INTRODUCTION
FUNDAMENTALS OF CT Current transformer isolates the measurement
and protection circuit from the primary high voltage level and
transforms the power system current to a smaller secondary current
which can be easily accessed by protection and measurement circuit.
Based on the application CTs are classified as 1. Protection CT2.
Metering CTIn protection application one is concerned about the
performance and error of the C.T. during fault and normal
condition, On the other hand metering application errors are of
concern under normal condition and not of concern when faulted or
abnormal condition occur. The performance of current transformer is
very vital to correct operation of a protection system. If current
transformer is not properly selected, saturation of CT core will
occur and leads to relay malfunctioning. Designing the CT for a
protection application has always been a challenge, over dimension
results in costly and over-sized CT, while under dimensioned CT
results in incorrect measurement of the power system current and
makes the protection system unreliable. Current transformer is an
instrument transformer used to lower down the value of current; it
basically has single turn primary and multi turn secondary. Like
voltage transformer it works on the principle of electromagnetic
induction, primary produces the changing flux which links with the
secondary winding to induce a voltage in secondary, but the
difference is that in voltage transformer secondary current is
varying and based on that the primary current flows that is primary
current is a function of secondary current, but in current
transformer the secondary current is a function of primary current,
primary current is a load current that line is carrying, this
current is stepped down and connected to the burden, here burden
can be measuring instrument (like ammeter) or it can be a
protective relay, CT basically consist of there parts1. Primary
winding 2. Secondary winding3. Core
1. Primary winding : primary winding is basically a line that
carries the load current that is to be stepped down, so the core of
a transformer is ring shaped which is placed over the line or
bulbar so that the line or bulbar acts as single turn primary, this
line current produces the flux that is carried to the secondary
winding by core, there is no of advantages of using a single turn
primary as compared to multi turn primary, in case of short
circuits the high value of current produces the forces on the
conductor, single turn has low effect of these forces as compare to
multiturns. 2. Secondary winding: secondary winding is multi turn
wound on a core, this winding is connected to the burden 3.
Core:core is a magnetic circuits used to carry flux, here core is
in the ring shape, primary conductor passes from the center of the
ring, core plays very important role in the accuracy of the CT,
both ratio error and phase error depends upon the material of the
core Based on the burden and there requirements, CTs are classified
as1. Metering CTs2. protection CTs3. special protection CTs 1.
Metering CTs: Measuring CTs are used to measure current under
normal condition, for metering CTs we defined both phase error and
ratio error by class of accuracy, these CTs are used to measure
current at normal conditions so there accuracy is defined from 100%
to 120%, for metering CT accuracy is given as class 1, class 2
etce.g. for class1 (IS:2075)
2. Potection CTs : these CTs are used to measure current at
abnormal condition (faults) and connected to relays to operate the
CBs, fault current is very high in magnitude so there accuracy is
defined at higher values of current, for metering CTs we defined
both phase error and ratio error by class of accuracy, for
Protection CTs the accuracy is given at 100% and ALF times normal
primary current, for Protection CT accuracy is given as class 5P,
class 10P etc, P stands for protection CT and 5 and 10 for accuracy
at Ip*ALF for e.g. class 5P(IS:2705) here accuracy at 100% is 1%
and at Ip*ALF is 5% 3. Special protection CT are used for special
protection systems like difference protection, here defined the
turns ratio error and Iexc at Vk, Vk/2, Vk/3
PROCEDURE
(i) study of various protection system(ii) study of various
parameters of CT(iii) collection of data base of various relay
requirements of different manufacturers(iv) CT sizing calculation
for each relay(v) Verification of calculation by using simulation
model(vi) Development of Automated sizing calculationStudy of
various protection system, different types of faults, abnormal
conditions during faults, ways to measure these abnormal
conditions, role of CTs in measuring these fault conditions.
Study of various parameters of CT. Formation of data base of
various relay requirements of different manufacturers, study the
market to find out relays available for various protection system,
collecting these relays details and forming a database of relays.
CT sizing calculation as per relay manufacture and verification of
this calculation by forming a simulation model. STEPS FOR DOING CT
SIZING CALCULATION FOR TRANSFORMER FEEDER OVER CURRENT PROTECTION
1.Select CT ratio based on max feeder current2. calculate fault MVA
of transformer fault MVA3. calculate through fault current IF(T)4.
calculate lead resistance Rlead5. calculate total secondary side
resistance RCT+Rlead+RB6. calculate knee point voltage as Vk(
IF(T)/N)(RCT+Rlead+RB)
EXAMPLE
Transformer detailsTransformer rating=100 MVAPrimary voltage
=220 kVSecondary voltage=33 kVMax feeder current on primary side= (
1.1xMVA rating x1000) / ( 1.73x Vp x min Tap) = (1.1 x120x1000)/ (
1.732x220x0.9) =385 AMax feeder current on secondary side= (
1.1xMVA rating x1000) / ( 1.73x Vs ) = (1.1 x120x1000) / (1.732x33)
=2309 A
CT Ratios Selected
IP(HV)500ASelected tap of main CT at HV side
IP(LV)2500ASelected tap of main CT at LV side
IS1ANominal secondary current
Primary side CTThrough fault MVA of transformer =MVA rating/%
impedance at ve tap =120/.1130=1062.417 MVAThrough fault current on
primary side IPF(T)=fault MVA/ 1.732xVP=1062.41/ 1.732 x 220=2.79
kACalculate total secondary resistance= RCT+Rlead+RB=2.5+ 1.916+0.2
ohmsCalculate knee point voltage Vk( IPF(T)/N)(RCT+Rlead+RB)Vk
(2790/500) (2.5+ 1.916+0.2) VVk 25.75 V
Secondary side CTThrough fault MVA of transformer =MVA rating/%
impedance at ve tap =120/.1130=1062.417 MVAThrough fault current on
primary side IPF(T)=fault MVA/ 1.732xVP=1062.41/ 1.732 x 33=18.71
kACalculate total secondary resistance= RCT+Rlead+RB=2.5+ 1.916+0.2
ohmsCalculate knee point voltage Vk ( IPF(T) /N)(RCT+Rlead+RB)Vk
(1871/500) (2.5+ 1.916+0.2) VVk 17.27 V
CT UNDER SYMMETRICAL FAULT CURRENT
INTRODUCTIONMost of the faults on the power system lead to
short-circuit condition. When such a condition occur, a heavy
current (called short circuit current) flows through the equipment,
causing considerable damage to the equipment and interruption of
service to the consumers. That fault on the power system which
gives rise to symmetrical currents (i.e. equal fault currents in
the line with 120 degree displacement) is called symmetrical fault.
This chapter will show how CT will behave when symmetrical fault
occurs and what should be the requirements of the CT for correctly
measuring the symmetrical fault currents.
KNEE-POINT VOLTAGE EQUATIONFigure 1 shows equivalent circuit of
CT
WHEN CT SECONDARY VOLTAGE IS LESS THAN CT KNEE POINT
VOLTAGEDetails of the CT used for simulationCT ratio-200/1CT knee
point voltage -34 voltsRCT+Rlead+RB=1CT primary current,
Ip=200AThere fore CT secondary current, Is=1ANow, CT secondary
voltage Vs=Is (RCT+Rlead+RB) =1(1) =1VFigure 2 shows the CT
secondary current (Is) and CT primary current referred to secondary
side (Ip).
Figure 2 shows that, when secondary voltage of CT is less than
Knee point voltage both CT secondary current(Is) and CT primary
current referred to secondary side (Ip) over lapping each other,
there is no saturation CT, secondary current is exact replica of
primary current. WHEN CT SECONDARY VOLTAGE IS MORE THAN CT KNEE
POINT VOLTAGEDetails of the CT used for simulationCT ratio-200/1CT
knee point voltage -34 voltsRCT+Rlead+RB=40CT primary current,
Ip=200AThere fore CT secondary current, Is=1ACT secondary voltage
Vs=Is (RCT+Rlead+RB) =1(40) = 40VNow CT knee point voltage (Vk) is
less than CT secondary voltage (Vs)Figure 3 shows the CT secondary
current (Is) and CT primary current referred to secondary side
(Ip)
Figure 3 shows that when secondary voltage of CT is more than CT
knee point voltage, there is saturation in the CT core and CT
secondary current is not exact replica of primary current. WHEN CT
SECONDARY VOLTAGE IS EQUAL TO CT KNEE POINT VOLTAGEDetails of the
CT used for simulationCT ratio-200/1CT knee point voltage -34
VRCT+RB+Rlead=34CT primary current, Ip=200AThere fore CT secondary
current, Is=1ACT secondary voltage Vs=Is (RCT+Rlead+RB) = 1(34)
=34VNow Ct knee point voltage (Vk) is equal to the CT secondary
voltage (Vs)Figure 34 shows the CT secondary current (Is) and CT
primary current referred to secondary side (Ip)
Figure 4 shows that the CT saturates after a quarter of a cycle
but will remain unsaturated after this. The reason for this is that
the flux in the CT is assumed to be zero when the fault starts. For
the first half cycle wave only half of the flux (+saturation
flux-0=saturation flux) is available compared to all the following
half wave forms (+saturation flux-(-saturation flux=2*saturation
flux). That means to avoid this condition we have to select the
knee-point voltage at least two times higher as the secondary
voltage generated
Figure 5 shows that when knee point voltage of CT is equal to
twice CT secondary voltage both CT secondary current(Is) and CT
primary current referred to secondary side over lapping each other,
there is no saturation CT secondary current is exact replica of
primary current. RESULTSFrom the above simulation following results
are obtained.
Table 1 Results for Symmetrical Fault Current
Relay saturation free timeKnee point voltage
Less than quarter cycle(Isc/N)(RCT+Rlead+RB)
In between quarter and half cycle2*(Isc/N) (RCT+Rlead+RB)
More than half cycle(Isc/N) (RCT+Rlead+RB)
CT UNDER TRANSIENT FAULT CURRENT Initially current transformer
(CT) sizing criteria were based on traditional symmetrical
calculations, assuming that fault current is sinusoidal but in
actual fault current is transient in nature. Transient fault
current during first few cycles contains both ac and dc components,
dc component decays over the time. Time constant for this transient
is proportional to X/R ratio of that primary system at the fault
location. DC component in the transient fault current will cause
the flux in the CT core to rise faster than it would have with
sinusoidal current. Thus it is necessary to over dimension the CT.
Now days fast numerical relays are used for protection systems
capable of detecting fault ranging from quarter cycle to cycle
time. If traditional symmetrical calculation approach for CT sizing
is used it will not provide correct value of power system current
to the relays making the protection system unreliable.
KNEE POINT VOLTAGE EQUATIONConsider primary fault current as
Putting this value of B (tsat) in equation above we get
TRANSFORMER FEEDER OVER CURRENT PROTECTION OVERCURRENT
PROTECTION:Figure shows block diagram of over current relay.
1.Primary CT: The purpose of the primary CT is to reproduce the
primary current waveform to the relay aux CT. 2. Low pass filter:
low pass filter to remove any high frequency content. 3. Sample and
Hold amplifier: Samples and hold analog signals at time interval
determine by the sampling clock to preserve the phase information4.
Multiplexer: Select one sample and hold sample at a time for
subsequent scaling and conversion for digital conversion5.
Programmable gain amplifier- used for amplification of signal6.
Analog to digital (A/D) conversion: An A/D converter converts
analog signal to a digital signal. 7. Microprocessor-Microprocessor
with appropriate software that provides the required protection
characteristic that are amplified to operate auxiliary units for
tripping, closing, alarms and so on.
Signal from CT is first brought into Low pass filter that remove
frequency content about 1/2 or 1/3 of the sampling frequency
(According to Nyquist criterion, A relay A/D converter needs to
sample faster than 2x per cycle to the highest frequency that is it
to monitor), for e.g. for 50 Hz system, A/D converter sampling rate
is 2 sample per cycle and cut of frequency of 50Hz. Modern Numeric
relay samples ranging from 16 to 24 samples per power system cycle.
In some relays, the entire sampled data is kept for oscillographic
records, but in the relay, only the fundamental component is needed
for most protection algorithms, unless a high speed algorithm is
used that uses sub cycle data to monitor for fast changing issues.
The sampled data is then passed through a low pass filter that
numerically removes the frequency content that is above the
fundamental frequency of interest (i.e., nominal system frequency),
and uses Fourier transform algorithms to extract the fundamental
frequency magnitude and angle. Next the microprocessor passes the
data into a set of protection algorithms, which are a set of logic
equations in part designed by the protection engineer, and in part
designed by the relay manufacturer, that monitor for abnormal
conditions that indicate a fault. If a fault condition is detected,
output contacts operate to trip the associated circuit breaker(s).
Over current relay are designed to operate under saturation
conditions, there is no need of considering the transient
dimensioning factor for calculating the knee-point voltage. Over
current relay take decision based on primary frequency component
only, it neglects the dc component, so there is no need of
considering transient over-dimensioning factorOver current relay
are designed to operate under saturation conditions, there is no
need of considering the transient dimensioning factor for
calculating the knee-point voltage. Over current relay take
decision based on primary frequency component only, it neglects the
dc component, so there is no need of considering transient
over-dimensioning factor.
OVER CURRENT PROTECTION OF TRANSFORMER FEEDER:Over current relay
cannot be used for primary protection without the risk of internal
faults causing extensive damage to the transformer. Differential
protection of transformer protects transformer from internal faults
but incase of external through fault current it does not protect
the transformer. A through fault external to transformer results in
an overload that can cause transformer failure if the fault is not
cleared promptly. It is widely reorganized that damage to
transformer. Over current protection for transformer feeder is used
to protect the transformer from external through faults and over
loads. It is basically a back up protection, which operates when
the unit protection of transformer and load side protection system
does not operate. Consider the fig given below.
Now if the overload protection system on load side and unit
protection system fails the transformer feeder over current
protection acts as back up protection. Protection relay used for
back up protection is either definite time or inverse time relay
and have there operating time much after the main protection
operating time, which can be calculated asBreker operating time=0.
1sRelay overshoot=.05sAllow.For error=.15sSafety
margin=0.1sTotal=0.4sSo operating time of backup relay starts much
after the main protection system and the dc transient components
delays to zero up to that point and also Over current relay
eliminates the dc component from the current, so there is no need
of considering the transient dimensioning factor, so the knee point
voltage of CT used for back up protection of transformer is given
as
Setting current of backup relay should be such that it protects
the transformer from damage without the protection device
operates.For transformer feeder over current protectionConsider a
simulation resultCT ratio 200/1Nominal CT primary current
Ipn-200ANominal CT secondary current-1AMaximum fault primary
current =20*Ipn=4000AKnee-point voltage is taken as
Rct+Rlead+Rrelay= 1.7 There fore Vk=34Now taking Ip=Ipn, CT
secondary current is given as
Taking Ip=20Ipn
Taking Ip=25Ipn
Taking Ip=30Ipn
Plotting curve between primary current /nominal primary current
and secondary current/nominal secondary current
This is clear from above waveforms and graphs that after
reaching the CT into saturation the rms value of CT secondary
current is still more than setting current and there is no need of
considering the transient over dimension factor.
STEPS FOR DOING CT SIZING CALCULATION FOR TRANSFORMER FEEDER
OVER CURRENT PROTECTION1. Select CT ratio based on max feeder
current 2. calculate fault MVA of transformer fault MVA3. calculate
through fault current IF(T)4. calculate lead resistance Rlead5.
calculate total secondary side resistance RCT+Rlead+RB6. calculate
knee point voltage as Vk( IF(T)/N)(RCT+Rlead+RB)EXAMPLETransformer
detailsTransformer rating=100 MVAPrimary voltage =220 kVSecondary
voltage=33 kVMax feeder current on primary side= ( 1.1xMVA rating
x1000) / ( 1.73x Vp x min Tap) = ( 1.1 x120x1000)/ ( 1.732x220x0.9)
=385 AMax feeder current on secondary side= ( 1.1xMVA rating x1000)
/ ( 1.73x Vs ) = ( 1.1 x120x1000) / ( 1.732x33) =2309 ACT ratio
selectedIP(HV)500ASelected tap of main CT at HV side
IP(LV)2500ASelected tap of main CT at LV side
IS1ANominal secondary current
Primary side CTThrough fault MVA of transformer =MVA rating/%
impedance at ve tap =120/.1130=1062.417 MVAThrough fault current on
primary side IPF(T)=fault MVA/ 1.732xVP=1062.41/ 1.732 x 220=2.79
kACalculate total secondary resistance= RCT+Rlead+RB=2.5+ 1.916+0.2
ohmsCalculate knee point voltage Vk( IPF(T)/N)(RCT+Rlead+RB)Vk
(2790/500) (2.5+ 1.916+0.2) VVk 25.75 VSecondary side CTThrough
fault MVA of transformer =MVA rating/% impedance at ve tap
=120/.1130=1062.417 MVAThrough fault current on primary side
IPF(T)=fault MVA/ 1.732xVP=1062.41/ 1.732 x 33=18.71 kACalculate
total secondary resistance= RCT+Rlead+RB=2.5+ 1.916+0.2
ohmsCalculate knee point voltage Vk ( IPF(T) /N)(RCT+Rlead+RB)Vk
(1871/500) (2.5+ 1.916+0.2) VVk 17.27 VPRACTICAL RESULTSFor
verifying the above results and CT model, practical experiments
were performed on the CT. Specifications of CT used are as follows
CT ratio-200:1Knee point voltage of CT-34 VWhen CT Secondary
Voltage is less than CT Knee Point VoltageRCT+Rlead= 1.1RB=10.1CT
primary current=50AStandard CT measured current=5A (50/5 ratio
used)CT under test measured current=.24 A (200/1 ratio)CT secondary
Voltage measured with CRO=2.432 V (RMS)
When CT Secondary Voltage is more than CT Knee Point
VoltageRCT+Rlead= 1.1RB=490CT primary current=50AStandard CT
measured current=5A (50/5 ratio used)CT under test measured
current=.1 A (200/1 ratio)CT secondary Voltage measured with CRO
=52 VFigure given below show practical and simulation result of CT
secondary current
WHEN CT SECONDARY VOLTAGE IS EQUAL TO CT KNEE POINT
VOLTAGERCT+Rlead= 1.1RB=135CT primary current=50AStandard CT
measured current=5A (50/5 ratio used)CT under test measured
current=.189 A (200/1 ratio)CT secondary Voltage measured with CRO
=25.56 VFigure given below show practical and simulation result of
CT secondary current
To avoid this first quarter cycle saturation, takingRCT+Rlead=
1.1RB=66CT primary current=50AStandard CT measured current=5A (50/5
ratio used)CT under test measured current=.22 A (200/1 ratio)CT
secondary Voltage measured =14.66 V
