Feeder Calculation Troubles, why?

[GFourteen]

I have a #3 THW Cu, 3wire 120/240v single phase feeder 219' in length. If I wanted to find the Ampacity per phase I should use this formula right?

I= VD*CM/D*K

CM- cmil of 52,620
D- 219'
K- (approx) 25.8
VD-(Voltage Drop) should I use 120v or 240v to get my 3% from 210.19(A)FPN:
why use one over the other? If this is a stupid question I'm sorry, I just don't understand

 

[david luchini]

If you want to find the ampacity of your #3 THW feeder, you should use Table 310.16.

 

[tryinghard]

I have a #3 THW Cu, 3wire 120/240v single phase feeder 219' in length. If I wanted to find the Ampacity per phase I should use this formula right?

I= VD*CM/D*K

CM- cmil of 52,620
D- 219'
K- (approx) 25.8
VD-(Voltage Drop) should I use 120v or 240v to get my 3% from 210.19(A)FPN:
why use one over the other? If this is a stupid question I'm sorry, I just don't understand

Single phase formula should be
VD = 2xKxIxD/CM
or
D = CMxVD/2xKxI
K=12.9 for cu & 21.2 for al (but you can calc these as well K=(RxCM/1000)). If you want to respect the 3% VD its the value of like 240x.03=7.2VD, so you can have up to 7.2 volts drop on a 240V circuit, your case is a 240v circuit not 120 so you'd use the 240.
? = 2x12.9xAmpsx219/52,620
Not stupid question at all.

 

[suemarkp]

Voltage drop is not a code requirement. When I've done long feeders, I do the calculations a few different ways. First you need to know what the typical load is how balanced it is. In workshops, perhaps there is a single large 120V tool. There may also be some normal 120V and 240V loads all the time (e.g. balanced lighting, heat), and some intermittent 240V ones.

So I'd calculate the 120V voltage drop with one ungrounded conductor having the sum of the 120V big load and the 240V baseline load. Calculate the drop in the neutral with just the 120V large load (remember to use a different CM value if the neutral conductor is smaller than the ungrounded). Treat as a % of 120V.

I'd then calculate the 240V voltage drop using the calculated load assuming it is evenly balanced. This is calculation most people do. Treat as a % of 240V.

Your "answer" is the worst of the two calculations.

 

[erickench]

You reduce voltage drop by increasing the size of the conductor thus reducing the resistance. You must use the formula to determine what size conductor you want to use and at what voltage drop.

 

[tryinghard]

I'm one who can be too hung up on the 3% voltage drop but in reality it doesn't matter most often. An example can be 240v - 3% = 233v, the 240 is usually nominal and the resulting 233v usually doesn't have any ill effect on what it supplies, most equipment 230v+/- rated can operate on 208v as well.

Check out this table iwire found from Table 1 National Steady State Voltage Regulation Standards