Output (kVA) equation of single phase transformer

The output equation relates the kVA rating of a transformer to its main dimensions (net core area $A_i$ and window area $A_w$) through the chosen magnetic loading $B_m$ and electric loading $\delta$. It is the starting point of transformer design.

Notation

• $f$ = frequency (Hz), $B_m$ = maximum flux density (Wb/m²), $\phi_m = B_mA_i$
• $A_i$ = net iron area of limb (m²), $A_w$ = window area (m²)
• $K_w$ = window space factor = copper area / window area
• $\delta$ = current density (A/m²), $T$ = turns, $I$ = current
• Subscripts 1, 2 for primary and secondary

Step 1: Voltage

EMF per turn and terminal voltage (neglecting drops):

$$E_t = 4.44 f\phi_m = 4.44fB_mA_i, \qquad V \approx E = 4.44fB_mA_iT$$

Step 2: Window (copper) area

In a single-phase core type transformer the window holds one primary and one secondary winding (half of each on each limb, but a full set passes through the window).

$$\begin{aligned} a_1 &= \frac{I_1}{\delta},\quad a_2 = \frac{I_2}{\delta}\\ \text{Copper area in window} &= T_1a_1 + T_2a_2 = \frac{T_1I_1 + T_2I_2}{\delta} = \frac{2AT}{\delta} \end{aligned}$$

since $T_1I_1 = T_2I_2 = AT$. This copper area is also $K_wA_w$, so

$$K_wA_w = \frac{2AT}{\delta} \;\Rightarrow\; AT = \frac{K_wA_w\delta}{2}$$

Step 3: Output

$$\begin{aligned} Q &= V_1I_1\times10^{-3} = 4.44fB_mA_iT_1I_1\times10^{-3}\\ &= 4.44fB_mA_i \cdot \frac{K_wA_w\delta}{2}\times10^{-3}\\ Q &= 2.22\,f\,B_m\,\delta\,K_w\,A_w\,A_i\times10^{-3}\ \text{kVA} \end{aligned}$$

Remarks

• The same equation applies to a single-phase shell type transformer, because there too the window carries one full set of windings.
• For a given rating, higher $B_m$ or $\delta$ gives a smaller core and window, but increases iron or copper losses and temperature rise.
• In design, $E_t = K\sqrt{Q}$ is used first to get $A_i$; the output equation then gives $A_w$.