印尼海德堡保护整定计算说明书Word文档格式.docx

1、YJV-26/35kV 2x3x（1x240），电缆长度为67m.2#、3#主变高压侧电缆阻抗参数一样，长度分别为52m和56m。2.4 主变压器阻抗主变型号为SFZ11-45000/33，40MVA（ONAN）/45MVA（ONAF），338x1.25%/11kV，Dyn11阻抗为：Ud%=10.5%2.5 主变电压电缆阻抗1#主变低压电缆型号为：YJV-8.7/15kV，4x3x（1x400），长度40m.2#、3#主变高压侧电缆阻抗参数一样，长度分别为26m和56m。2.6 N-0535-MAIN SUBSTATION至下级变电站电缆阻抗2.6.1 N-0535-MAIN SUBSTAT

2、ION至N-3030-Substation 电缆型号为：YJV-8.7/15kV 2x（3x185），长度： 350m2.6.2 N-0535-MAIN SUBSTATION至N-4022-Substation YJV-8.7/15kV 1x（3x185），长度： 420m2.6.3 N-0535-MAIN SUBSTATION至N-4031-Substation YJV-8.7/15kV 2x（3x120），长度： 510m2.6.4 N-0535-MAIN SUBSTATION至N-5051-Substation YJV-8.7/15kV 3x95，长度： 90m2.6.5 N-0535-

3、MAIN SUBSTATION至N-0661-Substation YJV-8.7/15kV 2x（3x95），长度：250m2.6.6 N-0535-MAIN SUBSTATION至N-5060-Substation 760m2.6.7 N-0535-MAIN SUBSTATION至N-6020-Substation YJV-8.7/15kV 3x185，长度：2.6.8 N-0535-MAIN SUBSTATION至N-2020-Substation 1100m2.6.9 N-0535-MAIN SUBSTATION至N- 3021-Substation YJV-8.7/15kV 3x15

4、0，长度：2.6.10 SUBSTATION中变压器电缆以一个为例，其他类似。N-3030-Substation RAW MILL T.R.YJV-8.7/15kV，3x95，长度50m2.6.11 SUBSTATION中电动机电缆N-3030-Substation RAW MILLYJV-8.7/15kV，3x120，长度170m3 保护整定计算whole substation 11kV zero sequence current TABCable type395120150185Cable length550912942507650A/kM 11.11.31.4Sub-total5.509

5、1.42340.32510.71sum18A3.1 35kV1#、2#INCOMING3.1.1 Basic parametersPhase CT ratioZero sequence CT ratioGap zero sequence CT ratioProtection relay type1500/1noneMICOM P143+P5433.1.2 Protection configuration and calculate rule protection configure time limit over-current，time delay over current, low vol

6、tage protection and optical differential protection。3.1.2.1 联络线差动保护定值整定 灵敏启动值Is1 According to avoid the maximum unbalance current setting under the condition of maximum load：按躲过最大负荷时最大不平衡电流整定。最大不平衡电流计算公式为：Ium1= Krel* （KapKccKer +m）*Imax 式中：Krel-可靠系数，取1.52.0Kap-为非周期分量系数，此处选1.0。Kcc-为互感器同型系数，一般选1。 Ker-

7、电流互感器的比误差，取0.1。Imax:最大负荷电流，按线路CT额定电流考虑。m-由于电流互感器变比未完全匹配产生的误差，初设时取0.05。Iunb.1= Krel* （KapKccKer +m）* Imax =1.5*（0.1+0.05）*In=0.225In,考虑1.5的可靠系数，Is2=1.5* Iunb.1=1.5*0. 225In=0.3375， setting for Is1=0.4In拐点制动电流整定值Is2按以下原理整定：1. 最大运行方式下的最大负荷。2. 保证基本无CT暂态误差的最大电流。进线CT等级为5P20，电流在20In以内误差在5%以内，综合考虑按2In整定，设置I

8、s2=2In斜率K1:在拐点制动电流下差动保护动作电流为：Id1=Krel* （KapKccKer +m）* Is2=1.5*0.15*2In=0.45In按上图计算公式为：K1=（ Id1- Is1）/ Is2=（0.45-0.4）/2=0.25，按0.3整定。斜率K2:According to avoid the maximum unbalance current setting under the condition of maximum outside short circuit：按躲过外部最大故障电流时最大不平衡电流整定。Ium2= Krel* （KapKccKer+m）*Ik.ma

9、x外部短路最大短路电流如下：Iunb.2=Krel*（ KapKccKer +m）*Ik/1.5=1.5*（0.1+0.05）*27/1.5=4.05In，考虑1.5可靠系数，在最大外部短路电流下差动保护动作电流为：Id2=1.5*4.05In=6.07InK2=（ Id2- Id1）/ （Iunb.2-Is2）=（6.07-0.45）/（27/1.5-2）=0.35，按0.5整定。High Set应与上面Id2值（6.07 In）接近，设置6 In。在P543保护装置中具体整定清单为： Current diff? Yes Is1 0.3In Is2 2In K1 30% K2 50% pha

10、se char DT phase time delay 0s Idiff Curve IEC SI phase TMS 1 phase Time Dial 1 PIT I Disable? No DIT Time 0.1s High Set 6In3.1.2.2 Time limit over-current进线在最大运行方式下电缆末端3 相短路的最大电流如下。进线在最大运行方式下电缆末端3 相短路的最大电流为23.4kA. 取灵敏系数为1.2，则电流速断保护整定值为1.2*23.4=28kA. 手动验证计算如下（标幺值）：系统最大阻抗为：Zs.max=100/1806=0.05537进线电缆

11、阻抗为：Zcab=/10.89=0.0128总阻抗为：Zm= Zs.max+2* Zcab=0.081短路电流为：Ik= （1/Zm）* 1.75=21.6kA,由于软件设置了一个助增系数为1.1，见下图：所以软件计算出短路电流为21.6kA*1.1=23.7kA，与软件仿真结果基本一致，所以以下不再手动计算短路电流，以软件计算为准。进线在最小运行方式下电缆首端的15%三相短路的短路电流如下。进线在最小运行方式下电缆首端的15%三相短路的短路电流17.5kA,两相短路电流为0.866*17.5=15kA，小于28kA, 故电流速断的有效保护范围0.0828Un）,考虑到主变高压母线及电缆故障引

12、起扩大故障，速断保护延时0.2s动作跳闸。time delay direction over current：According to the maximum load current, due to a single line can be run with three main transformer load, according to twice the rated current setting, action time cooperate with the superior over-current protection, and consider direction lockou

13、t, direction toward line. calculate：2In=2*1500A=3kA，delay time 0.5s。灵敏度校验： 按过流保护按线路末端两相短路来校验，在最小运行方式下，线路末端两相短路时，其0.866*16.17=14kA.保护灵敏度为：14/3=4.66，完全满足要求。low voltage protection： According to the maximum operating mode the three-phase short-circuit at the low voltage side of main transformer, high vo

14、ltage side busbar voltage value to calculate, consider 1.3 safety factor, action time 0.5 s。76%/1.3=58%Un，setting for 0.55 Un。3.1.3 Summary protection settingTime limit over-currentTime relay direction over-currentLow voltage17kA/0.5 Un /0.2s3kA /0.5s0.55 Un/0.5s3.2 35kV1#、2#、3#TRANSFORMER3.2.1 Basi

15、c parametersTransformer capacity（KVA）Transformer short-circuit impedanceThe transformer rated current （primary）Gap Zero sequence CT ratioProtection type4500010.5%787A100/133/1MICOM P632+P1433.2.2 Protection configuration and calculate rule Protection configure Compound voltage over current protectio

16、n （including low voltage and negative sequence voltage setting）, overload protection, differential protection （including the percentage differential pickup value, slope ratio, base point and Unrestrained differential protection） and zero-sequence differential protectionCompound voltage lockout over

17、current protection Considering protection cooperate relationship, compound voltage over current protection, in order to improve the protection sensitivity, action value according to maximum load current setting, consider 1.1 times of rated current, calculate: 1.1 * 45000/1.732/787 = 787A （primary va

18、lue）, 0.9 s trip bus coupler, 1.1 s trip low voltage side of the breaker. Low voltage setting Under main transformer low voltage side of minimum operation mode, the bus voltage is 23.98 KV, consider 0.9 safety factor, the calculation for : 23.98/0.9/33=0.8, according to 0.7Un setting. Negative seque

19、nce voltage setting According to avoid the biggest negative sequence voltage under the unbalanced load setting。 Normal running negative sequence voltage is not more than 2% of rated voltage, generally the whole 6 v。Overload protection Overload, in accordance with 1.05 times rated current transformer

20、 setting, action time 9 s, alarm.ratio different protection pickup value According to avoid flow the maximum unbalance current setting under the condition of maximum load：Iunb.max=Krel*（Ker+U+m）*Ie=1.5*（0.06+0.1+0.05）*Ie=0.112Ie, Generally setting for 0.2 ratio。The differential protection restraint

21、slope ratio： Largest busbar three-phase short-circuit current 6.86 kA, produced by the unbalance current Iunb.max = （Kap*Kcc*Ker+U+m）*Ik（3）=（2*0.5*0.1+0.05+0.05）*6.86kA=686A/787A=0.87Ie, The differential protection operating current is：Krel*Iunb.max=1.5*0.87Ie=1.3Ie,slope ratio：（1.3-0.2）/（6860/787-1

22、）=0.14,general setting for 0.3.different protection base point： base point according to max load current calculate，general setting for 1.5 the rated current。Unrestrained different protection： According to the minimum operation mode ,main transformer low voltage side two phase short circuit current s

23、etting calculation, reliable coefficient of 1.3, then 4700*0.866/1.3=3130A=4.0Ie.Zero-sequence differential protection According to avoid the external unbalance current of single-phase grounding fault setting :Iop=Krel*（fi+m）*Ik（3）=1.5*（0.1+0.05）*6860/787=0.29Ie,setting for 0.3Ie，slope ratio 0.5，bas

24、e point :1Ie，Unrestrained different protection :4Ie.3.2.3 1.2.3. Summary protection settingTime relay over-current with compound voltage lockoutOver loadDifferent pickupSlope rateBase pointUnrestrained different protection866A/0.7 Un /6V/0.5s1.05Ie，9s,signal0.2In0.31.5In7InPickup value: 0.3Ie,slope

25、ratio:0.5,base point: 1Ie, Unrestrained different protection :4Ie3.3 35kV1#、2#COUPLING3.3.1 base parametersMICOM P1433.3.2 Protection configuration and calculate ruleprotection configuration：Charging protection, over current protectionCharging protection According to the two phase short circuit curr

26、ent calculation under minimal busbar operation mode, reliable coefficient of 1.3, 1 s exit after charging, setting calculation is: 15.14/1.3 = 11.64 kA, setting for 11 KAOver current protection The over-current protection is backup for transformer differential protection after charging over and according to the two phase short