Conversion from Three Phase to Two Phase (Scott Connection)

Conversion from Three Phase to Two Phase (Scott Connection)

• With the help of Scott connection, proposed by C.F. Scott, it is possible to obtain 2 phase supply which is required for furnaces or even three phase load can be driven from the available 2 phase supply source. The Scott connection which is serving this purpose is shown in the Fig. 7.6.1.

• This connection uses two transformers with different ratings. But identical transformers with suitable tappings may also be used for the interchangeability and provision of spares. One of the transformers having 50 % tapping is called main transformer and other one having 86.6 % tapping is called teaser transformer. If the secondaries of the two transformers are connected as shown in the Fig. 7.6.1 then two phase, three wire system is obtained.

• One end of the primary winding of the teaser transformer is connected to the centre tapping provided on the primary winding of the main transformer. The two ends of the primary winding of the main transformer and 86.6 % tapping point on the teaser transformer is connected to a balanced three phase supply. The voltage per turn is same both in primary of both main and teaser transformer. With the equal number of turns on secondaries of both the transformers, the secondary voltage will be equal in magnitude which results in symmetrical 2 phase system.

• The same connection drawn slight differently is shown in the Fig. 7.6.2.

• The main transformer primary winding consists of N₁ turns connected between lines Y and B of a symmetrical three phase supply. V_RY' V_YB and V_BR are all line voltages. Hence V_RY = V_YB = V_BR = V_L But RO being the altitude of the equilateral triangle, the voltage V_RO is √3 / 2 V_L. The voltage per turn will be same in primaries of both the transformers if number of turns between R and O are (√3/2) N₁. With this then the terminal voltages on the secondary windings having same turns will be equal in magnitude and have a phase difference of 90° between them.

• The point O is not the neutral point of 3 phase supply voltage as its voltage with respect to any line is not V_L / √3. N is the neutral point shown in Fig. 7.6.3. Voltage V_RN is nothing but V_L / √3 whereas V_RO is (√3/2) V_L. Hence the voltage between N and O will be

• Since 0.29 is one third of 0.866, N divides the teaser winding RO in the ratio 2: 1.

• Now let us consider the unity power factor load. The teaser secondary is supplying a current of I_2T Neglecting the magnetizing current I₀ we have,

Transformation ratio,

'• Each half of the primary winding of the main transformer carries current of I_1m consisting of two parts.

i) First part balances the main secondary current I_2M

ii) The second part is equal to one half of the teaser primary current i.e. ½ I_1T

• The main transformer primary winding forms a return path for the teaser primary current which is divided into two halves at point O in either direction.

The current in each half is equal to

This current is shown in the Fig. 7.6.4.

• Thus the phase R supplies current I_1T which is divided into two equal parts and is flowing in the main transformer in the opposite directions. Thus the line currents on the primary side are vectorially given by,

• The currents in the lines Y and B are obtained vectorially. The teaser transformer currents flowing in the two halves of the primary winding of the main transformer in the opposite direction and have no magnetic effect on the core and does not take part in balancing the secondary ampere turns of the main transformer. Thus when two phase load of unity power factor is balanced then three phase side is also balanced.

• Now let us consider balanced two phase load at a lagging p.f. of cos ϕ. The corresponding phasor diagram is shown in the Fig. 7.6.5. The three phase side is again balanced as the currents drawn from the three phase system are equal balanced and lag by angle ϕ with respect to their respective phase voltages. This can be shown mathematically.

• Let us consider the equal currents at a power factor of cos ϕ lagging from secondary side,

Thus all the currents in the primary side are equal in magnitude.

We have from the phasor diagram,

• Since the power factors of the loads for the main transformer and the teaser transformer are equal, the phase angle between the secondary currents is also 90°. The angle between the primary currents I_1T and I_1M is also 90°. Thus the three line currents I_R', I_y and I_B are displaced from each other by 120° and lag the respective phase voltages by an angle . This proves that if the two phase load is balanced then the loading on the three phase side is also balanced.

• Now we will consider the case of unbalanced two phase load having different currents and different power factors. The phasor diagram for this case can be constructed in a similar manner. It is shown in the Fig. 7.6.6.

 

1. Applications of Scott Connection

• The Scott connection is used for:

1. Linking three phase system with two phase system with the flow of power in either vilah direction.

2. Used in electric furnace installation where two single phase furnaces are to be operated from three phase supply providing balanced load.

3. Used to supply single phase loads as electric traction power to keep the load on the three phase supply almost balanced.

4. To supply single phase loads from three phase system for better balancing of primary power system.

Ex. 7.6.1 Two transformers are required for a Scott connection operating from a 415 V, 3 phase supply for supplying two single phase furnaces 225 V on the two phase side. If the total output is 150 kVA, calculate the secondary to primary turns ratio and the winding currents of each transformer.

Sol. :

For the main transformer

Primary voltage = 415 V

Secondary voltage = 225 V

Transformation ratio, K = Secondary voltage / Primary voltage

Ex. 7.6.2 In a Scott connection, calculate the values of line currents on the 3 phase side, if the loads on 2 phase side are 300 kW and 400 kW, both at 125 V and 0.707 p.f. and the three phase line voltage is 3300 V. The 300 kW load is on the leading phase of the 2 phase side. Neglect the losses.

Sol. :

 Primary line voltage =3300 V

Secondary voltage = 125 V

Load carried by teaser secondary = 300 kW at 0.707 p.f.

Load carried by main secondary = 400 kW at 0.707 p.f.

Transformation ratio for main transformer,

Power supplied by teaser transformer = 300 kW at 0.707 p.f.

Secondary current in teaser transformer, I_2T = 300 × 10³ / 125 × 0.707= 3394.62 A

 

Review Questions

1. With the help of phasor diagram explain how 2 phase supply can be obtained from 3 phase supply using Scott connection.

2. Two electric furnaces are supplied with single-phase circuit at 100 V from a 3-phase, 13,750 V supply by means of two single-phase Scott-connected transformers with similar secondary windings when the load on main transformer is 625 kW and on the teaser is 1000 kW, what current will flow in each of 3-phase lines at unity power factor? 

[Ans.: I_R=83.9 A, l_y = l_B=61.83 A]