AC dissipation

TyPR124
This is probably a dumb question, but I only need to know for a program I would like to write.

Basically, on average (I realize there are a lot of factors involved, but a rough estimate is all I need), how much energy dissipates over a specific length of a conductor (power lines, whatever they are made of)? (I don't know what an appropriate length would be, as I have no idea at all how far AC can travel. I have a vague idea of DC, but that is only because I am into networking.)

Also, since W=V*I, and I'm assuming that I shouldn't decrease, then V is what decreases?

Thank you.

gnurf
You would need the resistance of the conductor

R = ρ * l / A

Where ρ is the specific resistance of the material (1.72×10^-8 Ωm for copper), l is the length of the conductor and A is the cross-sectional area.

This conductor will dissipate P = I²R watts if you send a current of I amperes through it. For time-varying current, replace I with the rms-value.

rorobee
" edison ac wiki "

This is probably a dumb question, but I only need to know for a program I would like to write.

Basically, on average (I realize there are a lot of factors involved, but a rough estimate is all I need), how much energy dissipates over a specific length of a conductor (power lines, whatever they are made of)? (I don't know what an appropriate length would be, as I have no idea at all how far AC can travel. I have a vague idea of DC, but that is only because I am into networking.)

Also, since W=V*I, and I'm assuming that I shouldn't decrease, then V is what decreases?

Thank you.

The voltage drop in a cable can be found (approximately) with this equation:

$$V = IRcos\phi + IXsin\phi$$

where,

V is the voltage drop in the circuit (line to neutral)
I is the current flowing in the conductor
R is the line resistance for one conductor, in ohms
X is the line reactance for one conductor, in ohms
$\phi$ is the angle whose cosine is the load power factor

This is the generally accepted approximate voltage drop formula from IEEE. There is an exact one but it's not usually necessary.

Note that the formula gives line to neutral voltage drop. Hence, if you have a single phase system you'll multiply by 2 for the total voltage drop; if it is a three phase system you multiply by 1.73.

CS

Bob S