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How to figure 3 phase voltage drop

How to figure 3 phase voltage drop

How to figure 3 phase voltage drop

(OP)
When figuring 3 phase voltage drop do you use the distance to the load or 2 times that?

RE: How to figure 3 phase voltage drop

In balanced three-phase circuits, you'd use the distance to the load to account for the voltage drop in the phase conductor.

In single-phase circuits, with the neutral conductor size equal to the phase conductor, you'd double the distance to account for the total voltage drop (phase plus neutral).

RE: How to figure 3 phase voltage drop

To add to the above, the key is to use correct voltage. The voltage drop in single line of a 3-phase (balanced) system is the drop in the single phase voltage.

For example, if you have 0.2 ohms resistance in each phase condutor of a 208/120V, 3-phase system and 15A of load. The VD per line will be 15A*0.2 ohm=3V, The percentage voltage drop will be 3V/120V=.025 or 2.5%. Line to line VD will be 5.2V, still 2.5% of 208V.

Same is true, it were a delta system, even if there is no physical neutral. The calcs are still done on per phase basis, using phase voltage=line voltage/1.732.

(For purists: Above example ignores reactance for simplicity.)

RE: How to figure 3 phase voltage drop

You still need to take a look at the catalog. They list the single phase and 3 phase values of the impedance, which is not the same.

Once you read the correct value from the table, you multiply by twice the distance in 1 phase and simply by the distance in 3 phase.

You should be aware enough to know that as you have more applianced branched in parallel on the same circuit, you need to calculate the voltage drop chunk by chunk; other easier methods exist of course but should be a nice exercise.

Also, the impedance values are given in some catalogs for cos phi = 0.8 as well as for cos phi = 1, which is the worst case (highest resistance).

RE: How to figure 3 phase voltage drop

where can you find these catalogs?

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