three phase transformers...the most elemental three-phase transformer consists on an arrangement...

THREE PHASE TRANSFORMERS

The most elemental three-phase

transformer consists on an

arrangement made of 3 single-

phase transformers

Three-phase bank of single-phase

transformers

Drawbacks of 3-phase banks:

Three units are needed

Much space is required

Much iron weight is required 1

INTRODUCTION


0

0

321

321

EEE

2

F1+F2+F3 = 0 The 3 legs can be joined

through a central leg

INTRODUCTION


The central leg can be

removed because the

flux is null

By removing the

central leg, material,

weight and space is

saved

The 3 legs can be joined

through a central leg

3

INTRODUCTION


4

INTRODUCTION


5


6


7


8

Yy

Yy

Dd

Dy

Yd

Upper case: primary side

Lower case: secondary side

WINDING CONNECTIONS


9

WINDING CONNECTIONS

What about the phase shift between the primary and secondary

voltages?

Dy


10

WINDING CONNECTIONS

Can we do that?


Yy0

11

WINDING CONNECTIONS


Yy6

12

WINDING CONNECTIONS


13

WINDING CONNECTIONS

Dy11


Yz1

14

WINDING CONNECTIONS


15

Other three-phase transformer connections see IEC 60076-1

Yy:

Dd:

Yd:

Dy:

Yz:

Dz:

l

l

n

l

n

l

nt

U

U

U

Ur

20

1

2

1

RATED VOLTAGE (IEC 60076-1)

The voltage assigned to be applied, or developed at no-load, between the

terminals of an untapped winding, or of a tapped winding connected on the

principal tapping .

For a three-phase winding it is the voltage between line terminals.

NOTE 1 The rated voltages of all windings appear simultaneously at no-load

when the voltage applied to one of them has its rated value.

2

1

2

1

2

1

3

3

N

N

U

U

U

UrYy

f

n

f

n

l

n

l

nt

2

1

2

1

2

1

N

N

U

U

U

UrDd

f

n

f

n

l

n

l

nt

2

1

2

1

2

1 33

N

N

U

U

U

UrYd

f

n

f

n

l

n

l

nt

2

1

2

1

N

N

U

Ur

f

n

f

nte

2

1

2

1

2

1

33 N

N

U

U

U

UrDy

f

n

f

n

l

n

l

nt


16

CONDITIONS FOR CONNECTING 2 TRANSFORMERS A AND B IN PARALLEL

1. Same turns-ratio: rtA = rtB (V1n,A = V1n,B and V2n,A = V2n,B)

2. Similar esc: escA ≈ escB (cAescA= cBescB )

3. Same angular displacement/clock number

Z2scA

I2B

Zload

Z2scB

I2A

2scB2BnB2scA2AnA2scB2B2scA2A ZIcZIc ZIZI

scBBscAA

2n

2scB2BnB

2n

2scA2AnA εcεc

V

ZIc

V

ZIc


17

Transformer tap: connection point along a transformer winding that allows a certain

number of turns to be selected.

It produces variable turns ratio and enables voltage regulation of the output.

The tap selection is made via a tap changer mechanism

In step-down distribution transformers, taps are usually found in HV windings

Taps

Taps

Taps

Windings

connections

TAPS


18

11000+5.0 %=11550 V

11000+2.5 %=11275 V

11000 V

11000-2.5 %=10275 V

11000-5.0 %=10450 V

385.15715

00011

20

1

2

1 l

l

n

l

n

l

nt

U

U

U

Ur

AII

AII

kVAS

n

l

n

n

l

n

n

57.2543

33.165

3150

22

11


19


20

TECHNICAL FEATURES OF 24 kV THREE-PHASE TRANSFORMERS

Power (kVA) 50 75 100 125 160 200 250 315 400 500 630 800 1000

Group

Yyn

0

Yyn

0

Yyn

0

Dyn

11

Dyn

11

Dyn

11

Dyn

11

Dyn

11

Dyn

11

Dyn

11

Dyn

11

Dyn

11

Dyn

11

P0 (kW) 0.24 0.33 0.40 0.48 0.58 0.69 0.82 0.98 1.17 1.38 1.64 1.96 2.15

Psc

(kW) 1.39 1.87 2.20 2.53 2.97 3.49 4.10 4.86 5.80 6.89 8.22 10.24 13.30

esc

(%) 4 4 4 4 4 4 4 4 4 4 4 6 6

I0 (%In) 4.7 4.1 3.3 3.0 2.7 2.4 2.2 2.1 2.0 2.0 1.9 1.8 1.6

Weight (kg) 385 481 570 655 731 834 976 1100 1422 1640 1930 2267 2645


A 900 kVA, 15000/3000-V, εsc = 8% three-phase transformer, has the primary and the

secondary windings connected as in the table. Determine the parameters of the table.

21 0.8 W 2.4 W 240/100 = 2.4 W 138.56/173.21 = 0.8 W Z2sc = U2,sc

f /I2f

138.56 V 240 V 0.08·3000 = 240 V 0.08·3000/3 = 138.56 V U2scf = esc·U2

f

173.21 A 100 A 173.21/3 = 100 A 173.21 A I2f

173.21 A 173.21 A 173.21 A 900000/(3·3000) = 173.21 A I2l

3000/3 V 3000 V 3000 V 3000/3 V U2f

3000 V 3000 V 3000 V 3000 V U2l

60 W 1200/20 = 60 W 20 W 692.8/34.64 = 20 W Z1sc = U1scf /I1

f

1200 V 0.08·15000 = 1200 V 692.8 V 0.08·15000/3 = 692.8 V U1scf = esc·U1

f

20 A 34.64/3 = 20 A 34.64 A 34.64 A Ilf

34.64 A 34.64 A 34.64 A 900000/(3·15000) = 34.64 A I1l

15000 V 15000 V 15000/3 V 15000/3 V U1f

15000 V 15000 V 15000 V 15000 V U1l

15000/(300/3) 5 (15000/3)/3000 (15000/3)/ 3000/3 = 5 Turns ratio: rte

5 5 5 15000/3000 = 5 Voltages ratio: rt

Dy Dd Yd Yy l: line f: phase


A 50 kVA Dy11 15000/380-V and three-phase transformer has relative voltage drops

εsc = 10% and εXsc = 8%. The transformer feeds through a line of 0.1+j0.2 W/wire a

balanced wye-connected load with 50º W per phase. Determine: a) R1sc, X1sc, Z1sc. b)

The voltage in the secondary, being the primary fed at 15000 V.

VUVU

VUVU

ff

ll

3

38015000

38015000

21

21

37.6847.392

1

2

1

2

1 N

N

U

Ur

U

Ur

f

n

f

ntel

n

l

nt

L3

N2

U1

L2

L1

N1

V1

W1

u2

N

v2

w2

N2

N2N1

N1

2.01.0 j W5

Solved from the point of view of the electrical machine Dy

W

W

W

810

108011.1

0001508.0

135011.1

000151.0

11.1000153

00050

33

2

1

2

1

1

11

1

11

1

11

1

11

1

111

scscsc

f

n

f

nXscscf

n

sc

f

nXsc

f

n

f

nscscf

n

sc

f

nsc

f

n

nf

n

f

n

f

nn

XZR

I

UX

U

ZI

I

UZ

U

ZI

AU

SIIUS

ee

ee º0150001

fU

scR1 scX 1 W 2.01.02

jrte

W52

ter

'21

ffII

'2

fU '

f

loadU






VUVr

UUZIU

VUVr

UUZZIU

ZZZII

l

load

te

f

loadf

loadload

ff

load

l

te

ff

loadline

ff

loadlinesc

ff

9.3582.20732.20737.68

8.16314'º67.48.16314'''

3.3665.21135.21137.68

2.45814'º42.22.45814''''

º67.4606.0''

º000015'

2

22

222

1

21

Solved from the point of view of the electrical machine Dy

º0150001 f

U

scR1 scX 1 W 2.01.02

jrte

W52

ter

'21

ffII

'2

fU '

f

loadU

W

W

W

6.23372537.68'

9.93445.4672.01.037.68'

1080810

2

2

1

load

line

sc

Z

jjZ

jZ


24

L3

N2

U1

L2

L1

N1

V1

W1

u2

N

v2

w2

N2

N2N1

N1

2.01.0 j W5





Solved from the point of view of the electrical system

VUVU

VUVU

ff

ll

3

38015000

38015000

21

21

47.3947.392

1

2

1,

2

1 N

N

U

Ur

U

Ur

f

n

f

nYtel

n

l

nt

3

810

3270

3

1080

3360

11.13

30001508.0

3

1350

345.450

11.13

3000151.0

925.111.133000153

00050

33

,2

1

2

1,

,1

,1

,1

,1

,1

,1,1

,

,1

,1

,1

,1

,1

,1,1

,

,1

,11,1

W

W

W

sc

scscYsc

sc

f

Yn

f

YnXsc

Yscf

Yn

Ysc

f

Yn

YXsc

sc

f

Yn

f

Ynsc

Yscf

Yn

Ysc

f

Yn

Ysc

f

Yn

nf

Yn

f

nY

f

Ynn

RXZR

X

I

UX

U

ZI

Z

I

UZ

U

ZI

AU

SIIUS

ee

ee


25





Solved from the point of view of the electrical system

VUVr

UUZIU

VUVr

UUZZIU

ZZZII

l

load

t

f

loadf

loadload

ff

load

l

t

ff

loadline

ff

loadlinesc

ff

9.3582.20732.20747.39

3.8180'º67.43.8180'''

3.3665.21135.21147.39

4.8350'º42.24.8350''''

º67.405.1''

º0300015'

2

22

222

1

21

W

W

W

3

6.23372547.39'

3

9.93445.4672.01.047.39'

3602703

1080810

2

2

1

load

line

sc

Z

jjZ

jj

Z

º03

150001

fU

scR1 scX 1 W 2.01.02

jrt

W 52

tr

'21

ffII

'2

fU '

f

loadU


A Dy, 320 kVA, esc=4%, Psc=3584 W, 10800/400-V three-phase transformer feeds a

balanced load which absorbs 300 A with PF = 0.8(i). In this situation U1 =11100 V.

Calculate the voltage in the load

From the statement it results: I1n = 17.11 A I2n = 461.88 A rt = 27

Y

2

1nY1scsc I·R·3P Ω4.0817.11·3

3584

I·3

PR

22

1nY

sc

Y1sc

Ω14.5817.11

3432/

I

UZ

1nY

1scY

Y1sc

Ω13.9974.0814.58RZX 222

1scY

2

1scYY1sc

73.74ºR

Xarctg

1scY

1scYsc

V43210800·0.04.UεU 1nsc1sc

'Z'IUU 1scYY2Y2Y1 73.74º14.5836.87º27

0030º'Uº

3

111002U1

V6278.28'U 2Y V10874.306278.28·3'Ul

2 V7540227

3010874.

.2 lU

26

PF = 0.8(i)


V1

R1

S1

T1

A=C'

C=B'

B=A'

W1

W2

V2

a

b c

R2

S2

T2

a’= b’= c’ Balanced

load

From the statement it results:

I1n = 3.007 A I2n = 189.92 A rt = 63.158

Y

2

2nY2scsc I·R·3P

Ω0.02338189.9178·3

2530

I·3

PR

22

2n

sc

Y2sc

Y

Ω0.0462189.9178

3

380·

100

4

I

UZ

Y2n

Y2sc

Y2sc

Ω0.039850.023380.0462RZX 22

Y

2

2scY

2

2scY2sc Ω0.03985j0.02338ZY2sc

2

t

Y1

2Yr

RR0.02338 Ω 53.37R

Y1 Ω 11160R3RY1Δ1 .

63000WW·3Q

84000WWP

21T

21T

º87.36)/Parctg(Q TTT A164.465

0.8·368.6·3

84000PI T

Y2

Y sc2,2Y2Y1Y ZIU'U V72.0219.8390.03985j0.02338·87.36164.465º03

368.6'U1Y

V824048.1 lU

a)

b)

A Dy, 125 kVA εsc = 4%, Po = 480 W, Psc = 2530 W, 24000/380-V, three-phase

transformer has a secondary winding with R2= 0.01 W. The wattmeter readings are

W1=60186.56 W, W2=23813.44 W, while V2=368.6 V. Compute: a) R1 b) V1 c) The

transformer efficiency d) The angular displacement of secondary with respect to

primary voltages.

27


sc

22

1n1o2

2

CuFe2

2

P·)/V(VPP

P

PPP

Pη

c

972.034.86379

84000

2530·866.024000

8.24048·48084000

84000η

2

2

c) c = I2/I2n = 164.465/189.92 = 0.866

d) Clock number = angular displacement between secondary and primary voltages

R1

S1

T1

A=C'

C=B'

B=A'

a

b c

R2

S2

T2

a’= b’= c’

A=C’

B=A’ C=B’

a b

c

N

Dy11

28

three phase transformers...the most elemental three-phase transformer consists on an arrangement...

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