El-Hawary M.E. Electrical Energy Systems

103

21eq

XXX

RRR

′

+=

′

+=

Example 4.1

A 100-kVA, 400/2000 V, single-phase transformer has the following parameters

R

1

= 0.01 R

2

= 0.25 ohms

X

1

= 0.03 ohms X

2

= 0.75 ohms

G

c

= 2.2 mS B

m

=6.7 mS

Note that G

c

and B

m

are given in terms of primary reference. The transformer

supplies a load of 90 kVA at 2000 V and 0.8 PF lagging. Calculate the primary

voltage and current using the equivalent circuits shown in Figure 4.5.Verify

your solution using MATLAB



.

Solution

Let us refer all the data to the primary (400 V) side:

R

1

= 0.01 ohm X

1

= 0.03 ohms

ohms01.0

2000

400

25.0

2

=













=

′

R

ohm03.0

2000

400

75.0

2

=













=

′

X

Thus,

ohm02.0

21eq

=

′

+=

RRR

ohm06.0

21eq

=

′

+=

XXX

The voltage V

2

= 2000 V; thus

V 400

2000

400

2000

2

=













=

′

V

The current

2

I

′

is thus

A 225

400

1090

3

2

=

×

=

′

I

The power factor of 0.8 lagging implies that

A 87.36225

2

$

−∠=

′

I

For ease of computation, we start with the simplest circuit of Figure

104

4.5(D). Let us denote the primary voltage calculated through this circuit by

d

V

1

.

It is clear then that

()

06.087.362250400

)(

eq221

$

−∠+∠=

′

+

′

=

j

XIjVV

d

Thus,

A 87.36225

V 516.1243.408

1

$

−∠=

∠=

d

I

V

Comparing circuits (C) and (D) in Figure 4.5, we deduce that

(

)

(

)

eq21eqeq221

RIVjXRIVV

dc

′

+=+

′

+

′

=

Thus,

()()

A 87.36225

V 127.178.411

04.087.36225516.1243.408

21

1

$

−∠=

′

=

∠=

−∠+∠=

II

V

c

Let us consider circuit (A) in Figure 4.5. We can see that

V 127.178.411

11

$

∠==

ca

VV

But

()

A 277.37418.227

127.178.411107.6102.287.36225

33

121

$

−∠=

∠×−×+−∠=

++

′

=

−−

j

VjBGII

aa

mc

Circuit (B) is a bit different since we start with

2

V

′

impressed on the

shunt branch. Thus,

()

$

277.3737.227

0400107.6102.287.36225

33

221

−∠=

∠×−×+−∠=

′

++

′

=

−−

j

VjBGII

mc

b

Now

105

(

)

$

1265.196.411

eqeq121

∠=

++

′

=

jXRIVV

bb

The following is a MATLAB



script implementing Example 4.1.

% Example 4-1

% To enter the data

R1=0.01;

X1=0.03;

Gc=2.2*10^(-3);

R2=0.25;

X2=0.75;

Bm=-6.7*10^(-3);

V2=2000;

N1=400;

N2=2000;

pf=0.8;

S=90*10^3;

% To refer all the data to the primary

side

R2_prime=R2*(N1/N2)^2;

X2_prime=X2*(N1/N2)^2;

Req=R1+R2_prime;

Xeq=X1+X2_prime;

%The voltage V2=2000 V; thus

V2_prime=V2*(N1/N2);

% To find I2' complex

% Power factor of 0.8 lagging

theta=acos(pf);

theta_deg=theta*180/pi;

I2_prime=abs(S/V2_prime);

I2_primecom=I2_prime*(cos(-

theta)+i*sin(-theta));

I2_primearg=-theta_deg;

% From the figure 5-5(d). The primary

voltage V1d is

V1d=V2_prime+i*I2_primecom*Xeq;

delta=angle(V1d);

delta_deg=delta*180/pi;

I1d=I2_primecom;

% Comparing circuit (c) and (d) in

figure 4.5 we have

V1c_compl=V2_prime+I2_primecom*(Req+i*X

eq);

V1c_mod=abs(V1c_compl);

V1c_arg=angle(V1c_compl);

106

MATLAB



con’t.

V1c_argdeg=V1c_arg*180/pi;

I1c=I2_primecom;

% Consider to the circuit (a) in figure

4.5

V1a_compl=V1c_compl;

I1a_compl=I2_primecom+(Gc+i*Bm)*V1a_com

pl;

I1a_mod=abs(I1a_compl);

I1a_arg=angle(I1a_compl);

I1a_argdeg=I1a_arg*180/pi;

% Consider to the circuit (b) in figure

4.5

I1b_compl=I2_primecom+(Gc+i*Bm)*V2_prim

e;

I1b_mod=abs(I1b_compl);

I1b_arg=angle(I1b_compl);

I1b_argdeg=I1b_arg*180/pi;

V1b_compl=V2_prime+I1b_compl*(Req+i*Xeq

);

V1b_mod=abs(V1b_compl);

V1b_arg=angle(V1b_compl);

V1b_argdeg=V1b_arg*180/pi;

% The exact equivalent circuit is now

considered

% as shown in Figure 4.4 (b. First we

calculate E1

E1_compl=V2_prime+I2_primecom*(R2_prime

+i*X2_prime);

E1_mod=abs(E1_compl);

E1_arg=angle(E1_compl);

E1_argdeg=E1_arg*180/pi;

% Now we calculate I1

I1_compl=I2_primecom+E1_compl*(Gc+i*Bm)

I1_mod=abs(I1_compl)

I1_arg=angle(I1_compl);

I1_argdeg=I1_arg*180/pi

% Thus, we have

V1_compl=E1_compl+I1_compl*(R1+i*X1)

V1_mod=abs(V1_compl)

V1_arg=angle(V1_compl);

V1_argdeg=V1_arg*180/pi

107

The solution is

Transformer Performance Measures

Two important performance measures are of interest when choosing

transformers. These are the voltage regulation and efficiency of the transformer.

The voltage regulation is a measure of the variation in the secondary voltage

when the load is varied from zero to rated value at a constant power factor. The

percentage voltage regulation (P.V.R) is thus given by

rated 2

rated 2)load no( 2

100P.V.R.

V

VV

−

=

(4.11)

If we neglect the exciting current and refer the equivalent circuit to the

secondary side, we have by inspection of Figure 4.6,

2

1

100P.V.R.

V

a

V

−













=

where a is the transformer ratio:

2

1

N

a

=

From the phasor diagram we have approximately in terms of transformer

constants:

EDU»

I1_compl = 1.8092e+002 - 1.3771e+002i

I1_mod = 227.3679

I1_argdeg = -37.2772

V1_compl = 4.1179e+002 + 8.1005e+000i

V1_mod = 411.8702

V1_argdeg = 1.1269

108

Figure 4.6

Transformer Approximate Equivalent Circuit and Associated Phasor Diagrams for

Voltage Regulation Derivation.

(

)

()



























−

+







+

≅

2

eqeq

sincos

2

1

sincos

100P.V.R.

222

V

RXI

V

XRI

LeqLeqL

LLL

φ

(4.12)

The efficiency of the transformer is the ratio of output (secondary)

power to the input (primary) power. Formally the efficiency is

η

:

1

2

P

=

η

(4.13)

Let I

L

be the load current.

l

PPP

+=

21

The power loss in the transformer is made of two parts: the I

2

R loss and the core

loss P

c

.

As a result, the efficiency is obtained as:

109

()

eq

2

cos

RIPIV

IV

LcLL

LL

++

=

φ

η

(4.14)

The following example utilizes results of Example 4.1 to illustrate the

computations involved.

Example 4.2

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

Solution

Let us apply the basic formula of Eq. (4.12). We have from Example 4.1:

ohm5.1

400

2000

06.0

ohm5.0

400

2000

02.0

A45

V2000

2

eq

2

eq

2

=













=













=

X

R

I

V

L

Thus substituting in Eq. (4.12), we get

[][]

percent9455.2

2000

)6.0(5.0)8.0(5.145

2

1

2000

)6.0(5.1)8.0(5.045

100P.V.R.

2

=

































−

+

=

To calculate the efficiency we need only to apply the basic definition.

Take the results of the exact circuit. The input power is

W000,72

8.01090

cos

W66.385,73

)404.38)(cos418.227)(77.411(

cos

3

2222

1111

=

××=

=

φ

IVP

Thus,

98112.0

66.385,73

000,72

==

η

The efficiency of a transformer varies with the load current I

L

. It

110

attains a maximum when

0

=

∂

L

I

η

The maximum efficiency can be shown to occur for

()

eq

2

RIP

Lc

= (4.15)

That is, when the I

2

R losses equal the core losses, maximum efficiency is

attained.

Example 4.3

Find the maximum efficiency of the transformer of Example 4.1 under the same

power factor and voltage conditions.

Solution

We need first the core losses. These are obtained from the exact equivalent

circuit of Figure 4.4

[]

()

$

57174.087.405

03.001.087.36225400

22221

∠=

+−∠+=

′

+

′′

+

′

=

j

XjRIVE

()

W407.362

)102.2()87.405(

32

2

1

=

×=

=

−

cc

GEP

For maximum efficiency,

()

eq

2

RIP

Lc

=

Referred to the primary, we thus have

)02.0(407.362

2

L

I

=

Thus for maximum efficiency,

111

98345.0

)407.362(2)8.0)(612.134)(400(

)8.0)(612.134)(400(

2cos

cos

A 612.134

2

max

=

+

=

+

′

=

cLL

LL

L

PIV

IV

I

φ

η

4.3 TRANSFORMER CONNECTIONS

Single-phase transformers can be connected in a variety of ways. To

start with, consider two single-phase transformers A and B. They can be

connected in four different combinations provided that the polarities are

observed. Figure 4.7 illustrates a series-series connection where the primaries of

the two transformers are connected in series whereas the secondaries are

connected in series. Figure 4.8 illustrates the series-parallel connection and the

parallel-series connection. Note that when windings are connected in parallel,

those having the same voltage and polarity are paralleled. When connected in

series, windings of opposite polarity are joined in one junction. Coils of unequal

voltage ratings may be series-connected either aiding or opposing.

Figure 4.7 Two Transformers with Primaries in Series and Secondaries in Series. (A) Connection

Diagram, and (B) Exact Equivalent Circuit.

112

Figure 4.8 Series-Parallel and Parallel-Series Connections for Single-Phase Transformers.

Three-Winding Transformers

The three-winding transformer is used in many parts of the power

system for the economy achieved when using three windings on the one core.

Figure 4.9 shows a three-winding transformer with a practical equivalent circuit.

The impedances Z

1

, Z

2

, and Z

3

are calculated from the three impedances

obtained by considering each pair of windings separately with

2

231312

1

ZZZ

Z

−+

=

(4.16)

2

132312

2

ZZZ

Z

−+

=

(4.17)

2

122313

3

ZZZ

Z

−+

=

(4.18)

El-Hawary M.E. Electrical Energy Systems

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