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TRANSCRIPT
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Current and voltage transformers
Current or voltage instrument transformers are necessary for isolating the protection,control and measurement equipment from the high voltages of a power system, and for
supplying the equipment with the appropriate values of current and voltage - generally
these are 1A or 5 for the current coils, and 120 V for the voltage coils.
The behavior of current and voltage transformers during and after the occurrence of a
fault is critical in electrical protection since errors in the signal from a transformer can
cause maloperation of the relays.
In addition, factors such as the transient period and saturation must be taken into
account when selecting the appropriate transformer.
When only voltage or current magnitudes are required to operate a relay then therelative direction of the current flow in the transformer windings is not important.
However, the polarity must be kept in mind when the relays compare the sum or difference
of the currents.
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1- Voltage transformers:
With voltage transformers (VTs) it is essential that the voltage from the secondary
winding should be as near as possible proportional to the primary voltage.
In order to achieve this, VTs are designed in such a way that the voltage drops in the
windings are small and the flux density in the core is well below the saturation value so
that the magnetization current is small; in this way magnetization impedance is obtained
which is practically constant over the required voltage range. The secondary voltage of aVTis usually 110 or120 Vwith corresponding line-to-neutral values. The majority of
protection relays have nominal voltages of110 or63.5 V, depending on whether their
connection is line-to-line or line-to-neutral.
Figure 1 Voltage transformer equivalent circuits
Figure 2 Vector diagram for voltage transformer
1.1 Equivalent circuits
VTs can be considered as small power transformers so that their equivalent circuit
is the same as that for power transformers, as shown in Figure 1a. The magnetization
branch can be ignored and the equivalent circuit then reduces to that shown in Fig 1b.
The vector diagram for a VTis given in Figure.2, with the length of the voltage
drops increased for clarity. The secondary voltage Vs lags the voltage Vp/n and is
smaller in magnitude. In spite of this, the nominal maximum errors are relatively small.
VTs have an excellent transient behaviour and accurately reproduce abrupt changes in.
the primary voltage.
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1.2 Errors
When used for measurement instruments, for example for billing and control
purposes, the accuracy of a VTis important, especially for those values close to the
nominal system voltage.
Notwithstanding this, although the precision requirements of a VTfor protection
applications are not so high at nominal voltages, owing to the problems of having to
cope with a variety of different relays, secondary wiring bu rd ens and the uncertainty
of system parameters, errors should he contained within narrow limits over a widerange of possible voltages under fault conditions.
This range should be between 5 and 173% of the nominal primary voltage for VTs
connected between line and earth.
Referring to the circuit in Figure 1a, errors in a VTare clue to differences in
magnitude and phase between Vp/n, and Vs. These consist of the errors under open-
circuit conditions when the load impedance B is infinite, caused by the drop in
voltage from the circulation of the magnetization current through the primary winding,
and errors due to voltage drops as a result of the load current ILflowing through both
windings. Errors in magnitude can be calculated fromErrorV T= {(n Vs - Vp) / Vp} x 100%. If the error is positive, then the secondary
voltage exceeds the nominal value.
1.3 Burden
The standard burden for voltage transformer is usually expressed in volt-amperes (V)
at a specified power factor.
Table 1 gives standard burdens based onANSIStandard C57.1 3. Voltage
transformers are specified inIECpublication 186 by the precision class, and the
value of volt-amperes (V).The allowable error limits corresponding to different class values are shown in
Table 2, where Vn is the nominal voltage. The phase error is considered positive when
the secondary voltage leads the primary voltage. The voltage error is the percentage
difference between the voltage at the secondary terminals, V2, multiplied by the
nominal transformation ratio, and the primary voltages V1.
1.4 Selection of VTs
Voltage transformers are connected between phases, or between phase and earth.
The connection between phase and earth is normally used with groups of three single-
phase units connected in star at substations operating with voltages at about 34.5 kV
or higher, or when it is necessary to measure the voltage and power factor of each
phase separately.
The nominal primary voltage of a VTis generally chosen with the higher nominal
insulation voltage (kV) and the nearest service voltage in mind. The nominal secondary
voltages are generally standardized at 110 and 120 V. In order to select the nominal
power of a VT, it is usualto acid together all the nominal V loadings of the apparatus
connected to
Table 1 Standard burdens for voltage TransformerStandard burden Characteristics for 120 V
and 60 Hz
Characteristics for 69.3 V
and 60 Hz
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design Volt-
amperes
power
factor
resistance() inductance
(H)
impedance
( )
resistance
( )
inductance
(H)
impedance
( )
W 12.5 0.10 115.2 3.040 1152 38.4 1.010 384
25.0 0.70 403.2 1.090 575 134.4 0.364 192
75.0 0.85 163.2 0.268 192 54.4 0.089 64
200.0 0.85 61.2 0.101 72 20.4 0.034 24
400.0 0.85 31.2 0.0403 36 10.2 0.0168 12
35.0 0.20 82.3 1.070 411 27.4 0.356 137
Table 2 Voltage transformers error limitsClass Primary
voltage
Voltage
error (
%)
Phase error
(min)
0.1
0.8 Vn , 1.0
Vn and 1.2
Vn
0.1 0.5
0.2 0.2 10.0
0.5 0.5 20.0
1.0 1.0 40.0
0.1
0.5 Vn
1.0 40.0
0.2 1.0 40.0
0.5 1.0 40.0
1.0 2.0 80.0
0.1
Vn
0.2 80.0
0.2 2.0 80.0
0.5 2.0 80.0
1.0 3.0 120.0
Vn = nominal voltageThe VT secondary winding. In addition, it is important to take account of the voltage
drops in the secondary wiring, especially if the distance between the transformers and the
relays is large.
1.5 Capacitor volt age transformers
In general, the size of an inductive VT is proportional to its nominal voltage and, for
this reason, the cost increases in a similar manner to that of a high voltage transformer. One
alternative, and a more economic solution, is to use a capacitor voltage transformer.
This device is effectively a capacitance voltage divider, and is similar to a resistivedivider in that the output voltage at the point of connection is affected by the load - in fact
the two parts of the divider taken together can be considered as the source impedance
which produces a drop in voltage when the load is connected.
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Figure 4 Capacitor VT equivalent circuit
The capacitor divider differs from the inductive divider in that the equivalent
impedance of the source is capacitive and the .fact that this impedance can be
compensated for by connecting a reactance in series at the point of connection.
With an ideal reactance there are no regulation problems - however, in an actual
situation on a network, some resistance is always present. The divider can reduce the
voltage to a value which enables errors to be kept within normally acceptable limits.
For improved accuracy a high voltage capacitor is used in order to obtain a biggervoltage at the point of connection, which can be reduced to a standard voltage using a
relatively inexpensive trans-former as shown in Figure 3.
simplified equivalent circuit of a capacitorVTis shown in Figure 4 in which Viis equal to the nominal primary voltage, Cis the numerically equivalent impedance equal
to ( C1 + C2 ),Lis the resonance inductance,Ri represents the resistance of the primary
winding of transformer plus the losses in CandL, andZe is the magnetization
impedance of transformer. Referred to the inter-mediate voltage, the resistance of the
secondary circuit and the load impedance are represented by and respectively,
while and represent the secondary voltage and current.
Figure 5 Capacitor VT vector diagram
It can be seen that, with the exception ofC, the circuit in Figure 4.4 is the same as the
equivalent circuit of a power transformer. Therefore, at the system frequency when CandL
are resonating and canceling out each other, under stable system conditions the capacitorVTacts like a conventional transformer.Ri andR's are not large and, in addition, Ie is
small compared to I' s, so that the vector difference between Vi and V's which constitutes
the error in the capacitorVT, is very small.
This is illustrated in the vector diagram shown in Figure 4.5 which is drawn for a
power factor close to unity. The voltage error is the difference in magnitude between Vi
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and V's, whereas the phase error is indicated by the angle . From the diagram it can be
seen that, for frequencies different from the resonant frequency, the values of EL and ECpredominate, causing serious errors in magnitude and phase.
CapacitorVTs display better transient behaviour than electro-magnetic VTs as the
inductive and capacitive reactance in series are large in relation to the load impedance
referred to the secondary voltage, and thus, when the primary voltage collapses, the
secondary voltage is maintained for some milliseconds because of the combination of the
series and parallel resonant circuits represented byL, Cand the transformerT.
2 Current transformers
Although the performance required from a current transformer(CT) varies with the type of
protection, high grade CTs must always be used. Good quality CTs are more reliable and
result in less application problems and, in general, provide better protection.
Figure 6 Current transformer equivalent circuits
The quality ofCTs is very important for differential protection schemes where the
operation of the relays is directly related to the accuracy of the CTs under fault
conditions as well as under normal load conditions.
CTs can become saturated at high current values caused by nearby faults; to avoid
this, care should be taken to ensure that under the most critical faults the CToperates on
the linear portion of the magnetization curve. In all these cases the CTshould be a bleto supply sufficient current so that the relay operates satisfactorily.
2.1 Equivalent circuit
An approximate equivalent circuit for a CTis given in Figure 4.6a,
Where n2ZH represents the primary impedanceZH referred to the secondary side, and
the secondary impedance is,ZL,Rm andXm represent the losses and the excitation of the
core.
The circuit in Figure 4.6a can be reduced to the arrangement shown in figure 4.6b
whereZH can be ignored, since it does not influence either the currentIH/nor the voltage
acrossXm. The current flowing throughXm is the excitation currente.The vector diagram, with the voltage drops exaggerated for clarity, is shown in Figure 4.7.
In general,ZL, is resistive andelags Vs by 90, so thatIe is the principal source of error.
Note that the net effect ofIeis to makeIlag and be much smaller thanH /n, the primary
current referred to the secondary side.
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Figure 7 Vector diagram for the CT equivalent
circuit
2.2 Errors
The causes of errors in a CTare quite different to those associated with VTs. In effect,
the primary impedance of a CTdoes not have the same influence
On the accuracy of the equipment it only adds an impedance in series with the line, which
can be ignored. The errors are principally due to the current which circulates through the
magnetizing branch.
The magnitude error is the difference in magnitude betweenH / n andIL and is equaltoIr the component ofIe in line with k (see Figure 7).
The phase error, represented by , is related toIq the component ofIe which is in
quadrature withIL. The values of the magnitude and phase errors depend on the relative
displacement betweenIe andIL, but neither of them can exceed the vectorial error it should
be noted that a moderate inductive load, withIe andILapproximately in phase, has a small
phase error and the excitation component results almost entirely in an error in the
magnitude.
2.3 AC saturationCerrors result from excitation current, so much so that, in order to check if a CTis
functioning correctly, it is essential to measure or calculate the excitation curve. The
magnetization current of a CTdepends on the cross section and length of the magnetic
circuit, the number of turns in the windings, and the magnetic characteristics of the material.
Thus, for a given CT, and referring to the equivalent circuit of Figure 4.6b, it can be
seen that the voltage across the magnetization impedance,Es, is directly proportional to the
secondary current. From this it can be concluded that, when the primary current and
therefore the secondary current is increased, these currents reach a point where the core
commences to saturate and the magnetization current becomes sufficiently high to producean excessive error.
When investigating the behaviour of a CT, the excitation current should he measured
at various values of voltage the so-called secondary injection test. Usually, it ismore
convenient to apply a variable voltage to the secondary winding, leaving the primary
winding open-circuited. Figure 4.8a shows the typical relationship between the secondary
voltage and the excitation current determined in this way.
In European standards the pointp on the curve is called the saturation or knee point
and is defined as the point at which an increase in the excitation voltage of ten per cent
produces an increase of 50 % in the excitation current. This point is referred to in the ANSI /
IEEE standards as the intersection of the excitation curves with a 45 tangent line, as
indicated in Figure 4.8b. The European knee point is at a higher voltage than the ANSI/IEEE
Knee point.
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2.4 Burden
The burden of a CTis the value in ohms-of the impedance on the secondary side of
the CTdue to the relays and the connections between the CTand the relays. By way of
example, the standard burdens forCTs with a nominal secondary current of 5 A are shown
in Table 3, based on ANSI Standard C57.13.
IEC Standard Publication 185(1987) specifies CTs by the class of accuracy followed by the
letter or P, which denotes whether the transformer is suitable for measurement or
protection purposes, respectively. The current and phase-error limits for measurement andprotection CTs are given in Tables 4a and 4.4b. The phase error is considered positive when
the secondary current leads the primary current.
The current error is the percentage deviation of the secondary current, multiplied by
the nominal transformation ratio, from the primary current, i.e. {(CTR x 2) I1} I1 (%),
whereI1 = primary current (A),I2 = secondary current (A) and CTR = current transformer
transformation ratio. Those CTclasses marked with `ext' denote wide range (extended)
current transformers with a rated continuous current of 1.2 or 2 times the nameplate current
rating.
2.5 Selection of CTsWhen selecting a CT, it is important to ensure that the fault level and normal load
conditions do not result in saturation of the core and that
CT magnetization curves
Figure 8a CT magnetization curves
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Figure 8b CT magnetization curves
a Defining the knee point in a CT excitation curve according to Europeanstandards
b Typical excitation curves for a multi ratio class C CT (From IEEE Standard
C57.13-1978; reproduced by permission of the IEEE).
Table 4.3 Standard burdens for protection
CTs with 5 secondary current
Designation Resistance
()
Inductance
(mH)
Impedance
()
Volt-
amps
(at 5 A)
Power
factor
B-10.5
2.3 1.0 25 0.5
B-2 1.0 4.6 2.0 50 0.5
B-4 2.0 9.2 4.0 100 0.5
B-8 4.0 18.4 8.0 200 0.5
The errors do not exceed acceptable limits. These factors can be assessed from: formulae;
CT magnetization curves;
CT classes of accuracy.
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The first two methods provide precise facts for the selection of the CT. The third only
provides a qualitative estimation. The secondary voltage in Figure 4.6U has to be
determined for all three methods. If the impedance of the magnetic circuit, Xm is high,
this can be removed from the equivalent circuit with little error' givingEs=Vs and thus:
Vs=IL (ZL+ZC+ZB) (1)
Where
Vs = r.m.s. voltage induced in the secondary winding=maximum secondary current in amperes;
this can be determined by dividing the maximum
Fault current on the system by the transformer
turns ratio selected
ZB = ex ter na l impedance connected
ZL = impedance of the secondary winding
ZC=impedance of the connecting wiring
Use of the formulaThis method utilizes the fundamental transformer equation:
Vs= 4.44.f. . N. Bmax.10 -8 V (2)
Where
f =frequency in Hz,
=cross-sectional area of core (cm2)
=number of turns
Bmax =flux density (lines/cm2)
Table 4 Error limits for measurement current transformers
Class % current error at the given
proportion of rated current shown
below
% phase error at the given proportion of the rated
current shown below
2.0
*
1.2 1.0
0
0.5
0
0.2
0
0.10
0.05
2.0
*
1.2 1.
0
0.5 0.
2
0.1 0.0
5
0.1
0.1 0.1
0.2 0.25 5 5
8 10
0.2 0.2 0.2 0.35
0.50 10 10 15 20
0.5 0.5 0.5 0.75
1.00 30 30 45 60
1.0 1.0 1.0 1.5 2.00 60 60 - 90 120
-
3.0 3.0 3.0 - - - - _ 120
- 120
- - -
0.1 0.1 0.1 0.2 0.25 0.4 5 - 5 8 10 15
0.2ext
0.2 0.2 0.35
0.50 0.75
10 - 10 15 20 30
0.5ext
0.5 0.5 0.75
1.00 1.5 30 - 30 45 60 90
1.0ext
1.0 1.0 1.5 2.00 60 - 60 - 90 120 -
3.0ext
3.0 - - 3.0 - - - 120 - - 120 - - -
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*ext = 200 %
Table 4b Error limits for protection current transformers
Accuracy
Class
+/- percentage
Current
ratio error
+/- Phase error
(minutes)
% Current 5 20 100 120 5 20 100 1200.1 0.4 0.2 0.1 0.1 15 8 5 5
0.2 0.75 0.35 0.2 0.2 30 15 10 10
0.5 1.5 0.75 0.5 0.5 90 45 30 30
1.0 3 1.5 1.0 1.0 180 90 60 60
Total error for nominal error limit current and nominal load is five per cent for 5P and 5
ext CTs and ten per cent for 10P and 10P ext CTs.The cross-sectional area of metal and the saturation flux density are sometimes
difficult to obtain.
The latter can be taken as equal to 100 000lines/Cm2, which is a typical value for
modern transformers. To use the formula, V is determined from eqn. 4.1 and Bmax. is then
calculated using eqn. 2. IfBmax.
Exceeds the saturation density, there could be appreciable errors in the secondary current
and the CT selected would not be appropriate.
Example 1.
Assume that a CT with a ratio of 2000/5 is available, having a steel core of highpermeability, a cross-sectional area of 3.25 In cm2 and a secondary winding with a
resistance of 0.31 . The impedance of the relays, including connections, is 2 . Determine
whether the CT would be saturated by a fault of 35 000 A at 50 Hz.
Solution
If the CT is not saturated, then the secondary current, IL, is
35 000x 5/2000=87.5 A. N= 2000/5 = 400 turns
And Vs=87.5x (0.31+2) =202.1 V. Using eqn. 4.2, Bmax, can now becalculated:Bmax = 202.1X108/4.44X50X3.25X400=70 030 lines/ cm2
Since the transformer in this example has a steel core of high permeability, this relativelylow value of flux density should not result in saturation.
Using the magnetization curve
Typical CT excitation curves which are supplied by manufacturers state the r.m.s.
current obtained on applying an r.m.s. voltage to the secondary winding, with the primary
winding open-circuited.
The curves give the magnitude of the excitation current required order to obtain a
specific secondary voltage.
The method consists of producing a curve which shows the relationship between the
primary and secondary currents for one tap and specified load conditions, such as shown in
Figure 4.9.
Starting with any value of secondary current, and with the help of the magnetisation
curves, the value of the corresponding primary current can be determined. The process is
summarized in the following steps:
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(a) Assume a value for IL.
(b) Calculate Vs in accordance with eqn. 4.1.
(c) Locate the value of Vs on the curve for the tap selected, and find the associated value of
the magnetization current, Ie.
d) Calculate I H / n (=IL + Ie) and multiply this value by n to refer it to the primary side of
the CT.
(e) This provides one point on the curve of IL against IH, and the process is then repeated
to obtain other values of IL and the resultant values of IH. By joining the points together
the curve of IL against IH is obtained.
Figure 4.9 using the
magnetization curve
a - assume a value for IL.
b - Vs = IL ( Z L + Z C + Z B )
c - find I e from the curve
d - IH=n(I1,+ I e )
e - draw the point on the curve
This method incurs an error in calculating IH /n by adding I e and IL togetherarithmetically and not vectorially, which implies not taking account of the load angle
and the magnetizations branch of the equivalent circuit. However, this error is not great
and the simplification snakes it easier to carry out the calculations.
After construction, the curve should be checked to confirm that the maximum
primary fault current is within the transformer saturation zone. If not, then it will be
necessary to repeat the process, changing the tap until the fault current is within the
linear part of the characteristic.
In practice it is not necessary to draw the complete curve because it is sufficient to
take the known fault current and refer to the secondary winding, assuming that there is nosaturation for the tap selected.
This converted value can be taken as IL initially for the process described earlier.
If the tap is found to be suitable after finishing the calculations, then a value of IH can be
obtained which is closer to the fault current.
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(I) limitation of size of CTs and more importantly
(II) the fact that the open circuit volts would be dangerously high for large CTs Primary
ratings, such as those encountered on large turbo alternators, e.g. 5,000 amperes. It is
standard practice in such applications to use a cascade arrangement of say 5,000/20A
together with 20/1A interposing auxiliary CTs
Instantaneous over current relays
Class P method of specification will a suffice. A secondary accuracy limit current greatly inexcess of the value t o cause relay operation serves no useful purpose and a rated accuracy
limit of5 will usually be adequate.
When such relays are set to operate at high values of over current, say from 5 to 15 times the
rated current o f the transformer, the accuracy limit factor must be at least as high as the
value of the setting current used in order to ensure fast relay operation.
Rated outputs higher than 15VA and rated accuracy limit factors higher than 10 are not
recommended for general purposes. It ispossible, however, to combine a higher rated
accuracy limit factor with a lower rated output and vice versa. But when the product of these
two exceeds 150 the resulting current transformer may be uneconomical, and/orofundulylarge dimensions.
Over current relays with Inverse and Definite Minimum Time
(IDMT) lag characteristic
In general, for both directional and non-directional relays class 10P current transformers
should be used
Earth fault relays with inverse timecharacteristic
(1) Schemes in which phase fault current stability and accurate time grading are not
required.
Class 10P current transformers are generally recommended in which the product of rated
output and rated accuracy limit fact or approaches 150 provided that the earth fault relay
is
not set below 20% of the rated current of the associated current transformer and that the
burden of the relay at its setting current does not exceed 4VA.
(2) Schemes in which phase fault stability and/or where time grading is critical.
Class 5P current transformers in which the product of rated output and accuracy
limit factor approaches 150 should be used.
They are in general suitable for ensuring phase fault stability up to 10 times the rated
primary current and for maintaining time grading of the earth f a u l t relays, up to current
values of the order of 10 times the earth fault setting provided t h a t the phase burden
effectively imposed on each current transformer does not exceed 50% of it s rated burden.
The rated accuracy limit factor is not less than 10 the earth fault relay is not set below 30 %The burden of the relay at its setting does not exceed 4VA
The use of a higher relay setting the use of an earth fault relay having a burden of less than
4VA at its setting The use of current transformers having a product of rated output and rated
accuracy factor in excess of 150.
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Class X Current Transformer
Protection current transformers specified in terms of complying with Class ' X I
Specification is generally applicable to unit systems where balancing of outputs from each
end of the protected plant is vital.
This balance, or stability during through fault conditions, is essentially of a transient nature
and thus the extent of the unsaturated (or linear) zone is of paramount importance. Hence a
statement of knee point voltage is the parameter of prime importance and it is normal toderive, from heavy current test results, a formula stating the lowest permissible value of VKif stable operation is to be guaranteed, e.g.
Vk = K In (RCT + 2RL + R0)
Where
K - Is a constant found by realistic heavy current tests?
In - rated current of C.T. and relay
RCT - secondary winding resistance of the line current transformersRL - lead burden (route length) in ohms
Ro - any other resistance (or impedance) in circuit
Formula for knee point voltage calculation of current transformer? Best Answer- Chosen by Voters
The formula for Knee Point Voltage is:
Vkp = K * If/CTR * (RCT + RL + RR)
in which,
K = Constant, conventionally taken as 2.0
Vkp = The minimum Knee Point Voltage
If = Maximum Fault Current at the location, in Amperes
CTR = CT Ratio
RCT = CT Secondary Winding Resistnace, in Ohms
RL = 2-way Lead Resistance, in Ohms
RR = Relay Burden, in Ohms
Motor & generator protection example settings DocumentTranscript
1. Motor protection exampleAssume you have a motor rated500 HP, .95 power factor & 90 % efficiency energized from a
4.16 Kvsource, using a microprocessor motor protectionrelay, provide the typical settings.The full load current canbe taken from the motor nameplate as well as the servicefactor. The motorfull load current can be calculated from the
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following: I = 500 (.746)/1.732 (4.16)(.85)(.9) =60 ampsMotor and Line Data FunctionsFunction DescriptionAdjustment / Display Range Setting IncrementsSettingMotor Nameplate FLA 1 2000A, adjustable between50-100% of Max Amp Rating.FLA must be programmedUpper limit of range automatically adjusts 1 60for relay tofunction downward as Service factor is increased.MotorNameplate 1.00 - 1.30 SF .05 1Overload Class During Start
NEMA / UL Class 5 20 5 Class 20Overload Class During RunNEMA / UL Class 5 30 5 Class 10Overload Reset 0 =Manual,1 = Auto,2 = Disabled Overload 1 1kV Voltage Input(nominal line, Medium Voltage) .60 15kV .01 4.16LineFrequency 50 or 60 Hz 60Acceleration Time 0-300 seconds[0=Disabled] 1 30Current Imbalance Trip % 0.1 - 30% of FLA[0=Disabled] 1(%) 15Current Imbalance Trip Delay 1 - 20seconds 1 (Second) 5Over Current Trip % 0,.50 300% ofFLA [0=Disabled] 1 (%) 200Over Current Trip Delay 1 - 20
seconds 1 8Under Current % 0, 10 90% of FLA [0=Disabled]1 (%) 35Under Current Trip Delay 1 - 60 seconds 1 15StallDetection Trip Level 0.100 600% of FLA [0=Disabled] 5 (%)600Stall Detection Trip Delay 1 - 10 seconds 1 4PeakCurrent Trip % 0.800 1400% [0=Disabled] 10(%) 1000(%)Peak Current Trip Delay 0..01 - .5 seconds .01 .0513/11/10 1
2. Ground Fault Current Trip Value 0.5 90% of CT Value[0=Disabled] 1 (%) 50Ground Fault Current Trip Delay 1 60seconds 1 5 Voltage Protection SettingsVoltage ImbalanceTrip % 0.1 30% [0=Disabled] 1 (%) 20Voltage ImbalanceTrip Delay 1 20 seconds 1 10Over Voltage Trip % 0.1 10%[0=Disabled] 1 (%) 5Over Voltage Trip Delay 1 20 seconds1 10Under Voltage Trip on Start % 0.1 20% [0=Disabled] 1(%) 20UV Trip on Start Delay 1 180 seconds 1 20UnderVoltage Trip on Run % 0.1 20% [0=Disabled] 1 (%) 20UVTrip Delay during Run 1 20 seconds 1 10 Phase andFrequency Protection SettingsPhase Rotation Trip 0.1 or 20=Disabled, 1=ABC, 2=ACB] 1 2Phase Rotation Trip Delay 1 20 seconds 1 2Phase Loss Trip and Delay 0.1-20 Seconds[0= Disabled] 1 5Over Frequency Trip Limit 0.1 10Hz[0=Disabled] 1 1Over Frequency Trip Delay 1 20 seconds 15Under Frequency Trip Limit 0.1 10Hz [0=Disabled] 11Under Frequency Trip Delay 1 20 seconds 1 5Motor KWTrip 0-2.0 = Disabled,1 = Over KW Trip,2 = Under KW Trip 10Motor KW Trip Point 20 100% of full load KW (disabled)1% 20(%)Motor KW Trip Delay Time 1 999 minutes
(disabled) 1 1Power Factor Trip Range 0.1 3 [0=Disabled,1=lag, 2=lead, 3= lead/lag] 1 2Power Factor Trip Point .01 1 .01 .10Power Factor Trip Delay Time 1 20 seconds 11013/11/10 2
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3. Power Factor Current Direction 0 - 1, [0=Normal, 1=Reversed] 1 1 Lockout / Inhibit SettingsCoast Down (BackSpin) Lockout Timer 0 = Disabled, or 1 - 60 minutes 1minute 1Maximum Starts per Hour 0 = Disabled, or 1 10starts 1 5Minimum Time Between Starts Inhibit 0 =Disabled, or 1 - 60 minutes 1 minute 12Note: NEMA Classtrip curves are based on a common tripping point of 600% ofmotor Full LoadAmps (FLA). Curves vary by the amount of
time before the unit trips. As an example, a Class 20curvewill trip in 20 seconds at 600% of FLA. Anotherexample, Class 10 will trip in 10 seconds at 600% ofFLA.PTValue:1-200 (: 1) 1:1 = direct voltage input, 2-200: 1 = kVVoltage Input 1 40 (4160/104) VCT Value: 5-2000 (:5) 560Number of Turns through CT: 1 5 (in 1 increments) 1Generator protection exampleRatings of generators:Ratedoutput (eg. 1,120 MVA), Maximum output (1230 MVA), Ratedrotation speed (300 rpm), Powerfactor (0.9), Number of
poles (2), Terminal voltage (27), Rated Armature current(23949), MaximumArmature current (26302), Short-circuitratio (> or = 0.5), Hydrogen gas pressure (0.52 MpaG),Insulation type (F), Temperature rise class (B), Coolingmethod (Stator: direct water), Efficiency(99 %), Hydrogenconsumption (12 m3/day).Functional Specifications ofgenerator protective relayNOMINAL SYSTEM FREQUENCYSETTING RANGE.............................................50 or 60HzRATED PRIMARY INPUT CURRENT OF PHASE AND NEUTRAL
CTS .........1 - 9999A in 1A stepsRATED PRIMARY SYSTEMPHASE-TO-PHASE VOLTAGE OF PTS......2 655 kV in .0.1 kVstepsRATED PT SECONDARY LINE-TO-LINEVOLTAGE ...............................50 125 V in 1 V stepsLOWSET OVERCURRENTELEMENTCharacteristic: .......................................................................................Definite time orinversePickup: ...........................................................................................1.0 2.5 pu of rated generator currentTime
delay: ...........................................................................................0.05 30.0 seconds (at 5pu Igen)HIGH SETOVERCURRENTELEMENTCharacteristic: .......................................................................................DefinitetimePickup: ............................................................................................1.0 9.9pu of rated generator currentTimedelay: ...........................................................................................0.05 3.0 seconds (at 5pu Igen)CURRENTUNBALANCE ELEMENTMaximum negative sequence currentrating;.................................0.05 0.5pu of rated generatorcurrent13/11/10 3
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4. Time multiplier of I2 tcurve ...................................................................5 80 secondsCooling time torated.............................................................................10 1800 secondsAlarm levelpickup.................................................................................0.03 0.5pu generatorcurrentAlarm level time
delay...........................................................................1 100 secondsREVERSE POWERELEMENT PICKUP ......................................0.02to 0.2pu rated generator currentTimedelay ...........................................................................................1 100 secondsLOSS OFFIELD ELEMENTMho circlesize ..........................................................................50 300% of rated generator
impedanceMhooffset .................................................................................5 50% of rated generatorimpedanceTimedelay ............................................................................................0.2 60 secondsIntegrationtime .....................................................................................0 10 secondsVOLTAGEELEMENTSCharacteristic ...................................
.....................................................Over, Underor Over+UnderPick-uplevel.......................................................................................1 50% change from ratedvoltageTimedelay ............................................................................................0.1 60.0secondsFREQUENCYELEMENTSCharacteristic ...................................
.....................................................Over, Underor Over+UnderPick-uplevel..........................................................................................0.05 9.99Hz fromnominalTimedelay ............................................................................................0.1 60.0 secondsTHERMALIMAGE ELEMENTTriplevel ...............................................................................................Fixed at 110%ratedThermal time constant ofalternator ......................................................1 400 minutesPre-alarmlevel.................................................................
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.....................50 110% of ratedUNDERPOWERELEMENT PICKUPLEVEL ..........................................................0.05 1.00 ofrated power outputTimedelay ............................................................................................0.1 60.0secondsUNDERIMPEDANCE ELEMENTSPickuplevel.................................................................
..........................01 1.0 pu ratedimpedanceTimedelay ............................................................................................0.02 9.99 secondsFIRSTLEVEL OVEREXCITATIONELEMENTCharacteristic ........................................................................................InversePickuplevel.................................................................
..........................1.0 - 2.0 puTimemultiplier.......................................................................................0.5 5.0SECOND LEVELOVEREXCITATIONELEMENTCharacteristic ........................................................................................DefinitetimePickuplevel...........................................................................................1.0 - 2.0 puTime
multiplier.......................................................................................0.1 60.0 seconds95%STATOR GROUND FAULT ELEMENTSPickuplevel.....................................................................................5 99% Rated zero sequencevoltageTimedelay ............................................................................................0.05 99.0 seconds100%STATOR GROUND FAULT ELEMENT3rd Harmonic
Pickuplevel ..............................................................1 30% Rated zero sequence voltageTimedelay ............................................................................................0.05 99.0 seconds TypicalSettings IEEE No. Function Typical Settings andRemarks24 Overexcitation PU: 1.1*VNOM/60;TD: 0.3; reset TD: 5 alarm P.U.:1.18*VNOM/6013/11/10 4
5. alarm delay: 2.5s25 Synchronism Check Max Slip: 6RPM;Max phase angle error: 10 Max VMAG error: 2.5% VNOM32Reverse Power (one stage) PU: turbine 1% of rated; 15 s .PU: Reciprocating engine: 10% of rated; 5 s32-1 Reverse
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Power (non-electrical, trip supervision) PU: same as 32; 3s40 Loss-of-field (VAR Flow approach) Level 1 PU: 60% VArating; Delay: 0.2s; Level 2 PU: 100% VA rating: 0.1s46Negative Sequence overcurrent I2 PU: 10% Irated; K=1049Stator Temperature (RTD) Lower: 95C; upper: 105C50/87Differential via flux summation Cts PU:10% INOM or less if1A relay may be used50/27 IE Inadvertent EnergizationOvercurrent with 27, 81 50: 0.5A (10% INOM) 27: 85%
Supervision VNOM (81: Similar)51N Stator Ground Over-current (Low, Med Z Gnd, PU: 10% INOM; curve: EI; TD: 4.Inst: Phase CT Residual) none. Higher PU required tocoordinate with load. No higher than 25% INOM.50/51NStator Ground Over- current (Low, Med Z Gnd, P.U.: 10%INOM; Curve EI**, TD4; Inst Neutral CT or Flux SummationCT) 100% INOM. Higher PU if required to coordinate withload. No higher than 25% INOM.51GN, 51N Stator GroundOver- current (High Z Gnd) PU: 10% IFAULT at HV Term.;
Curve: VI***; TD:4.51VC Voltage Controlled overcurrent PU:50% INOM; Curve: VI***; TD: 4. Control voltage:80%VNOM.51VR Voltage Restrained overcurrent PU: 175%INOM; Curve: VI***; TD: 4. Zero Restraint Voltage: 100%VNOM L-L59N, 27-3N, 59P Ground Overvoltage 59N: 5%VNEU during HV terminal fault; 27-3N: 25% V3rd duringnormal operation; TD: 10s 59P: 80% VNOM67IE DirectionalO/C for Inadvertent Energization PU: 75-100% INOM GEN;Definite Time (0.1- 0.25 sec.) ; Inst: 200% INOM
GEN13/11/10 5
6. 81 Over/under frequency For Generator protection: 57,62Hz, 0.5s; For Island detection condition: 59, 61Hz,0.1s87G Generator Phase Differential Fixed: 0.4A; orVariable: Min P.U.: 0.1 * Tap; Tap: INOM; Slope: 15%87NGenerator Ground Variable: Min P.U.: 0.1 times tap; Slope15%; Differential Time delay: 0.1s; choose low tap 67N:current polarization; time: 0.25A; Curve: VI***; TD: 2;Instantaneous: disconnect87UD 13 Unit Differential Min PU:0.35*Tap; Tap: INOM; Slope 30%**: EI: extremely inverse.***:VI: very inverse.13/11/10 1. Motor protection exampleAssume you have a motor rated 500HP, .95 power factor & 90 % efficiency energized from a 4.16 Kvsource, using a microprocessor motor protection
relay, provide the typical settings.The full load current can be taken from the motor nameplate as well as the
service factor. The motorfull load current can be calculated from the following: I = 500 (.746)/1.732 (4.16)(.85)(.9)
=60 amps Motor and Line Data FunctionsFunction Description Adjustment / Display Range Setting Increments
SettingMotor Nameplate FLA 1 2000A, adjustable between 50-100% of Max Amp Rating.FLA must be
programmed Upper limit of range automatically adjusts 1 60for relay to function downward as Service factor is
increased.Motor Nameplate 1.00 - 1.30 SF .05 1Overload Class During Start NEMA / UL Class 5 20 5 Class
20Overload Class During Run NEMA / UL Class 5 30 5 Class 10Overload Reset 0 = Manual,1 = Auto,2 = Disabled
Overload 1 1kV Voltage Input (nominal line, Medium Voltage) .60 15kV .01 4.16Line Frequency 50 or 60 Hz60Acceleration Time 0-300 seconds [0=Disabled] 1 30Current Imbalance Trip % 0.1 - 30% of FLA [0=Disabled]
1(%) 15Current Imbalance Trip Delay 1 - 20 seconds 1 (Second) 5Over Current Trip % 0,.50 300% of FLA
[0=Disabled] 1 (%) 200Over Current Trip Delay 1 - 20 seconds 1 8Under Current % 0, 10 90% of FLA
[0=Disabled] 1 (%) 35Under Current Trip Delay 1 - 60 seconds 1 15Stall Detection Trip Level 0.100 600% of FLA
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[0=Disabled] 5 (%) 600Stall Detection Trip Delay 1 - 10 seconds 1 4Peak Current Trip % 0.800 1400%
[0=Disabled] 10(%) 1000 (%)Peak Current Trip Delay 0..01 - .5 seconds .01 .0513/11/10 1
2. Ground Fault Current Trip Value 0.5 90% of CT Value [0=Disabled] 1 (%)
50Ground Fault Current Trip Delay 1 60 seconds 1 5 Voltage Protection SettingsVoltage
Imbalance Trip % 0.1 30% [0=Disabled] 1 (%) 20Voltage Imbalance Trip Delay 1 20
seconds 1 10Over Voltage Trip % 0.1 10% [0=Disabled] 1 (%) 5Over Voltage Trip Delay 1
20 seconds 1 10Under Voltage Trip on Start % 0.1 20% [0=Disabled] 1 (%) 20UV Trip on
Start Delay 1 180 seconds 1 20Under Voltage Trip on Run % 0.1 20% [0=Disabled] 1
(%) 20UV Trip Delay during Run 1 20 seconds 1 10 Phase and Frequency Protection
SettingsPhase Rotation Trip 0.1 or 2 0=Disabled, 1=ABC, 2=ACB] 1 2Phase Rotation Trip
Delay 1 20 seconds 1 2Phase Loss Trip and Delay 0.1-20 Seconds [0= Disabled] 1 5Over
Frequency Trip Limit 0.1 10Hz [0=Disabled] 1 1Over Frequency Trip Delay 1 20 seconds
1 5Under Frequency Trip Limit 0.1 10Hz [0=Disabled] 1 1Under Frequency Trip Delay 1
20 seconds 1 5Motor KW Trip 0-2.0 = Disabled,1 = Over KW Trip,2 = Under KW Trip 1
0Motor KW Trip Point 20 100% of full load KW (disabled) 1% 20(%)Motor KW Trip Delay
Time 1 999 minutes (disabled) 1 1Power Factor Trip Range 0.1 3 [0=Disabled, 1=lag,
2=lead, 3= lead/lag] 1 2Power Factor Trip Point .01 1 .01 .10Power Factor Trip Delay Time
1 20 seconds 1 1013/11/10 2
3. Power Factor Current Direction 0 - 1, [0=Normal, 1= Reversed] 1 1 Lockout /
Inhibit SettingsCoast Down (Back Spin) Lockout Timer 0 = Disabled, or 1 - 60 minutes 1minute 1Maximum Starts per Hour 0 = Disabled, or 1 10 starts 1 5Minimum Time
Between Starts Inhibit 0 = Disabled, or 1 - 60 minutes 1 minute 12Note: NEMA Class trip
curves are based on a common tripping point of 600% of motor Full LoadAmps (FLA).
Curves vary by the amount of time before the unit trips. As an example, a Class 20
curvewill trip in 20 seconds at 600% of FLA. Another example, Class 10 will trip in 10
seconds at 600% ofFLA.PT Value:1-200 (: 1) 1:1 = direct voltage input, 2-200: 1 = kV
Voltage Input 1 40 (4160/104) VCT Value: 5-2000 (:5) 5 60Number of Turns through CT: 1
5 (in 1 increments) 1 Generator protection exampleRatings of generators:Rated output (eg.
1,120 MVA), Maximum output (1230 MVA), Rated rotation speed (300 rpm), Powerfactor
(0.9), Number of poles (2), Terminal voltage (27), Rated Armature current (23949),
MaximumArmature current (26302), Short-circuit ratio (> or = 0.5), Hydrogen gas pressure(0.52 Mpa G),Insulation type (F), Temperature rise class (B), Cooling method (Stator: direct
water), Efficiency(99 %), Hydrogen consumption (12 m3/day).Functional Specifications of
generator protective relayNOMINAL SYSTEM FREQUENCY SETTING
RANGE.............................................50 or 60 HzRATED PRIMARY INPUT CURRENT OF PHASE
AND NEUTRAL CTS .........1 - 9999A in 1A stepsRATED PRIMARY SYSTEM PHASE-TO-PHASE
VOLTAGE OF PTS......2 655 kV in .0.1 kV stepsRATED PT SECONDARY LINE-TO-LINE
VOLTAGE ...............................50 125 V in 1 V stepsLOW SET OVERCURRENT
ELEMENTCharacteristic: .......................................................................................Definite
time or inversePickup: ...........................................................................................1.0 2.5
pu of rated generator currentTime
delay: ...........................................................................................0.05 30.0 seconds (at 5puIgen)HIGH SET OVERCURRENT
ELEMENTCharacteristic: .......................................................................................Definite
timePickup: ............................................................................................1.0 9.9pu of rated
generator currentTime delay: ...........................................................................................0.05
3.0 seconds (at 5pu Igen)CURRENT UNBALANCE ELEMENTMaximum negative sequence
current rating;.................................0.05 0.5pu of rated generator current13/11/10 3
4. Time multiplier of I2 t curve ...................................................................5 80
secondsCooling time to rated.............................................................................10 1800
secondsAlarm level pickup.................................................................................0.03 0.5pu
generator currentAlarm level time delay...........................................................................1 100 secondsREVERSE POWER ELEMENT PICKUP ......................................0.02 to 0.2pu rated
generator currentTime delay ...........................................................................................1
100 secondsLOSS OF FIELD ELEMENTMho circle
size ..........................................................................50 300% of rated generator
impedanceMho offset .................................................................................5 50% of rated
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generator impedanceTime
delay ............................................................................................0.2 60 secondsIntegration
time .....................................................................................0 10 secondsVOLTAGE
ELEMENTSCharacteristic ........................................................................................Over,
Under or Over+UnderPick-up level.......................................................................................1
50% change from rated voltageTime
delay ............................................................................................0.1 60.0
secondsFREQUENCY
ELEMENTSCharacteristic ........................................................................................Over,
Under or Over+UnderPick-up
level..........................................................................................0.05 9.99Hz from
nominalTime delay ............................................................................................0.1 60.0
secondsTHERMAL IMAGE ELEMENTTrip
level ...............................................................................................Fixed at 110%
ratedThermal time constant of alternator ......................................................1 400
minutesPre-alarm level......................................................................................50 110% of
ratedUNDERPOWER ELEMENT PICKUP LEVEL ..........................................................0.05
1.00 ofrated power outputTime
delay ............................................................................................0.1 60.0
secondsUNDERIMPEDANCE ELEMENTSPickup
level...........................................................................................01 1.0 pu rated
impedanceTime delay ............................................................................................0.02 9.99
secondsFIRST LEVEL OVEREXCITATION
ELEMENTCharacteristic
........................................................................................InversePickup
level...........................................................................................1.0 - 2.0 puTime
multiplier.......................................................................................0.5 5.0SECOND LEVEL
OVEREXCITATION
ELEMENTCharacteristic ........................................................................................Definite
timePickup level...........................................................................................1.0 - 2.0 puTime
multiplier.......................................................................................0.1 60.0 seconds95%
STATOR GROUND FAULT ELEMENTSPickup
level.....................................................................................5 99% Rated zero sequence
voltageTime delay ............................................................................................0.05 99.0
seconds100% STATOR GROUND FAULT ELEMENT3rd Harmonic Pickup
level ..............................................................1 30% Rated zero sequence voltageTime
delay ............................................................................................0.05 99.0 seconds Typical
Settings IEEE No. Function Typical Settings and Remarks24 Overexcitation PU:
1.1*VNOM/60; TD: 0.3; reset TD: 5 alarm P.U.: 1.18*VNOM/6013/11/10 4
5. alarm delay: 2.5s25 Synchronism Check Max Slip: 6RPM; Max phase angle error:
10 Max VMAG error: 2.5% VNOM32 Reverse Power (one stage) PU: turbine 1% of rated; 15
s . PU: Reciprocating engine: 10% of rated; 5 s32-1 Reverse Power (non-electrical, trip
supervision) PU: same as 32; 3 s40 Loss-of-field (VAR Flow approach) Level 1 PU: 60% VA
rating; Delay: 0.2s; Level 2 PU: 100% VA rating: 0.1s46 Negative Sequence overcurrent I2
PU: 10% Irated; K=1049 Stator Temperature (RTD) Lower: 95C; upper: 105C50/87
Differential via flux summation Cts PU:10% INOM or less if 1A relay may be used50/27 IE
Inadvertent Energization Overcurrent with 27, 81 50: 0.5A (10% INOM) 27: 85% Supervision
VNOM (81: Similar)51N Stator Ground Over- current (Low, Med Z Gnd, PU: 10% INOM;
curve: EI; TD: 4. Inst: Phase CT Residual) none. Higher PU required to coordinate with load.
No higher than 25% INOM.50/51N Stator Ground Over- current (Low, Med Z Gnd, P.U.: 10%
INOM; Curve EI**, TD4; Inst Neutral CT or Flux Summation CT) 100% INOM. Higher PU if
required to coordinate with load. No higher than 25% INOM.51GN, 51N Stator Ground Over-
current (High Z Gnd) PU: 10% IFAULT at HV Term.; Curve: VI***; TD:4.51VC Voltage
Controlled overcurrent PU: 50% INOM; Curve: VI***; TD: 4. Control voltage: 80%VNOM.51VR
Voltage Restrained overcurrent PU: 175% INOM; Curve: VI***; TD: 4. Zero Restraint
Voltage: 100% VNOM L-L59N, 27-3N, 59P Ground Overvoltage 59N: 5% VNEU during HV
terminal fault; 27-3N: 25% V3rd during normal operation; TD: 10s 59P: 80% VNOM67IE
Directional O/C for Inadvertent Energization PU: 75-100% INOM GEN; Definite Time (0.1-
0.25 sec.) ; Inst: 200% INOM GEN13/11/10 5
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6. 81 Over/under frequency For Generator protection: 57, 62Hz, 0.5s; For Island
detection condition: 59, 61Hz, 0.1s87G Generator Phase Differential Fixed: 0.4A; or
Variable: Min P.U.: 0.1 * Tap; Tap: INOM; Slope: 15%87N Generator Ground Variable: Min
P.U.: 0.1 times tap; Slope 15%; Differential Time delay: 0.1s; choose low tap 67N: current
polarization; time: 0.25A; Curve: VI***; TD: 2; Instantaneous: disconnect87UD 13 Unit
Differential Min PU: 0.35*Tap; Tap: INOM; Slope 30%**: EI: extremely inverse.***: VI: very
inverse.13/11/10
Basler Excitation Technical Papers
Excitation Replacement Solutions
Technical Paper Description Additional Resources
Alterrex Excitation Upgrades -
Convection-Cooled Bridges andReplacement of Automatic Voltage
Regulator
Bridge replacement, testing, installation and
commissioning plus expectations for voltageregulator replacement of Alterrex exciter
system.
EX-ALT_VR.pdf
EX-ALT_BRDG.pdf
EX-ALT2.pdf
EX-ALT3.pdf
Alternate Solutions to Replacing
Aged Static Exciter Systems
Replacing the analog portion of a static exciter
system with a digital front-end controller intothe existing power rectifier bridge/s.
EX-RETRO.pdf
EX-RETRO2.pdf
Application of Static Excitation
Systems for Rotating Exciter
Replacement
The static exciter system - including the
power control devices, power transformer and
automatic voltage regulator
Convection Cooled Bridges Offer
New Solution to Old Alterrex Water-
Cooled Bridges
Air-cooled retrofit solutions for water-cooled
Alterrex EX-ALT_VR.pdf
EX-ALT_BRDG.pdf
EX-ALT2.pdf
http://www.basler.com/php/download.php?file=alterrexexcupgrades.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=alterrexexcupgrades.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=alterrexexcupgrades.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=alterrexexcupgrades.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=EX-ALT_VR.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=EX-ALT_BRDG.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=EX-ALT2.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=Ex-alt3.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=AlternateSolutions.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=AlternateSolutions.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=EX-RETRO.pdf&id=364&type=reshttp://www.basler.com/php/download.php?file=EX-RETRO2.pdf&id=364&type=reshttp://www.basler.com/php/download.php?file=appofstat.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=appofstat.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=appofstat.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=AlterrexBridge.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=AlterrexBridge.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=AlterrexBridge.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=EX-ALT_VR.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=EX-ALT_BRDG.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=EX-ALT2.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=EX-ALT_VR.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=EX-ALT_BRDG.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=EX-ALT2.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=Ex-alt3.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=AlternateSolutions.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=AlternateSolutions.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=EX-RETRO.pdf&id=364&type=reshttp://www.basler.com/php/download.php?file=EX-RETRO2.pdf&id=364&type=reshttp://www.basler.com/php/download.php?file=appofstat.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=appofstat.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=appofstat.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=AlterrexBridge.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=AlterrexBridge.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=AlterrexBridge.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=EX-ALT_VR.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=EX-ALT_BRDG.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=EX-ALT2.pdf&id=365&type=reshttp://www.basler.com/php/download.php?file=alterrexexcupgrades.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=alterrexexcupgrades.pdf&id=403&type=reshttp://www.basler.com/php/download.php?file=alterrexexcupgrades.pdf&id=403&type=res -
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EX-ALT3.pdf
Front End Analog Conversion to
Digital Saves Cost for Upgrades ofExcitation System
Decision factors in purchasing a 6 SCR full-
wave bridge or reusing an existing 3 SCRhalf-wave
EX-SSE1.pdf
Retrofitting SCT/PPT Excitation
Systems with Digital Control
Retrofit of the SCT/PPT (Saturable CurrentTransformer/Power Potential Transformer)
Static Excitation System
EX-SCTPPT.pdf
EX-RETRO2.pdf
Selecting the Excitation System for
the Additional Turbine Generator atthe Port Wentworth Pulp Mill
Evaluating whether or not to reuse the
generator's original compound excitationsystem
Xcel Energy Northern RegionExperience with Excitation System
Upgrades and Retrofits
The scope of excitation replacement projects,
the reasoning behind them, and the resourcesrequired to plan, design, install and
commission the equipment
return to top
Excitation System Operation
Technical Paper Description Additional Resources
Avoiding Loss of Voltage SensingRunaway for Generator Excitation
Systems
Protective safeguards built into
excitation systems to manage loss of
voltage sensing events and prevent
damage
EX-DECSLOS.pdf
Coordination of Digital ExcitationSystem Settings for Reliable
Operation
Coordinating the various settings of theexcitation system as well as performance
checking the voltage regulator and limiter
Digital Excitation System ProvidesEnhanced Tuning Over Analog
Systems
Features in excitation control speedcommissioning and offer a greater variety of
tuning tools
The Effect of Reactive Compensatorsand Coordination with Volts/Hertz
Limiting
Three types of reactive compensators andVolts/Hertz Limiter effect the generator
excitation system.
Parallel Operation with a NetworkSystem
Techniques required to successfully parallelsynchronous generators
Tuning a PID Controller for a Digital
Excitation Control System
How the digital controller is tuned for
optimum generator performance
Self-tuning of the PID Controller for
a Digital Excitation Control System
An indirect method for self-tuning of the PID
controller gains
Voltage Regulator and ParallelOperation
Control of the voltage regulator whenoperating generators in parallel.
Voltage Versus Var/Power FactorRegulation on Synchronous
Generators
When paralleled to the utility bus,
synchronous generators can be controlledusing either terminal voltage or var/power
factor control
return to top
Excitation Basics and Design
Technical Paper Description Additional Resources
Designing a Voltage Regulation
System
Excitation system theory when generatorapplications are in the design and proposal
stages
Introduction to SynchronizingAutomatic synchronizing process and a guidefor selection of the proper synchronizer
Comparison BE1-25A and Eaton XMSynchronizer
Updated!Specifying Excitation
Systems for Procurement
Considerations involved in specifying the
excitation system for a project.
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8/2/2019 Model sht 1
26/30
New!Especificando Sistemas deExcitacin para Adquisicin
Spanish language translation of the
paper above
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Excitation System Testing
Technical Paper Description Additional Resources
Easing NERC Testing with NewDigital Excitation Systems North American Electric Reliability Councilpolicy involving generator testing.
Feature Enhancements in NewDigital Excitation System Speeds
Performance Testing
Technology built into the excitation systemcan ease the burden of performing required
tests
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Power System Stability
Technical Paper Description Additional Resources
Power System Stability Various types of power system instability
EX-PSS1.pdf
EX-PSS2.pdf
Power System Stabilizer Performancewith Summing Point type Var/PowerFactor Controllers
Power system stabilizer performance does nothave to deteriorate during a system transientwhen voltage support is needed
Understanding Power System
Stability
Types of power system instability and the
effects of system impedance and excitation.
Voltage Regulator with Dual PID
Controllers Enhances Power System
Stability
Voltage regulator with two PID controllers
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Synchronous Motor Controls
Technical Paper Description Additional Resources
Plant Efficiencies Benefit bySelection of Synchronous Motor
Application of a synchronous motor EX-SYNCH1.pdf
Synchronous Motor ControlBasic theory of operation and control of the
synchronous motor
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Basler Protective Relay Technical Papers
Protection Theory and Concepts
Technical Paper Description Additional Resources
Automatic Reclosing - Transmission
Line Applications and Considerations
Applying autoreclosing to transmission
circuits.
Commissioning Numerical RelaysExperienced-based changes to commissioningtests and revised documentation of settings as
a method for commissioning
A Derivation of SymmetricalComponent Theory and Symmetrical
Component Networks
Review symmetrical component analysis ElectricCalcs_R30.xls
Generator Protection Application
Guide
Assists in the selection of numerical
multifunction and single function relays toprotect a generator
Load Shedding for Utility and
Industrial Power System Reliability
Analysis of the conditions in which load
shedding may be needed and review of loadshed implementation strategies
PC-IPS01.pdf
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er.com/php/download.php?file=ElectricCalcs_R30.xls&id=262&type=reshttp://www.basler.com/php/download.php?file=genprotguidC.pdf&id=198&type=reshttp://www.basler.com/php/download.php?file=genprotguidC.pdf&id=198&type=reshttp://www.basler.com/php/download.php?file=loadshed.pdf&id=216&type=reshttp://www.basler.com/php/download.php?file=loadshed.pdf&id=216&type=reshttp://www.basler.com/php/download.php?file=PC-IPS01.pdf&id=334&type=res 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8/2/2019 Model sht 1
27/30
Negative Sequence Relaying
Applications in Ungrounded andHigh Impedance Grounded Industrial
Systems
Negative sequence overcurrent sensing todetect faults
Power Quality: Measurement of Sags
and Interruptions
Measurement of power quality use of meters
for residential and small commercialcustomers as a tool to track outage issues.
A Practical Guide for Detecting
Single-Phasing on a Three-PhasePower System
Reference for the loss of one or two phases of
a radial three phase system and includessuggested detection and protection methods
ElectricCalcs_R30.xls
Single_Phase_mc7_rev_1_1.mc
d
PC-95102
Protective Relaying Issues in Low
Voltage Systems as Addressed in the
National Electric Code
Low voltage (LV) design practices, as
mentioned in the National Electric Code
(NEC)
Reliability Considerations ofMultifunction Protection
Reliability of numerical multifunction
protection systems against predecessor
technologies.
A Review of Ferroresonance The basics of a ferroresonant condition Ferro_R1.xls
A Review of Negative SequenceCurrent
Negative sequence current and the damage itcan cause to rotating machines
PC-IPS01.pdf
A Review of System GroundingMethods and Zero Sequence Current
Sources
Grounding methods, sources of ground
current, ground current flow in the power
system, and how each impacts protectiverelaying
A Survey of Cold Load Pickup
PracticesCold load pickup, the causes and effects
Voltage Restrained Time Overcurrent
Relay Principles, Coordination, andDynamic Testing Considerations
Dynamic response of 51/27R relays in which
the overcurrent element conforms to the
dynamic characteristic defined in ANSI
C37.112
51V Timing Model Mathcad simulation
file(Must have Mathcad Professional Edition installed)
Zero Sequence Impedance of
Overhead Transmission Lines
Transmission line impedances, centering on
zero sequence impedances
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Motor Protection
Technical Paper Description Additional Resources
Extending Motor Life with UpdatedThermal Model Overload Protection
Updates in the 49 thermal model
provide better motor protection
Motor Protection Application GuideOverview of motor hazards anddetection and protection options, plus
typical setting value range for BE1-11m.
PC-11m.pdf
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Breaker Failure and Monitoring Protection
Technical Paper Description Additional Resources
Breaker Monitoring with NumericalRelays
Breaker Duty/Contact Wear monitoring
Fundamentals and Advances inBreaker Failure Protection
Deciding whether to apply dedicated breakerfailure protection.
PC-50BF1.pdfHow Can Current Dropout Affect
Breaker Failure Timing Margins
Current decay time in applying breaker failure
protection.
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