Basics of Current and Voltage Transformers Power Transmission and Distribution Author Page 241 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Equivalent current transformer circuit I1 jX1 R1 I 2′ = I1 ⋅ N1 N2 jX2 Im P1 N1 N2 U R2 I2 S1 Zm Zb 2 P2 Ideal CT S2 X1 = Primary leakage reactance R1 = Primary winding resistance X2 = Secondary leakage reactance Z0 = Magnetizing impedance R2 = Secondary winding resistance Zb = SecondaryAuthor load Note: Page 242 2007-08 Power Transmission and Distribution Normally the leakage fluxes X1 and X2 can be neglected Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Current transformer, simplified equivalent circuit jX2 1 : N2 i2 i1 i'1 = N2 1 R2 ⋅ i1 LW L2<< LW im ZB R2 i2 Power Transmission and Distribution RB Author Page 243 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Current transformer: Phase displacement (δ) and current ratio error (ε) I1 . I1 N1 N2 ε N1 I2 N2 I2 δ Power Transmission and Distribution Zb Author Im Page 244 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI CT classes to IEC 60044-1: 5P or 10P Pi = I sec .2 × R CT Specification: 300/1 A 5P10, 30 VA Ratio In -Prim / In -Sek. 5% accuracyat I= n x In Actual accuracy limit factor in operation is higher as the CT ALF ' = ALF × is normally underburdened : Operating ALF: ALF‘ RCT ≤ 5 Ohm Rated burden (nominal power) PNB Accuracy limit factor ALF Pi + PNB Pi + PBB Dimension criterium: I ALF ' ≥ SC − max × KTF Power Transmission I n and Distribution KTF (over-dimensioning factor) considers the single sided CT over-magnetising due to the d.c. Author component in short circuit current ISC. KTF values required in practice depend on relay type and design. Recommendations are provided by manufacturers (see Application Guides) Page 245 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Definition of the CT knee-point voltage U2 10 % IEC 60044-1 Class PX UKN formerly 50 % British Standard BS3938: Class X Power Transmission and Distribution Author Im Page 246 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Current transformer, Standard for steady-state performance IEC 60044-1 specifies the following classes: Accuracy class Current error at nominal current (In) 5P ±1% 10P ± 5% Angle error δ at rated current In ± 60 minutes Total error at n x In (rated accuracy limit) 5% 10 % Power Transmission and Distribution Author Page 247 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Current transformers, Standard for transient performance IEC 60044-6 specifies four classes: Class Error at rated current Ratio error TPX (closed iron core) TPY with anti-remanence air gap TPZ linear core TPS 2007-08 Remanence Angle error ± 0,5 % ± 30 min εˆ ≤ 10% ± 1,0 % ± 30 min εˆ ≤ 10% ± 1,0 % ± 180 ± 18 min εˆ ≤ 10% Author no limit < 10 % Power Transmission and Distribution (a.c. current only) Special version for high impedance protection Author (Knee point voltage, internal secondary resistance) closed iron core Page 248 Maximum error at rated accuracy limit negligible No limit Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Current transformer saturation Steady-state saturation with AC current Power Transmission and Distribution Author Transient saturation with offset current Page 249 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Transient CT saturation due to DC component ISC Short circuit current DC flux non-saturated ΦS Φ AC flux Power Transmission and Distribution Author Im Page 250 Magnetising current 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Course of CT-flux during off-set short-circuit current ΙP primary current t t − B ω ⋅TN ⋅TS − TN = 1+ (e − e TS ) TN − TS Bˆ d.c. component ~ TS N TS TN ⎛T ⎞ − BMax = 1+ ω ⋅TS ⋅ ⎜⎜ ⎟ B~ ⎝ TS ⎠ t TN total flux B t B Max = BMax transient d.c. flux B~ TS = a.c. flux Author t B~ Page 251 2007-08 Author TN ⋅ TS TS ⋅ ln TS − TN TN LW Power Transmission 1 and Distribution = Ri + RB ω ⋅ tanδ For 50 Hz: TS = 10900 δ [min ] [ms] Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI CT transient over-dimensioning factor KTF KTF 6 0 ⎛ TN ⎞ KTF = 1+ ωTS⎜⎜ ⎟⎟ ⎝ TS ⎠ TS TS − TN ∞ (KTF ≈1+ωTN) 5000 5 0 Closed iron core 1000 4 0 TS [ms] 500 3 0 250 2 0 100 Linear core 1 0 Power Transmission and Distribution TN = network time constant (short-circuit time constant) Author 50 Page 252 2007-08 100 150 200 Author TN [ms] TS = CT secondary time constant Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI CT with closed iron core, Over-dimensioning factor KTF‘ for specified time to saturation (tM) 15 14 tM tM → ∞ K‘td 50 ms 40 ms 10 30 ms 20 ms 6 5 10 ms Page 253 20 2007-08 40 60 80 Tp Author ⎞ ⎟ ⎟ ⎟⎟ ⎠ Power Transmission and Distribution tM → ∞ : KTF = 1+ ωTN = 1+ Author 0 K' TF tM ⎛ ⎜ = 1 + ωTN ⋅ ⎜1 - e T N ⎜⎜ ⎝ XN RN 100 [ ms] Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI CT over-dimensioning factor KTF (tM,TN) in the case of short time to saturation (tM) KTF tM 4 Θ (tM,TN) (el. degree) 160 10 ms 3.5 140 3 120 8 ms 2.5 tM 100 2 10 ms 8 ms 6 ms Power Transmission and Distribution 4 ms 2 ms 80 6 ms 1.5 1 60 40 4 ms 0.5 0 180 2 ms 0 Page 254 20 40 60 80 2007-08 T N Author 0 100 in ms 20 Author 0 20 40 60 80 100 Copyright © Siemens AG 2007. All rights reserved. TPTD msPTI N inSE Current transformer magnetising and de-magnetising ΙP B t BMax B B = BR + (BMax. − BR ) ⋅ e BR − BMax t TS BR Power Transmission and Distribution Ιm Author t Page 255 2007-08 Author Ιm Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Current transformer Course of flux in the case of non-successful auto-reclosure B Bmax tF1 = duration of 1st fault t tF1 tDT tF2 tDT dead time Power Transmission and Distribution tF2 =duration of 2nd fault tF1 tDT+tF2 ⎡ tF2 tF1 tF2 ⎡ − ⎤ Author − − ⎤ − ⋅ N ⋅TS Bmax ⎢ ω ⋅TN ⋅TS − TN T ω = 1+ − e TS ⎥ ⋅ e TS + ⎢1+ (e (e TN − e TS ⎥ ⎢ ⎢ ⎥ ⎥ TN −TS TN −TS Bˆ~ ⎢⎣ ⎢⎣ ⎥⎦ ⎥⎦ Page 256 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Current transformer magnetising curve and point of remanence I B II III up to 80% < 10% negligible H = im ⋅ w Power Transmission and Distribution I: closed iron core (TPX) II: core with anti-remanence air-gaps (TPY) Author Page 257 2007-08 Author III: Linearised core (TPZ) Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Current transformers TPX und TPY Course of the flux with non-successful auto-reclosure BR BR closed iron core (TPX) Power Transmission and Distribution core with antiremanence air-gaps (TPY) Author tDT tF1 Page 258 2007-08 tF2 Author t Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Standards of voltage transformers VT classes to IEC 60044-2 Class designation Permissible error at 0.05 ·UN and 1.0 · UN Voltage error FU Angle error δ 3P ± 3.0 % 120 minutes 5P ± 6.0 % 240 minutes All 3P and 5P protection CTs must additionally comply with one of the below VT metering classes! VT classes for measurement IEC 60044-2 Class designation Page 259 Permissible voltage error in % at 1.0 · UN 0.1 0.2 0.1 0.2 0.5 0.5 1 3 1 3 2007-08 Permissible angle error in minutes at 1.0·UN Power Transmission and Distribution 5 Author Author 10 20 30 Not determined Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Capacitive voltage transformer, Equivalent circuit 400 kV C1 = 5 nF C2 = 95 nF 3 RF 20 kV LF 3 CE = C 1 + C 2 C1 U*1 = UP ⋅ C1 + C2 Page 260 2007-08 100 V CF 3 LO RE C T Author Author ZB Power Transmission and Distribution RF RT LT LF CF U*2 R'B L'B Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Transient performance of CVTs, Recommendations acc. to IEC 60044-5 Ratio Measured voltage transients after fault at voltage maximum and zero-crossing (Example) Time Ts in ms t UP(t) Primary voltage 2 ⋅ Us ⋅100% Classes 3PT1 6PT1 3PT2 6PT2 3PT3 6PT3 10 --- ≤ 25 ≤4 20 ≤ 10 ≤ 10 ≤2 40 ≤ 10 ≤ 2 ≤2 60 ≤ 10 ≤ 0.6 ≤2 ≤ 10 ≤ 0.2 ≤2 90 Legend: Us (t) Power Transmission and Distribution Recommendations to IEC 60044-5 US(t) Secondary voltage Author Page 261 2007-08 (1) Fault inception (2) Aperiodic damping of US(t) (3) Periodic damping of US(t) Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI CT dimensioning K OD = KTF Pi + PBN RCT + RBN ALF' = ALF ⋅ = ALF ⋅ Pi + PB RCT + RB Actual connected burden : PB= RB ⋅ I2N2 P + PB RCT + RB = ALF'⋅ ALF = ALF'⋅ i Pi + PBN RCT + RBN with I ALF ' ≥ K OD ⋅ K IN Theory: K OD ≥ K TF ⋅ K Re m K Re m = 1 % Remanence 1− 100 Page 262 2007-08 rated CT burden: PBN internal burden of the CT: Pi=Ri ⋅ I2N2 RB=Rl+RR= burden resistance Rl= resistance of connecting cables RR= burden resistance of the relay No saturation for the total short-circuit duration: No saturation for the specified time tM: Practice:Author K' TF = BMax XN = 1 + ωTN = 1 + RN Bˆ ~ Power tM ⎤ ⎡ Transmission andtMDistribution − − ω T N ⋅ TS ⋅ (e TN − e TS ⎥ K ' 'TF = ⎢1 + TN − TS ⎥ ⎢ ⎣ ⎦ Remanence only considered in extra high voltage systems (EHV) KTF-values acc. to relay manufacturers‘ guides Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI Practical CT requirements Transient over-dimensioning factors KTF (AR not considered) Distance 7SA6 and 7SA522 Overcurrent 7SJ56 Close-in fault: KTF ≥ 5 Fault atbalance point: KTF ≥ 2 ( ≥ 1 if TN < 30ms) ALF‘≥ I>>setting/ IN, at least 20 Internal fault External fault Transformer Differential 7UT6 KTF≥ 0.75 (Saturation free time ≥ 4 ms) KTF≥ 1.2 (Saturation free time ≥ 5 ms) Line differential 7SD61 KTF≥ 0.5 (Saturation free time ≥ 3 ms) KTF≥ 1.2 Power Transmission and Distribution (Saturation free time ≥ 5 ms) Bus differential 7SS52 Page 263 Author KTF≥ 0.5 (Saturation free time ≥ 3 ms) 2007-08 Author KTF≥ 0.5 (Saturation free time ≥ 3 ms) Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI CT dimensioning for Example differential protection (1) 1. Calculation of fault currents 110 kV, 3 GVA F1 110/20 kV 40 MVA uT=12% F2 OH-line: l = 8 km, zL‘= 0,4 Ω/km F3 F4 Net 300/1A ∆ IT 7UT61 Transf. : ∆ IL 7SD61 ∆ IL Impedances related to 20 kV: Impedances related to 110 kV: Net : 200/1A 200/1A 1200/1A U N 2 ⎡ kV 2 ⎤ ⎥⎦ 110 2 ⎢⎣ ZN = = = 4.03 Ω SSC ' ' [MVA ] 3000 U N 2 ⎡ kV 2 ⎤ Author ⎥⎦ uT [% ] 110 2 12 % ⎢⎣ ZT = ⋅ = ⋅ = 36.3 Ω PN -T [MVA ] 100 40 100 Net : Transf. : U N 2 ⎡ kV 2 ⎤ ⎢⎣ Power⎥⎦ Transmission 20 2 and Distribution ZN = = = 0.13 Ω SSC ' ' [MVA ] 3000 U N 2 ⎡ kV 2 ⎤ ⎢⎣ ⎥⎦ uT [% ] 20 2 12 % ZT= ⋅ = ⋅ = 1.2 Ω PN -T [MVA ] 100 40 100 Line : Z L = l[km ]⋅ z L ' [Ω/km ] = 8 ⋅ 0,4 = 3,2 Ω Page 264 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI CT dimensioning for Example differential protection (2) 1.1 ⋅ U N / 3 1.1 ⋅ 110kV/ 3 = = 17.3 kA ZN 4.03Ω F1 I F1 = F2 1.1 ⋅ U N / 3 1,1 ⋅ 110kV/ 3 I F2 = = = 1.73 kA Z N + ZT 4.03Ω + 36.3Ω F3 F4 I F3 = I F4 = 1.1 ⋅ U N / 3 1.1 ⋅ 20kV/ 3 = = 9.55 kA Z N + ZT 0.13Ω + 1.2Ω 1.1 ⋅ U N / 3 1,1 ⋅ 20kV/ 3 = = 2.8 kA Z N + Z T + Z L 0.13Ω + 1.2Ω + 3.2 Ω Dimensioning of the 110 kV CTs for the transformer differential protection: Manufacturer recommends for relay 7UT61: The saturation free time of 3 ms corresponds to KTF≥ 0,75 See diagram, page 59 Power Transmission and Distribution Criterion 1) therefore reads: I 17300 ALF' ≥ K TF ⋅ F1 = 0,75 ⋅ = 43 IN 300 1) Saturation free time ≥ 4ms for internal faults 2) Over-dimensioning factor KTF ≥ 1,2 for through flowing currents (external faults) For criterion 2) we get: Author I 1730 ALF' ≥ K TF ⋅ F2 = 1,2 ⋅ =7 IN 300 The 110 kV CTs must be dimensioned according to criterion 1). Page 265 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI CT dimensioning for Example differential protection (3) We try to use a CT type: 300/1, 10 VA, 5P?, internal burden 2 VA. Pi + Poperation 2 + 2.5 ALF ≥ ⋅ ALF ' = ⋅ 43 = 16.1 Pi + Prated 2 + 10 (Connected burden estimated to about 2.5 VA) Chosen, with a security margin : 300 /1 A, 5P20, 10 VA, R2≤ 2 Ohm (Pi ≤ 2VA) Specification of the CTs at the 20 kV side of the transformer: It is good relaying practice to choose the same dimensioning as for the CTs on the 110 kV side: 1200/1, 10 VA, 5P20, R2≤ 2 Ohm (Pi ≤ 2VA) Dimensioning of the 20 kV CTs for line protection: 1‘) Saturation free time ≥ 3ms for internal faults 2‘) Over-dimensioning factor KTF ≥ 1.2 for through flowing currents (external faults) Power Transmission and Distribution For relay 7SD61, it is required: The saturation free time of 3 ms corresponds to KTF≥ 0.5 See diagram, page 59 Criterion 1‘) therefore reads: I 9550 ALF' ≥ K TF ⋅ F3 = 0.5 ⋅ = 24 IN 200 Page 266 2007-08 Author For criterion 2‘) we get: I 2800 ALF' ≥ KTF ⋅ F4 = 1.2 ⋅ = 16.8 IN 200 Author The 20 kV line CTs must be dimensioned according to criterion 1‘). Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI CT dimensioning for Example differential protection (4) For the 20 kV line we have considered the CT type: 200/5 A, 5 VA, 5P?, internal burden ca. 1 VA Pi + Poperation 1+1 ⋅ ALF ' = ⋅ 24 = 8 ALF ≥ 1+ 5 Pi + Prated (Connected burden about 1 VA) Specification of line CTs: We choose the next higher standard accuracy limit factor ALF=10 : Herewith, we can specify: CT Type TPX, 200/5 A, 5 VA, 5P10, R2≤ 0.04 Ohm ( Pi≤ 1 VA) Power Transmission and Distribution Author Page 267 2007-08 Author Copyright © Siemens AG 2007. All rights reserved. PTD SE PTI