(1)建立微分方程并求解 u1=1.414*U1sin(ωt+α) = Lk dik/dt + Rkik
t=0时,ik=0,则 ik=i'k+i''k=1.414*U1/Zk ×[sin(ωt+α-φk)-sin(α-φk)e-Rkt/Lk] =-1.414*Ik[cos(ωt+α)-cosαe-Rkt/Lk]
(2)分析 ikmax
Ikmax=1.414*IK[1+e-(Rkπ)/(ωLk)]=kk*1.414*IK
Ikmax=(1.5~1.8)*1.414*I1N/ZK* ZK*=0.06时,Ikmax/(1.414*I1N)=25~30 (3)暂态短路时的机械应力
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