Losses & efficiency formulas

$k_h$ is a material constant and $V$ is the core volume. The exponent (the Steinmetz coefficient) depends on the core material, and 1.6 is the classical value. A problem will state the exponent to use, or will say only that $P_h\propto f$ at constant $B_{max}$.

$t$ is the lamination thickness, and $k_e\propto 1/\rho$ (core resistivity). This form assumes sinusoidal flux and thin laminations. At the same $B_{max}$, $f$ and core volume, halving $t$ cuts $P_e$ to one quarter. This is why cores are laminated rather than solid.

Valid only when $B_{max}$ is held constant across the tests. A core-loss separation test is designed exactly this way.

$N$ is the number of turns of the winding and $A_c$ the net core area. The winding resistance and leakage drops are neglected, so $V\approx E$. If $V$ and $f$ are changed in the same ratio (constant $V/f$), $B_{max}$ stays fixed.

Follows from $B_{max}\propto V/f$. At constant voltage, the eddy current loss does not depend on frequency, and the hysteresis loss falls as frequency rises (with a Steinmetz exponent of 1.6).

Applies to any machine: motor, generator or transformer. Express $P_{out}$, $P_{in}$ and every loss in the same unit (W or kW). The last form is convenient for a motor, when the electrical input is known.

Assumes the output is proportional to the load current (fixed terminal voltage and power factor), a constant loss that does not depend on load, and a variable loss that scales as $I^2$. Setting $d\eta /dI=0$ then gives the condition shown. Any loss that is proportional to $I$ (such as the DC brush-drop loss) drops out of the condition. For a transformer, the load (kVA) at which efficiency is maximum does not depend on power factor. The value of ${\eta }_{max}$ does depend on it, and is highest at unity power factor.

$x$ is the fraction of rated (full) load, $P_i$ the constant loss, and $P_{cu,fl}$ the full-load variable ($I^2$) loss. Valid when the constant loss stays fixed and the variable loss scales as $x^2$. For a transformer, the load at maximum efficiency is $xS$ (kVA).

Output energy over input energy across 24 hours. $t_k$ is the time in hours spent at load fraction $x_k$, with output $P_{out,k}$. The core loss applies for all 24 h while the transformer is energised, even at no load. The copper loss applies only while the transformer carries load.

$s=(n_s-n)/n_s$ is the per-unit slip, with $n_s=120f/P$ rpm. $P_{airgap}$ is the power crossing the air gap: stator input minus stator copper and stator iron losses. The ratio holds for three-phase totals, or per phase if used consistently. So $P_{rotor,cu}=s{P}_{airgap}$ and $P_{mech,gross}=(1-s)P_{airgap}$. Shaft output = $P_{mech,gross}$ − friction and windage − stray load loss. Because of this chain, efficiency can never exceed $(1-s)$, and in practice is below it.