Power System Protection - Engineeringlorama/Week2b.pdfPower System Protection Dr. Lionel R. Orama...
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Power System Protection
Dr. Lionel R. Orama Exclusa, PEWeek 2
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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
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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 =
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Current Transformers (CT’s)
4002000:5
2001000:5
120600:5
80400:5
40200:5
20100:5
1050:5
TurnsCurrent
Common ratios Polarity dot convention
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Current Transformers (CT)
36-550KV4000A
12-40KV2500A
5-25KV4000A
600V2000A
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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 −=
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Jan 27/2006 Copyright L.R. Orama 2006 7
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
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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
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Jan 30/2006 Copyright L.R. Orama 2006 10
CT Accuracy Classification
• 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!)
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CT Accuracy Classification
( )
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.
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CT Accuracy Classification
Is the CT adequate?
%10
100,100
100,100
<
<=
≤=
≤
RE
then
AIA
VZ
orA
VZAI
or
VV
CTRATING
B
RATINGBCT
RATINGT
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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)
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Evaluating CT Performance
( )
2
1
1
2
1
21
0
0
K
KIE
KEI
ZRK
ZRN
IK
IKKE
ES
SE
BCT
BCTP
ES
=→=
=→=+=
+=
−=
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Evaluating CT Performance
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!
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Transient Performance
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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
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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.
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Transient Performance
* W.A. Elmore, Protective Relaying Theory & Application,Marcel Dekker, NY, 1982, pg. 80. dt
dNe
φ−=
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Transient Performance
* W.A. Elmore, Protective Relaying Theory & Application,Marcel Dekker, NY, 1982, pg. 80.
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Transient Performance• Offset fault current
– CT response
* W.A. Elmore, Protective Relaying Theory & Application,Marcel Dekker, NY, 1982, pg. 80.
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Transient Performance
• 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)
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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