Current Transformer Sizing
(August 19th, 2010) By: Siva Singupuram James 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 Hall Effect CT Magneto-Optical
Bus Bar
Typical Current Transformers
Introduction Application
Current Transformers (CTs) are instrument transformers that are used to supply a reduced value of current to metering, protective relays, and other instruments. CTs 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 contains
1) 2) 3) 4) Iron Core Secondary windings Primary conductor 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
VSTD ! 20 I N Z STD
For IN = 5A secondary C Class C100 C200 C400 C800 ZSTD () 1.0 2.0 4.0 8.0 VSTD (V) 100 200 400 800
Metering Accuracy Classifications
Metering Accuracy CTs 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: 0.1, 0.2, 0.5, 0.9, 1.8 Volt-amperes (va) equivalent: 2.5va, 5.0va, 12.5va, 22.5va, 45va Typical Number 0.3 B 0.2
Max Ratio Error %
Burden
Ohms (Burden)
Relaying Accuracy Classifications
Relaying Accuracy CTs 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 fullload amplitude.
Class C (C for calculated) is low leakage reactance type typical of donut units Formerly Class L (L for Low Leakage) Class T (T for tested) is high leakage reactance type typical of bar-type units Formerly Class H (H for High Leakage)
Typical Number 10 C 800
10% Max Ratio Error at 20x Rated Current Low Leakage Unit Max secondary voltage developed at 20x rated current without exceeding the +10% ratio error
Available secondary voltages: 100V, 200V, 400V, 800V Burden support (ZSTD): 1.0, 2.0, 4.0, 8.0
CT Knee point
ANSI Method
CT Knee point
ANSI Method
IEC Method
Shorting of CT when not used
CTs 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 CTs 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
Is
s
Ip
s
Ie
Is (Rs jXs ) Is Z LOAD
Hysteresis/Residual Flux
Referring 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)
Based on IEEE survey of 141 cts on a 230kV system Remanent Flux % 0-20 21-40 41-60 61-80 Percentage of cts 39 18 16 27
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))
CT terminal Voltage (Vgh)
JN ! BAN !
vdt !
0
2I F (Z B
RS )(cos [t e
R t L
)dt !
R 1 L Lt L sin[t e ! ! 2I Z RS
[ R R R 2I (Z RS ) [L [L[t [L sin[t e ! ! [ R R R 2I (Z RS ) X X [t X sin[t e ! ! [ R R R X X [t X AN [ ! 2I (Z RS ) sin[t R e R
CT terminal Voltage (Vgh)
BAN[
max
X ! 2 I F ( Z B RS )1 R
Re call : BAN[
max ! VS max
S max
X ! 2 I ( Z RS )1 R X ! 2 I F Z B 1 R X ! I F Z B 1 R
Vgh _ max
Vgh _ max 2
CT terminal Voltage (Vgh)
Vgh _ max 2 Vgh _ max
X ! I F Z B 1 R
VSTD ! 20 I N Z STD
2 X I F Z B 1 e VSTD R Re call : VSTD ! 20 I N Z STD
X I F Z B 1 e 20 I N Z S R
IF ZB I N Z STD
X 1 e 20 R
To Avoid CT saturation
IF ZB I N Z STD X 1 e 20 R
X 1 I f Z b e 20 R
I F ! Max Fault Current Z B ! Burden from the CTs point of view (CTwdg + Cable Leads + Relays) I N ! CT primary rating if IF is given in primary Amps or ! CT secondary rating if IF is given in secondary Amps
Z STD !
If !
Standard Burden = 1, 2, 4, 8 (at 60) Max Fault in per unit of CT secondary nominal Current Burden in per unit of standard burden
Zb !
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 C800 2000/5 (full ratio) If = 3.07 pu Zb = 0.5 pu X/R = 12
X I f Z b e 20 1 R
(1+12)(3.07)(0.5) 20 ? 19.955 20 CT OK!
Example #1 C800 2000/5 (full ratio) If = 3.07 pu Zb = 0.5 pu X/R = 12
Example #2 C800 2000/5 (full ratio) If = 7.69 pu Zb = 0.5 pu X/R = 12
X I f Z b e 20 1 R
X 1 I f Z b e 20 R
(1+12)(3.07)(0.5) 20 ? 19.955 20 CT OK!
(1+12)(7.69)(0.5) 20 ? 49.985 > 20 CT Saturates
Time to Saturate
s
K s 1 ! 1 ln1 X R
X 1 ! [R
Ts ! time to saturate ln ! Natural log function T1 ! primary system time constant K s ! the saturation factor Vx/Vs, where Vx is the excitation voltage at
10A and, Vs is the saturation voltage IF x ZB
[ ! 2f, where f is the system frequency X ! Reactance of the primary system to the point of the fault R ! Resistance of the primary system to the point of the fault
Waveforms
Waveforms
Multi-Ratio CTs
1200/5, C800 tapped at 600/5 C400 effective rating 4.0 effective standard burden
600 EffectiveRating ! 800 ! 400 1200
600 Effective _ Std _ Burden ! 8; ! 4 ; 1200
CT winding Resistance
Typical turn ratio resistance = 2.5m/turn For example a 2000:5 ratio CT; 2000 / 5 !
2000 2 .5 m ; ! 1 .0 ; 5
Lead Resistance Calculation
Formula to determine Copper lead resistance Lead resistance, /1000 = e0.232G-2.32 where, G = AWG number
For a 10AWG cable at (305m = 1000) R = e0.232(10) -2.32 = e0 = 1/1000 Similarly for 8AWG cable at (305m = 1000) R = e0.232(8) -2.32 = 0.629/1000 10AWG cable at (152m = 500) will have a resistance of R = 1/1000x500 = 0.5
CT Application
Type of Fault Connection Wye (connect at CT) Wye (connected at switchhouse) Delta (connected at switchhouse) Delta (connected at CT) 3 ph or ph-to-ph Z = RS + RL + ZR Z = RS + 2RL + ZR Z = RS + 2RL +3 ZR Z = RS + 3RL +3 ZR Ph-to-ground Z = RS + 2RL + ZR Z = RS + 2RL + ZR Z = RS + 2RL +3 ZR Z = RS + 2RL +2 ZR
Z is the effective impedance seen by the CT RS is the CT secondary winding resistance and CT lead resistance; also includes any relay impedance that is inside the delta connection () RL is the circuit one-way lead resistance () ZR is the relay impedance in the CT secondary current path ()
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 Parameters SCC = 7316A @ V=240kV X/R = 12 Transformer size = 50MVA; 240/138kV Zxmfr = 8.0% Solve for X/R at 138kV. Solve for SCC at 138kV.
Example for X/R
Step #1: Calculate the equivalent source impedance
% Z source ! KVAtransformer / KVAshortcircuit
x100
KVAshortcircuit ! 3 kVLL SCC
KVAshortcircu it ! 3 240 kV 7316 A ! 3,041,204.09 kVA % Z source ! 50,000kVA / 3,041,204.09kVA
x100 ! 1.644%
Example for X/R
Step #2: Source R and X
X 5 ! tan R
1
X ! Sin5 * Z R ! Cos5 * Z
%Zsource ! 1.644
X / R ! 12
5 ! tan 12
! 85.24
1 S
X ! Sin(85.24) * (1.644%) ! 1.638% R ! Cos (85.18) * (1.644%) ! 0.138%
Example for X/R
Step #3: Transformer R and X
% Z xmfr ! 8 . 0
X / R ! 15
(From IEEE typical value)
5 ! tan 1 15
! 86.19
X ! Sin(86.19) * (8.0%) ! 7.982% R ! Cos(86.19) * (8.0%) ! 0.531%
Example for X/R
Step #4: Total Resistance and Reactance at 138kV
Rtotal ! Rsource Rxmfr ! 0.138 % 0.531 % ! 0.669 % X total ! X source X xmfr ! 1.638 % 7.982 % ! 9.62%
Step #5: X/R at138kV
X X total 9 . 622 % ! ! ! 14 . 4 R R total 0 . 669 %
Example for SCC
Step #1: Calculate the equivalent source impedance
% Z source ! KVAtransformer / KVAshortcircuit
x100
KVAshortcircuit ! 3 kVLL SCC
KVAshortcircu it ! 3 240 kV 7316 A ! 3,041,204.09 kVA % Z source ! 50,000kVA / 3,041,204.09kVA
x100 ! 1.644%
Example for SCC
Step #2: Calculate the 138kV side full load current
FLA LV !
kVA xmfr 3 kV LL
FLA LV
50 ,000 kVA ! ! 209 .18 A 3 (138 kV )
Example for SCC
Step #3: Calculate the short circuit current on the 138kV side
SCC LV
FLALV 100 ! (% Z xmfr % Z source )
SCC LV
209 .18 A 100 ! ! 2,169 .06 A (8.0% 1.644 %)
Global Example
C200 400/5 (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 valid on the 138kV 240kV 138kV
Global Example
Case #1) From previous slide for X/R and SCC on the 138kV side.
X X 9 . 622 % ! total ! ! 14 . 4 R R total 0 . 669 %
For a C200: ZSTD = 2.0 Rone-way = 1/1000x500 = 0.5
ZB = RS + 2RL + ZR ZB = 0.2 + 2(0.5) + 0.1=1.3 Note: 0.1 = digital & 1.0 for electromechanical
Zb !
1 .3 ; ZB ! ! 0 . 7 pu 2 .0 ; Z STD
Global Example
From previous slide for X/R and SCC on the 138kV side.
SCC LV !
If !
209 .18 A 100 ! 2,169 .06 A (8.0% 1.644 %)
2 ,169 A IF ! ! 5 . 5 pu 400 A IN
X 1 I f Z b e 20 ? R
1 14.4
5.5 0.7 ! 60
CT Saturates
Global Example
K s 1 Ts ! T1 ln1 X R
T1 !
X 14.4 ! ! 0.038s [R 377 2169 A VS ! I F Z B ! (1.3;) ! 35.3V 400 / 5
Vx ! 230V Taken from excitation Curve at IE = 10A (full ratio)
Ks ! V X 230V ! ! 6.33 Vs 35.3V
8.5 1 X ! 19ms } (" 1cycle) Ts ! 0.038 ln1 R
Global Example
Case #2 Choosing C400 (2000/5 tapped at 600/5).
Effective _ Std _ Burden !
1 . 41 ; ZB ! ! 1 . 18 pu 1 .2 ; Z STD 2 ,169 A IF ! ! 3 . 62 pu 600 A IN
600 4 ; ! 1 .2 ; 2000
ZB = 1.4 due to 600/5 ratio CT winding resistance.
Zb !
If !
X 1 I f Z b e 20 ? R
1 14.4
3.62 1.18 ! 66
CT Saturates
Global Example
K s 1 Ts ! T1 ln1 X R
T1 !
X 14.4 ! ! 0.038s [R 377
2169 A VS ! I F Z B ! (1.41;) ! 25.5V 600 / 5
Vx ! 150V
Ks !
Taken from exication Curves and at IE = 10A (600/5)
V X 150V ! ! 5.8 Vs 25.5V
5.8 1 X ! 11ms 1cycle Ts ! 0.038 ln1 R
Global Example
Case #3 Choosing C400 (2000/5 full ratio).
For a C400: ZSTD = 4.0
Zb ! 2 . 1; ZB ! ! 0 . 525 pu ZB = 2.1 due to 2000/5 ratio CT winding resistance. 4 .0 ; Z STD 2 ,169 A IF ! ! 1 . 08 pu 2000 A IN
If !
X 1 I f Z b e 20 ? R
1 14.4
1.08 0.525 ! 8.73
CT OK
Global Example
Case #4 Choosing C400 (2000/5 full ratio with 80% remenant flux).
X 1 I f Z b e 20 * (1 80 %) ! 4 ? R
1 14.4
1.08 0.525 ! 8.73
CT Saturates
Case #4 Choosing C400 (2000/5 full ratio with 50% remenant flux).
X 1 I f Z b e 20 * (1 50 %) ! 10 ? R
1 14.4
1.08 0.525 ! 8.73
CT OK
IEEE PSRC CALCULATOR
Saturation Curve
Vs
mfgr's data log-log plot, equal decade spacing
slope = 1/S
Ve volts rms
Ie
amps rms
10
The Saturation Curve above is represented by the following equation
log V e ! 1 log I e log V i S
Where Vi is the value of Ve for Ie=1, that is for (log Ie=0)
Measuring the S
HOW TO MEASURE THE SLOPE Use a ruler to determine the slope. Do NOT read currents or voltages for this purpose. (We assume equal decade spacing of the axes)
10 0 1 2 3 4 5 6 7 8 9
Ruler, inches or centimeters slope = 0.7/9.3 = 1/S S = 9.3/0.7 = 13
IEEE PSRC CALCULATOR EXAMPLE
Case #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?
