To study the performance of the Scott connected transformers on no-load and full load.

- Sometimes two-phase electric equipment like an electric arc furnace or two-phase induction motors used in servo applications might need a two phase supply whereas only a three-phase supply may be available.
- This problem can be tackled by means of Scott connected transformer which converts a balanced three-phase supply to a balanced two-phase supply.
- Such a connection can also supply a single phase load by suitably connecting the secondary windings of the two phase system without creating an imbalance in the three phase system.

- The Scott-T Connection is the method of connecting two single phase transformer to perform the 3-phase to 2-phase conversion and vice-versa.
- The two transformers are connected electrically but not magnetically. One of the transformer is called the main transformer, and the other is called the auxiliary or teaser transformer.
- The figure below shows the Scott-T transformer connection
- The main transformer is center
tapped at D and is connected to the
line B and C of the 3-phase side. It
has primary BC and secondary
*a*_{1}*a*_{2} - The teaser transformer is
connected to the line terminal A
and the center tapping D. It has
primary AD and the secondary
*b*_{1}*b*_{2} - The identical, interchangeable transformers are used for Scott-T connection in which each transformer has a primary winding of
*T*_{p}turns and is provided with tapping at 0.289*T*_{p}, 0.5*T*_{p}, and, 0.866*T*_{p}

- The line voltages of the 3-phase system
*V*_{AB},*V*_{BC}, and*V*_{CA}which are balanced are shown in the figure below. The same voltage is shown as a closed equilateral triangle. - The figure below shows the primary winding of the main and the teaser transformer.
∣

*V*_{AB}∣ = ∣*V*_{BC}∣ = ∣*V*_{CA}∣ = ∣*V*_{L}∣ - The D divides the primary BC of the main transformers into
two halves and hence the number of turns in portion BD =
the number of turns in portion DC =
*T*_{p}/2. - The voltage
*V*_{BD}and*V*_{DC}are equal, and they are in phase with*V*_{BC}.$$V_{BD}=V_{DC}=\frac{1}{2}V_{BC}=\frac{1}{2}V_L <0^0$$

- The voltage between A and D is
$$\begin{aligned} &V_{BD}=V_{DC}=\frac{1}{2}V_{BC}=\frac{1}{2}V_{L} <0^{0} \\ &V_{AD}=V_{BD}+V_{AB} \\ &V_{AD}=V_{L} \left(\frac{-1}{2}+j \frac{\sqrt{3}}{2}\right)+\frac{1}{2} V_{L} \\ &V_{AD}=V_{L} \left(j \frac{\sqrt{3}}{2}\right)=0.866 V_{L}<90^{0} \end{aligned}$$

- The teaser transformer has the primary voltage rating that is \(\frac{\sqrt{3}}{2}\) of the voltage ratings of the main transformer.
- Voltage
*V*_{AD}is applied to the primary of the teaser transformer and therefore the secondary of the voltage*V*_{2t}of the teaser transformer will lead the secondary terminal voltage*V*_{2m}of the main transformer by 90^{o}as shown in the figure below.

$$\frac{V_{S_{1}}}{V_{AD}}=\frac{T_{S}}{T_{AD}}\\V_{2t}=\frac{T_{S}}{T_{AD}}V_{AD}=\frac{T_{S}}{\frac{\sqrt{3}}{2}T_{P}}\mathbf{x}\frac{\sqrt{3}V_{t}}{2}\\\frac{T_{S}}{T_{L}}V_{L}=V_{2m}$$

- For keeping the voltage per turn same in the primary of the main transformer and the primary of the teaser transformer, the number of turns in the primary of the teaser transformer should be equal to $${\frac{\sqrt{3}}{2T_{p}}}$$.
- Thus, the secondaries of both transformers should have equal voltage ratings. The
*V*_{2t}and*V*_{2m}are equal in magnitude and apart in time; they result in the balanced 2-phase system.

- Let V be the primary line to line voltage and
*N*_{1},*N*_{2}be the no. of turns in the primary and secondary respectively, then the secondary voltage will be \(V\cdot \left( \dfrac{N_2}{N_1}\right)\). - The active number of turns in the main transformer’s primary
*N*_{1}and the teaser transformer primary is \(\left(\dfrac{\sqrt{3}}{2}\right)N_1\).

- 3ɸAC Supply R to 3ɸ Auto Verica Input Terminal R
- 3ɸ AC Supply Y to 3ɸ Auto Verica Input Terminal Y
- 3ɸ AC Supply B to 3ɸ Auto Verica Input Terminal B
- 3ɸ Auto Verica Output R to Wattmeter(1) Terminal M
- Wattmeter Terminal L to LV Side Transformer(1) 86.6%
- 3ɸ Auto Verica Output Y to LV Side Transformer(2) Terminal 115 V
- 3ɸ Auto Verica Output B to Wattmeter(2) Terminal M
- Wattmeter(2) Terminal L to LV Side Transformer(2) Terminal 0
- Wattmeter(1) Terminal V to Wattmeter(2) Terminal V
- Wattmeter(2) Terminal V to LV Side Transformer(2) Terminal 115 V
- LV Side Transformer(1) Terminal 0 to LV Side Transformer(2) Terminal 50%
- HV Side Transformer(1) Terminal 230V to Input Multifunction meter(1) Terminal(R)
- HV Side Transformer(1) Terminal 0 to Input Multifunction meter(1) Terminal(B)
- Multifunction meter (1) Output Terminal(R) to Resistive Load Terminal(R)
- Multifunction meter (1) Output Terminal(B) to Resistive Load Terminal(B)
- HV Side Transformer(2) Terminal 230V to Input Multifunction meter(2) Terminal(R)
- HV Side Transformer(2) Terminal 0 to Input Multifunction meter(2) Terminal(B)
- Multifunction meter (2) Output Terminal(R) to Resistive Load Terminal(R)
- Multifunction meter (2) Output Terminal(B) to Resistive Load Terminal(B)

- Connections must be made tight.
- Before making or breaking the circuit, supply must be switched off.

- Connect the circuit.
- Keep the 3 phase auto-transformer in minimum output position and switch on the 3- phase supply.
- Apply rated voltage to the primary windings of Main and Teaser transformers with the help of the three phase auto - transformer.
- Apply load on the transformers up to full load current. By loading the
Transformers in steps note down the
*V**L*,*I*_{L},*W*from multifunctional meter and Input wattmeter readings. Also, take Input current reading on the 3-phase side. (it should be 1.15 times ILoad). - After taking the meter readings remove the load on the transformer.
- Bring back the auto transformer to its minimum output position and switch off the 3- phase supply.

$$\begin{aligned}&\mathrm{Input~Power~(~P_i~)=W_{lp}+W_{2p}}\\&\mathrm{Output~Power~(~P_0)=V_1*I_1+V_2*I_2}\\&\text{Efficiency}=\text{Output Power ( P}_0)/\text{Input Power ( P}_\mathrm{i})\end{aligned}$$

- Draw the graph Efficiency Vs Output

- What are the applications of Scott connection?
- If the currents in the 2-phase side are unbalanced what will be its effect on the 3-phase side?
- How will you find out the phase shift between the two secondary voltages in Scott connection (This was not done in your experiment)?