saturation ct because of dc offset
TRANSCRIPT
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Example 1:
A 1200/5, C400 CT with excitation curves shown on above figure, is connected to a
2.0 burden. Based on the accuracy classification, what is the maximum symmetrical
fault current that may be applied to this CT without exceeding a 10% ratio error?
Answer:
Based on the criteria that the CT can deliver 20 times rated secondary current
without exceeding a 10% ratio error, the maximum fault current will be 24000A.
However, with a 2.0 burden, this will result in a voltage below the knee point of the CT
and, as a practical matter, it will be within 10% accuracy at higher currents. This can onlybe accurately determined from excitation or ratio correction curves and not from the
accuracy classification. For example, a CT with characteristics shown in above figure
will produce between 180-240A without exceeding the 10% ratio error, depending on the
power factor of the 2.0 burden.
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Example 2:
A 1200/5, C400 CT is connected on the 1000/5 tap. What is the maximum secondary
burden that can be used and still maintain rated accuracy at 20 times rated symmetrical
secondary current?
Answer:
Since the secondary voltage capability is directly proportional to the connected tap, the
CT will support a voltage of 1000/1200400V or 333V. Twenty times the rated
secondary current is 100A. Therefore, the maximum burden is 333V/100A or 3.33
Example 3:
Assume that secondary burden in a relay circuit is 5. The relay setting is 2A and the CT
ratio is 300/5. Using above figure, calculate the primary current required to operate the
relay?
Answer:
VB=5 times 2A=10V
The secondary exciting current from above figure is approximately 0.04A.
)( STP INI
)( SE IIN
= 300/5(0.04+2) A=122A
Example 4:
A relay is expected to operate for a 7000A primary current. The CT ratio is 600/5.
Secondary burden is 3.5 . What is the error for the CT shown in above figure?
Answer:
The total secondary fault current is (7000/600)5=58A. Assume that exciting current is
negligible.
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)( SBSS RRIV
= 58(3.5+0.31)
= 221V
The exciting current will not be negligible, however, and the calculation will not be
iterated.
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CT saturation and DC- offset current
Role of DC off-set current
Typically fault current consists a symmetrical ac component and a dc offset current. To
understand this concept, consider a transmission line unloaded exited by an equivalent voltage
source. The fault strikes at time0tt . This can be simulated by closing the switch at 1tt
LjR
or
Z models the line impedance. The fault current in the line is given by
0)( ti 00 tt
0
0||
)sin()(
tt
m eIZ
tVti 0tt
Where is the time constant of the line =L/R. The faul t current has two components in
it. The first component models the steady state sinusoidal ac response while the second current
is the dc offset current due to the presence of inductive component in the circuit. Recall that
current in an inductance can not change instantaneously. As t, the instantaneous dc
current, a consequence of maintaining initial condition )()( 00 titi , decays exponentially to
zero and the current reaches the ac steady state values. While the dc offset current, would in
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theory persist till infinity, its trace in the actual wave form would not be seen beyond a certain
time constants. Table-I illustrates the values of t
e
up to 10 time constants.
Time t = 0 t = t = 2 t = 4 t = 6 t = 8 t = 10
t
e
1 0.3678 0.1353 0.0183 0.0024 0.0003 0.00004
It is more or less obvious that, dc offset is not seen in the waveform after 5 time constants.
The value ofI0 can be worked out by setting the current at ott to zero.
This implies that
)sin( 00 tZ
VI m
Thus
)( 0
)sin()sin()(
tt
mm etZ
Vt
Z
Vti
fig.2
Clearly, the peak value of dc offset current depends upon
Time at which fault strikesPhase angle of ac voltage
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Z & of transmission line
Figure 2 shows the waveforms of
a)
symmetrical ac componentb) dc offset currentc) total current for various values of , &
0t
It can be seen that severity of dc offset component in fault current is maximum when
a)
b)2
0
t
For example, if angle of transmission line is 800, then with = 80
0&
50220
t
=200
1sec = 5msec, the severity of dc offset current would equal
Z
VI m0 , which is also the
peak value of symmetrical ac component of the current. This leads us to an important
conclusion. Viz. peak value
1) dc offset current can be as high as the symmetrical ac peak2) The dc offset current can be positive or negative (see fig2)3) Dc offset current may be totally absent
eg. If , 00 t
4) While, in above analysis, we have considered a single phase current, a 3 fault on a 3transmission line would always induce dc offset current in atleast 2 phases.
In the remaining lecture, we analyze the effect of dc offset current on CT performance.
DC- offset current and CT saturation
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We now plan to show that CT can saturate on dc offset current. Also, we plan to show
that the resulting distortions in the CT secondary current can be un-acceptably high. While
doing this analysis, we will neglect ac symmetrical component. In other words, we rest our
belief in superposition theorem atleast qualitatively and will finally evaluate effect using it
Notice that the current that we are dealing with is non-linear, a rigorous application of super
position theorem is simply out of question.
First consider an ideal CT excited by the dc offset current source. An ideal CT will faithfully
replicates primary current waveform on the secondary side. Hence, the secondary current
would be given by
t
eN
Iti
02 )(
and the voltage developed across CT secondary would be given by
t
eN
RItv
02 )( where1
2
N
NN
Typical voltage waveform is shown in fig. (5)
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For simplicity, let us assume that the initial flux in the transformer core at t=0 is
zero 0)0( ; Then we can compute the flux in the transformer core by using faradays law
dtdNV 22 ---------(2)
t
dtvt0
2)0()(
t
eN
RI1
2
0
)1(2
0
t
eN
LI
)1()0()(2
0
t
eN
LIt
)1(2
0
t
eN
LI
------- (3)
as a consequence of dc offset current,
Thus, flux in the core increases exponentially to a peak value of
2
0max
N
LIcd as t
Z
V
N
L m Z
Vm maxcd
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Note that unlike ac voltage induced flux, which is sinusoidal, this flux is unidirectional. The ac
voltage induced flux has zero average value. However, dc offset induced does not have this
nice feature. The total flux in ideal CT core is a summation of ac flux and dc flux.
The ac flux in the CT core can be obtained by substituting operatordt
dby j . Hence
phasor relationship between phase2
V & ac is given by
2
2
Nj
V
If 2( ) sin( )mv t V t , then
)2
sin(2
t
N
Vmac
The peak value of ac flux is given by
2
max
N
Vmac
However max02IRVm
Hence2
max
02max
N
IRac
and peak value of the total flux is given by
2
max
0
2
maxmax
N
LI
N
Vmdcac
In practice, if this flux exceeds the knee-point flux in the core (see fig.), then the CT core
will saturate.
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As a consequence of CT core saturation, the secondary current would not faithfully replicate
the primary current. Infact, in practice it is observed that CT secondary current is clipped. The
clipping ofCT current leads to blinding of the relay which cannot function further. Hence,
CT saturation in presence of dc offset current is a serious problem which relay designers have
to face. Note that dc flux accumulates gradually. (Depends upon the transmission line time
constant ( ). It is apparent that saturation should not occur immediately after the inception of
the fault. Thus, if the relay is fast enough in decision making, it is likely that a relaying
decision would be undertaken before the CT fully saturates. This is another important reason
for increasing the speed of relaying system. For bus-fault protection, where the dc saturation
due to dc offset current can be a significant contributing factor, quarter cycle operations *****
specifically are imposed. Similarly, a distance relay is expected to operate within -1 cycle
time.
CT oversizing factors
Typically, an efficient design of transformer would correspond to choosing the core
cross section such that acm should be near the knee point of B-H curve. One obvious way of
avoiding the CT saturation on dc flux is to oversize the core so that for flux )(maxmax
dcac , the
corresponding B is below the knee-point. Hence, the factor max
maxmax)(
ac
dcac
is called core-
oversizing factor.
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Core-oversizing factor =max
max
1
ac
dc
20
21
NRI
NLIo
R
L1
R
X 1
Note that X/R in above equation is the transmission line X/R ratio. For a 220KV line X/R 10.
This would imply that transformer core should be oversized by a factor of 11. For a 400KV
line, typical value of X/R 20. This would imply an oversizing required of about 21 times the
usual design. Clearly this high amount of oversizing is not practical. Thus, the important
conclusion is that, protection engineers have to live with the saturation problem.
Cautions in CT selection:
While choosing a CT for a particular application, it is necessary to observe following
precautions.
1. The CT rating and continuous load current should match. For example, if max loadcurrent is 90A, a 100:5 Ct may be acceptable but 50:5 is not acceptable.
2. The maximum fault current should be less than 20 times the CT rated current. forexample 100:5 CT can be used, so long as burden on the CT & maximum primary fault
current is below 2000A.
3. The voltage rating of CT should be compatible. For example, 100:5 C100 would givelinear response, upto 20 times rated current provided CT burden is kept
below(100/20*5=1 ). With 2 burden, this CT can be used only if maximum current
is limited to 1000A.
4. Parallel of CTs e.g. in differential protection, or with SLG fault can create significanterrors in CT performance. One should in general ascertain that magnetizing current is
kept much below the pick up value.
Following example, illustrates this point
Exercise problems:
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If the current ratio is adequate for a protection, but CT burden is high; then the
performance of CT may deteriorate due to large magnetizing current and/or saturation problem.
The CT performance can be improved by connecting the CTs in series.
1) Show the dotted terminals for correct secondary series connection2) What is the VA of CT in fig (a) & (b) respectively?1) Electromechanical relays tend to saturate at high currents. This reduces the relay burden
on CT, and so that the CT performance at moderately high currents may be considered
better than at relays rated burden at 5A.
2) Use of instantaneous over current relays has the potential to overcome this problem ofsaturation of CTs
3) Differential protection can operate on external faults due to the un equal saturation ofCTs
Lecture-6
Examples
6. If a 300:5 class C CT is connected to a meter with resistance 1IR andsecondary current in the CT is 4.5A find out the primary current voltage
developed across the meter and % rate error. Lead wire resistance
02.0LR secondary resistance SR of a 300/5 CT 15.0
Diagram
1IR , 02.0LR 15.0SR AIS 5.4
Total secondary resistance SLIT RRRR
17.1 Secondary voltage TS RI
17.15.4
V265.5
From Fig 5.7,
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Exciting current IE for 5.265V
= 0.03A
Turns ratio N = 300/5 = 60
)( ESp IINI
= 60(4.5 + 0.03)
= 271.8AVoltage across meter IS RI
14.5 = 4.5V
Ratio error 1005.4
03.0100
S
E
I
I
= 0.67%
bR
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Lecture8
Examples
1. Design a CCVT for a 132kV transmission line using the following data.Resistive Burden VA150)3(
Hzf 3 , phase angle error= 40 minConsider 4 choices of V2 as 33kV, 11kV, 6.6kV and 3.3kV
Diagram 1 Diagram 2Transmission line voltage V = 132kV. Suppose V2 (P_N) be the voltage to be
produced by the capacitive potential divider with capacitance values C1 and C2and L the value of tuning inductor. The standardized VT secondary voltage is 110
volts (L-L).
Here specification for phase angle error is 40 minutes variation in frequency
can be upto Hzf 3 . Phase angle error for change in by in the above
equation circuit, is given by
)1
( 2CL
At tuning frequencyLC
12
Substituting 2 1LC
Phase angle error )( LL
L2
% phase angle error bRa
L2
2 --- (1)
Using this equation the value L for different values of V2 is found out.
1) Let V2 be 33kV (L - N)2
2
'
3150
b
V
R
52' 108.217bb RaR
322 f
rad01164.060180
40min40
From eqn (1)
322
108.21701164.0
2
5'
bRL
H2.6722
FFL
CC
39
2211051.11051.1
1
2) )(112 NLkVV 3 2
' 43 (11 10 ) 242 10150
bR
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' 40.01164 242 10
2 2 2 3
bRL
H2.747
2.747)4/3(
112221
L
CC
F21036.1
3) kVV 6.62 3 2
' 3 (6.6 10 )
150bR
41012.87 ' 4
0.01164 87.12 10
2 2 2 3
bRL
,269H
FCC 221 1077.3
4) kVV 3.32 3 2
' 3 (3.3 10 )
150b
R
41078.21 ' 40.01164 21.78 10
2 2 2 3
bR
L
H25.67
1 2 0.151C C F
The values of L, 21 CC for different values of V2 are tabulated below.
V2 L 21 CC 33kV 6722.2H 0.00151 F
11kV 747.2H 0.0136 F
6.6kV 269H 0.0377 F
3.3kV 67.25H 0.151 F
From the above table it is clear that smaller the value of V2, the smaller is thevalue of L and higher the value of C1 and C2 for tuning condition. If we select too low
value of V2 and L then capacitance values will be beyond available limits, and if we
select higher value of V2 and L, then CCVT and inductor will become bulky. So a
compromise is necessary and let us select V2 = 6.6kVFor V2 = 6.6kV
L = 269H
FCC 0377.021
Now,1
21
2 C
CC
V
V
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1
6
3
3 100377.0
106.63
10132
C
3
63
110132
10106.630377.0
F
C
F0033.0 FC 0344.02
In this design, we explained the basic concept for CCVT design and we assumed
the transformer to be ideal. But in actual design practice the value of magnetizing
impedance of transformer, resistance of reactor etc have to be taken into account,
as ratio error and phase angle error will also get affected by these values.
2. DiagramThe equivalent circuit of a CCVT is shown in fig 8.3. The values of C1 and C2 are
0.0018 F and 0.018 F respectively. Tuning inductor has an inductance of
497H and resistance of 4620 .Xm of the 6.6kV VT is 1M , core loss = 20 watts per phase, VA burden =150VA per phase. Value of Cm for compensating the current drawn by m is equal
to F910183.3 .(a)Verify the appropriateness of choice of L and Cm.Ans: If FC 0018.01 and FC 0186.02 then the value L of tuning inductor is
given by
)(
1
21
2 CCL
where 2 f and f = tuning frequency
2 6
1
(2 50) (0.0018 0.0186) 10L
= 496.7H which is equal to the given value of L. Now 61 10mX
1m
m
XC
6
1 1
(2 50) 1 10m
m
CX
93.183 10 F The value is also same as the selected value of Cm Hence the selection of both L
and Cm is appropriate.
(b)Find out the nominal value of V/V2Ans: 1 2
2 1
0.0018 0.0186
0.0018
C CV
V C
= 11.33
V = 11.33 x 6.6132
3kV
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(c) If the frequency drops from 50Hz to 47Hz, what would be the values of ratioerror and phase angle error?
Ans: Core loss = 20w
2
2 20m
VW
R
2 2
2 (6600)
20 20m
VR
62.18 10 VA burden = 150VA (resistive)
2
2 150b
V
R
2 2
2 (6600)150 150
bVR
52.904 10 The equivalent circuit can be represented as shown below.
Diagram 8.12610mX at f = 50Hz
62 10mf L 610
3183.12 50
mL H
The frequency of interest is 47Hz. Hence values of Xm and other impedance can
be calculated at 47Hz. The above circuit can be simplified asDiagram 8.13
Where1 1 1
m
m m b
jj C
Z R X R
9
6 5
1 12 47 3.183 10
2.18 10 2 47 3183.1 2.904 10
jj
6 6 6 60.459 10 1.064 10 0.94 10 3.44 10j j 6 6(3.902 0.124) 10 3.904 10 1.82j
6
1
3.904 10 1.82
Z
256147.5 1.82
256018.32 8135.15j
thth
VI
jR j L Z
C
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6
6600 0
4620 2 47 497 256018.32 8135.152 47 0.0204 10
jj j
6600 0
4620 146768.9 165994 256018.32 8135.15j j j
6600 0
260638.32 11089.84j
6600 0
260874.14 2.44A
T thV I Z
6600 0256147.5 1.82
260874.14 2.44
6480.42 4.26
Hence % ratio error(6600 6480.42)
1006600
=1.81%
Phase angle error = 4.26
2 6.6V kV
rM
2 2
0t t2
2
10
eq
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dt L dt LC
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d ii
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t
e
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| |
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VI t
Z
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