Principle of operation of CT
§ A current transformer
is defined as “as an instrument transformer in which the secondary current is
substantially proportional to the primary current (under normal conditions of
operation) and differs in phase from it by an angle which is approximately zero
for an appropriate direction of the connections.”
§ Current transformers
are usually either “measuring” or “protective” types.
Some Definitions used for CT:
1) Rated primary current:
§ The value of primary
current which appears in the designation of the transformer and on which the
performance of the current transformer is based.
2) Rated secondary current:
§ The value of
secondary current which appears in the designation of the transformer and on
which the performance of the current transformer is based.
§ Typical values of
secondary current are 1 A or 5 A. In the case of transformer differential
protection, secondary currents of 1/ root 3 A and 5/ root 3 A are also
specified.
3) Rated burden:
§ The apparent power of
the secondary circuit in Volt-amperes expressed at the rated secondary current
and at a specific power factor (0.8 for almost all standards)
4) Rated output:
§ The value of the
apparent power (in volt-amperes at a specified power (factor) which the current
transformer is intended to supply to the secondary circuit at the rated
secondary current and with rated burden connected to it.
5) Accuracy class:
§ In the case of
metering CT s, accuracy class is typically, 0.2, 0.5, 1 or 3.
§ This means that the
errors have to be within the limits specified in the standards for that
particular accuracy class.
§ The metering CT has
to be accurate from 5% to 120% of the rated primary current, at 25% and 100% of
the rated burden at the specified power factor.
§ In the case of
protection CT s, the CT s should pass both the ratio and phase errors at the
specified accuracy class, usually 5P or 10P, as well as composite
error at the accuracy limit factor of the CT.
6) Current Ratio Error:
§ The error with a
transformer introduces into the measurement of a current and which arises from
the fact that actual transformation ratio is not equal to the rated transformer
ratio. The current error expressed in percentage is given by the formula:
§ Current error in % =
(Ka(Is-Ip)) x 100 / Ip
§ Where Ka= rated
transformation ratio ,Ip= actual primary current, Is= actual secondary current
when Ip is flowing under the conditions of measurement
7) Accuracy limit factor:
§ The value of primary
current up to which the CT complies with composite error requirements. This is
typically 5, 10 or 15, which means that the composite error of the
CT has to be within specified limits at 5, 10 or 15 times the rated primary
current.
8) Short time rating:
§ The value of primary
current (in kA) that the CT should be able to withstand both thermally and
dynamically without damage to the windings, with the secondary circuit being
short-circuited. The time specified is usually 1 or 3 seconds.
9) Instrument security factor (factor of
security):
§ This typically takes
a value of less than 5 or less than 10 though it could be much higher if the
ratio is very low. If the factor of security of the CT is 5, it means that the
composite error of the metering CT at 5 times the rated primary current is equal
to or greater than 10%. This means that heavy currents on the primary are not
passed on to the secondary circuit and instruments are therefore protected. In
the case of double ratio CT’s, FS is applicable for the lowest ratio only.
10) Class PS X CT:
§ In balance systems of
protection, CT s with a high degree of similarity in their characteristics is
required. These requirements are met by Class PS (X) CT s. Their performance is
defined in terms of a knee-point voltage (KPV), the magnetizing current (Imag) at
the knee point voltage or 1/2 or 1/4 the knee-point voltage, and the resistance
of the CT secondary winding corrected to 75C. Accuracy is defined in terms of
the turn’s ratio.
11) Knee point voltage:
§ That point on the
magnetizing curve where an increase of 10% in the flux density (voltage) causes
an increase of 50% in the magnetizing force (current).
§ The ‘Knee Point
Voltage’ (Vkp) is defined as the secondary voltage at which an increase of 10%
produces an increase in magnetizing current of 50%. It is the secondary voltage
above which the CT is near magnetic saturation.
12) Core balance CT (CBCT):
§ The CBCT, also known
as a zero sequence CT, is used for earth leakage and earth fault protection.
The concept is similar to the RVT. In the CBCT, the three core cable or three
single cores of a three phase system pass through the inner diameter of the CT.
When the system is fault free, no current flows in the secondary of the CBCT.
When there is an earth fault, the residual current (zero phase sequence
current) of the system flows through the secondary of the CBCT and this
operates the relay. In order to design the CBCT, the inner diameter of the CT,
the relay type, the relay setting and the primary operating current need to be
furnished.
13) Phase displacement:
§ The difference in
phase between the primary and secondary current vectors, the direction of the
vectors being so chosen that the angle is zero for the perfect transformer. The
phase displacement is said to be positive when the secondary current vector
leads the primary current vector. It is usually express in minutes
14) Highest system voltage:
§ The highest rms line
to line voltage which can be sustained under normal operating conditions at any
time and at any point on the system. It excludes temporary voltage variations
due to fault condition and the sudden disconnection of large loads.
15) Rated insulation level:
§ That combination of
voltage values (power frequency and lightning impulse, or where applicable,
lightning and switching impulse) which characterizes the insulation of a
transformer with regard to its capability to withstand by dielectric stresses.
For low voltage transformer the test voltage 4kV, at power-frequency, applied
during 1 minute.
16) Rated short-time thermal current (Ith):
§ The rms value of the
primary current which the current transformer will withstand for a rated time,
with their secondary winding short circuited without suffering harmful effects.
17) Rated dynamic current (Idyn):
§ The peak value of the
primary current which a current transformer will withstand, without being
damaged electrically for mechanically by the resulting electromagnetic forces,
the secondary winding being short-circuited.
18) Rated continuous thermal current (Un)
§ The value of current
which can be permitted to flow continuously in the primary winding, the
secondary windings being connected to the rated burdens, without the
temperature rise exceeding the specified values.
19) Instrument security factor (ISF or Fs):
§ The ratio of rated
instrument limits primary current to the rated primary current. The times that
the primary current must be higher than the rated value, for the composite
error of a measuring current transformer to be equal to or greater than 10%,
the secondary burden being equal to the rated burden. The lower this number is,
the more protected the connected instrument are against.
20) Sensitivity
§ Sensitivity is
defined as the lowest value of primary fault current, within the protected
zone, which will cause the relay to operate. To provide fast operation on an in
zone fault, the current transformer should have a ‘Knee Point Voltage’ at least
twice the setting voltage of the relay.
21) Field Adjustment of Current Transformer Ratio:
§ The ratio of current
transformers can be field adjusted to fulfil the needs of the
application. Passing
more secondary turns or more primary turns through the window will
increase or decrease the turns ratio.
Actual Turns Ratio = (Name Plate Ration- Secondary Turns Added) /
Primary Turns.
Types of Current transformers (CT’s)
According
to Construction of CT:
1) Bar Type:
§ Bar types are
available with higher insulation levels and are usually bolted to the current
caring device.
§ Bar type current
transformers are insulated for the operating voltage of the system.
§ Bar-type CTs operate
on the same principle of window CTs but have a permanent bar installed as a
primary conductor
2) Wound CT’s:
§ Capacity: There are
designed to measure currents from 1 amp to 100 amps.
§ the most common one
is the wound type current transformer. The wound type provides excellent
performance under a wide operating range. Typically, the wound type is
insulated to only 600 volts.
§ Since the load
current passes through primary windings in the CT, screw terminals are provided
for the load and secondary conductors. Wound primary CT’s are available in
ratios from 2.5:5 to 100:5.
§ Wound CTs have a
primary and secondary winding like a normal transformer. These CTs are rare and
are usually used at very low ratios and currents, typically in CT secondary
circuits to compensate for low currents, to match different CT ratios in
summing applications, or to isolate different CT circuits. Wound CTs have very
high burdens, and special attention to the source CT burden should be applied
when wound CTs are used.
3) Window:
§ Window CTs are the most
common. They are constructed with no primary winding and are installed around
the primary conductor. The electric field created by current flowing through
the conductor interacts with the CT core to transform the current to the
appropriate secondary output. Window CTs can be of solid or split core construction.
The primary conductor must be disconnected when installing solid window CTs.
However, split core CTs can be installed around the primary conductor without
disconnecting the primary conductor
§ Ring Core CT’s :
§ Capacity: There are
available for measuring currents from 50 to 5000 amps
§ Size: with windows
(power conductor opening size) from 1″ to 8″ diameter.
§ Split Core CT’s:
§ Capacity: There are
available for measuring currents from 100 to 5000 amps.
§ Size: with
windows in varying sizes from 1″ by 2″ to 13″ by 30″.
§ Split core CT’s have
one end removable so that the load conductor or bus bar does not have to be
disconnected to install the CT.
4) Bushing
§ Bushing CTs are
window CTs specially constructed to fit around a bushing. Usually they cannot
be accessed, and their nameplates are found on the transformer or
circuit-breaker control cabinets.
§ The bushing type is
typically used around the bushing on circuit breakers and transformers and may
not have a hard protective outside cover.
§ Donut type current
transformers are typically insulated for 600 volts. To ensure accuracy, the
conductor should be positioned in the center of the current transformer
opening.
According
to Application of CT:
1) Measuring CT:
§ The principal
requirements of a measuring CT are that, for primary currents up to 120% or
125% of the rated current, its secondary current is proportional to its primary
current to a degree of accuracy as defined by its “Class” and, in the case of
the more accurate types, that a specified maximum phase angle displacement is
not exceeded.
§ A desirable
characteristic of a measuring CT is that it should “saturate” when
the primary current exceeds the percentage of rated current specified as the
upper limit to which the accuracy provisions apply. This means that at these
higher levels of primary current the secondary current is less than
proportionate. The effect of this is to reduce the extent to which any
measuring device connected to the CT secondary is subjected to current
Overload.
§ On the other hand the
reverse is required of the protective type CT, the principal purpose of which
is to provide a secondary current proportional to the primary current when it
is several, or many, times the rated primary current. The measure of this
characteristic is known as the “Accuracy Limit Factor” (A.L.F.).
§ A protection
type CT with an A.L.F. of 10 will produce a proportional current in the
secondary winding (subject to the allowable current error) with primary
currents up to a maximum of 10 times the rated current.
§ It should be
remembered when using a CT that where there are two or more devices to be
operated by the secondary winding, they must be connected in series across the
winding. This is exactly the opposite of the method used to connect two or more
loads to be supplied by a voltage or power transformer where the devices are
paralleled across the secondary winding.
§ With a CT, an
increase in the burden will result in an increase in the CT secondary output
voltage. This is automatic and necessary to maintain the current to the correct
magnitude. Conversely, a reduction in the burden will result in a reduction in
the CT secondary output voltage.
§ This rise in
secondary voltage output with an increase in burden means that, theoretically,
with infinite burden as is the case with the secondary load open circuit, an
infinitely high voltage appears across the secondary terminals. For practical
reasons this voltage is not infinitely high, but can be high enough to cause a
breakdown in the insulation between primary and secondary windings or between
either or both windings and the core. For this reason, primary current should
never be allowed to flow with no load or with a high resistance load connected
across the secondary winding.
§ When considering the
application of a CT it should be remembered that the total burden imposed on
the secondary winding is not only the sum of the burden(s) of the individual
device(s) connected to the winding but that it also includes the burden imposed
by the connecting cable and the resistance of the connections.
§ If, for example, the
resistance of the connecting cable and the connections is 0.1 ohm and the
secondary rating of the CT is 5A, the burden of the cable and connections (RI2)
is 0.1 x 5 x 5 = 2.5VA. This must be added to the burden(s) of the connected
device(s) when determining whether the CT has an adequately large burden rating
to supply the required device(s) and the burden imposed by the connections.
§ Should the burden
imposed on the CT secondary winding by the connected device(s) and the
connections exceed the rated burden of the CT the CT may partly or fully
saturate and therefore not have a secondary current adequately linear with the
primary current.
§ The burden imposed by
a given resistance in ohms [such as the resistance of a connecting cable] is
proportional to the square of the rated secondary current. Therefore, where
long runs of cable between CT and the connected device(s) are involved, the use
of a 1A secondary CT and a 1A device rather than 5A will result in a 25-fold
reduction in the burden of the connecting cables and connections. All burden
ratings and calculations are at rated secondary current.
§ Because of the
foregoing, when a relatively long [more than a very few meters] cable run is
required to connect a CT to its burden [such as a remote ammeter] a calculation
should be made to determine the cable burden. This is proportional to the
“round trip” resistance, i.e. twice the resistance of the length of twin cable
used. Cable tables provide information on the resistance values of different
sizes of conductors at 20o C per unit length.
2) Protective CT:
§ The calculated
resistance is then multiplied by the square of the CT secondary current rating
[25 for 5A, 1 for 1A]. If the VA burden as calculated by this method and added
to the rated burden(s) of the device(s) to be driven by the CT exceeds the CT
burden rating, the cable size must be increased [to reduce the resistance and
thus the burden] or a CT with a higher VA burden rating must be used, or a
lower CT secondary current rating [with matching change in the current rating
of the device(s) to be driven] should be substituted
Nomenclature of CT:
1. Ratio: input / output
current ratio
2. Burden (VA): total burden
including pilot wires. (2.5, 5, 10, 15 and 30VA.)
3. Class: Accuracy
required for operation (Metering: 0.2, 0.5, 1 or 3, Protection: 5, 10, 15, 20,
30).
4. Accuracy Limit
Factor:
5. Dimensions: maximum &
minimum limits
6. Nomenclature of CT:
Ratio, VA Burden, Accuracy Class, Accuracy Limit Factor.
7. Example: 2000/5, 15VA
5P10 (Ratio: 2000/5, Burden: 15VA, Accuracy Class: 5P, ALF: 10)
8. As per IEEE Metering
CT: 0.3B0.1 rated Metering CT is accurate to 0.3 percent if the
connected secondary burden if impedance does not exceed 0.1 ohms.
9. As per IEEE Relaying
(Protection) CT: 2.5C100 Relaying CT is accurate within 2.5 percent if the
secondary burden is less than 1.0 ohm (100 volts/100A).
1) Current Ratio of
CT:
§ The primary and
secondary currents are expressed as a ratio such as 100/5. With a 100/5 ratio
CT, 100A flowing in the primary winding will result in 5A flowing in the
secondary winding, provided the correct rated burden is connected to the
secondary winding. Similarly, for lesser primary currents, the secondary
currents are proportionately lower.
§ It should be noted
that a 100/5 CT would not fulfil the function of a 20/1 or a 10/0.5 CT as the
ratio expresses the current rating of the CT, not merely the ratio of the
primary to the secondary currents.
§ The rated secondary
current is commonly 5A or 1A, though lower currents such as 0.5A are not
uncommon. It flows in the rated secondary load, usually called the burden, when
the rated primary current flows in the primary winding.
§ Increasing or
Decreasing Turns Ratio of CT:
§ Increasing Number of
Turn: Increasing the number of primary turns can only decrease the turn’s
ratio. A current transformer with a 50 to 5 turn’s ratio can be changed to a 25
to 5 turn’s ratio by passing the primary twice through the window.
§ Increasing or
Decreasing Turns Ratio:
§ The turn’s ratio can
be either increased or decreased by wrapping wire from the secondary through
the window of the current transformer.
§ Increasing the turn’s
ratio with the secondary wire, turns on the secondary are essentially
increased. A 50 to 5 current transformer will have a 55 to 5 ratio when adding
a single secondary turn.
§ Decreasing the turn’s
ratio with the secondary wire, turns on the secondary are essentially
decreased. A 50 to 5 current transformer will have a 45 to 5 ratio when
adding a single secondary turn.
§ Decreasing the turn’s
ratio with the primary, accuracy and VA burden ratings are the same as the
original configuration.
§ Increasing the turn’s
ratio with the secondary will improve the accuracy and burden rating.
§ Decreasing the turn’s
ratio with the secondary will worsen the accuracy and burden rating.
§ When using the
secondary of a current transformer to change the turn’s ratio, the right hand
rule of magnetic fields comes into play. Wrapping the white lead or the
X1 lead from the H1 side of the transformer through the window to the H2 side
will decrease the turn’s ratio. Wrapping this wire from the H2 side to
the H1 side will increase the turn’s ratio.
§ Using the black or X2
lead as the adjustment method will do the opposite of the X1(white) lead.
Wrapping from the H1 to the H2 side will increase the turns ratio, and wrapping
from the H2 to the H1 side will decrease the turns ratio.
2) Burden
of CT:
§ Common burden ratings
of CT: 2.5, 5, 10, 15 and 30VA.
§ The external load
applied to the secondary of a current transformer is called the “burden”.
§ The burden of CT is
the maximum load (in VA) that can be applied to the CT secondary.
§ The burden can be
expressed in two ways.
§ The burden can be
expressed as the total impedance in ohms of the circuit or the total
volt-amperes (VA) and power factor at a specified value of current or voltage
and frequency.
§ Formerly, the
practice was to express the burden in terms of volt-amperes (VA) and power
factor, the volt-amperes being what would be consumed in the burden
impedance at rated secondary current (in other words, rated secondary
current squared times the burden impedance). Thus, a burden of 0.5Ωimpedance
may be expressed also as “12.5 VA at 5 amperes,” if we assume the usual
5-ampere secondary rating. The VA terminology is no longer standard, but it
needs defining because it will be found in the literature and in old data.
Burden for Measuring CT:
§ Total burden of
Measuring CT = Sum of Meters Burden in VA (Ammeter, Wattmeter, Transducer etc.)
connected in series to the CT secondary circuit + Connecting Secondary Circuit
Cable Burden in VA.
§ Cable burden = I2 x
R x2 L, where I = CT secondary current, R = cable resistance per length, 2L is
the tro &fro distance of cable length L from CT to metering circuits. If
the proper size and short length of wire is used, cable burden can be ignored.
§ The CT secondary
circuit load shall not be more than the CT VA rating. If the load is less than
the CT burden, all meters connected to the measuring CT should provide correct
reading.
§ In the case of
Measuring Current transformer, the burden depends on the connected meters and
quantity of meters on the secondary i.e. no of Ammeters, KWh meters, Kvar
meters, Kwh meters, transducers and also the connection cable burden (I2 x
R x2 L) to metering shall be taken into account.
§ Note Meters burden
can be obtained from manufacturer catalogue.
§ Selected CT burden
shall be more than the calculated burden
Burden for Protecting CT:
§ In the case of
Protection CTs the burden is calculated in the same way as above except the
burden of individual protective relays burden shall be considered instead of
meters. The connecting cable burden is calculated in the same way as metering
CT
§ Total burden of
Protection CT=Connecting cable Burden in VA + sum of Protective relays Burden
in VA.
§ All manufacturers can
supply the burden of their individual devices. Although not used very often
these days, induction disk over-current devices always gave the burden for the
minimum tap setting. To determine the impedance of the actual tap setting being
used, First Square the ratio of minimum divide by the actual tap setting used
and, second multiply this value by the minimum impedance.
§ Suppose an impedance
of 1.47 + 5.34j at the 1A tap. To apply the relay at the 4A tap the engineer
would multiply the impedance at the 1A taps setting by (1/4)2. The impedance at
the 4A tap would be 0.0919 + 0.3338j or 0.3462 Z at 96.4 power factor.
§ The CT burden
impedance decreases as the secondary current increases, because of
saturation in the magnetic circuits of relays and other devices. Hence, a given
burden may apply only for a particular value of secondary current. The old
terminology of volt-amperes at 5 amperes is most confusing in this respect
since it is not necessarily the actual volt amperes with 5 amperes flowing, but
is what the volt-amperes would be at 5 amperes
§ If there were no
saturation. Manufacturer’s publications give impedance data for several values
of over current for some relays for which such data are sometimes required.
Otherwise, data are provided only for one value of CT secondary current.
§ If a
publication does not clearly state for what value of current the burden
applies, this information should be requested. Lacking such saturation data,
one can obtain it easily by test. At high saturation, the impedance approaches
the DC resistance. Neglecting the reduction in impedance with saturation makes
it appear that a CT will have more inaccuracy than it actually will have. Of
course, if such apparently greater inaccuracy can be tolerated, further
refinements in calculation are unnecessary. However, in some applications neglecting
the effect of saturation will provide overly optimistic results; consequently,
it is safer always to take this effect into account.
§ It is usually
sufficiently accurate to add series burden impedances 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 impedances is necessary to prove that the CTÕs will be suitable,
such an application should be avoided.
§ If the impedance at
pickup of a tapped over current-relay coil is known for a given pickup tap, it
can be estimated for pickup current for any other tap. The reactance of a
tapped coil varies as the square of the coil turns, and the resistance varies
approximately as the turns. At pickup, there is negligible saturation, and the
resistance is small compared with the reactance. Therefore, it is usually
sufficiently accurate to assume that the impedance varies as the square of the
turns. The number of coil turns is inversely proportional to the pickup
current, and therefore the impedance varies inversely approximately as the
square of the pickup current.
§ Whether CT is
connected in wye or in delta, the burden impedances are always connected in
wye. With wye-connected CT the neutrals of the CT and of the burdens are
connected together, either directly or through a relay coil, except when a
so-called zero phase-sequence-current shunt is used.
§ It is seldom correct
simply to add the impedances of series burdens to get the total, whenever two
or more CT are connected in such a way that their currents may add or subtract
in some common portion of the secondary circuit. Instead, one must calculate
the sum of the voltage drops and rises in the external circuit from one CT
secondary terminal to the other for assumed values of secondary currents
flowing in the various branches of the external circuit. The effective CT
burden impedance for each combination of assumed currents is the calculated CT
terminal voltage divided by the assumed CT secondary current. This effective
impedance is the one to use, and it may be larger or smaller than the actual
impedance which would apply if no other CTÕs were supplying current to the
circuit.
§ If the primary of an
auxiliary CT is to be connected into the secondary of a CT whose accuracy is
being studied, one must know the impedance of the auxiliary CT viewed from its
primary with its secondary short-circuited. To this value of impedance must be
added the impedance of the auxiliary CT burden as viewed from the primary side
of the auxiliary CT; to obtain this impedance, multiply the actual burden
impedance by the square of the ratio of primary to secondary turns of the
auxiliary CT. It will become evident that, with an auxiliary CT that steps up
the magnitude of its current from primary to secondary, very high burden
impedances, when viewed from the primary, may result.
§ Burden is depending
on pilot lead length
§ For Metering Class
CTs burden is expressed as ohms impedance. For Protection-class CTs burden is express
as volt-amperes (VA).
|
VA |
Applications |
|
1 To 2 VA |
Moving iron ammeter |
|
1 To 2.5VA |
Moving coil
rectifier ammeter |
|
2.5 To 5VA |
Electrodynamics
instrument |
|
3 To 5VA |
Maximum demand
ammeter |
|
1 To 2.5VA |
Recording ammeter
or transducer |
§ Burden (VA) of copper
wires between instrument & current transformer for 1A and 5A secondary’s
|
Cross Section (mm2) |
CT 1 Amp Secondary Burden in VA
(Twin Wire) |
|||||
|
Distance |
||||||
|
10 meter |
20 meter |
40 meter |
60 meter |
80 meter |
100 meter |
|
|
1.0 |
0.35 |
0.71 |
1.43 |
2.14 |
2.85 |
3.57 |
|
1.5 |
0.23 |
0.46 |
0.92 |
1.39 |
1.85 |
2.31 |
|
2.5 |
0.14 |
0.29 |
0.57 |
0.86 |
1.14 |
1.43 |
|
4.0 |
0.09 |
0.18 |
0.36 |
0.54 |
0.71 |
0.89 |
|
6.0 |
0.06 |
0.12 |
0.24 |
0.36 |
0.48 |
0.6 |
|
Cross Section (mm2) |
CT 5 Amp Secondary Burden in VA
(Twin Wire) |
|||||
|
Distance |
||||||
|
1 meter |
2 meter |
4 meter |
6 meter |
8 meter |
10 meter |
|
|
1.5 |
0.58 |
1.15 |
2.31 |
3.46 |
4.62 |
5.77 |
|
2.5 |
0.36 |
0.71 |
1.43 |
2.14 |
2.86 |
3.57 |
|
4.0 |
0.22 |
0.45 |
0.89 |
1.34 |
1.79 |
2.24 |
|
6.0 |
0.15 |
0.30 |
0.60 |
0.89 |
1.19 |
1.49 |
|
10.0 |
0.09 |
0.18 |
0.36 |
0.54 |
0.71 |
0.89 |
CT Burden Calculation:
§ The Actual burden is
formed by the resistance of the pilot conductors and the protection relay(s).
The resistance of a conductor (with a constant cross-sectional area) can be
calculated from the equation:
§ R =ƿxL / A
§ where ƿ =
resistivity of the conductor material (given typically at +20°C) ,L= length of
the conductor , A = cross sectional area
§ If the resistivity is
given in μΩm, the length in meters and the area in mm2, the equation 1 will
give the resistance directly in ohms.
§ Resistivity:
Copper 0.0178 µΩm at 20 °C and 0.0216 µΩm at
75 °C
Burden
of CT for 4 or 6 wire connection:
§ If 6-wire connection
is used, the total length of the wire, naturally, will be two times the
distance between the CT and the relay. However, in many cases a common
return conductor is used as shown in figure then, instead of multiplying the
distance by two, a factor of 1.2 is typically used. This rule only applies to
the 3-phase connection only. The factor 1.2 allows for a situation, where
up to 20% of the electrical conductor length, including terminal resistances,
uses 6-wire connection and at least 80% 4-wire connection.
§ Example: the distance
between the CT and the relay is 5 meters the total length is 2 x 5 m = 10 meter
for 6-wire connection, but only 1.2 x 5 m = 6.0 meter when 4-wire connection is
used.
Burden
of the relay:
§ Example: The Distance between
the CTs and the protection relay is 15 meters, 4 mm2 Cu conductors in 4-wire
connection are used. The burden of the relay input is less than 20 mΩ (5 A
inputs). Calculate the actual burden of the CT at 75°C , the input impedance is
less than 0.020 Ω for a 5 A input (i.e. burden less than 0.5 VA) and less than
0.100 Ω for a 1 A input (i.e. less than 0.1 VA):
§ Solution:
§ ƿ = 0.0216 µΩm (75°C)
for copper conductor.
§ R =ƿxL / A ,R =
0.0216 µΩm x (1.2 x 15 m) / 4 mm2 = 0.097 Ω
§ Burden of CT = 0.097
Ω + 0.020 Ω = 0.117 Ω.
§ Using CTs of burden
values higher than required, is unscientific since it leads to inaccurate
reading (meter) or inaccurate sensing of fault / reporting conditions.
§ Basically, such high
value of design burden extends saturation characteristics of CT core leading to
likely damage to the meter connected across it under overload condition. e.g.
When we expect security factor (ISF) to be 5, the secondary current should be
restricted to less than 5 times in case primary current shoots to more than 5
times its rated value.
§ In such an overload
condition, the core of CT is desired to go into saturation, restricting the
secondary current thus the meter is not damaged. However, when we ask for
higher VA, core doesn’t go into saturation due to less load (ISF is much higher
than desired) which may damage the meter.
§ To understand the
effect on Accuracy aspect, let’s take an example of a CT with specified burden
of 15 VA, and the actual burden is 2.5 VA:15 VA CT with less than 5 ISF will
have saturation voltage of 15 Volts (15/5×5), and actual burden of 2.5 VA the
saturation voltage required shall be ( 2.5/5 x 5) 2.5 Volts against 15 Volts
resulting ISF = 30 against required of 5.
§ Example: Decide Whether 5A,20VA
CT is sufficient for following circuit
§ Total instrument
burden = 2 + 2 + 3 + 2 + 4 = 13V A.
§ Total pilot load
resistance = 2 x 0.1 = 0.2 ohm.
§ With 5A secondary
current, volt-drop in leads is 5 x 0.2 = 1 V.
§ Burden imposed by
both leads = 5A x 1 V = 5V A.
§ Total burden on CT =
13 + 5 = 18V A.
§ As the CT is rated
20V A, it has sufficient margin.
3) Accuracy Class of
CT:
§ The CT accuracy is
determined by its certified accuracy class which is stamped on nameplate. For
example, CT accuracy class of 0.3 means that the CT is certified by the
manufacturer to be accurate to within 0.3 percent of its rated ratio value for
a primary current of 100 percent of rated ratio.
§ CT with a rated ratio
of 200/ 5 with accuracy class of 0.3 would operate within 0.45 percent of its
rated ratio value for a primary current of 100 amps. To be more explicit, for a
primary current of 100A it is certified to produce a secondary current between
2.489 amps and 2.511 amps.
§ Accuracy is specified
as a percentage of the range, and is given for the maximum burden as expressed
in VA. The total burden includes the input resistance of the meter and
the loop resistance of the wire and connections between the current transformer
and meter.
§ Example: Burden = 2.0
VA. Maximum Voltage drop = 2.0 VA / 5 Amps = 0.400 Volts.
§ Maximum
Resistance = Voltage / Current = 04.00 Volts / 5 Amps =0.080 Ohms.
§ If the input
resistance of the meter is 0.010Ω, then 0.070Ω is allowed for loop resistance of
the wire, and connections between the current transformer and the meter. The
length and gauge of the wire must be considered in order to avoid exceeding the
maximum burden.
§ If resistance in the
5 amp loop causes the burden to be exceeded, the current will drop. This will
result in the meter reading low at higher current levels.
§ As in all
transformers, errors arise due to a proportion of the primary input current
being used to magnetize the core and not transferred to the secondary winding.
The proportion of the primary current used for this purpose determines the
amount of error.
§ The essence of good
design of measuring current transformers is to ensure that the magnetizing
current is low enough to ensure that the error specified for the accuracy class
is not exceeded.
§ This is achieved by
selecting suitable core materials and the appropriate cross-sectional area of
core. Frequently in measuring currents of 50A and upwards, it is convenient and
technically sound for the primary winding of a CT to have one turn only.
§ In these most common
cases the CT is supplied with a secondary winding only, the primary being the
cable or bus bar of the main conductor which is passed through the CT aperture
in the case of ring CTs (i .e. single primary turn) it should be noted that the
lower the rated primary current the more difficult it is (and the more
expensive it is) to achieve a given accuracy.
§ Considering a core of
certain fixed dimensions and magnetic materials with a secondary winding of say
200 turns (current ratio 200/1 turns ratio 1/200) and say it takes 2 amperes of
the 200A primary current to magnetize the core, the error is therefore only 1%
approximately. However considering a 50/1 CT with 50 secondary turns on the
same core it still takes 2 amperes to magnetize to core. The error is then 4%
approximately. To obtain a 1% accuracy on the 50/1 ring CT a much larger core
and/or expensive core material is required
§ Accuracy Class of
Metering CT:
|
Metering Class CT |
|
|
Class |
Applications |
|
0.1 To 0.2 |
Precision measurements |
|
0.5 |
High grade kilowatt hour meters for commercial grade kilowatt hour
meters |
|
3 |
General industrial measurements |
|
3 OR 5 |
Approximate measurements |
|
Protective System |
CT Secondary |
VA |
Class |
|
Per current for
phase & earth fault |
1A |
2.5 |
10P20 Or 5P20 |
|
5A |
7.5 |
10P20 Or 5P20 |
|
|
Unrestricted earth
fault |
1A |
2.5 |
10P20 Or 5P20 |
|
5A |
7.5 |
10P20 Or 5P20 |
|
|
Sensitive earth
fault |
1A or 5A |
|
Class PX use relay
manufacturers formula |
|
Distance protection |
1A or 5A |
|
Class PX use relay
manufacturers formula |
|
Differential
protection |
1A or 5A |
|
Class PX use relay
manufacturers formula |
|
High impedance
differential impedance |
1A or 5A |
|
Class PX use relay
manufacturers formula |
|
High speed feeder
protection |
1A or 5A |
|
Class PX use relay
manufacturers formula |
|
Motor protection |
1A or 5A |
5 |
5P10 |
§ Accuracy Class of
Letter of CT:
|
Metering Class CT |
|
|
Accuracy Class |
Applications |
|
B |
Metering Purpose |
|
Protection Class CT |
|
|
C |
CT has low leakage flux. |
|
T |
CT can have significant leakage flux. |
|
H |
CT accuracy is applicable within the entire range of secondary
currents from 5 to 20 times the nominal CT rating. (Typically wound CTs.) |
|
L |
CT accuracy applies at the maximum rated secondary burden at 20
time rated only. The ratio accuracy can be up to four times greater than the
listed value, depending on connected burden and fault current. (Typically
window, busing, or bar-type CTs.) |
§ Accuracy Class of
Protection CT:
|
Class |
Applications |
|
10P5 |
Instantaneous over
current relays & trip coils: 2.5VA |
|
10P10 |
Thermal inverse
time relays: 7.5VA |
|
10P10 |
Low consumption
Relay: 2.5VA |
|
10P10/5 |
Inverse definite
min. time relays (IDMT) over current |
|
10P10 |
IDMT Earth fault
relays with approximate time grading:15VA |
|
5P10 |
IDMT Earth fault
relays with phase fault stability or accurate time grading: 15VA |
§ Accuracy
Class: Metering Accuracy as per IEEE C37.20.2b-1994
|
Ratio |
B0.1 |
B0.2 |
B0.5 |
B0.9 |
B1.8 |
Relaying Accuracy |
|
50:5 |
1.2 |
2.4 |
– |
– |
– |
C or T10 |
|
75:5 |
1.2 |
2.4 |
– |
– |
– |
C or T10 |
|
100:5 |
1.2 |
2.4 |
– |
– |
– |
C or T10 |
|
150:5 |
0.6 |
1.2 |
2.4 |
– |
– |
C or T20 |
|
200:5 |
0.6 |
1.2 |
2.4 |
– |
– |
C or T20 |
|
300:5 |
0.6 |
1.2 |
2.4 |
2.4 |
– |
C or T20 |
|
400:5 |
0.3 |
0.6 |
1.2 |
1.2 |
2.4 |
C or T50 |
|
600:5 |
0.3 |
0.3 |
0.3 |
1.2 |
2.4 |
C or T50 |
|
800:5 |
0.3 |
0.3 |
0.3 |
0.3 |
1.2 |
C or T50 |
|
1200:5 |
0.3 |
0.3 |
0.3 |
0.3 |
0.3 |
C100 |
|
1500:5 |
0.3 |
0.3 |
0.3 |
0.3 |
0.3 |
C100 |
|
2000:5 |
0.3 |
0.3 |
0.3 |
0.3 |
0.3 |
C100 |
|
3000:5 |
0.3 |
0.3 |
0.3 |
0.3 |
0.3 |
C100 |
|
4000:5 |
0.3 |
0.3 |
0.3 |
0.3 |
0.3 |
C100 |
Important of accuracy & phase
angle
§ Current error is an
error that arises when the current value of the actual transformation ratio is
not equal to rated transformation ratio.
§ Current error (%) =
{(Kn x Is – Ip) x 100}/Ip
§ Kn = rated
transformation ratio, Ip = actual primary current, Is = actual secondary
current
§ Example: In case of a
2000/5A class 1 5VA current transformer
§ Kn = 2000/5 = 400
turn, Ip = 2000A, Is = 4.9A
§ Current error = ((400
x 4.9 – 2000) x100)/2000 = -2%
§ For protection class
current transformer, the accuracy class is designed by the highest permissible
percentage composite error at the accuracy limit primary current prescribed for
the accuracy class concerned.
§ Accuracy class
includes: 5P, 10P
By phase angle
§ Phase error is the
difference in phase between primary & secondary current vectors, the
direction of the vectors to be zero for a perfect transformer.
§ You will experience a
positive phase displacement when secondary current vector lead primary current
vector.
§ Unit of scale
expressed in minutes / cent radians.
§ Circular measure =
(unit in radian) is the ratio of the distance measured along the arc to the
radius.
§ Angular measure =
(unit in degree) is obtained by dividing the angle subtended at the center of a
circle into 360 deg equal division known as “degrees”.
§ Limits of current
error and phase displacement for measuring current transformer (Classes 0.1 To
1)
|
Accuracy Class |
+/- Percentage Current (Ratio) Error
at % Rated Current |
+/- Phase Displacement at % Rated
Current |
||||||||||
|
Minutes |
Centi radians |
|||||||||||
|
5 |
20 |
100 |
120 |
5 |
20 |
100 |
120 |
5 |
20 |
100 |
120 |
|
|
0.1 |
0.4 |
0.2 |
0.1 |
0.1 |
15 |
8 |
5 |
5 |
0.45 |
0.24 |
0.15 |
0.15 |
|
0.2 |
0.75 |
0.35 |
0.2 |
0.2 |
30 |
15 |
10 |
10 |
0.9 |
0.45 |
0.3 |
0.3 |
|
0.5 |
1.5 |
0.75 |
0.5 |
0.5 |
90 |
45 |
30 |
30 |
2.7 |
1.35 |
0.9 |
0.9 |
|
1.0 |
3 |
1.5 |
1 |
1 |
180 |
90 |
60 |
60 |
5.4 |
2.7 |
1.8 |
1.8 |
§ limits of
current error and phase displacement for measuring current transformer For
special application
|
Accuracy Class |
+/- Percentage Current (Ratio) Error
at % Rated Current |
+/- Phase Displacement at % Rated
Current |
|||||||||||||
|
Minutes |
Centi radians |
||||||||||||||
|
1 |
5 |
20 |
100 |
120 |
1 |
5 |
20 |
100 |
120 |
1 |
5 |
20 |
100 |
120 |
|
|
0.2S |
0.75 |
0.35 |
0.2 |
0.2 |
0.2 |
30 |
15 |
10 |
10 |
10 |
0.9 |
0.4 |
0.3 |
0.3 |
0.3 |
|
0.5S |
1.50 |
0.75 |
0.5 |
0.5 |
0.5 |
90 |
45 |
30 |
30 |
30 |
2.7 |
1.3 |
0.9 |
0.9 |
0.9 |
§ limits of
current error for measuring current transformers (classes 3 and 5)
|
Accuracy Class |
+/- Percentage Current (Ratio) Error
at % Rated Current |
|
|
|
|
|||
|
50 |
120 |
|
|
|
3 |
3 |
3 |
|
|
5 |
5 |
5 |
|
Class X Current Transformer:
§ Class X current
transformer is use in conjunction with high impedance circulating current
differential protection relay, eg restricted earth fault relay. As illustrated
in IEC60044-1, the class X current transformer is needed.
§ The following
illustrates the method to size a class X current transformer.
§ Step 1: calculating
knee point voltage Vkp
§ Vkp = {2 x Ift
(Rct+Rw)}/ k
§ Vkp = required CT
knee point voltage, Ift = max transformer through fault in ampere
§ Rct = CT secondary
winding resistance in ohms, Rw = loop impedance of pilot wire between CT and
the
§ K = CT transformation
ratio
§ Step 2: calculate
Transformer through fault Ift
§ Ift = (KVA x 1000)/
(1.732 x V x Impedance)
§ KVA = transformer
rating in kVA , V = transformer secondary voltage, Impedance = transformer
impedance
§ Step 3: How to obtain
Rct
§ To measure when CT is
produce
§ Step 4: How to obtain
Rw
§ This is the
resistance of the pilot wire used to connect the 5th class X CT at the
transformer star point to the relay
§ In the LV
switchboard. Please obtain this data from the Electrical contractor or
consultant. We provide a table to Serve as a general guide on cable
resistance.
§ Example:
§ Transformer Capacity:
2500kVA
Transformer
impedance: 6%
Voltage system : 22kV
/ 415V 3phase 4 wire
Current transformer
ratio: 4000/5A
Current transformer
type: Class X PR10
Current transformer
Vkp : 185V
Current transformer
Rct : 1.02½ (measured)
Pilot wire resistance
Rw : 25 meters using 6.0mm sq cable
= 2 x 25 x 0.0032 =
0.16½
Ift = (kVA x 1000) /
(1.732 x V x impedance) = (2500 x 1000) / (1.732 x 415 x 0.06)= 57,968 (Say
58,000A)
Vkp = {2 x Ift
(Rct+Rw)} / k= {2 x 58000 (1.02+0.16)} / 800= 171.1½.
4) Accuracy
Limit Factor:
§ Standard Accuracy
Limit Factors: 5, 10, 15, 20 and 30.
§ Accuracy of a CT is
another parameter which is also specified with CT class. For example, if a
measuring CT class is 0.5M (or 0.5B10), the accuracy is 99.5% for the CT, and
the maximum permissible CT error is only 0.5%.
§ Accuracy limit
Factor is defined as the multiple of rated primary current up to which the
transformer will comply with the requirements of ‘Composite Error’. Composite
Error is the deviation from an ideal CT (as in Current Error), but takes
account of harmonics in the secondary current caused by non-linear magnetic
conditions through the cycle at higher flux densities.
§ The electrical
requirements of a protection current transformer can therefore be defined as :
§ Selection of Accuracy
Class & Limit Factor.
§ Class 5P and 10P
protective current transformers are generally used in over current and
unrestricted earth leakage protection. With the exception of simple trip
relays, the protective device usually has an intentional time delay, thereby
ensuring that the severe effect of transients has passed before the relay is
called to operate. Protection Current Transformers used for such applications
are normally working under steady state conditions Three examples of such
protection is shown. In some systems, it may be sufficient to simply detect a
fault and isolate that circuit. However, in more discriminating schemes, it is
necessary to ensure that a phase to phase fault does not operate the earth
fault relay.
§ Calculation of the
Accuracy limit factor
§ Fa=Fn X ( (Sin+Sn)
/ (Sin+Sa) )
§ Fn = Rated Accuracy
Limit Factor, Sin = Internal Burden of CT secondary Coil
§ Sn= Rated Burden of
CT (in VA), Sa= Actual Burden of CT (in VA)
§ Example: The internal
secondary coil resistance of the CT(5P20) is 0.07 Ω, the secondary burden
(including wires and relay) is 0.117 Ω and the CT is rated 300/5, 5P20, 10 VA.
Calculate the actual accuracy limit factor.
§ Fn = 20 (CT data
5P20), Sin = (5A)2 × 0.07 Ω =1.75 VA, Sn = 10 VA (from CT data),
§ Sa = (5A)2 × 0.117 Ω
= 2.925 VA
§ Accuracy limit factor
ALF (Fa) = 20 X ((1.75+10) / (1.75+2.925)) =50.3
Selection of CT:
1) Indoors or Out Door:
§ Determine where CT
needs to be used. Indoor transformers are usually less costly than outdoor
transformers. Obviously, if the current transformer is going to be enclosed in
an outdoor enclosure, it need not be rated for outdoor use. This is a common
costly error in judgment when selecting current transformers.
2) What do We need:
§ The first thing we
need to know that what degree of accuracy is required. Example, if you simply
want to know if a motor is lightly or overloaded, a panel meter with 2 to 3%
accuracy will likely suit for needs. In that case the current transformer needs
to be only 0.6 to 1.2% accurate. On the other hand, if we are going to drive a
switchboard type instrument with 1% accuracy, we will want a current
transformer with 0.3 to 0.6 accuracy. We must keep in mind that the accuracy
ratings are based on rated primary current flowing and per ANSI standards may
be doubled (0.3 becomes 0.6%) when 10% primary current flows. As mentioned
earlier, the rated accuracies are at stated burdens. We must take into
consideration not only the burden of the load (instrument) but you must
consider the total burden. The total burden includes the burden of the current
transformers secondary winding, the burden of the leads connecting the
secondary to the load, and the burden of the load itself. The current
transformer must be able to support the total burden and to provide the
accuracy required at that burden. If we are going to drive a relay you must
know what relay accuracy the relay will require.
3) Voltage Class:
§ You must know what
the voltage is in the circuit to be monitored. This will determine what the
voltage class of the current transformer must be as explained earlier.
4) Primary Conductor:
§ If you have selected
a current transformer with a window you must know the number, type and size of
the primary conductor(s) in order to select a window size which will
accommodate the primary conductors.
5) Application:
§ The variety of
applications of current transformers seems to be limited only by ones
imagination. As new electronic equipment evolves and plays a greater role in
the generation, control and application of electrical energy, new demands will
be placed upon current transformer manufacturers and designers to provide new
products to meet these needs
6) Safety:
§ For personnel and
equipment safety and measurement accuracy, current measurements on conductors
at high voltage should be made only with a conducting shield cylinder placed
inside the CT aperture. There should be a low electrical impedance connection
from one end only to a reliable local ground. An inner insulating cylinder of
adequate voltage isolation should be between the shield cylinder and the conductor
at high voltage. Any leakage, induced or breakdown current between the high
voltage conductor and the ground shield will substantially pass to local ground
rather than through the signal cable to signal ground. Do not create a “current
loop” by connecting the shield cylinder to ground from both ends. Current
flowing in this loop will also be measured by the CT.
7) CT output signal termination:
§ The CT output coaxial
cable should preferably be terminated in 50 ohms. CT characteristics are
guaranteed only when CT is terminated in 50 ohms. The termination should
present sufficient power dissipation capability. When CT output is
terminated in 50 ohms, its sensitivity is half that when terminated in a
high-impedance load.
Installing of CT:
§ Measurements must
have the same polarity to keep the power factor and direction of power flow
measurements accurate and consistent.
§ Most CTs are labelled
that shows which side of the CT should face either the source or the load.
§ Primary Side : The Primary of CT is
marked with H1 and H2 ( or only marking dot on one side)
§ The label “H1” or dot
defines the direction as flowing current into the CT (H1 or the dot should face
the Power source side). H2 side to load facing direction
§ Secondary Side: The Secondary
(The output wires) of CT is marked with X1 and X2.
§ X1 corresponds to
H1, or the input side.The X1 secondary terminal is the polarity terminal. The
polarity marks of a current transformer indicate that when a primary current
enters at the polarity mark (H1) of the primary, a current in phase with the
primary current and proportional to it in magnitude will leave the polarity
terminal of the secondary (X1).
§ Normally CT’s should
not be installed on live services. The power should be disconnected when
the CT’s are installed. Many times this is not possible
because of critical loads such as computers, laboratories, etc. that cannot be
shut down. Split core CT’s should not be installed on live un
insulated bus bars under any conditions.
Modification of Primary &
Secondary Turns Ratio:
§ The nameplate current
ratio of the current transformer is based on the condition that the primary
conductor will be passed once through the transformer opening. If necessary,
this rating can be reduced in even multiples by looping this conductor two or
more times through the opening.
§ A transformer having
a rating of 300 amperes will be changed to 75 amperes if four loops or turns
are made with the primary cable.
§ The ratio of the
current transformer can be also modified by altering the number of secondary
turns by forward or back-winding the secondary lead through the window of the
current transformer.
§ By adding secondary
turns, the same primary amperage will result in a decrease in secondary output.
§ By subtracting
secondary turns, the same primary amperage will result in greater secondary
output. Again using the 300:5 example, adding two secondary turns will require
310 amps on the primary to maintain the 5 amp secondary output or 62/1p =
310p/5s.
§ Subtracting two
secondary turns will only require 290 amps on the primary to maintain the 5 amp
secondary output or 58s/5p = 290p/5s. The ratio modifications are achieved in
the following manner:
§ To add secondary
turns, the white lead should be wound through the CT from the side opposite the
polarity mark.
§ To subtract turns,
the white lead should be wound through the CT from the same side as the
polarity mark.
1) Modifications in Primary Turns Ratio of CT:
§ The ratio of the
current transformer can be modified by adding more primary turns to the
transformer. By adding primary turns, the current required to maintain five
amps on the secondary is reduced.
§ Ka = Kn X (Nn/Na)
§ Ka= Actual Turns
Ration.
§ Kn=Name Plate T/C
Ratio.
§ Nn=Name Plate Number
of Primary Turns.
§ Na=Actual Number of
Primary Turns.
§ Example: 100:5
Current Transformers.
2) Modifications in Secondary Turns Ratio of CT:
§ Formula : Ip/Is
= Ns/Np
§ Ip = Primary Current
, Is = Secondary Current , Np = No of Primary Turns, Ns = No of Secondary Turns
§ Example: A 300:5 Current
Transformer.
§ The ratio of the
current transformer can be modified by altering the number of secondary turns
by forward or back winding the secondary lead through the window of the current
transformer.
§ By adding secondary
turns, the same primary current will result in a decrease in secondary output.
By subtracting secondary turns, the same primary current will result in greater
secondary output.
§ Again using the 300:5
example adding five secondary turns will require 325 amps on the primary to
maintain the 5 amp secondary output or: 325 p / 5s = 65s / 1p
§ Deducting 5 secondary
turns will only require 275 amps on the primary to maintain the 5 amp secondary
output or: 275p / 5s = 55s / 1p
§ The above ratio
modifications are achieved in the following manner:
§ Current Transformer
Ratio Modification:
|
CT Ratio |
Number of Primary Turns |
Modified Ratio |
|
100:5A |
2 |
50:5A |
|
200:5A |
2 |
100:5A |
|
300:5A |
2 |
150:5A |
|
100:5A |
3 |
33.3:5A |
|
200:5A |
3 |
66.6:5A |
|
300:5A |
3 |
100:5A |
|
100:5A |
4 |
25:5A |
|
200:5A |
4 |
50:5A |
|
300:5A |
4 |
75:5A |
§ A primary turn is the
number of times the primary conductor passes through the CT’s window. The main
advantage of this ratio modification is you maintain the accuracy and burden
capabilities of the higher ratio. The higher the primary rating the better the
accuracy and burden rating.
§ You can make smaller
ratio modification adjustments by using additive or subtractive secondary
turns.
§ For example, if
you have a CT with a ratio of 100:5A. By adding one additive secondary turn the
ratio modification is 105:5A, by adding on subtractive secondary turn the ratio
modification is 95:5A.
§ Subtractive secondary
turns are achieved by placing the “X1” lead through the window from the H1 side
and out the H2 side. Additive secondary turns are achieved by placing the “X1” lead
through the window from the H2 and out the H1 side.
§ So, when there is
only one primary turn each secondary turn modifies the primary rating by 5
amperes. If there is more than one primary turn each secondary turn value is
changed (i.e. 5A divided by 2 primary turns = 2.5A).
§ The following
table illustrates the effects of different combination of primary and secondary
turns:
|
CT RATIO 100:5A |
||
|
PRIMARY TURNS |
SECONDARY TURNS |
RATIO ADJUSTMENT |
|
1 |
-0- |
100:5A |
|
1 |
1+ |
105:5A |
|
1 |
1- |
95:5A |
|
2 |
-0- |
50:5A |
|
2 |
1+ |
52.5:5A |
|
2 |
2- |
45.0:5A |
|
3 |
-0- |
33.3:5A |
|
3 |
1+ |
34.97:5A |
|
3 |
1- |
31.63:5A |
Advantages of using a CT having 1A
Secondary:
§ The standard CT
secondary current ratings are 1A & 5A,The selection is based on the lead
burden used for connecting the CT to meters/Relays.5A CT can be used where
Current Transformer & protective’s device are located within same
Switchgear Panel.
§ 1A CT is preferred if
CT leads goes out of the Switchgear.
§ For Example if CT is
located in Switch Yard & CT leads have to be taken to relay panels located
in control room which can be away.1A CT is preferred to reduce the load burden.
For CT with very High lead length, CT with Secondary current rating of 0.5 Amp
can be used.
§ In large Generator
Circuits, where primary rated current is of the order of few kilo-amperes
only,5A CTs are used, 1A CTs are not preferred since the turns rations becomes
very high & CT becomes unwieldy.
Danger with Current Transformer:
§ When a CT secondary
circuit is closed, current flows through it, which is an exact proportion of
the primary current, regardless of the resistance of the burden. In the CT have
a ratio of 1OOO/5A and to have 1OOOA flowing in the primary is carrying exactly
5A.
§ If the secondary
terminals S1 and S2 are short- circuited, there is no voltage between them.
§ If now the
short-circuit be replaced by a resistance of, say, 0.5 ohm the same 5A will
flow through, causing a volt-drop of 2.5V and a burden of 5 x 2.5 = 12.5V A. If
the resistance were increased to 5 ohms the terminal voltage with 5A flowing
would rise to 25V and the burden to 125V A.
§ The greater the
resistance, the greater would be the voltage and burden until, as it approached
infinity (the open-circuit condition), so also in theory would the voltage (and
burden) become infinite. This cannot of course happen in practice because the
CT would saturate or the terminals flash over due to the very high secondary
voltage between them. But it does show the danger of open-circuiting the
secondary of running CT. lethal voltages can be produced at the point of opening.
This is why CT secondaries are never fused.
§ The danger from an
open-circuited CT is twofold. It can produce lethal voltages and so is a very
real danger to personnel. The high voltage across the secondary winding could
also cause insulation failure in that winding, leading at best to inaccuracy
and at worst to burn- out or fire.
§ Before ever an
instrument or relay is removed from the secondary loop of a running CT (if such
a thing had to be done), the wires feeding that instrument must first be
securely short- circuited at a suitable terminal box or, better, at the CT
itself. Similarly, if a running CT is ever to be taken out of circuit, it must
first be firmly shorted. CTs with 1 A secondary’s are more dangerous than those
with 5A, as the induced voltages are higher.
§ Ammeter resistance is
very low ,the current transformer normally works short circuited.
§ If for any reason the
ammeter is taken out of secondary winding then the secondary winding must be
short circuited with the help of short circuit switch .
§ If this is not done,
then due to high m.m.f. will set up high flux in the core and it will produces
excessive core loss which produce heat and high voltage across the secondary
terminals
§ Hence the secondary of
current transformer is never left open
Sizing of CT for Building:
§ New construction: size the CT to
handle about 80% of the circuit breaker capacity. If the building is served by
a 2000 amp breaker, use 1600 amp (2000 x 0.8) CT’s.
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