发布时间 : 星期四 文章某电机修造厂全厂总降压变电所及配电系统设计更新完毕开始阅读1ecc9e04bdeb19e8b8f67c1cfad6195f312be8cc
relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review contengle of clant overview answon (slightly) 17, aers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \hen in-line solution. Category fraction multiplication word problem score w of soliDivision applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their nce in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle assificatind subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-revied content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of cirsignificacular cone is slightly solid surface area and volume 1, size 2, table ...和某电机修造厂全厂总降压变电所及配电系统设计
一、生产任务及车间组成 1．本厂产品及生产规模
本厂主要承担全国冶金工业系统矿山、冶炼和轧钢设备的配件生产，即以生产铸造、锻造、铆焊、毛坯件为主体，生产规模为：铸钢件1万吨、铸铁件3千吨、锻件1千吨、铆焊件2千5百吨。
2．本厂车间组成 （1）铸钢车间；（2）铸铁车间；（3）锻造车间；（4）铆焊车间；（5）木型圈车间及木型库；（6）机修车间；（7）砂库；（8）制材场；（9）空压站；（10）锅炉房；（11）综合楼；（12）水塔；（13）水泵房及污水提升站等。
二、设计依据
1． 厂区平面布置图(略)
2． 全厂各车间负荷计算表如下：各车间380伏负荷 cos?tg? 序车间或用设备 计 算 负 荷 变压器 备注 Kx 号 电单位名容量 台数及 Pjs Qjs Sjs 称 (KW) (KW) (Kvar) (KVA容量 ) No.1变电所 1 铸钢车间 2000 0.4 0.65 1.17 800.01231.2×936.00 0 30 1250 0.4 0.7 0.7 0.6 1.02 400.0571.3 408.00 0 7 1.33 Ynn0 No.2变电所 2 3 4 铸铁车间 1000 砂库 小计 110 128.1 3 477.0698.62×500 Ynn0 510.41 0 0 77.00 102.41 Ynn0 No.3变电所 5 6 7 铆焊车间 1200 1＃水泵房 小计 28 0.3 0.45 1.98 360.0798.5 712.80 0 5 0.75 0.8 0.75 21.00 15.75 26.25 381.0822.1900 728.55 0 6 No.4变电所 1 2 3 4 5 空压站 390 0.85 0.25 0.75 0.88 331.5441.5 291.72 0 8 0.65 1.17 机修车间 150 锻造车间 220 37.50 43.88 57.72 0.3 0.55 1.52 0.6 0.6 1.33 1.33 66.00 100.32 65.05 5.60 木型车间 185.85 0.35 题材场 20 0.2120.0 8 108.2 86.51 4 7.45 9.32 relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review contengle of clant overview answon (slightly) 17, aers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \hen in-line solution. Category fraction multiplication word problem score w of soliDivision applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their nce in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle assificatind subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-revied content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of cirsignificacular cone is slightly solid surface area and volume 1, size 2, table ...和8 6 7 综合楼 小计 20 0.9 1 1 18.00 18.00 25.46 523.6757.8800 547.88 5 8 Ynn0 No.5变电所 1 2 3 4 5 锅炉房 2＃水泵房 300 28 0.75 0.75 0.8 0.8 0.75 225.0281.2 168.75 0 5 0.75 Ynn0 21.00 26.44 9.10 15.75 26.25 30.93 40.69 6.83 11.38 仓库（1、88.12 2） 污水 提升站 小计 14 0.3 0.65 1.17 0.65 0.8 0.75 281.5358.6400 222.26 4 9 各车间6千伏高压负荷 1 2 3 4 电弧炉 工频炉 空压机 小计 2×1250 0.9 0.87 0.57 2250.1282.52589. 00 0 85 0.48 600.0665.5 288.00 0 4 2×300 0.8 0.9 2×250 0.85 0.85 0.62 500.0588.3 310.00 0 0 3250.1880.53754. 00 0 83 全厂合计 6600.4825.58175. 00 9 96 说明：No.1变电所和No.2变电所设置两台变压器外，其余设置一台变压器。 3． 供用电协议
工厂与电业部门所签订的供用电协议主要内容如下：
（1）工厂电源从电业部门某220/35千伏变电所，用35千伏双回架空线路引入本厂，其中一个为工作电源，一个作为备用电源，该变电所距离工厂东侧4.5km处，单位长度电抗值为0.4Ω/km。
（2）供电系统短路技术数据如下： 区域变电所35kV母线短路数据如下：
系统最大运行方式：Sdmax＝200MVA；系统最小运行方式：Sdmin＝175MVA （3）电部门对本厂提出的技术要求
①区域变电所35kV配出线路定时限过电流保护装置的整定时间为2秒，工厂总降不应大于1.5秒。
②该厂的总平均功率因数值应在0.9以上。 ③在企业总降压变电所高压侧进行计量。 三、设计范围与任务 1．负荷计算
全厂总降变电所负荷计算，是在车间负荷计算基础上进行的，考虑车间变电所变压器的功率损耗，从而求出全厂总降变电所高压侧计算负荷及总功率因数。列出负荷计算表，表达
relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review contengle of clant overview answon (slightly) 17, aers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \hen in-line solution. Category fraction multiplication word problem score w of soliDivision applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their nce in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle assificatind subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-revied content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of cirsignificacular cone is slightly solid surface area and volume 1, size 2, table ...和设计成果。
2． 总降变电所位置和各个变压器台数以及容量的选择 考虑电源进线方向，综合考虑设置各个变电所的有关因素，结合全厂计算负荷以及扩建备用的需要，确定主变台数容量。
3． 厂总降压变电所主接结线设计 根据变电所配电回路数，负荷要求可靠性级别的计算负荷值，确定高低压侧的接线形式。 4．厂区高压配电系统设计 根据厂内负荷情况，从技术、经济合理性确定厂区配电电压。择优选择配电网布置方案，按选定配电系统作线路结构与敷设方式设计。
5．工厂供配电系统短路电流计算
工厂用电，通常为电网末端负荷，其容量远远小于电网容量，均按无限容量系统供电进行短路电流计算。
6．改善功率因数装置设计
根据负荷计算要求本厂的高压配电所的COS?，通过查表和计算求出达到供电部门要求的数值所需补偿的无功功率。由产品样本选出需补偿电容器的规格和数量，并选用合适的电容器柜。
7．变电所高低压侧设备选择
参照短路电流计算数据和各回路计算负荷以及相应的额定制，选择各种电器设备、开关柜等。用主结线图、设备材料表等表达设计成果。
8．继电保护及二次结线设计
内容包括继电保护装置、监视及测量仪表、控制和信号装置及备自投，用二次回路原理图或展开图及元件材料表来表达设计成果。
9．变电所防雷、接地装置设计
参考本地气象、地质资料设计防雷装置，并进行接地装置设计计算。 10．总降变电所变、配电装置总体布置设计
综合前述设计计算成果，参照有关规程，进行室内、室外变配电装置的总体布置和施工设计。
11．车间(机加车间)变电所及低压配电系统设计
根据生产工艺要求，车间环境，用电设备容量、分布情况等进行设计，确定车间变电所所用变台数、容量。
四、本厂的负荷性质
本厂为三班工作制，年最大有功负荷利用小时数为6000小时。属于二级负荷。 五、工厂的自然条件 1．气象条件
（1）最热月平均最高温度为30℃；
（2）土壤中0.7～1米深处一年中最热月平均温度为20℃； （3）土壤冻结深度为1.10米； （4）夏季主导风向为南风； （5）年雷暴日数为31天。 2.地质及水文条件
根据地质工程勘探资料获悉，该厂区地址原为耕地，地势平坦，地层以砂质粘土为主，地质条件较好，地下水位为2.8～5.3米。地耐压力为20吨/平方米。
总有功功率 ∑P = 6600kw
relationship, established equivalent relationship 14, and subject: application problem (4)--scores and percentage application problem review contengle of clant overview answon (slightly) 17, aers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units \hen in-line solution. Category fraction multiplication word problem score w of soliDivision applications engineering problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their nce in rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly used time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)--line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle assificatind subject: geometry preliminary knowledge (2)--plane graphics review content triangle, and edges shaped, and round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-revied content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboid, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of cirsignificacular cone is slightly solid surface area and volume 1, size 2, table ...和 总无功功率 ∑Q = 4825.59kvar 总视在功率 ∑S = 8175.96 KV·A
?P6600??0.807?0.9 全厂功率因数 cos???S4825.59所以要进行无功补偿
?1?1QC??P30(tancos0.807?tancos0.9)?1636.8KVar
取QC?2400KVar 低压侧补偿后无功功率：
Q1??Q?Q?4825.59?240?0C低压侧补偿后视在功率：
22 S2?(?P30)?(Q2)?26600? a524K2V5.r924225.?59K7V0 3A1.6变压器损耗：
515 ?P?0.01S2?0.0?S6 ?Q?0.062?0.0?高压侧有功功率：
703?1.6k1W0 5.474k2V1.a r90703?1.6P6?600?105.4?7 P2??P??高压侧无功功率：
670kW5. 47 Q2?Q1??Q?2825.59?421.90?2841.49kVar 高压侧视在功率： S2?222P2?Q2?6706.55?22841.?497kV2 8A5.0补偿后的功率因数：
COS??P26705.47??0.92?0.9S27285.0
2.1.2 主变压器的选择
主变压器选择要求S?S2 ，故选择型号为SC10-8000/35的变压器两台。一台工作，一台备用。
表2-1主变型号及参数
品牌 型号 W W P0 Pk75? 总电压联UK 质组合 结长*宽*高 轨距 LPA 量 (kV) 组高低标dB(% % kg mm*mm*mm mm*mm 压 压 号 A) I0