Voltage drop in wire. How do YOU calculate it?

[fastline]

I have recently run into a discrepancy over voltage drop in large service entrance wire and curious how you guys are calculating it. DC power seems pretty basic but this would be 60hz AC current.

Example would be 4/0 Aluminum. Looking for the hard calculations. One that is indicated is (2*L * R *I)/1000. I don't think this is accounting for AC but rather DC current. Can someone sharper than me clarify correct calculations, not just an NEC table? Concerns with the wire are not the wire heating but rather the Vdrop due to the run length.

Thanks

 

[crutschow]

Whether it's AC or DC current makes no difference in calculating the voltage drop due to line resistance. Just multiply the AC current (rms value) times the line length times the resistance per unit length. This will give the ACrms voltage drop in the line (the value that an AC voltmeter would measure). Remember to multiply that by two since there are two wires carrying the current (one hot and one return).

 

[fastline]

Maybe I am off center here but whether it is a ph-N or ph-ph load, there is sort of a loop that is created as current passes "to" the load on one conductor and passes "back" to the transformer on the other? Why do I rarely see the same amount of load on the neutral and the NEC allows a reduced size N conductor in some installations? Is this because most of the load is just on the ph-ph loads and there is a calculation done to consider how much actual N load there is? Guess I am still a little confuses as to the distance being calculated.

 

[RCinFLA]

Maybe I am off center here but whether it is a ph-N or ph-ph load, there is sort of a loop that is created as current passes "to" the load on one conductor and passes "back" to the transformer on the other? Why do I rarely see the same amount of load on the neutral and the NEC allows a reduced size N conductor in some installations? Is this because most of the load is just on the ph-ph loads and there is a calculation done to consider how much actual N load there is? Guess I am still a little confuses as to the distance being calculated.

There are two reasons why neutral will not carry the current that L1 and L2 do. First and most obvious is high power loads (air conditioners, hot water heaters, cooking ranges) are usually 240 vac powered which have zero neutral current.

A breaker box that is fairly balanced has 120 vac loads on either side of the 240 vac L1, L2 lines. If there is a load of 10 amps from L1 to neutral and a load of 10 amps from L2 to neutral, the neutral line back to utility transformer has zero current. The net neutral return current for any given load situation is difference between side 1 and side 2 120vac loads. This is because the two are out of phase so their neutral current cancels out.

If the out of phase part confuses you think of putting two identical 100 watt 120v bulbs in series. You can run them on 240 vac with or without a neutral connection.

As an FYI, NEC wire size / amperage standards are based on wire heating per unit length, not voltage drop. It is up to the installer to ensure there is not excessive voltage drop for a given total run length. Wire heating is a safety issue. The number that is close to NEC wire table is roughly 2 watts per foot of wire. There is variation for type of insulation and conduit type which effects the ability of the wire inside to dissipate the heat. Larger diameter wire has greater surface area to dissipate heat. For example, SOOW cable with heavy insulation is derated for maximum current handling. Obviously the number of current carrying wires effect the amount of heat generated. This matters if more then one circuit pair is put in a conduit or if the circuit is a three phase system.

 

[KeepItSimpleStupid]

(2*L * R *I)/1000 particular formula is essentially derived from:

2 * L ; Hot and return as was said
R/1000; is from the table where R is ohms/1000 ft, thus (R/1000)/1000 is R

You have to use the table for aluminum and for the specific wire gage.

P = V * I works when the loads are resistive and thus V is an RMS value and I is an RMS value and it would be equlivelent to DC.

Remember that the NEC also classifies loads as Continuous such as space heating and they will be derated.

The service entrance calcs will be quite different. You wont have the AC and heat on at the same time, for instance. Your home stove won't be on for more than 3 hours at a time and would not be continuous, but in a restauraunt it would be.

The NEC stuff is all based on self-heating, so type of cable, in conduit, ambiant, motor, how long it's going to be on all enter in. Tolerable drops are on the order of 3 to 5% max depending on use and ratings.