BASIC ELECTRICAL TECHNOLOGY DET 211/3 Chapter 6: Single Phase Transformer.

BASIC ELECTRICAL TECHNOLOGY

DET 211/3

Chapter 6: Single Phase Transformer

A transformer is a device that changes ac electric energy at one voltage level to ac electric energy at another voltage level through the action of a magnetic field.

Introduction to Transformer

The most important tasks performed by transformers are:

• changing voltage and current levels in electric power systems.• matching source and load impedances for maximum power

transfer in electronic and control circuitry.• electrical isolation (isolating one circuit from another or

isolating dc while maintaining ac continuity between two circuits).

It consists of two or more coils of wire wrapped around a common ferromagnetic core. One of the transformer windings is connected to a source of ac electric power – is called primary winding and the second transformer winding supplies electric power to loads – is called secondary winding.

An ideal transformer is a lossless device with an input winding and output winding.

Ideal Transformer

1S

p

S

Slossless

aN

N

tv

tv

s

p

s

p )(

)(

a = turns ratio of the transformer

)t(iN)t(iN sspp

ati

ti

s

p 1

)(

)(

Power in ideal transformer

cosppinout IVPP

sinppinout IVQQ

sinppinout IVSS

Where is the angle between voltage and current

Impedance transformation through the transformer

The impedance of a device – the ratio of the phasor voltage across it in the phasor current flowing through it:

L

LL I

VZ

LL ZaZ 2'

The equivalent circuit of a transformer

• Copper (I2R) losses: Copper losses are the resistive heating in the primary and secondary windings of the transformer. They are proportional to the square of the current in the windings.

• Eddy current losses: Eddy current losses are resistive heating losses in the core of the transformer. They are proportional to the square of the voltage applied to the transformer.

• Hysteresis losses: Hysteresis losses are associated with the arrangement of the magnetic domain in the core during each half cycle. They are complex, nonlinear function of the voltage applied to the transformer.

• Leakage flux: The fluxes ΦLP and ΦLS which escape the core and pass through only one of the transformer windings are leakage fluxes. These escaped fluxes produce a self inductance in the primary and secondary coils, and the effects of this inductance must be accounted for.

The major items to be considered in the construction of such a model are:

Nonideal or actual transformer

Mutual flux

Nonideal or actual transformer

Nonideal or actual transformer

Ep = primary induced voltage Es = secondary induced voltageVp = primary terminal voltage Vs = secondary terminal voltageIp = primary current Is = secondary currentIe = excitation current IM = magnetizing currentXM = magnetizing reactance IC = core currentRC = core resistance Rp = resistance of primary windingRs = resistance of the secondary winding Xp = primary leakage reactanceXs = secondary leakage reactance

Transformer equivalent circuit, with secondary impedances referred to the primary side

Nonideal or actual transformer

Transformer equivalent circuit

Dot convention

1. If the primary voltage is positive at the dotted end of the winding with respect to the undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are the same with respect to the dots on each side of the core.

2. If the primary current of the transformer flows into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary winding.

Exact equivalent circuit the actual transformer

a) The transformer model referred to primary side

b) The transformer model referred to secondary side

Approximate equivalent circuit the actual transformer

a) The transformer model referred to primary side

b) The transformer model referred to secondary side

Exact equivalent circuit of a transformer

Ep = primary induced voltage Es = secondary induced voltageVp = primary terminal voltage Vs = secondary terminal voltageIp = primary current Is = secondary currentIe = excitation current IM = magnetizing currentXM = magnetizing reactance IC = core currentRC = core resistance Rp = resistance of primary windingRs = resistance of the secondary winding Xp = primary leakage reactanceXs = secondary leakage reactance

a/III sep

MCe III

ppppp E)jXR(IV

CCp RIE

)jX(IE MMp

)jX//R(IE MCep

Primary side Secondary side

ssssS V)jXR(IE

Lss ZIV

s

p

p

s

s

p

s

p

N

N

I

I

E

E

V

Va

R pI p X p

V p E p aV s

a 2 R sI s /aa 2 X s

I e

Exact equivalent circuit of a transformer referred to primary side

V p /a

aI pR p /a 2 X p /a 2 I s

E p /a = E sV s

R s X s

aI c

R c/a 2X M /a 2

aI m

aI e

Exact equivalent circuit of a transformer referred to secondary side

Approximate equivalent circuit of a transformer referred to primary side

Approximate equivalent circuit of a transformer referred to secondary side

Vp/a

aIp

+

-

ReqsIsjXeqs

Rc/a2 jXM/a2 Vs

+

-

Reqs=Rp/a2+Rs

Xeqs=Xp/a2+Xs

Vp

Ip

+

-

ReqpIs/ajXeqp

RcjXM

aVs

+

-

Reqp=Rp+a2Rs

Xeqp=Xp+a2Xs

Example

A single phase power system consists of a 480V 60Hz generator supplying a load Zload=4+j3 through a transmission line ZLine=0.18+j0.24 Answer the following question about the system.

a) If the power system is exactly as described below (figure 1(a)), what will be the voltage at the load be? What will the transmission line losses be?

b) Suppose a 1:10 step-up transformer is placed at the generator end of the transmission line and a 10:1 step down transformer is placed at the load end of the line (figure 1(b)). What will the load voltage be now? What will the transmission line losses be now?

Example

Figure 1 (a)

Figure 1 (b)

V=48000V

IG+

-VLoad

T1 T2

1:10 10:1ZLine=0.18+j0.24

ILoadILine

ZLoad=4+j3V=48000V

IG+

-

VLoad

ZLoad=0.18+j0.24

ILine

ILoad

Solution Example

(a) From figure 1 (a) shows the power system without transformers. Hence IG = ILINE = ILoad. The line current in this system is given by:

8.378.90

8.3729.5

0480

24.318.4

0480

)34()24.018.0(

0480

j

jj

V

ZZ

VI

loadlineline

Solution Example

Therefore the load voltage is:

9.0454

)9.365)(8.378.90(

)34)(8.378.90(

A

jA

ZIV loadlineload

and the line losses are

W

A

RIP linelineloss

1484

)18.0()8.90(

)(2

2

Solution Example

(b) From figure 1 (b) shows the power system with the transformers. To analyze the system, it is necessary to convert it to a common voltage level. This is done in two steps:

i) Eliminate transformer T2 by referring the load over to the transmission’s line voltage level.

ii) Eliminate transformer T1 by referring the transmission line’s elements and the equivalent load at the transmission line’s voltage over to the source side.

The value of the load’s impedance when reflected to the transmission system’s voltage is

300400

)34()1

10(

'

2

2

j

j

ZaZ loadload

Solution Example

The total impedance at the transmission line level is now:

88.363.500

24.30018.400

'

j

ZZZ loadlineeq

The total impedance at the transmission line level (Zline+Z’load) is now reflected across T1 to the source’s

voltage level:

88.36003.5

)340024.00018.0(

)30040024.018.0()10

1(

)'(

'

2

2

2

jj

jj

ZZa

ZaZ

loadline

eqeq

Solution Example

Notice that Z’’load = 4+j3 and Z’line =0.0018+j0.0024 . The resulting equivalent circuit is shown below. The

generator’s current is:

AV

IG

88.3694.95

88.36003.5

0480

a) System with the load referred to the transmission

system voltage level

b) System with the load and transmission line referred to the generator’s voltage level

Solution Example

Knowing the current IG, we can now work back and find Iline

and ILoad. Working back through T1, we get:

A

A

IN

NI

ININ

GS

pline

linesGp

88.36594.9

)88.3694.95(10

11

1

11

Solution Example

Working back through T2 gives:

A

A

IN

NI

ININ

lines

pload

loadslinep

88.3694.95

)88.36594.9)(1

10(

2

2

22

It is now possible to answer the questions. The load voltage is given by

V

A

ZIV loadloadload

01.07.479

)87.365)(88.36594.9(

Solution Example

and the line losses are given by:

W

A

RIP linelineloss

7.16

)18.0()594.9(

)(2

2

Notice that raising the transmission voltage of the power system reduced transmission losses by a factor of nearly 90. Also, the voltage at the load dropped much less in the system with transformers compared to the system without

transformers.