17

cos d dE V X I

3 3 cosaP V I

cos cos sina q dI ab de I I

sin q qV X I

sinq

q

VI

X

cosd

d

E VI

X

23 3 cos sin 3 sin 3 sin 2

2d q

q dd d q

X XE VP V I I V

X X X

3 3 sind

E VP

X Approximately, the second item can be ignored:

Consider a simple case ignoring Ra and Xl

d d q qE V jX I jX I

q d d

d q q

E V jX IV jX I

q axisd axis

Vq

Vd

Iq

Id

18

Power Transformer

F=IN=/P

Ideal Transformer Real TransformerWinding resistance 0 0

Leakage flux 0 0 (windings do not link the same flux)

Permeabilityof the core

Infinite (flux is produced even with magnetizing current IP=0)

Finite(magnetizing current IP0 is needed to

produce flux)

Core losses0

0 (including hysteresis losses and eddy

current losses due to time varying flux)

19

Real Transformers• Modeling the current under no-load

conditions (due to finite core permeability and core losses)When I2=I’2=0 I1=I0 0 I0=Im +Ic

Im is the magnetizing current to set up the core flux – In phase with flux and lagging E1 by 90o, modeled by E1/(jXm1)

Ic supplies the eddy-current and hysteresis losses in the core– A power component, so it is in phase with E1, modeled by E1/(Rc1)

• Modeling flux leakages– Primary and secondary flux leakage reactances: X1 and X2

• Modeling winding resistances– Primary and secondary winding resistances: R1 and R2

Ideal Transformer

1 2 1

2 2 2

E I N aE I N

20

22 2 2 2

2 2

1 12 2

2 2

Z R jX a Z

N NR j XN N

Referred to the primary

side.

Exact Equivalent Circuit Ideal Transformer

1 2 1

2 2 2

E I N aE I N

22 2 2

1

/ NI I a IN

a2 Z

a V

1/a I

1/a2Z

1/aV

aI

12 2 2

2

NV aV VN

21

Approximate Equivalent Circuits• Since V1E1, Z1 can be combined with Z’2 to become an equivalent Ze1

Referred to the primary side:

1 2 1 1 2e eV V R jX I

22 1

1 1 2 1 22

eNR R a R R RN

2

2 11 1 2 1 2

2e

NX X a X X XN

Referred to the secondary side:

1 2 2 2 2e eV V R jX I 2

2 22 1 2 1 2

1

/eNR R a R R RN

2

2 22 1 2 1 2

1

/eNX X a X X XN

V2

• If power transformers are designed with very high permeability core and very small core loss, the shunt branch can be ignored. Then, I1=I’2 and I’1=I2.

22

Determination of Equivalent Circuit Parameters

• Open-circuit (no-load) test– Neglect (R1+jX1)I0 (since |R1+jX1|<<|Rc1//jXm1|)– Measure input voltage V1, current I0, power P0 (core/iron loss)

21

10

| |c

VRP

1

1c

c

VIR

2 20| | | |m cI I I

11

| || |m

m

VXI

P0

23

• Short-circuit test– Apply a low voltage VSC to create rated current ISC

– Neglect the shunt branch due to the low core flux

1| || || |

sce

sc

VZI

1 2| |sc

esc

PRI

2 21 1 1| |e e eX Z R

ISC

VSC

PSC

24

Transformer Performance

• Efficiency: 95% - 99%• Given |V2,rated|, |S|=3|V2,rated||I2,rated| (full-load rated VA) and PF• Actual load is |I2|=n|I2,rated| where n<1 is the fraction of the full load power

Pout=n|S|PF

Pc, rated=3Re2|I2,rated|2 : full-load copper loss (current dependent)

Pc=3Re2|I2|2 =3Re2|I2,rated|2 n2

Pi, rated: core/iron loss at rated voltage (mainly voltage dependent, almost constant)

• Maximum efficiency (constant PF) occurs when

output powerinput power

2

0 0d dd I dn

,

,

when i rated

c rated

Pn

P

Learn Example 3.4

2, , , ,

| | | || | | | /c rated i rated c rated i rated

n S PF S PFn S PF n P P S PF n P P n

25

Voltage Regulation• % change in terminal voltage from no-load to full load (rated)

– Generator

– Transformer:Utilizing the equivalent circuit referred to one side:

VR= 100nl rated

rated

V VV

VR= 100rated

rated

E VV

2, 2,

2,

VR= 100nl rated

rated

V VV

1 2

2

VR= 100V V

V

1 2

2

VR= 100V V

V

1 2 1 1 2e eV V R jX I

1 2 2 2 2e eV V R jX I

+

V

-

I

V=Vnl when I=0

V=Vrated when I=Irated

Source (generator,

transformer, etc.)

( )a s aE V R jX I

Secondary

Primary

26

Three‐Phase Transformer Connections• A bank of three single-phase transformers connected in Y or arrangements• Four possible combinations: Y-Y, △-△, Y-△ and △-Y

– Y-connection: lower insulation costs, with neutral for grounding, 3rd harmonics problem (3rd harmonic voltages/currents are all in phase, i.e. van3=vbn3=vcn3=Vmcos3t)

– -connection: more insulation costs, no neutral, providing a path for 3rd harmonics (all triple harmonics are trapped in the loop), able to operate with only two phases (V-connection)

– Y-Y and - : HV/LV ratio is same for line and phase voltages; Y-Y is rarely used due to the 3rd harmonics problem.

– Y- : commonly used as voltage step-down transformers– -Y : commonly used as voltage step-up transformers

27

3rd harmonics problem with three‐phase transformers• △ configuration provides a closed path for 3rd harmonics, or in

other words, all triple harmonics are trapped in the △ loop.– The 120o phase difference between the fundamental harmonic of I1 an I2 and

I3 becomes 360o (in phase) for the 3rd harmonic– Then by KCL, Ia(3)=I1(3)-I3(3)=0; also Ib(3)=Ic(3)=0

Distorted waveform

Fundamental

3rd harmonic3

9

, 3, 5, 7 9

, 3, 5, 7, 9

, 3, 5, 7 9

, 5, 7

, 5, 7

, 5, 7

28

Y‐△ and △ ‐Y Connections• Y-△ and Y-△ connections result in a 30o phase shift between the primary and

secondary line-to-line voltages• According to the American Standards Association (ASA), the windings are

arranged such that the HV side line voltage leads the LV side line voltage by 30o

– E.g. Y- (HV-LV) Connection with the ratio of turns a= NH/NX>1

,

,

H P An H

X P ab X

V V N aV V N

,

,

3 30H L AB

H P An

V VV V

, ,X L X PV V

,

,

3 30H L AB

X L ab

V V aV V

HV Side (indicated by “H”) in Y connection:

LV Side (indicated by “X”) in △ connection:

=VH,P

=VX,P =VX,L

=VH,L

(Complex ratio)

29

Per‐Phase Model for Y‐△ or △ ‐Y Connection

• Neglect the shunt branch• Replace the △ connection by an equivalent Y connection• Work with only one phase (equivalent impedances are line-to-

neutral values ZeY=Ze/3)• All voltages are line-to-neutral voltages

E.g. for Y-△ Connection, V1 is the phase voltage of the Y side and V2 is the phase (line-to-neutral) voltage of the △ side

V1

30

Autotransformers• A conventional two-winding transformer can be changed into an autotransformer by

connecting its two coils in series. • The connection may use a sliding contact to providing variable output voltage.• An autotransformer has kVA rating increased but loses insulation between primary

and secondary windings

20MVA (115/69kV) McGraw-Edison Substation Auto-Transformer (Y-Y) (Source:

http://www.tucsontransformer.com)(Source: EPRI Power System Dynamic Tutorial)

31

Autotransformer Model

1 2 1

2 1 2

V I N aV I N

1 1

2 1 2 22 2

(1 )H L L LN NV V V V V V V a VN N

IL

11 2 1 1

2

(1 )L HNI I I I I a IN

21 2 1 1 1

1

2 2

(1 )

1 (1 )

auto

w w conducted

NS V V I V IN

S S Sa

Power rating advantage

Equivalent Circuit (if the equivalent impedance is referred to the HV side)

Transformed power (thru

EM induction)

Conducted power (V2I1)

• Apparent power:

32

+

600V

-

-

120V

+

+

480V

-

H1

H2

X2

X1

I2=125A

I2-I1=100A

I1=25A

+

600V

-

+

120V

-

+

720V

-

H1

H2

X1

X2

I2=125A

I2+I1=150A

I1=25A

+

120V

-

-

600V

+

-

480V

+

X1

X2

I1=25A

I2-I1=100A

I2=125A

H2

H1

|Si|=600150=90kVA |Si|=120100=12kVA|So|=720125=90kVA |So|=48025 =12kVA

The maximum apparent power |Smax|=max(|E1|,|E2|)(|I1|+|I2|)= (|E1|+|E2|) max(|I1|, |I2|)

600V 120V

H1

H2 X2

X1

|Si|=600100=60kVA |So|=480125=60kVA

Conventional transformer connected as an autotransformer (Example 11‐2 from ECE325 – Wildi’s book)