The Wye-Delta Center Tapped Transformer Bank

W. H. KerstingWayne Carr

Milsoft Utility Solutions

1

Introduction

• Loads to be served– Three-phase induction motor– 120/240 volt single-phase loads

• Transformer connection– Grounded wye-delta– Ungrounded wye-delta

• The question is should the neutral be grounded, grounded through an impedance or ungrounded?

2

General Thoughts on Grounding

• Always taught that the grounded wye-delta is preferred because when a primary phase is open the secondary will still provide three-phase voltages.

• Always taught that if the neutral is ungrounded upon the loss of a primary phase the secondary will become single-phase.

3

Concerns about Grounding

• Paper at the 2016 REPC demonstrated large “feedback” current for upstream ground faults when transformer is connected grounded wye-delta.

• Do both the grounded and ungrounded cases provide a dangerous backfeed voltage?

• Purpose of this paper is to investigate the magnitude of the “feedback” voltages for both connections.

4

The Thirty Degree Transformer Connection

N

+ +

- -

-

+

+

ANV

ANVT

CNVT

+0Z

+

-

+

-

-

+

+-

+-

n

+

-

+

- nanVt

nbVt

bcVt

caVnbV

bcV

cI

bI

nI

aI

1Z

yZ

zZ

2Z

cbIDacID

a

b

c

?NGV +-

BI+

AGV BNVT

CGV

BGV

CI

GI

anV

a

c

b

AInaID

bnID

+

AB

G

C

Note in the figure that there is a question mark between the neutral and ground. A primary purpose of this presentation is to answer the question on whether or not to ground the neutral. Note also that there are switches on the three primary phases. This is done so that various open switch positions can be studied in order to help answer the question about grounding.

5

Very Basic Transformer Theory

6

+_

+_

1Vt

2Vt

+

_

VT

0I :1tn1H

2H

1X

2X

3X

1I

2I

+ +

_ _VtVT

1I 2I:1tn1X

2X

1H

2H

1 2

Two Winding Transformer1

1t

t

Vt VTn

I In

= ⋅

= ⋅( )

1 2

1

0 1 2

Center Tap Transformer1

221

2

t

t

t

Vt Vt VTn

VT n Vt

I I In

= = ⋅⋅

= ⋅ ⋅

= ⋅ +⋅

Two Laws Must Always be Satisfied

N

+ +

- -

-

+

+

ANV

ANVT

CNVT

+0Z

+

-

+

-

-

+

+-

+-

n

+

-

+

- nanVt

nbVt

bcVt

caVnbV

bcV

cI

bI

nI

aI

1Z

yZ

zZ

2Z

cbIDacID

a

b

c

?NGV +-

BI+

AGV BNVT

CGV

BGV

CI

GI

anV

a

c

b

AInaID

bnID

+

AB

G

C

7

Kirchhoff's Voltage Law: Kirchhoff's Current Law0 0

Every Node: an nb bc ca a b c n

in out

V V V V I I I II I

+ + + = + + + =

=∑ ∑

Grounded Wye-Delta Test System

[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]

[ ] [ ] [ ]

Forward Sweep:2

3 2

4 3

Backward Sweep:

ABC ABC ABC ABC

anbc ABC abcn n

anbc anbc anbc anbc

ABC t abc

V LG ELG Z I

V At V LG Bt I

V V Zs I

I d I

= − ⋅

= ⋅ − ⋅

= − ⋅

= ⋅

8

Subtransmission Line

4.0'

2.5' 4.5'

3.0'

3

n

25.0

21Phase Conductors: "Linnett" 336/400/ 26/7 ACSR

Neutral Conductor: 4/0 ACSR

Z.ABC

2.2878 5.3902i+

0.7797 2.5084i+

0.7674 1.9247i+

0.7797 2.5084i+

2.3331 5.2408i+

0.79 2.1182i+

0.7674 1.9247i+

0.79 2.1182i+

2.3074 5.3253i+

=

9

Grounded Wye-Delta Transformer Bank

10

The transformer bank data are: Lighting transformer (Phase A): 25, 7200-240 volts, Zt 0.012 0.017pukVA j= = + Power transformers (Phases B and C) 10, 7200-240 volts, Zt 0.016 0.014pukVA j= = +

[ ]ABCVLG [ ]anbcV

[ ]ABCI [ ]abcI

Final Grounded Wye-Delta Model[ ]ABCVLG [ ]anbcV

[ ]ABCI [ ]abcI

11

[ ] [ ] [ ] [ ] [ ][ ] [ ] [ ]

[ ] [ ] [ ][ ] [ ] [ ] [ ] [ ] [ ]( ) [ ]( ) [ ][ ] [ ] [ ]

where:

0 12

anbc t ABC t abcn n

ABC t abcn

t nn

t anbcn n nn

t nn

V A VLG B I

I d I

A KVL BV

B KVL BV Z G AI Zt Z CI

d AI CI

= ⋅ − ⋅

= ⋅

= ⋅

= ⋅ ⋅ ⋅ + + ⋅

= ⋅

Quadraplex Secondary

Quadraplex

4

1

2

3

3 - Insulated 1/0 All Aluminum1 - Bare 1/0 ACSR

Insulation Thickness = 80 mil

Zsanbc

0.1012 0.1429j+

0.009 0.1102j+

0.009 0.1102j+

0.009 0.1163j+

0.009 0.1102j+

0.1012 0.1429j+

0.009 0.1102j+

0.009 0.1084j+

0.009 0.1102j+

0.009 0.1102j+

0.1012 0.1429j+

0.009 0.1163j+

0.009 0.1163j+

0.009 0.1084j+

0.009 0.1163j+

0.1151 0.1534j+

=

12

The Secondary System

13

The three phase induction motor is:

25 HP, 240 volts, slip = 0.3 Stator Z: 0.0366 0.08

Rotor Z: 0.0394 0.08

Magnitizing Z: 0 2.1

pu

pu

pu

Zs j

Zr j

Zm j

= +

= +

= +

The single-phase lighting loads are:

2.85 0.9367 kW+jkvar4.5 2.1794 kW+jkvar6.8 4.2143 kW+jkvar

an

nb

ab

S jS jS j

= +

= +

= +

1IS

2IS bIM3IS

cIM

aIM

1SL

2SL3SL

bcV

_anV

_ +

+ _

+

+

_ b

n

c

nbVcaV

aI

nI

bI

cI

av

bv

cv

a

Motor

Phase A Open, Motor Connected

14

Backfeed Voltages:1 6302.8/ 1.9 volts2 6313.7/ 1.8 volts

AG

AG

VV

= −

= −

[ ] [ ]

Connected Motor Voltages and Currents201.5/ 0.6 38.8/ 86.4

219.0/ 115.2 52.4/ 158.1226.6/118.5 73.3/51.2

6.6% 33.7%

abc abc

unbalance unbalance

VM IM

V I

− − = − = − = =

Phase A Open, Motor Off

15

[ ]

Backfeed Voltages1 6557.0/ 1.1 volts2 6565.5/ 1.0 volts

Secondary Currents66.4/ 26.618.0/ 39.284.0/150.7

0

AG

AG

a

nanbc

b

c

VV

II

III

= −

= −

− − = =

Single-Phase Load Voltages103.4/ 0.7108.1/ 0.8211.5/ 0.8

an

nb

bc

VLVLVL

− = − −

Phase B and Phase C Open

16

Phase B Open Phase C Open1 6794.5/ 120.4 1 6874.0/119.12 6806.1/ 120.3 2 6878.6/119.2

1.7 1.2%11

BG CG

BG CG

unbalance unbalance

unbalance

V VV VV VI

= − =

= − =

= =

= .7% 6.9%unbalanceI =

Because the voltage unbalance is below the 3% level, the motor will not be tripped

Summary for Grounded Wye-Delta

• No matter which phase is open, there will be a significant backfeed voltage at the open switch

• When phase A is open the voltage unbalance at the motor will be high and therefore the protective device will trip the motor off.

• When phase B or phase C is open the voltage unbalance is low so motor will continue to operate.

• In all cases the single-phase loads are served

17

Ungrounded Wye-Delta Circuit

18

N

+ +

- -

-

+

+

++

+

-- NGV

Cv

Bv

AvANVT BNVT

CNVT

0Z

Source NI

B

G

A

C

+

_+__

+

AGV

CGV

BGV

SW B

SW AAGE

BGE

+__

+CGE

SW C

+

12.47 kV

AI

BI

CI

a

+

-

+

-

+-

+

-

n

+

-

+

-

23V

av

bv

cv

Motor

anVt

nbVt

bcVtacI

bnI

naI

cbI+

_

caVt

c

b

bcV

anV

nbV nI

aIM

cIM

bIM

2Z

bZ

cZ

1Z1ZL

2ZL3ZL

aI

bI

cI

Phase A Open No-Load Operating Condition

• Motor and Single-Phase Loads are Off

• Single-Phase voltages so motor will not start• Without three-phase voltages the rotational

magnetic field needed to start the motor is not there.

19

[ ]0

207.8/ 90207.8/90

VM = −

Phase A Open with Motor and Loads On

20

[ ] [ ]117.3/ 20.9 71.7/ 168.6169.3/ 106.0 78.6/ 101.8

214.0/107.1 125.6/46.529.7 % 36.5 %

abc abc

unbalance unbalance

VM IM

V I

− − = − = − = =

Because of the high voltage unbalance the protective device for the motor will open turning the motor off.

Open Phase A, Motor Off

21

N

+ +

- -

-

+

+

++

+

a

+-

+-

n

+

-

-- NGV

ANVT BNVT

CNVT

anVt

nbVt

bcVt+

_

caVt

c

b

0Z

2Z

bZ

cZ

1Z1SL

2SL3SL

B

A

+

_+__

+

AGV

BGV

SW C

SW A

SW B

+

_

+

+

AGVS

BGVS

CGVS CGV

G

C

BI

AI

CI

naI

bnIcbIacI

aI

nI 1IL

2ILbI 3IL

0cI =

3600/1800, 6235.4/ 90, 6235.4/90

0, 207.8/ 90, 207.8/901 3600/180, , 1 7200/120

All currents primary, secondary and single-phase l1 7

oa200/ 1

ds = 20

NG

AN BN CN

an nb bc ca

AG B CGG

VVT VT VTVt Vt Vt VtV V V

=

= = −

=

=

= = =

= =−

− =

0

Phase A OpenCurrents

22

N

+ +

- -

-

+

+

++

+

a

+-

+-

n

+

-

-- NGV

ANVT BNVT

CNVT

anVt

nbVt

bcVt+

_

caVt

c

b

0Z

2Z

bZ

cZ

1Z1SL

2SL3SL

B

A

+

_+__

+

AGV

BGV

SW C

SW A

SW B

+

_

+

+

AGVS

BGVS

CGVS CGV

G

C

BI

AI

CI

naI

bnIcbIacI

aI

nI 1IL

2ILbI 3IL

0cI =

• 0 therefore 0A an bnI I I= + = • Path for and so B C C BI I I I= − • Because of above ac cbI I= − • With motor open 0cI = for all conditions • At node c 0 this is impossible since c ac cb ac cbI I I I I= − = = − • Because of the above 0ac cbI I= = • Therefore 0B CI I= =

Kirchhoff’s Current Law (KCL)

23

N

+ +

- -

-

+

+

++

+

a

+-

+-

n

+

-

-- NGV

ANVT BNVT

CNVT

anVt

nbVt

bcVt+

_

caVt

c

b

0Z

2Z

bZ

cZ

1Z1SL

2SL3SL

B

A

+

_+__

+

AGV

BGV

SW C

SW A

SW B

+

_

+

+

AGVS

BGVS

CGVS CGV

G

C

BI

AI

CI

naI

bnIcbIacI

aI

nI 1IL

2ILbI 3IL

0cI =

• In previous slide 0ac cb na nb cI I I I I= = + = = • Since 0 then na bn bn naI I I I+ = = − • KCL at node a: 0a na ac na naI I I I I= − = − = • KCL at node b: 0b cb bn bn naI I I I I= − = − = • KCL at node n: 2n bn na na na naI I I I I I= − = − − = − ⋅

• KCL on secondary: 0 0 2

2 0a b c n na bn na

na na na

I I I I I I II I I+ + + = = − + − ⋅

+ − ⋅ =

Phase A OpenVoltages

24

N

+ +

- -

-

+

+

++

+

a

+-

+-

n

+

-

- - NGV

ANVT BNVT

CNVT

anVt

nbVt

bcVt+

_

caVt

c

b

0Z

2Z

bZ

cZ

1Z1SL

2SL3SL

BI

AI

naI

bnIbcIacI

aI

nI 1IL

2ILbI 3IL

0cI =CI

B

A

+

_+__

+

AGV

BGV

SW C

SW A

SW B

+

_

+

+

AGVS

BGVS

CGVS CGV

G

C

• Voltage applied to primary windings: 12,470/ 90BC BN CNVT VT VT= − = − • Phase B winding voltage: 0.5 6235/ 90BN BCVT VT= ⋅ = − • Phase C winding voltage: 0.5 6235/90CN BCVT VT= − ⋅ = • The secondary ideal voltages are: 207.8/ 90 and 207.8/90bc caVt Vt= − = • KVL: 2 0 0an nb bc ca anVt Vt Vt Vt Vt+ + + = ⋅ + = • Since: 0 then 0an ANVt VT= = • Primary neutral voltage: 3600/180NG BG ANV V VT= − = • Phase A backfeed voltage: 0 3600/180AG AN NG NGV VT V V= + = + =

Phase B Open with Motor and Loads On

25

[ ] [ ]192.1/ 10.3 102.3/ 84.0

119.3/ 121.4 36.8/54.2186.0/133.0 78.8/114.2

28.0 % 49.3 %

abc abc

unbalance unbalance

VM IM

V I

− − = − = = =

Because of the high voltage unbalance the protective device for the motor will open turning the motor off.

Open Phase B, Motor Off

26

N

+ +

- -

-

+

+

++

+

a

+-

+-

n

+

-

-- NGV

ANVT BNVT

CNVT

anVt

nbVt

bcVt+

_

caVt

c

b

0Z

2Z

bZ

cZ

1Z1SL

2SL3SL

B

A

+

_+__

+

AGV

BGV

SW C

SW A

SW B

+

_

+

+

AGVS

BGVS

CGVS CGV

G

C

BI

AI

CI

naI

bnIcbIacI

aI

nI 1IL

2ILbI 3IL

0cI =

_ _

7200/00, 12470.8/ 30, 12470.8/150

0, 415.7/ 30, 415 / 7/1501 1 19049.4/ 19.7200/0, , 1 7200/120

All currents primary, secondary and single-phase load1

NG

AN BN CN

an nb bc ca

AG CG GB

VVT V

V

T VTVt Vt Vt VtV V

=

= = − =

=

=

= −

−

= =

= =

s = 0

Phase B OpenCurrents

27

N

+ +

- -

-

+

+

++

+

a

+-

+-

n

+

-

-- NGV

ANVT BNVT

CNVT

anVt

nbVt

bcVt+

_

caVt

c

b

0Z

2Z

bZ

cZ

1Z1SL

2SL3SL

B

A

+

_+__

+

AGV

BGV

SW C

SW A

SW B

+

_

+

+

AGVS

BGVS

CGVS CGV

G

C

BI

AI

CI

naI

bnIcbIacI

aI

nI 1IL

2ILbI 3IL

0cI =

_ _

• 0 therefore 0B cbI I= = • There is a path for and so C A A CI I I I= − • With 0 and 0 then KCL says 0cb c acI I I= = = • With 0 then 0 and 0ac C AI I I= = = • In short 0A B CI I I= = = • However: 0 or na bn bn naI I I I+ = = − • What about ?an nbVt Vt= =

Secondary and Load Currents

28

a

+-

+

-

nanVt

nbVt

b

2Z

1Z1SL

3SL

aI

aI

aIaI

aI

aI

2 aI⋅

02SL

• Look at the circuit and note that bn naI I= − • Shown in the circuit is only since a nb an aI I I I= − = • This makes the same currents flow in phase a and phase b • Now must have 2 aI⋅ flowing in the neutral as shown • Note the same current must flow thru the two 120 volt loads • This requires the voltages an bnVL VL= • This is impossible because of the polarities of and an nbVt Vt • Therefore the only thing that works is 0an nb ANVt Vt VT= = =

Computation of Open Switch B Line-to-ground Voltage

29

N

+

-

-+

+

a

- n

+

-

- - NGV

ANVT BNVT

CNVT

anVt

nbVt

bcVt+

_

caVt

c

bbZ

cZ

1SL

2SL

3SL

B

A

+

_+__

+

AGV

BGV

SW C

SW A

SW B

+

_

+

+

AGVS

BGVS

CGVS CGV

G

C

_ _-

• With the shorted voltages the only primary voltage is 12,470/150CN CAVT VT= =

• Now 1 415.7/150ca CAVt VTn

= ⋅ =

• To satisfy KVL: 415.7/ 30bc caVt Vt= − = − • Therefore: 12,470/ 30BN t bcVT n Vt= ⋅ = − • With the neutral and phase A shorted: 7200/0NG AGV V= = • Now: 19049.4/ 19.1BG BN NGV VT V= + = −

BGV

Phase C Open with Motor and Loads On

30

N

+ +

- -

-

+

+

++

+

a

+-

+-

n

+

-

-- NGV

ANVT BNVT

CNVT

anVt

nbVt

bcVt+

_

caVt

c

b

0Z

2Z

bZ

cZ

1Z1SL

2SL3SL

B

A

+

_+__

+

AGV

BGV

SW C

SW A

SW B

+

_

+

+

AGVS

BGVS

CGVS CGV

G

C

BI

AI

CI

naI

bnIcbIacI

aI

nI 1IL

2ILbI 3IL

0cI =

_ _

[ ] [ ]165.5/2.1 30.6/34.5

229.0/ 127.3 103.5/ 174.7178.2/98.6 78.2/ 5.6

20.0 % 56.7 %

abc abc

unbalance unbalance

VM IM

V I

= − = − − = =

Open Phase C, Motor Off

31

N

+ +

- -

-

+

+

++

+

a

+-

+-

n

+

-

-- NGV

ANVT BNVT

CNVT

anVt

nbVt

bcVt+

_

caVt

c

b

0Z

2Z

bZ

cZ

1Z1SL

2SL3SL

B

A

+

_+__

+

AGV

BGV

SW C

SW A

SW B

+

_

+

+

AGVS

BGVS

CGVS CGV

G

C

BI

AI

CI

naI

bnIcbIacI

aI

nI 1IL

2ILbI 3IL

0cI =

_ _

7200/00, 12470.8/ 150, 12470.8/30

0, 415.7/ 150, 415 / 7/301 7200/0, 1 7200/ 120, 1 19049.4/19.1

All currents primary, secondary and single-phase load

NG

AN BN CN

an nb bc ca

AG BG CG

VVT VT VTVt Vt Vt VtV V V

=

= = − =

= = = − =

= = − =

s = 0

Conclusions - 1

• The original question was whether the wye should be grounded. The answer is still ???

• Grounded Wye-Delta– Provides three-phase service with an open phase– High backfeed short circuit current for upstream

ground faults– Provides a backfeed voltage below rated LG

voltage but still dangerous

32

Conclusions - 2

• Ungrounded Wye-Delta– Does not provide three-phase service with an

open phase– Does not provide a backfeed current for upstream

ground faults– Provides a very high backfeed voltage (2.65 times

rated LG voltage) when phase B or C is open– Lower backfeed voltage when the center tap

phase is open

33

Conclusions-3

• So what is the answer?– Good question– Neither connection is perfect– What is known is that both connections will

provide a very dangerous backfeed voltage on the downstream side of the open switch

– Linemen must be trained to understand that just because the switch is open, it is not save to touch the line on either side of the switch

34