Calculation for Ct Selection
Short Description
Calculation for Ct Selection...
Description
西南电力设计院
计 算 书 首 页
工 程 名 称 TIRODA THERMAL POWER PROJECT 1980 MW (3x660 MW Phase – I) Tiroda, Gondia District, Maharashtra
设计阶段
印度 TIRODA(提隆达)3X660MW 燃煤电站 概 念 设 计 BASIC DESIGN
运用软件名称
无
专
业
电 二
计 算 书 名 称 CALCULATION FOR CT SELECTION
CT 选型计算
批
准 ：
月
日
审
核 ：
月
日
校
核 ：
月
日
设计计算 ：
月
日
2008 年 11 月 12 日
CONTENTS 1. SPECIFICATION OF CALCULATION ....................................................................1 2. PRINCIPLE ON CT SELECTION ..............................................................................1 3 CALCULATION FORMULA.......................................................................................1 3.1 SELECTION FOR RATIO OF CT........................................................................1 3.2 SELECTION OF ACCURACY CLASS OF CT...................................................1 4. SECONDARY LOAD VERIFICATION .....................................................................2 4.1 SECONDARY LOAD OF CT FOR MEASUREMENT METERS ......................2 4.2 SECONDARY LOAD OF CT FOR RELAY PROTECTION..............................3 5. ACCURACY LIMIT FACTOR FOR PROTECTION CT CLASS P......................4 6. CHECKING COMPUTATIONS FOR CT CLASS P................................................4 7. CALCULATION EXAMPLE FOR MAIN PROTECTION CT...............................5 8. CALCULATION EXAMPLE FOR MAIN MEASUREM ENT CT .......................13
1. SPECIFICATION OF CALCULATION This calculation shall be done according to the specifications as follows: «The Guide for selection and calculation of current transformers and voltage transformers» «Manual for designing electrical part of electric power engineering »(2)
2. PRINCIPLE ON CT SELECTION 2.1 It should meet the requirement of rated primary current, maximum load current and dynamic, thermal stable current while short circuit. 2.2 It should meet the requirement of accuracy class for measuring meters and automation equipments in secondary circuit and 10% error characteristic curve for protection requirement s.
3 CALCULATION FORMULA 3.1 SELECTION FOR RATIO OF CT 3.1.1 Selection for Ratio of CT for Measurement Meters Rated primary current of CT shall be: IN≥1.25Ie Where:
(1-1)
IN --- Primary current of CT; Ie--- Rated current of generator or transformer, for power lines it shall be
maximum load current. Secondary current of CT shall be 5A or 1A 3.1.2 Selection for Ratio of CT for Relay Protection Rated primary current of CT should be greater than maximum long-time load current of electrical equipments. Secondary current of CT shall be 5A or 1A.
3.2 SELECTION OF ACCURACY CLASS OF CT 3.2.1 Selection of Accuracy Class of CT for Measurement Meters Accuracy class of CT refers to permissible error under definite secondary loads. It meets the requirements in the following tables. Tab. 1-1 Accuracy Class of CT and Electrical Measuring Devices Accuracy class Accuracy class of CT of meters 0.5
0.5
1.0
0.5
1.5
1.0
2.5
1.0
1/16
Tab. 1-2 Accuracy Class of CT and Energy Metering Accuracy class of watthour Accuracy class of varhour Accuracy class of CT meters
meters
0.2S
2.0
0.2S or 0.2
0.5S
2.0
0.2S or 0.2
1.0
2.0
0.5S
2.0
3.0
0.5S
3.2.2 Selection of Accuracy Class of CT for Relay Protection According to the regulation of IEC, the accuracy class for relay protection shall be 5P and 10P. Its error refers to Tab. 1-3. Tab. 1-3 Error of 5P, 10P Specified by IEC Accuracy
Ratio
error
under Phase angle error under Composed error under
class
rated primary current rated (%)
primary
current rated accuracy primary
(‘)
current (%)
(rad×10²) 5P
±1
±60
±1.8
5
10P
±3
⁄
⁄
10
4. SECONDARY LOAD VERIFICATION 4.1 SECONDARY LOAD OF CT FOR MEASUREMENT METERS Rated secondary load of CT shall be derived from the formula as follows: Rbn= VA/I 22
(1-2)
Where: Rbn--- Rated secondary load of CT(Ω); VA--- Secondary capacity of CT(VA); I 2 --- Rated secondary current of CT. For measuring meters, secondary load of CT shall be calculated as follows: Rb= KcjZcj+KlcZl+Zc
(1-3)
Where: Rb--- Actual secondary load of CT; Zcj--- Resistance of winding for measuring meters(Ω); Zc--- Contacting resistance which is 0.05~0.1Ω; Zjx--- Resistance of connecting wires (Ω); Kcj--- Impedance conversion coefficient of measuring meters; 2/16
Kmcx--- Impedance conversion coefficient of connecting wires. Resistance of winding for measuring meters (Zcj) shall be calculated from the formula as follows: Zcj= Pl/ I 22
(1-4)
Where: Pl--- Consumed power of measuring meters; I 2 --- Rated secondary current of CT. Impedance conversion coefficient of CT under different wiring modes refers to Tab. 1-4. Tab. 1-4 Impedance Conversio n Coefficient of CT for Measuring Meters Wiring mode Single
Three-phase
of CT
star type
phase
Two-phase star type
two-phase
Triangle
diff. wiring Zcj = Zcj 0
Zcj =0 0
Klc
2
1
√3
√3
2√3
3
Kcj
1
1
√3
1
√3
3
Remarks
Zcj is load resistance 0
in zero line circuit
4.2 SECONDARY LOAD OF CT FOR RELAY PROTECTION Secondary load of CT shall be derived from the formula as follows: Rb= KmcZj+KlcZl+Zc
(1-5)
Where: Rb--- Actual secondary load of CT; Zj--- Resistance of winding for relay(Ω); Zc--- Contacting resistance which is 0.05~0.1Ω; Zl--- Resistance of connecting wires (Ω); Kmc--- Impedance conversion coefficient of relay; Klc--- Impedance conversion coefficient of connecting wires. Tab. 1-5 Impedance Conversion Coefficient of CT for Relay Protection Wiring mode of CT
Three-phase short circuit
Two-phase short circuit
single-phase short circuit
Klc
Kmc
Klc
Kmc
Klc
Kmc
Single phase
2
1
2
1
2
1
Three-phase star type
1
1
1
1
2
1
Two-phase
Zj = Zj
√3
√3
2
2
2
2
star type
Zj = 0
√3
1
2
1
2
1
two-phase diff. wiring
2√3
√3
4
2
Triangle
3
3
3
3
2
2
0
0
3/16
Resistance of relay (Zj ) shall be derived from the formulla as follows: Zj= P/I 2
(1-6)
Where: P--- Consumed active power of relay at the first setting; I--- Operating current of relay at the first setting. Resistance of connecting wires (Zl) shall be obtained from the following formula: Zl= ρL/S
(1-7)
Where: ρ--- Resistance coefficient which is 0.0184Ω•mm²/m for copper cable; L--- Length of cable(m); S--- Sectional cross-area of cable(mm²). When Rb is smaller than Rbn, then CT meets the requirement of secondary load.
5. ACCURACY LIMIT FACTOR FOR PROTECTION CT CLASS P According to the regulation of IEC, the ALF for 5P of CT shall be 5, 10, 15, 20, 30, 40 and more greater value. Where: ALF --- Rated accuracy limit factor;
6. CHECKING COMPUTATIONS FOR CT CLASS P For ensur ing the security and reliability of protection device, the checking computations for CT class P shall be achieved according as the formula below: Esl=ALF× I 2 (Rct+Rbn); (1-8) Es=K×Kpcf× I 2 (Rct+Rb),
Kpcf= Ipal/ IN
(1-9)
Where: Esl --- Rated Secondary limiting E.M.F Es --- Protective checking Secondary E.M.F ALF --- Rated accuracy limit factor; IN--- Rated primary current of CT I 2 --- Rated secondary current of CT; Rct--- secondary winding resistance of CT; K---Specified transient factor; for G-T unit differential protection, K=10; for others protection, K=1 or 2; Kpcf --- Protective checking factor Ipal --- For differential protection, Ipal is Maximum short circuit current flowing through CT while external short circuit; For overcurrent and distance protection, Ipal is maximum short circuit current flowing through CT while internal short circuit. When Es is smaller than Es1, then the choosen parameters of CT CLASS P are acceptable. 4/16
7. CALCULATION EXAMPLE FOR MAIN PROTECTION CT Notes: The data about system short-circuit current please refer to electrical primary short-circuit current calculation reports.
7.1 EXAMPLE 1 CT at Generator Bushing for Gen. differential protection Accuracy Class: 5P40 ALF=40 Ratio=25000/5A Rct= 6.05Ω Rated secondary load= 60VA; Rbn=2.4Ω P=4VA I 2 =5A L=150m S=6mm² Ipal=105.9kA(Maximum through fault current while Gen. terminal short circuit) Secondary load computations: Zj= P/I 2 = 4/5 2 = 0.16Ω Zl= ρL/S= 0.0184x150/6= 0.46Ω Rn= KmcZj+KlcZl+Zc= 1×0.16+1×0.46+0.1= 0.72Ω< Rbn Checking computations: Esl=ALF×In(Rct+Rbn)=40×5×(6.05+2.4)=1690 Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn) =10×105900/25000×5×(6.05+0.72)=1421.7 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
7.2 EXAMPLE 2 CT at Gen. transformer LV side(Generator Bushing and UT Bushing)for Gen. transformer differential protection Accuracy Class: 5P40 ALF=40 Ratio=25000/5A Rct= 6.05Ω Rated secondary load= 60VA; Rbn=2.4Ω 5/16
P=4VA I 2 =5A L=150m(Generator Bushing) or 250m(UT Bushing), herein select L=250m as input S=6mm² Ipal=112.7kA(Maximum through fault current while Gen. terminal short circuit) Secondary load computations: Zj= P/I 2 = 4/5 2 = 0.16Ω Zl= ρL/S= 0.0184x250/6= 0.767Ω Rn= KmcZj+KlcZl+Zc= 1×0.16+1×0.767+0.1= 1.03Ω< Rbn Checking computations: Esl=ALF×In(Rct+Rbn)=40×5×(6.05+2.4)=1690 Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn) =10×112700/25000×5×(6.05+1.03)=1595.8 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
7.3 EXAMPLE 3 CT at 400kV switchyard for Gen. transformer differential protection Accuracy Class: 5P40 ALF=40 Ratio=3000/1A Rct= 12Ω Rated secondary load= 15VA; Rbn=15Ω P=4VA I 2 =1A L=600m S=6mm² Ipal=112.7×22/400=6.2kA(Maximum through fault current while Gen. terminal short circuit) Secondary load computations: Zj= P/I 2 = 4/1 2 = 4Ω Zl= ρL/S= 0.0184x600/6= 1.84Ω Rn= KmcZj+KlcZl+Zc= 1×4+2×1.84+0.1= 7.78Ω< Rbn Checking computations: Esl=ALF×In(Rct+Rbn)=40×1×(12+15)=1080 6/16
Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn)Kpcf=10×6200/3000×1×(12+7.78)=408.8 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
7.4 EXAMPLE 4 TA at Gen. transformer HV side bushing for Gen. transformer complex-voltage lockout overcurrent protection. Accuracy Class: 5P40 ALF=40 Ratio=1250/5A Rct= 0.6Ω Rated secondary load= 60VA; Rbn=2.4Ω P=4VA I 2 =5A L=250m S=6mm² Ipal=3.6kA(Maximum short circuit current flowing through CT while 400kV busbar short circuit) Secondary load computations: Zj= P/I 2 = 4/5 2 = 0.16Ω Zl= ρL/S= 0.0184x250/6= 0.767Ω Rn= KmcZj+KlcZl+Zc= 1×0.16+1×0.767+0.1= 1.03Ω< Rbn Checking computations: Esl=ALF×In(Rct+Rbn)=40×5×(0.6+2.4)=600 Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn) =2×3600/1250×5×(0.6+1.03)=46.9 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
7.5 EXAMPLE 5 CT at unit transformer(UT) HV side bushing for UT differential protection Accuracy Class: 5P20 ALF=20 Ratio=1500/5A 7/16
Rct= 0.8Ω Rated secondary load= 60VA; Rbn=2.4Ω P=4VA I 2 =5A L=250m S=6mm² Ipal=27.9×11/22=13.95kA(Maximum through fault current while 11kV busbar short circuit) Secondary load computations: Zj= P/I 2 = 4/5 2 = 0.16Ω Zl= ρL/S= 0.0184x250/6= 0.767Ω Rn= KmcZj+KlcZl+Zc= 1×0.16+1×0.767+0.1= 1.03Ω< Rbn Checking computations: Esl=ALF×In(Rct+Rbn)=20×5×(0.8+2.4)=320 Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn) =2×13950/1500×5×(0.8+1.03)=170.2 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
7.6 EXAMPLE 6 CT at UT LV side for UT differential protection Accuracy Class: 5P20 ALF=20 Ratio=3000/5A Rct= 2Ω Rated secondary load= 30VA; Rbn=1.2Ω P=4VA I 2 =5A L=150m S=6mm² Ipal=27.9kA(Maximum through fault current while 11kV busbar short circuit) Secondary load computations: Zj= P/I 2 = 4/5 2 = 0.16Ω Zl= ρL/S= 0.0184x150/6= 0.46Ω Rn= KmcZj+KlcZl+Zc= 1×0.16+1×0.46+0.1= 0.72Ω< Rbn 8/16
Checking computations: Esl=ALF×In(Rct+Rbn)=20×5×(2+1.2)=320 Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn) =2×27900/3000×5×(2+0.72)=253 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
7.7 EXAMPLE 7 CT at 400kV switchyard for 400kV busbar differential protection and 400kV outgoing line protection. Accuracy Class: 5P40 ALF=40 Ratio=3000/1A Rct= 12Ω Rated secondary load= 15VA; Rbn=15Ω P=4VA I 2 =1A L=150m S=6mm² Ipal=40kA(which is maybe the maximum short circuit current flowing through CT while 400kV system short circuit.) Secondary load computations: Zj= P/I 2 = 4/1 2 = 4Ω Zl= ρL/S= 0.0184x150/6= 0.46Ω Rn= KmcZj+KlcZl+Zc= 1×4+1×0.46+0.1= 4.56Ω< Rbn Checking computations: Esl=ALF×In(Rct+Rbn)=40×1×(12+15)=1080 Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn) =2×40000/3000×1×(12+4.56)=441.6 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
7.8 EXAMPLE 8 CT at station transformer(ST) HV side bushing for ST differential protection 9/16
Accuracy Class: 5P20 ALF=20 Ratio=300/5A Rct= 0.25Ω Rated secondary load= 60VA; Rbn=2.4Ω P=4VA I 2 =5A L=250m S=6mm² Ipal=27.9×11/400=0.767kA(Maximum through fault current while 11kV busbar short circuit) Secondary load computations: Zj= P/I 2 = 4/5 2 = 0.16Ω Zl= ρL/S= 0.0184x250/6= 0.767Ω Rn= KmcZj+KlcZl+Zc= 1×0.16+1×0.767+0.1= 1.03Ω< Rbn Checking computations: Esl=ALF×In(Rct+Rbn)=20×5×(0.25+2.4)=265 Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn) =2×767/300×5×(0.25+1.03)=32.7 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
7.9 EXAMPLE 9 CT at station transformer(ST) LV side for ST differential protection Accuracy Class: 5P20 ALF=20 Ratio=3000/5A Rct= 2Ω Rated secondary load= 30VA; Rbn=1.2Ω P=4VA I 2 =5A L=150m S=6mm² Ipal=27.9 (Maximum through fault current while 11kV busbar short circuit) Secondary load computations: 10/16
Zj= P/I 2 = 4/5 2 = 0.16Ω Zl= ρL/S= 0.0184x150/6= 0.46Ω Rn= KmcZj+KlcZl+Zc= 1×0.16+1×0.46+0.1= 0.72Ω< Rbn Checking computations: Esl=ALF×In(Rct+Rbn)=20×5×(2+1.2)=320 Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn) =2×27900/3000×5×(2+0.72)=253 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
7.10 EXAMPLE 10 CT at 400kV switchyard for station transformer(ST) HV side T line differential protection Accuracy Class: 5P40 ALF=40 Ratio=3000/1A Rct=12Ω Rated secondary load= 15VA; Rbn=15Ω P=4VA I 2 =1A L=600m S=6mm² Ipal=40kA(which is maybe the maximum short circuit current flowing through CT while 400kV system short circuit.) Secondary load computations: Zj= P/I 2 = 4/1 2 = 4Ω Zl= ρL/S= 0.0184x600/6=1.84Ω Rn= KmcZj+KlcZl+Zc= 1×4+1×1.84+0.1= 5.94Ω< Rbn Checking computations: Esl=ALF×In(Rct+Rbn)=40×1×(8+15)=920 Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn) =2×40000/3000×1×(12+5.94)=478.4 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
11/16
7.11 EXAMPLE 11 CT at startup transformer(ST) HV side bushing for ST differential protection Accuracy Class: 5P20 ALF=20 Ratio=600/1A Rct=3Ω Rated secondary load= 15VA; Rbn=15Ω P=4VA I 2 =1A L=600m S=6mm² Ipal=22.4×11/220=1.12kA(Maximum through fault current while 11kV busbar short circuit) Secondary load computations: Zj= P/I 2 = 4/1 2 = 4Ω Zl= ρL/S= 0.0184x600/6= 1.84Ω Rn= KmcZj+KlcZl+Zc= 1×0.16+1×1.84+0.1= 2.1Ω< Rbn Checking computations: Esl=ALF×In(Rct+Rbn)=20×1×(3+15)=360 Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn) =2×1120/600×1×(3+2.1)=19 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
7.12 EXAMPLE 12 CT at startup transformer(ST) LV side for ST differential protection Accuracy Class: 5P20 ALF=20 Ratio=3000/5A Rct= 2Ω Rated secondary load= 30VA; Rbn=1.2Ω P=4VA I 2 =5A L=150m S=6mm² 12/16
Ipal=22.4 (Maximum through fault current while 11kV busbar short circuit) Secondary load computations: Zj= P/I 2 = 4/5 2 = 0.16Ω Zl= ρL/S= 0.0184x150/6= 0.46Ω Rn= KmcZj+KlcZl+Zc= 1×0.16+1×0.46+0.1= 0.72Ω< Rbn Checking computations: Esl=ALF×In(Rct+Rbn)=20×5×(2+1.2)=320 Es=K×Kpcf×In(Rct+Rn) =K× Ipal/ IN ×In(Rct+Rn) =2×22400/3000×5×(2+0.72)=203.1 Esl＞Es Therefore, CT meets the requirement of checking computations for CT class P.
8. CALCULATION EXAMPLE FOR MAIN MEASUREMENT CT 8.1 EXAMPLE 1 CT at Generator Bushing for me tering/measurement /AVR Accuracy Class: 0.2S(metering)/ 0.2 (measurement) / 0.5 (AVR) Ratio=25000/5A Rated secondary load=60VA; Rbn=2.4Ω Pl=2×6VA(metering)/ 20×0.2 (measurement) / 6 (AVR), herein select Pl=12VA as input I 2 =5A L=150m S=6mm² Zl= ρL/S= 0.0184×150/6= 0.46Ω Zcj= Pl/ I 22 = 12/5 2 = 0.48Ω Rn= KcjZcj+KlcZl+Zc= 1×0.48+1×0.46+0.1= 1.04Ω Rn < Rbn Therefore, CT meets the requirement of secondary load.
8.2 EXAMPLE 2 CT at UT HV side bushing for me tering and measurement Accuracy Class: 0.2 Ratio=1500/5A Rated secondary load=60VA; Rbn=2.4Ω
13/16
Pl=6VA(metering)+5×0.2 (measurement) =7VA I 2 =5A L=250m S=6mm² Zl= ρL/S= 0.0184×250/6= 0.77Ω Zcj= Pl/ I 22 = 7/5 2 = 0.28Ω Rn= KcjZcj+KlcZl+Zc= 1×0.28+1×0.77+0.1= 1.15Ω Rn < Rbn Therefore, TA meets the requirement of secondary load.
8.3 EXAMPLE 3 TA at Gen. transformer HV side bushing for measurement Accuracy Class: 0.2 Ratio=1250/5A Rated secondary load=60VA; Rbn=2.4Ω Pl=10×0.2 (measurement) =2VA I 2 =5A L=250m S=6mm² Zl= ρL/S= 0.0184×250/6= 0.77Ω Zcj= Pl/ I 22 = 2/5 2 = 0.08Ω Rn= KcjZcj+KlcZl+Zc= 1×0.08+1×0.77+0.1= 0.95Ω Rn < Rbn Therefore, CT meets the requirement of secondary load.
8.3 EXAMPLE 3 CT at Gen. transformer HV side bushing for measurement Accuracy Class: 0.2 Ratio=1250/5A Rated secondary load=60VA; Rbn=2.4Ω Pl=10×0.2 (measurement) =2VA I 2 =5A L=250m S=6mm² 14/16
Zl= ρL/S= 0.0184×250/6= 0.77Ω Zcj= Pl/ I 22 = 2/5 2 = 0.08Ω Rn= KcjZcj+KlcZl+Zc= 1×0.08+1×0.77+0.1= 0.95Ω Rn < Rbn Therefore, CT meets the requirement of secondary load.
8.4 EXAMPLE 4 CT at station transformer HV side bushing for metering/measurement Accuracy Class: 0.2S(metering)/ 0.2 (measurement) Ratio=300/5A Rated secondary load=60VA; Rbn=2.4Ω Pl=2×6VA(metering)/ 10×0.2 (measurement) , herein select Pl=12VA as input I 2 =5A L=250m S=6mm² Zl= ρL/S= 0.0184×250/6= 0.77Ω Zcj= Pl/ I 22 = 12/5 2 = 0.48Ω Rn= KcjZcj+KlcZl+Zc= 1×0.48+1×0.77+0.1= 1.35Ω Rn < Rbn Therefore, CT meets the requirement of secondary load.
8.5 EXAMPLE 5 CT at startup transformer HV side for metering/measurement Accuracy Class: 0.2S(metering)/ 0.2 (measurement) Ratio=300/1A Rated secondary load=15VA; Rbn=15Ω Pl=2×6VA(metering)/ 10×0.2 (measurement) , herein select Pl=12VA as input I 2 =1A L=600m S=6mm² Zl= ρL/S= 0.0184×600/6= 1.84Ω Zcj= Pl/ I 22 = 12/5 2 = 0.48Ω Rn= KcjZcj+KlcZl+Zc= 1×0.48+1×1.84+0.1= 2.42Ω
15/16
Rn < Rbn Therefore, CT meets the requirement of secondary load.
16/16
View more...
Comments
