ct 2ndry resistance

8/12/2019 Ct 2ndry Resistance

1/19

Example 1:

A 1200/5, C400 CT with excitation curves shown on above figure, is connected to a

2.0 burden. ased on the accurac! c"assification, what is the #axi#u# s!##etrica"

fau"t current that #a! be a$$"ied to this CT without exceeding a 10% ratio error&

Answer:

ased on the criteria that the CT can de"iver 20 ti#es rated secondar! current

without exceeding a 10% ratio error, the #axi#u# fau"t current wi"" be 24000A.

'owever, with a 2.0 burden, this wi"" resu"t in a vo"tage be"ow the (nee $oint of the CT

and, as a $ractica" #atter, it wi"" be within 10% accurac! at higher currents. This can on"!be accurate"! deter#ined fro# excitation or ratio correction curves and not fro# the

accurac! c"assification. )or exa#$"e, a CT with characteristics shown in above figure

wi"" $roduce between 1*0+240A without exceeding the 10% ratio error, de$ending on the

$ower factor of the 2.0 burden.

1


2/19

Example 2:

A 1200/5, C400 CT is connected on the 1000/5 ta$. hat is the #axi#u# secondar!

burden that can be used and sti"" #aintain rated accurac! at 20 ti#es rated s!##etrica"

secondar! current&

Answer:

-ince the secondar! vo"tage ca$abi"it! is direct"! $ro$ortiona" to the connected ta$, the

CT wi"" su$$ort a vo"tage of 1000/1200400 or . Twent! ti#es the ratedsecondar! current is 100A. Therefore, the #axi#u# burden is /100A or .

Example 3:

Assu#e that secondar! burden in a re"a! circuit is 5. The re"a! setting is 2A and the CT

ratio is 00/5. sing above figure, ca"cu"ate the $ri#ar! current reuired to o$erate the

re"a!&

Answer:

5 ti#es 2A10

The secondar! exciting current fro# above figure is a$$roxi#ate"! 0.04A.

3 STP INI =

3 SE IIN +=

00/50.0423 A122A

Example 4:

A re"a! is ex$ected to o$erate for a 6000A $ri#ar! current. The CT ratio is 700/5.

-econdar! burden is .5 . hat is the error for the CT shown in above figure&

Answer:

The tota" secondar! fau"t current is 6000/700355*A. Assu#e that exciting current isneg"igib"e.

2


3/19

3SBSS

RRIV +=

5*.50.13

221

The exciting current wi"" not be neg"igib"e, however, and the ca"cu"ation wi"" not be

iterated.

8ec 9 6

#ax 1 0

2

2

dc

N RI

N =

: :

m

line

V

Z

1

2

2 : :

m

line

VN R

N Z 11 2

2: :

m

line

V NN R

Z N= 1 0

2

2

3dct

N RI t e

N

= 1 022

tN I

i eN

=

1 02 2

2 20

1 3 03 1 3

t tN RI

t dt eN N

= = 1 022

3 03 1 3t

N RIt e

N

= +

8ec+ 61

2

2

3 sin 3: :

ac m

line

V N t R t

Z N = + d

dt! 2

2

ac

V

! N

=

1

2

2

3 sin 3: : 2

mac

line

RV Nt t

Z N

= + #ax 1

2

2: :

mac

line

RV N

Z N

=

#ax #ax 1 1

2 2

2 2: : : :

m mac dc

line line

RV RVN N

Z N N Z

+ = +

1

2

2

1

2

2

: :1 1

: :

m

line

m

line

V NR

Z N

V N R

Z N

= + = + 1 line

line

"

R

= + 1 line

line

#

R= +

1 #R= +

8ec+7

20 5 100SI A= = = 71.0SR 1000

4001200

SV = #ax

400; 15

2.71A= =

120015

5=

VVS .10315.052 =+= 6000

5 5*.700

SI A= =7*00

5 7*500

SI A= =

*1.5*31.05.5* =+=SV 7

100 1007*

E

S

I

I = = 5.2BR

002.04

5

7*2.5 0.253SV = + 1IR = 0.51SR = < =sin 3 =sin > 3tpi I t I e $ t

= + < =sin cos 3 =cos sin > 3tpi I t e I t $ t = + +

= cos sinI t =cos < sin cos cos sin 3>tI e t + + =sin cosI t

2=sin < sin cos cos 3>tI e t + + =sin tI e =sin

3t tI

e e

8ec 9 5


4/19

100

p

S

S

II

N

I

p

S E

II I

N = 100

E

S

I

I

100

p

S

S

II

N

I

?*45

8ec + ?

100t% T

t%

V V

V

3 2

0

1 3"

= 2

0

1

"& = 20 1"&

= < ' () =

1 3

t

e*

di t Ri idt "

& dt= + +

2

2

1

e*

d di d iR i "

dt dt & dt = + +

0

1 3" &

e*

dV V !I "

d &

= 2

0

12 3

e*

I&

= + 201

e*"& = 3I " " = +

2! "I 3" &V V 2 2: : 2 2

tan: :

" &

R + +

V V "I "I

V a R I a R

= = =

2

2

+

"

a R

72 10m' " =

710 1*.12

m" (

= =

4


5/19

CT saturation and @C+ offset current

o"e of @C off+set current

T!$ica""! fau"t current consists a s!##etrica" ac co#$onent and a dc offset current. To

understand this conce$t, consider a trans#ission "ine un"oaded exited b! an euiva"ent vo"tage

source. The fau"t stri(es at ti#e 0tt= . This can be si#u"ated b! c"osing the switch at 1tt=

"!R + or Z #ode"s the "ine i#$edance. The fau"t current in the "ine is given b!

03 =ti 00 tt

++

=

0

0::

3sin3

tt

meI

Z

tVti 0tt

here B is the ti#e constant of the "ine B 8/. The fau"t current has two co#$onents in

it. The first co#$onent #ode"s the stead! state sinusoida" ac res$onse whi"e the second current

is the dc offset current due to the $resence of inductive co#$onent in the circuit. eca"" that

current in an inductance can not change instantaneous"!. As t , the instantaneous dc

current, a conseuence of #aintaining initia" condition 33 00+

= titi , deca!s ex$onentia""! to

Dero and the current reaches the ac stead! state va"ues. hi"e the dc offset current, wou"d in

theor! $ersist ti"" infinit!, itEs trace in the actua" wave for# wou"d not be seen be!ond a certain

ti#e constants. Tab"e+; i""ustrates the va"ues of t

e

u$ to 10 ti#e constants.

5


6/19

Ti#e t 0 t B t 2B t 4B t 7B t *B t 10B

t

e

1 0.76* 0.15 0.01* 0.0024 0.000 0.00004

;t is #ore or "ess obvious that, dc offset is not seen in the wavefor# after 5 ti#e constants.

The va"ue ofI,can be wor(ed out b! setting the current at -tt= to Dero.

This i#$"ies that

3sin 00 += tZ

VI m

Thus

3 0

3sin3sin3

tt

mmet

Z

Vt

Z

Vti

++=

fig.2

C"ear"!, the $ea( va"ue of dc offset current de$ends u$on

Ti#e at which fau"t stri(es

Fhase ang"e of ac vo"tage

Z G of trans#ission "ine

)igure 2 shows the wavefor#s of

a3 s!##etrica" ac co#$onent

7


7/19

b3 dc offset current

c3 tota" current for various va"ues of , G 0t

;t can be seen that severit! of dc offset co#$onent in fau"t current is #axi#u# when

a3 =

b32

0

=t

)or exa#$"e, if ang"e of trans#ission "ine is *00, then with *00G5022

0

=

t

200

1

sec 5#sec, the severit! of dc offset current wou"d eua"Z

VI m=0 , which is a"so the $ea(

va"ue of s!##etrica" ac co#$onent of the current. This "eads us to an i#$ortant

conc"usion. iD. $ea( va"ue

13 dc offset current can be as high as the s!##etrica" ac $ea(

23 The dc offset current can be $ositive or negative see fig23

3 @c offset current #a! be tota""! absent

eg. ;f = , 00 =t

43 hi"e, in above ana"!sis, we have considered a sing"e $hase current, a fau"t on a

trans#ission "ine wou"d a"wa!s induce dc offset current in at"east 2 $hases.

;n the re#aining "ecture, we ana"!De the effect of dc offset current on CT $erfor#ance.

@C+ offset current and CT saturation e now $"an to show that CT can saturate on dc offset current. A"so, we $"an to show

that the resu"ting distortions in the CT secondar! current can be un+acce$tab"! high. hi"e

doing this ana"!sis, we wi"" neg"ect ac s!##etrica" co#$onent. ;n other words, we rest our

be"ief in su$er$osition theore# at"east ua"itative"! and wi"" fina""! eva"uate effect using it

6


8/19

Hotice that the current that we are dea"ing with is non+"inear, a rigorous a$$"ication of su$er

$osition theore# is si#$"! out of uestion.

)irst consider an idea" CT excited b! the dc offset current source. An idea" CT wi"" faithfu""!

re$"icates $ri#ar! current wavefor# on the secondar! side. 'ence, the secondar! current

wou"d be given b!

t

eN

Iti

= 02

3

and the vo"tage deve"o$ed across CT secondar! wou"d be given b!

t

eN

RI

t

= 0

2 3 where 1

2

N

NN

=

T!$ica" vo"tage wavefor# is shown in fig. 53

*


9/19

)or si#$"icit!, "et us assu#e that the initia" f"ux in the transfor#er core at t0 is Dero

030 = I Then we can co#$ute the f"ux in the transfor#er core b! using farada!Es "aw

dt

dNV

22 = +++++++++23

= t

dtt0 2303

=

t

eN

RI1

2

0

312

0

t

eN

"I

=

313032

0

t

eN

"It

+=

312

0

t

eN

"I

= +++++++ 3

as a conseuence of dc offset current,

Thus, f"ux in the core increases ex$onentia""! to a $ea( va"ue of

2

0#ax

N

"Icd = as t

Z

V

N

" m

Z

Vm

#ax

cd

Hote that un"i(e ac vo"tage induced f"ux, which is sinusoida", this f"ux is unidirectiona". The ac

vo"tage induced f"ux has Dero average va"ue. 'owever, dc offset induced does not have this

nice feature. The tota" f"ux in idea" CT core is a su##ation of ac f"ux and dc f"ux.

The ac f"ux in the CT core can be obtained b! substituting o$eratordt

db! ! . 'ence

$hasor re"ationshi$ between $hase 2V G ac is given b!

?


10/19

2

2

N!

V

=

;f 2 3 sin 3m t V t = + , then

32

sin2

+= tNV

mac

The $ea( va"ue of ac f"ux is given b!

2

#ax

N

Vm

ac

=

'owever#ax

02IRVm=

'ence2

#ax

02#ax

N

IRac

=

and $ea( va"ue of the tota" f"ux is given b!

2

#ax

0

2

#ax#ax

N

"I

N

Vm

dcac +=+

;n $ractice, if this f"ux exceeds the (nee+$oint f"ux in the core see fig.3, then the CT core

wi"" saturate.

As a conseuence of CT core saturation, the secondar! current wou"d not faithfu""! re$"icate

the $ri#ar! current. ;nfact, in $ractice it is observed that CT secondar! current is c"i$$ed. The

c"i$$ing of CT current "eads to Jb"indingK of the re"a! which cannot function further. 'ence,

CT saturation in $resence of dc offset current is a serious $rob"e# which re"a! designers have

to face. Hote that dc f"ux accu#u"ates gradua""!. @e$ends u$on the trans#ission "ine ti#e

10


11/19

constant 3. ;t is a$$arent that saturation shou"d not occur i##ediate"! after the ince$tion of

the fau"t. Thus, if the re"a! is fast enough in decision #a(ing, it is "i(e"! that a re"a!ing

decision wou"d be underta(en before the CT fu""! saturates. This is another i#$ortant reason

for increasing the s$eed of re"a!ing s!ste#. )or bus+fau"t $rotection, where the dc saturation

due to dc offset current can be a significant contributing factor, uarter c!c"e o$erations LLLLL

s$ecifica""! are i#$osed. -i#i"ar"!, a distance re"a! is ex$ected to o$erate within M+1 c!c"e

ti#e.

CT oversiDing factors

T!$ica""!, an efficient design of transfor#er wou"d corres$ond to choosing the core

cross section such thatac

m shou"d be near the (nee $oint of +' curve. Nne obvious wa! of

avoiding the CT saturation on dc f"ux is to oversiDe the core so that for f"ux 3 #ax#ax

dcac + , the

corres$onding is be"ow the (nee+$oint. 'ence, the factor #ax

#ax#ax 3

ac

dcac

+is ca""ed core+

oversiDing factor.

Core+oversiDing factor #ax

#ax

1ac

dc

+

20

21NRI

N"I-

+=

R

"+=1

R

#+=1

Hote that O/ in above euation is the trans#ission "ine O/ ratio. )or a 220P "ine O/ 10.

This wou"d i#$"! that transfor#er core shou"d be oversiDed b! a factor of 11. )or a 400P

"ine, t!$ica" va"ue of O/ 20. This wou"d i#$"! an oversiDing reuired of about 21 ti#es the

usua" design. C"ear"! this high a#ount of oversiDing is not $ractica". Thus, the i#$ortant

conc"usion is that, $rotection engineers have to "ive with the saturation $rob"e#.

Cautions in CT se"ectionQ

11


12/19

hi"e choosing a CT for a $articu"ar a$$"ication, it is necessar! to observe fo""owing

$recautions.

1. The CT rating and continuous "oad current shou"d #atch. )or exa#$"e, if #ax "oad

current is ?0A, a 100Q5 Ct #a! be acce$tab"e but 50Q5 is not acce$tab"e.

2. The #axi#u# fau"t current shou"d be "ess than 20 ti#es the CT rated current. for

exa#$"e 100Q5 CT can be used, so "ong as burden on the CT G #axi#u# $ri#ar! fau"t

current is be"ow 2000A.

. The vo"tage rating of CT shou"d be co#$atib"e. )or exa#$"e, 100Q5 C100 wou"d give

"inear res$onse, u$to 20 ti#es rated current $rovided CT burden is (e$t

be"ow100/20L51 3. ith 2 burden, this CT can be used on"! if #axi#u# current

is "i#ited to 1000A.

4. Fara""e" of CTEs e.g. in differentia" $rotection, or with -8R fau"t can create significant

errors in CT $erfor#ance. Nne shou"d in genera" ascertain that #agnetiDing current is

(e$t #uch be"ow the $ic( u$ va"ue.

)o""owing exa#$"e, i""ustrates this $oint

Exercise pr-+lems:

;f the current ratio is adeuate for a $rotection, but CT burden is highI then the

$erfor#ance of CT #a! deteriorate due to "arge #agnetiDing current and/or saturation $rob"e#.The CT $erfor#ance can be i#$roved b! connecting the CTEs in series.

13 -how the dotted ter#ina"s for correct secondar! series connection

23 hat is the A of CT in fig a3 G b3 res$ective"!&

12


13/19

13 S"ectro#echanica" re"a!s tend to saturate at high currents. This reduces the re"a! burden

on CT, and so that the CT $erfor#ance at #oderate"! high currents #a! be considered

better than at re"a!Es rated burden at 5A.

23 se of instantaneous over current re"a!s has the $otentia" to overco#e this $rob"e# of

saturation of CTEs

3 @ifferentia" $rotection can o$erate on externa" fau"ts due to the un eua" saturation of

CTEs

8ecture+7

Sxa#$"es

7. ;f a 00Q5 c"ass C CT is connected to a #eter with resistance =1IR and

secondar! current in the CT is 4.5A find out the $ri#ar! current vo"tage

deve"o$ed across the #eter and % rate error. 8ead wire resistance = 02.0"Rsecondar! resistance SR of a 00/5 CT = 15.0

@iagra#

=1IR , = 02.0"R = 15.0SR AIS 5.4=Tota" secondar! resistance S"IT RRRR ++=

= 16.1

-econdar! vo"tage TS RI =

16.15.4 =

V275.5=

)ro# )ig 5.6,Sxciting current ;S for 5.275

0.0ATurns ratio H 00/5 70

34 ESp IINI +=

704.5 0.03 261.*A

o"tage across #eter IS RI =

14.5= 4.5

atio error 1005.4

0/.0100 ==

S

E

I

I

0.76%

+R

1


14/19

8ecture 9 *

Sxa#$"es1. @esign a CCT for a 12( trans#ission "ine using the fo""owing data.

esistive urden VA1503/4 =

()' /=

, $hase ang"e error 40 #inConsider 4 choices of 2as (, 11(, 7.7( and .(

@iagra# 1 @iagra# 2Trans#ission "ine vo"tage 12(. -u$$ose 2FH3 be the vo"tage to be

$roduced b! the ca$acitive $otentia" divider with ca$acitance va"ues C1 and C2and 8 the va"ue of tuning inductor. The standardiDed T secondar! vo"tage is 110vo"ts 8+83.

'ere s$ecification for $hase ang"e error is 40 #inutes variation in freuenc!

can be u$to ()' /= . Fhase ang"e error for change in b! in the aboveeuation circuit, is given b!

+= 31

2

&

"

At tuning freuenc!"&

12=

-ubstituting2 1

"& =

Fhase ang"e error += 34 ""

= "2

% $hase ang"e error+Ra

"2

2 = +++ 13

sing this euation the va"ue 8 for different va"ues of 2 is found out.

13 8et 2be ( 8 + H3

22

=

150+

VR

=

== 52= 10*.216++ RaR/22 == '

rad01174.0701*0

40#in40 =

==

)ro# en 13

22

10*.21601174.0

2

5=

=

=

+R"

(2.7622=

.."

&&

?

221 1051.11051.11 ===+

23 34112 N"/VV = 2

= 4 11 10 3 242 10150

+R

= =

= 40.01174 242 10

2 2 2

+R"

= =

14


15/19

(2.646=

2.64634/

112221

==+"

&&

.210/7.1 =

3 /VV 7.72=

2= 7.7 10 3

150+R

=

= 41012.*6= 40.01174 *6.12 10

2 2 2

+R"

= =

,27?(=

.&& 221 1066. =+

43 /VV /./2= 2

= . 10 3

150

+R

=

= 4106*.21= 40.01174 21.6* 10

2 2 2

+R"

= =

(25.76=

1 2 0.151& & .+ =

The va"ues of 8, 21 &&+ for different va"ues of 2 are tabu"ated be"ow.2 8 21 &&+

( 7622.2' 0.00151 .

11( 646.2' 0.017 .

7.7( 27?' 0.066 ..( 76.25' 0.151 .

)ro# the above tab"e it is c"ear that s#a""er the va"ue of 2, the s#a""er is the

va"ue of 8 and higher the va"ue of C 1and C2for tuning condition. ;f we se"ect too "ow

va"ue of 2 and 8 then ca$acitance va"ues wi"" be be!ond avai"ab"e "i#its, and if wese"ect higher va"ue of 2 and 8, then CCT and inductor wi"" beco#e bu"(!. -o a

co#$ro#ise is necessar! and "et us se"ect 2 7.7(

)or 2 7.7(

8 27?'

.&& 066.021 =+

How,1

21

2 &&&

VV +=

1

7

/

/ 100/66.0

107.7/

101/2

&

=

7

11012

10107.7066.0

=

.&

.00//.0=

15


16/19

.& 044.02=;n this design, we ex$"ained the basic conce$t for CCT design and we assu#ed

the transfor#er to be idea". ut in actua" design $ractice the va"ue of #agnetiDing

i#$edance of transfor#er, resistance of reactor etc have to be ta(en into account,as ratio error and $hase ang"e error wi"" a"so get affected b! these va"ues.

2. @iagra#

The euiva"ent circuit of a CCT is shown in fig *.. The va"ues of C 1and C2are

0.001* . and 0.01* . res$ective"!. Tuning inductor has an inductance of4?6' and resistance of 4720.O# of the 7.7( T is 1U, core "oss 20 watts $er $hase, A burden

150A $er $hase. a"ue of C# for co#$ensating the current drawn b! # is eua"

to .?101*/./ .

a3 erif! the a$$ro$riateness of choice of 8 and C#.

AnsQ ;f .& 001*.01= and .& 01*7.02= then the va"ue 8 of tuning inductoris given b!

3

1

21

2 &&"

+=

where 2 ' = and ' tuning freuenc!

2 7

1

2 503 0.001* 0.01*73 10"

=

+

4?7.6' which is eua" to the given va"ue of 8. How71 10m# =

1m

m

#&

=

7

1 1

2 503 1 10m

m

= =

?.1* 10 .=

The va"ue is a"so sa#e as the se"ected va"ue of C# 'ence the se"ection of both 8

and C# is a$$ro$riate.

b3 )ind out the no#ina" va"ue of /2

AnsQ1 2

2 1

0.001* 0.01*7

0.001*

& &V

V &

+ += =

11.

11. x 7.7

12

/V=

c3 ;f the freuenc! dro$s fro# 50'D to 46'D, what wou"d be the va"ues of ratio

error and $hase ang"e error&AnsQ Core "oss 20w

2

2 20m

V0

R=

17


17/19

2 2

2 77003

20 20m

VR = =

72.1* 10=

A burden 150A resistive32

2

150+

V

R =2 2

2 77003

150 150+

VR = =

52.?04 10= The euiva"ent circuit can be re$resented as shown be"ow.

@iagra# *.12710m# = at f 50'D

72 10m' " =710

1*.12 50m" (= = The freuenc! of interest is 46'D. 'ence va"ues of O #and other i#$edance can

be ca"cu"ated at 46'D. The above circuit can be si#$"ified as@iagra# *.1

here1 1 1

m

m m +

!! &

Z R # R= + +

?

7 5

1 12 46 .1* 10

2.1* 10 2 46 1*.1 2.?04 10

!!

= + +

7 7 7 70.45? 10 1.074 10 0.?4 10 .44 10! ! = + + 7 7.?02 0.1243 10 .?04 10 1.*2! = =

7

1

.?04 10 1.*2Z

=

257146.5 1.*2=

25701*.2 *15.15!= +

t%t%

VI

!R ! " Z

&

=+ +

7

7700 0

4720 2 46 4?6 25701*.2 *15.152 46 0.0204 10

!! !

= + + +

[ ]

7700 0

4720 14767*.? 175??4 25701*.2 *15.15! ! !=

+ + +

7700 0

2707*.2 110*?.*4!=

16


18/19

7700 0

270*64.14 2.44A=

T t%V I Z=

7700 0257146.5 1.*2

270*64.14 2.44

=

74*0.42 4.27=

'ence % ratio error7700 74*0.423

1007700

=

1.*1%

Fhase ang"e error 4.27

2 7.7V /V=

r

= 2 2

+

0t t2

2

10

e*

d i R dii

dt " dt "& + = + =

12 50nw

"&= = 2 n

RIw

"=

22

22 0n n

d iw w i

dt+ + = 1

ct 2ndry resistance

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