A load with a low power factor draws more load current than necessary, so the supply conductors need to have a greater cross-sectional area than would otherwise be necessary.
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A 400 MCM copper conductor with an insulation temperature factor of 90 degrees C is rated at 380 amps.A 500 MCM copper conductor with an insulation temperature factor of 75 degrees C is rated at 380 amps.These conductors can have either 600 or 1000 volt insulation rating depending on what the system voltage in use is.
Yes. The two main reasons are:For a given size of power line more power can be carried by the line at higher voltage and lower amperage with less resistive heating. The heating due to resistance is a waste of energy. This due to Power = Volts X Amps. The higher the Volts the lower the Amps for some amount of Power. The more Amps carried by a conductor the more heat it will produce due to resistance.Since we can use a smaller size of power line to carry a given amount of power we will use less copper or aluminum to make that power line. The less material we use to make the conductor the less energy we have to use to make that conductor. Also the less energy it takes to transport, install, and support that conductor.
The largest conductor listed in the National Electrical Code is a 2000 kcmil.
When power factor is at unity, the voltage and current waves are aligned or in phase with one another. Since power is the product of voltage and current, power transfer is maximized at unity power factor. When power is transmitted at a lower power factor, greater current is required to deliver the same amount of power. When current is increased, the size of the transmission, distribution and generation systems, all have to be increased accordingly, along with the price of the killowatt-hour at the meter.
Use this formula to find amperage when kW is shown. Amps = kW x 1000/1.73 x Volts x pf. Use a power factor of .85 for the formula. Once the current is calculated re-ask the question for cable size for the calculated current.