Ungrounded Wye-Delta Transformer Bank Backfeed Voltage
W. H. Kersting Milsoft Utility Solutions
Introduction • Loadstobeserved– Three-phaseinduc4onmotor– 120/240voltsingle-phaseloads
• Transformerconnec4on– Groundedwye-delta
• Developsbackfeedshortcircuitcurrentsforupstreamgroundfaults
– Ungroundedwye-delta• Purposeofthispaper
TypicalSystemOne-LineDiagram
4-wire, gr. wye 1/0 QuadraplexInfinite Bus
12,470 VSingle-Phase Loads
M Induction Motor
1 2 3 4
The Three-Phase System
30 Degree Step-Down Connection
N
+ +
- -
-
+
+
++
+
a
+
-
+
-
+
-
+
-
n
+
-
+
-
-
23V
- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVtacI
bnI
naI
cbI+
_
caVt
c
b
bcV
anV
nbV
cI
bI
aI
0Z
2Z
bZ
cZ
1Z
BI
AI
NI
CI
ANV
CNV ABV
BNV
abV
bcV
caVRated Primary
Rated Secondaryt
kVLNn
kVLL=
30DegreeStep-UpConnec4on
N
+ +
- -
-
+
+
++
+
a
+
-
+
-
+
-+
-
n
+
-
+
-
-
23V
- NGV
anVt
nbVt
bcVtacI
bnI
naI
cbI+
_
caVt
c
b
bcV
anV
nbV
cI
bI
aI
0Z
2Z
bZ
cZ
1Z
BI
AI
NI
CI
BNVT
CNVT
ANVT
ANV
BNV
CNV ABV
acVcaV
abVcbV
bcVbaV
Voltage Equations for NMSU Connection
[ ] [ ] [ ]
[ ]
2 0 0 00 0 1 00 0 0 1
2 0 0 0where: 0 0 1 0
0 0 0 1
anAN
nbBN t
bcCN
ca
ABC anbc
t
VtVT
VtVT n
VtVT
Vt
VTLN AV Vt
AV n
⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⋅ ⋅⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
⎣ ⎦
= ⋅
⎡ ⎤⎢ ⎥= ⋅ ⎢ ⎥⎢ ⎥⎣ ⎦
[ ] [ ] [ ]
[ ]
0.5 0 00.5 0 010 1 00 0 1
0.5 0 00.5 0 01where: 0 1 00 0 1
anAN
nbBN
bc tCN
ca
anbc ABC
t
VtVT
VtVT
Vt nVT
Vt
Vt BV VTLN
BVn
⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= ⋅ ⋅ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ ⎦
⎣ ⎦⎣ ⎦
= ⋅
⎡ ⎤⎢ ⎥⎢ ⎥= ⋅⎢ ⎥⎢ ⎥⎣ ⎦
N
+ +
- -
-
+
+
++
+
a
+
-
+
-
+
-+
-
n
+
-
+
-
-
23V
- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt
acIbnI
naI
cbI+
_
caVt
c
b
bcV
anV
nbV
cI
bI
aI
0Z
2Z
bZ
cZ
1Z
BI
AI
NI
CI
Current Equations for NMSU Connection
[ ] [ ] [ ]
[ ]
0.5 0.5 0 01 0 0 1 0
0 0 0 1
0.5 0.5 0 01where: 0 0 1 0
0 0 0 1
naA
bnB
cbtC
ac
ABC anbc
t
II
II
InI
I
I AI ID
AIn
⎡ ⎤⎡ ⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⋅ ⋅⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
⎣ ⎦
= ⋅
⎡ ⎤⎢ ⎥= ⋅ ⎢ ⎥⎢ ⎥⎣ ⎦
[ ] [ ] [ ]
[ ]
5 1 31 5 311 1 361 1 3
5 1 31 5 31where: 1 1 361 1 3
naa
bnb
cbc
ac
anbc abc
II
II
II
I
ID Dd I
Dd
⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥− − − ⎢ ⎥⎢ ⎥ ⎢ ⎥= ⋅ ⋅ ⎢ ⎥⎢ ⎥ ⎢ ⎥− −⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ ⎦−⎣ ⎦⎣ ⎦
= ⋅
⎡ ⎤⎢ ⎥− − −⎢ ⎥= ⋅⎢ ⎥− −⎢ ⎥−⎣ ⎦
N
+ +
- -
-
+
+
++
+
a
+
-
+
-
+
-+
-
n
+
-
+
-
-
23V
- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt
acIbnI
naI
cbI+
_
caVt
c
b
bcV
anV
nbV
cI
bI
aI
0Z
2Z
bZ
cZ
1Z
BI
AI
NI
CI
Ini6alStudyofNMSUTransformers
█NMSU Power Lab Transformers@Each rated: 0.5 kVA, 120−120/240 Volts, 𝑍↓𝑡 =0+1𝑗0.036 PU
N
+ +
- -
-
+
+
++
+
a
+
-
+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
B
G
A
C
_+__
+
AGV
CGV
BGV
SW C
SW B
Source120 volt
SW A
AGE
CGE
BGE
+__
+
+ +
No-LoadStudiesofNMSUSystem
• Standard30degreeconnec4onssothesecondarywillhavebalancedthree-phasevoltages
• Developequa4onsforsteadystate• Modifyfor– SW-Aopen– SW-Bopen– SW-Copen
The NMSU System
N
+ +
- -
-
+
+
++
+
a
+
-+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
120 LGSource
B
G
A
C
+
_+__
+
AGV
CGV
BGV
SW A
SW C
SW B
Turns ratio: 120 0.5240tn = =
Source LG Voltages: [ ]120/0
120/ 120120/120
ABCELG⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
Source LL Voltages: [ ]207.8/30
207.8/ 90207.8/90
ABCELL⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦
Assumption: With one switch open the two remaining primaries will share the line-to-line voltage equally.
KVL On Secondary Terminals
( )
( )
on secondary 0
1 1 1 1 02 21 0
0 0
an nb bc ca
AN AN BN CNt t t t
AN BN CNt
AN BN CN t
kVLVt Vt Vt Vt
VT VT VT VTn n n n
VT VT VTnVT VT VT n
+ + + =
⋅ + ⋅ + ⋅ + ⋅ =⋅ ⋅
⋅ + + =
+ + = ⋅ =
N
+ +
- -
-
+
+
++
+
a
+
-
+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
B
G
A
C
_+__
+
AGV
CGV
BGV
SW C
SW B
Source120 volt
SW A
AGE
CGE
BGE
+__
+
+ +
All Switches Closed
1. 1. 2. 2. 3.
A AN NG AN
B BN NG BN
C CN NG
ELG VT V VTELG VT V VTELG VT V
= +
= +
= + 3. 14. 0 4.
25. 0 5.
CN
an AN ant
nb an
VT
V VT Vtn
Vt Vt
= − + ⋅⋅
= − +
.
16. 0 6.
17. 0 7.
8. 0 +
nb
bc BN bct
ca CN cat
AN BN CN
Vt
Vt VT Vtn
Vt VT Vtn
VT VT V
= − + ⋅
= − + ⋅
= + 8. NGV
All Switches Closed - Results
[ ] [ ]
[ ]
[ ] [ ] [ ]
120/0120/0
120/0120/ 120
240/ 120120/120
240/120
00 0
0
120/0120/ 120120/120
ABC anbc
NG AG AN
ABC ABC
VTLN Vt
V E VT VNG
VSLG VTLN VNG
⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥= − =⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎣ ⎦
⎣ ⎦
⎡ ⎤⎢ ⎥= − = = ⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤⎢ ⎥= + = −⎢ ⎥⎢ ⎥⎣ ⎦
SW-AOpen
N
+ +
- -
-
+
+
++
+
a
+
-+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
B
G
A
C
_+__
+
AGV
CGV
BGV
SW C
SW B
Source120 volt
SW A
AGE
CGE
BGE
+__
+
+ +
SW-A open and B and C closed:
( )
207.8/ 901 103.9/ 902
103.9/900
0.5 0 0 00.5 0 0 010 1 0 207.8/ 900 0 1 207.8/90
BC BC
BN BC
CN BN
AN AN CN
anAN
nbBN
bc tCN
ca
NG
VT ELL
VT VT
VT VTVT VT VT
VtVT
VtVT
Vt nVT
VtV EL
= = −
= ⋅ = −
= − =
= − + =
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⋅ ⋅ =⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦
⎣ ⎦ ⎣ ⎦⎣ ⎦=
[ ] [ ] [ ]
60/18060/180
120/ 120120/120
BG CN
ABC
L VT
VNG VTLN VNG
− =
⎡ ⎤⎢ ⎥= − = −⎢ ⎥⎢ ⎥⎣ ⎦
SW-B Open
N
+ +
- -
-
+
+
++
+
a
+
-+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
B
G
A
C
_+__
+
AGV
CGV
BGV
SW C
SW B
Source120 volt
SW A
AGE
CGE
BGE
+__
+
+ +
( )
207.8/1501 103.9/1502
103.9/ 300
0.5 0 0 103.9/ 300.5 0 0 103.9/ 3010 1 0 00 0 1 207.8/150
CA CA
CN CA
AN CN
BN CN AN
anAN
nbBN
bc tCN
ca
VT ELL
VT VT
VT VTVT VT VT
VtVT
VtVT
Vt nVT
Vt
= =
= ⋅ =
= − = −
= − + =
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ −⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢= ⋅ ⋅ =⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦
⎣ ⎦ ⎣ ⎦⎣ ⎦
[ ] [ ] [ ]
60/60120/060/60
120 /120
NG BC BN
ABC
V ELG VT
VLG VTLN VNG
⎥⎥⎥⎥
= − =
⎡ ⎤⎢ ⎥= + = ⎢ ⎥⎢ ⎥⎣ ⎦
SW-C Open
N
+ +
- -
-
+
+
++
+
a
+
-+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
B
G
A
C
_+__
+
AGV
CGV
BGV
SW C
SW B
Source120 volt
SW A
AGE
CGE
BGE
+__
+
+ +
( )
207.8/301 103.9/302
103.9/ 1500
0.5 0 0 103.9/300.5 0 0 103.9/3010 1 0 207.8/ 1500 0 1 0
AB AB
AN AB
BN AN
CN AN BN
anAN
nbBN
bc tCN
ca
VT ELL
VT VT
VT VTVT VT VT
VtVT
VtVT
Vt nVT
Vt
= =
= ⋅ =
= − = −
= − + =
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⋅ ⋅ =⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ −⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦
⎣ ⎦ ⎣ ⎦⎣ ⎦
[ ] [ ] [ ]
60/ 60120 / 0
120 / 12060 / 60
NG AB AN
ABC
V ELG VT
VLG VTLN VNG
⎥⎥
= − = −
⎡ ⎤⎢ ⎥= + = −⎢ ⎥⎢ ⎥−⎣ ⎦
New Mexico State Power Lab
This circuit was set up in the New Mexico State University (NMSU) power lab. The purpose was to verify the Mathcad results. Understand that the Mathcad routine assumes “ideal” transformers while the Lab uses real transformers which have shunt impedances and capacitance.
N
+ +
- -
-
+
+
++
+
a
+
-
+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
B
G
A
C
_+__
+
AGV
CGV
BGV
SW C
SW B
Source120 volt
SW A
AGE
CGE
BGE
+__
+
+ +
Mathcad vs. Lab Results
MathcadResults NMSULabResults
SW-A SW-B SW-C ANVT 8.5 104 102
BNVT 106 2.4 105
CNVT 101 102 6.3
NGV 55 55 55
anVt 8.5 104 104
nbVt 8.5 104 104
bcVt 213 4.7 209
caVt 201 204 12.8
AGV 55 120 120
BGV 120 61 120
CGV 120 120 63
SW-A SW-B SW-C ANVT 0 103.9 103.9
BNVT 103.9 0 103.9
CNVT 103.9 103.9 0
NGV 60 60 60
anVt 0 103.9 103.9
nbVt 0 103.9 103.9
bcVt 207.8 0 207.8
caVt 207.8 207.8 0
AGV 60 120 120
BGV 120 60 120
CGV 120 120 60
NMSU System with Single-Phase Load
25 Unknowns 12 voltages 13 currents
N
+ +
- -
-
+
+
++
+
a
+
-+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
120 LGSource
B
G
A
C
+
_+__
+
AGV
CGV
BGV
1SL
2SL3SL
aI
nI
bI
1IL
2IL 3IL
AI
BI
CI
naI
cbIacI
bnI
SW C
SW A
SW B
_+
_
+
_AGE
CGE
+BGE
Single-Phase Loading WindMil and Mathcad
• WindMil PQ single-phase loads
• Using WindMil results the constant impedance loads computed for Mathcad
[ ]0.5/18.20.45/25.80.55/31.8
SL⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
[ ]25.4 8.326.8 13.082.7 51.2
jZL j
j
+⎡ ⎤⎢ ⎥= +⎢ ⎥⎢ ⎥+⎣ ⎦
WindMil and Mathcad Full-Load Results
WindMil Mathcad AI 8.59/-29.3 8.59/-29.3
BI 4.29/150.7 4.29/150.7
CI 4.29/150.7 4.29/150.7
aI 6.64/-26.9 6.64/-26.9
bI 6.24/148.1 6.24/148.1
cI 0 0
anV 116.0/-3.9 116.0/-3.9
nbV 115.8/-3.8 115.8/-3.8
abV 231.8/-3.8 231.8/-3.8
NGV 0 0
Open Switch Studies • Majorpurposeofthetotalstudywastodeterminewhathappenswhenoneofthethreeupstreamswitchesisopen
• Studiestobecompletedwiththesingle-phaseloads– SW-Aopen– SW-Bopen– SW-Copen
SW-A Open - Currents
SW-A Open - Voltages
N
+ +
- -
-
+
+
++
+
a
+
-+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
120 LGSource
B
G
A
C
+
_+__
+
AGV
CGV
BGV
1SL
2SL
3SL
aI
nI
bI
1IL
2IL 3IL
AI
BI
CI
naI
cbIacI
bnI
SW C
SW A
SW B
_+
_
+
_AGE
CGE
+BGE
1) Only voltage on transformers is BCV
a) 12BN BCVT V= ⋅
b) 12CN BCVT V= − ⋅
i) 1bc BN
tVt VT
n= ⋅
ii) 1ca BN
tVt VT
n= − ⋅
iii) ( )
KVL: 0 0 0 0
an nb bc ca
t AN BN BN
AN BN BN
AN
Vt Vt Vt Vtn VT VT VTVT VT VTVT
+ + + =
⋅ + − =
+ − =
=
SW-A Open – Mathcad Results
N
+ +
- -
-
+
+
++
+
a
+
-+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
120 LGSource
B
G
A
C
+
_+__
+
AGV
CGV
BGV
1SL
2SL
3SL
aI
nI
bI
1IL
2IL 3IL
AI
BI
CI
naI
cbIacI
bnI
SW C
SW A
SW B
_+
_
+
_AGE
CGE
+BGE
0, 103.9/ 90, 103.9/9060/180
0, 207.8/ 90, 207.8/9060/180120/ 120120/120
60 0.5120
AN BN CN
NG
an nb bc ca
AG AN NG
BG BN NG
CG CN NG
VT VT VTVVt Vt Vt VtVS VT VVS VT VVS VT V
Ratio
= = − =
=
= = = − =
= + =
= + = −
= + =
= =
SW-B Open - Currents
N
+ +
- -
-
+
+
++
+
a
+
-+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
120 LGSource
B
G
A
C
+
_+__
+
AGV
CGV
BGV
1SL
2SL
3SL
aI
nI
bI
1IL
2IL 3IL
AI
BI
CI
naI
cbIacI
bnI
SW C
SW A
SW B
_+
_
+
_AGE
CGE
+BGE
1) 0 alwayscI =
a) 0BI = i) 0cbI =
ii)
Because 00
Therefore: 0 and 0 0 0
c
ac
C A
ac an nb cb
a b c
II
I II I I II I I
=
=
= =
= = = =
= = =
SW-B Open - Voltages
N
+ +
- -
-
+
+
++
+
a
+
-+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
120 LGSource
B
G
A
C
+
_+__
+
AGV
CGV
BGV
1SL
2SL
3SL
aI
nI
bI
1IL
2IL 3IL
AI
BI
CI
naI
cbIacI
bnI
SW C
SW A
SW B
_+
_
+
_AGE
CGE
+BGE
1) All currents are zero
a) Since the load currents are zero the center tap voltages and an nbVt Vt must be zero i) Therefore: 0ANVT =
(1) With 0AI = the primary neutral is shifted up to the terminal A (2) This causes and CN CA AN CAVT V VT V= = −
SW-B Open – Mathcad Results
N
+
-
-+
+
a
+
-+
-
n
+
-
- NGV
BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
2Z
bZ
cZ
1Z
120 LGSource
B
G
A
C
+
_+__
+
AGV
CGV
BGV
1SL
2SL
3SL
aI
nI
bI
1IL
2IL 3IL
AI
BI
CI
naI
cbIacI
bnI
SW C
SW A
SW B
_+
_
+
_AGE
CGE
+BGE
0, 207.8/ 30, 207.8/150120/0
0, 415.7/ 30, 415.7/150120/0317.5/ 19.1120/120
317.5 2.65120
AN BN CN
NG
an nb bc ca
AG AN NG
BG BN NG
CG CN NG
VT VT VTVVt Vt Vt VtVS VT VVS VT VVS VT V
Ratio
= = − =
=
= = = − =
= + =
= + = −
= + =
= =
SW-C Open – Mathcad Results
N
+ +
- -
-
+
+
++
+
a
+
-+
-
n
+
-
-- NGV
ANVT BNVT
CNVT
anVt
nbVt
bcVt+
_
caVt
c
b
0Z
2Z
bZ
cZ
1Z
120 LGSource
B
G
A
C
+
_+__
+
AGV
CGV
BGV
1SL
2SL
3SL
aI
nI
bI
1IL
2IL 3IL
AI
BI
CI
naI
cbIacI
bnI
SW C
SW A
SW B
_+
_
+
_AGE
CGE
+BGE
0, 207.8/ 150, 207.8/30120/0
0, 415.7/ 150, 415.7/30120/0120/ 120317.5/19.1
317.5 2.65120
AN BN CN
NG
an nb bc ca
AG AN NG
BG BN NG
CG CN NG
VT VT VTVVt Vt Vt VtVS VT VVS VT VVS VT V
Ratio
= = − =
=
= = = − =
= + =
= + = −
= + =
= =
Summary of Open Switches
12.47 kV System Data
5 mile
4-wire, gr. wye
500 ft.
1/0 quadraplex
Single-Phase Loads
Induction MotorInfinite Bus
12,470 V
Four Wire Centered Tapped Ungrd. Y-D Transformer System
3 single-phasetransformers
Primary Line: 336,400 26/7 ACSR and 4/0 ACSR Secondary Line: 1/0 AA Quadraplex Lighting Transformer: 25 kVA, 7200-120/240 volts, 0.012 0.017LZt j= + per-unit Power Transformers: 10 kVA, 7200-240 volts, 0.16 0.014PZ j= + per-unit
Single-Phase 120 volt loads: 1 1
2 2
3.0, 0.95 lag5.0, 0.90 lag
kVA PFkVA PF
= =
= =
Single-Phase 240 volt load: 3 38.0, 0.85 lagkVA PF= =
Induction Motor:
25, Line-to-line voltage = 240 volt0.0336 0.08 per-unit0.0394 0.08 per-unit
2.1 per-unit
s
r
m
kVAZ jZ jZ j
=
= +
= +
=
12.47 kV System Unknowns
█For Mathcad Routine@ 39 unknowns
System Voltages: , , , , , , , , , NG an nb bc ca an nb ab bc caV V V V V VL V VL VL VL Transformer Voltages: , , , , , , AN BN CN an nb bc caVT VT VT Vt Vt Vt Vt Voltage Drop: , , , , , A B C a b cv v v v v v Line Currents: , , , , , A B C a b cI I I I I I Transformer Currents: , , , na bn cb acI I I I Load Currents: 1 2 3, , , , , a b cIL IL IL IM IM IM
N
+ +
- -
-
+
+
++
+
a
+
-
+
-
+
-+
-
n
+
-
+
-
-
2 3V
- NGV
Cv
Bv
AvANVT BNVT
CNVT
av
bv
cv
Motor
anVt
nbVt
bcVtacI
bnI
naI
cbI+
_
caVt
c
b
bcV
anV
nbVn I
aIM
cIM
bIM
0Z
2 Z
bZ
cZ
1 Z1ZL
2ZL3ZL
SourceN I
B
G
A
C
+
_+__
+
AGV
CGV
BGV
SW B
12.47 kV
AI
BI
CI
aI
bI
cI
SW A
AGE
BGE
+__
+
CGE
SW C
+__
+
+
WindMil and Mathcad • Beforecheckingfortheopenswitchcondi4onstheMathcadrou4neneededtobeverifiedbyrunningWindMiltogetthecorrectanswersunderafull-loadcondi4on.
• WindMilwasfirstrun.TheMathcadrou4neassumesalinearsystemsotheresultsofWindMilwereusedtocomputethesingle-phaseloadequivalentimpedances.
The Three-Phase System – Full Load WindMil and Mathcad Results
N
+ +
- -
-
+
+
++
+
a
+
-
+
-
+
-+
-
n
+
-
+
-
-
2 3V
- NGV
Cv
Bv
Av
ANVT BNVT
CNVT
av
bv
cv
Motor
anVt
nbVt
bcVt
acIbnI
naI
cbI+
_
caVt
c
b
bcV
anV
nbVnI
cI
bI
aI
aIM
cIM
bIM
0Z
2Z
bZ
cZ
1Z1ZL
2ZL3ZL12.5 kV
Source
BI
AI
NI
CI
B
G
A
C
+
_+__
+
AGV
CGV
BGV
SW A
SW C
SW B
[ ] [ ] [ ]
[ ]
7200/0 7074.7/ 0.4 109.7/2.07200/ 120 , 7228.3/ 120.4 57.7/5.1, 103.1/ 1.57200/120 7221.5/120.4 212.7/0.3
2.51/ 27.61.77/ 172.11.48/108.9
ABC ABC NG
ABC
VS VT V VL
I
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = − = = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
−⎡ ⎤⎢= −⎢⎢⎣
[ ] [ ]
[ ] [ ]
101.0/ 43.7 27.4/ 16.2, 132.4/164.7 , 48 / 5/ 27.4
62.4 / 52.5 27.6/ 31.5
212.7/0.3 44.9/ 71.0221.4/ 117.7 , 53.1/ 172.4223.9/119.4 62.4/52.5
V
abc anbI IL
VM IM
− −⎡ ⎤ ⎡ ⎤⎥ ⎢ ⎥ ⎢ ⎥= = −⎥ ⎢ ⎥ ⎢ ⎥⎥ ⎢ ⎥ ⎢ ⎥−⎦ ⎣ ⎦ ⎣ ⎦
−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
oltage Unbalance = 3.04%, Current Unbalance = 16.69%
• SinceWindMilandMathcadresultsareiden4cal– ConfidencethattheMathcadrou4neiscorrect– WillusetheMathcadrou4netostudytheopenswitchcondi4ons• SW-AOpenwithmotorandsingle-phaseloads• Allotherswitchesopenwithmotoroff
SW-A Open with Motor and Single-Phase Loads
[ ] [ ]117.3/ 21.0 71.7/ 168.6169.3/ 106.0 78.6/ 101.8214.0/107.1 125.6/46.5
Voltage Unbalance = 29.7% Current Unbalance = 36.5%
Negative sequence relay on motor will disco
VM IM− −⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥= − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
nnect motor
SW-A Open With No Motor
[ ] [ ]0 3600/ 180
6235.4/ 90 , 3600/ 180, 7200/ 1206235.4/ 90 7200/120
Note that the LG voltage on the transformer side of the open phase A will be 3600 volts. It is no
ABC NG ABCVT V VS−⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥= − = − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦
[ ]
t zero.
00
207.8/ 90207.8/90
With no 120/240 voltages across the single-phase 120/240 volt loads, the load currents will be zero.
abcVLN
⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥−⎢ ⎥⎣ ⎦
SW-B Open With No Motor
SW-C Open With No Motor
NMSU System • Thesystemwascreatedinordertoanalyzehowtheungroundedwye-deltabankbehavesundernormalandabnormalcondi4ons– Thesystemwasstudiedunderano-loadcondi4onusingaMathcadrou4ne
– ThesystemwassetupintheNMSUpowerlaboratorytochecktheresultsfortheno-loadcondi4on
– Theresultswereverifiedtakingintoaccountthe“ideal”vs.realtransformers
12.47 kV System
• Ungrounded Wye-Delta – Can serve 120/240 volt single-phase loads and a
three-phase induction motor – Problems when one phase is open
• Voltage unbalance on motor will cause the motor relay to trip the motor off • If the open phase is serving the center tap transformer
there will be a nominal line-to-ground voltage of 3600 volts at transformer side of the open switch
12.47 System Conculations (cont.)
– Whentheopenswitchisservingoneofthepowertransformerstheline-to-groundvoltageatthetransformersideoftheopenswitchis19,049volts
– The19,049voltsis2.654mestheratedline-to-groundvoltageofthesystem
Conclusions
• Thecombina4on120/240voltsingle-phaseloadandthree-phaseinduc4onmotorcanbeservedbyagroundedwye-deltaorungroundedwye-deltabank– Groundedwye-deltabank
• Willcon4nuetoservebothloadswithanopenphase• Problemswithbackfeedshortcircuitcurrentonupstreamgroundfaults
Conclusions (cont.) – Ungroundedwye-deltabank
• Cannotservecombina4onloadswithanopenphase• Anopenphaseresultsinaveryhighline-to-groundvoltageatthetransformersideoftheopenswitch
