a toy factory needs a maximum power supply of 500kw at a mains voltage of 220v. If it is transmitted at 220v,the total resistance of the cables is 0.01.
Find the power loss in the cables and the available voltage reaching the fastory.
請問點計?
唔識計...
2007-04-08 12:52 pm
回答 (2)
2007-04-08 2:42 pm
✔ 最佳答案
Power supply, Psupply= 500 kW
= 500 000 W
Voltage, V
= 220 V
Current in the cable, I
= P / V
= 500 000 / 220
= 2272.7 A
Total resistance of the cables, R
= 0.01 Ω
So, power loss in the cables, Ploss
= I^2R
= (2272.7)^2(0.01)
= 51 652.9 W
= 51 700 W
Potential difference across the cable, Vdiff
= IR
= (2272.7)(0.01)
= 22.7 V
So, available voltage reaching the factory, Vavail
= V - Vdiff
= 220 - 22.7
= 197.3 V
參考: Myself~~~
2007-04-08 8:10 pm
Since the 500 kW is the maximum power supply,
the power dissipated in the factory may be < 500 kW.
Operating resistance of the machine in toy factory R(t) = (220^2)/500*1000 = 0.0968 Ω
Potential difference across the cables = 220*0.01/(0.01+0.0968) = 20.599 V
Power loss in the cables = (20.599^2)/0.01 = 42400 W
Available voltage reaching the factory = 220-20.599 = 199 V
the power dissipated in the factory may be < 500 kW.
Operating resistance of the machine in toy factory R(t) = (220^2)/500*1000 = 0.0968 Ω
Potential difference across the cables = 220*0.01/(0.01+0.0968) = 20.599 V
Power loss in the cables = (20.599^2)/0.01 = 42400 W
Available voltage reaching the factory = 220-20.599 = 199 V
收錄日期: 2021-04-14 18:59:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070408000051KK00647