ct sizing
TRANSCRIPT
Current Transformer Sizing
(August 19th, 2010)
By: Siva SingupuramJames Trinh
Agenda
• Current Transformer (CT) introduction• Mathematical Modeling• Saturated Waveforms• CT saturation (Hysteresis Curve)• Remnant Flux, High Fault Currents, DC offset• Saturation Voltage, • CT ratings, Selection and Applications• ElectroMagnetic Transients Program (EMTP)/
PSCAD
Introduction CT types
Donut
Bus Bar
Hall Effect CT Magneto-Optical
Typical Current Transformers
Introduction Application
• Current Transformers (CT’s) are instrument transformers that are used to supply a reduced value of current to metering, protective relays, and other instruments. CT’s provide galvanic isolation from the high voltage primary, permit grounding of the secondary for safety, and step-down the magnitude of the measured current to a value that can be safely handled by the instruments (high fault currents).
• To achieve the above goals CT contains1) Iron Core2) Secondary windings3) Primary conductor4) External insulation
CT Symbols• When current flows in the CT from the H1 lead polarity (±) lead through,
the burden (load), and return to the secondary X2 non-polarity lead. The next half cycle the current will reverse.
ANSI IEC
Current Transformer (CT) Principle
Electrical Schematic 3D-Representation
CT Connections
ANSI Standard Letter Designation• B-The CT is a metering type• C-The CT has low leakage flux and the excitation characteristic can be
used directly to determine performance. (Accuracy can be calculated before manufacturing)
• K-Same as the C rating, but the knee point voltage must be at least 70% of the secondary terminal voltage rating.
• T-The CT can have significant leakage flux (Accuracy must be determined by testing at the factory)
• H-The CT accuracy is applicable within the entire range of secondary currents from 5 to 20 time the nominal CT rating. (Typically wound CTs)
• L-Same as C class except there were two accuracy classes 2.5% and 10%. The ratio accuracy can be up to four times greater than the listed value, depending on connected burden and fault current. (Typically window, busing, or bar-type CTs)
• L and H ANSI rating are applicable to old CT that where manufactured before 1954
ANSI Standard Terminal Voltage Class C CTs
C Class ZSTD (Ω) VSTD (V)
C100 1.0 100
C200 2.0 200
C400 4.0 400
C800 8.0 800
STDNSTD ZIV 20
For IN = 5A secondary
Metering Accuracy Classifications• Metering Accuracy CT’s are used where a high degree of accuracy is
required from low-load values up to full-load of a system. These are utilized by utility companies for revenue metering.
Available in Maximum Ratio Error Classes of: ±0.3%, ±0.6%, ±1.2%, ±2.4%
For Burdens (Loads) of: 0.1Ω, 0.2Ω, 0.5Ω, 0.9Ω, 1.8Ω
Volt-amperes (va) equivalent: 2.5va, 5.0va, 12.5va, 22.5va, 45va
Typical Number
0.3 B 0.2
Burden Ohms (Burden)Max Ratio Error ±%
Relaying Accuracy Classifications• Relaying Accuracy CT’s are used for supplying current to protective relays.
In this application, the relays do not normally operate in the normal load range, but they must perform with a reasonable degree of accuracy at very high overload and fault-current levels which may reach 20x the full-load amplitude.
Class C (C for calculated) is low leakage reactance type – typical of donut units – Formerly Class L (L for Low Leakage)
Typical Number
10 C 800
Low Leakage UnitMax secondary voltage developed at 20x rated current without exceeding the +10% ratio error
10% Max Ratio Error at 20x Rated Current
Class T (T for tested) is high leakage reactance type – typical of bar-type units – Formerly Class H (H for High Leakage)
Available secondary voltages: 100V, 200V, 400V, 800V
Burden support (ZSTD): 1.0Ω, 2.0Ω, 4.0Ω, 8.0Ω
CT Knee point
ANSI Method
CT Knee point
ANSI MethodIEC Method
Shorting of CT when not used
• CT’s can be very dangerous devices! Stepping down the current means that the voltage is automatically stepped up the same ratio. The secondary's of the CT’s are, therefore; always kept shorted when not in use and/or when no low impedance ammeter is connected! On the open-circuited secondary of a CT, just a few volts on the primary voltage would become a lethal thousands of volts on the secondary.
CT Equivalent Circuit
LOADsssss ZIjXRIV )(
es
ps I
N
II
Hysteresis/Residual FluxReferring to figure on the left, if H is gradually increased from zero to H1, the magnetization follows path oa. Decreasing H back to zero will leave the material magnetized at a B-value corresponding to point b. Distance ob on the B axis is called the “Residual Flux Density”. To bring B back to zero requires a negative value of H corresponding to point c on the cure. Distance oc on the H-axis is called the Coercive Force. Decreasing H to H2=-H1 results in the B-value of point d. Finally increasing H back to the original H1 now causes the magnetization to traverse path defa. The latter path is nearly equivalent to path abcd, but flipped about the B and H axes. Path abcdefa is called a hysteresis loop which occurs when an AC voltage is applied to a coil on a ferromagnetic core. As the voltage and current cycle between their max and min values, the core magnetization repeatedly traverses a hysteresis loop.
Remanent Flux (Information Only)
Remanent Flux % Percentage of cts
0-20 39
21-40 18
41-60 16
61-80 27
Based on IEEE survey of 141 cts on a 230kV system
DC offset
Saturation Occurs
CT Errors impact
Errors to in currents magnitude and angle will have a significant effect on protection relays particularly (directional (67), distance (21), and differentials (87))
t tL
R
SBF
tdtetRZIvdtBANN
00))(cos(2
R
Le
R
LtRZI
tL
R
SBF
sin1
2
R
Xe
R
Xt
RZI tX
RSBF
sin
)(2
R
Xe
R
XtRZIBAN
tX
R
SBF
sin)(2
R
Le
R
Lt
RZI tL
RSBF
sin)(2
CT terminal Voltage (Vgh)
maxmax SVBAN
R
XRZIBAN SBF 1)(2max
:Recall
R
XRZIV SBFS 1)(2max
R
XZIV BFgh 12max_
R
XZI
VBF
gh 12max_
CT terminal Voltage (Vgh)
R
XZI
VBF
gh 12max_
STDNSTDgh ZIV
V20
2max_
STDBF VR
XZI
1
STDNSTD ZIV 20:Recall
STDNBF ZIR
XZI 201
201
R
X
Z
Z
I
I
STD
B
N
F
CT terminal Voltage (Vgh)
201
R
X
Z
Z
I
I
STD
B
N
F 201
bf ZIR
X
CT primary rating if IF is given in primary Amps orNI
Standard Burden = 1, 2, 4, 8 Ω (at 60°)STDZ
Max Fault in per unit of CT secondary nominal CurrentfI
Burden in per unit of standard burdenbZ
CT secondary rating if IF is given in secondary Amps
Max Fault CurrentFI Burden from the CT’s point of view (CTwdg + Cable Leads + Relays)BZ
To Avoid CT saturation
Limitation of CT Selection Criterion
• We face high X/R ratio and high fault currents near generation sites. This becomes impractical to size the CT to avoid saturation during a asymmetrical fault. The formula derived earlier should not be used for this study and we must size the CT with reasonable sensitivity for line end faults
• For this situation we should analyze in further detail via EMTP simulation.
Example #1
C8002000/5 (full ratio)If = 3.07 puZb = 0.5 puX/R = 12
19.955 ≤ 20
201
bf ZIR
X
(1+12)(3.07)(0.5) ≤ 20 ?
CT OK!
Example #1
C8002000/5 (full ratio)If = 3.07 puZb = 0.5 puX/R = 12
201
bf ZIR
X
19.955 ≤ 20
Example #2
C8002000/5 (full ratio)If = 7.69 puZb = 0.5 puX/R = 12
49.985 > 20
201
bf ZIR
X
(1+12)(3.07)(0.5) ≤ 20 ? (1+12)(7.69)(0.5) ≤ 20 ?
CT OK! CT Saturates
RX
K
TT
s
s
1
1ln1 R
XT
1
primary system time constant1T the saturation factor Vx/Vs, where Vx is the excitation voltage at 10A and, Vs is the saturation voltage IF x ZB
sK
Resistance of the primary system to the point of the faultR
Natural log functionln time to saturatesT
Time to Saturate
Reactance of the primary system to the point of the faultX2πf, where f is the system frequency
Waveforms
Waveforms
Multi-Ratio CTs
1200/5, C800 tapped at 600/5C400 effective rating4.0 Ω effective standard burden
4008001200
600atingEffectiveR
481200
600__ BurdenStdEffective
CT winding Resistance
Typical turn ratio resistance = 2.5mΩ/turn
0.15.25
20005/2000 mFor example a 2000:5 ratio CT;
Lead Resistance CalculationFormula to determine Copper lead resistanceLead resistance, Ω/1000’ = e0.232G-2.32
where, G = AWG number
For a 10AWG cable at (305m = 1000’)R = e0.232(10) -2.32 = e0 = 1Ω/1000’
R = 1Ω/1000’x500’ = 0.5Ω10AWG cable at (152m = 500’) will have a resistance of
Similarly for 8AWG cable at (305m = 1000’)R = e0.232(8) -2.32 = 0.629Ω/1000’
CT Application
ConnectionType of Fault
3 ph or ph-to-ph Ph-to-ground
Wye (connect at CT) Z = RS + RL + ZR Z = RS + 2RL + ZR
Wye (connected at switchhouse) Z = RS + 2RL + ZR Z = RS + 2RL + ZR
Delta (connected at switchhouse) Z = RS + 2RL +3 ZR Z = RS + 2RL +3 ZR
Delta (connected at CT) Z = RS + 3RL +3 ZR Z = RS + 2RL +2 ZR
Z is the effective impedance seen by the CTRS is the CT secondary winding resistance and CT lead resistance; also
includes any relay impedance that is inside the delta connection (Ω)RL is the circuit one-way lead resistance (Ω)ZR is the relay impedance in the CT secondary current path (Ω)
IEEE typical X/R Ratios
IEEE typical X/R Ratios
IEEE typical X/R Ratios
IEEE typical X/R Ratios
IEEE typical X/R Ratios
Parameters for X/R & SCC Example Given ParametersSCC = 7316A @ V=240kVX/R = 12Transformer size = 50MVA; 240/138kVZxmfr = 8.0%
Solve for X/R at 138kV.Solve for SCC at 138kV.
Example for X/R
%644.110009.204,041,3/000,50% xkVAkVAZ source
Step #1: Calculate the equivalent source impedance
SCCkVKVA LLitshortcircu 3
100/% xKVAKVAZ itshortcircurtransformesource
kVAAkVKVA itshortcircu 09.204,041,373162403
Example for X/RStep #2: Source R and X
644.1% sourceZ
12/ RX
24.8512tan 1
%638.1%)644.1(*)24.85( SinX
%138.0%)644.1(*)18.85( CosR
R
X1tan ZSinX *ZCosR *
Example for X/RStep #3: Transformer R and X
0.8% xmfrZ
15/ RX
19.8615tan 1
%982.7%)0.8(*)19.86( SinX
%531.0%)0.8(*)19.86( CosR
(From IEEE typical value)
Example for X/RStep #4: Total Resistance and Reactance at 138kV
%669.0%531.0%138.0 xmfrsourcetotal RRR
%62.9%982.7%638.1 xmfrsourcetotal XXX
Step #5: X/R at138kV
4.14%669.0
%622.9
total
total
R
X
R
X
Example for SCC
%644.110009.204,041,3/000,50% xkVAkVAZ source
Step #1: Calculate the equivalent source impedance
SCCkVKVA LLitshortcircu 3
100/% xKVAKVAZ itshortcircurtransformesource
kVAAkVKVA itshortcircu 09.204,041,373162403
Example for SCCStep #2: Calculate the 138kV side full load current
LL
xmfrLV
kV
kVAFLA
3
AkV
kVAFLALV 18.209
)138(3
000,50
Example for SCCStep #3: Calculate the short circuit current on the 138kV side
)%(%
100
sourcexmfr
LVLV ZZ
FLASCC
AA
SCCLV 06.169,2%)644.1%0.8(
10018.209
Global Example
C200400/5 (full ratio)IF = 7,316A (HV side)X/R = 15 (HV side faults)One-way RL length = 152m (500’)Determine what CT Class and Ratio is validon the 138kV
240kV 138kV
Global Example
4.14%669.0
%622.9
total
total
R
X
R
X
Case #1) From previous slide for X/R and SCC on the 138kV side.
Rone-way = 1Ω/1000’x500’ = 0.5Ω
For a C200: ZSTD = 2.0Ω
ZB = RS + 2RL + ZR
ZB = 0.2Ω + 2(0.5Ω) + 0.1Ω=1.3
Note: 0.1Ω = digital & 1.0 Ω for electromechanical
puZ
ZZ
STD
Bb 7.0
0.2
3.1
Global Example
AA
SCCLV 06.169,2%)644.1%0.8(
10018.209
?201
bf ZIR
X
From previous slide for X/R and SCC on the 138kV side.
puA
A
I
II
N
Ff 5.5
400
169,2
607.05.54.141 CT Saturates
Global Example
RX
K
TT
s
s
1
1ln1
sR
XT 038.0
377
4.141
33.63.35
230
V
V
V
VK
s
Xs
VVx 230
VA
ZIV BFS 3.35)3.1(5/400
2169
Taken from excitation Curve at IE = 10A (full ratio)
)1(19
15.8
1ln038.0 cyclemsRXTs
Global ExampleCase #2 Choosing C400 (2000/5 tapped at 600/5).
puZ
ZZ
STD
Bb 18.1
2.1
41.1
2.142000
600__ BurdenStdEffective
puA
A
I
II
N
Ff 62.3
600
169,2
?201
bf ZIR
X
6618.162.34.141 CT Saturates
ZB = 1.4Ω due to 600/5 ratio CT winding resistance.
Global Example
RX
K
TT
s
s
1
1ln1
sR
XT 038.0
377
4.141
8.55.25
150
V
V
V
VK
s
Xs
VVx 150
VA
ZIV BFS 5.25)41.1(5/600
2169
Taken from exication Curves and at IE = 10A (600/5)
cyclemsRXTs 111
18.5
1ln038.0
Global ExampleCase #3 Choosing C400 (2000/5 full ratio).
puZ
ZZ
STD
Bb 525.0
0.4
1.2
puA
A
I
II
N
Ff 08.1
2000
169,2
?201
bf ZIR
X
73.8525.008.14.141 CT OK
For a C400: ZSTD = 4.0Ω
ZB = 2.1Ω due to 2000/5 ratio CT winding resistance.
Global ExampleCase #4 Choosing C400 (2000/5 full ratio with 80% remenant flux).
?4%)801(*201
bf ZIR
X
73.8525.008.14.141 CT Saturates
Case #4 Choosing C400 (2000/5 full ratio with 50% remenant flux).
?10%)501(*201
bf ZIR
X
73.8525.008.14.141 CT OK
IEEE PSRC CALCULATOR
Vs
a m p s rm s
slope= 1/S
lo g - lo g p lo t ,e q u a l
d e c a d es p a c in g
vo l tsrm s
I e
mf gr 'sd a taV e
1 0
Saturation Curve
iee VIS
V loglog1
log
The Saturation Curve above is represented by the following equation
Where Vi is the value of Ve for Ie=1, that is for (log Ie=0)
Measuring the ‘S’
0
1
2
3
4
5
6
7
8
9
1
0
slope = 0.7/9.3 = 1/S
S = 9.3/0.7 = 13
Ruler, inches or centimeters
Use a ruler to determine the slope.Do NOT read currents or voltages for this purpose.(We assume equal decade spacing of the axes)
HOW TO MEASURE THE SLOPE
IEEE PSRC CALCULATOR EXAMPLECase #1) From previous slide for X/R and SCC on the 138kV side.
EMTP/PSCAD SIMULATION
• The CTs are subject to saturation during DC transient of fault current , hence there is growing interest in simulations like EMTP/PSCAD.
• The IEEE-PSRC CT calculator is only accurate at high fault currents and burden , hence to get better idea of CT performance at low fault currents/ burden we need to model in PSCAD/EMTP.
• The primary aim of PSCAD simulation is to obtain digitized records, which can be recognized as secondary analog signal using D/A conversion and amplification, for the purpose of relay testing.
EMTP/PSCAD SIMULATION
PSCAD -CT Modeling
QUESTIONS?