Parallel Operation Of Single Phase Transformers - Electrical Machines -

Introduction

Load on a distribution or other practical transformer usually grows over time. When a transformer approaches or exceeds its rated load during its life, two practical responses are possible. One option is to replace the transformer by a larger rating; this has limits because the new transformer must be planned for future expansion and will often operate at light load for long periods, reducing utilisation efficiency. The second option is to connect another transformer in parallel with the existing one. Parallel operation has advantages: it allows growth in load without a single very large transformer, permits one transformer to be left out of service when loading is light, and simplifies maintenance because a single spare transformer of similar rating suffices.

Basic connection and essential checks

When two single-phase transformers are connected in parallel, their primaries are connected to the same source and their secondaries supply the same load. The following conditions must be observed before paralleling:

• Polarity check: The instantaneous polarities (phasing) of the transformer terminals must be the same before paralleling. The corresponding instantaneous positive terminals on both secondaries must be connected together and the other terminals likewise. If polarity is reversed on one transformer, closing the paralleling switch will produce a direct short between windings and cause large short-circuit currents.
• No-load (open-circuit) secondary voltages: The no-load secondary voltages must be very close in magnitude. If the no-load voltages differ significantly, large circulating currents flow at the instant the paralleling switch is closed; these currents are limited only by the series impedances of the transformers (typically small, 5-10%), and can be dangerously large.
• Total KVA rating: The combined KVA ratings of the transformers must be at least equal to the expected load KVA; otherwise paralleling will not meet demand.

Equivalent circuits and current sharing (identical no-load voltages)

Assume two single-phase transformers with their secondaries connected in parallel and that their induced (no-load) secondary voltages are identical. With the secondary side as reference, each transformer can be represented by its series equivalent impedance.

Let the series impedances be Z₁ = R1 + jX1 and Z₂ = R2 + jX2. Let the currents supplied by the two transformers be I1 and I2, and the total load current be IL. The series impedances appear effectively in parallel as seen by the load, and the following relations hold:

• Z₁·I1 = Z₂·I2 (voltage drops across each series impedance are equal because secondaries are tied together)
• IL = I1 + I2
• Solving gives I1 = (Z₂ / (Z₁ + Z₂)) · IL and I2 = (Z₁ / (Z₁ + Z₂)) · IL

Phasor relations: if V_L is the load voltage phasor and IL its current phasor, then V_L = I1·Z₁ = I2·Z₂. The phasor diagram shows I1 and I2 such that their vector sum is IL. From the relations above we see:

• Phase condition for maximum utilisation: For the transformer currents to be in phase (so that magnitudes add algebraically), the ratio I1/I2 must be a real scalar. That requires the impedance angles to be equal: angle(Z₁) = angle(Z₂), i.e. R1/X1 = R2/X2. This is a desirable (but not strictly essential) condition because when the currents are in phase the pair supplies the greatest possible IL for given transformer current limits.
• KVA sharing: Multiply the current expressions by V_L* (complex conjugate) to obtain complex KVA (apparent power). If SL denotes the complex KVA demanded by the load, then S1 = (Z₂ / (Z₁ + Z₂)) · SL and S2 = (Z₁ / (Z₁ + Z₂)) · SL.
• Proportional sharing by rated KVA: It is desirable that the transformers share load KVA in proportion to their individual KVA ratings, i.e. S1 / S2 = S1,rated / S2,rated. From the sharing formula this implies |Z₂| / |Z₁| = S1,rated / S2,rated.
• Practical rule (per-unit equality): If both transformers have the same rated secondary voltage, then equality of per-unit series impedances on their own bases ensures proportional sharing: Z₁ (pu) = Z₂ (pu).

Consequences if sharing condition is not met

If the condition |Z₂| / |Z₁| = S1,rated / S2,rated is not satisfied, one transformer will reach rated KVA before the other. Suppose S2 reaches its rated KVA first. Let SL,max denote the total load KVA when S2 = S2,rated. Then

• S2,rated = (|Z₁| / (|Z₁| + |Z₂|)) · SL,max
• Hence SL,max = ((|Z₁| + |Z₂|) / |Z₁|) · S2,rated
• The load on transformer 1 at that instant is S1 = (|Z₂| / (|Z₁| + |Z₂|)) · SL,max = (|Z₂| / |Z₁|) · S2,rated

If (|Z₂| / |Z₁|) is less than (S1,rated / S2,rated) then S1 < s1,rated when s2 reaches s2,rated. thus the parallel group cannot be further loaded even though one transformer still has spare capacity, and the combined capacity of the two is not fully exploited. for maximum combined usable kva it is therefore desirable to ensure equal per-unit impedances and equal impedance 

Parallel connection when no-load voltages differ (slightly different turns ratios)

In practice, two transformers to be paralleled may have slightly different turns ratios and hence slightly different induced secondary (no-load) voltages. When the induced voltages differ, the simple parallel equivalent with impedances directly in parallel no longer strictly applies; there will be internal voltage differences and circulating currents between transformers even before external load is applied.

Represent the two transformer equivalent circuits on the common secondary (load) side. Let the induced (open-circuit) secondary phasors referred to the secondary terminals be E1′ and E2′ respectively. Let the series impedances referred to the secondary be Z₁ and Z₂. The load admittance is Y_L = 1/ZL, and transform admittances are Y₁ = 1/Z₁ and Y₂ = 1/Z₂. If the load terminal voltage phasor is V_L, node KCL at the load node gives:

• V_L·Y_L = (E1′ - V_L)·Y₁ + (E2′ - V_L)·Y₂
• Rearrange to obtain V_L·(Y_L + Y₁ + Y₂) = E1′·Y₁ + E2′·Y₂
• Therefore V_L = (Y₁·E1′ + Y₂·E2′) / (Y_L + Y₁ + Y₂)
• Transformer currents are I1 = Y₁·(E1′ - V_L) and I2 = Y₂·(E2′ - V_L)

From these equations individual currents and the KVA supplied by each transformer can be determined for any given load. When E1′ ≠ E2′, circulating magnetising or short-circuit currents flow between transformers; their magnitude is determined by the difference E1′ - E2′ and the series impedances.

Worked numerical example (step-by-step solution)

Problem statement (as given): Two single-phase transformers have identical open circuit secondary voltage rating 11 kV/3.3 kV. Short-circuit test (HV winding shorted) data are:

• Transformer 1: applied voltage 200 V, circulating current 400 A, power dissipated (on short-circuit) 15 kW.
• Transformer 2: applied voltage 100 V, circulating current 400 A, power dissipated (on short-circuit) 20 kW.

These two transformers are connected in parallel and supply a load current IL = 750 A at 0.8 power factor lagging. Find the KVA supplied by each transformer and the load terminal voltage.

Solution (computations given stepwise; each numerical step on a separate line followed by a short explanation):

Modulus of series impedance of transformer 1: 0.5 Ω = 200 V ÷ 400 A

This is |Z₁| from the short-circuit test.

Series resistance of transformer 1: 0.094 Ω = 15 000 W ÷ (400 A)²

R1 is obtained from the measured power loss during the short-circuit test.

Leakage reactance of transformer 1: 0.491 Ω = √(0.5² - 0.094²)

X1 is found from |Z₁|² = R1² + X1².

Modulus of series impedance of transformer 2: 0.25 Ω = 100 V ÷ 400 A

This is |Z₂| from transformer 2 short-circuit test.

Series resistance of transformer 2: 0.125 Ω = 20 000 W ÷ (400 A)²

R2 is obtained from the short-circuit power loss for transformer 2.

Leakage reactance of transformer 2: 0.2165 Ω = √(0.25² - 0.125²)

X2 computed from |Z₂|² = R2² + X2².

Impedance angles: θ1 = cos⁻¹(R1 / |Z₁|) = 79.1°

θ1 is the impedance angle of Z₁.

Impedance angles: θ2 = cos⁻¹(R2 / |Z₂|) = 60°

θ2 is the impedance angle of Z₂.

Complex impedances: Z₁ = 0.5 ∠79.1° Ω and Z₂ = 0.25 ∠60° Ω

Express Z₁ and Z₂ in polar form for phasor calculations.

Sum of impedances: Z₁ + Z₂ = 0.7407 ∠72.8° Ω

Compute Z₁ + Z₂ by phasor addition (or convert to rectangular and add, then convert back to polar).

Load current phasor: IL = 750 ∠-36.87° A (0.8 p.f. lagging corresponds to cos⁻¹(0.8) = 36.87°)

Reference chosen so that IL angle is -36.87°.

Current supplied by transformer 1: I1 = (Z₂ / (Z₁ + Z₂)) · IL = 253 ∠-49.6° A

Use I1 = (Z₂ / (Z₁ + Z₂)) IL and multiply phasors.

Current supplied by transformer 2: I2 = (Z₁ / (Z₁ + Z₂)) · IL = 506 ∠-30.5° A

Use I2 = (Z₁ / (Z₁ + Z₂)) IL and multiply phasors.

Check: I1 + I2 = IL (vector sum: 253∠-49.6° + 506∠-30.5° = 750∠-36.87°)

The phasor sum recovers the given IL (small differences may be due to rounding).

Determine the secondary terminal voltage magnitude. The referred (induced) secondary voltage magnitude corresponding to 11 kV primary and 3.3 kV secondary is 3.3 kV (given transformer rating). Let V1′ denote the induced secondary phasor magnitude 3300 V at angle δ. Then V_L = V1′ - Z₁·I1 (phasor equation V_L = V1′ - I1·Z₁).

Compute phasor drop across Z₁: I1·Z₁ (numerical phasor multiplication gives magnitude and angle).

The angle difference θ1 - ∠I1 = 79.1° - (-49.6°) = 128.7°; use phasor multiplication to find the real and imaginary components of the drop.

Using the geometry of the phasor diagram and solving for V_L and δ yields V_L ≈ 3190 V

Detailed phasor algebra gives the load terminal voltage magnitude as about 3.19 kV.

Apparent power supplied by transformer 1: S1 = V_L · I1 = 3190 V × 253 A = 807.5 kVA

Compute complex S1 and determine its power factor (S1 shows a lagging power factor of approximately 0.648).

Apparent power supplied by transformer 2: S2 = V_L · I2 = 3190 V × 506 A = 1615 kVA

Compute complex S2 and determine its power factor (S2 shows a lagging power factor of approximately 0.862).

Discussion of example and practical points

• Because the impedance angles θ1 and θ2 differ, the currents I1 and I2 are not in phase; their algebraic magnitudes do not add but phasor sum equals IL. This reduces the usable combined current compared with the ideal case where impedance angles match.
• Transformer 2 in the example supplies approximately twice the KVA of transformer 1 because its series impedance magnitude is smaller; the per-unit impedances are not equal on their own bases, so sharing is not proportional to rated KVA.
• When paralleling transformers in practice, engineers match polarities, ensure secondary voltages are very close, and prefer equal per-unit impedances and similar X/R ratios to obtain predictable and fair load sharing.

Summary

Parallel operation of single-phase transformers is a practical method to increase available KVA and to provide flexibility and redundancy. Essential preconditions are correct polarity and closely matched no-load secondary voltages. Desirable additional conditions are equal per-unit series impedances (on own bases) and similar impedance angle (X/R ratio) so that currents are nearly in phase and KVA is shared proportionally to ratings. When induced voltages differ slightly, node equations with admittances yield the exact terminal voltage and individual currents; circulating currents may then exist even with no external load. Practical design and tests (open-circuit and short-circuit tests) provide the impedances and losses needed for correct paralleling and for predicting load sharing.