Power System Protection - Engineeringlorama/Week2b.pdfPower System Protection Dr. Lionel R. Orama Exclusa, PE Week 2

Power System Protection

Dr. Lionel R. Orama Exclusa, PEWeek 2

Jan 25/2006 Copyright L.R. Orama 2006 2

CT Performance

Readings-Mason Chapter 7,pgs 99-115• Steady State (Symmetrical)

– Basics– Accuracy Classification– Effects on Accuracy– Evaluating CT performance

• Transient (Asymmetrical)– Time to Saturation

Copyright L.R. Orama 2006 3

Current Transformers (CT)

• CT produces current proportional to the system conductor current

• Ideal CT has:

• Rated ICT typically 5A (called the CT 2ry

current)

CT

SYSCT N

II =


Current Transformers (CT’s)

4002000:5

2001000:5

120600:5

80400:5

40200:5

20100:5

1050:5

TurnsCurrent

Common ratios Polarity dot convention


Current Transformers (CT)

36-550KV4000A

12-40KV2500A

5-25KV4000A

600V2000A


CT Steady State PerformanceEquivalent circuit of a

bushing CT• CT is a toroid, leakage

flux is minimum– No leakage reactance

in circuit equivalent

• ZB is called the Burden (load)– Impedance in the

internal circuits of the relay

100% ×=CT

E

I

IRatioError

BCTT ZIV =

EP

CT IN

II −=


CT Accuracy ClassificationWith 20 times rated 2RY current

(5A X 20 = 100A), CT is classified on a basis of the VMAXit can sustain without exceeding a specified Ratio Error (RE)

• Example using line 2 on the table (10T20):– 10 is the %RE (at 20 times

Rated Current-100A)– 20 is the VMAX it can maintain

without exceeding the 10% RE (VRATING)

• T is for tested device• C for calculated device• RE < 10% for ICT < 100A and

ZB< VT/100A

100% ×=CT

E

I

IRatioError

CalcTest

10C80010T800

10C40010T400

10C20010T200

10C10010T100

10C5010T50

10C2010T20

10C1010T10



CT Accuracy Classification

• Determination of CT accuracy class– From CT Saturation Curve (120 turns)

• ICT=100A, IE=10A, ES=500V• VT=ES-100RCT (sat curve determines the RCT)

– Ex., if RCT=0.372ohm, then

• VT=500V-100A(0.372ohm)=463V

– CT Accuracy Class would be C400• Since from last table class jumps from 400 to 800

– This is true assuming all I’s, V’s & Z’s are in phase



• True error at VT = RatingES=VT+ICTRCT

• At VT=800V & ICT=100A• RCT=(.0031ohm/turn)X240turns=0.774ohm• ES=800+100(0.744)=874V (on the curve)• IE=0.4A• %RE= IE /ICT X100=0.4%<10% (OK!)



( )

100×=

+

−=

−==

−=

+=

CT

E

BCTEP

S

BEP

BCTT

EP

CT

TCTCTS

I

IRE

ZRIN

IE

ZIN

IZIV

IN

II

VRIE

Accu.REIEESN

Accu.REIEESZB

Accu.REIEESRCT

Accu.REIEES

upIP

Accu.REIEES

The higher the turns ratio of the CT, the better the accuracy.



Is the CT adequate?

%10

100,100

100,100

<

<=

≤=

≤

RE

then

AIA

VZ

orA

VZAI

or

VV

CTRATING

B

RATINGBCT

RATINGT


Evaluating CT Performance

( )

( ) ( )

ES

BCTEBCTP

S

BCTEP

S

EP

CTBCTCTCTS

BCTTTCTCTS

IKKE

ZRIZRN

IE

ZRIN

IE

IN

IIZIRIE

ZIVVRIE

21 −=

+−+=

+

−=

−=+=

=+=

L

L

Given IP and a relay IPU, calculate weather or not the relay will operate(ICT>IPU)



( )

2

1

1

2

1

21

0

0

K

KIE

KEI

ZRK

ZRN

IK

IKKE

ES

SE

BCT

BCTP

ES

=→=

=→=+=

+=

−=



else

II

if

IN

II

puCT

EP

CT

≥

−=

IE is the operating point of the relay atthe above conditions, then

Try another IPU setting or another N

IPU setting is OK!


Transient Performance


Transient Performancev t( )

V

10sin ω t⋅ θ+( )⋅:= i t( )

V

Zsin ω t⋅ θ+ φ−( ) sin θ φ−( ) exp

t−τ

⋅−

⋅:=

0 0.01 0.02 0.03 0.04 0.05 0.06

1000

500

500

1000Short Circuit Current (5mi dist. line)

time (s)

curr

ent (

A),

tent

h of

vol

tage

(V

)

i t( )

v t( )

t

3 Feb 2006 Copyright L.R. Orama 2006 18

Transient Performance• Offset fault current

* W.A. Elmore, Protective Relaying Theory & Application,Marcel Dekker, NY, 1982, pg. 80.



* W.A. Elmore, Protective Relaying Theory & Application,Marcel Dekker, NY, 1982, pg. 80. dt

dNe

φ−=





Transient Performance• Offset fault current

– CT response




• CT Flux density for theasymmetrical fault current

−=

−

−⋅=

=

−

−

)cos(2

)sin(1

110

2

1

2

8

teIi

teAN

IRB

AB

DC

DC

t

t

DC

ω

ωω

τ

φ

τ

τ

B-Flux Density (lines/in2)

Total CT Burden (resistive)


Transient Performance• Time to Saturation

2

8102

1ln

ANk

kIR

Bt

DC

SDC

⋅=

−−=

ττ BS-Saturation Flux Density

125,000 to 130,000 lines/in2

•States how long the CT will accurately reproduces the fault current•Time constant is that of the power system•R is the resistance of the CT + wiring

Power System Protection - Engineeringlorama/Week2b.pdfPower System Protection Dr. Lionel R. Orama Exclusa, PE Week 2

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