Voltage drop calculations

[Jpflex]

I don’t understand entirely why, when I solve algebraic for unknown CM size (NEC chapter 9 table 8) of conductor able to yield less than 3 percent voltage drop from source to final breaker, I get a CM/ AWG size too small for the ampacity required


For a 3 phase Delta delta, 4800 volt primary 480 volt secondary, secondary current 360.85 amperes I get an answer of 1/0 aluminum wire good for only 120 amperes with a voltage drop of 14.4 volts

14.4 volts/ 480 volt is within target 3% voltage drop but the answer yields a conductor less than allowed for its ampacity

Is this to say this is the minimum size wire to yield this VD?

14.4 VD = 1.732 (3 phase) x 21.2 (copper K resistance per foot mill) x 360.08 (Amperes) / divided by unknown CM

SOLVE FOR CM

 

[Dennis Alwon]

I can't explain it but the formula is not perfect as I have run into that problem also.

 

[Rock86]

Well I think one factor that needs to be considered is that the ampacity rating is not based on the conductor but by the insulation being used. Therefore if you are only looking for CM it would be very likely that you would come up with a lower wire gauge than the NEC would allow because you are solving an equation based on only what will happen to (in this case) the copper conductor.

 

[Jpflex]

Ye

Well I think one factor that needs to be considered is that the ampacity rating is not based on the conductor but by the insulation being used. Therefore if you are only looking for CM it would be very likely that you would come up with a lower wire gauge than the NEC would allow because you are solving an equation based on only what will happen to (in this case) the copper conductor.

Yes I came to realize this later. Since I am making equation = to voltage drop, the result will be exact size needed to drop achieve this voltage drop regardless if conductors insulation can handle this amount of current on a small conductor

 

[Dennis Alwon]

If you study the formula you will see you will get bad results when the distance is less than the amperage. For example--- 60 feet @ 100 amps will give bad results however 100 feet @ 60 amps will give an accurate number. I believe this is correct

 

[ggunn]

I don’t understand entirely why, when I solve algebraic for unknown CM size (NEC chapter 9 table 8) of conductor able to yield less than 3 percent voltage drop from source to final breaker, I get a CM/ AWG size too small for the ampacity required


For a 3 phase Delta delta, 4800 volt primary 480 volt secondary, secondary current 360.85 amperes I get an answer of 1/0 aluminum wire good for only 120 amperes with a voltage drop of 14.4 volts

14.4 volts/ 480 volt is within target 3% voltage drop but the answer yields a conductor less than allowed for its ampacity

Is this to say this is the minimum size wire to yield this VD?

14.4 VD = 1.732 (3 phase) x 21.2 (copper K resistance per foot mill) x 360.08 (Amperes) / divided by unknown CM

SOLVE FOR CM

There is nothing wrong. Minimum ampacity and voltage drop calculations are done with different variables; you need to consider both and use the larger of the two results. For long conductors voltage drop will usually dictate the minimum conductor size and for short conductors minimum ampacity will usually prevail.

 

[Jpflex]

There is nothing wrong. Minimum ampacity and voltage drop calculations are done with different variables; you need to consider both and use the larger of the two results. For long conductors voltage drop will usually dictate the minimum conductor size and for short conductors minimum ampacity will usually prevail.

How many feet do you need until voltage drop tends to matter, considered a long run or exceeds 3 to 5 percent?

I noticed so far that a few hundred feet didn’t seem to drop voltage much.

As I was saying before, I realize now that placing the formula equal to a 3 percent drop will yield a conductor cm size needed to achieve this drop but may not be large enough for the amperes put into the equation or going through the conductor realistically

 

[Tulsa Electrician]

I would always start with VD based on the minum circuit conductor needed. Then once that falls out of percentage then go to CM. This will yield accurate results.

 

[Dennis Alwon]

How many feet do you need until voltage drop tends to matter, considered a long run or exceeds 3 to 5 percent?

The VD, of course, varies depending on the ampacity also. If you consider the formula at full ampacity, ie, 20 amps for #12 and 40 amps for #8 then 150' would be pushing the limit but again that will vary.

I noticed so far that a few hundred feet didn’t seem to drop voltage much.

Again that depends on the amperage

As I was saying before, I realize now that placing the formula equal to a 3 percent drop will yield a conductor cm size needed to achieve this drop but may not be large enough for the amperes put into the equation or going through the conductor realistically

I think this may be true only when the distance is short and the amperage is high. For instance a 50' run at 100 amps with #6 wire will result in a 2% drop but we know #6 isn't large enough for the 100 amps.

 

[wwhitney]

Is this to say this is the minimum size wire to yield this VD?

14.4 VD = 1.732 (3 phase) x 21.2 (copper K resistance per foot mill) x 360.08 (Amperes) / divided by unknown CM

SOLVE FOR CM

Your formula above doesn't include the length in feet. So you would be solving for the VD for 1 foot of conductor.

As for the broader question, we know that ampacity/CM is not constant but is decreasing with increasing CM (since heat rejection goes by the surface area, or sqrt(CM)). The voltage drop formula, however, is inversely proportional to CM.

So the distance at which a voltage drop criterion will control wire size, rather than an ampacity criterion, will (a) depend linearly on voltage (since the criterion is a % of the system voltage (b) depend on the circuit configuration (2-wire vs 3-wire balanced single phase vs 3 phase etc) via the coefficient in the voltage drop formula and (c) be larger for larger currents.

Cheers, Wayne

 

[ggunn]

How many feet do you need until voltage drop tends to matter, considered a long run or exceeds 3 to 5 percent?

I noticed so far that a few hundred feet didn’t seem to drop voltage much.

As I was saying before, I realize now that placing the formula equal to a 3 percent drop will yield a conductor cm size needed to achieve this drop but may not be large enough for the amperes put into the equation or going through the conductor realistically

As my dad was fond of saying, how long is a piece of string?

What you are seeing is completely normal. Voltage drop percentage depends on conductor resistance and length, current, and voltage a la Ohm's Law. Ampacity depends on the temperature stability of conductor insulation for a given amount of current. The two calculations have very little to do with each other, and either can set the lower bound on conductor size. Run them both and select the higher minimum conductor size.