What is the power loss due to resistance in the line?

Question

34. Power transmission lines often use a form of electric current called alternating current, but in many regions, such as the Province of Quebec, high-voltage direct-current lines are used instead. Direct current is the kind of electric current you are studying in this chapter. A certain direct-current power transmission line has a resistance of 0.255 Ω/km. 812 kV of potential drives the current from the generating station to a city located 125 km from the plant.

Answer 1

The IR power losses in a transmission line depend upon the resistance and the current. If you have a resistance of .255 ohm/km and 125km to go to town, then you have 125 x .255 ohms loss between the power station and the city. The voltage does not matter. You only need to find the current in the wire and multiply it, squared, by the resistance you just resolved. That will be the loss in the wire.

DC is used for long distance transmission to avoid skin effect losses from the AC field pushing the current to the outside of the wire, leaving little current in the middle, which increases losses due to decreased current path volume.

Answer 2

You can apply ohms law to calculate the energy loss. Most of the energy is lost through heat that radiates from the wire.

Edison was a proponent of direct current because he thought it would be safer. However, most power agencies use alternating current and transformers to boost the voltage (reducing the amperage) and allow the power to be tranferred over greater distances.

delta V = I x R

Electric Power:
P = I x V

P = I^2 x R

Answer 3

To calculate specific power loss under these conditions you would need to know the current flowing through the line, or know the load resistance in order to calculate load current.

WK

Answer 4

P= IV; I = V/R
so P = V^2/R (rearange to this as you know V and R, I unknown)
so P/km = (812,000)^2/0.255
= 2.59x10^12
so total Power loss is 323x10^12W (or 323TeraW)