毕业设计论文：某电机制造厂总降压变电所及高压配电

1、毕业设计论文：某电机制造厂总降压变电所及高压配电 毕业设计(论文) 题 目 总降压变电所-XXXXXX毕业设计 教学点 专 业 年 级 姓 名 指导教师 定稿日期： 2021 年 6月 1 日 I摘 要为使工厂供电工作很好地为工业生产效劳，切实保证工厂生产和生活用电的需要，并做好节能工作，本设计在大量收集资料，并对原始资料进行分析后，做出35kV变电所及变电系统电气局部的选择和设计，使其到达以下根本要求：1、平安 在电能的供给、分配和使用中，不发生人身事故和设备事故。2、可靠 满足电能用户对供电可靠性的要求。3、优质 满足电能用户对电压和频率等质量的要求4、经济 供电系统的投资少，运行费用低，

2、并尽可能地节约电能和减少有色金属的消耗量。 此外，在供电工作中，又合理地处理局部和全局、当前和长远等关系，既照顾局部的当前的利益，又要有全局观点，顾全大局，适应开展。按照国家标准GB50052-95 ?供配电系统设计标准?、GB50059-92 ?35110kV变电所设计标准?、GB50054-95 ?低压配电设计标准?等的规定，工厂供电设计遵循以下原那么：1、遵守规程、执行政策；遵守国家的有关规定及标准，执行国家的有关方针政策，包括节约能源，节约有色金属等技术经济政策。2、平安可靠、先进合理；做到保障人身和设备的平安，供电可靠，电能质量合格，技术先进和经济合理，采用效率高、能耗低和性能先进电

3、气产品。3、近期为主、考虑开展；根据工作特点、规模和开展规划，正确处理近期建设与远期开展的关系，做到远近结合，适当考虑扩建的可能性。4、全局出发、统筹兼顾。按负荷性质、用电容量、工程特点和地区供电条件等，合理确定设计方案。工厂供电设计是整个工厂设计中的重要组成局部。工厂供电设计的质量直接影响到工厂的生产及开展。 关键词：节能、配电、平安、合理、开展II目 录摘要 ···················

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127、183;············································· 28 致谢 ···&

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138、····································· 33IV1绪论1.1设计题目某电机制造厂总降压变电所及高压配电系统设计1.2设计依据1.2.1工厂总平面布置图略1.2.2全厂各车间负荷情况汇总表。11.

139、2.3供用电协议。1当地供电部门可提供两种电源：从某220/35KV区域变电站提供电源，该站距离厂南5公里；从某35/10KV变电所，提供10KV备用电源，该所距离厂南5公里。2配电系统技术数据。1区域变电站35KV母线短路数据为：2配电系统错误！未找到引用源。 3供电部门对工厂提出的技术要求：区域变电站35KV馈电线路定时限过流保护装置的整定时间为1.8秒，要求厂总降压变电所的保护动作时间不大于1.3秒。工厂在总降压变电所35KV侧计量。功率因素值应在0.9以上。1.2.4工厂的负荷性质本工厂大局部车间为一班制，少数车间为两班或三班制，年最大负荷利用小时数为2500小时。锅炉房提供高压蒸汽，

140、停电会使锅炉发生危险。由于距离市区较远，消防用水需要厂方自备。因此，锅炉房要求较高的可靠性。21.2.5工厂的自然条件1年最高气温为40，年最低气温5，年平均气温为10。2站所选地址地质以粘土为主，地下水位3-5米。3风向以东南风为主。 1.3设计任务及设计大纲1.3.1高压供电系统设计根据供电部门提供的资料，选择本厂最优供电电压等级。1.3.2总变电所设计1主结线设计：根据设计任务书，分析原始资料与数据，列出技术上可能实现的多个方案，经过概略分析比拟，留下2-3个较优方案进行详细计算和分析比拟经济计算分析时，设备价格、使用综合投资指标，确定最优方案。2短路电流计算：根据电气设备选择和继电保护

141、的需要，确定短路计算点，计算三相短路电流，计算结果列出汇总表。3主要电气设备选择：主要电气设备的选择，包括断路器、隔离开关、互感器、导线截面和型号、绝缘子等设备的选择及效验。选用设备型号、数量、汇总设备一览表。4主要设备继电保护设计：包括主变压器、线路等元件的保护方式选择和整定计算。5配电装置设计：包括配电装置形式的选择、设备布置图。6防雷、接地设计：包括直击雷保护、进行波保护和接地网设计。 31.4设计成果1.4.1设计说明书包括对各种设计方案分析比拟的扼要说明，并附有必要的计算及表格。1.4.2设计图纸1降压变电所电气主结线图。2变电所平面布置图3主变压器保护原理接线图。42供电电压等级选

142、择2.1电源电压等级选择根据供电协议可知，当地供电部门可提供电源为两种：主电源为厂南的某220/35KV区域变电站提供，10kV备用电源为厂南的某35/10KV变电所提供，因此可以考虑本厂总降压变电所主电源采用35kV电压等级，经过变压后采用10kV输送至各个车间变电房降压至0.4kV直供负荷。同时采用10kV作为保安电源，为锅炉房等一级负荷提供备用电源。3全厂负荷计算3.1变电所的负荷计算 3.1.1用电设备的负荷计算根据设计任务书的要求，按照需要系数法及以下计算公式 错误！未找到引用源。错误！未找到引用源。错误！未找到引用源。 错误！未找到引用源。得各项数据列表如下下表数据均为35kV侧：

143、 53.1.2变压器损耗估算Pb=1%Sj=0.01×10527.37=105.27kw Qb=5%Sj=0.05×10527.37=526.37kvar 63.1.3无功功率补偿计算从设计任务书的要求可知，工厂35kV高压侧进线在最大负荷时，其功率因素不应小于0.9，考虑到变压器的无功功率损耗Qb，远远大于有功功率损耗Pb，因此，在变压器的10kV侧进行无功功率补偿时，其补偿后的功率因素应稍大于0.9，现设cos=0.95，那么10kV侧在补偿前的功率因素为:错误！未找到引用源。 错误！未找到引用源。错误！未找到引用源。 错误！未找到引用源。错误！未找到引用源。因此，所需

144、要的补偿容量为：错误！未找到引用源。错误！未找到引用源。错误！未找到引用源。选取错误！未找到引用源。35kV侧在补偿后的负荷及功率因素计算：错误！未找到引用源。错误！未找到引用源。错误！未找到引用源。错误！未找到引用源。错误！未找到引用源。 错误！未找到引用源。错误！未找到引用源。错误！未找到引用源。满足了设计任务书的要求，其计算数据如下：7根据设计任务书的要求以及以上计算结果，选取：并联补偿电容为 BWF10.5-100-1型电容器50只。补偿总容量为 100kvar×50=5000kvar。3.1.4变压器选择根据补偿后的总计算负荷8425kVA，同时考虑工厂5-10年的负荷增长

145、，变压器容量考虑一定的预留，本期工厂负荷能保证变压器运行在60-70%经济负荷区内即可，因此选择型号为：SFZ7-10000-35±3*2.5%/10.5kV YN,d11的变压器。 84系统主接线方案的选择 4.1方案1：单回路高压线路变压器组、低压单母线分段主接线 4.2方案2：双回路高压线路变压器组、低压单母线分段主接线 4.3方案的比拟与选择根据设计任务书的要求，本厂根本负荷为一班制，少数负荷为两班或三班制，属于二级负荷；同时锅炉房供电可靠性要求高，属于一级负荷。主接线的设计必须满足工厂电气设备的上述要求，因此：方案1：该方案35kV侧为单回路线路-变压器组接线、10kV单母

146、线，与10kV备用电源通过母联连接，正常运行时母联合闸，由主电源供给锅炉房；当主电源故障或主变等设备停电检修退出运行时，母联分闸，由10kV备用电源直供锅炉房及其他重要负荷。由于本厂根本负荷为二级负荷，对供电可靠性要求不高，采用单回路进线和1台主变根本可满足对二级负荷供电的要求，对于锅炉房等重要负荷采用10kV备用电源作为备用，以保证工厂的重要用电设备不会出现长时间断电，即在任何时候都能满足对二级负荷的供电要求。方案2：该方案35kV侧采用从220/35kV变电站出双回路电源、高压线路变压器组接线、10kV侧为单母线分段接线。方案2的特点就是采用双电源、可靠性高。其缺点就是设备投资大、运行维护

147、费用高，同时本厂最大负荷利用小时仅为2600小时，相对来说，变压器的利用率低，2台主变的空载损耗将大大超过1台主变的选择。选择结果：从上述分析可知，方案1能满足供电要求，同时设备投资、运行维 9护费用和占地面积、建筑费用等方面均由于方案2，技术和经济的综合指标最优，因此，在本设计中，选用方案1作为本设计的主接线方案。方案详细的图纸见?35/10kV降压变电所电气主接线图?。 105变电所位置及变压器、配电装置选择5.1变电所位置根据变电所选址原那么：a.变电所尽量选择在负荷中心，可减少低压损耗；b.便于维修；c.便于进出线；d.节约费用；e.便于运行平安的原那么，将35/10kV总降压变电所设

148、置在木工车间后侧。具体位置见附件3?工厂总平面布置图?。5.2变压器选择根据设计方案的选择结果，本期只设计1台主变压器即可满足需要，因此变压器选择结果不变，即为：额定容量 35：电压等级5.3所用变压器选择为保证变电所正常运行，需要设置所用变压器。根据常规，本所所用变压器可以选择为：SC9-30/10 10±5%/0.4kV Y,y11 阻抗电压4%，布置在10kV柜内。5.3配电装置选择根据供电电压等级选择的结果：进线电源采用35kV，经过变压器降为10kV供给各车间配电所，从经济性和运行维护等方面考虑，35kV配电装置采用户外常规布置，10kV采用户内配电装置。116短路电流计算

149、6.1确定计算电路及计算电抗6.1.1计算电路图 错误！未找到引用源。 错误！未找到引用源。 错误！未找到引用源。 错误！未找到引用源。 错误！未找到引用源。错误！未找到引用源。 设基准容量错误！未找到引用源。基准电压 错误！未找到引用源。 错误！未找到引用源。6.1.2归算前的等值电路图 6.1.3计算电抗将所有电抗归算到35kV侧：系统电抗X1*=Xsmax*=SB/Sdmax=100/580=0.172最大运行方式下X1*=Xsmin*=SB/Sdmin=100/265=0.377最小运行方式下 架空线路电抗 X2*=XL*=XOLSB/VB12=0.4×5×100/

150、372=0.146 12变压器电抗 X3*=XT1*=SS%/100×SB/ST1=(7.5/100) ×(100/10)=0.756.1.4归算后的等值电路图 6.2最大运行方式下的短路点计算6.2.1 d1点的短路电流计算10kV母线侧没有电源，无法向35kV侧提供短路电流，即可略去不计，那么d1点短路电流标幺值为：I d1*错误！未找到引用源。=错误！未找到引用源。=3.145换算到35kV侧0秒钟短路电流有名值I = I d1*×错误！未找到引用源。 =3.145×错误！未找到引用源。 = 4.908kA 根据?电力工程电气设计手册?的相关规定，

151、Kch = 1.8，当不计周期分量的衰减时， 短路电流全电流最大有效值Ich = 错误！未找到引用源。×I =错误！未找到引用源。×4.908=7.41kA当不计周期分量衰减时，短路电流冲击电流ich =2 Kch× I =2 ×1.87×I = 2.55× I = 2.55×4.908= 12.515 kA短路容量 S = 3 UB × I = 3 ×37×4.908 =314.52MVA6.2.2 d2点的短路电流计算10kV母线侧没有电源，无法向35kV侧提供短路电流，即可略去不计，那么d

152、2点短路电流标幺值为：I d1*错误！未找到引用源。=错误！未找到引用源。=0.936换算到10kV侧0秒钟短路电流有名值I = I d1*×错误！未找到引用源。 =0.936×错误！未找到引用源。 =5.15kA根据?电力工程电气设计手册?的相关规定，Kch = 1.8，当不计周期分量的衰减时， 短路电流全电流最大有效值Ich = 错误！未找到引用源。×I =错误！未找到引用源。×5.15=7.78kA当不计周期分量衰减时，短路电流冲击电流ich =2 Kch× I =2 ×1.87×I = 2.55× I =

153、2.55×5.15= 13.133kA13短路容量 S = 3 UB × I = 3 ×10.5×5.15 =93.66MVA 6.3最小运行方式下的短路点计算6.3.1 d1点的短路电流计算同上所得，那么d1点短路电流标幺值为：I d1*错误！未找到引用源。=错误！未找到引用源。=1.912换算到35kV侧0秒钟短路电流有名值I = I d1*×错误！未找到引用源。 =1.912×错误！未找到引用源。 =2.983KA 根据?电力工程电气设计手册?的相关规定，远离发电厂的地点变电所取电流冲击系数Kch = 1.8，当不计周期分量的衰

154、减时，Ich = 错误！未找到引用源。×I =错误！未找到引用源。×2.983=4.505kA当不计周期分量衰减时，短路电流冲击电流ich =2 Kch× I =2 ×1.87×I = 2.55× I = 2.55×2.983= 7.61kA短路容量 S = 3 UB × I = 3 ×37×2.983=191.16MVA6.2.2 d2点的短路电流计算10kV母线侧没有电源，无法向35kV侧提供短路电流，即可略去不计，那么d2点短路电流标幺值为：I d1*错误！未找到引用源。=错误！未找到引用

155、源。=0.785换算到10kV侧0秒钟短路电流有名值I = I d1*×错误！未找到引用源。 =0.785×错误！未找到引用源。 =4.317KA 根据?电力工程电气设计手册?的相关规定，Kch = 1.8，当不计周期分量的衰减时， 短路电流全电流最大有效值Ich = 错误！未找到引用源。×I =错误！未找到引用源。×4.317=6.519kA当不计周期分量衰减时，短路电流冲击电流ich =2 Kch× I =2 ×1.87×I = 2.55× I = 2.55×4.317= 11.008kA短路容量 S

156、 = 3 UB × I = 3 ×10.5×4.317=78.51MVA以上计算结果列表如下：14 157高压电气设备的选择7.1 35kV架空线的选择考虑到变压器在电压降低5%时其出力保持不变，所以35kV架空线相应的Igmax=1.05Ie即：Igmax =1.05 ×错误！未找到引用源。=1.05 ×错误！未找到引用源。=0.173kA 根据设计条件 Tmax=2500h 取J=1.3那么导体经济截面面积 S=Igmax/J=173/1.3=133.08mm2。7.1.1选择导线按照经济电流密度：选择LGJ-150/20钢芯铝绞线，其室外载流量为Ij1=306A，面积为S=145.68mm2,导线最高允许温度为70，根据工作环境温度为30，查综合修正系数K=0.94， 306=287.64AIgmax，满足电流的要求。7.1.2热稳定校验按最大运行方式d2点短路：根据设计任务书的条件，变电所的继保动作时限不能大于1.3秒，即top=1.3s，断路器开短时间tos=0.2s，非周