transformer

Transformer

Professor Mohamed A. El-Sharkawi

El-Sharkawi@University of Washington

2

Why do we need transformers?

• Increase voltage of generator’s output– Transmit high power at low current– Reduce cost of transmission system

• Adjust voltage to a usable level

• Create electrical isolation

• Match load impedance

• Filters


3

220kV-750kV220kV-750kV

15 kV- 25kV15 kV- 25kV

208V- 416V208V- 416V

TransmissionTransmissionTransformerTransformer

DistributionDistributionTransformerTransformer

ServiceServiceTransformerTransformer


4

Transmission Transformer


5

Distribution Transformer


6

Distribution Transformer


7

Service Transformer


8

Service Transformer bank


9

Service Transformer bank


10

Service Transformer


11

Service Transformer


12

Service Transformer


13

Service Transformer


14

Service Transformer


15

Low power Transformer


16

Basic Components

Iron Core Insulated Copper Wire


17

Basic Components

Laminated iron core

Insulated copper wire


18


19


20


21

i1 i2

N1e1 N2

e2

dteN

1dt

dNe

11

11

dt

dNe

22

Primary Secondary


22

2

1

2

1

2

1

N

N

dtd

N

dtd

N

te

te

2

1

2

1

N

N

E

E

2

2

1

1

N

E

N

E

Basic Analysis:Voltagei1

i2

N1e1+_

N2e2

+_

• Volts/turn is constant• Voltages are in phase (no phase shift)• Voltage magnitudes vary with turns ratio.


23

Basic Analysis: Power and current

21 SS

*22

*11 IEIE

1

2

1

2*2

*1

N

N

E

E

I

I

1

2

2

1

N

N

I

I 2211 ININ

i1i2

N1e1+_

N2e2

+_

• Currents are in phase.

• Current ratio is opposite to the voltage ratio


24

Basic Analysis: Reflected impedance

I1I2

N1E1 N2

E2

Primary Secondary

Source

Load

Flux

Zload

2

2

I

EZload


25

Basic Analysis: Reflected impedance

'loadZ

I1

E1

Primary

Source

1

1

I

EZ '

load

2

2

1'

2

2

1

1

2

2

1'

N

NZZ

N

N

I

I

E

E

Z

Z

loadload

load

load


26

Single-Phase, Ideal Transformer Ratings

2 KVA, 120/240 V

Apparent Power

Primary Voltage Secondary Voltage

N2N1

+

-

V1

I1 I2

+

-V2


27

Rated Values

• Rated voltage: The device can continuously operate at the rated voltage without being damaged due to insulation failure

• Rated current: The device can continuously operate at the rated current without being damaged due to thermal destruction


28

ExampleTransformer rating:

2 KVA, 240/120 V

Compute the currents

KVAIVIVS 2 2211

A 33.8V240

KVA 2

11

V

SI

A 67.16V 120

KVA 2

22

V

SI

N2N1

+

-

V1

I1 I2

+

-V2


29

Multi-Secondary Transformer


30

Multi-secondary windings


31

I1

I2N1E1

N2E2

Primary

I3

N3E3

3

1

3

1

2

1

2

1

N

N

E

E

N

N

E

E


32

Current ratio: superposition

1

2212 N

NII I12

I2N1E1

N2E2

Primary


33

Current ratio: superposition

1

3313 N

NII I13

N1E1

Primary

I3

N3E3


34

Superposition

I1

I2N1E1

N2E2

Primary

I3

N3E3

332211

1

33

1

2213121

NININI

N

NI

N

NIIII


35

Superposition

I1

I2N1E1

N2E2

Primary

I3

N3E3

331 SSS *33

*22

*11 IEIEIE


36

Example

• The transformer consists of one primary winding and two secondary windings. The number of turns is each winding is

A voltage source of 120V is applied to the primary winding, and purely resistive loads are connected across the secondary windings. A wattmeter placed in the primary circuit measures 300W. Another wattmeter placed in the secondary winding N2 measures 90W. Compute the following:

• The voltages of the secondary windings• The currents in N3• The power consumed by the load connected across N3

500;1000;4000 321 NNN


37

Solution

304000

1000120

1

212

N

NEE

154000

500120

1

313

N

NEE

5.2120

300

cos 11

11

E

PI

330

90

cos 22

22

E

PI


38

Solution

3

332211

5003*10004000*5.2 I

NININI

143 I

3

332211

153*305.2*120 I

IEIEIE

143 I


39

Solution

3

321

90300 P

PPP

2103P


40

Autotransformer


41

I1 I2

N1V1N2

E2

A1

A2

B1

B2

E1 V2


42

Autotransformer: Voltage and current

21 III load

I2

N2

V1

N1I1

Is

Iload

V2

A1

A2

B1

B2

E1

E2

211 EEV


43

Autotransformer

2211 IEIESA

loadsB IVIVS 21

N2N1

+

-

E1

I1 I2

+

-

E2

I2

N2

V1

N1I1

Is

Iload

V2

A1

A2

B1

B2

E1

E2


44

Autotransformer: Power

12

12111211 )(

IESS

IEIEIEEIVS

AB

sB

AB SS


45

ExampleRatings of regular transformer: 10 kVA, 400/200 V

New voltage ratio: 600/200 V

Compute the new ratings

Solution

AI

AI

254.0

10

502.0

10

1

2

kVAIVSB 152560011

I2

N2

V1

N1I1

Is

Iload

V2

A1

A2

B1

B2

E1

E2


46

VARIC: Variable Auto-Transformer

I2N2

V1

N1

I1

Is

Iload

V2

N3

Z

Y

Sliding terminal


47

Output Voltage

21

2

NN

NVV sload

ssload VNN

NNVV

21

21

21

321

NN

NNNVV sload

I2N2

V1

N1

I1

Is

Iload

V2

N3

Z

Y

Sliding terminal

At Y

At Z


48

Three-Phase Transformer


49

3-phase transformer


50

3-phase transformer Y-Y connection. Also known as star-star connection

a

c b

N1

A

BC

N2

2

1

N

N

V

V

AN

an 2

1

3

3

N

N

V

V

V

V

AN

an

AC

ac

n N

Rat

io o

f P

has

e V

olta

ge

Rat

io o

f L

ine

Vol

tage


51

3-phase transformer ( -)

A

B

C

N2

a

bc

N1

2

1

N

N

V

V

AC

ac

Rat

io o

f P

has

e V

olta

ge a

nd

lin

e vo

ltag

e


52

3-phase transformer (Y-)

Also known as star-delta connectionA

B

C

N2

N1

2

1

N

N

V

V

AC

an 2

133

N

N

V

V

V

V

AC

an

AC

ac

a

c b

n

Rat

io o

f P

has

e V

olta

ge

Rat

io o

f L

ine

Vol

tage


53

3-phase transformer bank (Y-)

N2N1Van

N2N1

N2N1

a

c

A

B

C

VABVab

1

2

an

AB

ab

AB

N3

N

V3

V

V

V

1

2

an

AB

N

N

V

V

b


54


55

Ratings of Ideal 3-phase Transformer

100 MVA, 13.8/138 KV

Apparent Power (3-phase)

Primary Voltageline-to-line

Secondary Voltageline-to-line


56

Example

• Three single-phase transformers are used to form a three-phase transformer bank. Each single-phase transformer is rated at 10 kVA, 13.8 KV/240 V. One side of the transformer bank is connected to a three-phase, 13.8 kV transmission line. The other side of the transformer is connected to a three-phase residential load of 415.7V, 9kVA at 0.8 power factor lagging. Determine the connection of the transformer bank, the voltage ratio of the transformer bank, and the line current of the bank at the 13.8 kV side


57

Solution• Secondary voltage (Low voltage side) should be in

Y to provide the needed residential voltage

• The high voltage side must be Delta-connection

– The line-to-line voltage of the supply is 13.8 kV. Same as the transformer rating of the primary.

– If the primary is connected in Y, the voltage of the load would be lower than 240 V.

V2403

7.415


58

Solution

N2N1Van

N2N1

a

c

A

B

C

VAB= 13.8 kV Vab

b

Van= 240 V


59

Solution2403

800,13

3

an

AB

ab

AB

V

V

V

V

N2N1Van

N2N1

a

c

A

B

C

VAB= 13.8 kV Vab

b

Van= 240 V

AI 5.12240

39000

2

Phase current of the load

Phase current of the Transformer primary

AN

NII 2174.0

800,13

2405.12

1

221

Line Current in primary

AII A 377.03 1


60

Actual Transformer

• Windings:– Resistance– Inductance

• Core:– Eddy Current– Hysteresis

i1i2

N1e1+_

N2e2

+_


61

R1X1

N1 N2

R2 X2

Ideal Transformer

Windings Impedance


62

Core Hysteresis

B

H

i

Ne+

_

dtefB

ifH


63

i

e


64

Core Model

e

i

RLet

tEe sinmax

tE

dte

cosmax

tR

E

R

ei sinmax i

e


65

Core Model

LettEe sinmax

tE

dte

cosmax

tL

Edte

Li

dt

diLe

cos1 max

i

e

i

e Xl


66

e

i

R Xl

i

e

i

e


67

Ro Xo

R1X1 R2 X2

V1

I1

Io

'2I

I2V2E1 E2

N1 N2

Equivalent Circuit

2

1

2

1

2

1

V

V

N

N

E

E

2

1

1

2

2

'2

I

I

N

N

I

I

load


68

Referred impedance

Ro Xo

R1X1 R2 X2

V1

I1

Io

'2I

I2E1 E2

N1 N2

V2

22222

12

2

11 VjXRI

N

NE

N

NE


69

Referred impedance

Ro Xo

R1X1 R2 X2

V1

I1

Io

'2I

I2E1 E2

N1 N2

V2

222

2

1'2

2

12

2

11 VjXR

N

NI

N

NE

N

NE


70

222

2

1'2

2

12

2

11 VjXR

N

NI

N

NE

N

NE

22

1'2

2

2

2

1'2

2

2

2

1'2

VN

NV

XN

NX

RN

NR

Define:

'2

'2

'2

'21 VjXRIE

Then:

22

122

2

2

1'21 V

N

NjXR

N

NIE


71

'2

'2

'2

'21 VjXRIE

Ro Xo

R1X1 R2 X2

V1

I1

Io

'2I

I2E1 E2

N1 N2

V2

Ro Xo

R1X1

V1

I1

Io '2I

E1

'2R

'2X

'2V

Equivalent Circuit Referred to Source Side


72

o'21 III

1o'2 III

Ro Xo

V1

I1

Io '2I

E1'2V

R1X1

'2R

'2X

'2o1 RRR '2o1 XXX

Practical Considerations


73

V1

I1

Ro Xo

Io '2I '

2V

R1X1 '

2R'2X

V1

I1

Ro Xo

Io

'2I

'2V

R1

X1 '2R

'2X


74

V1

'21 II '

2V

R1

X1 '2R

'2X

Define:'21eq

'21eq

XXX

RRR


75

Analysis of Transformer

'21 II

V1'2V

eqR eqX

eqeq'2

'21 jXRIVV

Z


76

22

1'2 V

N

NV

TerminologiesTerminologies

2

2

2

1'2

2

2

2

1'2

XN

NX

RN

NR

2I

21

2'2 I

N

NI

2V Load VoltageLoad Voltage

Load Voltage referred to Source sideLoad Voltage referred to Source side

Impedance referred to Source sideImpedance referred to Source side

Load CurrentLoad Current

Load current referred to Source sideLoad current referred to Source side


77

Analysis of Transformer

'21 II

V1'2V

eqR eqX

eqeq'2

'21 jXRIVV

Z’


78

eqX

'21 II

V1'2V

eqR

Z’

'2I

eq'2 RI

eq'2 XI


79

'21 II

V1'2V

eqR eqX

Z

eq'2eq

'2

'21 XIjRIVV

eq'2 RI

eq'2 XI

'2I

eq'2 ZI

'2V

1V


80

Ratings of Actual 3-phase Transformer

100 MVA, 13.8/138 KV

Apparent Power (3-phase)

line-to-line

'V2

line-to-line2V


81

Example

102

1 N

N

5000;1000;10;1 0'21

'21 XRXXXRRR oeqeq

ol .Z 3050

A transformer has the following parameters:

The rated voltage of the primary winding is 1000V. Compute the load voltage.


82

Solution

502

2

1

N

NZZ L

'L

A..jZXjR

VI o

o'Leqeq

' 31387173050101

01000 01

2

'21 II

V1'2V

eqR eqX

Z’

V...ZIV oo'L

'' 022 31888530503138717

V.N

NVV ' 588

10

1885

1

222


83

Voltage Regulation VR

loadfull

loadfullloadno

V

VVVR

'2

'21

V

VVVR

Measured at the load side

'21 II '

2V

eqR eqX

V1 Loa

d


84

ExampleCalculate the voltage regulation of the transformer in the previous problem.

Solution:

%14.01007200

72005.7209VR

V

VVVR

'2

'21


85

Efficiency

V1

I1

Ro Xo

Io'2I '

2V

ReqXeq

in

out

P

P

PowerInput

PowerOutput

LossesPP outin

ironculosses PPP

eq2'

2cu RIP

o

21

iron R

VP

cosIVP '2

'2out


86

Example

A 10 kVA, 2300/230 V, single phase distribution transformerhas the following parameters:

k4.69X;k6.75R

05.6R;12XX;8.5R

oo

'2

'211

At full load and 0.8 power factor lagging, compute the efficiencyof the transformer.


87

Solution

W23.22405.68.5)35.4(RIP 2eq

2'2cu

W

R

V

R

VP

ooiron 70

600,75

2300 22'2

21

kW88.010cosScosIVP '2

'2out

A35.42300

000,10

V

SI

'2

'2

%45.961007023.2248000

8000

P

P

in

out

transformer

Documents

univers of washington

universit of washington

service transformerelsharkawi

voltage ratioelsharkawi

secondary windingselsharkawi

n n nelsharkawi

thermal destructionelsharkawi

e s si e i e i e e i