Transformer losses, efficiency and maximum efficiency condition

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

1. Core or iron loss ($P_i$)

• Hysteresis loss: energy lost in repeatedly magnetising the core in opposite directions. $P_h = \eta_h B_m^{1.6} f V$ (Steinmetz). Reduced by using silicon steel (CRGO) with a narrow hysteresis loop.
• Eddy current loss: circulating currents induced in the core. $P_e = k_e B_m^2 f^2 t^2 V$, where $t$ is lamination thickness. Reduced by thin, insulated laminations (0.35-0.5 mm).
• Since $V$ and $f$ are constant, $B_m$ is constant, so iron loss is constant at all loads. It is measured by the open-circuit test.

2. Copper loss ($P_{cu}$)

• $I^2R$ loss in primary and secondary windings: $P_{cu} = I_1^2R_1 + I_2^2R_2 = I_1^2R_{01} = I_2^2R_{02}$.
• Varies with the square of load: at fraction $x$ of full load, $P_{cu} = x^2P_{cu,FL}$.
• Measured by the short-circuit test.

Minor losses: stray loss (leakage flux in tank and clamps) and dielectric loss in insulation, usually neglected.

Calculating efficiency

$$\eta = \frac{\text{Output}}{\text{Output} + \text{Losses}} = \frac{x\,S\cos\phi}{x\,S\cos\phi + P_i + x^2P_{cu,FL}}\times 100$$

where $S$ = rated VA. Because efficiency is very high (95-99%), direct measurement of input and output is inaccurate; it is therefore found indirectly from the OC and SC test losses.

All-day efficiency (for distribution transformers) = output energy in kWh / input energy in kWh over 24 hours.

Condition for maximum efficiency

With $V_2$ and $\cos\phi$ constant:

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

Dividing by $I_2$:

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

$\eta$ is maximum when the denominator is minimum, so

$$\begin{aligned} \frac{d}{dI_2}\left(\frac{P_i}{I_2} + I_2R_{02}\right) &= 0 \\ -\frac{P_i}{I_2^2} + R_{02} &= 0 \\ P_i &= I_2^2R_{02} = P_{cu} \end{aligned}$$

So efficiency is maximum when copper loss equals iron loss.

The load and current at maximum efficiency:

$$x = \sqrt{\frac{P_i}{P_{cu,FL}}}, \qquad I_2 = I_{2,FL}\sqrt{\frac{P_i}{P_{cu,FL}}}$$ $$\eta_{max} = \frac{x\,S\cos\phi}{x\,S\cos\phi + 2P_i}$$

For a given load current, efficiency is highest at unity power factor.

Example: $S = 100$ kVA, $P_i = 500$ W, $P_{cu,FL} = 2000$ W: $x = \sqrt{0.25} = 0.5$, so $\eta_{max}$ at upf $= \frac{50000}{50000 + 1000} = 98.04\%$.