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T-08: Scott (T-T) Connection

ECE 2207 — Explanation Answers (Sorted by Topic)

Topic Overview: This document compiles all semester final questions and explanation answers on Scott (T-T) Connection from 7 years of exams (2017–2024). Repeated questions appear once with all exam appearances noted.

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T-08: Scott (T-T) Connection: Conversion Between 3-Phase and 2-Phase

Appears in: 2018 Q4(a), 2018 Q4(b), 2019 Q4(b)

Why phase conversion is needed

Standard electrical power generation and transmission are universally 3-phase. However, certain heavy industrial loads: such as large electric arc furnaces, induction heating equipment, and two-phase AC servomotors: require balanced two-phase power (two equal voltages displaced by 90° in time).

Directly connecting a single-phase or two-phase load across a 3-phase line causes severe voltage unbalance, overheating nearby generators and motors. The Scott connection (invented by Charles F. Scott) provides a balanced conversion: it draws balanced 3-phase currents from the supply while delivering balanced 2-phase power to the load (or vice versa).

Construction and connection details

The scheme uses two single-phase transformers:

1. Main transformer ($T_M$):

   • The primary winding has $N_1$ turns with a center tap $D$ at exactly 50% turns ($N_1/2$).
   • Connected across two line terminals of the 3-phase supply: lines $A$ and $B$.
   • Primary voltage is the full line-to-line voltage $V_{AB}=V_L$.

2. Teaser transformer ($T_T$):

   • The primary winding has $N_T=\frac{\sqrt{3}}{2}{N}_1\approx 0.866N_1$ turns.
   • Connected between the center tap $D$ of the main transformer primary and the third line terminal $C$.
   • Voltage across the teaser primary is the median of the equilateral voltage triangle: $V_{DC}=\frac{\sqrt{3}}{2}{V}_L\approx 0.866V_L$.

3. Secondaries:

   • Both main and teaser secondaries have an equal number of turns $N_2$.
   • This ensures both output phase voltages have identical magnitude: $V_{2M}=V_{2T}=V_2$.

Mathematical proof of 90° phase displacement

In a balanced 3-phase system, the line voltages form an equilateral triangle $ABC$ with sides equal to $V_L$:

• Let line voltage $V_{AB}=V_L\angle 0°$ (horizontal reference).
• Center tap $D$ divides $AB$ into two equal halves: $V_{AD}=V_{DB}=\frac{1}{2}{V}_L\angle 0°$.
• Vertex $C$ is at distance $V_L$ from both $A$ and $B$.

From equilateral triangle geometry, the altitude $DC$ from the midpoint $D$ to vertex $C$ is perpendicular to base $AB$: $$V_{DC}=V_C-V_D$$ $$|V_{DC}|=\sqrt{V_L^2-{\left(\frac{V_L}{2}\right)}^2}=\sqrt{\frac{3}{4}{V}_L^2}=\frac{\sqrt{3}}{2}{V}_L\approx 0.866V_L$$

Because $DC\perp AB$, the phasor $V_{DC}$ is shifted by exactly 90° in time relative to $V_{AB}$: $$V_{AB}=V_L\angle 0°$$ $$V_{DC}=\frac{\sqrt{3}}{2}{V}_L\angle 90°$$

Secondary induced EMFs

The secondary voltage of the main transformer is: $$V_{2M}=V_{AB}\times \frac{N_2}{N_1}=V_L\frac{N_2}{N_1}\angle 0°$$

The secondary voltage of the teaser transformer is: $$V_{2T}=V_{DC}\times \frac{N_2}{N_T}=\left(\frac{\sqrt{3}}{2}{V}_L\angle 90°\right)\times \frac{N_2}{\frac{\sqrt{3}}{2}{N}_1}=V_L\frac{N_2}{N_1}\angle 90°$$

Notice how the factor $\frac{\sqrt{3}}{2}$ cancels out perfectly. The two output voltages have identical magnitudes and are in exact time quadrature (90° phase shift): $$\boxed{|V_{2M}|=|V_{2T}|\text{and}{V}_{2T}\text{ leads }{V}_{2M}\text{ by }90°}$$

This constitutes a true, balanced 2-phase electrical supply.

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[2018 Q4(b)]: Worked Numerical Problem

📋 Appeared in: 2018 Q4(b)

Problem: Two T-connected transformers supply a 440V, 33 kVA balanced load from a 3300V balanced 3-phase supply. Find: (i) voltage and current rating of each coil, (ii) kVA rating of main and teaser.

Step-by-step physical solution

Given:

• 3-phase supply line voltage: $V_L=3300\text{ V}$.
• 2-phase balanced load: $S_{total}=33\text{ kVA}$, $V_2=440\text{ V}$.
• Power per phase in 2-phase system: $S_{phase}=33/2=16.5\text{ kVA}$.

1. Secondary coil ratings (identical for both transformers):

• Secondary voltage: $V_2=440\text{ V}$.
• Secondary current: $$I_2=\frac{S_{phase}}{V_2}=\frac{16500\text{ VA}}{440\text{ V}}=37.5\text{ A}$$

2. Main transformer primary coil ratings:

• Main primary is connected across lines $A$ and $B$: $$V_{1,\text{main}}=V_L=3300\text{ V}$$
• Primary current: $$I_{1,\text{main}}=\frac{S_{phase}}{V_{1,\text{main}}}=\frac{16500}{3300}=5.0\text{ A}$$

3. Teaser transformer primary coil ratings:

• Teaser primary has $0.866N_1$ turns and is connected from center-tap $D$ to phase $C$: $$V_{1,\text{teaser}}=\frac{\sqrt{3}}{2}\times 3300=0.866\times 3300=2857.8\text{ V}\approx 2858\text{ V}$$
• Primary current (carries line current from phase $C$): $$I_{1,\text{teaser}}=\frac{S_{phase}}{V_{1,\text{teaser}}}=\frac{16500}{2857.8}=5.77\text{ A}$$

4. kVA rating of main and teaser transformers: $${\text{kVA}}_{\text{main}}=V_{1,\text{main}}\times I_{1,\text{main}}=3300\text{ V}\times 5.0\text{ A}=\boxed{16.5\text{ kVA}}$$ $${\text{kVA}}_{\text{teaser}}=V_{1,\text{teaser}}\times I_{1,\text{teaser}}=2857.8\text{ V}\times 5.77\text{ A}=\boxed{16.5\text{ kVA}}$$

Conclusion: Both transformers operate at identical apparent power ratings ($16.5\text{ kVA}$ each), perfectly sharing the total $33\text{ kVA}$ load.

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