Transformer A loading
80%
Transformer Parallel Operation Calculator
Verify whether two transformers can run in parallel (vector group, ratio, impedance, capacity ratio) and see how the load splits between them in inverse proportion to their short-circuit impedance.
Result
Enter your load → get a full design & BOM in seconds
Sample outputParallel load sharing
798 / 502 kVA
1300 kVA total across 1000 kVA (uk 6%) + 630 kVA (uk 6%) · Feasible for parallel operation ✓
Transformer B loading
80%
Combined capacity
1630 kVA
Overall load factor
80%
Parallel-operation conditions (IEC 60076)
| Condition | Value | Verdict |
|---|---|---|
| Same vector group | Matched (e.g. Dyn11 = Dyn11) | PASS ✓ |
| Same voltage ratio | 0% difference (≤ 0.5% recommended) | PASS ✓ |
| Impedance within tolerance | |uk_A − uk_B| = 0.0% of mean (≤ 10%) | PASS ✓ |
| Capacity ratio ≤ 3:1 | 1.59 : 1 | PASS ✓ |
Load sharing calculation
| Weight A | Sa ÷ uk_A = 1000 ÷ 6 = 166.7 |
| Weight B | Sb ÷ uk_B = 630 ÷ 6 = 105.0 |
| Load A | 1300 × 166.7 ÷ 271.7 = 798 kVA (80%) |
| Load B | 1300 × 105.0 ÷ 271.7 = 502 kVA (80%) |
Conclusion
The two units can operate in parallel; load splits in inverse proportion to their short-circuit impedance.
Bill of materials
| Equipment | Spec | Qty | Subtotal |
|---|---|---|---|
| Transformer A (oil-immersed S11) | 1000 kVA · uk 6% | 1 | POA |
| Transformer B (oil-immersed S11) | 630 kVA · uk 6% | 1 | POA |
| Total equipment | POA |
For power-system and plant engineers sizing parallel transformer banks for redundancy, capacity expansion or N−1 reliability.
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Worked example
A pre-computed reference example (crawlable — no JavaScript required). Enter your own parameters above for a live result.
Example 1 — 1000 kVA + 630 kVA, matched 6% impedance
Parallel load sharing
798 / 502 kVA
1300 kVA total across 1000 kVA (uk 6%) + 630 kVA (uk 6%) · Feasible for parallel operation ✓
Transformer A loading
80%
Transformer B loading
80%
Combined capacity
1630 kVA
Overall load factor
80%
Parallel-operation conditions (IEC 60076)
| Condition | Value | Verdict |
|---|---|---|
| Same vector group | Matched (e.g. Dyn11 = Dyn11) | PASS ✓ |
| Same voltage ratio | 0% difference (≤ 0.5% recommended) | PASS ✓ |
| Impedance within tolerance | |uk_A − uk_B| = 0.0% of mean (≤ 10%) | PASS ✓ |
| Capacity ratio ≤ 3:1 | 1.59 : 1 | PASS ✓ |
Load sharing calculation
| Weight A | Sa ÷ uk_A = 1000 ÷ 6 = 166.7 |
| Weight B | Sb ÷ uk_B = 630 ÷ 6 = 105.0 |
| Load A | 1300 × 166.7 ÷ 271.7 = 798 kVA (80%) |
| Load B | 1300 × 105.0 ÷ 271.7 = 502 kVA (80%) |
Conclusion
The two units can operate in parallel; load splits in inverse proportion to their short-circuit impedance.
Bill of materials
| Equipment | Spec | Qty | Subtotal |
|---|---|---|---|
| Transformer A (oil-immersed S11) | 1000 kVA · uk 6% | 1 | POA |
| Transformer B (oil-immersed S11) | 630 kVA · uk 6% | 1 | POA |
| Total equipment | POA |
Example 2 — mismatched impedance and ratio
Parallel load sharing
883 / 417 kVA
1300 kVA total across 1000 kVA (uk 6%) + 630 kVA (uk 8%) · Not recommended — check conditions below ✗
Transformer A loading
88%
Transformer B loading
66%
Combined capacity
1630 kVA
Overall load factor
80%
Parallel-operation conditions (IEC 60076)
| Condition | Value | Verdict |
|---|---|---|
| Same vector group | Matched (e.g. Dyn11 = Dyn11) | PASS ✓ |
| Same voltage ratio | 1.5% difference (≤ 0.5% recommended) | FAIL ✗ |
| Impedance within tolerance | |uk_A − uk_B| = 28.6% of mean (≤ 10%) | FAIL ✗ |
| Capacity ratio ≤ 3:1 | 1.59 : 1 | PASS ✓ |
Load sharing calculation
| Weight A | Sa ÷ uk_A = 1000 ÷ 6 = 166.7 |
| Weight B | Sb ÷ uk_B = 630 ÷ 8 = 78.8 |
| Load A | 1300 × 166.7 ÷ 245.4 = 883 kVA (88%) |
| Load B | 1300 × 78.8 ÷ 245.4 = 417 kVA (66%) |
Circulating current
A 1.5% no-load ratio difference drives a circulating current of ≈ 80 kVA between the two units (adds to the load, does not supply the load).
Bill of materials
| Equipment | Spec | Qty | Subtotal |
|---|---|---|---|
| Transformer A (oil-immersed S11) | 1000 kVA · uk 6% | 1 | POA |
| Transformer B (oil-immersed S11) | 630 kVA · uk 8% | 1 | POA |
| Total equipment | POA |
How it was calculated
- · Load sharing weight: w = S_nom ÷ uk for each unit.
- · Load of unit A: S_A = S × w_A ÷ (w_A + w_B); unit B likewise.
- · Loading: S_A ÷ S_nom,A (must stay ≤ 100%).
- · Parallel conditions: same vector group, same voltage ratio (≤0.5%), impedance within ±10% of each other, capacity ratio ≤ 3:1.
- · Circulating current (ratio mismatch): I_c ≈ ΔU ÷ (Z_A + Z_B) in per-unit.
Referenced standards
| Standard | Scope |
|---|---|
| IEC 60076-1 | Power transformers — general requirements |
| IEC 60076-8 | Power transformers — application guide |
| IEEE C57.12.00 | Standard for general requirements for liquid-immersed distribution and power transformers |
Frequently asked questions
What conditions must transformers meet to run in parallel?
Per IEC 60076, parallel transformers must match: voltage ratio (within ~0.5%), Vector Group, impedance voltage uk (within ~10%), and preferably power rating ratio under 3:1. Mismatched vector groups cause circulating current and severe overheating. The calculator checks these conditions and reports pass/fail.
How is load shared between parallel transformers?
Load shares inversely with impedance: the Transformer with lower uk carries more load. Two transformers of equal rating with uk 6% and 4.5% split load in the ratio 4.5:6, so the 4.5% unit carries 57%. The calculator computes the load share by impedance and flags overloading.
Why run transformers in parallel?
Parallel operation provides redundancy (one can be taken out for maintenance without outage), allows capacity to be added incrementally, and can improve efficiency at light load by running only one unit. The trade-off is higher fault level at the bus. The calculator models the parallel arrangement and its Fault Current.
What is circulating current between parallel transformers?
Circulating current flows when parallel transformers have unequal voltage ratios or different vector groups, even with no load, causing extra losses and heating. A 1% ratio mismatch can drive 5-10% circulating current. IEC 60076 limits the ratio difference to ~0.5%. The calculator checks for this.
What fault level results from paralleling transformers?
Paralleling transformers roughly sums their short-circuit contributions. Two 1000 kVA, 6% uk transformers give ~48 kA at the LV bus versus ~24 kA for one, so the switchgear must be uprated. The calculator computes the combined fault level so you can check panel ratings.
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Results are engineering estimates for preliminary design and reference only. Always verify with the applicable standards, the equipment nameplate and a licensed engineer before procurement or installation.
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