印尼海德堡保护整定计算说明书Word文档格式.docx

上传人:b****7 文档编号:22408342 上传时间:2023-02-03 格式:DOCX 页数:96 大小:337.53KB
下载 相关 举报
印尼海德堡保护整定计算说明书Word文档格式.docx_第1页
第1页 / 共96页
印尼海德堡保护整定计算说明书Word文档格式.docx_第2页
第2页 / 共96页
印尼海德堡保护整定计算说明书Word文档格式.docx_第3页
第3页 / 共96页
印尼海德堡保护整定计算说明书Word文档格式.docx_第4页
第4页 / 共96页
印尼海德堡保护整定计算说明书Word文档格式.docx_第5页
第5页 / 共96页
点击查看更多>>
下载资源
资源描述

印尼海德堡保护整定计算说明书Word文档格式.docx

《印尼海德堡保护整定计算说明书Word文档格式.docx》由会员分享,可在线阅读,更多相关《印尼海德堡保护整定计算说明书Word文档格式.docx(96页珍藏版)》请在冰豆网上搜索。

印尼海德堡保护整定计算说明书Word文档格式.docx

YJV-26/35kV2x3x(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),33±

8x1.25%/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。

2.6N-0535-MAINSUBSTATION至下级变电站电缆阻抗

2.6.1N-0535-MAINSUBSTATION至N-3030-Substation

电缆型号为:

YJV-8.7/15kV2x(3x185),长度:

350m

2.6.2N-0535-MAINSUBSTATION至N-4022-Substation

YJV-8.7/15kV1x(3x185),长度:

420m

2.6.3N-0535-MAINSUBSTATION至N-4031-Substation

YJV-8.7/15kV2x(3x120),长度:

510m

2.6.4N-0535-MAINSUBSTATION至N-5051-Substation

YJV-8.7/15kV3x95,长度:

90m

2.6.5N-0535-MAINSUBSTATION至N-0661-Substation

YJV-8.7/15kV2x(3x95),长度:

250m

2.6.6N-0535-MAINSUBSTATION至N-5060-Substation

760m

2.6.7N-0535-MAINSUBSTATION至N-6020-Substation

YJV-8.7/15kV3x185,长度:

2.6.8N-0535-MAINSUBSTATION至N-2020-Substation

1100m

2.6.9N-0535-MAINSUBSTATION至N-3021-Substation

YJV-8.7/15kV3x150,长度:

2.6.10SUBSTATION中变压器电缆

以一个为例,其他类似。

N-3030-SubstationRAWMILLT.R.

YJV-8.7/15kV,3x95,长度50m

2.6.11SUBSTATION中电动机电缆

N-3030-SubstationRAWMILL

YJV-8.7/15kV,3x120,长度170m

3保护整定计算

wholesubstation11kVzerosequencecurrentTAB

Cabletype

95

120

150

185

Cablelength

5509

1294

250

7650

A/kM

1

1.1

1.3

1.4

Sub-total

5.509

1.4234

0.325

10.71

sum

18A

3.135kV1#、2#INCOMING

3.1.1Basicparameters

PhaseCTratio

ZerosequenceCTratio

GapzerosequenceCTratio

Protectionrelaytype

1500/1

 none

MICOMP143+P543

3.1.2Protectionconfigurationandcalculaterule

protectionconfiguretimelimitover-current,timedelayovercurrent,lowvoltageprotectionandopticaldifferentialprotection。

3.1.2.1联络线差动保护定值整定

灵敏启动值Is1

Accordingtoavoidthemaximumunbalancecurrentsettingundertheconditionofmaximumload:

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1.5~2.0

Kap---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---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,settingforIs1=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Is2=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:

Accordingtoavoidthemaximumunbalancecurrentsettingundertheconditionofmaximumoutsideshortcircuit:

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.max

外部短路最大短路电流如下:

Iunb.2=Krel*(KapKccKer+Δm)*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=6.07In

K2=(Id2-Id1)/(Iunb.2-Is2)=(6.07-0.45)/(27/1.5-2)=0.35,按0.5整定。

HighSet

应与上面Id2值(6.07In)接近,设置6In。

在P543保护装置中具体整定清单为:

Currentdiff?

Yes

Is10.3In

Is22In

K130%

K250%

phasecharDT

phasetimedelay0s

IdiffCurveIECSI

phaseTMS1

phaseTimeDial1

PITIDisable?

No

DITTime0.1s

HighSet6In

3.1.2.2Timelimitover-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<

15%,应改用电流闭锁电压速断保护。

电流闭锁电压速断按正常运行时电缆的75%处三相短路电流来整定,正常运行最小运行方式考虑,仿真计算如下.

流经电缆故障电流为16.55kA,母线电压为8.28%。

所以速断保护整定为:

17kA,闭锁电压整定为0.5Un(>

0.0828Un),考虑到主变高压母线及电缆故障引起扩大故障,速断保护延时0.2s动作跳闸。

timedelaydirectionovercurrent:

Accordingtothemaximumloadcurrent,duetoasinglelinecanberunwiththreemaintransformerload,accordingtotwicetheratedcurrentsetting,actiontimecooperatewiththesuperiorover-currentprotection,andconsiderdirectionlockout,directiontowardline.

calculate:

In=2*1500A=3kA,delaytime0.5s。

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。

lowvoltageprotection:

Accordingtothemaximumoperatingmodethethree-phaseshort-circuitatthelowvoltagesideofmaintransformer,highvoltagesidebusbarvoltagevaluetocalculate,consider1.3safetyfactor,actiontime0.5s。

76%/1.3=58%Un,settingfor0.55Un。

3.1.3Summaryprotectionsetting

Timelimitover-current

Timerelaydirectionover-current

Lowvoltage

17kA/0.5Un/0.2s

3kA/0.5s

0.55Un/0.5s

3.235kV1#、2#、3#TRANSFORMER

3.2.1Basicparameters

Transformercapacity(KVA)

Transformershort-circuitimpedance

Thetransformerratedcurrent(primary)

GapZerosequenceCTratio

Protectiontype

45000

10.5%

787A

100/1

33/1

MICOMP632+P143

3.2.2Protectionconfigurationandcalculaterule

ProtectionconfigureCompoundvoltageovercurrentprotection(includinglowvoltageandnegativesequencevoltagesetting),overloadprotection,differentialprotection(includingthepercentagedifferentialpickupvalue,sloperatio,basepointandUnrestraineddifferentialprotection)andzero-sequencedifferentialprotection

Compoundvoltagelockoutovercurrentprotection

Consideringprotectioncooperaterelationship,compoundvoltageovercurrentprotection,inordertoimprovetheprotectionsensitivity,actionvalueaccordingtomaximumloadcurrentsetting,consider1.1timesofratedcurrent,calculate:

1.1*45000/1.732/787=787A(primaryvalue),0.9stripbuscoupler,1.1striplowvoltagesideofthebreaker.

Lowvoltagesetting

Undermaintransformerlowvoltagesideofminimumoperationmode,thebusvoltageis23.98KV,consider0.9safetyfactor,thecalculationfor:

23.98/0.9/33=0.8,accordingto0.7Unsetting.

Negativesequencevoltagesetting

Accordingtoavoidthebiggestnegativesequencevoltageundertheunbalancedloadsetting。

Normalrunningnegativesequencevoltageisnotmorethan2%ofratedvoltage,generallythewhole6v。

Overloadprotection

Overload,inaccordancewith1.05timesratedcurrenttransformersetting,actiontime9s,alarm.

ratiodifferentprotectionpickupvalue

Accordingtoavoidflowthemaximumunbalancecurrentsettingundertheconditionofmaximumload:

Iunb.max=Krel*(Ker+ΔU+Δm)*Ie=1.5*(0.06+0.1+0.05)*Ie=0.112Ie,Generallysettingfor0.2ratio。

Thedifferentialprotectionrestraintsloperatio:

Largestbusbarthree-phaseshort-circuitcurrent6.86kA,producedbytheunbalancecurrentIunb.max=(Kap*Kcc*Ker+ΔU+Δm)*Ik(3)

=(2*0.5*0.1+0.05+0.05)*6.86kA=686A/787A=0.87Ie,Thedifferentialprotectionoperatingcurrentis:

Krel*Iunb.max=1.5*0.87Ie=1.3Ie,sloperatio:

(1.3-0.2)/(6860/787-1)=0.14,generalsettingfor0.3.

differentprotectionbasepoint:

basepointaccordingtomaxloadcurrentcalculate,generalsettingfor1.5theratedcurrent。

Unrestraineddifferentprotection:

Accordingtotheminimumoperationmode,maintransformerlowvoltagesidetwophaseshortcircuitcurrentsettingcalculation,reliablecoefficientof1.3,then4700*0.866/1.3=3130A=4.0Ie.

Zero-sequencedifferentialprotection

Accordingtoavoidtheexternalunbalancecurrentofsingle-phasegroundingfaultsetting:

Iop=Krel*(fi+Δm)*Ik(3)=1.5*(0.1+0.05)*6860/787=0.29Ie,settingfor0.3Ie,sloperatio0.5,basepoint:

1Ie,Unrestraineddifferentprotection:

4Ie.

3.2.31.2.3.Summaryprotectionsetting

Timerelayover-currentwithcompoundvoltagelockout

Overload

Differentpickup

Sloperate

Basepoint

Unrestraineddifferentprotection

866A/0.7Un/6V/0.5s

1.05Ie,9s,signal

0.2In

0.3

1.5In

7In

Pickupvalue:

0.3Ie,sloperatio:

0.5,basepoint:

1Ie,Unrestraineddifferentprotection:

4Ie

3.335kV1#、2#COUPLING

3.3.1baseparameters

MICOMP143

3.3.2Protectionconfigurationandcalculaterule

protectionconfiguration:

Chargingprotection,overcurrentprotection

Chargingprotection

Accordingtothetwophaseshortcircuitcurrentcalculationunderminimalbusbaroperationmode,reliablecoefficientof1.3,1sexitaftercharging,settingcalculationis:

15.14/1.3=11.64kA,settingfor11KA

Overcurrentprotection

Theover-currentprotectionisbackupfortransformerdifferentialprotectionafterchargingoverandaccordingtothetwophaseshort

展开阅读全文
相关资源
猜你喜欢
相关搜索

当前位置:首页 > 小学教育 > 数学

copyright@ 2008-2022 冰豆网网站版权所有

经营许可证编号:鄂ICP备2022015515号-1