Selection of Current Transformers and Wire Sizing in Substations Sethuraman Ganesan ABB Inc. Allentown, PA
Presented to:
59th Conference for Protective Relay Engineers Texas A&M University College Station, Texas April 4-6, 2006 1
Discussion Paper
Characteristics of CT Metering and Protection Class Specifications of CTs CT Wiring and other issues IEEE Std C57.13, Guide C37.110 IEC Std 60044-6
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CT Simplified Circuit and Phasor
IP
IP /n
1:n a
IS
e RCT
c Xm
RB
IE
d
b
f Vef ISRCT
IS
Vcd=n. Vab
IP n
IE 3
Metering
Metering class Typical Spec 0.3 B-0.1
Meters can be off Protection CTs • Thermal stress • Auxiliary CTs • Burdens of auxiliary CTs, accuracy
Summation CTs
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Protection Class CTs
Ratings, Ratio
Polarity
Class, Knee point voltage, Excitation characteristics Vx
g n i z i e t e g a n t l g a o V M
Vk
10A(10%) Secondary Current 5
AC Saturation
Severe Saturation Too large CT secondary burden, currents Ideal
Actual
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CT ratings to avoid AC saturation
Vx > If (RCT+RL+RB) Vx = Saturation Voltage If
= CT secondary current during fault
RCT= CT Secondary Resistance- Ohms RL = CT lead Resistance- Ohms RB = CT Connected burden ResistanceOhms
7
CT Transient Saturation
Caused by DC Transients in the power system 2 1 t n e r r 0 u C
DC AC 1
2
-1
-2
Cycles
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CT Transient Saturation (Minimum Math!)
i
= current , v = voltage Φ = Flux in the magnetic core, all instantaneous;
i
α v α (d Φ /d t)
where d Φ /d t represents the rate of change of flux.
i α (d Φ /d t) Integrating,
∫i α Φ Rewriting,
Φ α ∫i (Flux is decided by area under the time function ‘i’) 9
Flux during AC currents v α i α dΦ/dt
Φ Φα
∫i
10
Flux during DC Transients v α i α dΦ/dt
Φ Φα
∫i
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CT ratings to avoid Transient saturation
Vx > If (1+X/R) (RCT+RL+RB) Where, X, R= Primary system reactance and resistances Avoiding CT saturation may not always be possible.
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Waveforms during AC+DC Transients Ideal CT secondary current
I
Actual CT secondary current
Time DCΦ (Ideal CT)
Φ
Saturation Φ
AC+DC Actual Φ in CT
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Typical ANSI Class C CT
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Specifications for ANSI CTs
Classification Letter C, K or T
C – Performance can be ‘C’alculated, low leakage reactance K- Same as C but with Knee point 70% of secondary terminal voltage T- Performance to be ‘T’ested
Recommended maximum secondary current 100A
Error max: 10% at 100A, so 10A error
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Class C CTs
C800- Develops >800V
At 100A current
At connected burden of 8 Ohms
Internal voltage > 800 + 100RCT
Burden 1,2,4,8 Ohms for C100, C200, C400,C800 etc (RBx100A = C Volt Rating)
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Class C CTs
To avoid AC Saturation, in C800,
100(RCT+ 8) > If (RCT+RL+RB) Typically
If < 100A
Connected burden RL+RB < 8 Ohms
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Class C CTs
To avoid DC Saturation, in C800,
100(RCT+ 8) > If (1+X/R)(RCT+RL+RB) Normally If < 100A, Connected burden is less than design burden; Define Ni = 100/ If ( Ideally >1) Define Nr = (RCT+ 8) / (RCT+RL+RB ) (Ideally >1)
The equation above becomes Ni. Nr > (1+X/R)
In other words CT saturation is avoided if (1+X/R) < Ni. Nr 18
Remanence
Remanence, Residual flux
Similar to permanent magnetism
Reduces available ‘excursion’ of flux to translate currents
If ψ is the per unit of maximum flux remaining as residual flux, CTs have to be oversized by a factor 1/(1- ψ) If ψ = 0.9, the above factor is 10, that bigger CT is required!!! 19
Remanence
Reduce
Gap in the steel core
Different core materials
Biased core
Account for remanence
Increase the CT size- Not an option always
Reduce the burdens, leads etc.
Make the relay faster- to operate before CT saturation starts
Increased slope
Special relays with algorithms 20
CT- Time to saturate
2
t/T 1
0
0
1
2
3
4
5
6
Vx / (IRT) Vx = Saturation Volts I = Symm. Secy Current, A R = Secy. Circuit Resist, Ω Ie = Exciting Current, A T = Primary Circuit Time Constant, Cycles t = Time to saturate in Cycles
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Calculating Lead Resistance - Example
Data
CT C400, 1000/5A, RCT = 0.25 Ohms
Fault Primary = 10kA at X/R = 15
Relay burden = Negligible
Calculations: If = 10000/CTR = 10000/200 = 50A Ni= 100/ 50 = 2 Nr = 4.25/(0.25+RL) Checking for adequacy, (1+X/R) > Ni.Nr (1+15) > 2 x 4.25 /(0.25 + RL) RL
< 0.28 Ohms 22
CT Lead wires
AWG Numbers are logarithmic
Numbers ‘increase’ with resistance (= ‘decrease’ with thicker wires)
AWG #10 has “1” Ohms for “1000”feet wire
Note: AWG # 13 has double the resistance
AWG #10 is most popular (easier to calculate the resistances!)
AWG#12 is adequate in most of the applications
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Case Study Fig 1
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Case Study Fig 2
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Case Study Fig 3
I
t
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Case Study Fig 4
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Conclusion
A thorough understanding of the application of CT is required
Previous experience of CT wire sizing may not always be correct in a newer application
More than adequate CT sizes and cable sizes waste resources
Application check is recommended, always for critical applications
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