110kv变电所课程设计计算书

第1页共9页110/10kV变电所电气部分设计计算书一、回路数变电所CP3=14MW有55%的重要负荷负荷采用双回路供电，45%非重要负荷采用单回路供电。由最大负荷Pmax=P3=14MW,每回10kV馈线功率为2MW，则重要负荷回路数：85.3214%552P3%55（回）根据对称性，重要负荷回路数为4。因为双回路供电，则为8回。非重要负荷回路数：15.3214%452P3%45（回）根据对称性，非重要负荷回路数选为4所以回路数目：8+4=12（回）根据对称性，选择N=12回。二、主变的选择变电所C最大负荷MVAPKSKSm11.319.0281cosm每台变压器容量MVASSmN67.1811.316.06.0变电所重要负荷MVASS11.179.02855.055.0maximpMVASSN11.171-211.171-nimp选择SN中较大者作为，mS，即MVAS67.18,m，SN=16000kVA过负荷校验：SN/Sm’=16000/18670=0.857，故在负荷曲线高于此值时均为过负荷，即欠负荷系数：8771.016000186706668.067.0221K过负荷系数：0905.11600018670484189.0222K查图可知，对应于K1允许的K2的值比实际算出的K2=1.0905大，所以SN=16000kVA满足正常负荷，所以选择SFZ7-16000/110型号。三、短路电流计算1、电抗标么值的计算为了计算方便选取如下基准值：基准容量：Sd=1000MVA基准电压：Ud（kV）10.5115基准电流：Id（kA）54.995.02系统归算额定电流：110kV侧：kAUSIddNS02.5115310003第2页共9页10kV侧：kAUSIddNS99.545.1031000'3'发电机归算额定电流：110kV侧：kAUPIdNGNG628.08.01153100cos310kV侧：kAUPIdNGNG873.68.05.103100cos'3'电抗标么值计算（归算至110kV侧）：发电机：984.18.0/501000124.0cos/2*1*NGddNGddGGPSXSSXXX变压器：625.24010001005.10100%2*1*NTdkTTSSXXu4.85.1210001005.10100%4*3*NdkTTSSXXu线路：665.01151000224.0220*0ddLUSLXX575.01151000194.0221*1ddLUSLXX817.01151000274.0222*2ddLUSLXX786.01151000264.0223*3ddLUSLXX544.01151000184.0224*4ddLUSLXX2、短路点选择及kI计算等值电路图如下：短路点的选择表格见设计说明书表6.2。第3页共9页（1）求1k点的kI。图中，665.00*1*LXX3045.22625.2984.1211**2*TGXXX967.0544.0423.0)]//([4321****3*LLLLXXXXX911.13045.2967.0665.0967.0665.02*3*1*3*1*4*XXXXXX623.6665.0967.03045.2967.03045.21*3*2*3*2*5*XXXXXX系统分支计算电抗：911.14**XXsj系统供给短路电流标么值：52329.0911.111**2/*'*'sjtstssXIIIkk系统供给短路电流有名值：tktkSSNSssIIkAIII2/6269.202.552329.0'*'电厂分支计算电抗标么值：8279.010008.0/100623.6/cos5**dNGGjSPXX电厂供给短路电流标么值：由tk=3+0.1=3.1s及8.0*GjX查汽轮发电机运算曲线得tk0s21.3s3.1s'*'GI1.31.2951.415有名值：kAIIINGGG8164.0628.03.1'*'kAIIINGGGtktk8133.0628.0295.12/2/*kAIIINGGGtktk8886.0628.0415.1*总的短路电流：kAIIIsG4433.36269.28164.0''''kAIIItktktkSG4402.36269.28133.02/2/2/第4页共9页kAIIItktktkSG5155.36269.28886.0热效应：skAIIItQtktkkk222222/228292.36)5155.34402.3104433.3(121.3)10(12短路冲击电流：kAIish7804.84433.355.255.2（2）求2k点的kI。图中，665.00*1*LXX3045.22625.2984.1211**2*TGXXX695.0272.0423.0)]//([2/4321****3*LLLLXXXXX561.13045.2695.0665.0695.0665.02*3*1*3*1*4*XXXXXX408.5665.0695.03045.2695.03045.21*3*2*3*2*5*XXXXXX系统分支计算电抗：561.14**XXsj系统供给短路电流标么值：641.0561.111**2/*'*'sjtstssXIIIkk系统供给短路电流有名值：tktkSSNSssIIkAIII2/2178.302.5641.0'*'电厂分支计算电抗标么值：676.010008.0/100408.5/cos5**dNGGjSPXX电厂供给短路电流标么值：由tk=2.5+0.1=2.6s及68.0*GjX查汽轮发电机运算曲线得tk0s26.2s2.6s'*'GI1.511.4951.65有名值：kAIIINGGG9483.0628.051.1'*'第5页共9页kAIIINGGGtktk9389.0628.0495.12/2/*kAIIINGGGtktk0362.1628.065.1*总的短路电流：kAIIIsG1661.42178.39483.0''''kAIIItktktkSG1567.42178.39389.02/2/2/kAIIItktktkSG254.42178.30362.1热效应：skAIIItQtktkkk222222/221175.45)254.41567.4101661.4(126.2)10(12短路冲击电流：kAIish6236.101661.455.255.2（3）求3k点的kI。图中，665.00*1*LXX3045.22625.2984.1211**2*TGXXX095.94.8272.0423.0//)]//([444321******3*TLLLLLXXXXXXX3845.123045.2095.9665.0095.9665.02*3*1*3*1*4*XXXXXX9174.42665.0095.93045.2095.93045.21*3*2*3*2*5*XXXXXX系统分支计算电抗：3845.124**XXsj系统供给短路电流标么值：0807.03845.1211**2/*'*'sjtstssXIIIkk系统供给短路电流有名值：tktkSSNSssIIkAIII2/4373.499.540807.0'*'电厂分支计算电抗标么值：3647.510008.0/1009174.42/cos5**dNGGjSPXX电厂供给短路电流标么值：由33647.5*GjX可得：第6页共9页1864.03647.511***'*'2/GjGGGXIIItktk有名值：kAIIItktkGGG2812.1873.61864.02/总的短路电流：kAIIItktk7185.54373.42812.12/短路冲击电流：kAIish5822.147185.555.255.2（4）求4k点的kI。图中，665.00*1*LXX3045.22625.2984.1211**2*TGXXX895.42.4272.0423.0////)]//([4344321*******3*TTLLLLLXXXXXXXX9725.63045.2895.4665.0895.4665.02*3*1*3*1*4*XXXXXX1627.24665.0895.43045.2895.43045.21*3*2*3*2*5*XXXXXX系统分支计算电抗：9725.64**XXsj系统供给短路电流标么值：1434.09725.611**2/*'*'sjtstssXIIIkk系统供给短路电流有名值：tktkSSNSssIIkAIII2/8856.799.541434.0'*'电厂分支计算电抗标么值：0203.310008.0/1001627.24/cos5**dNGGjSPXX电厂供给短路电流标么值：由30203.3*GjX可得：3311.00203.311***'*'2/GjGGGXIIItktk有名值：kAIIItktkGGG2756.2873.63311.02/总的短路电流：kAIIItktk1612.102756.28856.72/短路冲击电流：kAIish9111.251612.1055.255.2第7页共9页3、短路电流计算结果表（详见设计说明书）四、断路器的选择1、各回路的最大持续工作电流axmI的计算（1）110kV进线上的QF、QS应考虑当L2断开时L1线路运行时通过的最大功率AUSSSNCBAax91.3491011039.0149.0189.028103I33m（2）变电所C高压侧，即主变110kV侧，考虑变压器过负荷1.3倍AUSNNTax17.109101103163.11033.1I333m（3）变电所C主变低压侧10kV，考虑变压器过负荷1.3倍AUSNNTax89.120010103163.11033.1I333m（4）10kV主母线、分段，按1.05倍变压器额定电流考虑AUSNNTax95.969101031605.110305.1I333m（5）10kV出线电器（计算双回路供电的重要用户，每回按2MW考虑）考虑2倍一回线的正常最大负荷电流AUPNax6.2569.010320002cos32Im2、修正系数k的计算对于电器：40＜θ≤60℃时，k=1—(θ—40)×0.0180≤θ≤40℃时，k=1+(40—θ)×0.005θ≤0℃时，k=1.2所以屋外：k1=1—(40—40)×0.018=1屋内：k2=1+(40—37)×0.005=1.0153、110kV进线上的断路器选择（屋外，k=1）①额定电压的选择UN≥UNs=110kV，所以UN=110kV≥110kV②额定电流的选择IN≥Imax=349.91A，所以IN=2000A349.91A③额定开断电流的选择INb