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企业10KV变电所及低压配电系统的设计(10)

时间:2017-05-07 20:41来源:毕业论文
1、T1、T2变压器:Uk % = 4.5 1)高压侧: ⑴ SB=100 MVA, UB1=10.5 KV, l=5 km, x0 =0.08 /km ,r0 =0.09/km(任务书给出) 确定基准电流:IB1= SB/ UB1 =5.5 KA (5-1) (1) 电缆线路电


1、T1、T2变压器:Uk % = 4.5
1)高压侧: ⑴ SB=100 MVA, UB1=10.5 KV, l=5 km, x0 =0.08 Ω/km ,r0 =0.09Ω/km(任务书给出)
确定基准电流:IB1= SB/ UB1 =5.5 KA                                    (5-1)
(1) 电缆线路电抗:XL*=I×x0 /(UB12/ SB)=0.36                             (5-2)
其等值电路为图:
 
图5-1 高压侧短路计算等效电路图
(2)求k-1点得短路总阻抗及三相短路电流及短路容量:
① 总阻抗:X*Σ(K-1) = X*1=0.36 mΩ                                       (5-3)
I(3)(K-1)= IB1/X*Σ(K-1)=5.5/0.36=15.27 KA                                   (5-4)
③ 其他三相短路电流:
对L较大的中、高压系统,取Ksh=1.8,则
i(3)sh= IpKsh=2.55* I(3)(K-1)=2.55*15.27=38.87 KA                         (5-5)
I(3)sh=1.51I(3)(K-1)=1.51*15.27=23.06 KA                                   (5-6)
I’’(3)=I(3) =I(3)(K-1)=15.27 KA                                             (5-7)
④三相短路容量:
S(3)(K-1)=SB/X*(K-1)=100MVA/0.36=277.8MVA                                (5-8)
2)低压侧:
(1)电力电缆的阻抗(需折算至低压侧)
X0=0.08 Ω/km, r0 =0.09Ω/km,l=5km.
RL’=lr0/(UB1/UB2)2                                                                                                              (5-9)
=0.09 5 (0.4/10.5)2
=0.7(mΩ)
XL’= lx0/(UB1/UB2)2                                                                                                          (5-10) 企业10KV变电所及低压配电系统的设计(10):http://www.youerw.com/zidonghua/lunwen_6642.html
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