when calculating the max voltage drop of 3% for feeder (or branch circuit) for a 120/208V, do I want the Vd to be 3% of 120V or 208V? And does that change for the 5% of the total circuit?
Last edited by unsaint33; 02-12-19 at 04:24 PM.
Reason: Claritying
when calculating the max voltage drop of 3% for feeder (or branch circuit) for a 120/208V, do I want the Vd to be 3% of 120V or 208V? And does that change for the 5% of the total circuit?
if the feeder has the same current on all three legs, the VD will be the same L-L as L-N.
if the feeder has the same current on all three legs, the VD will be the same L-L as L-N.
So, in order to select the feeder conductors big enough to limit the Vd under 3%, should my 3% Vd be 3.6V or 6.24V? (3.6V is 3% of 120V & 6.24V is 3% of 208V)
So, in order to select the feeder conductors big enough to limit the Vd under 3%, should my 3% Vd be 3.6V or 6.24V? (3.6V is 3% of 120V & 6.24V is 3% of 208V)
If you have 3% VD on the L-N side, that would be 116.4 V. That would make the L-L voltage 201.6 V, or a 3% VD.
Originally Posted by petersonra
if the feeder has the same current on all three legs, the VD will be the same L-L as L-N.
I should have said the VD in % is the same. Obviously the voltage drop in volts is not the same.
So, in order to select the feeder conductors big enough to limit the Vd under 3%, should my 3% Vd be 3.6V or 6.24V? (3.6V is 3% of 120V & 6.24V is 3% of 208V)
They are all telling you straight.
3% Voltage drop on each 120V leg gets you 3% Voltage drop on the phase to phase 208V
It is a vector (trigonometry) problem. Vphase-neutral X sqrt(3) = Vphase-phase
120V X 1.732 = 207.8V
Works the same for the Voltage Drops
3% X 120V = 3.6V
3.6V X 1.732 = 6.24V
.03 X 207.8V = 6.23V
The only difference we are seeing is the round off error.
The attached sketches show the angles.
Without data you’re just another person with an opinion – Edwards Deming
They are all telling you straight.
3% Voltage drop on each 120V leg gets you 3% Voltage drop on the phase to phase 208V
It is a vector (trigonometry) problem. Vphase-neutral X sqrt(3) = Vphase-phase
120V X 1.732 = 207.8V
Works the same for the Voltage Drops
3% X 120V = 3.6V
3.6V X 1.732 = 6.24V
.03 X 207.8V = 6.23V
The only difference we are seeing is the round off error.
The attached sketches show the angles.
I deal with balanced lines (3 phase PV inverters), so for me the answer is simple; it's just the one way drop on an individual conductor @120V. There's no current on the neutral, so no voltage drop.
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