用380V的电压,电线长500米,要带动400千瓦的电机,问要多少平方的铜线?

2024-10-31 10:18:19
推荐回答(2个)
回答(1):

1、先求400千瓦的电机的电流 Ie,设功率因数为0.8:
Ie = 400KW / (根3 x 380V x:0.8)= 760A
760A就是电缆通过的电流。
2、根据 Ie,确定电缆截面积:
经查《电缆安全载流量》知,0.6/1kV三芯 + 一芯(3x185+95)PVC绝缘电力铜电缆的载流量280A,可以使用3根并联,载流量为280Ax3=840A>760A。
3、500米电缆的长度对电压的影响:
经查,3x185+95电缆的导体直流电阻是0.091Ω/km。
电缆500米,在正常额定电流下的压降 = 0.091Ω/km x 0.5km x 280A =12.7V
12.7V是380V的3.4%,在允许范围之内。
电机输入端电压约为380V-12.7V=367.3V
4、结论:
380V的电压,电线长500米,要带动400千瓦的电机,可选用三芯 + 一芯(3x185+95)的电缆,满负荷时压降约为12.7V。

回答(2):

这也太远了,,为什么不装就近装设变压器。
计算压降:△U=PL/CS 400KW*500M/77*360=200000/27720=7.2%
380*7=26V 使用360平方的三相铜芯线,压降在26V左右,这样才可以正常启动。
这样用线比安变压器贵多了。

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