El-Hawary M.E. Electrical Energy Systems

113

Figure 4.9 Three-Winding Transformer and Its Practical Equivalent Circuit.

The I

2

R or load loss for a three-winding transformer can be obtained from

analysis of the equivalent circuit shown.

The Autotransformer

The basic idea of the autotransformer is permitting the interconnection

of the windings electrically. Figure 4.10 shows a two-winding transformer

connected in an autotransformer step-up configuration. We will assume the

same voltage per turn, i.e.,

2

1

N

V

N

V

=

The rating of the transformer when connected in a two-winding configuration is

2211rated

IVIVS

==

(4.19)

In the configuration chosen, the apparent power into the load is

()













+=

2

1

22

2210

1

N

IV

IVVS

(4.20)

The input apparent power is

()













+=

2

1

11

211

1

N

IV

IIVS

i

Thus the rating of the autotransformer is higher than the original rating of the

two-winding configuration. Note that each winding passes the same current in

both configurations, and as a result the losses remain the same. Due to the

increased power rating, the efficiency is thus improved.

114

Autotransformers are generally used when the ratio is 3:1 or less. Two

disadvantages are the lack of electric isolation between primary and secondary

and the increased short-circuit current over that the corresponding two-winding

configuration.

Example 4.4

A 50-kVA, 2.4/0.6-kV transformer is connected as a step-up autotransformer

from a 2.4-kV supply. Calculate the currents in each part of the transformer and

the load rating. Neglect losses. Verify your solution using MATLAB



.

Solution

With reference to Figure 4.10, the primary winding rated current is

A 83.20

4.2

50

1

==

I

The secondary rated current is

A 33.83

6.0

50

2

==

I

Thus the load current is

A 33.83

=

L

I

The load voltage is

kV 3

21

=+=

VVV

L

As a result, the load rating is

kVA250

==

LLL

IVS

Note that

Figure 4.10 Step-Up Autotransformer.

115

kV4.2

A16.104

1

21

==

=

+=

VV

III

i

Thus,

kVA150)16.104)(4.2(

==

i

S

A MATLAB



script implementing Example 4.4 is shown here

The solution is obtained as

Three-Phase Transformer Connections

For three-phase system applications it is possible to install three-phase

transformer units or banks made of three single-phase transformers connected in

the desired three-phase configurations. The latter arrangement is advantageous

from a reliability standpoint since it is then possible to install a single standby

% Example 4-4

% Autotransformer

KVA=30;

KVp=2.4;

KVs=0.6;

% The primary winding rated current is

I1=KVA/KVp

% The secondary rated current is

I2=KVA/KVs

% The load current is

IL=I2

% the load voltage is

VL = KVp+KVs

% The load rating is

SL=VL*IL

EDU»

I1 = 12.5000

I2 = 50

IL = 50

VL = 3

SL = 150

116

single-phase transformer instead of a three-phase unit. This provides a

considerable cost saving. We have seen that there are two possible three-phase

connections; the Y-connection and the

∆

-connection. We thus see that three-

phase transformers can be connected in four different ways. In the Y/Y

connection, both primary and secondary windings are connected in Y. In

addition, we have

∆

/

∆

, Y/

∆

, or

∆

/Y connections. The Y-connected windings

may or may not be grounded.

The Y/

∆

configuration is used for stepping down from a high voltage to

a medium or low voltage. This provides a grounding neutral on the high-voltage

side. Conversely, the

∆

/Y configuration is used in stepping up to a high voltage.

The

∆

/

∆

connection enables one to remove one transformer for maintenance

while the other two continue to function as a three-phase bank (with reduced

rating) in an open-delta or V-connection. The difficulties arising from the

harmonic contents of the exciting current associated with the Y/Y connection

make it seldom used.

In Figure 4.11, the four common three-phase transformer connections

are shown along with the voltage and current relations associated with the

transformation. It is important to realize that the line-to-ground voltages on the

∆

side lead the corresponding Y-side values by 30

°

and that the line currents on

the

∆

side also lead the currents on the Y side by 30

°

.

Consider the Y/

∆

three-phase transformer shown in Figure 4.12. We

can show that

$

303

2

1

−∠=

anAn

V

N

V

That is the

∆

-side line-to-ground secondary voltage V

an

leads the Y-side line-to-

ground primary voltage V

An

by 30

°

.

Turning our attention now to the current relations, we have

$

303

2

1

∠=

Aa

I

N

I

Thus the secondary line current leads the primary current by 30

°

.

Three-phase autotransformers are usually Y-Y connected with the

neutral grounded. A third (tertiary)

∆

-connected set of windings is included to

carry the third harmonic component of the exciting current. A schematic

diagram of a three-phase autotransformer with a

∆

-tertiary is shown in Figure

4.13.

117

Figure 4.11 Three-Phase Transformer Connections.

Control Transformers

Transformers are used not only to step up or step down bulk power

voltages but also as a means for controlling the operations of the power system.

Two examples of control transformer applications involve (1) tap changing

under load (TCUL) transformers, and (2) the regulating transformer.

Load Tap Changing

The TCUL transformer maintains a prescribed voltage at a point in the

system by changing the transformation ratio by increasing or decreasing the

number of active turns in one winding with respect to another winding. This is

performed while not interfering with the load. In practice, a voltage measuring

device actuates the motor that drives the tap changer. If the actual voltage is

higher than a desired upper limit, the motor will change to the next lower tap

118

Figure 4.12 A Y-∆ Transformer and a Phasor Diagram.

Figure 4.13 Schematic Diagram of a Three-Winding Autotransformer.

119

voltage; similarly, a voltage lower than the desired will cause a change to the

next higher up.

The Regulating Transformer

The regulating transformer changes (by a small amount) the voltage

magnitude and phase angle at a certain point in the system. Figure 4.14 shows

the arrangement of a regulating transformer. Assume that:

$

120

0

+∠=

−∠=

∠=

VV

cn

bn

an

The primary windings of the transformers A, B, and C are connected in

∆

. The secondary windings 1, 3, and 5 are connected in Y with their voltages

adjustable. From the phase-shift property in

∆

-Y transformers, we have

$

150

3

90

3

30

3

∠=

−∠=

∠=

m

mo

m

lo

m

ko

V

The magnitude of V

m

can be controlled in a small range and is utilized for

adjusting the magnitude of the three-phase voltage

a

V

′

,

b

V

′′

, and

c

V

′

. The

tertiary windings 2, 4, and 6 have voltages

$

30

∠=

φ

VV

rl

(4.21)

$

90

−∠=

φ

VV

sm

(4.22)

$

150

∠=

φ

VV

tk

(4.23)

The magnitude V

φ

is adjustable and is used for control of the phase angle of the

voltages

a

V

′

,

b

V

′′

, and

c

V

′

.

We can derive the voltages V

km

, V

lk

, V

ml

from V

ko

, V

lo

, V

mo

as

120

Figure 4.14 Schematic of Regulating Transformer.

$

120

0

+∠=

−∠=

∠=

mml

mlk

mkm

VV

Note that V

km

, V

lk

, and V

ml

are in phase with the system voltages V

an

, V

bn

, and

V

cn

. The voltages V

rl

, V

sm

, and V

tk

are 90

°

out of phase with the same voltages.

The incremental voltages

∆

V

a

,

∆

V

b

, and

∆

V

c

are given by

mrc

ltb

ksa

VV

=∆

or

$

900

−∠−∠=−=∆

φ

VVVVV

msmkma

(4.24)

$$

150120

∠−−∠=−=∆

φ

VVVVV

mtklkb

(4.25)

$$

30120

∠−+∠=−=∆

φ

VVVVV

mrlmlc

(4.26)

The

∆

V values are added in series in each phase to give

aanna

VVV

∆+=

′

(4.27)

121

Figure 4.15 Output Voltages of Regulating Transformer.

bbnnb

VVV

∆+=

′

(4.28)

ccnnc

VVV

∆+=

′

(4.29)

A phasor diagram of the voltages in the system is shown in Figure 4.15.

PROBLEMS

Problem 4.1

A 50-kVA, 400/2000 V, single-phase transformer has the following parameters:

R

1

= 0.02 ohm R

2

= 0.5 ohm

X

1

= 0.06 ohm X

2

= 1.5 ohm

G

c

= 2 mS Bm = -6 mS

Note that G

c

and B

m

are given in terms of primary reference. The transformer

supplies a load of 40 kVA at 2000 V and 0.8 PF lagging. Calculate the primary

voltage and current using the equivalent circuits shown in Figure 4.5 and that of

Figure 4.4. Verify your solution using MATLAB



.

Problem 4.2

Find the P.V.R. and efficiency for the transformer of Problem 4.1.

Problem 4.3

Find the maximum efficiency of the transformer of Problem 4.1, under the same

conditions. Verify your solution using MATLAB



.

122

Problem 4.4

The equivalent impedance referred to the primary of a 2300/230-V, 500-kVA,

single-phase transformer is

Z = 0.2 + j0.6 ohm

Calculate the percentage voltage regulation (P.V.R.) when the transformer

delivers rated capacity at 0.8 power factor lagging at rated secondary voltage.

Find the efficiency of the transformer at this condition given that core losses at

rated voltage are 2 kW.

Problem 4.5

A 500/100 V, two-winding transformer is rated at 5 kVA. The following

information is available:

A. The maximum efficiency of the transformer occurs when the output

of the transformer is 3 kVA.

B. The transformer draws a current of 3 A, and the power is 100 W

when a 100-V supply is impressed on the low-voltage winding with

the high-voltage winding open-circuit.

Find the rated efficiency of the transformer at 0.8 PF lagging. Verify your

solution using MATLAB



.

Problem 4.6

The no-load input power to a 50-kVA, 2300/230-V, single-phase transformer is

200 VA at 0.15 PF at rated voltage. The voltage drops due to resistance and

leakage reactance are 0.012 and 0.018 times rated voltage when the transformer

is operated at rated load. Calculate the input power and power factor when the

load is 30 kW at 0.8 PF lagging at rated voltage. Verify your solution using

MATLAB



.

Problem 4.7

A 500 KVA, 2300/230 V single phase transformer delivers full rated KVA at

0.8 p.f. lagging to a load at rated secondary voltage. The primary voltage

magnitude is 2400 V under these conditions and the efficiency is 0.97. Find the

equivalent circuit parameters of this transformer neglecting the no load circuit.

Problem 4.8

Solve Problem 4.7 using MATLAB



for a power factor of 0.7.

Problem 4.9

A single phase transformer has a turns ratio of 2:1, and an equivalent reactance

X

eq

= 4 ohms. The primary voltage is 2020 V at 0.75 p.f. lagging. The voltage

regulation for this power factor is found to be 0.09, and the efficiency is 95%

under these conditions. Neglect the no load circuit.

A. Find the transformer’s equivalent circuit resistance R

eq

referred to

El-Hawary M.E. Electrical Energy Systems

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