An engineering company is hired to improve the efficiency of a transmission line. In their report, they discuss ways that will allow them to reduce the current by a factor of four. Assuming that this is done, by what factor will the power losses be reduced?
a. four
b. eight
c. sixteen
d. two
If you can explain the answer that would be very helpful! :)
Physics help: electric power transmission?
2018-01-12 3:29 am
回答 (3)
2018-01-13 8:26 am
The answer would be 16 based on the following assumptions: The transmission line losses are only dependent on the resistance of the conductors. All other forms of loss due to corona, leakage, inductive coupling, etc. are ignored or remain exactly the same and only the current level changes.
Based on this, Power loss = I^2*R, lets assume a total resistance for the line of 5 ohms. Starting current is 100 amps.
Power Loss = (100*100*5) = 50,000 watts
A current reduction to 25 amps (100/4) will give Power Loss = (25*25*5) = 3,125 watts
The ratio of change = 50,000/3,125 = 16, which if you notice (4)^2 = 16.
Hope this helps.
Based on this, Power loss = I^2*R, lets assume a total resistance for the line of 5 ohms. Starting current is 100 amps.
Power Loss = (100*100*5) = 50,000 watts
A current reduction to 25 amps (100/4) will give Power Loss = (25*25*5) = 3,125 watts
The ratio of change = 50,000/3,125 = 16, which if you notice (4)^2 = 16.
Hope this helps.
2018-01-12 5:52 am
In fact this is not the case for power transmission lines.
There are losses from the increased volts. Which is why there is a limit to how far we can take this process.
At the moment a million volts is about the limit. After which the losses from sparking, ionization, electric fields etc all lose more than is gained through lowering the current.
Randy has already explained that the power loss is I^2 R where R is the resistance of the lines.
If you don't alter the lines in any way then halving the current is going to alter the power loss by a factor of ?
There are losses from the increased volts. Which is why there is a limit to how far we can take this process.
At the moment a million volts is about the limit. After which the losses from sparking, ionization, electric fields etc all lose more than is gained through lowering the current.
Randy has already explained that the power loss is I^2 R where R is the resistance of the lines.
If you don't alter the lines in any way then halving the current is going to alter the power loss by a factor of ?
2018-01-12 3:30 am
P = I^2 * R. You haven't changed the resistance of the transmission lines, so P is proportional to I^2.
What's the square of a factor of four?
What's the square of a factor of four?
收錄日期: 2021-05-01 20:57:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20180111192918AARfQfO