2×300MW机组火电厂电气部分的设计毕业设计 下载本文

内容发布更新时间 : 2024/5/19 10:38:19星期一 下面是文章的全部内容请认真阅读。

2*300MW火力发电厂电气部分设计

单相短路 20.829 135.214 71.561 31.976 25.573 31.166 两相短路 15.857 37.905 41.241 两相接地短路 20.117 78.048 35.742 第二部分计算结果如下:

(单位为KA)

第二节 短路电流计算

一、计算电路图和各元件电抗标么值的计算

取SB=1000MVA;Ub=Uav;XC=0.27Ω 发电机和变压器:

1000?0.8?1.152 1F、2F: XG*=UG%×SB=18×

SN10010012517.11000?0.85??0.4845 3F、4F: XG*=UG%×SB=

SN100100300变压器: 1#、2# :XT*=UK%?SN=13?1000=0.867

100SB1001503#、4# :XT*=UK%?SN=14?1000=0.378

100SB100370联络自耦变压器: Uk12%?13.43 Uk13%?11.74 Uk23%?18.66

11Uk1%=(Uk12%?Uk13%?Uk23%)?(13.43?11.74?18.66)?3.255

22

Uk2%?11(Uk12%?Uk23%?Uk13%)?(13.43?18.66?11.74)?10.175 22 25

2*300MW火力发电厂电气部分设计

Uk3%?12(U1k13%?Uk23%?Uk12%)?2(11.74?18.66?13.43)?8.485 X*%L1?Uk100?SNS?3.255?1000?0.109 B100300X*Uk%L2?100?SNS?10.175?1000300?0.339 B100X*k%SNL3?U100?S?8.45?1000?0.566 B100300高压厂用分裂变压器: X1-2= X1-2ˊ/(1+Kf/4)=0.066 XT1*=(1-3.744)X1-2×1000/40=0.107 XT2*=12KfX1-2×1000/40=3.086

厂高备用变压器: X1-2= X1-2ˊ/(1+Kf)=0.099

XT1*=(1-.3.54)X1-2×1000/31.5=0.392

XT2*=12KfX1-2×1000/31.5=4.4

三. 正序网络变换及三相短路电流计算 1.短路点dX1 :X1?X31.152?13?2?0.8672?1.010 :XX7?X914?2?0.389?0.4852?0.437 Y/Δ变换:(由X11、X13、X15、转化为X16、X17)

X0.3?0.44816=0.3+0.448+

1.01?0.881 X1.010?0.44817=1.010+0.448+

0.3?2.966 26

2*300MW火力发电厂电气部分设计

图1.各设备正序阻抗图

图2

X1Σ=

11111???0.8812.9660.270.437?0.134

27

2*300MW火力发电厂电气部分设计

三相短路计算电抗: XS1JS=40.881×

1000300?2?0.85=0.881 X2JS= 0.437×=0.308 10001000125?2?0.8X1JS=2.966×=0.927 XSC2JS=0.27×1000?0.27

10001000查汽轮发电机运算曲线得:

1F、2F: I*″=1.12 I*(0.1)=01.09 I*(0.2)=1.02 I*(4)=1.36 3F、4F: I*″=3.56 I*(0.1)=3.05 I*(0.2)=2.7 I*(4)=2.7 系统C1:Ic1*″=1.3 Ic1(0.1)*=1.23 Ic1(0.2)*=1.15 Ic1(4)*=0.1.46 系统C2:Ic2*″=4.1 Ic2(0.1)*=3.4 Ic2(0.2)*=3.0 Ic2(4)*=2.37 求短路点各个时刻短路电流: I”=1.12×

250+ 1.3.56×3?231600+1.3×3?2311000+4.1 ×3?2311000=19.534 3?231I0.1=1.09×

250+3.05×3?231600+1.23×3?231600+3.4×3?2311000=16.828 3?231I0.2=1.02×

250+2.7×3?231600+1.15×3?231600+3.0×3?231600=15.059 3?231I04=1.36×

250+2.31×3?231600+1.46×3?231600+2.37×3?231600=13.026 3?231 28

2*300MW火力发电厂电气部分设计

冲击电流:ich=1.85×2×19.534=51.108KA 2.短路点d2三相短路电流计算(按图1所示) X106 =

11111???0.9593.2880.270.8735?1?0.161Ω 6.195×107=×106+×10=0.161+0.389=0.55 电流分布系数:C1?C2?X1060.161X0.161??0.596 C3?106??0.168 X80.27X1030.959X106X0.1610.161??0.184 C4?106??0.049 X1050.8735X1043.228X107X0.550.55??0.923 X109?107??2.989 C10.596C20.184X107X0.550.55??3.274 X111?107??11.22 C30.168C40.049其转移阻抗为:X108?X110?图3

换算成计算阻抗:

125?2300XJS111?11.22?0.8?3.506 XJS12?0.4845?0.85?0.171

10001000300 XJS105?2.989?0.85?1.055 XJS110?3.274 XJS108?0.923

1000 29