电力系统短路电流计算(20)
时间：2025-04-08
电力系统短路电流计算。
2
=0.15×380×0.76
120
=137.18 mΩ
RS=rSUrM ηM cos M
PrM
=0.034×914.533
= 31.094 mΩ
2
U ηM cos M R=rrM
2
RRPrM
= 0.021×914.533
= 19.205 mΩ
2 电动机至主汇流排电缆阻抗
Re=rl/a =0.279×35÷2
= 5.198 mΩ
Xe=xl/a =0.081×35÷2 = 1.437 mΩ
A.4.1.2 等效电动机M2阻抗
XU2
rM ηM cos MM′′=x′M
′P rM
= 2
0.188×380×0.76
408
= 50.568 mΩ
2
RUrM ηM cos MS=rSP rM
= 0.043×268.98
= 11.566 mΩ
2 RUrM ηM cos MR=rRP rM
= 0.027×268.98
= 7.262 mΩ
A.4.2 大电动机M1时间常数计算
T1000X′Me
′Me′′=
2πfR R
=1000(X′M
′+Xe) 2πfRR
=1000×(137.18+1.437)100π×19.205
= 22.975 ms
TX′M′
+Xe)dcMe=1000(2πf(R)
S+Re =
10×138.617π ×(31.094+5.198)
= 12.158 ms
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