Transformer A loading
80%
Calculadora de Operación en Paralelo de Transformadores
Verifique si dos transformadores pueden operar en paralelo (grupo vectorial, relación, impedancia, relación de capacidad) y vea cómo se divide la carga entre ellos en proporción inversa a su impedancia de cortocircuito.
Resultado
Ingrese su carga → obtenga un diseño completo y BOM en segundos
Salida de ejemploParallel 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 |
Para ingenieros de sistemas de potencia y plantas que dimensionan bancos de transformadores en paralelo para redundancia, expansión de capacidad o confiabilidad N−1.
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Ejemplo resuelto
Un ejemplo de referencia precalculado (indexable — no requiere JavaScript). Introduzca sus propios parámetros arriba para obtener un resultado en vivo.
Ejemplo 1 — 1000 kVA + 630 kVA, impedancia igualada al 6%
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 |
Ejemplo 2 — Impedancia y relación desajustadas
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 |
Cómo se calculó
- · Peso de reparto de carga: w = S_nom ÷ uk para cada unidad.
- · Carga de la unidad A: S_A = S × w_A ÷ (w_A + w_B); unidad B igualmente.
- · Carga: S_A ÷ S_nom,A (debe mantenerse ≤ 100%).
- · Condiciones de paralelo: mismo grupo vectorial, misma relación de tensión (≤0.5%), impedancia dentro de ±10% entre sí, relación de capacidad ≤ 3:1.
- · Corriente de circulación (desajuste de relación): I_c ≈ ΔU ÷ (Z_A + Z_B) en por unidad.
Normas referenciadas
| Standard | Scope |
|---|---|
| IEC 60076-1 | Transformadores de potencia — requisitos generales |
| IEC 60076-8 | Transformadores de potencia — guía de aplicación |
| IEEE C57.12.00 | Estándar para requisitos generales de transformadores de distribución y potencia sumergidos en líquido |
Preguntas frecuentes
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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