Parallel transformers and per-unit impedance load sharing

Conditions for parallel operation

1. Same voltage ratio (same primary and secondary voltage ratings), otherwise a circulating current flows even at no load.
2. Same polarity, otherwise a dead short circuit occurs through the secondaries.
3. Same per-unit (percentage) impedance, so that load is shared in proportion to kVA ratings.
4. Same X/R ratio, so that both operate at the same power factor as the load.
5. For three-phase units: same phase sequence and same phase displacement (same vector group).

Conditions 1, 2 and 5 are essential; 3 and 4 are desirable for proper load sharing.

Justification: load sharing depends on per-unit impedance

Consider transformers A and B with equal voltage ratios in parallel, with equivalent impedances $Z_A$ and $Z_B$ (referred to secondary), supplying load current $I$ at voltage $V$.

        I_A    Z_A
   +---->----[====]---+
   |                  |      I
 E +                  +---->---- load (V)
   |    I_B    Z_B    |
   +---->----[====]---+

Both have the same EMF $E$ and same terminal voltage $V$, so the impedance drops are equal:

$$I_A Z_A = I_B Z_B \quad\Rightarrow\quad \frac{I_A}{I_B} = \frac{Z_B}{Z_A}$$

With $I = I_A + I_B$:

$$I_A = I\,\frac{Z_B}{Z_A + Z_B}, \qquad I_B = I\,\frac{Z_A}{Z_A + Z_B}$$

Multiplying by $V$ gives kVA shares:

$$S_A = S\,\frac{Z_B}{Z_A + Z_B}, \qquad S_B = S\,\frac{Z_A}{Z_A + Z_B}$$

So the current shared by each is inversely proportional to its ohmic impedance.

Now write each impedance in per unit on its own rating. With rated currents $I_{A,r}$, $I_{B,r}$:

$$z_A = \frac{I_{A,r}Z_A}{V}, \qquad z_B = \frac{I_{B,r}Z_B}{V}$$

Substitute $Z_A = \frac{z_A V}{I_{A,r}}$ and $Z_B = \frac{z_B V}{I_{B,r}}$ into $I_A Z_A = I_B Z_B$:

$$\begin{aligned} I_A\frac{z_A V}{I_{A,r}} &= I_B\frac{z_B V}{I_{B,r}} \\ \frac{I_A/I_{A,r}}{I_B/I_{B,r}} &= \frac{z_B}{z_A} \end{aligned}$$

$I_A/I_{A,r}$ and $I_B/I_{B,r}$ are the fractional loadings of A and B. For both to carry the same fraction of their rating (load shared in proportion to kVA capacity):

$$\frac{I_A}{I_{A,r}} = \frac{I_B}{I_{B,r}} \iff z_A = z_B$$

Hence the per-unit impedances, each on its own base, must be equal. If they are not, the transformer with the smaller per-unit impedance takes more than its share and may be overloaded while the other is under-used. If also $X/R$ is equal, the impedance angles match and both currents are in phase with the load current, so no transformer works at a worse pf.

Example: a 500 kVA unit with $z = 0.05$ pu and a 250 kVA unit with $z = 0.05$ pu share 600 kVA as 400 kVA and 200 kVA (both 80% loaded).