井下电钳工 如何计算俩相短路线电流
1个回答

Id5=Ue/2√(∑R)2+(∑X)2

∑R =Rg/kb2+Rb1=0.429/52 +0.0113=0.02846Ω

∑X=Xx+Xg/ kb2+Xb1=0.0288+0.15/52+0.1072=0.142Ω

Xx—系统电抗,取0.0288

6KV电缆取1000M,查表的R0=0.412Ω/KM

X0=0.15Ω/KM

Rg =0.412Ω

Xg=0.075Ω

Rb1 =0.0135Ω Xb1=0.1072Ω

Id5= Ue/2√(∑R)2+(∑X)2=4142A

变压器短路电流4142A

橡套电缆70平方,长1500米短路电流为:813A

∑R=0.02846+0.42=0.7046Ω

∑X=0.142+0.07614=0.21814Ω

Id6= Ue/2√(∑R)2+(∑X)2=813A