which are almost constant, so it is a constant loss (the same at all loads). It is measured by the open-circuit test.
   • Hysteresis loss: $P_h = \eta B_m^{1.6} f V$; reduced by using silicon steel / CRGO.
   • Eddy current loss: $P_e = K_e B_m^2 f^2 t^2 V$; reduced by thin insulated laminations.
2. Copper loss ($P_{cu}$) – $I_1^2R_1 + I_2^2R_2 = I_2^2 R_{02}$ in the windings. It varies with the square of the load current, so it is a variable loss. The full-load value is measured by the short-circuit test. At a fraction x of full load, $P_{cu} = x^2 P_{cu,fl}$.
3. Stray loss – from leakage flux causing eddy currents in the tank, clamps and conductors. Usually small and included in copper loss.
4. Dielectric loss – in the insulation (oil, paper); important only at very high voltage.

Efficiency

$$\eta = \frac{\text{output}}{\text{output} + \text{losses}} = \frac{V_2 I_2\cos\phi_2}{V_2 I_2\cos\phi_2 + P_i + I_2^2 R_{02}}$$

Condition for maximum efficiency

Take $V_2$ and $\cos\phi_2$ as constant. Dividing the numerator and denominator by $I_2$:

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

η is maximum when the varying part of the denominator, $D = \dfrac{P_i}{I_2} + I_2R_{02}$, is minimum:

$$\begin{aligned} \frac{dD}{dI_2} &= -\frac{P_i}{I_2^2} + R_{02} = 0 \\ I_2^2 R_{02} &= P_i \end{aligned}$$

($d^2D/dI_2^2 = 2P_i/I_2^3 > 0$, so it is a minimum of D.)

Efficiency is maximum when variable copper loss equals constant iron loss.

Load at maximum efficiency

$$x = \frac{I_2}{I_{2,fl}} = \sqrt{\frac{P_i}{P_{cu,fl}}}, \qquad \text{kVA at } \eta_{max} = \text{rated kVA}\times\sqrt{\frac{P_i}{P_{cu,fl}}}$$ $$\eta_{max} = \frac{x S\cos\phi}{x S\cos\phi + 2P_i}$$

Also, for a given kVA, η is highest at unity power factor.

Example: a transformer with $P_i = 1$ kW and $P_{cu,fl} = 4$ kW has maximum efficiency at $x = \sqrt{1/4} = 0.5$, i.e. half load. Distribution transformers are designed with low $P_i$ so that η is maximum at about 50–70% load, as they are lightly loaded for much of the day.