Three Phase Transformer               

 

26.1 Goals of the lesson

Three phase system has been adopted in modern power system to generate, transmit and distribute power all over the world. In this lesson, we shall first discuss how three number of single phase transformers can be connected for 3-phase system requiring change of voltage level. Then we shall take up the construction of a 3-phase transformer as a single unit. Name plate rating of a three phase transformer is explained. Some basic connections of a 3-phase transformer along with the idea of vector grouping is introduced.

Key Words: bank of three phase transformer, vector group.

After going through this section students will be able to answer the following questions.

 

• Point out one important advantage of connecting a bank of 3-phase transformer.

 

• Point out one disadvantage of connecting a bank of 3-phase transformer.

 

• Is it possible to transform a 3-phase voltage, to another level of 3-phase voltage by using two identical single phase transformers? If yes, comment on the total kVA rating obtainable.

 

• From the name plate rating of a 3-phase transformer, how can you get individual coil rating of both HV and LV side?

 

• How to connect successfully 3 coils in delta in a transformer?

 

26.2 Three phase transformer

It is the three phase system which has been adopted world over to generate, transmit and distribute electrical power. Therefore to change the level of voltages in the system three phase transformers should be used.

Three number of identical single phase transformers can be suitably connected for use in a three phase system and such a three phase transformer is called a bank of three phase transformer. Alternatively, a three phase transformer can be constructed as a single unit.

26.3 Introducing basic ideas

In a single phase transformer, we have only two coils namely primary and secondary. Primary is energized with single phase supply and load is connected across the secondary. However, in a 3-phase transformer there will be 3 numbers of primary coils and 3 numbers of secondary coils. So these 3 primary coils and the three secondary coils are to be properly connected so that the voltage level of a balanced 3-phase supply may be changed to another 3-phase balanced system of different voltage level.

Suppose you take three identical transformers each of rating 10 kVA, 200 V / 100 V, 50 Hz and to distinguish them call them as A, B and C. For transformer-A, primary terminals are marked as A₁A₂ and the secondary terminals are marked as a₁a₂. The markings are done in such a way that A₁ and a₁ represent the dot (•) terminals. Similarly terminals for B and C transformers are marked and shown in figure 26.1.

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It may be noted that individually each transformer will work following the rules of single phase transformer i.e, induced voltage in a₁a₂ will be in phase with applied voltage across A₁A₂ and the ratio of magnitude of voltages and currents will be as usual decided by a where a = N₁/N₂ = 2/1, the turns ratio. This will be true for transformer-B and transformer-C as well i.e., induced voltage in b₁b₂ will be in phase with applied voltage across B1_B2 and induced voltage in c₁c₂ will be in phase with applied voltage across C₁C₂.

Now let us join the terminals A₂, B2 _{and C}2 _{of the 3 primary coils of the transformers and no inter connections are made between the secondary coils of the transformers. Now to the free terminals A}1_{, B1} and C₁ a balanced 3-phase supply with phase sequence A-B-C is connected as shown in figure 26.2. Primary is said to be connected in star.

If the line voltage of the supply is 2003 VLLV=, the magnitude of the voltage impressed across each of the primary coils will be 3 times less i.e., 200 V. However, the phasors 12AAV, A1 A2 Transformer-Aprimary B1 B2 Transformer-Bprimary C1 C2 Transformer-Cprimary a1 a2 Transformer-A secondary b1 b2 Transformer-B secondary c1 c2 Transformer-C secondary Figure 26.1: Terminal markings along with dots Figure 26.2: Star connected primary with secondary coils left alone. Balanced 3-phase supply A1 A2 B1 B2 C1 C2 a1 a2 b1 b2 c1 c2 A B C A1 B1 C1 A2B2C2 a1 b1 c1 a2 b2 c2

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12BBVand 12CCVwill be have a mutual phase difference of 120º as shown in figure 26.2. Then from the fundamental principle of single phase transformer we know, secondary coil voltage 12aaVwill be parallel to 12AAV; 12bbVwill be parallel to 12BBVand 12ccVwill be parallel to 12CCV. Thus the secondary induced voltage phasors will have same magnitude i.e., 100 V but are displaced by 120º mutually. The secondary coil voltage phasors 12aaV, 12bbVand 12ccV are shown in figure 26.2. Since the secondary coils are not interconnected, the secondary voltage phasors too have been shown independent without any interconnections between them. In contrast, the terminals A₂, B2 _{and C}2 _{are physically joined forcing them to be equipotential which has been reflected in the primary coil voltage phasors as well where phasor points A}2_{, B2} and C₂ are also shown joined. Coming back to secondary, if a voltmeter is connected across any coil i.e., between a₁ and a₂ or between b₁ and b₂ or between c₁ and c₂ it will read 100 V. However, voltmeter will not read anything if connected between a₁ and b₁ or between b₁ and c₁ or between c₁ and a₁ as open circuit exist in the paths due to no physical connections between the coils.

Imagine now the secondary coil terminals a₂, b₂ and c₂ are joined together physically as shown in figure 26.3. So the secondary coil phasors should not be shown isolated as a₂, b₂ and c₂ become equipotential due to shorting of these terminals. Thus, the secondary coil voltage phasors should not only be parallel to the respective primary coil voltages but also a₂, b₂ and c₂ should be equipotential. Therefore, shift and place the phasors 12aaV, 12bbVand 12ccVin such a way that they remain parallel to the respective primary coil voltages and the points a₂, b₂ and c₂ are superposed.

Here obviously, if a voltmeter is connected between a₁ and b₁ or between b₁ and c₁ or between c₁ and a₁ it will read corresponding phasor lengths a₁b₁ or b₁ c₁ or c₁a₁ which are all equal to 2003V. Thus, 11abV, 12bcV and 21caVare of same magnitude and displaced mutually by 120º to form a balanced 3-phase voltage system. Primary 3-phase line to line voltage of 2003V is therefore stepped down to 3-phase, 1003V line to line voltage at the secondary. The junction of A₂, B2 _{and C}2 _{can be used as primary neutral and may be denoted by N. Similarly the junction of a}2_{, b}2 _{and c}2 _{may be denoted by n for secondary neutral.} Figure 26.3: Both primary & secondary are star connected. Balanced 3-phase supply A1 A2 B1 B2 C1 C2 a1 a2 b1 b2 c1 c2 A B C A1 B1 C1 A2B2C2 a1 b1 c1 a2 b2 c2

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26.3.1 A wrong star-star connection

 

In continuation with the discussion of the last section, we show here a deliberate wrong connection to highlight the importance of proper terminal markings of the coils with dots (•). Let us start from the figure 26.2 where the secondary coils are yet to be connected. To implement star connection on the secondary side, let us assume that someone joins the terminals a₂, b₁ and c₂ together as shown in figure 26.4.

The question is: is it a valid star connection? If not why? To answer this we have to interconnect the secondary voltage phasors in accordance with the physical connections of the coils. In other words, shift and place the secondary voltage phasors so that a₂, b₁ and c₂ overlap each other to make them equipotential. The lengths of phasors 12aaV, 12bbVand 12ccV are no doubt, same and equal to 100 V but they do not maintain 120º mutual phase displacement between them as clear from figure 26.4. The magnitude of the line to line voltages too will not be equal. From simple geometry, it can easily be shown that

122111abbccaVVV=≠

Thus both the phase as well as line voltages are not balanced 3-phase voltage. Hence the above connection is useless so far as transforming a balanced 3-phase voltage into another level of balanced 3-phase voltage is concerned.

Appropriate polarity markings with letters along with dots (•) are essential in order to make various successful 3-phase transformer connections in practice or laboratory.

26.3.2 Bank of three phase transformer

In the background of the points discussed in previous section, we are now in a position to study different connections of 3-phase transformer. Let the discussion be continued with the same three single phase identical transformers, each of rating 10kVA, 200V / 100V, 50Hz,. These Figure 26.4: Both primary & secondary are star connected. Balanced 3-phase supply A1 A2 B1 B2 C1 C2 a1 a2 b1 b2 c1 c2 A B C A1 B1 C1 A2B2C2 a1 b1 c1 a2 b2 c2 120º 120º 60º 60º

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transformers now should be connected in such a way, that it will change the level of a balanced three phase voltage to another balanced three phase voltage level. The three primary and the three secondary windings can be connected in various standard ways such as star / star or star / delta or delta / delta or in delta / star fashion. Apart from these, open delta connection is also used in practice.

Star-star connection

We have discussed in length in the last section, the implementation of star-star connection of a 3-phase transformer. The connection diagram along with the phasor diagram are shown in figure 26.5 and 26.6.

As discussed earlier, we need to apply to the primary terminals (A₁B1_C1_{) a line to line voltage of 200V so that rated voltage (200 V) is impressed across each of the primary coils of the individual transformer. This ensures 100 V to be induced across each of the secondary coil and the line to line voltage in the secondary will be 100V. Thus a 3-phase line to line voltage of 200V is stepped down to a 3-phase line to line voltage of 100V. Now we have to calculate how much load current or kVA can be supplied by this bank of three phase transformers without over loading any of the single phase transformers. From the individual rating of each transformer, we know maximum allowable currents of HV and LV windings are respectively I}3 3 3 3 HV _{= 10000/200 = 50A and I}LV _{= 10000/100 = 100A. Since secondary side is connected in star, line current and the winding currents are same. Therefore total kVA that can be supplied to a balanced 3-phase load is kVA. While solving problems, it is not necessary to show all the terminal markings in detail and a simple and popular way of showing the same star-star connection is given in figure 26.7.} ()33310010030LLLVI==

B A B C A1 B1 C1 A2 B2 C2 Balanced 3-phase supply Balanced 3-phase load c2 b2 a2 c1 b1 a1 Figure 26.5: Star/star Connection. A1 B1 C1 a2 b2 c2 Figure 26.6: Phasor diagram. 200V50A50AIHV = 50 A A

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Star-delta connection

To connect windings in delta, one should be careful enough to avoid dead short circuit. Suppose we want to carry out star / delta connection with the help of the above single phase transformers. HV windings are connected by shorting A₂, B2 _{and C}2 _{together as shown in the figure 26.8. As we know, in delta connection, coils are basically connected in series and from the junction points, connection is made to supply load. Suppose we connect quite arbitrarily (without paying much attention to terminal markings and polarity), a}1 _{with b}2 _{and b}1 _{with c}1_{. Should we now join a}2 _{with c}2 _{by closing the switch S, to complete the delta connection? As a rule, we should not join (i.e., put short circuit) between any two terminals if potential difference exists between the two. It is equivalent to put a short circuit across a voltage source resulting into very large circulating current. Therefore before closing S, we must calculate the voltage difference between a}2 _{with c}2_{. To do this, move the secondary voltage phasors such that a}1 _{and b}2 _{superpose as well as b}1 _{with c}1 _{superpose - this is because a}1 _{and b}2 _{are physically joined to make them equipotential; similarly b}1 _{and c}1 _{are physically joined so as to make them equipotential. The phasor diagram is shown in figure 26.9. If a voltmeter is connected across S (i.e., between a}2 _{and c}2_{), it is going to read the length of the phasor . By referring to phasor diagram of figure 26.9, it can be easily shown that the voltage across the switch S, under this condition is. So this connection is not proper and the switch S should not be closed.} 22acVo100+2cos60100200V22acV==

nced 3

Another alternative way to attempt delta connection in the secondary could be: join a₁ with b₂ and b₁ with c₂. Before joining a₂ with c₁ to complete delta connection, examine the open circuit voltage 21acV. Following the methods described before it can easily be shown that 21acV= 0, which allows to join a₂ with c₁ without any circulating current. So this, indeed is a correct delta connection and is shown in figure 26.10 where a₁ is joined with b₂, b₁ is joined with c₂ and c₁ is joined with a₂. The net voltage acting in the closed delta in this case is zero. Although voltage exists in each winding, the resultant sum becomes zero as they are 120° mutually apart. The output terminals are taken from the junctions as a, b and c for supplying 3-phase load. The corresponding phasor diagram is shown in figure 26.11. c1 b1 a1A B C A1 B1 C1 A2 B2 C2 Bala

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A

B

C

Here also we can calculate the maximum kVA this star / delta transformer can handle without over loading any of the constituents transformers. In this case the secondary line to line voltage is same as the winding voltage i.e., 100V, but the line current which can be supplied to the load is1003. Because it is at this line current, winding current becomes the rated 100A. Therefore total load that can be supplied is()33 1003 100 VA=30kVALLLVI=. Here also total kVA is 3 times the kVA of each transformer. The star-delta connection is usually drawn in a simplified manner for problem solving and easy understanding as shown in figure 26.12.

Another valid delta connection on the LV side is also possible by joining a₂ with b₁, b₂ with c₁ and c₂ with a₁. It is suggested that the reader tries other 3-phase connections and verify that the total KVA is 3-times the individual KVA of each transformer. However, we shall discuss about delta / delta and open delta connection.

Delta / delta and open delta connection

Here we mention about the delta/delta connection because, another important and useful connection namely open delta connection can be understood well. Valid delta connection can be implemented in the usual way as shown in the figure 26.13. The output line to line voltage will be 100V for an input line voltage of 200V. From the secondary one can draw a line current of 1003A which means a total of 30 kVA can be supplied without overloading any of the 200V50A50AIHV = 50 A A B C 2003 V NBalanced 3-phase supply Balanced 3-phase load 100VLVI=1003 A 100A100Aa b c Figure 26.12: Simplified way of showing star-star connection 100V100 A LVI=1003 A

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individual transformers. A simplified representation of the delta-delta connection is shown in figure 26.15 along with the magnitude of the currents in the lines and in the coils of HV and LV side.

Let us now imagine that the third transformer C be removed from the circuit as shown in the second part of the figure 26.13. In effect now two transformers are present. If the HV sides is energized with three phase 200V supply, in the secondary we get 3-phase balanced 100V supply which is clear from the phasor diagram shown in figure 26.14. Although no transformer winding exist now between A₂ & B1 _{on the primary side and between a}2 _{& b}1 _{on the secondary side, the voltage between A}2 _{& B1} on the primary side and between a₂ & b₁ on the secondary side exist. Their phasor representation are shown by the dotted line confirming balanced 3-phase supply. But what happens to kVA handling capacity of the open delta connection? Is it 20 kVA, because two transformers are involved? Let us see. The line current that we can allow to flow in the secondary is 100A (and not 1003 as in delta / delta connection). Therefore total maximum kVA handled is given by()33 100 100 VA=17.32 kVALLLVI=, which is about 57.7% of the delta connected system. This is one of the usefulness of using bank of 3-phase transformers and connecting them in delta-delta. In case one of them develops a fault, it can be removed from the circuit and power can be partially restored.

 

A

 

B

 

C

A₁

 

B1

 

C₁

 

A₂

B2

 

C₂

 

c₂

 

b₂

a₂

 

Delta-delta Connection.

 

c₁

 

b₁

a₁

 

a

 

b

 

c

A₁

 

A₂

 

B1

 

B2

C₁

 

C₂

 

a₁

 

a₂

 

b₁

 

b₂

 

c₁

 

c₂

 

Delta-delta

 

A

 

B

 

C

 

A₁

 

B1

 

A₂

 

B2

 

b₂

 

a₂

 

Open delta Connection.

 

b₁

 

a₁

 

a

 

b

 

c

 

A₁

 

A₂

 

B1

 

B2

 

a₁

 

a₂

 

b₁

 

b₂

 

Delta-delta Figure 26.13: Delta/delta and open delta connection.Figure 26.14: Phasor diagram

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26.3.3 3-phase transformer- a single unit

Instead of using three number of single phase transformers, a three phase transformer can be constructed as a single unit. The advantage of a single unit of 3-phase transformer is that the cost is much less compared to a bank of single phase transformers. In fact all large capacity transformers are a single unit of three phase transformer.

A1

A2a

a

To understand, how it is constructed let us refer to figure 26.16. Here three, single phase transformers are so placed that they share a common central limb. The primary and the secondary windings of each phase are placed on the three outer limbs and appropriately marked. If the primary windings are connected to a balanced 3-phase supply (after connecting the windings in say star), the fluxes φ_A(t), φ_{B(t) and φ}C_{(t) will be produced in the cores differing in time phase mutually by 120°. The return path of these fluxes are through the central limb of the core structure. In other words the central limb carries sum of these three fluxes. Since instantaneous sum of the fluxes, φ}A_{(t)+ φ}B(t)+ φ_C(t) = 0, no flux lines will exist in the central limb at any time. As such the central common core material can be totally removed without affecting the working of the transformer. Immediately we see that considerable saving of the core material takes place if a 3-phase transformer is constructed as a single unit. The structure however requires more floor area as the three outer limbs protrudes outwardly in three different directions. φA φA φB φC φC φB φA φB φC A1A2a1a2C1C2c1c2B1B2b1b2Figure 26.16: A conceptual three phase transformer. Figure 26.17: A practical core type three phase transformer. 200V50A50AIHV = 50 A A B C 2003 V N Balanced 3-phase supply Balanced 3-phase load 100VLVI=1003 A100A100Aa b c Figure 26.15: Simplified way of showing delta-delta connection 100V100 A LVI=1003 A

A further simplification of the structure can be obtained by bringing the limbs in the same plane as shown in the figure 26.17. But what do we sacrifice when we go for this simplified structure? In core structure of figure 26.16, we note that the reluctance seen by the three fluxes are same, Hence magnetizing current will be equal in all the three phases. In the simplified core structure of figure 26.17, reluctance encountered by the flux φ_{B is different from the reluctance encountered by fluxes φ}A _{and φ}C_{, Hence the magnetizing currents or the no load currents drawn will remain slightly unbalanced. This degree of unbalanced for no load current has practically no influence on the performance of the loaded transformer. Transformer having this type of core structure is called the core type transformer.}

26.4 Vector Group of 3-phase transformer

The secondary voltages of a 3-phase transformer may undergo a phase shift of either +30° leading or -30° lagging or 0° i.e, no phase shift or 180° reversal with respective line or phase to neutral voltages. On the name plate of a three phase transformer, the vector group is mentioned. Typical representation of the vector group could be Yd1 or Dy11 etc. The first capital latter Y indicates that the primary is connected in star and the second lower case latter d indicates delta connection of the secondary side. The third numerical figure conveys the angle of phase shift based on clock convention. The minute hand is used to represent the primary phase to neutral voltage and always shown to occupy the position 12. The hour hand represents the secondary phase to neutral voltage and may, depending upon phase shift, occupy position other than 12 as shown in the figure 26.18.

The angle between two consecutive numbers on the clock is 30°. The star-delta connection and the phasor diagram shown in the figures 26.10 and 26.11 correspond to Yd₁. It can be easily seen that the secondary a phase voltage to neutral n (artificial in case of delta connection) leads the A phase voltage to neutral N by 30°. However the star delta connection shown in the figure 26.19 correspond to Yd₁₁. A N a n 30º lag 12 123456 11 10 9 8 7 ANan30º lead 1212 3 4 561110987A N a n 0º shift 12 123456 11 10 9 8 7 ANan180ºshift 1212 3 4 561110987Figure 26.18: Clock convention representing vector groups.

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26.5 Tick the correct answer

 

1. The secondary line to line voltage of a star-delta connected transformer is measured to be 400 V. If the turns ratio between the primary and secondary coils is 2 : 1, the applied line to line voltage in the primary is:

 

(A) 462 V (B) 346 V (C) 1386 V (D) 800 V

 

2. The secondary line to line voltage of a delta-delta connected transformer is measured to be 400 V. If the turns ratio between the primary and secondary coils is 2 : 1, the applied line to line voltage in the primary is:

 

(A) 462 V (B) 346 V (C) 1386 V (D) 800 V

 

3. The secondary line to line voltage of a delta-star connected transformer is measured to be 400 V. If the turns ratio between the primary and secondary coils is 2 : 1, the applied line to line voltage in the primary is:

 

(A) 800 V (B) 500 V (C) 1386 V (D) 462 V

 

4. The secondary line current of a star-delta connected transformer is measured to be 100 A. If the turns ratio between the primary and secondary coils is 2 : 1, the line current in the primary is:

 

(A) 50 A (B) 28.9 A (C) 57.7 A (D) 60 A

 

5. The secondary line current of a delta-star connected transformer is measured to be 100 A. If the turns ratio between the primary and secondary coils is 2 : 1, the line current in the primary is:

 

(A) 86.6 A (B) 50 A (C) 60 A (D) 57.7 A A B C A1 B1 C1 A2 B2 C2 Balanced 3-phase supply c2 b2 a2 c1 b1 a1 a b c a1 a2 b2 b1c1 c2 A1 B1 C1 A2 B2 C2 N n30ºFigure 26.19: Connection and phasor diagram for Y d11.

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6. The primary line current of an open delta connected transformer is measured to be 100 A. If the turns ratio between the primary and secondary coils 2 : 1, the line current in the primary is:

 

(A) 173.2 A (B) 200 A (C) 150 A (D) 50 A

 

7. Two single-phase transformers, each of rating 15 kVA, 200 V / 400 V, 50 Hz are connected in open delta fashion. The arrangement can supply safely, a balanced 3-phase load of:

 

(A) 45 kVA (B) 25.9 kVA (C) 30 kVA (D) 7.5 kVA

 

8. In figure 26.20 showing an incomplete 3-phase transformer connection, the reading of the voltmeter will be:

 

(A) 0 V (B) 173.2 V (C) 346.4 V (D) 300 V

26.6 Problems

 

1. Three number of single phase ideal transformers, each of rating. 10kVA, 200V / 100V, 50Hz is connected in star/delta fashion to supply a balanced three phase 20 kW, 0.8 power factor load at 100V(line to line). Draw a circuit diagram for this. Calculate (i) what line to line voltage should be applied to the primary side? (ii) Calculate the line and phase currents on the secondary and primary sides and indicate them on the diagram.

 

 

2. How two identical single phase transformers each of rating 5kVA, 200V/100V, 50Hz be used to step down a balanced 3-phase, 200V supply to a balanced 3-phase, 100V supply? Explain with circuit and phasor diagrams. Calculate also the maximum kVA that can be supplied from this connection.

 

 

3. A balanced 3-phase load of 20kW, 0.8 power factor lagging is to be supplied at a line to line voltage of 110V. However, a balanced 3-phase voltage of 381V (line to line) is available. Using three numbers of identical single phase ideal transformers each of rating 10kVA, 220V/110V, 50Hz make an arrangement such that the above load can be supplied. Draw the circuit diagram and show the magnitude of currents in the lines and in the windings of the transformers on both LV and HV side.

Figure 26.20:Balanced 3-phase supply A1 A2 B1 B2 C1 C2 a1 a2 b1 b2c1 c2 A B C S N2 = 75 V 400 V line to line N1 = 100

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4. Refer to the following figure 26.21 which shows the windings of a 3-phase transformer. Primary turns per phase is 250. Each phase has got two identical secondary windings each having 100 turns. The primary windings are connected in star by shorting A₂, B2 _{and C}2 _{and supplied from a balanced 3-phase 1000 V (line to line), 50 Hz source.}

 

 

a) If the secondary coils are connected by joining a₂ with b₃ and b₄ with c₁ then calculate11acV.

 

b) All the 6 coils are connected in series in the following way:

 

a₂ joined with b₂ b₁ joined with c₂

c₁ joined with c₄ c₃ joined with b₄

b₃ joined with a₃

Draw the phasor diagram and calculate the voltage 14aaV A1 A2 B1 B2 C1 C2 a1 a2 b1 b2 c1 c2 a3 a4 b3 b4 c3 c4 Figure 26.21: 3-phase transformer with two secondary coils per phase

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