题目内容
某电站输送电压为U=5000V,输送功率为P=500kW,这时安装在输电线路的起点和终点的电能表一昼夜里读数相差4800 kW·h(即4800度电),求:
(1)输电线的电阻;
(2)若要使输电损失功率降到输送功率的1.6 %,电站应使用多高的电压向外输电?
(1)输电线的电阻;
(2)若要使输电损失功率降到输送功率的1.6 %,电站应使用多高的电压向外输电?
(1)20 Ω (2)2.5×104 V
试题分析:(1)因为两地的电能表读数之差即为线路损失的电能,由此可求线路损失的功率:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004538021595.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004538052936.png)
又由
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004538083803.png)
得:输电线的电阻
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004538114666.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004538146565.png)
(2)要使输电损失的功率降到输送功率的1.6%,则有
P损′=500 kW×1.6%=8 kW (2分)
而
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004538161708.png)
得输送电压U′=500000×
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004538177617.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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