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 CT 2. Metering CT In 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 oversized 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 parts 1. Primary winding 2. Secondary winding 3. Core
1.
Primary winding : primary winding is basically a line that carries the load current that is 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 as 1. Metering CTs 2. protection CTs 3. 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 etc e.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 calculation Study 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 current 2. calculate fault MVA of transformer fault MVA 3. calculate through fault current I F(T) 4. calculate lead resistance R lead 5. calculate total secondary side resistance R CT+Rlead+RB 6. calculate knee point voltage as V k≥( IF(T)/N)(RCT+Rlead+RB)
EXAMPLE Transformer details Transformer rating=100 MVA Primary voltage =220 kV Secondary voltage=33 kV Max feeder current on primary side= ( 1.1xMVA rating x1000) / ( 1.73x V p x min Tap) = (1.1 x120x1000)/ ( 1.732x220x0.9) =385 A Max feeder current on secondary side= ( 1.1xMVA rating x1000) / ( 1.73x V s ) = (1.1 x120x1000) / (1.732x33) =2309 A CT Ratios Selected IP(HV) IP(LV) IS
500 2500 1
A A A
Selected tap of main CT at HV side Selected tap of main CT at LV side Nominal secondary current
Primary side CT Through fault MVA of transformer =MVA rating/% impedance at – ve tap =120/.1130=1062.417 MVA Through fault current on primary side IPF(T)=fault MVA/ 1.732xVP =1062.41/ 1.732 x 220=2.79 kA Calculate total secondary resistance= RCT+Rlead+RB =2.5+ 1.916+0.2 ohms Calculate knee point voltage Vk≥( IPF(T)/N)(RCT+Rlead+RB) Vk≥ (2790/500) (2.5+ 1.916+0.2) V Vk≥ 25.75 V Secondary side CT Through fault MVA of transformer =MVA rating/% impedance at – ve tap =120/.1130=1062.417 MVA Through fault current on primary side IPF(T)=fault MVA/ 1.732xVP =1062.41/ 1.732 x 33=18.71 kA Calculate total secondary resistance= RCT+Rlead+RB =2.5+ 1.916+0.2 ohms Calculate knee point voltage Vk≥ ( IPF(T) /N)(RCT+Rlead+RB) Vk≥ (1871/500) (2.5+ 1.916+0.2) V Vk≥ 17.27 V CT UNDER SYMMETRICAL FAULT CURRENT INTRODUCTION Most 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 EQUATION Figure 1 shows equivalent circuit of CT
WHEN CT SECONDARY VOLTAGE IS LESS THAN CT KNEE POINT VOLTAGE Details of the CT used for simulation CT ratio-200/1 CT knee point voltage -34 volts RCT+Rlead+RB=1Ω CT primary current, I p=200A There fore CT secondary current, I s=1A Now, CT secondary voltage V s=I s (RCT+Rlead+RB) =1(1) =1V Figure 2 shows the CT secondary current ( Is) and CT primary current referred to secondary side ( I p ). ’
Figure 2 shows that, when secondary voltage of CT is less than Knee point voltage both CT secondary current( I s) and CT primary current referred to secondary side ( I p ) 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 VOLTAGE Details of the CT used for simulation CT ratio-200/1 CT knee point voltage -34 volts RCT+Rlead+RB=40Ω CT primary current, I p=200A There fore CT secondary current, I s=1A CT secondary voltage V s=I s (RCT+Rlead+RB) =1(40) = 40V Now CT knee point voltage (V k) is less than CT secondary voltage (V s) Figure 3 shows the CT secondary current ( Is) and CT primary current referred to secondary side ( I p ) ’
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 VOLTAGE Details of the CT used for simulation CT ratio-200/1 CT knee point voltage -34 V RCT+RB+Rlead=34Ω CT primary current, I p=200A There fore CT secondary current, I s=1A CT secondary voltage V s=I s (RCT+Rlead+RB) = 1(34) =34V Now Ct knee point voltage ( V k ) is equal to the CT secondary voltage ( V s) Figure 34 shows the CT secondary current ( I s) and CT primary current referred to secondary side ( I p ) ’
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(I s) 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. RESULTS From the above simulation following results are obtained. Table 1 Results for Symmetrical Fault Current Relay saturation free time Less than quarter cycle In between quarter and half cycle More than half cycle
Knee point voltage ≥(I sc /N )(RCT+Rlead+RB) ≥2*(I sc /N ) (RCT+Rlead+RB) ≥(I sc /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 day’s 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 EQUATION Consider primary fault current as
Putting this value of B ( t sat ) 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 information 4. Multiplexer: Select one sample and hold sample at a time for subsequent scaling and conversion for digital conversion 5. Programmable gain amplifier- used for amplification of signal 6. 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 factor 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 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 as Breker operating time=0. 1s Relay overshoot=.05s Allow.For error=.15s Safety margin=0.1s Total=0.4s So 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 protection Consider a simulation result CT ratio 200/1 Nominal CT primary current Ipn-200A Nominal CT secondary current-1A Maximum fault primary current =20*Ipn=4000A Knee-point voltage is taken as
Rct+Rlead+Rrelay= 1.7 There fore Vk=34 Now 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 PROTECTION 1. Select CT ratio based on max feeder current 2. calculate fault MVA of transformer fault MVA 3. calculate through fault current IF(T) 4. calculate lead resistance R lead 5. calculate total secondary side resistance R CT+Rlead+RB 6. calculate knee point voltage as Vk≥( IF(T) /N)(RCT+Rlead+RB) EXAMPLE Transformer details Transformer rating=100 MVA Primary voltage =220 kV Secondary voltage=33 kV Max feeder current on primary side= ( 1.1xMVA rating x1000) / ( 1.73x V p x min Tap) = ( 1.1 x120x1000)/ ( 1.732x220x0.9) =385 A Max feeder current on secondary side= ( 1.1xMVA rating x1000) / ( 1.73x V s ) = ( 1.1 x120x1000) / ( 1.732x33) =2309 A CT ratio selected IP(HV) 500 A Selected tap of main CT at HV side IP(LV) 2500 A Selected tap of main CT at LV side IS 1 A Nominal secondary current Primary side CT Through fault MVA of transformer =MVA rating/% impedance at – ve tap =120/.1130=1062.417 MVA
Through fault current on primary side I PF(T) =fault MVA/ 1.732xVP =1062.41/ 1.732 x 220=2.79 kA Calculate total secondary resistance= R CT+Rlead+RB =2.5+ 1.916+0.2 ohms Calculate knee point voltage V k≥( IPF(T) /N)(RCT+Rlead+RB) Vk≥ (2790/500) (2.5+ 1.916+0.2) V Vk≥ 25.75 V Secondary side CT Through fault MVA of transformer =MVA rating/% impedance at – ve tap =120/.1130=1062.417 MVA Through fault current on primary side I PF(T) =fault MVA/ 1.732xVP =1062.41/ 1.732 x 33=18.71 kA Calculate total secondary resistance= R CT+Rlead+RB =2.5+ 1.916+0.2 ohms Calculate knee point voltage V k≥ ( IPF(T) /N)(RCT+Rlead+RB) Vk≥ (1871/500) (2.5+ 1.916+0.2) V Vk≥ 17.27 V PRACTICAL RESULTS For verifying the above results and CT model, practical experiments were performed on the CT. Specifications of CT used are as follows CT ratio-200:1 Knee point voltage of CT-34 V When CT Secondary Voltage is less than CT Knee Point Voltage RCT+Rlead= 1.1Ω RB=10.1Ω CT primary current=50A Standard 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 Voltage RCT+Rlead= 1.1Ω RB=490Ω CT primary current=50A Standard 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 V Figure given below show practical and simulation result of CT secondary current
WHEN CT SECONDARY VOLTAGE IS EQUAL TO CT KNEE POINT VOLTAGE RCT+Rlead= 1.1Ω RB=135Ω CT primary current=50A Standard 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 V Figure given below show practical and simulation result of CT secondary current
To avoid this first quarter cycle saturation, taking RCT+Rlead= 1.1Ω RB=66Ω CT primary current=50A Standard 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
