# Conductor and insulator losses

1. Jun 8, 2013

### anhnha

Hi,
I need help to understand these statements:
The generators at the power plant can be operated as either, constant voltage source (CVS), or constant current source (CCS). To obtain a CVS, you spin the generator at constant speed. To obtain a CCS, you spin at constant torque. The CVS is used exclusively in power distribution because losses are lower. Conductors lose more power than insulators, so they generate at 100% voltage all the time, & the current varies with load, & it is usually well below 100% capacity.

Can you help me explain the bold part? I don't understand why insulator losses are taken into acount.
Where does these losses come from? With conductor losses, I think it is caused by resistance of the conductor but I am totally lost about insulator loss.

2. Jun 9, 2013

### PhanthomJay

There is no such thing as a perfect insulator, so as long as there is a path to ground there will be small currents passing along its surface. These currents are called leakage currents, and are typically very small compared to the line currents, and thus power losses through the insulators are very small compared to the wire losses through (i^2)R heating losses. If the insulators get contaminated however, now you've got more than losses to worry about.

3. Jun 9, 2013

### anhnha

Thanks,
Losses in conductor is: L = i^2*R
How can I calculate the loss in insulator?
And can you explain more about the bold sentences? I am not quite understand it.

4. Jun 9, 2013

### PhanthomJay

It's a bit beyond my expertise, but insulator power losses are dependent upon a number of variables, like leakage distance, air temperature, and humidity, and losses are best determined through lab testing. Insulator power losses are orders of magnitude less than conductor losses: for a long transmission line, conductor losses might be a few MW, whereas insulator losses might be a few KW . Power lines have conductors sized for peak amperage during short term high temperature conditions; on a regular day to day basis, the conductors normally carry well below, on average, 40 percent of their current rating, so i^2*R losses are not as drastic as they would be if carrying full rated current all the time.