More Than You Ever Wanted to Know About Electrical Engineering: Power Transmission Losses
We’ve been talking about step-down transformers that can convert between a high line voltage and more moderate local voltages. You might be asking yourself why we need to do such a thing in the first place. Why can’t we just transmit utility power at 120 V or 240 V?
All right, let’s try it. Say we need to transmit 20 MW of power over a 100 km. If we do it at 240 Vrms, we’ll have a current of 83.3 kArms flowing through the conductor.
Some of the 20 MW we’re transmitting is going to be lost - that is, dissipated as heat. We’d like to minimize losses, so let’s say we’re aiming for 97% efficiency - no more than 3% of that 20 MW lost. Since power dissipated is a function of current and resistance, we know we’ll need a conductor with an overall resistance of 8.64 x 10^-5 Ohms over a distance of 100 km.
The resistance of a conductor is a function of its length, l, its cross-sectional area, A, and the conductor’s resistivity, ρ, which is a property of the material it’s made of. We’ll assume a resistivity here of 8 x 10^-8 Ohm-meters. From here, we can figure out what kind of cross-sectional area our conductor needs - in other words, how large the cable’s diameter has to be.
To make this work, you’d need a conductor with a radius of 5.4 m. Clearly, this isn’t going to happen.
If you want a conductor that’s actually practical, that means you need to transmit power at high voltage - since a higher voltage at the same power will result in a lower current, this means your losses will be lower, allowing you to use a much smaller conductor. Here are the same calculations done with a transmission voltage of 240 kVrms instead of 240 Vrms.
In this case, you need a conductor with a radius of about 0.5 cm, which is much more reasonable.