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Current Transformers
GRID Technical Institute
This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.
Current Transformer Function
X Reduce power system current to lower value for measurement. X Insulate secondary circuits from the primary. X Permit the use of standard current ratings for secondary equipment.
REMEMBER : The relay performance DEPENDS on the C.T which drives it !
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Instrument Transformer Standards IEC
IEC 185:1987
CTs
IEC 44-6:1992
CTs
IEC 186:1987
VTs
BS 7625
VTs
BS 7626
CTs
BS 7628
CT+VT
BS 3938:1973
CTs
BS 3941:1975
VTs
AMERICAN
ANSI C51.13.1978
CTs and VTs
CANADIAN
CSA CAN3-C13-M83
CTs and VTs
AUSTRALIAN
AS 1675-1986
CTs
EUROPEAN
BRITISH
3
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Current Transformer Construction
BAR PRIMARY
WOUND PRIMARY
Primary
Secondary
4
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Bar Primary Type Current Transformer(s)
Primary Conductor Primary Insulation
Core Secondary Winding
5
s
Ring Type Current Transformer
Typical Protection Bar-Primary Current Transformer Laminated ‘strip’ wound steel toroidal core Insulation to stop flash-over from HV primary to core & secondary circuit
1000 turns sec. ? ‘Feeder’ or ‘Bus-bar’ forming 1 turn of primary circuit
RELAY
1000A ? Generator, or system voltage source
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1A ? Insulation covered wire, giving inter-turn insulation & secondary to core insulation
Polarity
Is P2
P1 Ip
S1
S2
Inst. Current directions :P1 Î P2 S1 Î S2 Externally 7
s
Flick Test
P1 Is
Ip
FWD kick on application,
S1 +
REV kick on removal of test lead.
-
Battery (6V) + to P1 AVO +ve lead to S1
V
S2
P2
8
s
Directional Protection P1
P2
S1
S2
IOP VPO L
67
IOP = Current direction for operation P1
P2
S2
S3
IFP
IFS
Operation for Forward Faults 9
s
IFP
IFS
No Operation for Reverse Faults
Differential Protection (1)
Operation for Internal Faults P1
P2
IFP1
IFP2
P2
P1
S1
S2
IFS1
IFS2
S2
S1
IFS
R
10
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Differential Protection (2)
Stability for External Faults P1
P2
S1
S2
IFP
IFS
IFS
R
11
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P2
P1
S2
S1
Differential Protection (3)
Maloperation due to Incorrect Connections P1
P2
S1
S2
IFP
IFS
P2
S2
P1
S1
IFP IFS
R 2IFS
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Basic Theory
13
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Basic Theory (1) IS IP
R
1 Primary Turn N Secondary Turns
For an ideal transformer :PRIMARY AMPERE TURNS = SECONDARY AMPERE TURNS ⇒ IP = N x IS
14
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Basic Theory (2) IS IP
ES
R
For IS to flow through R there must be some potential ES = the E.M.F. ES = IS x R ES is produced by an alternating flux in the core. ES ∝ dØ dt 15
s
Basic Theory (3) NP IP NS IS
EK ZCT
ZB
VO/P 16
s
=
ISZB = EK - ISZCT
Basic Formulae
Circuit Voltage Required : ES = IS (ZB + ZCT + ZL) Volts where :IS
=
Secondary Current of C.T. (Amperes)
ZB
=
Connected External Burden (Ohms)
ZCT
=
C.T Winding Impedance (Ohms)
ZL
=
Lead Loop Resistance (Ohms)
Require EK > ES
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Basic Formulae (1) Maximum Secondary Winding Voltage : EK = 4.44 x B A f N Volts …. 1 where :-
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EK
=
Secondary Induced Volts (knee-point voltage)
B
=
Flux Density (Tesla)
A
=
Core Cross-sectional Area (square metres)
f
=
System Frequency (Hertz)
N
=
Number of Turns
Simple Selection Example Example Calculation : C.T. Ratio = 2000 / 5A
Max Flux Density = 1.6 T
RS = 0.31 Ohms
Core C.S.A = 20 cm2
IMAX Primary = 40 kA Find maximum secondary burden permissible if no saturation is to occur. Solution :
N = 2000 / 5 = 400 Turns IS MAX = 40,000 / 400 = 100 Amps From equation 1 the knee point voltage is :VK = (4.44 x 1.6 x 20 x 50 x 400) = 284 Volts 104 Therefore Maximum Burden = 284 / 100 = 2.84 Ohms Hence Maximum CONNECTED burden : 2.84 - 0.31 = 2.53 Ohms
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Mag Curves Flux Density (B) Tesla (Wb/m2)
Material : Cross
1.8 1.6
Knee Point
1.4 1.2 1.0 0.8 0.6 0.4 0.2 0
0
0.02
0.04
0.06
0.08
0.1
0.12
Magnetizing Force (H) Ampere-Turns / mm (Magnetic Field Strength : H, A/m) 20
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Mag Curves Cont...
ES ES = 4.44 N f A B = Kv B
(Secondary E.M.F.)
where :- Kv = 4.44 N f A (B in Wb/m2 ; A in m2) Ie = H . L = K i . H N where :Ki = L/N L = mean magnetic path in metres H = amp. Turns / metre Ie = Amps
(Magnetising Current) Ie 21
s
Mag Curves Cont... B
Multiply By KV to obtain Volts
Units :
KV = E = 4.44 N f A (m2 x turns) B Ki = Ie = L (m / turns) H
Note :
N
L = Mean Magnetic Path
ro
Mean Magnetic Path = 2πr r ri
r = r o - ri 2
H Multiply By Ki to obtain Amps
KV x B = Volts (E) Ki x H = Amps 22
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+ ri
Magnetisation Characteristic (1)
C.T Ratio 100/5 A connected to CDG11. Relay Burden = 2 VA (at 10% rated I) Problem :Calculate the necessary values of Kv and Ki to provide the necessary output at ten times setting (Assume maximum flux density before saturation = 1.6 Tesla)
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Magnetisation Characteristic (2) (i)
Bar Type Primary CDG current setting = 10% = 0.5 A Volts required to operate relay = 2 / 0.5 = 4 V Therefore voltage required at 10 times setting = 10 x 4 = 40 V Hence for a flux density of 1.6T the C.T. should have an output of at least 40 V. With bar primary, number of turns = 20. Ek = 4.44 B A f N = 4.44 x 1.6 x A x 10-4 x 20 x 50 A = 40 / 0.71 = 56.3 cm2 Assume a stacking factor of 0.92 => Total C.S.A. = 61.2 cm2 Assume I.D. = 18cm, O.D. = 30cm & Depth = 10.2cm KV = AN / 45 = (56.3 x 20) / 45 = 25 Ki = L / N = 24TT / 20 = 3.77 cm / turn
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Magnetisation Characteristic (3)
(ii)
Wound Type Primary Assume primary has 5 turns - Therefore secondary has 100 turns Ek = 4.44 B A f N = 4.44 x 1.6 x A x 10-4 x 100 x 50 A = 40 / 3.55 = 11.26 cm2 Assume a stacking factor of 0.92 => Total C.S.A. = 12.24 cm2 Assume I.D. = 18cm, O.D. = 30cm & Depth = 2.04cm KV = AN / 45 = (11.26 x 100) / 45 = 25 Ki = L / N = 24TT / 100 = 0.754 cm/turn
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Low Reactance Design
With evenly distributed winding the leakage reactance is very low and usually ignored. Thus ZCT ~ RCT
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Exciting Voltage (VS)
Knee-Point Voltage Definition
+10% Vk Vk +50% Iek
Iek Exciting Current (Ie) 27
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C.T. Equivalent Circuit Ip
ZCT Is
P1 Ip/N
Ie
S1 N
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Ze
Vt
Es
Ip = Primary rating of C.T.
Ie
= Secondary excitation current
N
Is
= Secondary current
= C.T. ratio
Zb = Burden of relays in ohms
Es = Secondary excitation voltage
(r+jx) ZCT = C.T. secondary winding impedance in ohms (r+jx) Ze = Secondary excitation impedance in ohms (r+jx)
Vt = Secondary terminal voltage across the C.T. terminals
s
Zb
Phasor Diagram Φ
Ip/N Ie Ie
Is Es
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Ep
Im Ic
Ep = Es =
Primary voltage Secondary voltage
Im = Ie =
Magnetising current Excitation current
Φ = Ic =
Flux Iron losses (hysteresis & eddy currents)
Ip = Is =
Primary current Secondary current
s
Saturation
30
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Steady State Saturation (1)
E= 100V
100A
100A
1A
1A 100/1
E
100/1
1 ohm
E
100 ohm
E= 1V
100A
1A
1A 100/1
31
E=?
100A E
s
10 ohm
E= 10V
100/1
E
1000 ohm
Steady State Saturation (2) Φ
+V
0V
A
-V
Time
Assume :- Zero residual flux Switch on at point A 32
s
Steady State Saturation (3) +V
0V
C
Φ -V
Time
Assume :- Zero residual flux Switch on at point C 33
s
Steady State Saturation (4) +V Φ
0V
B
-V
Time
Assume :- Zero residual flux Switch on at point B 34
s
Steady State Saturation (5)
+V Mag Core Saturation 0V Mag Core Saturation -V
Time 35
s
Steady State Saturation (6) +V
Prospective Output Voltage Φ
Mag Core Saturation
0V Mag Core Saturation
-V
+V Output lost due to steady state saturation
0V Actual Output Voltage
-V Time 36
s
Transient Saturation v = VM sin (wt + σ) L1
R1 Z1
i1
v = VM sin (wt + σ) i1 = +
VM V sin (wt + σ - ∅ ) = M sin (σ - ∅ ) . e Z1 Z1
= + Ιˆ1 sin (wt + σ - ∅ ) - Ιˆ1 sin (σ - ∅ ) . e =
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STEADY STATE
s
+
-R1t / L1
-R1t / L1
TRANSIENT
where : -
Z1 =
R12 + w 2L12
∅ = tan-1 V Ιˆ1 = M Z1
wL1 R1
Transient Saturation : Resistive Burden
Required Flux ØSAT
FLUX Actual Flux Mag Current
0
Primary Current Secondary Current CURRENT 0
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10
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20 30
40 50
60 70
80
M
CT Types
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Current Transformer Function
Two basic groups of C.T. X
Measurement C.T.s
Limits well defined X
Protection C.T.s
Operation over wide range of currents Note : They have DIFFERENT characteristics
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Measuring C.T.s Measuring C.T.s X Require good accuracy up to approx 120% rated current. X Require low saturation level to protect instruments, thus use nickel iron alloy core with low exciting current and knee point at low flux density.
B Protection C.T.
Protection C.T.s X Accuracy not as important as above. X Require accuracy up to many times rated current, thus use grain orientated silicon steel with high saturation flux density. 41
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Measuring C.T.
H
CT Errors
42
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Current Transformer Errors ZS
Is
Ip Ie Ze
Zb
Es
Φ Ip/N Ie
I Iq p/N Ie Ic’ Is Es 43
Ie Is Ep
s
Current Transformer Error Transformer Error vs Primary Current
Error
Current Error Phase Error Primary Current
Rated Current
Errors can be reduced by :1. 2. 3. 44
s
Using better quality magnetic material Shortening the mean magnetic path Reducing the flux density in the core
Current Transformer Errors Current Error Definition. Error in magnitude of the secondary current, expressed as a percentage, given by :Current error % = 100 (kn IS - Ip) Ip kn Ip
= =
Rated Transmission Ratio Actual Primary Current
Is
=
Actual Secondary Current
Current Error is :+ve : When secondary current is HIGHER than the rated nominal value. -ve : When secondary current is LOWER than the rated nominal value. 45
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Current Transformer Errors
Phase Error Definition. The displacement in phase between the primary and secondary current vectors, the direction of the vectors being chosen so the angle is zero for a perfect transformer. Phase Error is :+ve : When secondary current vector LEADS the primary current vector. -ve : When secondary current vector LAGS the primary current vector.
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Current Transformer Ratings
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Current Transformer Ratings (1) Rated Burden X Value of burden upon which accuracy claims are based X Usually expressed in VA X Preferred values :2.5, 5, 7.5, 10, 15, 30 VA
Continuous Rated Current X Usually rated primary current
Short Time Rated Current X Usually specified for 0.5, 1, 2 or 3 secs X No harmful effects X Usually specified with the secondary shorted
Rated Secondary Current X Commonly 1, 2 or 5 Amps 48
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Current Transformer Ratings (2) Rated Dynamic Current Ratio of :IPEAK : IRATED (IPEAK = Maximum current C.T. can withstand without suffering any damage). Accuracy Limit Factor - A.L.F. (or Saturation Factor) Ratio of :IPRIMARY : IRATED up to which the C.T. rated accuracy is maintained. e.g. 200 / 1A C.T. with an A.L.F. = 5 will maintain its accuracy for IPRIMARY < 5 x 200 = 1000 Amps 49
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Choice of Ratio Clearly, the primary rating IP ≥ normal current in the circuit if thermal (continuous) rating is not to be exceeded. Secondary rating is usually 1 or 5 Amps (0.5 and 2 Amp are also used). If secondary wiring route length is greater than 30 metres - 1 Amp secondaries are preferable. A practical maximum ratio is 3000 / 1. If larger primary ratings are required (e.g. for large generators), can use 20 Amp secondary together with interposing C.T. e.g. 5000 / 20 - 20 / 1 50
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Current Transformer Designation
Class “P” Specified in terms of :i) Rated burden ii) Class (5P or 10P) iii) Accuracy limit factor (A.L.F.) Example :15 VA 10P 20 To convert VA and A.L.F. into useful volts Vuseful ≈ VA x ALF IN
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BS 3938 Classes :-
5P, 10P. ‘X’
Designation (Classes 5P, 10P) (Rated VA)
(Class)
(ALF)
Multiple of rated current (IN) up to which declared accuracy will be maintained with rated burden connected. 5P or 10P. Value of burden in VA on which accuracy claims are based. (Preferred values :- 2.5, 5, 7.5, 10, 15, 30 VA) ZB = rated burden in ohms = Rated VA IN2 52
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Protection Current Transformers
Table 3 - Limits of Error for Accuracy Class 5P and 10P
Accuracy Current Error at Phase Displacement at Composite Error Class rated primary rated primary current (%) at rated current (%) Minutes Centiradians accuracy limit primary current 5P 5 ±1 ±60 ±1.8 10P
53
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±3
10
Interposing CT
54
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Interposing CT
LINE CT
NP
NS
ZCT
Burden presented to line CT = ZCT + ZB x NP2 NS2 55
s
ZB
NEG.
5A
1A
0.5Ω
R 500/5
0.1Ω
1VA @ 1A ≡ 1.0Ω
0.4Ω
‘Seen’ by main ct :- 0.1 + (1)2 (2 x 0.5 + 0.4 + 1) = 0.196Ω (5)2 Burden on main ct :- I2R = 25 x 0.196 = 4.9VA Burden on a main ct of required ratio :0.5Ω
R 500/1
1.0Ω
Total connected burden = 2 x 0.5 + 1 = 2Ω Connected VA = I2R = 2 ∴ The I/P ct consumption was about 3VA. 56
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Current Transformer Designation
57
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Current Transformer Designation Class “X” Specified in terms of :-
58
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i)
Rated Primary Current
ii)
Turns Ratio (max. error = 0.25%)
iii)
Knee Point Voltage
iv)
Mag Current (at specified voltage)
v)
Secondary Resistance (at 75°C)
Choice of Current Transformer X Instantaneous Overcurrent Relays
Class P Specification A.L.F. = 5 usually sufficient For high settings (5 - 15 times C.T rating) A.L.F. = relay setting
X IDMT Overcurrent Relays
Generally Class 10P Class 5P where grading is critical Note : A.L.F. X V.A < 150 X Differential Protection
Class X Specification Protection relies on balanced C.T output 59
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Selection Example
60
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Burden on Current Transformers
1. Overcurrent : RCT + RL + Rr
2. Earth : RCT + 2RL + 2Rr
RCT
RCT
RCT RCT
Rr
61
Rr
s
RCT
IF RL
RL
RCT
IF
RL
Rr
RL
Rr
RL Rr
IF
IF
RL
Rr
RL
Rr
RL
Rr
Overcurrent Relay VK Check Assume values :
If max C.T
= =
7226 A 1000 / 5 A 7.5 VA 10P 20
RCT = Rr = RL =
0.26 Ω 0.02 Ω 0.15 Ω
Check to see if VK is large enough : Required voltage = VS = IF (RCT + Rr + RL) = 7226 x 5 (0.26 + 0.02 + 0.15) = 36.13 x 0.43 = 15.54 Volts 1000 Current transformer VK approximates to :VK ~ VA x ALF + RCT x IN x ALF In = 7.5 x 20 + 0.26 x 5 x 20 = 56 Volts 5 VK > VS therefore C.T VK is adequate 62
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Earth Fault Relay VK Check Assume values : As per overcurrent. Note
For earth fault applications require to be able to pass 10 x relay setting.
Check to see if VK is large enough : Total load connected = 2RL + RCT + 2Rr = 2 x 0.15 + 0.26 + 2 x 0.02 ∴
Maximum secondary current = 56 = 93.33A 0.6
Typical earth fault setting
= =
30% IN 1.5A
Therefore C.T can provide > 60 x setting C.T VK is adequate 63
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VK = 56 Volts
Voltage Transformers
64
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Voltage Transformers
65
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X
Provides isolation from high voltages
X
Must operate in the linear region to prevent accuracy problems - Do not over specify VT
X
Must be capable of driving the burden, specified by relay manufacturer
X
Protection class VT will suffice
Typical Working Points on a B-H Curve Flux Density ‘B’
Saturation
1.6
Tesla 1.0
0.5
Cross C.T.’s & Power Transformers
V.T.’s
Protection C.T. (at full load) ‘H’ 1000
2000
3000 Magnetising Force AT/m
66
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Types of Voltage Transformers
Two main basic types are available: X Electromagnetic VT`s
Similar to a power transformer May not be economical above 132kV X Capacitor VT`s (CVT)
Used at high voltages Main difference is that CVT has a capacitor divider on the front end.
67
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Electromagnetic Voltage Transformer
NP / NS = Kn
LP
RP IP
EP = ES
68
s
IS
Ie LM
VP
LS
RS
IM
Re IC
VS
ZB
(burden)
Basic Circuit of a Capacitor V.T.
C1 L T
VP C2
69
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ZB VC2
Vi
VS
Ferro-resonance
X The exciting impedance of auxiliary transformer T and the capacitance of the potential divider form a resonant circuit. X May oscillate at a sub normal frequency X Resonant frequency close to one-third value of system frequency X Manifests itself as a rise in output voltage, r.m.s. value being 25 to 50 per cent above normal value X Use resistive burden to dampen the effect
70
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VT Earthing
X Primary Earthing
Earth at neutral point Required for phase-ground measurement at relay X Secondary Earthing Required for safety Earth at neutral point When no neutral available - earth yellow phase (VERY COMMON) No relevance for protection operation
71
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VT Construction
X
5 Limb
Used when zero sequence measurement is required (primary must also be earthed)
X
Three Single Phase
Used when zero sequence measurement is required (primary must also be earthed)
X
3 Limb
Used where no zero sequence measurement is required
X
V Connected (Open Delta)
72
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No yellow phase Cost effective Two phase-phase voltages No ground fault measurement
VT Connections
Broken Delta A
B
da
a
73
s
C
N
V Connected a
b
c
dn
b
c n
a
b
c
VT Construction - Residual
X Used to detect earthfault X Useful where current operated protection cannot be used X Connect all secondary windings in series X Sometimes referred to as `Broken Delta` X Residual Voltage is 3 times zero sequence voltage X VT must be 5 Limb or 3 single phase units X Primary winding must be earthed
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Voltage Factors Vf
X Vf is the upper limit of operating voltage.
75
X
Important for correct relay operation.
X
Earthfaults cause displacement of system neutral, particularly in the case of unearthed or impedance earthed systems.
s
Protection of VT’s
76
X
H.R.C. Fuses on primary side
X
Fuses may not have sufficient interrupting capability
X
Use MCB
s
GRID Technical Institute
This document is the exclusive property of Alstom Grid and shall not be transmitted by any means, copied, reproduced or modified without the prior written consent of Alstom Grid Technical Institute. All rights reserved.
