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CT applications guidelines

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Title:
CT applications guidelines utilities point of view
Creator:
Stone, Gerald Wayne
Place of Publication:
Denver, CO
Publisher:
University of Colorado Denver
Publication Date:
Language:
English
Physical Description:
vi, 135 leaves : illustrations ; 29 cm

Subjects

Subjects / Keywords:
Current transformers (Instrument transformer) ( lcsh )
Transients (Electricity) ( lcsh )
Electric power systems -- Protection ( lcsh )
Current transformers (Instrument transformer) ( fast )
Electric power systems -- Protection ( fast )
Transients (Electricity) ( fast )
Genre:
bibliography ( marcgt )
theses ( marcgt )
non-fiction ( marcgt )

Notes

Bibliography:
Includes bibliographical references (leaves 133-135).
Thesis:
Submitted in partial fulfillment of the requirements for the degree, Master of Science, Electrical Engineering
General Note:
Department of Electrical Engineering
Statement of Responsibility:
by Gerald Wayne Stone.

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Source Institution:
|University of Colorado Denver
Holding Location:
|Auraria Library
Rights Management:
All applicable rights reserved by the source institution and holding location.
Resource Identifier:
30674273 ( OCLC )
ocm30674273
Classification:
LD1190.E54 1993m .S76 ( lcc )

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Full Text
n
CT APPLICATIONS GUIDELINES:
UTILITIES POINT OF VIEW
by
Gerald Wayne Stone
B.S., University of Nebraska, 1976
A thesis submitted to the
Faculty of the Graduate School of the
University of Colorado at Denver
in partial fulfillment
of the requirements for the degree of
Master of Science
Electrical Engineering
1993
i1 v a


This thesis for the Master of Science
degree by
Gerald Wayne Stone
has been approved for the
Department of
Electrical Engineering
by
William R. Roemish
Tamal Bose
/- /C~?3
Date


Stone, Gerald Wayne (M.S., Electrical Engineering)
CT Applications Guidelines: Utilities Point of View
Thesis directed by Professor Pankaj K. Sen
ABSTRACT
Historically Current Transformers (CTs) have
provided power system current information to various
monitoring devices such as protective relays. The
application of CTs in the utility industry has followed
several standard practices. This thesis will review the
steady state application methods and transient estimation
techniques used in the past, develop a computer model to
simulate current transformer performance under steady
state and transient conditions, and compare the results.
The computer simulations will include a three-phase
radial feeder, a three-phase bus differential scheme, and
a three-phase power transformer differential scheme.
This abstract accurately represents the content of the
candidate's thesis. I recommend its publication.
Signed
Pankaj K. Sen


ACKNOWLEDGEMENTS
This thesis is submitted to the University of
Colorado at Denver as the final requirement for the
Master of Science degree in Electrical Engineering. The
work could not have been completed without the
cooperation and assistance of several people and
organizations.
I especially wish to thank my academic and thesis
advisor, Dr. Pankaj K. Sen, Professor of Electrical
Engineering and Computer Science at the University of
Colorado at Denver, whose guidance and encouragement have
been invaluable during this endeavor.
I am also grateful to Jeff Selman of Tri-State
Generation and Transmission Association for his support
and assistance with EMTP/ATP.
I would also like to thank George Laughlin of Public
Service Company of Colorado for his countless
conversations and support.
Finally, I wish to thank Public Service Company of
Colorado for the use of their computers and word
processing equipment.
iv


CONTENTS
ACKNOWLEDGEMENTS................................... iv
CHAPTER
I. INTRODUCTION ..................................... 1
II. STEADY STATE PERFORMANCE ........................ 4
Classification "C" or "T"...................... 6
Burden/Accuracy ................................. 7
Equivalent Circuit ............................. 10
CT Performance Evaluation ...................... 14
Steady State Mathematical Modeling and
Simulation................................... 16
III. TRANSIENT PERFORMANCE ........................... 21
BH Loop/Nonlinearity ........................... 26
Remanence....................................... 26
Materials....................................... 28
Equivalent Circuit ............................. 29
Transient Mathematical Modeling and
Simulation................................... 33
IV. COMPARISON WITH EXISTING TECHNIQUES AND RESULTS 58
V. CONCLUSIONS...................................... 62
v


APPENDIX......................................... 64
APPENDIX A: SHORT CIRCUIT INFORMATION .... 65
APPENDIX B: CT CHARACTERISITCS .................. 67
APPENDIX C: CT STEADY STATE CALCULATIONS ... 69
APPENDIX D: CT TRANSIENT CALCULATIONS .... 100
APPENDIX E: EMTP SIMULATION............... 107
GLOSSARY ............................................. 125
REFERENCES...................................... 133
Vi


CHAPTER I
INTRODUCTION
The protection of major equipment in a power utility
requires the application of protective devices. One of
the common links in the changing world of relay equipment
from existing electromagnetic relays to the new
generation of solid state and microprocessor multi-
function relays is the quality of the source of
information presented to the protective relay equipment.
Historically Current Transformers (CTs) and
Potential Transformers (PTs) have been the primary
equipment used to supply this information to the
protective relays. These instrument transformers
proportionally reduce the system voltage and current
values to safe magnitudes.
The performance of a current transformer is a
measure of its ability to reproduce accurately the
primary current in secondary amperes both in wave shape
and magnitude. There are two components of this which
will affect this reproduction. The symmetrical ac
component under steady state conditions and the
asymmetrical dc offset component which is a result of a
transient condition.
1


The behavior of protective relays is usually
specified using sinusoidal voltages and currents but dc
offsets in the primary current of CTs and remanence in
the CT cores frequently result in secondary currents that
have substantial distortion. This distortion is the
result of CT core saturation which can affect the
performance of the relay protection scheme. The time-
to-saturation, during which time the CT output current is
a faithful replica of the primary current, can be
determined from generally available power-system and CT
parameters by using graphical solutions [l].
Standard ratings and application guides deal with
the steady-state sine wave behavior of a CT. Despite the
use of recommended rules to avoid saturation, transient
saturation always occurs in critical relay applications.
After the primary fault current has ceased, the CT can
produce unidirectional decaying output current and can
have a high level of flux trapped in the core, both of
which can affect the performance of relay protection
schemes. Consequently, the relay engineer needs an
understanding of the non-linear characteristics of CTs,
CT accuracy ratings, and the transient behavior of CTs.
This thesis will do the following:
1. Review existing CT application methods
2. Develop and use a CT computer model
2


3. Compare the results of existing CT
application methods with the computer
model simulation.
3


CHAPTER II
STEADY STATE PERFORMANCE
The initial application of CTs for use by protection
relays requires the knowledge of the normal power circuit
load current, the location and configuration of the CTs,
the CT secondary burden, and the power circuit short
circuit current.
The CT secondary current can be calculated by
dividing the primary current by the CT ratio. Usually
CTs which monitor the power flowing in the electric
system have a ratio selected so that the secondary
circuit will produce a predetermined current magnitude
for a projected maximum power system load under normal
conditions. In the United States, standard CTs and
relays are rated at 5 amps while European CTs and relays
are rated for 1 amp secondary current.
The location and configuration of the CTs are of
importance. If a power transformer is to be protected,
then the CT ratios of the high voltage side and low
voltage side of the power transformer will be different.
CT ratios selected for a transformer differential current
protection scheme would have to provide secondary
currents from both CTs that would have to be matched to
4


within a certain limit. High voltage bus CTs would be
paralleled in a ring bus or breaker and a half bus
scheme. And feeder CTs are usually used by themselves.
The CT is designed to deliver to a secondary
standard burden from one to twenty times rated secondary
current without exceeding a 10% ratio error. For a
particular voltage class of CT the larger the burden
connected to the CT secondary the lower the magnitude of
secondary current that can be reproduced without
distortion.
Accuracy calculations for maximum CT secondary
currents usually use three phase and single line to
ground fault current magnitudes. If these calculations
are satisfactory then the phase to phase and two phase to
ground fault calculations would also be satisfactory.
Application of CTs require the specification of many
parameters including the mechanical construction, type of
insulation (dry or liquid), ratio in terms of primary and
secondary currents or voltages, continuous thermal
rating, short time thermal and mechanical ratings,
insulation class, impulse level, service conditions,
accuracy, and connections.
5


Classification "C11 or "T11
There are two basic types of CT construction used in
relay applications which are grouped into the categories
of Toroidal and Wound. Bushing, Window, or Bar CTs use
the Toroidal construction. The Bushing CTs are generally
multi-ratio and are commonly used in circuit breakers,
power transformers, generators and switchgear. Window or
Bar CTs are used at medium or lower voltages. This type
of construction has the primary conductor passing through
the center of the core. The secondary windings are
uniformly distributed around the toroidal core. All the
flux which links the primary conductor also links the
secondary winding. The leakage flux, and thus the
leakage reactance, is negligible.
The C classification per ANSI C57.13 is usually
applied to Bushing, Window, or Bar CTs. The uniformly
distributed windings produce negligible leakage flux and
therefore the excitation characteristics can be used
directly to determine the CT performance. The C
classification indicates that ratio correction at any
current can be calculated adequately if the burden,
secondary winding resistance, and the excitation
characteristics are known. The C classification applies
to all tap sections of the current transformer winding.
Wound CTs have a primary winding of one or more
6


turns and secondary winding on a common core similar to
power transformers. This type of transformer is usually
used for medium and low voltage applications. The
secondary windings are not evenly distributed due to
physical space reguired for insulation and bracing of the
primary winding. This results in fringing effects of the
nonuniformly distributed windings, and therefore flux is
present which does not link both primary and secondary
windings. Auxiliary current transformers are of the
Wound construction.
The T classification per ANSI C57.13 is applied to
Wound current transformers whose core leakage flux
affects the ratio appreciably. Appreciably is defined as
one percent difference between the actual ratio
correction and the ratio correction calculated. The T
classification indicates that the CT has to be tested to
identify its operating characteristics.
Burden/Accuracy
The external load applied to the secondary of a
current transformer is called the "burden". The burden
is either expressed in total ohms impedance with
effective resistance and reactance components, or as the
total volt-amperes and power factor at the specified
value of current or voltage, and freguency.
7


Accuracy classes are indicated by the number
following the C or T designations. This designation
indicates the secondary terminal voltage which the CT can
deliver to a standard burden at 20 times rated secondary
current without exceeding 10% ratio correction. This 10%
will not be exceeded at any current from 1 to 20 times
rated secondary current at the standard burden or any
lower standard burden. For relaying the voltages classes
are C100, C200, C400, and C800 corresponding to standard
burdens of l ohm (B-l), 2 ohms (B2), 4 ohms (B-4) and 8
ohms (B-8) respectively. For a CT rated C200 with a
secondary 5 amp rating, a voltage of 200 volts would be
developed across the CT secondary terminals for a 2 ohm
burden at 100 amps (20 5 amps). Standard burdens have
a .5 power factor.
The performance of a CT used in protective relaying
is largely dependent on the total burden or impedance in
the secondary circuit of the CT. The CT core flux
density (and thus the amount of saturation) is directly
proportional to the voltage that the CT secondary must
produce. So for a given amount of secondary current, the
larger the burden impedance becomes, the greater is the
tendency of the CT to saturate. The total burden
includes the burdens of the individual relays, meters,
and other equipment plus the resistance of
8


interconnecting leads and the internal resistance of the
CT.
Sufficient accuracy results if series burden
impedances are added arithmetically. The results will be
slightly pessimistic, indicating slightly greater than
actual CT ratio inaccuracy. But, if a given application
is so borderline that vector addition of impedance is
necessary to prove that the CTs will be suitable, such an
application should be avoided.
The CT burden impedance of most electromechanical
relays decrease as the secondary current increases
because of saturation in the magnetic circuits of the
devices. At high saturation, the electromechanical relay
burden impedance approaches its DC resistance.
Therefore, a given burden may apply only for a particular
value of current.
The physical connections of CTs effects the burden
as seen by each individual CT. When CT primaries and
secondaries are connected in series, the burden on each
individual CT is decreased in proportion to the number of
CTs in use. When CTs are connected in parallel the
effect is to increase the burden on each individual CT.
The amount of increase is dependent upon the number of
CTs and distribution of current between the CTs. This is
the case in breaker and half, ring, and double bus
9


arrangements.
In three phase CT connections, the burden on an
individual CT can vary with the type of connection (wye
or delta) and the type of fault on the system (1 or 2-
line to ground or multi-phase).
Also beware of attempts to "step-up" current from
the main CT to the relay with the use of an auxiliary CT.
The auxiliary CT may be adequate, but remember that main
CT will see the burden impedance multiplied by the
auxiliary CT ratio squared.
Equivalent Circuit
In an ideal current transformer the primary ampere-
turns are equal to the secondary ampere-turns. However,
every core material requires some energy to produce the
magnetic flux in the core. Thus, in an actual current
transformer, the secondary ampere-turns are equal to the
primary ampere-turns minus the exciting ampere-turns.
An approximate equivalent circuit is shown in Figure
1. The primary current is stepped down through a perfect
transformer (no-loss). The primary leakage impedance
(ZH) is referred to the secondary by turns ratio (n2)
squared. The CT secondary consists of the secondary
impedance (ZL) and the core loss (Rm) and exciting
impedances (Xm). (ZB) is the CT burden.
10


Figure 1 The Equivalent Circuit and Phasor Diagram
of a Current Transformer [2].
The generalized circuit can be further reduced. ZH
can be neglected, since it influences neither the
perfectly transformed current IH/n nor the voltage across
Xm. The current through Xm, the magnetizing branch, is
Ie, the exciting current. The phasor diagram, with
exaggerated voltage drops is also shown. In general, ZL
is resistive, Ie lags Vcd by 90 degrees and is the prime
source of error. Note that the net effect of Ie is to
cause IL to lead and to be smaller than the perfectly
transformed current IH/n.
11


noo
s 100
001
100
scoonomt mis Excmwc amps
Figure 2 Typical Excitation Curves for Multi-Ratio C
Class Current Transformer [3].
Figure 2 above, extracted from ANSI C57.13, shows
the typical excitation curves for a multi-ratio C class
current transformer. The maximum tolerance of excitation
values above and below the knee are also specified.
These curves define the relationship of the secondary
exciting current (Ie) to the secondary voltage (Ee) The
unsaturated slope is determined by the magnetic core
material. The saturated region is the air-core
reactance.
When the current transformer core is unsaturated,
12


the error due to the exciting current is normally
negligible. When the voltage is above the knee of the
excitation curve, operating in its saturated region the
exciting current is no longer negligible, therefore, the
ratio error of the current transformer becomes much
greater.
For T class current transformer, the leakage flux
could be appreciable. The exciting flux should be
considered, along with the leakage flux, in determining
Figure 3 Typical T Class Overcurrent Ratio Curves [3].
13


current transformer accuracy. Although a test should be
done, Figure 3, also extracted from ANSI C57.13, shows
typical overcurrent ratio curves for a T class current
transformer.
Any simple equivalent diagram for a CT is, at best,
crude. Exciting current is accompanied by harmonics
which, in turn, produce harmonic relay currents. An
analysis for application purposes is usually made on the
basis of sinusoidal fundamental quantities. While this
approach is highly simplified, the equivalent diagram is
an excellent tool for picturing the phenomenon and
estimating the approximate performance to be expected.
CT Performance Evaluation
The symmetrical component of CT performance is
determined by the highest current that can be reproduced
without saturation resulting in large errors. Two types
of errors introduced by CTs are ratio errors and phase
angle errors. For relaying applications the phase angle
error is neglected. This is because the load on the
secondary of a CT during a fault is generally of such a
high lagging power factor that the secondary current is
practically in phase with the exciting current and
therefore the effect of the exciting current on the phase
angle accuracy is negligible. Also relays can usually
14


tolerate a phase angle error without much effect on
operation, unlike in a metering application. If the CTs
do not saturate then the excitation current is negligible
and can be ignored.
Although the performance required of CTs varies with
the relay application, high quality CTs should always be
used. The better-quality CTs reduce application
problems, present fewer hazards, and generally provide
better relaying. The quality of the CTs is most critical
for differential schemes, where the performance of all
the transformers must match. In these schemes, relay
performance is a function of the accuracy of
reproduction, not only at load currents, but at all fault
currents as well.
Some differences in performance can be accommodated
in the relays. In general, the performance of CTs is not
so critical for transmission line protection. The CTs
should reproduce reasonably faithfully for faults near
the remote terminal, or at the balance point for
coordination or measurement.
For large magnitude, close-in faults, the CTs may
saturate; however, the magnitude of the fault current is
not usually critical to the relay. For example, an
induction overcurrent relay may be operating on the flat
part of the curve for a large magnitude, close-in fault.
15


Here it is relatively unimportant whether the CT
secondary current is accurate, since the timing is
essentially identical. The same is true for
instantaneous or distance-type relaying for a nearby
internal fault well inside the cut-off or balance point.
In all cases, however, the current transformer should
provide sufficient current to operate the relay
components.
The performance and/or application of CTs under
steady state conditions has been estimated with the help
of three basic methods.
1. Calculation of CT performance using CT
design dimensions and materials
(Formula Method).
2. Calculations based on CT secondary
excitation curves (Excitation Method).
3. Calculations based on ANSI Standard:
Current Transformer Accuracy Classes
Examples of each of these estimation methods are
discussed next.
Steady State Mathematical Modeling and Simulation
The existing Dillion 115/25kv Substation (Figure 4)
was chosen to compare the three steady state calculation
methods. The relay protection schemes included three
16


areas or zones of protection. The first zone consisted
of differentially connected CTs around a 115/25kV Power
Distribution Transformer. The second zone consists of
the 25kV bus differential protection. And the third zone
is the 25kV feeder.
The Power Distribution Transformer is protected by a
General Electric STD15C transformer differential relay.
The 25kV bus is protected by a General Electric PVD11
differential voltage relay. The 25kV feeders are
protected by General Electric IAC77B phase and IFC53B
ground time overcurrent relays.
The substation has a system three-phase fault duty
of 1152 MVA (5783 amps) on the 115kV bus and a 162 MVA
(3741 amps) on the 25kV bus. The power distribution
transformer has a 15/20/25 MVA, 115/25kV rating connected
delta-wye with an impedance of 7.95%. An initial load of
12.5 MVA was served by the substation.
The CT characteristics including the excitation
curve and CT internal resistances are listed in Appendix
B. The CT connection leads (#10 CU) from the CT location
to the relay have an estimated length of 500 feet.
The estimated performance of the CTs with the tap
ratios as indicated using the Formula, Excitation and
ANSI methods are listed in the Table 1 below.
A three-phase fault on the power transformer high
17


voltage terminal will saturate the CT with a maximum
current of approximately 145% of rated CT secondary
CT Steady State Calculation Summary
25kV Feeder Current Transformer
C800, 400/5 Tap of 1200/5 MRCTS
1.55 ohms CT Secondary Burden
Three-Phase Fault Level at the feeder breaker
3744 Primary Amps, 47 CT Secondary Amps
Formula Method Excitation Method ANSI Method
Will Not Will Not Will Not
Saturate Saturate Saturate
25kV Bus Differential Main Breaker
Current Transformer
C800, 1200/5 Tap of 1200/5 MRCT
Three-Phase Fault Level on the 25kV bus
4.1 ohm CT Secondary Burden
3741 Primary Amps, 16 CT Secondary Amps
Formula Method Excitation Method ANSI Method
Will Not Will Not Will Not
Saturate Saturate Saturate
115kV Power Transformer Differential
Current Transformer
C400, 200/5 Tap Of 600/5 MRCT
2.5 ohms CT Secondary Burden
Three-Phase Fault Level at the power high voltage
terminals
5784 Primary Amps, 144 CT secondary Amps
Formula Method Excitation Method ANSI Method
Will Saturate Will Saturate Will Saturate
Table 1 Estimated Performance of CTs
current. All three steady state methods indicated that
saturation would occurr.
The CT secondary current magnitudes for the bus and
feeder faults were less than the 100 amps maximum
18


secondary current and did not saturate the CTs for the
secondary impedance used.
19


DILLIDN SUBSTATIDN
CURRENT TRANSFORMER CONFIGURATION
Figure 4 Dillion Substation One-Line
20


CHAPTER III
TRANSIENT PERFORMANCE
In modern HV and EHV power systems more and more
emphasis is placed on the ability of CTs to provide an
accurate reproduction of the primary electric system
current as input to the protective relay systems. Relay
decisions need to be made while an asymmetrical dc
transient component of fault current is still present.
This component can cause some degree of CT saturation and
thus operating current distortion.
CTs generally produce a faithful replica of the
primary current from the instant of fault current
initiation until saturation begins. This interval,
called the "time-to-saturation"[1], can be calculated for
most ring-core CTs using a series of generalized curves
and readily obtained CT and power system parameters.
This method was used to developed a set of curves from
which the time-to-saturation can be obtained for a
calculated value of saturation factor (Ks).
Distortion of the CT secondary output current begins
whenever conditions are such that the core flux density
enters the "region of saturation". The factors
influencing the core flux density are the physical
parameters of the CT, the magnitude, duration, and
21


waveform of the primary currents and the nature of the
secondary burden. Saturation of the core can be produced
by excessive symmetrical fault currents as well as by
lower magnitude asymmetrical (dc offset) fault currents.
The latter are far more important in determining the
transient response of CTs.
When a fully offset fault current of the form shown
in Figure 5 below
is impressed on the primary of a current transformer, the
dc offset will, in general, cause a rise in flux in the
core several times greater .than that required to
transform the 60-Hz component of current. Figure 6 below
shows the rise in flux in a CT core when a fully offset
current is applied to a CT with a resistive burden. In
22


this diagram the quantity 0ac plotted on the abscissa
represents the flux required to reproduce the 60-Hz
component of fault current while the quantity represents the flux required to reproduce the transient
(dc) component of current. The variation in the
transient component of flux will be a function of both
the primary circuit dc time constant and the time
constant of the CT secondary circuit. The latter time
constant is a function of CT secondary resistance, burden
resistance, burden inductance, and the magnetizing
impedance. The total flux, Figure 6 Rise of Flux in the Core of a Current
Transformer [1].
23


offset current is considerably greater than that required
to reproduce a symmetrical current.
In the above illustration, the CT will reproduce the
offset fault current completely as long as the induction
does not reach the saturation-flux-density level. In
modern CTs, this saturation flux density is between 1.9
and 2.0 teslas (20 kilogausses). If the induction
greatly exceeds saturation flux density, the CT may
produce a severely distorted output current similar to
that shown in Figure 7. In this illustration a 1200:5 CT
with a 2.6-ohm resistive burden is subjected to a fully
Figure 7 Distortion in Secondary Current due to
Saturation [1].
offset primary current of 24,000 amperes (20 times
24


rated). As can be seen, the CT begins to saturate in the
first cycle. With pure resistance burden, a sharp drop
off of secondary current occurs in each cycle. Addition
of inductance to the burden causes a more gradual drop-
off of secondary current.
In many instances, CTs will be capable of accurately
reproducing offset currents for a number of cycles before
starting to saturate. However, under some conditions,
saturation cannot be avoided and CTs will produce
distorted output currents.
The time-to-saturation of a CT is determined by the
following parameters:
1. Fault current magnitude
2. Degree of fault current offset
3. Time constant of the dc component of fault
current
4. Residual flux in the core (Remanence)
5. Secondary circuit resistance and power
factor, including the effect of the
secondary winding resistance.
6. Secondary excitation impedance of CT at
power system frequency (from standard
exciting current measurements)
7. CT turns ratio
25


BH Loop/Nonlinearitv
Usually published magnetic characteristics of
various materials which are used in the core of CTs are
presented as plots of flux density B as a function of the
magnetic field intensity H. The property of the magnetic
material traces out a BH curve with a different path of
flux density B for an increasing and decreasing magnetic
field intensity H. When equal magnitude of positive and
negative magnetic field intensity are applied to a magnet
material the result is a symmetrically cyclically BH path
or loop called hysteresis. The flux density B lags
behind the magnetic field intensity H. The value of flux
density B when the magnetic field intensity H is removed
is call the residual flux density.
The nonlinear properties of magnetic materials as
characterized by the hysteresis loop are the result of
the magnetic materials used, the physical construction of
the magnetic circuit, and the placement of the windings
on the magnetic material. Nonlinearity in CT application
is undesirable because it will distort the secondary
current waveform.
Remanence
Remanence is commonly found in the cores of relaying
CTs operating in service. Remanence or Residual flux is
26


a result of the amount of flux in the core of a CT at the
moment of primary circuit interruption. In fact, any
value of flux between zero and the CTs' saturation level
may be retained in the core. With remanence the core may
be operating indefinitely at a high peak flux density
under normal system load conditions. Then, when a power
system fault occurs, the time-to-saturation may be much
shorter than without remanence.
The remanent flux in the core depends primarily on
the magnitude of primary current, the impedance of the
secondary circuit and the amplitude and time constant of
any offset transient. Since the impedance of the
secondary circuit is generally fixed, the magnitude of
remanent flux is governed by the magnitude of the
symmetrical component of the primary current and the
magnitude of the offset transient prior to the primary
current interruption. Maximum remanent flux can be
obtained under conditions whereby the primary current is
interrupted while the transformer is in a saturated
state.
The performance of both C and T class CTs is
influenced by this remanence or residual magnetism.
Relay action could be slow or even incorrect.
Remanence can be substantially eliminated from CTs
by the use of small air gaps in the core. A CT with an
27


air gap in the core has a fairly low residual flux,
approximately 10% of saturation density. Residual flux
or current transformers with no intentional air gap is
approximately 90% of saturation density.
Materials
Magnetic circuits have been made from many types of
magnetic materials. Included in the list are certain
forms of iron and its alloys in combination with cobalt,
nickel, aluminum, and tungsten. These are known as
ferromagnetic materials and are easy to magnetize since
they have a high value of relative permeability ur.
Figure 8 Normal Magnetization Curves for Common Magnetic
Materials [1].
28


The magnetic permeability of ferromagnetic
materials varies with magnetic field intensity starting
at a relative low value and increasing to a maximum then
falling off with increasing saturation. Typical
magnetization curves are as shown in Figure 8 above.
The permeability is also generally different for
increasing flux density that for decreasing flux density
at the same value of magnetic field intensity as
exhibited by the hysteresis loop when magnetization is
carried through a complete cycle.
Magnetic cores that operate at low frequencies (in
the audio-frequency range) are comprised of silicon steel
laminations. Modern CTs are made from silicon steel.
silicon steels 1.2 to 1.94 Tesla 1.55 Tesla avg
older materials, >1.55 Tesla avg
prior 1947
Equivalent Circuit
A digital computer model of the Dillion Substation
electrical system was developed. A program designed to
simulate an electrical network for the study of
transients, Electromagnetic Transients Program (EMTP),
was used to develop a model of a three-phase power system
including CT secondary circuits. This program was
originally developed by Dr. Hermann Dommel in the late
29


1960's. Since that time the Bonneville Power
Administration (BPA) along with many universities and
other utilities have developed a small computer version
called the Alternative Transients Program. These two
programs are quite similar and were used in this study.
The Dillion Substation power transformer and the
associated CTs were modeled using the Saturable
Transformer model. This model represents a two winding
or single phase transformer with the advantage that a
saturable component can be included in the internal
connections of the model. The Saturable Transformer
model can be represented as shown in Figure 9 below.
There is one single-phase two-winding ideal transformer
where the correct transformation ratios for winding 2 can
be modeled with respect to winding 1. Both windings have
an associated leakage-impedance, characterized by
resistance R and inductance L. Leakage inductance of the
high voltage winding has to be non-zero, but the low
voltage winding can have a zero value. The saturation
effect is confined to a single non-linear reactor in the
circuit of the low voltage winding. As a consequence,
the saturation branch is connected to the star point,
which is not always the best connection point. Ideally
the nonlinear inductance should be connected to
30


BUS11
IDEAL
BUSTOP N1 : N2
Figure 9 Single-Phase Two-Winding [5].
that point of the equivalent star circuit where the
integrated voltage equals the iron-core flux. Normally,
this is only true for the winding which is closest to the
iron core.
In the case of saturation, the Type-98 pseudo-
nonlinear reactor model is used internally. For a CT the
(Vrms,Irms) characteristic is obtained from the CT
excitation curves published by the CT manufacturer. An
auxiliary routine "SATURATION" is used to convert the
Vrms,Irms curve information into (Flux, Current) peak
value information for the EMTP program. Excitation
losses (iron core) are confined to a linear resistance
(RMAG) which is in parallel to the saturation branch.
The specification of remanent magnetism requires a
(flux current) hysteresis curve. Recall that the shape
of the hysteresis loop for an inductor depends primarily
on the material of the core, while the scaling of the
hysteresis loop depends on the geometry, the number of
31


turns and other construction factors. The auxiliary
routine "HYSDAT" is used to create a Type-96 hysteretic
inductor characteristic which will model the
(flux,current) characteristic of a transformer using the
curve shape corresponding to ARMCO Mn steel. Without
proper field testing to obtain the CT hysteresis loop
this is a good estimate for the CT hysteresis behavior.
The Saturable Transformer Model could not model
hysteresis internally. To model the hysteresis and
remanence characteristics the Type-96 hysteretic inductor
was modeled on the CT secondary terminals.
The impedance values used in the CT model were
obtained from the published voltage-current excitation
curves. The secondary resistance can be obtained in a
ohmic value per turn. The secondary leakage impedance is
negligible due to the uniformly distributed secondary
windings and has a value of zero. The primary value of R
and L is not easy to obtain because there is not a
primary winding as such but only a terminal or wire that
the secondary winding is placed around. The impedance
values of the CT primary and secondary circuits were
estimated in the EMTP model by transforming 10% of the
secondary impedance to the primary side of the CT. The
values to calculate the flux-current saturation curve
were read from the published voltage-excitation current
32


curve for the CT and converted to flux-current values by
the "SATURATION" auxiliary program. The value of RMAG
was estimated by dividing the voltage squared at the knee
of the voltage-excitation current by the associated
excitation power loses. The CT model input data
calculations are shown in Appendix D.
Transient Mathematical Modeling and Simulation
The Dillion Substation model consisted of three
areas or zones of protection that can be found in most
relay protection schemes. The first zone consisted of
differentially connected CTs around a 115/25kV Power
Distribution Transformer. The second zone consists of
the 25kV bus differential protection. And the third zone
is the 25kV feeder.
In addition to the CT model the substation was
modeled with a system three-phase fault duty of 1152 MVA
on the 115kV bus and a 162 MVA on the 25kV bus. The
power distribution transformer was a 15/20/25 MVA,
115/25kV rating connected delta-wye with an impedance of
7.95%. All bus work was modeled with minimal resistance.
An initial load of 12.5 MVA was served by the substation.
The CT secondary circuit or burdens included both
connecting wires and relay burdens. The connecting leads
were estimated at 500 feet of #10 Copper from the CT
33


location to the relay location. The relay impedances
were estimated at the expected value of secondary current
under a three phase bus fault condition.
Three separate phase faults were initiated (Figure
10) at the power transformer high voltage terminals, the
25kV bus, and on the 25kV feeder. The EMTP program was
written to initiate a fault at a time when the largest
dc offset would occur.
The 25kV feeder fault was simulated with the ATP
program. The program requirements were small enough for
a solution on a personal computer (Figure 11). As can be
seen in Figures 12 and 13, although while a secondary
fault current of only 49 amps is expected, there is some
secondary current distortion due to saturation of the CT
core with the transient dc offset. In Figures 14 and 15
the effects of hysteresis increase the amount of current
distortion and saturation of the CT core. Remanence in
the CT core as modeled in Figures 16 and 17 reduced the
level of current distortion and CT core saturation.
The 25kV bus fault was simulated with the EMTP
program (Figure 18). The program requirements were too
large for a personal computer and were simulated on a
mainframe IBM computer. Although the EMTP program would
solve, numerical oscillations were encountered and could
not be eliminated. The results are shown in Figures 19,
34


20, 21, 22, 23, and 24. The transient dc offset is
present and reduces the secondary current magnitude.
Although there is little distortion.
The H5kV power transformer high voltage terminal
fault was simulated with the EMTP program (Figure 25).
The program requirements were too large for a personal
computer and were simulated on a mainframe IBM computer.
Again the EMTP program would solve, but numerical
oscillations were encountered and could not be
eliminated. The results are shown in Figures 26, 27, 28,
29, 30, and 31. The CT current is extremely distorted
when using the saturation model. While the addition of
hysteresis and remanence did not result in distortion but
a reduction in current magnitude.
35


DILLIDN SUBSTATION
CURRENT TRANSFORMER CONFIGURATION
Figure 10 Dillion Substation One-Line
36


DILLION SUBSTATION
25kV FEEDER SIMULATION
Zc = .65 + J.0185 OHM
Zp = .0000038 + J.0000067 DHM Zs = .223 + j.000 DHM
Figure 11 25kV Feeder EMTP Model
37


CT Primary and Secondary Current Comparision (peak)
taJ
u
oo
t [ms]
Figure 12 25kV Feeder EMTP Model, Phase A CT with Saturation


CT Flux Density
[ IIW-K J
t [ms]
Figure 13 25kV Feeder EMTP Model, Phase A CT with Saturation


CT Primary and Secondary Current Comparision (peak)
1*1
Figure 14 25kV Feeder EMTP Model, Phase A CT with Hysteresis


CT Flux Density
Cmv-sl
Figure 15 25kV Feeder EMTP Model, Phase A CT with Hysteresis


CT Primary and Secondary Current Comparision (peak)
Cal
Figure 16 25kV Feeder EMTP Model, Phase A CT with Remanence


CT Flux Density
lj
Emv-sD
t [ms]
Figure 17 25kV Feeder EMTP Model, Phase A CT with Remanence


DILLIDN SUBSTATION
25kV BUS SIMULATION
Zc = .65 +J.0185 DHM
Zp = .0000012 + J.0000022 DHM Zs = .670 + J.000 DHM
Figure 18 25kV Bus EMTP Model
44


UFPENT
CT Primary and Secondary Current Comparision (peak)
o
o
Figure 19 25kV Bus EMTP Model, Phase A CT with Saturation


VC:.
CT Flux Density
£
0>
o
0
pi'
u
D
N
D
O
I
Figure 20 25kV Bus EMTP Model, Phase A CT with Saturation


CT Primary and Secondary Current Comparision (peak)
o
o
Figure 21
25kV Bus EMTP Model, Phase A CT with Hysteresis


CT Flux Density
Figure 22 25kV Bus EMTP Model, Phase A CT with Hysteresis


CT Primary and Secondary Current Comparision (peak)
Figure 23 25kv Bus EMTP Model, Phase A CT with Remanence


CT Flux Density
ui
o
HIlllSECOOS
Figure 24 25kV Bus EMTP Model, Phase A with Remanence


DILLIDN SUBSTATIDN
115/25kV TRANSFORMER DIFFERENTIAL SIMULATION
Zp = .0000073 + J.0000013 OHM Zs = .104 + J.000 OHM
SOURCE
Zc = .65 = J/0185 OHM
600/5MR
@200/5
Zr = .044 OHM
Zr = .044 OHM
LOAD
Zc = .65 = J/0185 OHM
1200/5MR
@1200/5
Zp = .0000013 +J.0000022 OHM 2s = .744 + JdOO OHM
Figure 25
]
115kV Power Transformer EMTP Model
51


-JRRE
CT Primary and Secondary Current Comparision (peak)
Figure 26 115kV Power Transformer EMTP Model, Phase A CT with Saturation


CT Flux Density
Figure 27 115kV Power Transformer EMTP Model, Phase A CT with Saturation


OT
CT Primary and Secondary Current Comparision (peak)
Figure 28 115kV Power Transformer EMTP Model, Phase A CT with Hysteresis


CT Flux Density
Figure 29 115kV Power Transformer EMTP Model, Phase A CT with Hysteresis


CT Primary and Secondary Current Comparision (peak)
o
o
Figure 30 115kV Power Transformer EMTP Model, Phase A CT with Remanence


SEC
CT Flux Density
ui
8 _
(in'
I V
.iu:
On
> i
MILllStOTCS
(i
n
Figure 31 115kV Power Transformer EMTP Model, Phase A CT with Remanence


CHAPTER IV
COMPARISON WITH EXISTING TECHNIQUES AND RESULTS
A comparison of the steady state, time-to-saturation
and the EMTP simulation shows that the different
techniques can give different results. Again the steady
state techniques do not take into consideration the dc
offset at the time of a fault occurrence. The time-to-
saturation method uses the various electrical system
constants to estimate the transient dc offset and can
account for remanence. The EMTP simulation that was
developed can model saturation, hysteresis, and
remanence.
For a three-phase fault on the 25kV feeder with the
following conditions:
C800 CT, 400/5 Tap of 1200/5 MRCT.
Three-Phase Fault Level on the 25kV Feeder
3741 Primary amps, 47 Secondary amps,
Total Burden of approximately 1.55 ohms
All three steady state estimations indicated that the CT
would not saturate. The time-to-saturation calculation
method indicated that the CT would saturate in seven
msec. The EMTP simulation with a fault initiation at
zero degrees for A phase current showed that the addition
58


of the dc transient component in the secondary circuit
resulted in saturation within approximately 4-8msec.
For a three-phase fault on the 25kV bus with the
following conditions:
C800 CT, 1200/5 Tap of 1200/5 MRCT.
Three-Phase Fault Level on the 25kV Feeder
3741 Primary amps, 47 Secondary amps,
Total Burden of approximately 4.1 ohms
The bus relay has a burden of 2600 ohms when a fault
occurs. This burden will be placed in parallel with other
feeder CTs for a bus fault feed through the main breaker.
With a burden of approximately 4.1 ohms, the steady state
estimations indicate that the CT would not saturate. The
time-to-saturation method indicated that the CT would
saturate in approximately 75msec. The EMTP simulation
showed a reduction in the magnitude of ct secondary
current but no distortion. Due to the numeric
oscillation a definite saturation period was hard to
distinguish.
For a three-phase fault on the 115kV power
transformer high voltage terminals with the following
conditions:
59


C400 CT, 200/5 Tap of 600/5 MRCT.
Three-Phase Fault Level at 115kV Bus
5784 Primary amps, 144 Secondary amps,
Total Burden of approximately 2.5 ohms
All three steady state estimation methods indicated that
there would be saturation. The time-to-saturation method
had immediate saturation in less that 1 msec. The EMTP
simulation indicates that saturation occurs with in the
first quarter cycle.
The primary purpose of attempting an EMTP simulation
of the Dillion Substation was to investigate the
interaction of a three-phase relay system connected under
normal operation and compare the results with existing
methods of CT performance estimation. The CT model that
was used in the EMTP simulation (Saturation Transformer
Model) developed some numerical oscillations due to the
small reactances and high currents in the CT model. It
seems that the EMTP model provided a good approximation
but that another type of model in the EMTP program may be
more suited for the CT application. The addition of
hysteresis could only be connected to the secondary
terminals and not the internal model of the CT. The
effects of hysteresis on the EMTP simulation were to
reduce the distortion of the secondary current as
compared to the saturation effects and to lengthen the
60


time that the dc component took to increase the flux
density in the secondary circuit.
Harmonics are a result of current distortion.
During transient conditions odd and even harmonics can be
developed. This was not addressed.
61


CHAPTER V
CONCLUSIONS
The application of current transformers requires the
understanding of the non-linear characteristics of CTs,
CT accuracy ratings, and the transient behavior of CTs.
Where the application does not require a great amount of
accuracy such as an overcurrent relay on a radial feeder,
the CT saturation may not be of great concern. The
protection of equipment with differential relay schemes
which will require that CTs which are paralleled to be
matched within a certain limit will need more study. CT
applications on extra high voltage lines will require a
thorough understanding of CT saturation and relay
response times. Again the higher quality CTs will reduce
application problems.
The comparison of steady state estimations methods,
the time-to-saturation, and the EMTP computer model
simulations indicated different results. The steady
state and time-to-saturation methods have been used quite
sucessfully. Application of each method will depend on
the project under design.
The EMTP computer model simulation was adjusted for
a maximum transient dc offset current at the time of the
62


three-phase fault. Naturally this is a realistic
condition but a condition which does not occur very
often. Therefore this computer model indicated a worst
case situation. The results of the simulation will need
to be refined because of the numerical oscillations
encountered. Another type of EMTP model may give better
results.
This investigation was primarly concerned with
current transformers. The transient effects of potential
transformers also need to be investigated. Both
instrument transformers provide the necessary information
to protection relays for proper operation.
Other areas for future study include relay response
to transient currents and voltages, the application of
the many relay schemes when exposed to transient current
and voltages, and the effects of harmonics generated by
saturated CTs.
63


APPENDIXES
A: Short Circuit Information
B: CT Characteristics
C: CT Steady State Calculations
D: CT Transient Calculations
E: EMTP Simulation
64


1. 3LC dots-In fault on:
PILLION 25 25. kV BUS X 25. kV IL
FAULT CURRENT (A 3 DEG)
SCO - SEO 0 SCO A PHASE B PHASE C PHASE
3740.73 -84.9 0.03 0.0 0.03 0.0 3740.73 -B4.9 3740.73 155.1 3740.73 35.1
THEVENIN IMPEDANCE (OHM)
0.34053*J3.84356 0.34053* J3.84356 9.8E-11*J3.23568
SHORT CIRCUIT MVA> 162.0 X/R RATIO- 11.2869 R0/X1- 2.5E-11 X0/X1 - 0.84185
MONITORED BRANCH: 0 DILL ION 25 25 a K < V 0 BUS X 25.0KV IL
SEQ - SEO 0 SEO A PHASE B PHASE C PHASE
RELAY CURRENT (A) 3744.33 84.9 o.oa o.o o.oa 0.0 3744.33 - 84.9 3744.33 155.1 3744.33 35.1
BUS VOLTAGES (KV, LG)
0 PILLION 25 25.0 0.0003 0.0 0.0003 0.0 0.0003 0.0 0.0003 0.0 0.0003 0.0 0.0003 0.0
0 BUS X 2S.0 o.oooa 0.0 o.oooa o.o o.oooa 0.0 o.oooa 0.0 0.0003 0.0 0.0003 0.0
3lo- o.oa c 1.0 A Va/Ia- 5.05.-0163 80.7 ChM (Vs-Vb)/(la-lb). 5. 05e-016a 80.7 Ohm
(Zo-ZD/321 0.0000 3 0.0
3'
APPENDIX A: SHORT CIRCUIT INFORMATION


1. 3LG Close-in fault on:
DILLON 115 115. kV DILL ION 25 25. kV 1T
FAULT CURRENT (A 3 DEG)
* SEO - SEO 0 SEO A PHASE B PHASE C PHASE
57B3.1S -BA.6 0.03 0.0 0.03 0.0 5783.13 -BA.6 57B3.13 155.A 5783.13 35.A
THEVENIN IMPEDANCE (OWO
1.082A3+J11.A298 1.0B2A3.J11.A298 1.19715.J11.7867
SHORT CIRCUIT MVA* 1151.9 X/R RATIO* 10.559A RO/X1* 0.10A7A XO/XI* 1.03122
Or
Or ....................................................................................................................................
MONITORED BRANCH: 3AA DILLON 115 115.0XV-> ODILLION25 25.0XV1T
+ SEO * SEO 0 SEO A PHASE B PHASE C PHASE
RELAY CURRENT (A) 5783.9a -84.6 O.oa o.o o.oa 0.0 5783.9a -84.6 5783.93 155.A 5783.9a 35. A
BUS VOLTAGES (XV, L -C)
544 DILLON 115 115.0 o.oooa 0.0 0.0009 o.o o.oooa 0.0 o.oooa o.o 0.0003 0.0 0.0009 0.0
0 DILL ION 25 25.0 7.4749 75.1 o.oooa 0.0 o.oooa 0.0 7.4749 75.1 7.A7A3 16A.9 7.4749 AA.9
3lo= o.oa 0.0 A Vs/las 1.22e-0163 -57.1 Ohms (Va-Vb)/(la-lb)= 1 ,22e-0163 -57.1 Ohms


APPENDIX B: CT CHARACTERISTICS
MARKED RATIO SEC. TURNS SEC. TAPS
50:5 10 X2-X3
100:5 20 X1-X2
150:5 30 X1-X3
200:5 40 X4-X5
250:5 50 X3-X4
300:5 60 X2-X4
400:5 80 X1-X4
450:5 90 X3-X5
500:5 100 X2-X5
600:5 120 . X1-X5
TYPICAL
EXCITING CURRENT CURVES
TYPE: BUSHING
FREQUENCY BO HZ.
MAXIMUM RATIO 600:5
TOTAL SEC. TURNS 120
SEC. RES. 0.0029OHMS/TURN @85 CENTIGRADE
ANSI ACCURACY <*00
BURDEN 4 0 ( 4.0OHMS. .50 PERCENT P. F. LAG)
67


a\
oo
TYPICAL SEC. EXCITATION CORVES
PART NO._______________ FREQUENCY _60. HERTZ
ANSI ACCURACY- C-800
RESISTANCE 7SC .0031 OHMS/ TURN


APPENDIX C: CT STEADY STATE CALCULATIONS
25kV Feeder CT
C800, 400/5 Tap on a 1200/5 MRCT
Formula Method
Excitation Method
ANSI CT Accuracy Method
25kV BUS CT
C800, 1200/5 Tap on a 1200/5 MRCT
Formula Method
Excitation Method
ANSI CT Accuracy Method
Power Transformer High Voltage CT
C400/ 200/5 Tap on a 600/5 MRCT
Formula Method
Excitation Method
ANSI CT Accuracy Method
69


Formula Method
25kV Feeder CT
C800, 400/5 Tap on a 1200/5 MRCT
CT ratio, maximum
CT ratio, tap
ct
ratio
ct
tap
1200
5
400
5
CT secondary resistance
ohm
Z : = 0031 ---ct
ct_sec_full turn ratio
Z
ct_sec_full
ohm
Z : = 0031 ---ct
ct_sec_tap turn tap
Z
ct_sec_tap
0.74ohm
0.25-ohm
the rated secondary load impedance
Z := 8.00ohm
rated
ct
tap
Z : ----------Z
ct_rated_tap ct rated
ratio
Z =2.67-ohm
ct_rated_tap
70


the rated secondary current
I := 5amp
rated
I := 20I I = 100-amp
sec cal rated sec cal
Maximum electrical system fault currents
I
25kV fault
3744amp
the cross-sectional area of the iron core
A := 12.in
the frequency in Hz
f := 60Hz
Voltage of CT full winding
V := 800volt
rated
I = 100amp
sec cal
Vs := I
\Z + Z 1
sec_cal l ct_sec_full rated J
rated
Vs = 874.4-volt
Vs
P : =
P = 1.77-tesla
-8
4.44- fA1ct -10
ratio
71


Voltage of CT tap winding
V := 800'
rated
ct
tap
ct
ratio
volt
V = 266.67-volt
rated
Vs : =
2 5kV fault
ct
tap
[Z=t_
sec_tap ct_rated_tap
Vs = 136.41-volt
Vs
J3 : =
/B = 0.83- tesla
-8
4.44fAct 10
tap
/3 is less than the Bmax rating for silicon steel
which can range between 1.2 to 1.9 tesla,
therfore there is no saturation for this application.
Note:
that the Burden modeled is approximately 1.55 ohms
This is less than the maximum of 2.67 ohms.

72



Excitation Curve Method
2 5kV Feeder CT
C800, 400/5 Tap on the 1200/5 MRCT
CT secondary rated voltage V := 800volt
rated 1200
CT ratio, maximum ct : =
ratio 5
400
CT ratio, tap ct : =
tap 5
Secondary Burden Voltage
ct
tap
V :=----------V V = 266.667-volt
tap_rated ct rated tap_rated
ratio
Secondary Burden Impedance
Z := 8ohm
rated
ct
tap
Z : =---------Z
ct_tap_rated ct rated
ratio
Z =2.667-ohm
ct_tap_rated
Secondary Current
I := 5amp
sec_rated
I := 20-1 I = 100-amp
sec cal sec rated sec cal
73


Maximum electrical sysem fault currents
I := 3744amp
2 5kV fault
Secondary burden (leads + relay)
Standard burden power factor
0 : = 5
b
Z := 1.31-ohm
acos|0 1 = 60-deg
i acos f0 1
Lb J
M
Z : = Z e
br b
Z = 0.655 + 1.134i-ohm
br
CT impedance
. 0031ohm
Z : =----------ct
ct_tap turn tap
CT impedance power factor
0 : = 1
ct
= 0.248-ohm
ct tap
M
acos f |=0- deg
i acos T I
L ct J
Z : = Z e
ctr ct_tap
= 0.248-ohm
ctr
CT excitation voltage
I
25kV_fault
V :=--------------I Z
cte ct [_ ct_tap
tap
p + z i
[_ ct_tap b J
V = 72.914-volt
cte
74


I
100amp
Maximum Secondary current
I := I
max sec_cal max
Maximum Secondary excitation current
I
max
I := -----
e 9
I = 11.111-amp
e
I : = I + I
st max e
by definition
I = 111.Illamp
st
I
e
----- = 10-%
I
st
at a maximum secondary excitation current of 11.1 amps,
1200/5 ratio, a tap of 400/5, 2.67 ohms, the
secondary excitation voltage is read from Es vs Ie graph
V := 280-volt
e
From Secondary excitation curve
for a excitation voltage of 72.9V. I := .085-amp
eg
I
25kV_fault
I := .------------ + I
st ct eg
tap
by definition
I
eg
----- = 0.181-%
I
st
For a C800 class CT, 1200/5 ratio on a tap of 400/5,
with a burden of 1.55 ohms the excitation current
is .085 amps or .181% error which is less than 10%.
75


ANSI CT Accuracy Method
2 5kV Feeder CT
C800, 400/5 Tap on a 1200/5 MRCT
CT secondary rated voltage
rated
s = 800volt
CT ratio, maximum
CT ratio, tap
ct
ratio
ct
tap
1200
5
400
5
Secondary Current
I := 5amp
sec rated
I := 20-1 I = 100'amp
sec cal sec rated sec cal
Maximum electrical system fault currents
I := 3744amp
25kV fault
Secondary Burden Voltage
ct
tap
V :=----------V V = 266.667-volt
tap rated ct rated tap_rated
ratio
76


Secondary burden (leads + relay)
Standard burden power factor
0 : = .5
b
i acos To 1
L b J
Z : = Z e
br b
CT impedance
Z := 1.31-ohm
b
Z = 0.655
br
60deg
+ 1.134iohm
.0031-ohm
Z : =---------ct
ct_tap turn tap
CT impedance power factor
Q : = 1
ct
Z : = Z
i
e
acos
]
ctr ct_tap
Maximum Burden current
Maximum ct secondary current
Z
ct_tap
Z
ctr
0.248-ohm
0 deg
0.248ohm
I
sec max
Z
max
I I = 100amp
sec cal sec max
ct V
tap rated
---------:--------- Z = 2.667-ohm
ct I max
ratio sec max
At a maximum fault current of 100 amps the CT will
saturate for an impedance greater than
2.667 ohm total burden.
77


Maximum Burden fault current
Maximum ct secondary fault current
I
2 5kV_fault
:= ------------- I
ct aec_max
tap
ct V
tap rated
Z :=--------------------Z
max ct I max
ratio sec max
I
sec max
46.8* amp
5.698-ohm
At a maximum fault current of 46.8 amps the CT will
not saturate for total burden less than 5.6 ohms.
78


Formula Method
25kV Bus CT
C800, 1200/5 Tap on a 1200/5 MRCT
CT ratio, maximum
CT ratio, tap
ct
ratio
ct
tap
1200
5
1200
5
CT secondary resistance
ohm
Z : = 0031 --ct
ct_sec_full turn ratio
Z
ct_sec_full
ohm
Z : = 0031 --ct
ct_sec_tap turn tap
Z
ct_sec_tap
0.74-ohm
0.741ohm
the rated secondary load impedance
Z s = 8.00ohm
rated
ct
tap
Z : = ----------Z
ct_rated_tap ct rated
ratio
Z = 8ohm
ct_rated_tap
79


the connecting lead and relay burden
Z : = 3.4 ohm
lr
the rated secondary current
I := 5amp
rated
I := 20- I I = 100-amp
sec cal rated sec cal
Maximum electrical system fault currents
I := 3744amp
25kV fault
the cross-sectional area of the iron core
A s = 12.in
the frequency in Hz
f := 60Hz
Voltage of CT full winding
V := 800-volt
rated
I = 100amp
sec cal
Vs := I
sec cal
\z + 2 1
L ct_sec_full rated J
rated
Vs = 874.4-volt
Vs
fi : =
(3 = 1.77 tesla
-8
4.44-f-A-ct -10
ratio
80


Voltage of CT tap winding
ct
V s= 800'
rated
tap
ct
ratio
volt
V = 800volt
rated
Vs s =
25kV fault
ct
tap
T
ct_sec_tap
+ Z 1
ct_rated_tap J
P
vs
-8
4.44- f-A-ct 10
tap
Vs = 136.41-volt
/3 = 0.28- tesla
/3 is less than the Bmax rating for silicon steel
which can range between 1.2 to 1.9 tesla,
therfore there is no saturation for this application.
Voltage of CT with a
large differential burden of 2600
f ct
V
rated
800-
tap
ct
ratio
volt
ohms
V
rated
Vs
I
25kV fault
ct
tap
[2
ct_sec_tap
+
Vs
P
Vs
-8
4.44- fA-ct 10
P
tap
CT
= 800volt
= 64.65-volt
= 0.13-tesla
not saturate
/3 is less than the Bmax rating for silicon steel
which can range between 1.2 to 1.9 tesla,
therfore there is no saturation for this application.
81


Excitation Curve Method
25kV Bus CT
C800, 1200/5 Tap on a 1200/5 MRCT
CT secondary rated voltage V := 800volt
rated
CT ratio, maximum
ct
ratio
1200
5
1200
CT ratio, tap ct :=
tap 5
Secondary Burden Voltage
ct tap = 800volt
V tap_rated ct rated tap rated ratio
Secondary Burden Impedance ct tap Z rated := 8ohm
Z : ct_tap_rated ct ratio z rated
z ct_tap_rated = 8ohm
Secondary Current I sec_rated := 5amp
I sec_cal := 20- I sec_rated I sec cal = 100amp
82


Maximum electrical sysem fault currents
I
2 5kV fault
3744amp
Secondary burden (leads + relay)
Standard burden power factor
: = .5
b
i acos ["
L b .
Z : = Z e
br b
Z := 3.4ohm
b
Z
acos
M
= 1.7 +
br
= 60deg
2.9 44 iohm
CT impedance
.0031-ohm
Z : -----------ct
ct_tap turn tap
CT impedance power factor
: = 1
ct
i acos ["
L ct
Z : = Z e
ctr ct_tap
CT excitation voltage
Z
ct_tap
Z
ctr
I
25kV_fault
V :=--------------[z + Z 1
cte ct |_ ct_tap b J
tap
V
cte
0.744'ohm
0 deg
0.744-ohm
64.646volt
83


Maximum Secondary current
I : = I
max sec cal
= 100amp
max
Maximum Secondary excitation current
I
max
= 11.111-amp
I := I + I
st max e
by definition
I = 111.111-amp
st
I
e
----- = 10-%
I
st
at a maximum secondary excitation current of 11.1 amps,
1200/5 ratio, 8 ohms the secondary excitation voltage
is read from Es vs Ie graph
V := 850volt
e
From Secondary excitation curve
for a excitation voltage is 64.6V.
I := .04amp
eg
I
25kV_fault
------------ + I
ct eg
tap
I
eg
by definition ---- = 0.256-%
I
st
For a C800 class CT, 1200/5 ratio on a tap of 1200/5,
with a burden of 3.4 ohms the excitation current is
.04 amps or .256% error. The CT will not saturate.
I
st
84


ANSI CT Accuracy Method
2 5kV Bus CT
C800, 1200/5 Tap on a 1200/5 MRCT
CT secondary rated voltage V := 800-volt rated
CT ratio, maximum 1200 ct : = ratio 5
CT ratio, tap 1200 ct : = tap 5
Secondary Current I := 5amp sec_rated
I := 20-I sec cal sec rated I = 100amp sec cal
Maximum electrical system fault currents
I : = 3744amp 25kV fault
Secondary Burden Voltage ct tap V := V tap_rated ct rated ratio V = 800-volt tap_rated
85


Secondary burden (leads + relay)
Standard burden power factor
0 : = .5
b
i acos T 1
L b J
Z : = Z e
br b
CT impedance
Z := 3.4ohm
b
acos To "I = 60-deg
L b J
Z = 1.7 + 2.944i-ohm
br
.0031-ohm
Z : =----------ct Z
ct_tap turn tap ct_tap
CT impedance power factor
0 : = 1
ct
]
Z := Z e
ctr ct_tap
Maximum Burden current
Maximum ct secondary current
acosIQ
s
Z
ctr
I
sec max
I
sec cal
I
sec max
0.744-ohm
0 deg
0.744-ohm
100amp
ct V
tap rated
Z :=------------------ Z = 8-ohm
max ct I max
ratio sec_max
At a maximum fault current of 100 amps the CT
will not saturate for a less than 8 ohm total burden.
86



Maximum Burden fault current
Maximum ct secondary fault current
I
25kV_fault
I := ------------
sec_max ct
tap
ct tap V rated
max ct ratio I sec max z max
At a maximum fault current of 15.5 amps the CT
will not saturate for a total burden
less than 51.282 ohms.
I =15
sec max
6 amp
282 ohm
87


Formula Method
Power Transformer High Voltage CT
C400, 200/5 Tap on a 600/5 MRCT 600
CT ratio, maximum ct ; =
ratio 5
CT ratio, tap 200
ct : =
tap 5
CT secondary resistance
ohm
Z := .0029------ct
ct_sec_full turn ratio
Z
ct_sec_full
ohm
Z := .0029------ct
ct_sec_tap turn tap
Z
ct_sec_tap
0.35* ohm
0.12-ohm
the rated secondary load impedance
Z := 4.00ohm
rated
ct
tap
Z : ----------Z
ct_rated_tap ct rated
ratio
Z =1.33-ohm
ct_rated_tap
88


the rated secondary current
I : = 5amp
rated
I := 20-1 I = 100-amp
sec cal rated sec cal
Maximum electrical system fault currents
Primary 115kV current
for a transformer
2 5kV fault
I := 814amp
lowt fault
Primary 115kV current
for a transformer
115kV fault
I := 5784amp
hight_fault
the cross-sectional area of the iron core
A
2
12.in
the frequency in Hz
f := 60Hz
Voltage of CT full winding
V := 400volt
rated
I = 100amp
sec cal
Vs := I
sec cal
rz + z ]
[_ ct_sec_full rated J
Vs = 434.8-volt
Vs
P : =
/3 = 1.76- tesla
-8
4.44-f-A-ct 10
ratio
89


Power Transforemer Low Side Fault
ct
tap
V : = 400 --------volt
rated ct
ratio
VB
I
lowt fault
ct |_ ct_sec_tap
tap
P
0
Vs
-a
4.44fA-ct 10
tap
V = 133.33-volt
rated
Z
ct_rated_tap
Vs = 29.49-volt
P = 0.36-tesla
P is less than the Bmax rating for silicon steel
which can range between 1.2 to 1.9 tesla,
therfore there is no saturation for this application.
Note: If the Secondary Load Burden is less than 1.33 Ohms
there CT will not saturate.
Power Transformer High Side Fault
ct
tap
V := 400 volt
rated ct
ratio
I
hight_fault
Vs := --------------[z
ct [_ ct_sec_tap
tap
V = 133.33-volt
rated
ct_rated_tap ]

Vs
-8
4.44-fA-ct -10
tap
Vs = 209.57-volt
P = 2.54-tesla
p is greater than the Bmax rating for silicon steel
which can range between 1.2 to.1.9 tesla,
therfore there CT is saturated for this application.
Note: The CT will saturate for a burden
greater that 1.33 ohms
90


Excitation Curve Method
Power Transformer High Voltage CT
C400 , 200/5 Tap on a 600/5 MRCT
CT secondary rated voltage V = 400volt
rated 600
CT ratio, maximum ct s
ratio 5
200
CT ratio, tap ct : =
tap 5
Maximum electrical system fault currents
transformer primary current
for a 25kV fault
I := 3 7 41amp
25kV_fault
25- kV
I : ------I
lowt_fault 115-kV 25kV_fault
I = 813.261-amp
lowt_fault
transformer primary current
for a 115kV fault
I := 5784-amp
115kV_fault
I : = I
hight_fault 115kV_fault
3
I = 5.78410 amp
hight_fault
91


Secondary Current
I := 5amp
sec rated
I := 20-1 I = 100-amp
sec cal sec rated sec cal
Secondary Burden Impedance
ct
tap
Z : ---------Z
ct_tap_rated ct rated
ratio
Z := 4ohm
rated
Z = 1.333-ohm
ct tap rated
Secondary burden (leads + relay)
Standard burden power factor
0 : = 5
b
i acos [" "I
Lb J
Z : = Z e
br b
Z := 1.4-ohm
b
acos| | = 60-deg

Z = 0.7 + 1.212i-ohm
br
CT impedance
ohm
Z : = 0029 --ct
ct tap
turn tap
Z =0.116-ohm
ct_tap
CT impedance power factor
0 : = 1
ct
i acos [" "I
L ct J
M
acos[ |=0-deg
Z : = Z e
ctr ct_tap
Z =0.116-ohm
ctr
92


133.333-volt
Secondary burden voltage
ct
tap
V :=----------V V
tap_rated ct rated tap_rated
ratio
I
lowt_fault
V :=--------------Z V
lowt_b ct b lowt_b
tap
I
hight_fault
V s =--------------Z
hight_b ct b
tap
CT excitation voltage
I
lowt_fault
V :=--------------fz + Z I
ct_lowt_e ct |_ ct_tap b J
tap
v
hight_b
28.464volt
202.44volt
V
ct_hight_e
V = 30.823-volt
ct lowt e
hight fault
ct
rz m i
L ct_tap b J
tap
V
ct_hight_er
V = 219.214-volt
ct hight e
hight. fault
ct
tap
-ReTz + Z "I
|_ ctr br J
V = 117.994-volt
ct_hight_er
93


Maximum Secondary current
I : = I
max sec_cal
Maximum Secondary excitation
I
max
I := --------
e 9
I : = I + I
st max e
I = 100amp
max
current
I = 11.111-amp
e
I = 111.111-amp
st
I
e
by definition ----- = 10-%
I
st
at a maximum secondary excitation current of 11.1 amps,
200/5 ratio, 1.333 ohms the secondary excitation voltage
is read from Es vs Xe graph
V := 135volt
e
Transformer low side fault
I := .08amp
eg
lowt_fault
------------ + I
ct eg
tap
I
eg
by definition ---- = 0.392-%
I
st
For a C400 class CT, 600/5 ratio on a tap of 200/5,
with a burden of 1.4 ohms the excitation current is
.08 amps or 0.392% error which is less than 10%
From Secondary excitation curve
for a excitation voltage of 30.3V.
I
I := -
st
94


Full Text

PAGE 1

CT APPLICATIONS GUIDELINES: UTILITIES POINT OF VIEW by Gerald Wayne Stone B.S., University of Nebraska, 1976 A thesis submitted to the Faculty of the Graduate School of the University of Colorado at Denver in partial fulfillment of the requirements for the degree of Master of Science Electrical Engineering 1993 ........... ..... 1

PAGE 2

This thesis for the Master of Science degree by Gerald Wayne Stone has been approved for the Department of Electrical Engineering by William R. Roemish Tamal Bose 1-lb-Cf3 Date

PAGE 3

Stone, Gerald Wayne (M.S., Electrical Engineering) CT Applications Guidelines: Utilities Point of View Thesis directed by Professor Pankaj K. Sen ABSTRACT Historically Current Transformers (CTs) .have provided power system current information to various monitoring devices such as protective relays. The application of CTs in the utility industry has followed several standard practices. This thesis will review the steady state application methods and transient estimation techniques used in the past, develop a computer model to simulate current transformer performance under steady state and transient conditions, and compare the results. The computer simulations will include a three-phase radial feeder, a three-phase bus differential scheme, and a three-phase power transformer differential scheme. This abstract accurately represents the content of the candidate's thesis. I recommend its publication. Sign

PAGE 4

ACKNOWLEDGEMENTS This thesis is submitted to the University of Colorado at Denver as the final requirement for the Master of Science degree in Electrical Engineering. The work could not have been completed without the cooperation and assistance of several people and organizations. I especially wish to thank my academic and thesis advisor, Dr. Pankaj K. Sen, Professor of Electrical Engineering and Computer Science at the University of Colorado at Denver, whose guidance and encouragement have been invaluable during this endeavor. I am also grateful to Jeff Selman of Tri-State Generation and Transmission Association for his support and assistance with EMTP/ATP. I would also like to thank George Laughlin of Public Service Company of Colorado for his countless conversations and support. Finally, I wish to thank Public Service Company of Colorado for the use of their computers and word processing equipment. iv

PAGE 5

CONTENTS ACKNOWLEDGEMENTS CHAPTER I. INTRODUCTION II. STEADY STATE PERFORMANCE Classification "C" or "T" Burden/Accuracy Equivalent Circuit iv 1 4 6 7 10 CT Performance Evaluation . . . . 14 Steady State Mathematical Modeling and Simulation III. TRANSIENT PERFORMANCE IV. v. BH Loop/Nonlinearity . Remanence Materials Equivalent Circuit Transient Mathematical Simulation Modeling and COMPARISON WITH EXISTING TECHNIQUES AND RESULTS CONCLUSIONS . v 16 21 26 26 28 29 33 58 62

PAGE 6

APPENDIX . . . . . . 64 APPENDIX A: SHORT CIRCUIT INFORMATION . . 65 APPENDIX B: CT CHARACTERISITCS . . 67 APPENDIX C: CT STEADY STATE CALCULATIONS 69 APPENDIX D: CT TRANSIENT CALCULATIONS . 100 APPENDIX E: EMTP SIMULATION GLOSSARY REFERENCES vi 107 125 133

PAGE 7

CHAPTER I INTRODUCTION The protection of major equipment in a power utility requires the application of protective devices. One of the common links in the changing world of relay equipment from existing electromagnetic relays to the new generation of solid state and microprocessor multifunction relays is the quality of the source of information presented to the protective relay equipment. Historically current Transformers (CTs) and Potential Transformers (PTs) have been the primary equipment used to supply this information to the protective relays. These instrument transformers proportionally reduce the system voltage and current values to safe magnitudes. The performance of a current transformer is a measure of its ability to reproduce accurately the primary current in secondary amperes both fn wave shape and magnitude. There are two components of this which will affect this reproduction. The symmetrical ac component under steady state conditions and the asymmetrical de offset component which is a result of a transient condition. 1

PAGE 8

The behavior of protective relays is usually specified using sinusoidal voltages and currents but de offsets in the primary current of CTs and remanence in the CT cores frequently result in secondary currents that have substantial distortion. This distortion is the result of CT core saturation which can affect the performance of the relay protection scheme. The timeto-saturation, during which time the CT output current is a faithful replica of the primary current, can be determined from generally available power-system and CT parameters by using graphical solutions [1]. Standard ratings and application guides deal with the steady-state sine wave behavior of a CT. Despite the use of recommended rules to avoid saturation, transient saturation always occurs in critical relay applications. After the primary fault current has ceased, the CT can produce unidirectional decaying output current and can have a high level of flux trapped in the core, both of which can affect the performance of relay protection schemes. Consequently, the relay engineer needs an understanding of the non-linear characteristics of CTs, CT accuracy ratings, and the transient behavior of CTs. This thesis will do the following: 1. Review existing CT application methods 2. Develop and use a CT computer model 2

PAGE 9

3. Compare the results of existing CT application methods with the computer model simulation. 3

PAGE 10

CHAPTER II STEADY STATE PERFORMANCE The initial application of CTs for use by protection relays requires the knowledge of the normal power circuit load current, the location and configuration of the CTs, the CT secondary burden, and the power circuit short circuit current. The CT secondary current can be calculated by dividing the primary current by the CT ratio. Usually CTs which monitor the power flowing in the electric system have a ratio selected so that the secondary circuit will produce a predetermined current magnitude for a projected maximum power system load under normal conditions. In the United States, standard CTs and relays are rated at 5 amps while European CTs and relays are rated for 1 amp secondary current. The location and configuration of the CTs are of importance. If a power transformer is to be protected, then the CT ratios of the high voltage side and low voltage side of the power transformer will be different. CT ratios selected for a transformer differential current protection schemewould have to provide secondary currents from both CTs that would have to be matched to 4

PAGE 11

within a certain limit. High voltage bus CTs would be paralleled in a ring bus or breaker and a half bus scheme. And feeder CTs are usually used by themselves. The CT is designed to deliver to a secondary standard burden from one to twenty times rated secondary current without exceeding a 10% ratio error. For a particular voltage class of CT the larger the burden connected to the CT secondary the lower the magnitude of secondary current that can be reproduced.without distortion. Accuracy calculations for maximum CT secondary currents usually use three phase and single line to ground fault current magnitudes. If these calculations are satisfactory then the phase to phase and two phase to ground fault calculations would also be satisfactory. Application of CTs require the specification of many parameters including the mechanical construction, type of insulation (dry or liquid), ratio in terms of primary and secondary currents or voltages, continuous thermal rating, short time thermal and mechanical ratings, insulation class, impulse level, service conditions, accuracy, and connections. 5

PAGE 12

Classification "C" or "T" There are two basic types of CT construction used in relay applications which are grouped into the categories of Toroidal and Wound. Bushing, Window, or Bar CTs use the Toroidal construction. The Bushing CTs are generally multi-ratio and are commonly used in circuit breakers, power transformers, generators and switchgear. Window or Bar CTs are used at medium or lower voltages. This type of construction has the primary conductor passing through the center of the core. The secondary windings are uniformly distributed around the toroidal core. All the flux which links the primary conductor also links the secondary winding. The leakage flux, and thus the leakage reactance, is negligible. The C classification per ANSI C57.13 is usually applied to Bushing, Window, or Bar CTs. The uniformly distributed windings produce negligible leakage flux and therefore the excitation characteristics can be used directly to determine the CT performance. The C classification indicates that ratio correction at any current can be calculated adequately if the burden, secondary winding resistance, and the excitation characteristics are known. The C classification applies to all tap sections of the current transformer winding. Wound CTs have a primary winding of one or more 6

PAGE 13

turns and secondary winding on a common core similar to power transformers. This type of transformer is usually used for medium and low voltage applications. The secondary windings are not evenly distributed due to physical space required for insulation and bracing of the primary winding. This results in fringing effects of the nonuniformly distributed windings, and therefore flux is present which does not link both primary and secondary windings. Auxiliary current transformers are of the Wound construction. The T classification per ANSI C57.13 is applied to Wound current transformers whose core leakage flux affects the ratio appreciably. Appreciably is defined as one percent difference between the actual ratio correction and the ratio correction calculated. The T classification indicates that the CT has to be tested to identify its operating characteristics. Burden/Accuracy The external load applied to the secondary of a current transformer is called the "burden". The burden is either expressed in total ohms impedance with effective resistance and reactance components, or as the total volt-amperes and power factor at the specified value of current or voltage, and frequency. 7

PAGE 14

Accuracy classes are indicated by the number following the C or T designations. This designation indicates the secondary terminal voltage which the CT can deliver to a standard burden at 20 times rated secondary current without exceeding 10% ratio correction. This 10% will not be exceeded at any current from 1 to 20 times rated secondary current at the standard burden or any lower standard burden. For relaying the voltages classes are C100, C200, C400, and C800 corresponding to standard burdens of 1 ohm (B-1), 2 ohms (B-2), 4 ohms (B-4) and 8 ohms (B-8) respectively. For a CT rated C200 with a secondary 5 amp rating, a voltage or 200 volts would be developed across the CT secondary terminals for a 4 ohm burden at 100 amps (20 5 amps). Standard burdens have a .5 power factor. The performance of a CT used in protective relaying is largely dependent on the total burden or impedance in the secondary circuit of the CT. The CT core flux density (and thus the amount of saturation) is directly proportional to the voltage that the CT secondary must produce. So for a given amount of secondary current, the larger the burden impedance becomes, the greater is the tendency of the CT to saturate. The total burden includes the burdens of the individual relays, meters, and other equipment plus the resistance of 8

PAGE 15

interconnecting leads and the internal resistance of the CT. Sufficient accuracy results if series burden impedances are added arithmetically. The results will be slightly pessimistic, indicating slightly greater than actual CT ratio inaccuracy. But, if a given application is so borderline that vector addition of impedance is necessary to prove that the CTs will be sui table, such an application should be avoided. The CT burden impedance of most electromechanical relays decrease as the secondary current increases because of saturation in the magnetic.circuits of the devices. At high saturation, the electromechanical relay burden impedance approaches its DC resistance. Therefore, a given burden may apply only for a particular value of current. The physical connections of CTs effects the burden as seen by each individual CT. When CT primaries and secondaries are connected in series, the burden on each individual CT is decreased in proportion to the number of CTs in use. When CTs are connected in parallel the effect is to increase the burden on each individual CT. The amount of increase is dependent upon the number of CTs and distribution of current between the CTs. This is the case in breaker and half, ring, and double bus 9

PAGE 16

arrangements. In three phase CT connections, the burden on an individual CT can vary with the type of connection (wye or delta) and the type of fault on the system (1 or 2line to ground or multi-phase). Also beware of attempts to "step-up" current from the main CT to the relay with the use of an auxiliary CT. The auxiliary CT may be adequate, but remember that main CT will see the burden impedance multiplied by the auxiliary CT ratio squared. Eguivalent Circuit In an ideal current transformer the primary ampereturns are equal to the secondary ampere-turns. However, every core material requires some energy to produce the magnetic flux in the core. Thus, in an actual current transformer, the secondary ampere-turns are equal to the primary ampere-turns minus the exciting ampere-turns. An approximate equivalent circuit is shown in Figure 1. The primary current is stepped down through a perfect transformer (no-loss). The primary leakage impedance (ZH) is referred to the secondary by turns ratio (n2 ) squared. The CT secondary consists of the secondary impedance (ZL) and the core loss (Rm) and exciting impedances (Xm). (Z8 ) is the CT burden. 10

PAGE 17

a) bl c) Figure 1 The Equivalent Circuit and Phasor Diagram of a Current Transformer [2]. The generalized circuit can be further reduced. ZH can be neglected, since it influences neither the perfectly transformed current IH/n nor the voltage across Xm. The current through Xm, the magnetizing branch, is Ie, the exciting current. The phasor diagram, with exaggerated voltage drops is also shown. In g:eneral, ZL is resistive, Ie lags Vcd by 90 degrees and is the prime source of error. Note that the net effect of Ie is to cause IL to lead and to be smaller than the perfectly transformed current IH/n. 11

PAGE 18

CURRENT TURN SEC ....!!!!.2.. RATIO RES o\BOVE THIS LINE '04 VOl.. TAG 100 20'1 o:o5 FOR A GIVEN EXCrnNG C:UAROIT 200 cor 0.10 FOR ANY UNIT WILL NOT BE LESS THAN I 500!1 flOor 0 .15 9ll'r. OF THE CURVE VALUE 1000 .... "' .J 0 ::> "' z ;: 100 u ... "' :I Ill: Ill: 400!1 eon 0.20 I 50Ct-5 100'1 O.Z!I 600 5 0.31 L .00!1 r&Oor I o.r v .L !I I I I 100 liOn 0 . 1000 ZOOor I 0.!11 1200 !1 zo" 0 .&1 OttiiiS AT 75 C J I ' i IIELOw THIS LINE THE 100"5-EICITING C:UARENT FOR A 1---" v. 1-::%: i i I &00"5 GIVEN 'IIOL TAGE FOR ANY.,.... !100-UNIT WILL NOT EXCEED b?: vy V: I : 400 IT / !-""" i !00!1 I z Vy-l 200!1 I v V; r; v k' 100 c Q u "' .. 10 I Ill II l{ f_ / v V;/, V; v v v v v ' V; w VI v v ll v J 1/ vh I Vf v J L J II: v VI v / v v r I/ Vj 1/ v I OJ 1.0 10 100 SECCINDAJn' ltMS EXCfTII5 AIIPS /e Figure 2 Typical Excitation Curves for Multi-Ratio c Class current Transformer (3]. Figure 2 above, extracted from ANSI C57.13, shows the typical excitation curves for a multi-ratio c class current transformer. The maximum tolerance of excitation values above and below the knee are also specified. These curves define the relationship of the secondary exciting current (Ie) to the secondary voltage (Ee). The unsaturated slope is determined by the magnetic core material. The saturated region is the air-core reactance. When the current transformer core is unsaturated, 12

PAGE 19

the error due to the exciting current is normally negligible. When the voltage is above the knee of the excitation curve, operating in its saturated region the exciting current is no longer negligible, therefore, the ratio error of the current transformer becomes much greater. For T class current transformer, the leakage flux could be appreciable. The exciting flux should be considered, along with the leakage flux, in determining u 20 II z .., a: 16 a: ::> u ,.. 14 a: c o z 12 0 u .., en 10 ..1 c 2 a: 0 z en 6 .., 2 4 2 I ..../ 8.1 THROUGH 8.0 'LL_ 8.0 / w / / 8\..4. ) v v lP v L v / v v 0 0 2 4 6 s 10 12 14 16 18 20 22 TIMES NORMAL PRIMARY CURRENT Figure 3 Typical T Class Overcurrent Ratio Curves (3]. 13

PAGE 20

current transformer accuracy. Although a test should be done, Figure 3, also extracted from ANSI C57.13, shows typical overcurrent ratio curves for a T class current transformer. Any simple equivalent diagram for a CT is, at best, crude. Exciting current is accompanied by harmonics which, in turn, produce harmonic relay currents. An analysis for application purposes is usually made on the basis of sinusoidql fundamental quantities. While this approach is highly simplified, the equivalent diagram is an excellent tool for picturing the phenomenon and estimating the approximate performance to be expected. CT Performance Evaluation The symmetrical component of CT performance is determined by the highest current that can be reproduced without saturation resulting in large errors. Two types of errors introduced by CTs are ratio errors and phase angle errors For relaying applications the phase angle error is neglected. This is because the load on the secondary of a CT during a fault is generally of such a high lagging power factor that the secondary current is practically in phase with the exciting current and therefore the effect of the exciting current on the phase angle accuracy is negligible. Also relays can usually 14

PAGE 21

tolerate a phase angle error without much effect on operation, unlike in a metering application. If the CTs do not saturate then the excitation current is negligible and can be ignored. Although the performance required of CTs varies with the relay application, high quality CTs should always be used. The better-quality CTs reduce application problems, present fewer hazards, and generally provide better relaying. The quality of the CTs is most critical for differential schemes, where the performance of all the transformers must match. In these schemes, relay performance is a function of the accuracy of reproduction, not only at load currents, but at all fault currents as well. Some differences in performance can be accommodated in the relays. In general, the performance of CTs is not so critical for transmission line protection. The CTs should reproduce reasonably faithfully for faults near the remote terminal, or at the balance point for coordination or measurement. For large magnitude, close-in faults, the CTs may saturate; however, the magnitude of the fault current is not usually critical to the relay. For example, an induction overcurrent relay may be operating on the flat part of the curve for a large magnitude, close-in fault. 15

PAGE 22

Here it is relatively unimportant whether the CT secondary current is accurate, since the timing is essentially identical. The same is true for instantaneous or distance-type_ relaying for a nearby internal fault well inside the cut-off or balance point. In all cases, however, the current transformer should provide sufficient current to operate therelay components. The performance and/or application of CTs under steady state conditions has been estimated with the help of three basic methods. 1. Calculation of CT performance using CT design dimensions and materials (Formula Method) 2. Calculations based on CT secondary excitation curves (Excitation Method). 3. Calculations based on ANSI Standard: Current Transformer Accuracy Classes Examples of each of these estimation methods are discussed next. Steady State Mathematical Modeling and Simulation The existing Dillion 115/25kv Substation (Figure 4) was chosen to compare the three steady state calculation methods. The relay protection schemes included three 16

PAGE 23

areas or zones of protection. The first zone consisted of differentially connected CTs around a 115/25kV Power Distribution Transformer. The second zone consists of the 25kV bus differential protection. And the third zone is the 25kV feeder. The Power Distribution Transformer is protected by a General Electric STD15C transformer differential relay. The 25kV bus is protected by a General Electric PVD11 differential voltage relay. The 25kV feeders are protected by General Electric IAC77B phase and IFC53B ground time overcurrent relays. The substation has a system three-phase fault duty of 1152 MVA (5783 amps) on the 115kV bus and a 162 MVA (3741 amps) on the 25kV bus. The power distribution transformer has a 15/20/25 MVA, 115/25kV rating connected delta-wye with an impedance of 7.95%. An initial load of 12.5 MVA was served by the substation. The CT characteristics including the excitation curve and CT internal resistances are listed in Appendix B. The CT connection leads (#10 CU) from the CT location to the relay have an estimated length of 500 feet. The estimated performance of the CTs with the tap ratios as indicated using the Formula, Excitation and ANSI methods are listed in the Table 1 below. A three-phase fault on the power transformer high 17

PAGE 24

voltage terminal will saturate the CT with a maximum current of approximately 145% of rated CT secondary CT steady state caiculation Summary 2SkV Feeder current Transformer C800, 400/5 Tap of 1200/5 MRCTS 1.55 ohms CT Secondary Burden Three-Phase Fault Level at the feeder breaker 3744 Primary Amps, 47 CT Secondary Amps Formula Method Excitation Method ANSI Method Will Not Will Not Will Not Saturate saturate Saturate 2SkV Bus Differential Main Breaker current Transformer C800, 1200/5 Tap of 1200/5 MRCT Three-Phase Fault Level on the 25kV 4.1 ohm CT Secondary Burden bus 3741 Primary Amps, 16 CT Secondary Formula Method Excitation Method Amps Will Not Will Not Saturate Saturate 11SkV Power Transformer Differential current Transformer C400, 200/5 Tap of 600/5 MRCT 2.5 ohms CT Secondary Burden ANSI Method Will Not Saturate Three-Phase Fault Level at the power high voltage terminals 5784 Primary Amps, 144 CT secondary Formula Method Excitation Method Will Saturate Will Saturate Amps ANSI Method Will Saturate Table 1 Estimated Performance of CTs current. All three steady state methods indicated that saturation would occurr. The CT secondary current magnitudes for the bus and feeder faults were less than the 100 amps maximum 18

PAGE 25

secondary current and did not saturate the CTs for the secondary impedance used. 19

PAGE 26

DILLION SUBSTATION CURRENT TRANSFORMER CONFIGURATION 1151 = .75 J:HI Zci = 0.74 IH4 21> = 3.4:5 IH4 rEEDER 87T 878 Figure 4 Dillion Substation one-Line 20

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CHAPTER III TRANSIENT PERFORMANCE In modern HV and EHV power systems more and more emphasis is placed on the ability of CTs to provide an accurate reproduction of the primary electric system current as input to the protective relay systems. Relay decisions need to be made while an asymmetrical de transient component of fault current is still present. This component can cause some degree of CT saturation and thus operating current distortion. CTs generally produce a faithful replica of the primary current from the instant of fault current initiation until saturation begins. This interval, called the "time-to-saturation"[!], can be calculated for most ring-core CTs using a series of generalized curves and readily obtained CT and power system parameters. This method was used to developed a set of curves from which the time-to-saturation can be obtained for a calculated value of saturation factor (Ks). Distortion of the CT secondary output current begins whenever conditions are such that the core flux density enters the "region of saturation". The factors influencing the core flux density are the physical parameters of the CT, the magnitude, duration, and 21

PAGE 28

waveform of the primary currents and the nature of the secondary burden. Saturation of the core can be produced by excessive symmetrical fault currents as well as by lower magnitude asymmetrical (de offset) fault currents. The latter are far more important in determining the transient response of CTs. When a fully offset fault current of the form shown in Figure 5 below TOTAL &S'fiiiiETRICAL GUIUtfiiT Figure 5 TIME Primary current waves [4]. is impressed on the primary of a current transformer, the de offset will, in general, cause a rise in flux in the core several times greater than that required to transform the 60-Hz component of current. Figure 6 below shows the rise in flux in a CT core when a fully offset current is applied to a CT with a resistive burden. In 22

PAGE 29

this diagram the quantity ac plotted on the abscissa represents the flux required to reproduce the 60-Hz component of fault current while the quantity tc represents the flux required to reproduce the transient (de) component of current. The variation in the transient component of flux will be a function of both the primary circuit de time constant and the time constant of the CT secondary circuit. The latter time constant is a function of CT secondary resistance, burden resistance, burden inductance, and the magnetizing impedance. The total flux, required to reproduce the 0 TOTAL Fl..UX TRANSIENT Fl..UX tc ALTERNATING FLUX. c TIME-----Figure 6 Rise of Flux in the Core of a Current Transformer [1). 23

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offset current is considerably greater than that required to reproduce a symmetrical current. In the above illustration, the CT will reproduce the offset fault current completely as long as the induction does not reach the saturation-flux-density level. In modern CTs, this saturation flux density is between 1.9 and 2.0 teslas (20 kilogausses). If the induction greatly exceeds saturation flux density, the CT may produce a severely distorted output current similar to that shown in Figure 7. In this illustration a 1200:5 CT with a 2.6-ohm resistive burden is subjected to a fully REFERRED TO SECONDARY 1&1 a: 1&1 < 1z 1&1 a: a: :l RESIDUAL F\..UX-0 TOTAL BURDEN = (2.6 ... J 01 OHMS POWER SYSTEM TC = 0.05 T1ME---------. Figure 7 Distortion in Secondary Current due to Saturation [1]. offset primary current of 24,000 amperes (20 times 24

PAGE 31

rated). As can be seen, the CT begins to saturate in the first cycle. With pure resistance burden, a sharp drop off of secondary current occurs in each cycle. Addition of inductance to the burden causes a more gradual dropoff of secondary current. In many instances, CTs will be capable of accurately reproducing offset currents for a number of cycles before starting to saturate. However, under some conditions, saturation cannot be avoided and CTs will produce distorted output currents. The time-to-saturation of a CT is determined by the following parameters: 1. Fault current magnitude 2. Degree of fault current offset 3. Time constant of the de component of fault current 4. Residual flux in the core (Remanence) 5. Secondary circuit resistance and power factor, including the effect of the secondary winding resistance. 6. Secondary excitation impedance of CT at power system frequency (from standard exciting current measurements) 7. CT turns ratio 25

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BH Loop/Nonlinearity Usually published magnetic characteristics of various materials which are used in the core of CTs are presented as plots of flux density B as a function of the magnetic field intensity H. The property of the magnetic material traces out a BH curve with a different path of flux density B for an increasing and decreasing magnetic field intensity H. When equal magnitude of positive and negative magnetic field intensity are applied to a magnet material the result is a symmetrically cyclically BH path or loop calleo hysteresis. The flux density B lags behind the magnetic field intensity H. The value of flux density B when the magnetic field intensity H is removed is call the residual flux density. The nonlinear properties of magnetic materials as characterized by the hysteresis loop are the result of the magnetic materials used, the physical construction of the magnetic circuit, and the placement of the windings on the magnetic material. Nonlinearity in CT application is undesirable because it will distort the secondary current waveform. Remanence Remanence is commonly found in the cores of relaying CTs operating in service. Remanence or Residual flux is 26

PAGE 33

a result of the amount of flux in the core of a CT at the moment of primary circuit interruption. In fact, any value of flux between zero and the CTs' saturation level may be retained in the core. With remanence the core may be operating indefinitely at a high peak flux density under normal system load conditions. Then, when a power system fault occurs, the time-to-saturation may be much shorter than without remanence. The remanent flux in the core depends primarily on the magnitude of primary current, the impedance of the secondary circuit and the amplitude and time constant of any offset.transient. Since the impedance of the secondary circuit is generally fixed, the magnitude of remanent flux is governed by the magnitude of the symmetrical component of the primary current and the magnitude of the offset transient prior to the primary current interruption. Maximum remanent flux can be obtained under conditions whereby the primary current is interrupted while the transformer is in a saturated state. The performance of both C and T class CTs is influenced by this remanence or .residual magnetism. Relay action could be slow or even incorrect. Remanence can be substantially eliminated from CTs by the use of small air gaps in the core. A CT with an 27

PAGE 34

air gap in the core has a fairly low residual flux, approximately 10% of saturation density. Residual flux or current transformers with no intentional air gap is approximately 90% of saturation density. Materials Magnetic circuits have been made from many types of magnetic materials. Included in the list are certain forms of iron and its alloys in combination with cobalt, nickel, aluminum, and tungsten. These are known as ferromagnetic materials and are easy to magnetize since they have a high value of relative permeability ur. 1.7 5 CD 0. 'f t'f.Ni / tfCILeHtOfl olio I { .. .,.,.-1 ( I j!\: 75 #I V". I l I r! ._/ .., .. rl 8 i/ v .. / / j 25 /__ ca'1 iron 1.50 1.25 ] 1.00 "' !! 0. 0. 0 0 50 100 150 200 250 300 350 400 1000 2000 3000 H (ampere-turM/IIIItlr) Figure 8 Normal Magnetization curves for Common Magnetic Materials [1]. 28

PAGE 35

The magnetic permeability of ferromagnetic materials varies with magnetic field intensity starting at a relative low value and increasing to a maximum then falling off with increasing saturation. Typical magnetization curves are as shown in Figure 8 above. The permeability is also generally different for increasing flux density that for decreasing flux density at the same value of magnetic field intensity as exhibited by the hysteresis loop when magnetization is carried through a complete cycle. Magnetic cores that operate at low frequencies (in the audio-frequency range) are comprised of silicon steel laminations. Modern CTs are made from silicon steel. silicon steels older materials, prior 1947 1.2 to 1.94 Tesla Equivalent Circuit 1.55 Tesla avg >1.55 Tesla avg A digital computer model of the Dillion Substation electrical system was developed. A program designed to simulate an electrical network for the study of transients, Electromagnetic Transients Program (EMTP), was used to develop a model of a three-phase power system including CT secondary circuits. This program was originally developed by Dr. Hermann Dommel in the late 29

PAGE 36

1960's. Since that time the Bonneville Power Administration (BPA) along with many universities and other utilities have developed a small computer version called the Alternative Transients Program. These two programs are quite similar and were used in this study. The Dillion Substation power transformer and the associated CTs were modeled using the Saturable Transformer model. This model represents a two winding or single phase transformer with the advantage that a saturable component can be included in the internal connections of the model. The Saturable Transformer model can be represented as shown in Figure 9 below. There is one single-phase two-winding ideal transformer where the correct transformation ratios for winding 2 can be modeled with respect to winding 1. Both windings have an associated leakage-impedance, characterized by resistance R and inductance L. Leakage inductance of the high voltage winding has to be non-zero, but the low voltage winding can have a zero value. The saturation effect is confined to a single non-linear reactor in the circuit of the low voltage winding. As a consequence, the saturation branch is connected to the star point, which is not always the best connection point. Ideally the nonlinear inductance should be connected to 30

PAGE 37

SUS1-1 R2 8US2-1 BUS2-2 Figure 9 Single-Phase Two-Winding [5]. that point of the equivalent star circuit where the integrated voltage equals the iron-core flux. Normally, this is only true for the winding which is closest to the iron core. In the case of saturation, the Type-98 pseudononlinear reactor model is used internally. For a CT the (Vrms,Irms) characteristic is obtained from the CT excitation curves published by the CT manufacturer. An auxiliary routine "SATURATION" is used to convert the Vrms,Irms curve information into (Flux, current) peak value information for the EMTP program. Excitation losses (iron core) are confined to a linear resistance (RMAG) which is in parallel to the saturation branch. The specification of remanent magnetism requires a (flux current) hysteresis curve. Recall that the shape of the hysteresis loop for an inductor depends primarily on the material of the core, while the scaling of the hysteresis loop depends on the geometry, the number of 31

PAGE 38

turns and other construction factors. The auxiliary routine "HYSDAT" is used to create a Type-96 hysteretic inductor characteristic which will model the (flux,current) characteristic of a transformer using the curve shape corresponding to ARMCO Mn steel. Without proper field testing to obtain the CT hysteresis loop this is a good estimate for the CT hysteresis behavior. The Saturable Transformer Model could not model hysteresis internally. To model the hysteresis and remanence characteristics the Type-96 hysteretic inductor was modeled on the CT secondary terminals. The impedance values used in the CT model were obtained from the published voltage-current excitation curves. The secondary resistance can be obtained in a ohmic value per turn. The secondary leakage impedance is negligible due to the uniformly distributed secondary windings and has a value of zero. The primary value of R and L is not easy to obtain because there is not a primary winding as such but only a terminal or wire that the secondary winding is placed around. The impedance values of the CT primary and secondary circuits were estimated in the EMTP model by transforming 10% of the secondary impedance to the primary side of the CT. The values to calculate th. e flux-current saturation curve were read from the published voltage-excitation current 32

PAGE 39

curve for the CT and converted to flux-current values by the "SATURATION" auxiliary program. The value of RMAG was estimated by dividing the voltage squared at the knee of the voltage-excitation current by the associated excitation power loses. The CT model input data calculations are shown in Appendix D. Transient Mathematical Modeling and Simulation The Dillion Substation model consisted of three areas or zones of protection that can be found in most relay protection schemes. The first zone consisted of differentially connected CTs around a 115/25kV Power Distribution Transformer. The second zone consists of the 25kV bus differential protection. And the third zone is the 25kV feeder. In addition to the CT model the substation was modeled with a system three-phase fault duty of 1152 MVA on the 115kV bus and a 162 MVA on the 25kV bus. The power distribution transformer was a 15/20/25 MVA, 115/25kV rating connected delta-wye with an impedance of 7.95%. All bus work was modeled with minimal resistance. An initial load of 12.5 MVA was served by the substation. The CT secondary circuit or burdens included both connecting wires and relay burdens. The connecting leads were estimated at 500 feet of #10 Copper from the CT 33

PAGE 40

location to the relay location. The relay impedances were estimated at the expected value of secondary current under a three phase bus fault condition. Three separate phase faults were initiated (Figure 10) at the power transformer high voltage terminals, the 25kV bus, and on the 25kV feeder. The EMTP program was written to initiate a fault at a time when the largest de offset would occur. The 25kV feeder fault was simulated with the ATP program. The program requirements were small enough for a solution on a personal computer (Figure 11). As can be seen in Figures 12 and 13, although while a secondary fault current of only 49 amps is expected, there is some secondary current distortion due to saturation of the CT core with the transient de offset. In Figures 14 and 15 the effects of hysteresis increase the amount of current distortion and saturation of the CT core. Remanence in the CT core as modeled in Figures 16 and 17 reduced the level of current distortion and CT core saturation. The 25kV bus fault was simulated with the EMTP program (Figure 18). The program requirements were too large for a personal computer and were simulated on a mainframe IBM computer. Although the EMTP program would solve, numerical oscillations were encountered and could not be eliminated. The results are shown in Figures 19, 34

PAGE 41

20, 21, 22, 23, and 24. The transient de offset is present and reduces the secondary current magnitude. Although there is little distortion. The 115kV power transformer high voltage terminal fault was simulated with the EMTP program (Figure 25) The program requirements were too large for a personal computer and were simulated on a mainframe IBM computer. Again the EMTP program would solve; but numerical oscillations were encountered and could not be eliminated. The results are shown in Figures 26, 27, 28, 29, 30, and 31. The CT current is extremely distorted when using the saturation model. While the addition of hysteresis and remanence did not result in distortion but a reduction in current magnitude. 35

PAGE 42

DILLION SUBSTATION CURRENT TRANSFORMER CONFIGURATION 1151
PAGE 43

SOURCE 1200/SMR @400/5 LOAD DILLION SUBSTATION 25kV FEEDER SIMULATION Zc = .65 + J.018S CHH Zc = .65 + J.OlBS CHH Zp = .59 + J.43 [Hf Zrg = .26 7 + J 730 DHM Zp = .0000038 + J.000006 7 DHH Zs = .223 + j.OOO DHH Figure 11 25kV Feeder EMTP Model 37

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w ()) CT Primary and Secondary current Comparision {peak) [a] 11)1) !'51l n .... ... ..... .,._,.. -50 -too 0 20 10 60 eo 100 t [ml Figure 12 25kV Feeder EMTP Model, Phase A CT with Saturation

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w CT Flux Density 1000 \) 500 -5()0 -1000 0 20 10 60 eo 100 t [mb] Figure 13 25kV Feeder EMTP Model, Phase A CT with Saturation

PAGE 46

0 CT Primary and Secondary Current Comparision (peak) 100 eo .... -50 -100 0 20 10 60 eo 100 t [mal Figure 14 25kV Feeder EMTP Model, Phase A CT with Hysteresis

PAGE 47

.... CT Flux Density [mv-s] HiUU woo1\ 1-\ /\\ /" 'J ., I \ 5()0 ., ....... -5ll0 -1000 -lbUU () 20 10 60 eo 100 l lnn. J Figure 15 25kV Feeder EMTP Model, Phase A CT with Hysteresis

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1\) CT Primary and Secondary Current Comparision (peak) [a] 100 !50 -bu -100 0 20 10 60 eo 100 t [msl Figure 16 25kV Feeder EMTP Model, Phase A CT with Remanence

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w CT Flux Density [mv-!il 1600 lliUU-':501) 500 -tooo-0 20 iO 60 80 100 t [mi>J Figure 17 25kV Feeder EMTP Model, Phase A CT with Remanence

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DILLION SUBST A TIDN 25kV BUS SIMULATION SOURCE 1200/5MR @1200/5 Fa.ult Zc = .65 + J.Ol85 OHM 25kV BUS 1200/SMR @1200/5 LDADg LOAD Zp = .0000012 + J.0000022 DHH Zr = 2600 OHM 1 Zc = .65 + J.Ol85 OHM Zs = .6 70 + J.OOO OHM Figure 18 25kV Bus EMTP Model 44

PAGE 51

U1 CT Primary and Secondary Current Comparision (peak) .... .. 8 .; 8 r. 0 0 I! 0 UJI) Q.ll Q.' tl' 0 0 .; N I :;! I Figure 19 25kV Bus EMTP Model, Phase A CT with Saturation

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,c:.. 0\ CT Flux Density II 0 0 ... N I) I) 0 "f -,--.C..co iii.oo a.oo u--l. mliii b t.m l&l.m -: : HllliSEcmJS 0 1 I I II I '! U> 0 '! " : Figure 20 25kV Bus EMTP Model, Phase A CT with Saturation

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-.J CT Primary and Secondary Current Comparision (peak) D D ,; Figure 21 25kV Bus EMTP Model, Phase A CT with Hysteresis

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co CT Flux Density 0 1: ,, .. '! .. u g (I -lll 1.00 10.00 "' Ul I .J'! 0-::- g ': u :: A . . '--.------------. U M U U Figure 22 25kV Bus EMTP Model, Phase A CT with Hysteresis

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\D CT Primary and Secondary Current Comparision (peak) ll 0 D 0 '! 0 N u 0 0 fz I!Jo ltll It' if )I D D N I u 0 I Figure 23 25kV Bus EMTP Model, Phase A CT with Remanence

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CT Flux Density u 0 D , u " nl g: ,i oi -;:oo ir.liJ-'1\:w-l.oolt.oo JJ,IIJ--l.oo .. a.oo "' "llliSltmlS "' I '" " {) : ;.o .. 'I " u Figure 24 25kV Bus EMTP Model, Phase A with Remanence

PAGE 57

DILLION SUBST ATIDN 115/25kV TRANSFORMER DIFFERENTIAL SIMULATION SOURCE 600/5MR @200/5 1200/5MR @1200/5 LOAD Zp = .0000073 + J.0000013 OHM Zs = .104 + J.OOO OHM Zc = .65 = J/0185 OHM Zo = .096 OHM L Zc = .65 = J/0185 OHM 600/5MR @200/5 Zr = .044 OHM Zr = .044 OHM 1200/5MR @1200/5 Zp = .0000013 + J.0000022 OHM Zs = .744 + J.OOO OHM Figure 25 115kV Power Transformer EMTP Model 51

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CT Primary and Secondary Current Comparision (peak) () I u 0 II ,, 0 '! 0 u I g U1 il;---r-N s.ro J. Z" ur it' Q: ::1 1.1 Figure 26 115kV Power Transformer EMTP Model, Phase A CT with Saturation

PAGE 59

U1 w CT Flux Density ,., I ll II .. 0 II u 2 1.00 ID.IIl hi Ul 1-0 r .. ,,: > ll I) I < .. \ r. J '.' .'\_, I ' o I I I 'tJi J !1 ft. m J b \ b i/oo .. Figure 27 115kV Power Transformer EMTP Model, Phase A CT with Saturation

PAGE 60

U1 CT Primary 0 o. 0 0 0 "' 0 D N 0 0 0 o I 0 0 en and Secondary Current Comparision (peak) ID.ill .. I'! ZJ bJn n -J l l u ; Figure 28 115kV Power Transformer EMTP Model, Phase A CT with Hysteresis

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CT U1 U1 Flux Density D m ,.; D " . : D q u'\too 5.11) ID.m w U1 I '" .Ju! Uo > D '? I i g 't l.ro i.ro u-so:oo -u HILLJS(ClHIS il.oo t.m . il:iii oo.oo e5.11J t:iiJ!w.IIJ ...ur...oo Figure 29 llSkV Power Transformer EMTP Model, Phase A CT with Hysteresis

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CT Primary and Secondary current Comparision (peak) 0 ,, '! [) .. 2 0 rr I g '1-,--y-I I \ II U1 '\ 00 \,IJl IU.IJl 1 1 1 I I I f 1--.\ 0\ zo 111 o D. : ) u () () I ,, I Figure 30 115kV Power Transformer EMTP Model, Phase A CT with Remanence

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Ul ...J CT Flux Density g .. .. IJ IJ 11 u H w Ill I I '' > 0 IJ I u .. I " !L.111 LJ llll ... "ILL ISECIHJS Figure 31 llSkV Transformer EMTP Model, Phase A CT with Remanence

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CHAPTER IV COMPARISON WITH EXISTING TECHNIQUES AND RESULTS A comparison of the steady state, time-to-saturation and the EMTP simulation shows that the different techniques can give different results. Again the steady state techniques do not take into consideration the de offset at the time of a fault occurrence. The time-tosaturation method uses the various electrical system constants to estimate the transient de offset and can account for remanence. The EMTP simulation that was developed can model saturation, hysteresis, and remanence. For a three-phase fault on the 25kV feeder with the following conditions: caoo CT, 400/5 Tap of 1200/5 MRCT. Three-Phase Fault Level on the 25kV Feeder 3741 Primary amps, 47 Secondary amps, Total Burden of approximately 1.55 ohms All three steady state estimations indicated that the CT would not saturate. The time-to-saturation calculation method indicated that the CT would saturate in seven msec. The EMTP simulation with a fault initiation at zero degrees for A phase current showed that the addition 58

PAGE 65

of the de transient component in the secondary circuit resulted in saturation within approximately 4-Smsec. For a three-phase fault on the 25kV bus with the following conditions: CSOO CT, 1200/5 Tap of 1200/5 MRCT. Three-Phase Fault Level on the 25kV Feeder 3741 Primary amps, 47 Secondary amps, Total Burden of approximately 4.1 ohms The bus relay has a burden of 2600 ohms when a fault occurs. This burden will be placed in parallel with other feeder CTs for a bus fault feed through the main breaker. With a burden of approximately 4.1 ohms, the steady state estimations indicate that the CT would not saturate. The time-to-saturation method indicated that the CT would saturate in approximately 75msec. The EMTP simulation showed a reduction in the magnitude of ct secondary current but no distortion. Due to the numeric oscillation a definite saturation period was hard to distinguish. For a three-phase fault on the 115kV power transformer high voltage terminals with the following conditions: 59

PAGE 66

C400 CT, 200/5 Tap of 600/5 MRCT. Three-Phase Fault Level at 115kV Bus 5784 Primary amps, 144 Secondary amps, Total Burden of approximately 2.5 ohms All three steady state estimation methods indicated that there would be saturation. The time-to-saturation method had immediate saturation in less that 1 msec. The EMTP simulation indicates that saturation occurs with in the first quarter cycle. The primary purpose of attempting an EMTP simulation of the Dillion Substation was to investigate the interaction of a three-phase relay system connected under normal operation and compare the results with existing methqds of CT performance estimation. The CT model that was used in the EMTP simulation (Saturation Transformer Model) developed some numerical oscillations due to the small reactances and high currents in the CT model. It seems that the EMTP model provided a good approximation but that another type of model in the EMTP program may be more suited for the CT application. The addition of hysteresis could only be connected to the secondary terminals and not the internal model of the CT. The effects of hysteresis on the EMTP simulation were to reduce the distortion of the secondary current as compared to the saturation effects and to lengthen the 60

PAGE 67

time that the de component took to increase the flux density in the secondary circuit. Harmonics are a result of current distortion. During transient conditions odd and even harmonics can be developed. This was not addressed. 61

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CHAPTER V CONCLUSIONS The application of current transformers requires the understanding of the non-linear characteristics of CTs, CT accuracy ratings, and the transient behavior of CTs. Where the application does not require a great amount of accuracy such as an overcurrent relay on a radial feeder, the CT saturation may not be of great concern. The protection of equipment with differential relay schemes which will require that CTs which are paralleled to be matched within a certain limit will need more study. CT applications on extra high voltage lines will require a thorough understanding of CT saturation and relay response times. Again the higher quality CTs will reduce application problems. The comparison of steady state estimations methods, the time-to-saturation, and the EMTP computer model simulations indicated different results. The steady state and time-to-saturation methods have been used quite sucessfully. Application of each method will depend on the project under design. The EMTP computer model simulation was adjusted for a maximum transient de offset current at the time of the 62

PAGE 69

three-phase fault. Naturally this is a realistic condition but a condition which does not occur very often. Therefore this computer model indicated a worst case situation. The results of the simulation will need to be refined because of the numerical oscillations encountered. Another type of EMTP model may give better results. This investigation was primarly concerned with current transformers. The transient effects of potential transformers also need to be investigated. Both instrument transformers provide the necessary information to protection relays for proper operation. Other areas for future study include relay response to transient currents and voltages, the application of the many relay schemes. when exposed to transient current and voltages, and the effects of harmonics generated by saturated CTs. 63

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APPENDIXES A: Short Circuit Information B: CT Characteristics C: CT Steady State Calculations D: CT Transient Calculations E: EMTP Simulation 64

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0'1 l11 ................................................................................................................................. I. 3LG Cloce In fault on: DILLION 25 25. I RElA T CURRENT (A) BUS VOLTAGES !IV, lli) 0 OILLIOH 25 25.0 0 BUS X 25.0 + SEQ 3144.311 .9 0 .00011 0 0 o.oooa o.o 0 BUS X SEQ o.oa o.o o .oooa o.o O.ODOiil 0.0 25.0KY IL 0 SEQ o.oa o.o 0.00011 0 0 0.00011 0 0 A PHASE 3744.3iil 9 O .OOOiil 0.0 O.OOOiil 0 0 B PHASE 3744.311 155., o.oooa o o 0.00011 0 0 C PHASE 3744.311 35., O.OOOiil 0 0 0.00011 0.0 31o O.Oiil 0.0 A Valla 5 D5eiil 1!10.7 et-5 !YaVb)/(1 albl 5 .05eOI6iil 1!10. 7 Ohfll8 (loZU/321 0.0000 il 0.0 ........................................................... -.. -............ .. ......... -.......... --... -...................................................... ............. -........ --..... --.................................................. "' '1j .r 0 H > .. tn :I: 0 n H n c::: H H :z: 0 H 0 2:

PAGE 72

0'\ 0'\ 1. 3LG Close-In fault on: DILLON 115 115. kV DILLION 25 25. kV 1T SEQ 5781. ltl 6 SEQ o.oa o o FAULT CURRENT (A a OEGl 0 SEQ o.oa o.o A PHASE 5781. 1a -84. 6 THEVUIN IMPEDANCE (OIM) 8 PHASE 578l.1a 155. 4 C PHASE 5783 .101 35.4 1.08243 J 11.4298 1.08243+)11.4298 I. 19715+ J 11. 1'1167 SHORT CIRCUIT MVA 1151.9 IIOIIIIOREO BRANCH: 344 DILLON 115 115.0KY RElAY CURRENT (A) BUS VOLT ACES u : v L C l 344 DilLON 115 115.0 0 DILLION 25 25.0 + SEll 578l.9a -84. 6 o.oooa o.o 7.474a -75 1 X/R RATIO= 10.5594 RO/X1 0.10474 XO/X1 1.03122 0 D JLLIOII 25 SEQ o.oa o o 0 .00001 0.0 0.00001 0.0 25 .OKV 1T 0 SEQ o.oa o o o.oooa o.o O.OOOiil 0.0 A PHASE 5783.901 -84.6 0.00001 0.0 7.474il -75.1 B PHASE 5783.901 155.4 o.ooo;i o o 7.474a 164.9 C PHASE 5783.901 35.4 0 000;} 0.0 7 474il 44.9 31o= o.oa o.o A Vella= 1.22ea -57.1 Ohio& (VaVb)/(lelb) 1.22ea .1 Olwns .r

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APPENDIX B: CT CHARACTERISITCS TYPICAL MARKm RATIO SEC. TURNS SEC. TAPS EX CITING CURRENT CURVES 50:5 10 X2-X3 100:5 20 Xt-X2 TYPE: BUSHING FREQUENCY 60 HZ. 150:5 30 X1-X3 200:5 40 X4-X5 250:5 50 X3-X4 MAXIMUM RATIO 6oo:5 TOTAL SEC. TURNS 120 300:5 60 X2-X4 400:5 80 X1-X4 450:5 90 X3-X5 SEC. RES. o.oo29 OHMS/TURN @850 CENTIGRADE 500:5 100 X2-XS ANSI ACCURACY c4oo. 600:5 120 xi-x5 BURDEN 4 0 ( 4 .oOHMS . soPERCENT P. F.LAGl 1,000! 500. I : ; -_ . tr ... ___ J-+-+, i . -=--:;..:..r: ---.!,.__;_--. :> .. 6o o 5 t---:-++-++iH+T--+--T/1 sao s 30i-' ... :,o:-' I Eacll .... .._ .,11 10 zo 30 67 450 5 400 5 300 5 250 5 zoo 5 150 5 100 5 50 5

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0\ 00 100 10 .00, .01 .05 0.1 0.5 SECONDARY EXCITING AMPERES -lc rYPICAL SEC. EXCITATION CUR\IES PART NO. _______ FREQUENCY ANSI RCSISrAtiCE@ , 10 20 30

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APPENDIX C: CT STEADY STATE CALCULATIONS 25kV Feeder CT caoo, 400/5 Tap on a 1200/5 MRCT Formula Method Excitation Method ANSI CT Accuracy Method. 25kV Bus CT caoo, 1200/5 Tap on a 1200/5 MRCT Formula Method Excitation Method ANSI CT Accuracy Method Power Transformer High Voltage CT C4001 200/5 Tap on a 600/5 MRCT Formula Method Excitation Method CT Accuracy Method 69

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Formula Method 25kV Feeder CT CSOO, 400/5 Tap on a 1200/5 MRCT 1200 CT ratio, maximum ct := ratio 5 CT ratio, tap 400 ct := tap 5 -----------------------------------------------------------CT secondary resistance ohm z : = 0031 -ct ct sec full turn ratio z = 0.74ohm ct sec full ohm z := .0031-ct ct sec_tap turn tap z = 0.25ohm ct _sec_tap -----------------------------------------------------------the rated secondary load impedance ct tap z := -----z ct_rated_tap ct ratio z z rated rated ct_rated_tap 70 := 8.00ohm = 2.67ohm

PAGE 77

the rated secondary current I := Samp rated I : = 20 I I = lOOamp sec cal rated sec cal -----------------------------------------------------------Maximum electrical system fault currents I := 3744amp 25kV fault -----------------------------------------------------------the cross-sectional area of the iron core 2 A : = 12. in -----------------------------------------------------------the frequency in Hz f := 60Hz -----------------------------------------------------------Voltage of CT full winding v : = BOO volt rated vs := I [z sec cal ct...:_sec_;full 0 Vs /3 := 4.44fAct -a ratio 71 I = lOOamp sec cal + z J rated Vs = 874.4volt {J = 1. 77 tesla

PAGE 78

-----------------------------------------------------------Voltage of CT tap winding v Vs := BOO [ ct ] tap rated := [I 25kV_fault ct tap Vs ] [' ct_sec_tap f3 : = -8 4.44fAct 10 tap v = 266.67volt rated + z J ct_rated_tap Vs = 136.4lvolt f3 = O.B3tesla f3 is less than the Bmax rating for silicon steel which can range between 1.2 to 1.9 tesla, therfore there is no saturation for this application. Note: that the Burden modeled is approximately 1.55 ohms This is less than the maximum of 2.67 ohms. 72

PAGE 79

Excitation Curve Method 25kV Feeder CT caoo, 400/5 Tap on the 1200/5 MRCT CT secondary rated voltage v := BOO volt rated 1200 CT ratio, maximum ct := ratio 5 400 CT ratio, tap ct : = tap 5 -----------------------------------------------------------Secondary Burden Voltage v := tap_rated ct tap -----V ct ratio v = 266.667volt rate. d tap_rated -----------------------------------------------------------Secondary Burden Impedance z : = ct_tap_rated ct tap -----z ct ratio z := 8ohm rated rated z = 2.667ohm ct_tap_rated -----------------------------------------------------------Secondary Current I := Samp sec rated I := 20 I I = lOOamp sec cal sec rated sec cal -----------------------------------------------------------73

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-----------------------------------------------------------Maximum electrical sysem fault currents I := 3744amp 25kV fault -----------------------------------------------------------Secondary burden (leads + relay) Standard burden power factor e := 5 b i acos[eb J z : = z e br b Z : = 1. 31 ohm b acos [e b ] = 60 deg z = 0.655 + 1.134iohm br -----------------------------------------------------------CT impedance .0031ohm z := -----ct z = 0.248ohrn ct_tap turn tap ct_tap CT impedance power factor e := 1 acos [e ct J = Odeg ct i acos [e ct J z := z e z = 0.248ohrn ctr ct_tap ctr CT excitation voltage v := cte I 25kV fault ct tap [z ct_tap + z b ] V = 72.914volt cte 74

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-----------------------------------------------------------Maximum Secondary current I := I I = lOOamp max sec cal max Maximum Secondary excitation current I max I := I = ll.l11amp e 9 e I := I + I I = 111. 111 amp st max e st I e by definition = 10% I st at a maximum secondary excitation current of 11.1 amps, 1200/5 ratio, a tap of 400/5, 2.67 ohms, the secondary excitation voltage is read from Es vs Ie graph V := e -----------------------------------------------------------From Secondary excitation curve for a excitation voltage of 72.9V. I := .OSSamp I := st by definition eg I 25kV fault I eg I st ct tap = 0.181\ For a C800 class CT, 1200/5 ratio on a tap of 400/5, with a burden of 1.55 ohms the excitation current is .085 amps or .181\ error which is less than 10\. 75 + I eg

PAGE 82

ANSI CT Accuracy Method 25kV Feeder CT caoo, 400/5 Tap on a 1200/5 MRCT CT secondary rated voltage v := SOOvolt rated 1200 CT ratio, maximum ct := ratio 5 400 CT ratio, tap ct := tap 5 -----------------------------------------------------------Secondary current I := 5amp sec rated I := 20I I = lOOamp sec cal sec rated sec cal -----------------------------------------------------------Maximum electrical system fault currents I := 3744amp 25kV fault -----------------------------------------------------------Secondary Burden Voltage ct tap v := -----v tap_rated ct ratio v 266.667volt rated tap_rated -----------------------------------------------------------76

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-----------------------------------------------------------Secondary burden (leads +relay) Standard burden power factor e := 5 b i aces [e b J z := z e br b Z := 1.3lohm b acos [e b J = 60 deq z = 0.655 + 1.134iohm br -----------------------------------------------------------CT impedance .003: ohm z := ------ct z = 0.248ohm ct_tap tap ct_tap CT impedance power factor e := 1 ct i acos [e ct J acos [e ct J = Odeq z := z e z = 0.248ohm ctr ct_tap ctr -----------------------------------------------------------Maximum Burden current Maximum ct secondary current I := I I lOOamp sec max sec cal sec max z := ct tap max ct ratio v rated z = 2.667ohm I max sec max At a maximum fault current of 100 amps the CT will saturate for an impedance greater than 2.667 ohm total burden. -----------------------------------------------------------77

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-----------------------------------------------------------Maximum Burden fault current Maximum ct secondary fault current I 25kV fault I := I = 46.8amp sec max ct sec max tap ct v tap rated z :s: z = 5.698ohm max ct I max ratio sec max At a maximum fault current of 46.8 amps the CT will not saturate for total burden less than 5.6 ohms. -----------------------------------------------------------78

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Formula Method 25kV Bus CT caoo, 1200/5 Tap on a 1200/5 MRCT 1200 CT ratio, maximum ct := ratio 5 CT ratio, tap 1200 ct := tap 5 -----------------------------------------------------------CT secondary resistance ohm z := .0031--ct ct sec full turn ratio z = 0.74ohm ct sec full -ohm z := .0031--ct ct_sec_tap turn tap z = 0.74ohm ct_sec_tap -----------------------------------------------------------the rated secondary load impedance ct tap z := ....... ----z ct_rated_tap ct ratio z z rated rated ct_rated_tap 79 := a.ooohm = Sohm

PAGE 86

the connecting lead and relay burden Z : = 3. 4 ohm lr the rated secondary current I := Samp rated I := 20I I = lOOamp sec cal rated sec cal Maximum electrical system fault currents I := 3744amp 25kV fault the cross-sectional area of the iron core 2 A:= 12.in the frequency in Hz f := 60Hz Voltage of CT full winding v := SOOvolt I = lOOamp rated sec cal Vs := I [z + z ] sec cal ct sec .full rated -0 Vs = 874.4volt Vs /3 : = /3 = 1.77tesla -a 4.44fAct 10 ratio 80

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-----------------------------------------------------------Voltage of CT tap winding := BOO [c:\ap ] volt rat1.o v rated = SOOvolt v rated Vs := {3 : = [I 2SkV_fault ] [ ct_sec_tap ct tap vs -8 4.44fAct tap + z J ct_rated_tap Vs = 136.41volt {3 = 0.2Bteela {3 is less than the Bmax rating for silicon steel which can range between 1.2 to 1.9 tesla, therfore there is no saturation for this application. -----------------------------------------------------------ohms v = 800vo1t rated [I 2SkV_fault ] [ ct_sec_tap J vs := + z ct lr tap vs = 64.6Svolt Vs {3 := -8 {3 = 0.13tee1a 4.44fAct 10 tap CT not saturate {3 is less than the Bmax rating for silicon steel which can range between 1.2 to 1.9 tesla, therfore there is no saturation for this application. 81

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Excitation Curve Method CT 25kV Bus CBOO, 1200/5 Tap on a 1200/5 MRCT CT secondary rated voltage v := BOO volt rated 1200 CT ratio, maximum ct := ratio 5 1200 CT ratio, tap ct : = tap 5 -----------------------------------------------------------secondary Voltage v := tap_rated ct tap -----v ct ratio v = BOOvolt rated tap_rated -----------------------------------------------------------Secondary Burden Impedance z := ct tap -----Z ct_tap_rated ct ratio z := Bohm rated rated z = Bohm ct_tap_rated -----------------------------------------------------------Secondary Current I := 5amp sec rated I := 20 I I = lOOamp sec cal sec rated sec cal -----------------------------------------------------------82

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-----------------------------------------------------------Maximum electrical sysem fault currents I := 3744amp 2SkV fault -----------------------------------------------------------Secondary burden (leads + relay) Standard burden power factor z e b br : = 5 i aces [0 b J : = Z e b Z := 3.4ohm b aces [0 b J = 60 deg Z = 1.7 + 2.944iohm br -----------------------------------------------------------CT impedance .003lohm z := ------ct Z = 0.744ohm ct_tap turn tap ct_tap CT impedance power factor e := 1 aces [0 ct ] = Odeg ct i aces [e ct ] z := z e z = 0.744ohm ctr ctr -----------------------------------------------------------CT excitation voltage v := cte I 25kV fault ct tap [z ct_tap + z b ] V = 64.646volt cte -----------------------------------------------------------83

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-----------------------------------------------------------Maximum Secondary current I := I I = lOOamp max sec cal max Maximum Secondary excitation current I max I := I = 11.111amp e 9 e I := I + I I = 111.11lamp st max e st I e by definition = 10\ I st at a maximum secondary excitation current of 11.1 amps, 1200/5 ratio, 8 ohms the secondary excitation voltage is read from Es vs Ie graph v := 850volt e From Secondary excitation curve for a excitation voltage is 64.6V. I st by definition I I := := .04amp eg 25kV fault + ct tap I eg = 0.256\ I st For a CSOO class CT, 1200/5 ratio on a tap of 1200/5, with a burden of 3.4 ohms the excitation current is .04 amps or .256\ error. The CT will not saturate. 84 I eg

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ANSI CT Accuracy Method 2SkV Bus caoo, 1200/5 Tap on CT secondary rated voltage CT ratio, maximum CT ratio, tap CT a 1200/5 v rated ct ratio ct tap MRCT := SOOvolt 1200 := 5 1200 := 5 -----------------------------------------------------------Secondary Current I := Samp sec rated I := 20 I I = 100arnp sec cal sec rated sec cal -----------------------------------------------------------Maximum electrical system fault currents I := 3744amp 25kV fault -----------------------------------------------------------Secondary Burden Voltage ct tap v := -----V tap_rated ct ratio v = SOOvolt rated tap_rated -----------------------------------------------------------85

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-----------------------------------------------------------Secondary burden (leads + relay) Standard burden power factor e := 5 b i acos[eb J z := z e br b Z := 3.4ohm b aces [e b J = 60 deg Z = 1.7 + 2.944iohm br -----------------------------------------------------------CT impedance .003lohm z : = -----ct z 0.744ohm ct_tap turn tap ct_tap CT impedance power factor e : = 1 aces [e J = Odeg ct ct i aces e:l J ct z := z e z = 0.744ohm ctr ct_tap ctr -----------------------------------------------------------Maximum Burden current Maximum ct secondary current I := I sec max sec cal ct tap v rated I = 100amp sec max z := Z = Bohm max ct I max ratio sec max At a maximum fault current of 100 amps the CT will not saturate for a less than B ohm total burden. -----------------------------------------------------------86

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-----------------------------------------------------------Maximum Burden fault current Maximum ct secondary fault current I I := sec max z : = 2SkV fault ct tap ct tap v rated I = 15.6amp sec max Z = 51.2B2ohm max ct I max ratio sec max At a maximum fault current of 15.5 amps the CT will not saturate for a total burden less than 51.282 ohms. -----------------------------------------------------------87

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CT ratio, CT ratio, Formula Method Power Transformer High Voltage CT C400, 200/5 Tap on a 600/5 MRCT maximum ct : = ratio tap ct := tap 600 5 200 5 -----------------------------------------------------------CT secondary resistance ohm z := .0029--ct ct sec full turn ratio z = 0. 35 ohm ct sec full ohm z := .0029--ct ct_sec_tap turn tap z = 0.12ohm ct_sec_tap -----------------------------------------------------------the rated secondary load impedance z := 4.00ohm rated ct tap z := z ct rated _tap ct rated -ratio z = 1.33ohm ct rated _tap 88

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-----------------------------------------------------------the rated secondary current I := Samp rated I := 20 I I = lOOamp sec cal rated sec cal -----------------------------------------------------------Maximum electrical system fault currents Primary llSkV current for a transformer 2SkV fault I := 814amp lowt fault Primary llSkV current for a transformer llSkV fault I := 5784amp hight_fault -----------------------------------------------------------the cross-sectional area of the iron core 2 A:= 12.in -----------------------------------------------------------the frequency in Hz f := 60Hz -----------------------------------------------------------Voltage of CT full winding v := 400volt I = lOOamp rated sec cal Vs := I [z + z ] sec cal ct_sec_full rated Vs = 434.8volt Vs f3 : = f3 = 1.76tesla -8 4.44fAct ratio 89

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-----------------------------------------------------------Power Transforemer Low Side Fault ct v rated I ve := f3 : = tap := 400---------volt ct lowt fault ct tap ratio [z ct sec tap --vs -8 4.44fAct 10 0 tap v = 133.33volt rated + z J ct_rated_tap Vs = 29.49volt f3 = 0.36tesla f3 is less than the Bmax rating for silicon steel which can between 1.2 to 1.9 tesla, therfore there is no saturation for this application. Note: If the Secondary Load Burden is less than 1.33 Ohms there CT will not saturate. -----------------------------------------------------------Power Transformer High Side Fault ct v rated I Vs : = {J : = tap := 400---------volt ct ratio hight_fault [z ct tap VB ct sec tap ---8 4.44fAct tap v = 133.33volt rated + z ] ct_rated_tap Vs = 209.57volt f3 = 2.54tesla f3 is greater than the Bmax rating for silicon steel which can range between 1.2 to 1.9 tesla, therfore there CT is saturated for this application. Note: The CT will saturate for a burden greater that 1.33 ohms 90

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Excitation Curve Method ?ower Transformer High Voltage CT C400, 200/5 on a 600/5 MRCT v := 400volt rated CT secondary rated voltage 600 CT ratio, maximum ct := ratio 5 200 CT ratio, tap ct := tap 5 ----------------------------------------------------------electrical system fault currents transformer primary current for a 25kV fault I 25kV_fault I := 374lamp 2SkV : = I lowt fault 115kV 25kV fault transformer primary current for a llSkV fault I := 5784amp llSkV fault .I := I hight fault llSkV fault 91 I lowt fault I hight_fault = 813.26lamp 3 = 5.78410 amp

PAGE 98

-----------------------------------------------------------Secondary Current sec rated I := S amp I := 20I I = lOOamp sec cal sec rated sec cal ----------------------------------------------------------Secondary Burden Impedance z : = 4 ohm ct rated tap z := ----z ct_tap_rated ct rated ratio z = 1.333ohm -----------------------------------------------------------secondary burden (leads + relay) Z : = 1. 4 ohm b Standard burden power factor a := 5 b i acos [e b ] z := z e acos [a b J = 60deq Z = 0.7 + 1.212iohm br b br -----------------------------------------------------------CT impedance ohm z := .0029-ct z = O.ll6ohm ct_tap turn tap ct_tap CT impedance power factor 0 := 1 ct i acos [e ] ct acos [e ct J = Odeq z := z e ctr ct_tap Z = O.ll6ohm ctr -----------------------------------------------------------92

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-----------------------------------------------------------secondary burden voltage ct tap v : = -----V tap_rated ct rated v := lowt b v : = hight_b ratio I lowt fault -------z I ct tap hight fault b --------z ct tap b v = 133.333volt tap_rated V = 28.464volt lowt b v hight_b = 202.44volt -----------------------------------------------------------CT excitation voltage I lowt fault v := ct lowt e ct tap [z + zb J ct_tap I v : = ct_hight_e v : = ct_hight_er hight fault I ct tap tap v = 30.823volt ct lowt e [z ct_tap + z b ] v = 219.214volt ct_hight_e Re [z ctr + z ] br v = 117.994volt ct_hight_er -----------------------------------------------------------93

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-----------------------------------------------------------Maximum Secondary current I := I I = lOOamp max sec cal max Maximum Secondary excitation current I max I 11.111amp I := e e 9 I := I + I I = 111.111amp st max e st I e by definition = 10% I st at a maximum secondary excitation current of 11.1 amps, 200/5 ratio, 1.333 ohms the secondary excitation voltage is read from Es vs Ie graph V := 135volt e Transformer low side fault From secondary excitation curve for a excitation voltage of 30.3V. I := .OBamp eg I lowt fault I := + I st ct tap I eg by definition = 0.392% I st For a C400 class CT, 600/5 ratio on a tap of 200/5, with a burden of 1.4 ohms the excitation current is .08 amps or 0.392\ error which is less than 10\ eg -----------------------------------------------------------94

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-----------------------------------------------------------Transformer high side fault From Secondary excitation curve for a excitation voltage of 219V. I st I .-I := 30amp eg hight_ fault + I ct tap I eg by definition = 17.182% I st For a C400 class CT, 600/5 ratio on a tap of 200/5, with a burden of 1.4 ohms the excitation current is 30 amps or 17% error which is more than 10%. This voltage is above the saturation voltage and any impedance greater than 1.33 ohms will will saturate the CT 95 eg

PAGE 102

ANSI CT Accuracy Method Power Transformer High Voltage CT C400, 200/5 Tap_ on a 600/5 MRCT CT secondary rated voltage v := 400volt rated 600 CT ratio, maximum ct := ratio 5 200 CT ratio, tap ct := tap 5 -----------------------------------------------------------Maximum electrical system fault currents transformer primary current for a 25kV fault I := 374lamp 25kV fault 2SkV I : = I lowt fault llSkV 25kV fault transformer primary current. for a llSkV fault I := 5784amp llSkV fault I := I hight_fault llSkV fault I lowt fault I hight_fault 96 = 813.261amp 3 = 5.784 amp

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-----------------------------------------------------------Secondary Current I := Samp sec rated I : = 20 I I = lOOamp sec cal sec rated sec cal -----------------------------------------------------------Secondary Burden Impedance Z := 4ohm ct rated tap z := -----z ct_tap_rated ct rated ratio z = .1.333ohm ct_tap_rated -----------------------------------------------------------secondary burden (leads + relay) Standard burden power factor 9 := .s b i aces [e b J z := z e br b Z := 1.4ohm b aces [9 b J = 60 deg Z = 0.7 + 1.212iohm br -----------------------------------------------------------CT impedance ohm Z := .0029---ct Z = 0.116ohm ct_tap turn tap ct_tap CT impedance power factor 9 := 1 aces [e ct J = Odeg ct i aces [9 ct J z := z e ctr ct_tap z = O.ll6ohm ctr -----------------------------------------------------------97

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-----------------------------------------------------------Secondary burden voltage ct v tap_rated v := lowt_b v := hight_b tap := -----v ct rated ratio I lowt fault I ct tap z b hight_fault ct tap z b v = 133.333volt tap_rated V = 28.464volt lowt b v hight_b = 202.44volt -----------------------------------------------------------Maximum Secondary current I := I I = lOOamp max sec cal max -----------------------------------------------------------98

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Maximum Burden power transformer low side fault current Maximum ct secondary current I lowt fault I := ct lowt fault ct tap I = 20.332amp ct lowt fault z : = max ct tap ct ratio I v rated ct lowt fault Z = 6.558ohm max At a maximum fault current of 20.35 amps the CT will not saturate for a 6.553 ohm total burden. -----------------------------------------------------------Maximum Burden power transformer high side fault current Maximum ct secondary current I I := ct_hight_fault hight_fault ct tap I = 144.6amp ct_hight_fault z : = max ct tap ct ratio I v rated ct_hight_fault z = 0.922ohm max At a maximum fault current of 144.6 amps the CT will saturate for a .922 ohm total burden. 99

PAGE 106

APPENDIX D: CT TRANSIENT CALCULATIONS 25kV Feeder CT csoo, 400/5 Tap on a 1200/5 MRCT Calculation of Time-To-Saturation 25kV Bus CT csoo, 1200/5 Tap on a 1200/5 MRCT Calculation of Time-To-Saturation Power Transformer High Voltage CT C400, 200/5 Tap on a 600/5 MRCT Calculation of Time-To-saturation 100

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calculation of Time-To-saturation 25kV Feeder CT caoo, 400/5 Tap on a 1200/5 MRCT X ----------------------------------------------------------Given: CT secondary saturation voltage (Vx), turns in CT secondary winding (N2), ac component of primary fault current (I1), resistance of CT secondary burden (R2) (including winding resistance). vx : = 240 volts N2 := 80 turns Il := 3741 amps R2 := 1. 55 ohm ----------------------------------------------------------First, calculate ct saturation factor (Ks). (assuming neither ct remanence or burden reactance) VxN2 Ks := I1R2 Ks = 3.3112 ---------------------------------------------------------Second, calculation of Tl and T2 Given: power-system short-circuit loop reactance (Xl), power-system short-circuit loop resistance (Rl), CT secondary magnetizing reactance (X2) plus burden reactance, power system angular frequency (W). Xl := 3.84356 ohm Rl := 0.34053 ohm X2 := 1000. ohm w := 377 radfsec 101

PAGE 108

X1 X2 T1 := T2 : = T1 = 0.0299 sec R1W R2W T2 = 1.7113 sec ----------------------------------------------------------Last, using calculated values of Ks, T1, and T2, read value of Time-to-Saturation from chart. Time-to-Saturation is n := 1,2 1000 n t : = n 1000 WT1T2 [ :::: "] e -e + 1 Ks := n T2 -Tl 20 Ks n / ./ 0 -3 t 1 10 n ms Ks = 3.046 6 Ks = 3.3482 7 max(Ks) = 11.5024 102

PAGE 109

Calculation of Time-To-Saturation 25kV Bus CT CBOO, 400/5 Tap on a 1200/5 MRCT ----------------------------------------------------------Given: CT secondary saturation voltage (Vx), turns in CT secondary winding (N2), ac component of primary fault current (I1), resistance of CT secondary burden (R2) (including winding resistance). Vx := 720 volts N2 := 240 turns I1 := 3744 amps R2 := 4.1 ohm ----------------------------------------------------------First, calculate ct saturation factor (Ks). (assuming neither. ct remanence or burden reactance) vxN2 Ks := IlR2 Ks = 11.257 ----------------------------------------------------------Second, calculation of Tl and T2 Given: power-system short-circuit loop reactance (Xl), power-system short-circuit loop resistance (Rl), CT secondary magnetizing reactance (X2) plus burden reactance, power system angular frequency (W). Xl := 3.84356 ohm Rl := 0.34053 ohm X2 := 7333. ohm w := 377 rad/sec 103

PAGE 110

X1 X2 Tl := T2 : = T1 = 0.0299 sec R1 W R2 W T2 = 4 .7441 sec Last, using calculated values of Ks, Tl, and T2, read value of Time-to-Sat.uration from chart. Time-to-Saturation is n : = 1,2 . 1000 n t : = n 1000 W T1 T2 [ -t -t ] n n --T2 T1 e -e + 1 Ks := n T2 T1 20 Ks n f ./ v 0 -3 t 1 10 n ms Ks = 11.2529 75 Ks = 11.281 76 max(Ks) = 11.9297 104

PAGE 111

Calculation of Time-To-Saturation Power Transformer High Voltage CT C400, 200/5 Tap on a 600/5 MRCT ----------------------------------------------------------Given: CT secondary saturation voltage (Vx), turns in CT secondary winding (N2), ac component of primary fault current (Il), resistance of CT secondary burden (R2) (including winding resistance). Vx := 130 volts N2 := 40 turns Il := 5784 amps R2 : = 2.5 ohm ----------------------------------------------------------First, calculate ct saturation factor (Ks). (assuming neither ct remanence or burden reactance) VxN2 Ks := IlR2 Ks = 0.3596 ----------------------------------------------------------second, calculation of T1 and T2 Given: power-system short-circuit loop reactance (Xl), power-system short-circuit loop resistance (Rl), CT secondary magnetizing reactance (X2) plus burden reactance, power system angular frequency (W). Xl Rl X2 w := := := := 11.4294 ohm 1. 08243 ohm 600.0 ohm 377 radfsec 105

PAGE 112

Xl X2 T1 := T2 := Tl = 0.028 sec RlW R2W T2 = 0.6366 sec ----------------------------------------------------------Last, using calculated values of Ks, T1, and T2, read value of Time-to-saturation from chart. Time-to-Saturation is n := 1,2 1000 n Ks t : = n 1000 : = [ :> :: "] e -e + 1 n T2 -T1 20 Ks n 0 10 -3 t n ms Ks = 1.3701 1 Ks = 10.0747 110 max(Ks) = 10.1451 106 ..... f", 1

PAGE 113

APPENDIX E: EMTP SIMULATION 25kV Feeder CT csoo, 400/5 Tap on a 1200/5 MRCT CT EMTP Input Calculations EMTP Saturation Conversion Program EMTP Hysteresis Conversion Program 25kV Bus CT csoo, 1200/5 Tap on a 1200/5 MRCT CT EMTP Input Calculations EMTP Saturation Conversion Program EMTP Hysteresis Conversion Program Power Transformer High Voltage CT C400, 200/5 Tap on a 600/5 MRCT CT EMTP Input Calculations EMTP Saturation Conversion Program EMTP Hysteresis Conversion Program Typical EMTP Input cards EMTP Feeder Input Fault Study with Saturation 107

PAGE 114

CT EMTP Input Calculations 25 kV Bus CT caoo, 1200/5 Tap on a 1200/5 MRCT Ct Voltage vs Excitation current calculations ----------------------------------------------------------1200 CT ratio, maximum ct := ratio 5 1200 CT ratio, tap ct := tap 5 -----------------------------------------------------------Secondary Resistance ohm ct := .0031---res turn R := .OO ohm ctleads R := ct ct + 2R R = 0.744ohm sec res tap ctleads sec -----------------------------------------------------------rated secondary current I := Samp rated I := 20I I = lOOamp sec cal rated sec cal -----------------------------------------------------------rated secondary voltage v sec cal ct tap v := SOOvolt rated := V V = 800 volt rated ct sec cal ratio ----------------------------------------------------------1.08

PAGE 115

-----------------------------------------------------------rated secondary MVA ct cal : = [V sec_cal I sec cal ] MVA MVA = O.OSMVA ct cal -----------------------------------------------------------From ct Saturation curves i := 0,1 3 ie := 00.200amp 20.000amp iepu := i ie i I sec cal 1 10 vpu i -1 10 iepu i 0.00004 0.00046 0.002 0.2 10 -4 109 v : = vpu i iepu i [010.0volt] 400.0volt 780.0volt 900.0volt := v i v sec cal 0 10 vpu i 0.0125 0.5 0.975 1.125

PAGE 116

BEGIN NEW DATA CASE c c c c c c c c c IBM MAIN FILE: PC FILE: INPUT: SCT80C SCTSOC.DAT SCTSOC.HCD current Transformer Saturation Calculations VBASE 266.67 VOLTS !BASE m 100.00 AMPS PBASE = 26667.00 WATTS c c Calculation of the current vs flux saturation curves from the knowledge c of the RMS magnetization current of the transformer. c c C SERASE SATURATION C --freq<-KVbase
PAGE 117

BEGIN NEW DATA CASE c c c c c c FILE NAME: HCTBOC.dat REFERENCE PILE: HYS. OAT c computation of the magnetic hysteresis curve of single-phase transformer, c using the HYSTERESIS routine. c saturation shape is currently available only for ARMCO Mh oriented c silicon steel. c Input data required is the positive current-flux point (amps and v-aec) c where the hysteresis curve becomes c output is for the third and first quadrants (amps and V-sec) in the format c required by TYPE-96 Hysteretic Inductor. c $ERASE HYSTERESIS c C itype><-level> l 2 { initialize punch buffer { special request word type 1 shape ARMCO si steel, level 2 10 data points c isat> .790623 .9378412 c positive single-valued saturation point, amps va v-aec same as point 4 of cts60a.dat'curve c l 2 3 4 5 6 7 8 c 345678901234567890123456789012345678901234567890123456789012345678901234567890 c $PUNCH, HCTBOC.pch BLANK END OF DATA FILE BEGIN NEW DATA CASE BLANK END OF ALL CASES { flush punch buffer into file hys2400.pch 111

PAGE 118

CT EMTP Input Calculations 25 kV Feeder CT caoo, 400/5 Tap on a 1200/5 MRCT Ct Voltage vs Excitation Current calculations -----------------------------------------------------------1200 CT ratio, maximum ct := ratio 5 400 CT ratio, tap ct := tap 5 -----------------------------------------------------------Secondary Resistance ohm ct := .0031---res turn R := .OOohm ctleads R := ct ct + 2 R R = 0.248ohm sec res tap ctleads sec -----------------------------------------------------------rated secondary current I := Samp rated I 20I I = lOOamp sec cal rated sec cal -----------------------------------------------------------rated secondary voltage v sec cal ct tap v := BOOvolt rated := V V = 266.667volt rated ct sec cal ratio ----------------------------------------------------------

PAGE 119

-----------------------------------------------------------rated secondary MVA MVA ct cal MVA = 0.026667MVA ct cal -----------------------------------------------------------From Ct saturation Curves i := 0,1 3 ie = = 20.00amp iepu := i ie i I sec cal 1 10 vpu i -1 10 iepu i 0.0003 0.0015 0.0048 0.2 10 -3 113 v := 250.0volt 300.0volt vpu := i v v i sec cal iepu i 10 0 vpu i 0.0525 0.5625 0.9375 1.125

PAGE 120

BEGIN NEW DATA CASE c c c c c c c c c c IBM MAIN FILE: SCT240C SCT240C.DAT SCT240C.MCO PC FILE: INPUT: current Transformer Saturation Calculations VBASE = 800.00 VOLTS IBASE = 100.00 AMPS PBASE = 50000.00 WATTS c Calculation of the current vs flux saturation curves from the knowledge C of the RMS magnetization current of the transformer. c c C $ERASE SATURATION C --freq<-KVbase
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BEGIN NEW DATA CASE c c c FILE NAME: HCT240C.dat c c REFERENCE FILE: HYS.DAT c c c c c c c c Computation of the magnetic hysteresis curve of single-phase transformer, using the HYSTERESIS routine. saturation shape is currently available only for ARMCO Mh oriented silicon steel. Input data required is the positive current-flux point (amps and V-sec) where the hysteresis curve becomes single-valued. c output is for the third and first quadrants (amps and V-sec) in the format required by TYPE-96 Hysteretic Inductor. c SERASE HYSTERESIS c C itype><-level> 1 2 C isat> initialize punch buffer special request word type l shape ARMCO si steel, level 2 10 data points .321464 2.926028 { pbsitive single-valued saturation point, amps vs V-sec c same as point 4 of cts60a.dat curve c 1 2 3 4 5 6 7 8 c 345678901234567890123456789012345678901234567890123456789012345678901234567890 c SPUNCH, HCT240C.pch BLANK END OF DATA FILE BEGIN NEW DATA CASE BLANK END OF ALL CASES { flush punch buffer into file hys2400.pch 115

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CT EMTP Input Calculations Power Transformer High Voltage CT C400, 200/5 Tap on a 600/5 MRCT Ct Voltage vs Excitation Current calculations CT ratio, maximum CT ratio, tap Secondary Resistance R := ct ct + 2R sec res tap ctleads rated secondary current I := 20I sec cal rated rated secondary voltage v := v ct tap sec cal rated ct ratio l.l.6 600 ct := ratio 5 200 ct := tap 5 ohm ct := .0029---res turn R := .OOohm ctleads R = 0.116ohm sec I := 5amp rated I = lOOamp sec cal v := 400volt rated v = 133.333volt sec cal

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-----------------------------------------------------------rated secondary MVA ct cal : = [V sec_cal I sec cal ] MVA MVA = 0.013333MVA ct cal -----------------------------------------------------------From Ct Saturation Curves i := 0,1 3 ie == 30.000amp iepu := i I ie i sec cal 1 10 vpu i/ -1 10 iepu i 0.00005 0.0013 0.004 0.3 -3 10 117 v : = [000. 50 volt] 070.00volt 130.00volt 138.00volt v i vpu := i v iepu i sec cal 0 10 vpu i 0.00375 0.525 0.975 1.035

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BEGIN NEW DATA CASE c c c c c c c c c c IBM MAIN FILE: SCT40C SCT40C.DAT SCT40C.MCD PC FILE: INPUT: current Transformer Saturation Calculations VBASE = 133.33VOLTS IBASE 100.00 AMPS PBASE 13333.00 WATTS c Calculation of the current va flux saturation curves from the knowledge C of the RMS magnetization current of the transformer. c c C $ERASE SATURATION C --freq<-KVbaee
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BEGIN NEW DATA CASE c c c c c c FILE NAME: HCT40C.dat REFERENCE FILE; HYS. DAT c Computation of the magnetic hysteresis curve of single-phase transformer, c using the HYSTERESIS routine. c saturation shape is currently available only for ARMCO Mh oriented c silicon steel. c Input data required is the positive current-flux point (amps and V-sec) c where the hysteresis curve becomes single-valued. c output is for the third and first quadrants (amps and v-sec) in the format c required by TYPE-96 Hysteretic Inductor. c $ERASE HYSTERESIS c C itype><-level> 1 2 C isat> .629409 .4876591 c c 1 initialize punch buffer special request word type 1 shape ARMCO si steel, level 2 10 data points positive single-valued saturation point, amps vs v-sec same as point 4 of cts60a.dat curve 2 3 4 5 6 7 8 c 345678901234567890123456789012345678901234567890123456789012345678901234567890 c $PUNCH, HCT40C. pch BLANK END OF DATA FILE BEGIN NEW DATA CASE BLANK END OF ALL CASES { flush punch buffer into file hys2400.pch 119

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BEGIN NEW DATA CASE c c ****** Public Service Company of Colorado ****** c c c c c c c c c c c c c c IBM MAIN: FILE NAME: CONDITION: DATE: DIAGNOSTIC 9 Miscellaneous Data Card 1-8 9-16 17-24 25-32 Del taT Tmax X opt Copt SO.E-6 .20 60.0 o.o SFEED3H SFEED.DAT BASE CASE 12/11/92 33-40 Epsiln 1.0E-9 c --+----l----+----2----+----3--------4----+----s----+----6----+----7----+----8 c 1-8 9-16 17-24 25-32 33-40 41-48 49-56 57-64 65-72 73-80 C IOUT IPLOT IDOUBL KSSOUT MAXOUT IPUN MEMSAV ICAT NENERG IPRSUP 101 5 0 1 1 0 2 c TACS HYBRID C /TACS c c OUTPUT + + + + + fixed fixed named named c -< in1> -< in2> -< in3> -< in4> -< inS> < -lo>< -hi>< -lo>< -hi> c no >< nl >< n2 >< n3 >< n4 >< nS >< n6 >< n7 > c do >< d1 >< d2 >< d3 >< d4 >< dS >< d6 >< d7 > lFLUXA +TFCTAA 1.0 o.o 1.0 c C Resident Sources c < A >< B >< c > < t-start>< t-stop > 90TFCTAA 91CBA 91B25A c c Fortran Expression c 99SCURRA =<************** free-format FORTRAN expression ********************> 1.0* CBA 99PCURRA c B25A* 1/80 C output variable c J3FLUXA SCURRA PCURRA c c ------------------------------> c c c c c c OUTPUT + -< in1> no >< nl do >< dl lFLUXB +TFCTBB 1.0 o.o + -< >< >< 1.0 C Resident Sources + in2> -< inJ> n2 >< d2 >< PHASE B <----------------------------------+ + fixed fixed named named -< in4> -< inS> < -lo>< -hi>< -lo>< -hi> nJ >< n4 >< nS >< n6 >< n7 > d3 >< d4 >< dS >< d6 >< d7 > 120

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c < 90TFCT88 9lC88 918258 c A >< c Fortran Expression 8 >< c > < t-stop > c 99SCURR8 =<************** free-format FORTRAN C88 expression > 99PCURR8 c l.O* 325B* C output variable l/80 33FLUX8 SCURR8 PCURR8 c C --------------------------------> PHASE C <---------------------------------0 C OUTPUT + C -< inl> c no >< nl c do >< dl c lFLUXC +TFCTCC l.O 0.0 + -< >< >< l.O c Resident Sources c < 90TFcrcc 9lC8C 9lB25C c A >< c Fortran Expression + in2> -< n2 d2 8 + in3> -< in4> >< n3 >< >< d3 >< >< c > + -< inS> ::14 d4 < >< nS >< dS fixed named named -lo>< -hi>< -lo>< -hi> >< n o >< n7 > >< d6 >< d7 > < t-start>< t-stop > c 99SCURRC 99PCURRC c =< free-format FORTRAN expression > l. 0* C8C B25C* l/80 c output variable 33FLUXC SCURRC PCURRC c BLANK card terminates all TACS data c C =====m============= => SOURCE EQUIVALENT AT THE 25KV BUS <=================== c C 8US-->BUS-->8US-->BUS-->--R-->--L(OHM)---> SlSOURCAB25A 3.23568 52SOURCBB25B .3405 3.84356 53SOURCCB25C c C =================> CURRENT TRANSFORMER, FEEDER OVERCURRENT <================= c c c c request C ---word----> TRANSFORMER c CURRENT TRANSFORMER TFCTAA 1-steady-stll busil <--I-><-op-> .0424 .0525 TFCTAA1250. C $INCLUDE, SCT80C.PCH, 4.24264069E-02 5.2519l083E-02 2.04676353E-Ol S.62704731E-Ol 7.90623061E-01 9.37841219E-Ol 121. 0

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c 4.4B277724E+Ol 1.12S40946E+OO 9999 c !NODE NAMES! !LEAKAGE z I C ____________ <-RL-><-LL-> ___________________________________ 3 1TFCTSATFCTNA .2221 0.000 .266 2TFCTPATFCTA 3.8E-66.6E-6.00333 c c c CURRENT TRANSFORMER TFCTBB c request C ---word----> TRANSFORMER TFCTAA 1-steady-stl I bus! I <--I-><-op-> TFCTBB c C $INCLUDE, SCTSOC.PCH, C 4.24264069E-02 S.25191083E-02 c 2.04676353E-01 S.62704731E-Ol C 7.90623061E-01 9.37841219E-01 C 4.48277724E+Ol 1.12S40946E+OO c 9999 c 1TFCTSBTFCTNB 2TFCTPBTFCTB c c c CURRENT TRANSFORMER TFCTCC c request C ---word----> TRANSFORMER TFCTAA c 1-steady-st II bus II <--I-><-op-> TFCTCC C $INCLUDE, SCTBOC.PCH, C 4.24264069E-02 5.25191083E-02 C 2.04676353E-Ol 5.62704731E-Ol C 7.90623061E-01 9.37841219E-Ol C 4.48277724E+Ol 1.12540946E+OO c 9999 c c 1TFCTSCTFCTNC 2TFCTPCTFCTC C =================-========a=> LOADS <================================= c c c FEEDER CURRENT TRANSFORMER TO BREAKER C BUS-->BUS-->BUS-->BUS--><-ohmR<-ohmL<----c c c c TFCTA BKRFlA .00001 TFCTB BKRFlB TFCTC BKRFlC .00001 .00001 BREAKER LOAD IMPEDANCE C BUS-->BUS-->BUS-->BUS--><-ohmR<-ohmL<----C c c c BKRF2A SO. BKRF2B SO. BKRF2C SO. FAULT IMPEDANCE C BUS-->BUS-->BUS-->BUS--><-ohmR<-ohmL<----c FAULTA .01000.0029S 122 f f 0 1 1 1 0 3 3 3 0 3

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c c c FAULTS FAULTC .01000.00295 .01000.00295 CT LOAD IMPEDANCE C BUS-->BUS-->BUS-->BUS--><-ohmR<-ohmL<----C c CBA RELAY A CBB RELAYS CBC RELAYC c RELAYARELAYN RELAYBRELAYN RELAYCRELAYN RELAYNRELAYG c TFCTNATCTG TFCTNBTCTG TFCTNCTCTG c TCTG RELAYG c RELAYG c c TFCTNA C TFCTNB C TFCTNC c .65 .65 .65 .59 .59 .59 .267 .0001 .0001 .0001 .65 .0001 .0185 .0185 .0185 .43 .43 .43 .73 .0185 lO.E-6 lO.E-6 lO.E-6 BLANK CARD ENDING system impedance data c 0 3 3 3 3 C CT SECONDARY MEASURING SWITCH c c Secondary Relay Current c C BUS-->BUS--><---TCLOSE<----TOPEN<-------:E c c c TFCTSACBA TFCTSBCBB TFCTSCCBC Primary Electrical System Cur:ent C BUS-->BUS--><---TCLOSE<----TOPEN<-------:E c B25A TFCTPA B25B TFCTPB B25C TFCTPC MEASURING MEASURING MEASURING MEASURING MEASUR::::m MEASURING 0 l 1 1 0 l l l C FEEDER BREAKER SWITCH c C BUS-->BUS--><---TCLOSE<----TOPEN<-------IE 0 BKRFlABKRF2A -.001 99. 1 c c BKRFlBBKRF2B BKRFlCBKRF2C -.001 -.001 99. 99. C ==============================> FAULT sw:TCH <=============================== c C BUS-->BUS--><---TCLOSE<----TOPEN<-------:E BKRFlAFAULTA 0. 0 .12400 1000 8KRF1BFAULTB 0.012400 BKRFlCFAULTC 0.012400 .1000 .1000 123 1 1 0 1

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c BLANK card ending switch cards c c ------1----+----2--------3--------4--------s--------6--------7---------8 c 3-8 11-20 21-30 31-40 61-70 c sus v PEAK L-G FREQUENCY PHS ANGLE T-Sl'ART 14SOURCA 20412.0 60.0 0.0 -1.0 3 14SOURC8 20412.0 60.0 -120.0 -1.0 3 14SOURCC 20412.0 60.0 120.0 -1.0 3 c BLANK card ending Voltage source cards c c Specify output Node Voltages c --+----l--------2--------3--------4----+----s--------6----+----7--------8 C 3-8 9-14 15-20 21-26 27-32 33-38 39-44 45-50 51-56 57-62 63-68 69-74 75-80 C 8US1->8US2->BUS3->BUS4->BUS5->BUS6->BUS7->BUS8->BUS9->BUS10>BUS1l>BUS12>BUS13> 825A B25B B25C BLANK card ending output node voltages c c c c CALCOMP PLOT 2CASE: DILLION 25KV FEEDER CT WITH SATURATION c Plot request Data .. C Graph Type: 4(volts) S(branch volts) 9(currents) N(# node volt plots) C Units: 1(deq) 2(cyc) 3(sec) 4(msec) 5(microsec) c Units per inch c __ Plot starting time c _Plot stopping time c __ Vmin vertical scale c Vmax scale VV<-<--<--<--<--I:US-->BUS-->BUS-->BUS-->HEADING-------->VERT AXIs------> c .. + 1 .. + 2 + 3 + 4 + 5 .. + 6 . + 7 ... + > C 1445.0 0.0110. B25A 8258 B25C 25KV BUS VOLTAGEVOLTAGE C 1945.0 0.0110. BRANCH 25KV PRI CURRENTCURRENT C B25A TFCTPAB25B TFCTPBB25C TFCTPC C 1945.0 0.0110. BRANCH 25KV SEC CURRENTCURRENT C TFCTSACBA TFCTSBCBB TFCTSCCBC c 1945.0 0.0110. c c 1945.0 0.0110. c c 1945.0 0.0110. c c 1945.0 0.0150. c BRANCH PCURRATACS BRANCH PCURRBTACS BRANCH PCURRCTACS BRANCH FLUXA TACS BLANK card ending Plot cards BEGIN NEW DATA CASE BLANK card SCURRATACS SCURRBTACS SCURRCTACS FLUXB TACS 124 25KV FEEDER 25KV FEEDER 25KV FEEDER CT FLUX A B FLUXC TACS A CURRENT B CURRENT c CURRENT c VOLT-SEC

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GLOSSARY accuracy. The extent to which the current in the secondary circuit reproduces the current in the primary circuit in the proportion stated by the marked ratio and represents the phase relationship of the primary current. accuracy classes for relaying (instrument transformers). Limits in terms of percent ratio error that have been established. accuracy ratings for relaying. The relay accuracy class is described by a letter denoting whether the accuracy can be obtained by calculation or must be obtained by test, followed by the maximum secondary terminal voltage that the transformer will produce at 20 times secondary current with one of the standard burdens without exceeding the relay accuracy class limit. ( This is usually taken as 10%.) bar-type current transformer. One that has a fixed, insulated straight conductor in the form of a bar, rod, or tube that is a single primary turn passing through the magnetic circuit and that is assembled to the secondary, core, and winding. 125

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burden of an instrument transformer. That property of the circuit connected to the secondary winding that determines the active and reactive power at the secondary terminals. The burden is expressed either as total ohms impedance with the effective resistance and reactance components, or as the total volt-amperes and power factor of the secondary devices and leads at the specified value of current or voltage, and frequency. bushing-type current transformer. One that has an annular core and a secondary winding insulated from and permanently assembled on the core but has no primary winding or insulation for a primary winding. This type of current transformer is for use with fully insulated conductor as the primary winding. A bushing-type current transformer usually is used in equipment where the primary conductor is a component part of other apparatus. (Note: This type of CT has very low leakage flux and is also known as a Low Inductance Type Current Transformer.) continuous-thermal-current rating factor CRF). The specified factor by which the rated primary current of a current transformer can be multiplied to obtain the maximum primary current that can be carried continuously without exceeding the limiting temperature rise from 30 C ambient air temperature. (When current transformers are incorporated internally as parts of larger transformers 126

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or power circuit breakers, they shall meet allowable average winding and hot spot temperatures under specific conditions and requirements of the larger apparatus.) current transformer. An instrument transformer intended to have its primary winding connected in series with the conductor carrying the current to be measured or controlled. (In window type current transformers, the primary winding is provided by the line conductor and is not an integral part of the transformer.) double-secondary current transformer. One that has two secondary coils each on a separate magnetic circuit with both magnetic circuits excited by the same primary winding. double-secondary voltage transformer. One that has two secondary windings on the same magnetic circuit insulated from each other and the primary. excitation losses for an instrument transformer. The watts required to supply the energy necessary to excite the transformer, which include the dielectric watts, the core watts, and the watts in the excited winding due to the excitation current. instrument transformer. A transformer that is intended to reproduce in its secondary circuit, in a definite and known proportion, the current or voltage of its primary circuit with the phase relations substantially preserved. 127

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marked ratio. The ratio of the rated primary value to the rated secondary value as stated on the nameplate. multiple-secondary current transformer. One that has three or more secondary coils each on a separate magnetic circuit with all magnetic circuits excited by the same primary winding. multi-ratio current transformer. One from which more than one ratio can be obtained by the use of taps on the secondary winding. percent ratio. The true ratio expressed in percent of the marked ratio. percent ratio correction of an instrument transformer. The difference between the ratio correction factor and unity expressed in percent. phase angle correction factor. The ratio of the true power factor to the measured power factor. It is a function of both the phase angles of the instrument transformer and the power factor of the primary circuit being measured. phase angle of an instrument transformer. The phase displacement, in minutes, between the primary and secondary value. polarity. The designation of the relative instantaneous directions of the currents entering the primary terminals 128

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and leaving the secondary terminals during most of each half cycle. rated current. The primary current selected for the basis for performance specifications of a current transformer. rated secondary current. The rated current divided by the marked ratio. rated secondary voltage. The rated voltage divided by the marked ratio. rated voltage of a voltage transformer. The primary voltage selected for the basis of performance specifications of a voltage transformer. ratio correction factor (RCF) The ratio of the true ratio to the marked ratio. The primary current or voltage is equal to the secondary current or voltage multiplied by the marked ratio times the ratio correction factor. secondary winding of an instrument transformer. The winding that is intended to be connected to the measuring or control devices. three-wire type current transformer. One that has two primary windings, each completely insulated for the rated insulation level of the transformer. This type of current transformer is for use on a three-wire, singlephase service. 129

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transformer correction factor (TCF). The ratio of true watts or watt-hours to the measured watts or watt-hours, divided by the marked ratio. true ratio. The ratio of the root-mean-square (rms) primary value to the rms secondary value under specified conditions. turn ratio of a current transformer. The ratio of the secondary winding turns to the primary winding turns. window-type current transformer. One that has a secondary winding insulated from and permanently assembled on the core, but has no primary winding as an integral part of the structure. Complete insulation is provided for a primary winding in the window through which one turn of the line conductor can be passed to provide the primary winding. wound-type current transformer. One that has a primary winding consisting of one or more turns mechanically encircling the core or cores. The primary and secondary windings are insulated from each other and from the core(s) and are assembled as an integral structure. Definitions from sources other that IEEE standards. knee point voltage (ANSI/IEEE C57.13-1978, 6.10.2). for class C transformers the knee point voltage is the point on the excitation curve where the tangent is at 45 to 130

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the abscissa. The excitation curve shall be plotted on log-log paper with square decades. knee point voltage
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Footnote: CCVT,CVT, and PT definitions were not given. 132

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REFERENCES [1] Power Systems Relaying Committee of the IEEE PES, "Transient Response of Current Transformers", IEEE Publication 76 CH 1130-4 PWR, January 1976. [2] Westinghouse Electric Corporation,"Applied Protective Relaying", 1976. [3] Transformers Committee of the IEEE PES, "IEEE Standard Requirements for Instrument Transformers", ANSI/IEEE C57.13-1978. [4] James G. Frame, Narendra Mohan, Tsu-huei Liu, "Hysteresis Modeling in an Electro-Magnetic Transients Program", IEEE Transactions on Power Apparatus and Systems, Vol PAS-101, No. 9, pp. 34033412, September 1982. [5] Electromagnetic Transients Program (EMTP) Rule Book, Bonneville Power Administration, Revised, September 1980. (6] c. Russell Mason,"The Art and Science of Protective Relaying", John Wiley & Sons, 1956. (7] Power system Relaying Committee of the IEEE PES, "Guide for the Application of Current Transformers Used for Protective Relaying Purposes", IEEE Publication C37.110/D6, January 1991. 133

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[8] Relay Work Group of the Western Systems Coordinating Council, "Relaying Current Transformer Application Guide", Preliminary, June 1989. [9] Leander W. Matsch, "Electromagnetic and Electromechanical Machines", Intext Education Publishers, 1972. [10] J. Lewis Backburn, "Protective Relaying, Principles and Applications", Marcel Dekker, Inc., 1987. [11] A.E. Fitzgerald, Charles Kingsley, Jr., Stephen D. Umans, "Electric Machinery", McGraw Hill, 1983. [12] Vincent Del Toro, "Electrical Engineering Prentice-Hall, Inc., 1972. [13] GEC Measurments, "Protective Relays Application Guide" The Genera 1 Electric Company p. 1. c. of England, 1975. [14] Allan Greewood, "Electrical Transients in Power Systems", John Wiley & Sons, Inc. 1991. [15] Harold Pender, William A. Del Mar, "Electrical Engineers' Handbook, Electric Power", John Wiley & Sons, Inc., 1949. [16] Alex Y. Wu, "The Analysis of Current Transformer Transient Response and Its Effect on Current Relay Performance", IEEE Transactions on Industry Applications, Vol. IA-21, No.4, pp. 793-802, MayfJune 1985. 134

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[17] H.W. Dommel, "Electromagnetic Transients Program Reference Manual(EMTP Theory Book)", 1986. 135