一个使用电压为380伏,电流为54.48安的潜水泵一小时走多少度电

2024-11-27 10:18:49
推荐回答(4个)
回答1:

下面给你两个计算方法:

如果54.45A是某一条线的电流则:

P=U*I*3相=220V*54.48A*3=约36KW

电量Q=Pt即功率和时间的乘积,一度电就是一千瓦每小时,36KW的功率一小时自然也就是36度电.

如果测量的是三线的线电流而非上面的那相电流则:

P=√3(根号三)*U*I=1.732*380V*54.48A=约36KW

这两种计算方式适当了解一下吧,结果是基本一样的,如果是非纯阻性负载还需要计算功率因数,即所计算出来的结果后乘上COSφ.

回答2:

P=IU*功率因数=380*54.48*0.8=16552.8W=16.5528KW
1小时耗电16.5528度电

回答3:

不管你选不选我,我告诉你
你的电机是30KW的。
你的电流是54.48A
54.48/2=27.24KWh
你的电机每小时大概就是27度电
不信,你可以看电表算一下。

回答4:

消耗总功率:相电压*相电流*3=220*54.48*3=?

注意不用乘COSφ。如果乘COSφ,就是有功功率,只有装有功表的情况下才会显示有功功率

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