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CT VT

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Basics of
Current and Voltage Transformers
Power Transmission and Distribution
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Page 241
2007-08
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Copyright © Siemens AG 2007. All rights reserved.
PTD SE PTI
Equivalent current transformer circuit
I1
jX1 R1
I 2′ = I1 ⋅
N1
N2
jX2
Im
P1
N1
N2 U
R2
I2
S1
Zm
Zb
2
P2
Ideal CT
S2
X1 = Primary leakage reactance
R1 = Primary winding resistance
X2 = Secondary leakage reactance
Z0 = Magnetizing impedance
R2 = Secondary winding resistance
Zb = SecondaryAuthor
load
Note:
Page 242
2007-08
Power Transmission and Distribution
Normally the leakage fluxes X1 and X2 can be neglected
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Copyright © Siemens AG 2007. All rights reserved.
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Current transformer, simplified equivalent circuit
jX2
1 : N2
i2
i1
i'1 =
N2
1
R2
⋅ i1
LW
L2<< LW
im
ZB
R2
i2
Power Transmission and Distribution
RB
Author
Page 243
2007-08
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PTD SE PTI
Current transformer:
Phase displacement (δ) and current ratio error (ε)
I1 .
I1
N1
N2
ε
N1
I2
N2
I2
δ
Power Transmission and Distribution
Zb
Author
Im
Page 244
2007-08
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PTD SE PTI
CT classes to IEC 60044-1: 5P or 10P
Pi = I sec .2 × R CT
Specification:
300/1 A
5P10, 30 VA
Ratio In -Prim / In -Sek.
5% accuracyat I= n x In
Actual accuracy limit factor
in operation is higher as the CT
ALF ' = ALF ×
is normally underburdened :
Operating ALF: ALF‘
RCT ≤ 5 Ohm
Rated burden
(nominal power) PNB
Accuracy
limit factor ALF
Pi + PNB
Pi + PBB
Dimension criterium:
I
ALF ' ≥ SC − max × KTF
Power Transmission
I n and Distribution
KTF (over-dimensioning factor) considers the single sided CT over-magnetising due to the d.c.
Author
component in short circuit current ISC.
KTF values required in practice depend on relay type and design.
Recommendations are provided by manufacturers (see Application Guides)
Page 245
2007-08
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PTD SE PTI
Definition of the CT knee-point voltage
U2
10 %
IEC 60044-1 Class PX
UKN
formerly
50 %
British Standard BS3938:
Class X
Power Transmission and Distribution
Author
Im
Page 246
2007-08
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Copyright © Siemens AG 2007. All rights reserved.
PTD SE PTI
Current transformer, Standard for steady-state performance
IEC 60044-1 specifies the following classes:
Accuracy class
Current error
at nominal
current (In)
5P
±1%
10P
± 5%
Angle error δ
at rated current
In
± 60 minutes
Total error at n x In
(rated accuracy limit)
5%
10 %
Power Transmission and Distribution
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Page 247
2007-08
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PTD SE PTI
Current transformers, Standard for transient performance
IEC 60044-6 specifies four classes:
Class
Error at rated current
Ratio error
TPX
(closed iron core)
TPY
with anti-remanence
air gap
TPZ
linear core
TPS
2007-08
Remanence
Angle error
± 0,5 %
± 30 min
εˆ ≤ 10%
± 1,0 %
± 30 min
εˆ ≤ 10%
± 1,0 %
± 180 ± 18 min
εˆ ≤ 10%
Author
no limit
< 10 %
Power Transmission and Distribution
(a.c. current only)
Special version for
high impedance protection
Author
(Knee point voltage, internal secondary resistance)
closed iron core
Page 248
Maximum error at
rated accuracy limit
negligible
No limit
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Current transformer saturation
Steady-state saturation with AC current
Power Transmission and Distribution
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Transient saturation with offset current
Page 249
2007-08
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Transient CT saturation due to DC component
ISC
Short circuit current
DC flux non-saturated
ΦS
Φ
AC
flux
Power
Transmission and Distribution
Author
Im
Page 250
Magnetising current
2007-08
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Course of CT-flux during off-set short-circuit current
ΙP primary current
t
t
−
B
ω ⋅TN ⋅TS − TN
= 1+
(e
− e TS )
TN − TS
Bˆ
d.c. component
~
TS
N TS TN
⎛T ⎞ −
BMax
= 1+ ω ⋅TS ⋅ ⎜⎜ ⎟
B~
⎝ TS ⎠
t
TN
total flux
B
t B Max =
BMax
transient
d.c. flux
B~
TS =
a.c. flux
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t
B~
Page 251
2007-08
Author
TN ⋅ TS
TS
⋅ ln
TS − TN TN
LW Power Transmission
1
and Distribution
=
Ri + RB ω ⋅ tanδ
For 50 Hz:
TS =
10900
δ [min ]
[ms]
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CT transient over-dimensioning factor KTF
KTF
6
0
⎛ TN ⎞
KTF = 1+ ωTS⎜⎜ ⎟⎟
⎝ TS ⎠
TS
TS − TN
∞ (KTF ≈1+ωTN)
5000
5
0
Closed iron core
1000
4
0
TS [ms]
500
3
0
250
2
0
100
Linear core
1
0
Power Transmission and Distribution
TN = network time constant
(short-circuit time constant)
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50
Page 252
2007-08
100
150
200
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TN [ms]
TS = CT secondary time constant
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CT with closed iron core,
Over-dimensioning factor KTF‘ for specified time to saturation (tM)
15
14
tM
tM → ∞
K‘td
50 ms
40 ms
10
30 ms
20 ms
6
5
10 ms
Page 253
20
2007-08
40
60
80
Tp
Author
⎞
⎟
⎟
⎟⎟
⎠
Power Transmission and Distribution
tM → ∞ :
KTF = 1+ ωTN = 1+
Author
0
K' TF
tM
⎛
⎜
= 1 + ωTN ⋅ ⎜1 - e T N
⎜⎜
⎝
XN
RN
100 [ ms]
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CT over-dimensioning factor KTF (tM,TN)
in the case of short time to saturation (tM)
KTF
tM
4
Θ (tM,TN)
(el. degree) 160
10 ms
3.5
140
3
120
8 ms
2.5
tM
100
2
10 ms
8 ms
6 ms
Power Transmission and Distribution
4 ms
2 ms
80
6 ms
1.5
1
60
40
4 ms
0.5
0
180
2 ms
0
Page 254
20
40
60
80
2007-08 T
N
Author
0
100
in ms
20
Author
0
20
40
60
80
100
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TPTD
msPTI
N inSE
Current transformer
magnetising and de-magnetising
ΙP
B
t
BMax
B
B = BR + (BMax. − BR ) ⋅ e
BR
−
BMax
t
TS
BR
Power Transmission and Distribution
Ιm
Author
t
Page 255
2007-08
Author
Ιm
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Current transformer
Course of flux in the case of non-successful auto-reclosure
B
Bmax
tF1 = duration of 1st fault
t
tF1
tDT
tF2
tDT dead time
Power Transmission and Distribution
tF2 =duration of 2nd fault
tF1
tDT+tF2 ⎡
tF2
tF1
tF2
⎡
− ⎤ Author
−
− ⎤
−
⋅
N ⋅TS
Bmax ⎢ ω ⋅TN ⋅TS − TN
T
ω
= 1+
− e TS ⎥ ⋅ e TS + ⎢1+
(e
(e TN − e TS ⎥
⎢
⎢
⎥
⎥
TN −TS
TN −TS
Bˆ~
⎢⎣
⎢⎣
⎥⎦
⎥⎦
Page 256
2007-08
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Current transformer
magnetising curve and point of remanence
I
B
II
III
up to 80%
< 10%
negligible
H = im ⋅ w
Power Transmission and Distribution
I: closed iron core (TPX)
II: core with anti-remanence air-gaps (TPY)
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Page 257
2007-08
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III: Linearised core (TPZ)
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Current transformers TPX und TPY
Course of the flux with non-successful auto-reclosure
BR
BR
closed iron core (TPX)
Power Transmission and Distribution
core
with antiremanence air-gaps (TPY)
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tDT
tF1
Page 258
2007-08
tF2
Author
t
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Standards of voltage transformers
VT classes to IEC 60044-2
Class
designation
Permissible error at 0.05 ·UN and 1.0 · UN
Voltage error FU
Angle error δ
3P
± 3.0 %
120 minutes
5P
± 6.0 %
240 minutes
All 3P and 5P protection CTs must
additionally comply with one of the
below VT metering classes!
VT classes for measurement IEC 60044-2
Class
designation
Page 259
Permissible voltage error in %
at 1.0 · UN
0.1
0.2
0.1
0.2
0.5
0.5
1
3
1
3
2007-08
Permissible angle error in minutes
at 1.0·UN
Power Transmission and Distribution
5
Author
Author
10
20
30
Not determined
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Capacitive voltage transformer,
Equivalent circuit
400 kV
C1 = 5 nF
C2 = 95 nF
3
RF
20 kV
LF
3
CE = C 1 + C 2
C1
U*1 = UP ⋅
C1 + C2
Page 260
2007-08
100 V
CF
3
LO
RE
C
T
Author
Author
ZB
Power Transmission and Distribution
RF
RT
LT
LF
CF
U*2
R'B
L'B
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Transient performance of CVTs,
Recommendations acc. to IEC 60044-5
Ratio
Measured voltage transients after fault at
voltage maximum and zero-crossing
(Example)
Time
Ts
in ms
t
UP(t) Primary voltage
2 ⋅ Us
⋅100%
Classes
3PT1
6PT1
3PT2
6PT2
3PT3
6PT3
10
---
≤ 25
≤4
20
≤ 10
≤ 10
≤2
40
≤ 10
≤ 2
≤2
60
≤ 10
≤ 0.6
≤2
≤ 10
≤ 0.2
≤2
90
Legend:
Us (t)
Power Transmission and Distribution
Recommendations to IEC 60044-5
US(t) Secondary voltage
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Page 261
2007-08
(1)
Fault inception
(2)
Aperiodic damping of US(t)
(3)
Periodic damping of US(t)
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CT dimensioning
K OD = KTF
Pi + PBN
RCT + RBN
ALF' = ALF ⋅
= ALF ⋅
Pi + PB
RCT + RB
Actual connected burden : PB= RB ⋅ I2N2
P + PB
RCT + RB
= ALF'⋅
ALF = ALF'⋅ i
Pi + PBN
RCT + RBN
with
I
ALF ' ≥ K OD ⋅ K
IN
Theory:
K OD ≥ K TF ⋅ K Re m
K Re m =
1
% Remanence
1−
100
Page 262
2007-08
rated CT burden:
PBN
internal burden of the CT: Pi=Ri ⋅ I2N2
RB=Rl+RR= burden resistance
Rl= resistance of connecting cables
RR= burden resistance of the relay
No saturation
for the total
short-circuit duration:
No saturation for
the specified time tM:
Practice:Author
K' TF =
BMax
XN
= 1 + ωTN = 1 +
RN
Bˆ ~
Power
tM ⎤
⎡ Transmission andtMDistribution
−
−
ω
T
N ⋅ TS
⋅
(e TN − e TS ⎥
K ' 'TF = ⎢1 +
TN − TS
⎥
⎢
⎣
⎦
Remanence only considered in extra high voltage systems (EHV)
KTF-values acc. to relay manufacturers‘ guides
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Practical CT requirements
Transient over-dimensioning factors KTF (AR not considered)
Distance
7SA6 and 7SA522
Overcurrent
7SJ56
Close-in fault:
KTF ≥ 5
Fault atbalance point:
KTF ≥ 2 ( ≥ 1 if TN < 30ms)
ALF‘≥ I>>setting/ IN, at least 20
Internal fault
External fault
Transformer Differential
7UT6
KTF≥ 0.75
(Saturation free
time ≥ 4 ms)
KTF≥ 1.2
(Saturation free
time ≥ 5 ms)
Line differential
7SD61
KTF≥ 0.5
(Saturation free
time ≥ 3 ms)
KTF≥ 1.2
Power Transmission and Distribution
(Saturation free
time ≥ 5 ms)
Bus differential
7SS52
Page 263
Author
KTF≥ 0.5
(Saturation free
time ≥ 3 ms)
2007-08
Author
KTF≥ 0.5
(Saturation free
time ≥ 3 ms)
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CT dimensioning for
Example differential protection (1)
1. Calculation of fault currents
110 kV, 3 GVA
F1
110/20 kV
40 MVA
uT=12%
F2
OH-line:
l = 8 km,
zL‘= 0,4 Ω/km
F3
F4
Net
300/1A
∆ IT
7UT61
Transf. :
∆ IL
7SD61
∆ IL
Impedances related to 20 kV:
Impedances related to 110 kV:
Net :
200/1A
200/1A
1200/1A
U N 2 ⎡ kV 2 ⎤
⎥⎦ 110 2
⎢⎣
ZN =
=
= 4.03 Ω
SSC ' ' [MVA ] 3000
U N 2 ⎡ kV 2 ⎤
Author
⎥⎦ uT [% ] 110 2 12 %
⎢⎣
ZT =
⋅
=
⋅
= 36.3 Ω
PN -T [MVA ] 100
40 100
Net :
Transf. :
U N 2 ⎡ kV 2 ⎤
⎢⎣ Power⎥⎦ Transmission
20 2
and Distribution
ZN =
=
= 0.13 Ω
SSC ' ' [MVA ] 3000
U N 2 ⎡ kV 2 ⎤
⎢⎣
⎥⎦ uT [% ] 20 2 12 %
ZT=
⋅
=
⋅
= 1.2 Ω
PN -T [MVA ] 100
40 100
Line : Z L = l[km ]⋅ z L ' [Ω/km ] = 8 ⋅ 0,4 = 3,2 Ω
Page 264
2007-08
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CT dimensioning for
Example differential protection (2)
1.1 ⋅ U N / 3 1.1 ⋅ 110kV/ 3
=
= 17.3 kA
ZN
4.03Ω
F1
I F1 =
F2
1.1 ⋅ U N / 3 1,1 ⋅ 110kV/ 3
I F2 =
=
= 1.73 kA
Z N + ZT
4.03Ω + 36.3Ω
F3
F4
I F3 =
I F4 =
1.1 ⋅ U N / 3 1.1 ⋅ 20kV/ 3
=
= 9.55 kA
Z N + ZT
0.13Ω + 1.2Ω
1.1 ⋅ U N / 3
1,1 ⋅ 20kV/ 3
=
= 2.8 kA
Z N + Z T + Z L 0.13Ω + 1.2Ω + 3.2 Ω
Dimensioning of the 110 kV CTs for the transformer differential protection:
Manufacturer recommends for relay 7UT61:
The saturation free time of 3 ms
corresponds to KTF≥ 0,75
See diagram, page 59
Power Transmission and Distribution
Criterion 1) therefore reads:
I
17300
ALF' ≥ K TF ⋅ F1 = 0,75 ⋅
= 43
IN
300
1) Saturation free time ≥ 4ms for internal faults
2) Over-dimensioning factor KTF ≥ 1,2
for through flowing currents (external faults)
For criterion 2) we get:
Author
I
1730
ALF' ≥ K TF ⋅ F2 = 1,2 ⋅
=7
IN
300
The 110 kV CTs must be dimensioned according to criterion 1).
Page 265
2007-08
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PTD SE PTI
CT dimensioning for
Example differential protection (3)
We try to use a CT type: 300/1, 10 VA, 5P?, internal burden 2 VA.
Pi + Poperation
2 + 2.5
ALF ≥
⋅ ALF ' =
⋅ 43 = 16.1
Pi + Prated
2 + 10
(Connected burden estimated to about 2.5 VA)
Chosen, with a security margin : 300 /1 A, 5P20, 10 VA, R2≤ 2 Ohm (Pi ≤ 2VA)
Specification of the CTs at the 20 kV side of the transformer:
It is good relaying practice to choose the same dimensioning as for the CTs on the 110 kV side:
1200/1, 10 VA, 5P20, R2≤ 2 Ohm (Pi ≤ 2VA)
Dimensioning of the 20 kV CTs for line protection:
1‘) Saturation free time ≥ 3ms for internal faults
2‘) Over-dimensioning factor KTF ≥ 1.2
for through flowing currents
(external
faults)
Power
Transmission
and Distribution
For relay 7SD61, it is required:
The saturation free time of 3 ms corresponds
to KTF≥ 0.5
See diagram, page 59
Criterion 1‘) therefore reads:
I
9550
ALF' ≥ K TF ⋅ F3 = 0.5 ⋅
= 24
IN
200
Page 266
2007-08
Author
For criterion 2‘) we get:
I
2800
ALF' ≥ KTF ⋅ F4 = 1.2 ⋅
= 16.8
IN
200
Author
The 20 kV line CTs
must be dimensioned
according to criterion 1‘).
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CT dimensioning for
Example differential protection (4)
For the 20 kV line we have considered the CT type: 200/5 A, 5 VA, 5P?, internal burden ca. 1 VA
Pi + Poperation
1+1
⋅ ALF ' =
⋅ 24 = 8
ALF ≥
1+ 5
Pi + Prated
(Connected burden about 1 VA)
Specification of line CTs:
We choose the next higher standard accuracy limit factor ALF=10 :
Herewith, we can specify: CT Type TPX, 200/5 A, 5 VA, 5P10, R2≤ 0.04 Ohm ( Pi≤ 1 VA)
Power Transmission and Distribution
Author
Page 267
2007-08
Author
Copyright © Siemens AG 2007. All rights reserved.
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