ece320 chapter 3.pdf

8/10/2019 ECE320 Chapter 3.pdf

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ECE 320

Energy Conversion and Power ElectronicsDr. Tim Hogan

Chapter 3: Transformers (Textbook Chapter 2)

Chapter Objectives

In this chapter you will be able to:

Choose the correct rating and characteristics of a transformer for a specific application

Calculate the losses, efficiency, and voltage regulation of a transformer under specific operating

conditions.

Experimentally determine the transformer parameters given its ratings.

Utilize the per unit system.

3.1 Introduction

Transformers do not have moving parts, nor are they energy conversion devices, however theirability to modify the current-voltage characteristics of a given load or source, make them invaluable

components in energy conversion systems. They are utilized for power applications and in lowpower signal processing systems. One application in power transmission is the use of transformers

on a transmission line utility pole commonly seen as a cylinder with a few wires sticking out. These

wires enter the transformer through bushings that provide isolation between the wires and the tank.Inside the tank there is an iron core commonly made of silicon-steel laminations that are 14 mils

(0.014) thick. The insulation often used is paper with the whole coil system immersed in insulating

oil. The oil increases the dielectric strength of the paper and helps to transfer heat from the core/coilassembly. An drawing of one such distribution transformer is shown in Figure 2.2 in your textbook.

Connection of the transformer to the transmission lines can take several electrical configurations. A

relatively simple connection to a 2.4 kV three phase transmission line is shown in Figure 1.2400 voltthree-phasethree-wire

primary system

A

B

C

2400

24002400

H1 H2

X1X2X3

120 voltone-phasetwo-wire service

Figure 1. Example configuration of a distribution pole transformer connection to

three phase power lines to provide 120 (V) service to your home.

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2400 voltthree-phasethree-wire

primary system

A

B

C

2400

24002400

H1 H2

X1X

2

X3

120 (V)

120 (V)240 (V)

Figure 2. Example configuration of a distribution pole transformer connection tothree phase power lines and provides 120 (V) and 240 (V) service.

If a neutral line is also part of the three phase transmission line (perhaps between the substationand your home), then the connection could be made as shown in Figure 3.

4160Y/2400 volt

three-phasefour-wire Ywith neutral

B

C

4160

4160

4160

H1 H2

X1X3

120 (V)

120 (V)240 (V)

A

NN2400

2400

2400

X2

Figure 3. Distribution transformer connection to provide 120 (V) and 240 (V)

service from a 4160Y/2400 (V) four-wire transmission line.

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For a three phase line at the service end, a system could be connected to a four wire three phase

transmission line source as shown in Figure 4below.

4160Y/2400 voltthree-phasefour-wire Ywith neutralprimary service

B

C

4160

41604160

H1

H2

X1X3

208 (V)

208 (V)208 (V)

A

NN2400

2400

2400

X2

H1

H2

X1X3 X2

H1

H2

X1X3 X2

B

C

A

N

three-phasefour-wire Ygrounded neutralsecondary service

120 (V)

120 (V)

120 (V)

Figure 4. Three phase to three phase distribution transformer connection providing afour-wire three phase distribution of 120 (V) and 208 (V) service from a

4160Y/2400 (V) four-wire transmission line.

Transformers are very common place in society, and have had significant impact at many power

levels. They also continue to be improved on today, with such studies as amorphous metals to furtherdecrease core losses. For more understanding, we begin with the ideal transformer.

3.2 Ideal Transformer

Under ideal conditions, the iron core has infinite permeability and the coils have zero electrical

resistance. Coil 1 has N1turns, and coil 2 has N2turns, and all of the magnetic flux is maintained in

the iron (no flux leakage).

i i21

e1 e2

F2F1

Figure 5. Transformer and equivalent magnetic circuit.

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The electromotive force, or emf, is represented with the symbol e. For an ideal transformer, this

is the voltage at the terminals of a given coil. The flux linkage in each coil is 11 N= and

22 N= . The electromotive force induced in each coil is then the time derivative of the flux

linkage or

dtdN

dtde 111 == (V) (3.1)

dt

dN

dt

de

2

22 == (V) (3.2)

The ratio of the voltage at the terminals of coil 1 to the voltage at coil 2 is then

2

1

2

1

N

N

e

e= (3.3)

Using an equivalent circuit shown in Figure 5with the magnetomotive force F1for coil 1 and F2

for coil 2 we would write:

0221121 == iNiNFF (3.4)

or

1

2

2

1

N

N

i

i= (3.5)

Transformers are often used with a voltage source connected to one coil and a load connected tothe other coil. The dots above the coils help to indicate the voltage reference marks for the coils such

that for an ideal transformer a positive voltage,dt

dNev

111 == in coil 1 (primary coil) results in a

positive voltage,dt

dNev

222 == in the coil 2 (secondary coil) as shown in Figure 6.

i i21

v1 v2

LoadN

1

N2

Figure 6. Circuit utilizing a transformer.

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The circuit symbol for the transformer is shown in Figure 7. Since the voltages across the coilstransform in the direct ratio of the turns and the current transforms in the inverse ratio of the turns,

then the impedance can also be transformed through the transformer.

i 1 i2

N1

N2

Figure 7. Transformer circuit symbol.

For a voltage applied to the primary coil, and a load impedance,ZL, connected to the secondary asshown in Figure 8.

ZL

+

V1 V2

+

N1 N2

I1 I2

Figure 8. Transformer with a load connected to the secondary.

The voltages and currents from secondary to primary are related by (3.3) and (3.5), or

22

11

VN

NV = (3.6)

21

21

IN

NI = (3.7)

Then the impedance measured at the terminals of the primary is

2

22

1

21

1

1

I

VNNZ

I

V

== (3.8)

Thus the following circuits are equivalent

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ZL

+

V1 V2

+

N1 N2

I1 I2

ZL

+

V1

N1 N2

I1

I2

N1( )N22

ZLN1( )N2

2

+

V1

I1

(a) (b) (c)

Figure 9. Three equivalent circuits assuming an ideal transformer.

3.3 Non-Ideal Transformer

There are several non-ideal properties that we can take into account by modifying the equivalentcircuit shown in Figure 7. These include a finite core permeability (c< ), leakage flux, core

losses, and coil resistance. The contribution from each of these is discussed below.

Finite Core Permeability

For the core of the transformer to have a finite permeability, then the circuit in Figure 5is

modified to include the reluctance of the core.

F2F1

R

Figure 10. Transformer shown in Figure 5, but with a core of finite permeability.

Then we can write

RFF == 221121 iNiN (3.9)

Defining a magnetomotive force that is equal to the drop across the reluctance of the core as:

RF == exc iN ,11 (3.10)

Solving for the flux gives

R

exiN ,11= (3.11)

The rate of change in the flux is proportional to the induced voltage as

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dt

diN

dt

iNd

Ndt

dNe

ex

ex

,121

,11

111

=

==

R

R

(3.12)

This induced voltage is proportional to the time derivative of the current i1,exwhich can be

represented by an inductor in our equivalent circuit with a value ofR

21NLm = . The equivalent circuit

for the ideal transformer is then modified to account for the finite permeability of the core by placingan additional inductor across the primary coil as shown in Figure 11.

+

e 1 e2

+

N1 N2

i'1

L

i2

m

i1

i1,ex

Ideal transformer

Figure 11. Equivalent circuit for a transformer with finite core permeability.

We can also see from Figure 11that 1,11 iii ex = which is in agreement with equations (3.9) and

(3.10).

Leakage Flux

With finite core permeability, not all of the flux will be confined to the metal core, but some will

leak outside the core in the surrounding air. The influence of this leakage flux can also be included

in the equivalent circuit, by considering an additional reluctance associated with the leakage flux, l1,

for coil 1, and l2for coil 2 such that

2

222

1

111

l

l

ll

iN

iN

R

R

=

=

(3.13)

These reduce the magnetizing flux, m, of the core, and modify the induced voltages for the primary(coil 1) and secondary (coil 2) to be

dt

diNe

dt

dN

dt

dN

dt

dv

dt

diNe

dt

dN

dt

dN

dt

dv

l

lm

l

lm

2

2

22

22

222

2

1

1

21

11

111

1

+=

+

==

+=

+

==

R

R

(3.14)

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The last term for v1can be represented by an inductor,

=

1

21

1l

lN

LR

and the last term for v2can be

represented by

=

2

22

2l

lN

LR

. These can be incorporated into the equivalent circuit as shown in

Figure 12.

+

e 1 e2

+

N1 N2

i'1

L

i2

m

i1

i1,ex

Ideal transformer

Ll1 Ll2

Figure 12. Equivalent circuit for a transformer with finite core permeability and

leakage flux.

Core Losses

The dissipation of heat in the core due to the flux through it can be represented by a resistor in the

equivalent circuit. It is placed in parallel toLmsince the power loss in the core is proportional to the

flux through the core squared, and thus is proportional to .21e

Resistance of the Coils

The resistance of copper is approximately 16.810-9

(m), however with hundreds to thousandsof turns for the primary and secondary coils it can lead to an appreciable resistance. Losses to these

resistances are proportional to the current through the coils squared.

Adding the core losses and resistance of the coils to the equivalent circuit we obtain our

completed model of the transformer as show in Figure 13.

+

e 1 e2

+

N1 N2i'1

L

i2

m

i1

i1,ex

Ideal transformer

Ll1 Ll2

Rc

R1 R2

V2

++

V1

Figure 13. Equivalent circuit for a non-ideal transformer.

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Example

Consider a transformer with a turns ratio of 4000/120, with a primary coil resistanceR1= 1.6 (),

a secondary coil resistance ofR2= 1.44 (m), leakage flux that corresponds toLl1= 21 (mH), and

Ll2= 19 (H), and a realistic core characterized byRc= 160 (k) andLm= 450 (H). The low voltageside of the transformer is at 60 (Hz), and V2= 120 (V), and the power there is P2= 20 (kW) at

pf= 0.85 lagging. Calculate the voltage at the high voltage side and the efficiency of the transformer.

7.169450602 === mm LX (k)

92.7021.060211 === lLX ()

16.71019602 622 === lLX (m)

( )pfVP

I

=2

22

/-31.8 = 196.08/ -31.8 (A)

( ) 045.198.120 22222 jjXRIVE +=++= (V)

83.347.4032 22

11 jE

N

NE +=

= = 4032.8/0.49 (V)

1017.3001.5 21

21 jI

NNI =

= (A)

0236.00254.011

1,1 jjXR

EImc

ex =

+= (A)

==+= 125.30255.5 1,11 jIII ex 5.918/ -31.87 (A)

( ) 2.695.4065 11111 jjXRIEV +=++= (V) = 4066/0.9 (V)

The power losses in the windings and core are:

39.550144.008.196 22222 === RIPR (W)

04.566.1918.5 21211 === RIPR (W)

64.10110160

8.40323

221 =

==

cc

R

EP (W)

08.21321 =++= cRRloss PPPP (W)

( )9895.0

08.2131020

10203

3

2

2 =+

=

+==

lossin

out

PP

P

P

P

3.4 Losses and Ratings

The impedances shown in Figure 13can be reflected into either the primary side, or the secondaryside of the transformer as shown in Figure 14.

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N1 N2

Lm

Ll1

Rc

R1 L l2N1( )N2

2

R2N1( )N2

2

+

V1 V2

+

V'2

+

N1 N2

V'1

+

V1

+

Lm

Ll2

Rc

R2L l1

N2( )N12

R1N2( )N1

2

V2

+

N2( )N1

2 N2( )N12

Figure 14. Reflection of the impedances to the primary or secondary side of the transformer.

For a given frequency, the power losses in the core (iron losses) increase with the voltage e1(ore2). If these losses exceed a particular limit, a hot spot in the transformer will reach a temperature

that dramatically reduces the life of the insulation. Limits are therefore put onE1andE2which arethe voltage limits for the transformer. The current limits for the transformer limitI1andI2to avoid

excessive Joule heating due to winding resistances.Typically a transformer is described by its rated voltages,E1NandE2N, that give both the limits

and turns ratio. The ratio of the rated currentsN

N

I

I

2

1 , is the inverse of the ratio of the voltages when

the magnetizing current is neglected. Instead of listing these rated current limits, a transformer is

described by its rated apparent power as:

NNNNN IEIES 2211 == (3.15)

When a transformer is operated at its rated conditions, that is at its maximum current and

maximum voltage, the magnetizing current,I1,ex, typically does not exceed 1% of the current in thetransformer. Its effect on the voltage drop on the leakage inductance and on the winding resistance is

therefore negligible.Under maximum (rated) current, the total voltage drops on the winding resistances, and leakage

inductances typically do not exceed 6% of the rated voltage. Their effect onE1andE2is small and

their effect on the magnetizing current can be neglected.Because these effects are small, we can modify the equivalent circuits shown in Figure 14to a

slightly inaccurate, but much more useful on shown in Figure 15.

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11/23

N1 N2

Lm

Ll1

Rc

R1

L l2N1( )N2

2

R2N1( )N2

2+

V1 V2

+

V'2

+

N1 N2

V'1

+

V1

+

Lm

Ll2

Rc

R2

L l1N2( )N1

2

R1N2( )N1

2

V2

+

N2( )N12

N2( )N12

Figure 15. Slightly inaccurate, but highly simplified equivalent circuits for the transformer.

When we utilize this simplification and work with the reflected voltages, the transformer

equivalent circuits can be shown as:

cR mL

Req + R'2

+

V1 V'2

+

= R1

Leq + L'l2= Ll1i'1

i1

i1,ex

V'1

+

L'mR'c V2

+Req + R2= R'1

Leq + L l2= L'l1

Figure 16. Working with V2or V1the above approximate circuits for the

transformers can simplify the analysis.

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Example (repeat with simplified equivalent circuits)

Consider a transformer with a turns ratio of 4000/120, with a primary coil resistanceR1= 1.6 (),

a secondary coil resistance ofR2= 1.44 (m), leakage flux that corresponds toLl1= 21 (mH), andLl2= 19 (H), and a realistic core characterized byRc= 160 (k) andLm= 450 (H). The low voltage

side of the transformer is at 60 (Hz), and V2= 120 (V), and the power there is P2= 20 (kW) atpf= 0.85 lagging. Calculate the voltage at the high voltage side and the efficiency of the transformer.

Using the simplified equivalent circuits of Figure 16, we can first findReq, andXeq

2.32

2

2

11 =

+= R

N

NRReq ()

876.15602 2

2

2

11 =

+= lleq L

N

NLX ()

then

( )pfVP

I

=2

22

/-31.8 = 196.08/ -31.8 (A)

22 VE =

102.35

1

221 j

N

NII =

= = 5.884/-31.8 (A)

4000

2

121 =

=

N

NEE (V)

0235.00258.0

111,1 jjXREI mcex

=

+=

(A)

==+= 125.30259.5 1,11 jIII ex 5.918/-31.87 (A)

( ) 79.694065 111 jjXRIEV eqeq +=++= (V) = 4066/0.98 (V)

The power losses in the windings and core are:

79.1102.3884.522

1 === eqR RIP eq (W)

28.10310160

4065

3

221 =

==cc R

VP (W)

07.214Re =+= cqloss PPP (W)

( )9894.0

07.2141020

10203

3

2

2 =+

=

+==

lossin

out

PP

P

P

P

These values agree well with the previous analysis using the more accurate model.

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14/23

b

bb

I

VZ

1

11 = (3.19)

1b

b

b

bb

I

V

N

N

I

VZ 1

2

1

2

2

22

== (3.20)

or

bb ZN

NZ 1

2

1

22

= (3.21)

To move impedances from one side of the transformer to the other, they get multiplied or divided

by the square of the turns ratio,

2

1

2

N

N, but so does the base impedance, hence the pu value of an

impedance stays the same regardless of which side of the transformer it is on.Through our choice of the bases in (3.16) and (3.17), we also see that for an ideal transformer

pupu

pupu

IIEE

,2,1

,2,1

=

=

Thus when using the per unit system, an ideal transformer has voltages and currents on one side that

are identical to the voltages and currents on the other side, and the ideal transformer can be

eliminated.

Example (repeat and solved using pu system)

Consider a 30 (kVA) rated transformer with a turns ratio of 4000/120, with a primary coil

resistanceR1= 1.6 (), a secondary coil resistance ofR2= 1.44 (m), leakage flux that corresponds

toLl1= 21 (mH), andLl2= 19 (H), and a realistic core characterized byRc= 160 (k) andLm= 450 (H). The low voltage side of the transformer is at 60 (Hz), and V2= 120 (V), and the power

there is P2= 20 (kW) atpf= 0.85 lagging. Calculate the voltage at the high voltage side and theefficiency of the transformer.

1. First calculate the impedances of the equivalent circuit.

40001 =bV (V)

301 =bS (kVA)

5.7

104

1030

3

3

1 =

=bI (A)

5331

21

1 ==b

bb

S

VZ ()

1202 =bV (V)

3012 == bb SS (kVA)

2502

22 ==

b

bb

V

SI (A)

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15/23

48.02

22 ==

b

bb

I

VZ ()

2. Convert everything to per unit.

003.0

1

1,1 ==

b

pu

Z

RR (pu)

003.02

2,2 ==

bpu

Z

RR (pu)

3001

, ==b

cpuc

Z

RR (pu)

0149.0602

1

1,1 =

=

b

lpul

Z

LX

(pu)

0149.0602

2

2,2 =

=

b

lpul

Z

LX

(pu)

3186021

, ==b

mpum

ZLX (pu)

12

2,2 ==

bpu

V

VV (pu)

6667.02

2,2 ==

bpu

S

PP (pu)

3. Solve in the per unit system.

)(

,2

,2,2

pfV

PI

pu

pupu

= /arccos(pf) = 0.666 j0.413(pu)

( ) 0.017361.0163 ,2,2,1 jjXRIVV eqeqpupupu +=++= (pu)

0031.00034.0

,

,1

,

,1, j

jX

V

R

VI

pum

pu

puc

pupum =+= (pu)

4163.06701.0 ,2,,1 jIII pupumpu =+= (pu)

00369.0,2,2 == pueqpuR RIP

eq

(pu)

00344.0,

2,1

, ==puc

pupuc

R

VP (pu)

( )9894.0

0034.00037.06667.0

6667.0

)( ,,2

,2=

++=

+=

pulosspu

pu

PP

P

1-15


16/23

4. Convert back to SI units. The efficiency is unitless and thus stays the same. Converting

gives1V

( ) =+=+== 46.698.40650.017361.01634000 ,111 jjVVV pub 4065.8/0.979 (V)

3.5 Testing TransformersPurchased transformers often give information on the frequency, winding ratio, power, and

voltage ratings, but not the impedances. These impedances are important in calculating the voltage

regulation, efficiency, etc. Use of open circuit and short circuit tests we will determineReq,Leq,Rc,andLm.

Open Circuit Test

Leaving one side of the transformer open circuited, while the other has the rated voltage

{Vin-oc= 1(pu)} applied to it, we measure the current and power. The current that flows into the

transformer is mostly determined by the impedancesXmandRc, and it is much lower than the rated

current for the transformer. It is often the case that the rated voltage for the low voltage side (lowtension side) of the transform since for the above example it would be easier applying 120 (V) instead

of 4000 (V). Since the units we use indicate if we are using the per unit system, the followingcalculations will drop the subscriptpu. Using the following equivalent circuit with the primary open

circuited, and V2= Voc= 120 (V) applied to the secondary.

V'1

+

L'mR'

c

V2

+Req + R2= R'1

Leq + L l2= L'l1

Figure 17. Open circuit testing of a 4000/120 rated transformer. With 120 (V)applied to the low voltage side {V2= 120 (V)}, the primary voltage for

this open circuit test is 4000 (V).

For this open circuit test, the following can be compared to the measured values of current and

power as:

cc

ococ

RR

VP

12== (pu) (3.22)

m

oc

c

ococ

jX

V

R

VI += (3.23)

1-16


17/23

22

111

mc

ocXR

I += (pu) (3.24)

Then from the open circuit test with measurements of the current and the power, we determine RcandXm.

Short Circuit Test

For the above open circuit test, the low voltage side of the transformer was chosen since it is

easier to apply the lower voltage during that test. Similarly, the rated current is lower on the high

voltage side of the transformer. With the short circuit test, the rated current is commonly applied tothe high voltage side of the transformer (since with a short circuited secondary, the applied voltage

required to reach the rated current is relatively low). With the rated current applied to the high

voltage side, we measure the voltage, Vscwhich is V1for this example, and the power, Psc.

(pu) (3.25)eqeqscsc RRIP == 12

( )eqeqscsc jXRIV += (3.26)

221 eqeqsc XRV += (pu) (3.27)

cR mL

Req + R'2

+

V1 V'2

+

= R1

Leq + L'l2= Ll1i'1

i1

i1,ex

= 0

Figure 18. Short circuit testing of a 4000/120, 30 (kVA) rated transformer. This

corresponds to the rated current on the high voltage side of 7.5 (A).

Then from the short circuit test with measurements of the current and the power, we determine

and21 RRReq += ( )21602 llm LLX += .

Example

A 60Hz transformer is rated 30 (kVA), 4000(V)/120(V). The open circuit test, with the highvoltage side open, gives Poc= 100 (W),Ioc= 1.1455 (A). The short circuit test, measured with the

low voltage side shorted, gives Psc= 180 (W), Vsc= 129.79 (V). Determine the equivalent circuit for

this transformer by using the per unit system.

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18/23

1. Bases:

40001 =bV (V)

301 =bS (kVA)

5.7104

10303

3

1 =

=bI (A)

5331

21

1 ==b

bb

S

VZ ()

1202 =bV (V)

3012 == bb SS (kVA)

250120

1030 3

2

12 =

==

b

bb

V

SI (A)

48.0250

120

222 ===bbb

I

VZ ()

2. Convert to (pu):

006.01030

1803,

=

=puscP (pu)

0324.0104

79.1293,

=

=puscV (pu)

00333.01030

1003,

=

=puocP (pu)

00458.0250

1455.1, ==puocI (pu)

3. Calculate the components of the equivalent circuit {dropping the (pu) subscripts}

eqscsc RIP2= or 006.01

2 === sc

sc

sceq P

I

PR (pu)

221 eqeqeqeqscscsc XRjXRIVV +=+== or 0318.0

22 == eqsceq RVX (pu)

c

ococ

R

VP

2

= or 3001

2

===ococ

occ

PP

VR (pu)

1-18


19/23

22

11

mcm

oc

c

ocococ

XRjX

V

R

VII +=+== or 318

1

1

22

=

=

c

oc

m

RI

X (pu)

With this knowledge, we can address the more common example of:

A 60Hz transformer is rated 30 (kVA), 4000(V)/120(V). The short circuit impedance is0.0324 (pu) and the open circuit current is 0.0046 (pu). The rated core losses are 100 (W) and the

rated winding losses are 180 (W). Calculate the efficiency and the necessary primary voltage when

the load at the secondary is at rated voltage, 20 (kW), and at 0.8pflagging.

Using (pu) system:

0324.0=scZ (pu)

006.0

1030

1801

3

2 =

=== eqeqscsc RRIP (pu)

017.022 == eqsceq RZX (pu)

coc

RP

1= or 300

1030

100

11

3

=

==oc

cP

R (pu)

22

11

mc

ocXR

I += or 3181

1

22

=

=

c

oc

m

RI

X (pu)

Now we have the equivalent circuit components of the transformer and can work on the full

circuit which includes the load that has a power of 20 (kW) or:

6667.01030

10203

3

2 =

=P (pu)

but the power at the load is

pfIVP = 222 or 8.016667.0 2= I thus (pu)8333.02 =I

as a phasor: 8333.02=

I / -

36.87 = 0.6667 j0.5 (pu)

( ) =+=++= 008334.00125.1 221 jjXRIVV eqeq 1.0125/0.472 (pu)V1= 1.0125 (pu)

0062.022 == eqR RIP eq (pu)

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034.02

==c

cc

R

VP (pu)

986.02

2 =++

=cR PPP

P

eq

converting V1 to SI units gives

405040000125.11,11 === bpu VVV (V)

3.6 Three-Phase Transformers

If we consider three-phase transformers as consisting of three identical one-phase transformers,

then we have an accurate representation as far as equivalent circuits and two-port models are

concerned, but it does not give insight into the magnetic circuit of the three-phase transformers.

The primaries and the secondaries of the one-phase transformers can be connected either in or

in Y configurations. In either case, the rated power of the three-phase transformer is three times thatof the one-phase transformers.

For the connection:

1VV ll = (3.28)

13IIl = (3.29)

For the Y connection:

13VV ll = (3.30)

1IIl = (3.31)

Connections to the three-phase transformer that are Y connected in the primary are shown in Figure

19.

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3.7 Autotransformers

An autotransformer is a transformer where the two windings (of turnsN1andN2) are not isolated

from each other, but are connected as shown in Figure 21.

I1

V1 I2

V2

N2

N1

I1

I2

Figure 21. An autotransformer.

From this figure that the voltage ratio in an autotransformer is:

2

21

2

1

N

NN

V

V += (3.32)

and the current ratio is:

2

21

1

2

N

NN

I

I += (3.33)

An interesting note on autotransformers is that the coil of turnsN1carries currentI1, while the coil of

turnsN2carries the (vectorial) sum of the two currents, . So if the voltage ratio were 1, no

current would flow through theN21

II

2coil. This characteristic leads to a significant reduction in the size

of the autotransformer compared to a similarly rated transformer, especially if the primary and

secondary voltages are of the same order of magnitude. These savings come at a seriousdisadvantage of the loss of isolation between the two sides.

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Chapter Notes:

To understand the operation of transformers we have to use both the Gausss law for magnetism (or

Biot-Savart law) and Faradays law.

Most transformers operate under or near rated voltage. The voltage drop in the winding resistanceand leakage reactance are usually small.

In both transformers and in rotating machines, the net mmf of all the currents must accordinglyadjust itself to create the resultant flux required by this voltage balance.

Leakage fluxes induce voltage in the windings that are accounted for in the equivalent circuit asleakage reactance (elementsLl1andLl2). The leakage-flux paths are dominated by paths through

air, and are thus almost linearly proportional to the currents producing them. The leakage

reactances therefore are often assumed to be constant (independent of the degree of saturation of

the core material).

The open- and short-circuit tests provide the parameters for the equivalent circuit of the

transformer.

Three-phase transformers can be considered to be made of three single-phase transformers for the

purpose of this course. The main issue is then to calculate the ratings, voltages, and currents ofeach.

Autotransformers are used mostly to vary the voltage a little. It is seldom that an autotransformer

will have a voltage ratio greater than two.

ece320 chapter 3.pdf

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