Re: 3 phase voltage drop
We can do better than Holt on this one, he said optimistically.
Originally Posted by ggunn
1. Make sure you distinguish between a circuit supplied by a delta or wye transformer and a load which is wired as delta or wye. If you do not keep this in mind, the confusion will be hard to recover from.
2. In a balanced wye source, there will be no current in the neutral. But you can either connect loads or sources from phase to phase or from phase to neutral.
In a balanced delta source, there will be no neutral. So your loads must connect from phase to phase only. (or you can derive a neutral, given a guaranteed balanced condition, but we will ignore that....)
3. The voltage drop is easily calculated as IxR once you know I. (No help there, I know....)
4. Your inverters are using three phases plus EGC or three phases plus neutral? If nominally wye, it is the latter. Each inverter is delivering a balanced output over all three phases or driving just one phase?
5. Assume for the purposes of this calculation that the 480 volt output of the inverters really is wye. That means that the phase to neutral voltage on each leg will be 480 volts.
To get the per-phase current, divide the output power by 3 and then calculate I from P=IV.
The voltage drop which concerns you will be in each of the phase wires only, as long as the source and load remain balanced. A calculator set up for a two wire, single phase circuit will automatically figure a voltage drop in both phase and neutral, giving you the wrong answer.
6. You can now get voltage drop in each single wire after finding the resistance per foot of each wire size. Note that getting the voltage drop you want will probably give you oversize conductors compared to the NEC minimum. I assume that if the wires will be very long you will also be calculating for using Aluminum as an alternative.
7. For a balanced delta, the load current is specified as the current from one phase wire to another, and the power to current relationship is based on that voltage. This is really where the wye versus delta differences all derive from, and the distinction between the transformer type and the load/source connected to it becomes critical.
In a 480 volt delta, the phase to (neutral) voltage will not be 480 when the phase to phase voltage is 480 (totally independent of whether the neutral exists or is carrying current. ) That is where the square root of three comes in, when you draw the triangle vector diagrams with a 120 degree angle at the apex. The circuit is the same when considered as a black box, but the nominal voltage of the circuit is measured at a different place.
In a 480 volt wye, you have both the 480 phase to neutral voltage and the 480 time root 3 voltage from phase to phase available for use.
Did that do it, or are the fish not biting yet?
Last edited by inetdog; September 5th, 2012 at 16:31 PDT.
Reason: expanded slightly.
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