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T-07b: Open-Delta (V-V) Connection

Section: A | Priority: 🔴 MUST | Exam Frequency: 6/7 years Sources: Theraja Ch-33 (Art. 33.8), VK Mehta Ch-7 (Art. 7.36), Slides L-11 S15–S18

Why This Topic Matters

Open-Delta appeared in 6 out of 7 papers (2017, 2018, 2019, 2020, 2021, 2023). It is the single most repeated question in the 3-phase transformer section. The question is almost always the same: "One transformer in a Δ-Δ bank fails. Show that 3-phase power can still be supplied. Prove the capacity reduces to 57.7%." This is a guaranteed 4-8 marks.

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📝 Key Definitions

Open-Delta (V-V) connection: "If one of the transformers in a Δ-Δ bank is damaged or removed, the remaining two transformers continue to supply 3-phase power. This is known as open-delta or V-V connection. The total kVA capacity reduces to $\sqrt{3}/3=57.7\%$ of the original closed-delta capacity." — VK Mehta, Art. 7.36

Utilization factor: "In open-delta, each transformer operates at a power factor of $\cos 30°=0.866$, not at unity. The utilization of each transformer is only $\sqrt{3}/2=86.6\%$ of its rated capacity." — VK Mehta

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Why 3-Phase Power Still Works

Start with three transformers in Δ-Δ: $T_{AB}$, $T_{BC}$, $T_{CA}$. Suppose $T_{CA}$ fails and is removed.

On the primary side: The 3-phase supply maintains all three line voltages. By KVL in the delta loop: $V_{CA}=-(V_{AB}+V_{BC})$. This voltage exists at the open terminals even without $T_{CA}$.

On the secondary side: $T_{AB}$ produces $V_{ab}=K\cdot V_{AB}$. $T_{BC}$ produces $V_{bc}=K\cdot V_{BC}$. By KVL: $V_{ca}=-(V_{ab}+V_{bc})=K\cdot V_{CA}$. All three secondary line voltages are present and balanced.

Result: Balanced 3-phase power is delivered using only two transformers.

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The 57.7% Capacity Proof

Let each single-phase transformer be rated $S=VI$ (voltage $V$, current $I$).

Closed Δ (3 transformers): $S_{\text{closed}}=3S$

Open Δ (2 transformers): Each transformer still operates at its rated $V$ and $I$. But for a balanced load, each transformer operates at $\cos 30°$ (not unity). The proof:

In open delta, the two remaining transformers must handle the full line current. The angle between each transformer's voltage and the current it carries is not 0° but ±30°.

$$S_{\text{open}}=2\times V\times I\times \cos 30°=2VI\times \frac{\sqrt{3}}{2}=\sqrt{3}\times VI=\sqrt{3}S$$

$$\frac{S_{\text{open}}}{S_{\text{closed}}}=\frac{\sqrt{3}S}{3S}=\frac{1}{\sqrt{3}}=\boxed{0.577=57.7\%}$$

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🏆 Golden Questions (Past Exam Archive)

🎯 Q1: One transformer in a Δ-Δ bank is damaged. Show that 3-phase power can still be supplied and prove the capacity reduces to 57.7%.

Appeared: 2017 Q7(a), 2018 Q3(c), 2019 Q4(a), 2020 Q4(b), 2021 Q3(b), 2023 Q4(a) — (4–8 marks)

Full Answer:

Part 1: Why 3-phase power still reaches the load

Consider three transformers $T_{AB}$, $T_{BC}$, $T_{CA}$ in Δ-Δ. Suppose $T_{CA}$ fails and is removed.

On the primary side: The 3-phase supply maintains $V_{AB}$ and $V_{BC}$. By KVL in the delta loop: $V_{CA}=-(V_{AB}+V_{BC})$. Even without $T_{CA}$, this voltage is present at the open terminals because it is imposed by the supply.

On the secondary side: $T_{AB}$ produces $V_{ab}=K\cdot V_{AB}$. $T_{BC}$ produces $V_{bc}=K\cdot V_{BC}$. By KVL: $V_{ca}=-(V_{ab}+V_{bc})=K\cdot V_{CA}$. All three secondary line voltages exist and are balanced.

Three-phase balanced power is delivered by just two transformers. This configuration is called the open-delta (V-V) connection.

Part 2: Capacity reduces to 57.7%

Let each single-phase transformer be rated $S=VI$ kVA.

Closed-Δ total: $S_{\text{closed}}=3\times VI=3S$

Open-Δ: Each transformer operates at rated $V$ and $I$. But for a balanced 3-phase load at unity pf, each transformer operates at effective power factor $\cos 30°=\sqrt{3}/2$ (due to the 30° phase displacement between transformer voltage and line current in the open configuration).

$$S_{\text{open}}=2\times VI\times \cos 30°=2VI\times \frac{\sqrt{3}}{2}=\sqrt{3}\times VI=\sqrt{3}S$$

$$\text{Ratio}=\frac{S_{\text{open}}}{S_{\text{closed}}}=\frac{\sqrt{3}S}{3S}=\frac{1}{\sqrt{3}}=0.577=\boxed{57.7\%}$$

Utilization factor per transformer: $\sqrt{3}/2=86.6\%$ (each transformer delivers only 86.6% of its rated capacity).

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🎯 Q2: Two 25 kVA transformers in Open-Δ: find maximum load without overloading, and load when third transformer closes the delta.

Appeared: 2020 Q4(c) — 4 marks

Full Answer:

(i) Open-Δ capacity:

$$S_{\text{open}}=\sqrt{3}\times S_{\text{each}}=\sqrt{3}\times 25=\boxed{43.3\text{ kVA}}$$

Check: Each transformer handles 25 kVA. Two transformers at $\cos 30°$: $2\times 25\times 0.866=43.3$ kVA. ✓

(ii) Closed-Δ capacity (third transformer added):

$$S_{\text{closed}}=3\times S_{\text{each}}=3\times 25=\boxed{75\text{ kVA}}$$

Ratio: $43.3/75=0.577$ (confirms 57.7%).

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⚡ Exam Tips & Common Mistakes

1. Don't write $2S$ for Open-Δ capacity. The capacity is $\sqrt{3}S$, not $2S$.
2. The 57.7% is the ratio to the CLOSED delta capacity. $\sqrt{3}S/3S=57.7\%$.
3. KVL argument is essential. Examiners want you to show WHY the third voltage still exists.
4. This is an emergency configuration. Not designed for permanent use.

🔗 Related Topics

• T-07a: 3-Phase Connections — The closed Δ-Δ configuration
• T-09: Vector Groups — Phase shifts in connections

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