一台电动机的额定电压是220V,额定功率5.5Kw,它正常工作时的电流有多大?连续工作2h耗电多少

2025-03-15 04:29:25
推荐回答(4个)
回答1:

你好:

——★1、单相电动机是感性负载,它的额定电流Ⅰ= P ÷ (U × 功率因数 × 效率)。【注意:公式:I=P/U 为电阻性负载的计算公式用这个公式计算电动机的电流,属于基本概念错误


——★2、额定电压220V、额定功率5.5Kw的单相电机,< 以单相电机的功率因数效率都为0.7计算 > 电流为5500W ÷(220V × 0.7 × 0.7) = 51A。


——★3、电动机的耗电很好计算:(电度为:千瓦* 时)5.5KW x 2h = 11度电

回答2:

正常工作电流I=5500/220=25A........连续工作2h耗电W=25*25*3600*2=4500000

回答3:

根据功率公式:I=P/U=5500/220=25A,连续工作2h耗电=5.5kw*2h=11kwh(度).

回答4:

额定电流是18.7安,,不是满负荷的话要低于这个数。连续两小时如果满负荷的话耗电11度,希望对你有帮助。

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