KTV总功率200KW,需要多大电缆

2025-01-23 11:43:33
推荐回答(5个)
回答1:

总功率200KW可选用150平方的铜芯电缆,但已无裕量,建议改用185平方的铜芯电缆200KW的负荷,大致电流400A。 YJV-3x185+1x150。 185平方的电缆,空气中载流375A,土中载流380A。

电压等级为380v情况下:总功率200KW,功率因数取0.8,线电流:I=P/√3*380*0.8=379.85A。查表:铝芯电缆配185平方毫米。铜芯电缆配120平方毫米。

P=1.732UIcosφ,I=P/(1.732×U×cosφ)I=300/(1.732×380V×0.85)=545A

P:功率;U:电压;I:电流;cosφ:功率因数,电阻负载取1,感性(比如电机)负载取0.85。

300KW负载至少要选择240平方的铜芯电线,240平方的铜芯电线明线敷设安全截流量约570A。


扩展资料:

零线和地线的根本差别在于一个构成工作回路,一个起保护作用叫做保护接地,一个回电网,一个回大地,在电子电路中这两个概念是要区别开来的。

真有效值/直流转换检测功率法的最大优点是测量结果与被测信号的波形无关,这就是“真正有效值”的含义。因此,它能准确测量任意波形的真有效值功率。测量真有效值功率的第一种方法是采用单片真有效值/直流转换器,首先测量出真有效值电压电平,然后转换成其真有效值功率电平。

回答2:

三相200KW,电流达到400A,选用240平方电缆可以保证供电万无一失。如果200KW不是同时使用,错开峰值,那么使用185平方电缆勉强,但是不能100%满负荷使用。
1、功率是指物体在单位时间内所做的功的多少,即功率是描述做功快慢的物理量。功的数量一定,时间越短,功率值就越大。
2、求功率的公式为功率=功/时间。功率表征作功快慢程度的物理量。单位时间内所作的功称为功率,用P表示。故功率等于作用力与物体受力点速度的标量积。

回答3:

三相四线制。200KW。它 的满负荷工作电流大约在400A左右距离变压器也不算远,所以不用考虑电压降。120平方的电缆的安全载流量在500左右。所以用120的挺合适的。240的有些浪费了。

回答4:

负载电流=105(kw)÷1.732÷380(V)=160(A)
总电流在100A以上的导线,每平方毫米可通过2A电流,所以建议用320平方的电缆。

回答5:

1、估计现场自然功率因数为0.75。
2、计算负载电流:
(1)总负载电流I=200/1.732/0.38/0.75
=405(A)
(2) 空调负载电流I=105/1.732/0.38/0.75
=212(A)
3、查电缆的长期允许截流量:
240平方毫米铜芯电缆在空气中的长期允许截流量为440A,建议主电缆选择此规格。
95平方毫米铜芯电缆在空气中的长期允许截流量为245A,建议空调电缆选择此规格。
4、为了人身安全,建议低压配电系统的接地型式采用TN-S系统或TN-C-S系统,即人们说的三相五线制,即加一根PE线,截面为火线电缆的1/2(规范规定)。

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