Output equation of three phase transformer

The output equation gives the kVA rating of a three-phase core type transformer in terms of the net core area $A_i$, window area $A_w$, flux density $B_m$ and current density $\delta$.

   +-------+-------+-------+
   |       |       |       |
  [L|H] W1 [H|L|H] W2 [H|L]
   |  R    |  Y    |  B    |
   +-------+-------+-------+
  3 limbs, 2 windows; each window holds
  LV + HV of two different phases

Step 1: Voltage per phase

$$E_t = 4.44f\phi_m = 4.44fB_mA_i, \qquad V_p \approx 4.44fB_mA_iT_p$$

Step 2: Copper in a window

A three-phase core type transformer has three limbs and two windows. Each window holds one coil side from each of two adjacent limbs. Each coil side carries the full cross-section of one primary and one secondary winding:

$$\text{Copper area in a window} = 2(a_pT_p + a_sT_s)$$

With $a_p = I_p/\delta$, $a_s = I_s/\delta$ and $I_pT_p = I_sT_s = AT$:

$$K_wA_w = 2\left(\frac{AT}{\delta} + \frac{AT}{\delta}\right) = \frac{4AT}{\delta} \;\Rightarrow\; AT = \frac{K_wA_w\delta}{4}$$

Step 3: Output

$$\begin{aligned} Q &= 3V_pI_p\times10^{-3} = 3\times4.44fB_mA_iT_pI_p\times10^{-3}\\ &= 3\times4.44fB_mA_i\times\frac{K_wA_w\delta}{4}\times10^{-3}\\ Q &= 3.33\,f\,B_m\,\delta\,K_w\,A_w\,A_i\times10^{-3}\ \text{kVA} \end{aligned}$$

where $Q$ is the total three-phase rating in kVA.

Use in design

• First, $E_t = K\sqrt{Q}$ (with $K \approx 0.45$ for distribution and 0.6–0.7 for power transformers) gives $\phi_m$ and $A_i = \phi_m/B_m$.
• Then the output equation gives $A_w$, from which window width and height are found with a chosen $H_w/W_w$ ratio (2 to 4).