Voltage drop motor load

Status
Not open for further replies.

Dsg319

Senior Member
Location
West Virginia
Occupation
Wv Master “lectrician”
If I were trying to figure voltage drop for a motor load using this formula 2xkxIxD/vd. For the amount of current do i just use the full load current at 100percent? Or Do I use full load current And 125percent?. My brain is telling me just the full load current rating at 100 percent but I want to make sure
 
There is no code requirement to answer this. You need to consider what your target for this VD is. Many cases will be "normal load current" or rated full load current, but also consider that even though the starting period is usually brief in most cases, you will have a VD associated with that higher current during starting and it can result in voltage sag for everything on same service/feeder.

A high inertia load that takes time to accelerate can also have longer term impact from the voltage drop during starting/acceleration.

All that said amount of use of the motor and other factors may come into play as well. A motor that is rated 208-240 volts, you may look at VD and decide that even though you are operating on 240 nominal you may find a little higher VD is still acceptable because the motor is still operating at a higher voltage then it would if operating on 208 nominal, and especially if it only runs for limited duty cycles. Something that runs long cycles or even nearly always on - you need to consider that VD also equates to heat loss in the conductors and whether or not it is worth reducing those losses even if it doesn't have much impact on the load itself.
 
During startup for small motors it’s 10 times the FLA for a couple seconds. On larger ones it’s 5.5-6xFLA. The load doesn’t matter it just changes the timing. The rest of the time unless the load fluctuates it stays at or below FLA.

That’s why when sizing for industrial use usually we add up motor horsepower and add five times the biggest motor. It’s at around 0.85 power factor worst case. So horsepower x .746 / .85 = kw. Then convert to amps from there.
 
Status
Not open for further replies.
Top