Voltage drop and wire size

[n8586m]

While I am just sitting here waiting for spring so that I can empty out my storage building which will allow me to take delivery of the new tractor I have picked out, I have seen several threads asking about wire size to run for added circuits. It appears that there is some confusion on just how to arrive at the correct wire size. Maybe I can shed some light on how to find the right wire size.

We will use the following formula:
This method is a bit more involved that the Ohms Law method, but the big advantage is you can use it for three-phase or single-phase. Here are some additional items to observe:

* Single-phase VD = 2 x K x I x D/CM.
* Three-phase VD = 1.732 x K x I x D/CM. The difference between this and the single phase formula is you replace the 2 with 1.732.
* K = Direct-Current Constant. K represents the dc resistance for a 1,000-circular mils conductor that is 1,000 ft long, at an operating temperature of 75ｺC. K is 12.9 ohms for copper and 21.2 ohms for aluminum.
* I = Amperes: The load in amperes at 100% (not at 125% for motors or continuous loads).
* D = Distance: The distance the load is from the power supply. When calculating conductor distance, use the length of the conductor溶ot the distance between the equipment connected by the conductor. To arrive at this length, add distance along the raceway route to the amount of wire sticking out at each end. An approximation is good enough. Where we specify distances here, we are referring to the conductor length.
* CM = Circular-Mils: The circular mils of the circuit conductor as listed in NEC Chapter 9, Table 8. I will include some standard sizes later.

Now we all remember 8th grade algebra so we can cross multply the formula and solve for the CM or wire size by picking the voltage drop we can stand for the application, usually not more than 5%.

CM= 2 x K x I x D/VD single phase, CM= 1.732 x K x I x D/VD three phase
(continued)

 

[n8586m]

Here are some of the most common wire sizes in circular mills (CM):

Wire Cir.Mills
Size
12 6530
10 10380
8 16510
6 26240
4 41740
3 52620
2 66360
1 83690
1/0 105600
2/0 133100
3/0 167800
4/0 211600

Recently someone wanted to know what wire size to run for four lights at a distance of about 300 feet, the total amp load was about 4. So let's do the math. Since he was using 120 volts, let's allow a voltage drop of 5% or 6 volts. The lights will run just fine on 114 volts.

CM= 2 x 12,9 x 4 x 300/6= 5,160 From the table, a #12 wire will be just fine.

If you got any questions, I will try to answer them. Hope this helps.

 

[SPYDERLK]

Actual conductor length is twice the distance from source to load. Im guessing that is accounted for by the "2" and the "3^1/2" for the 2 and 3 phase formulae?
larry

 

[Wayne County Hose]

Eeesh. I have a degree in electronics and you guys lost me. Actually I can comprehend it, but it's easier to look in a book.

 

[n8586m]

SPYDERLK

It's not do to the length out and back but I will have to look into it, the 1.732 is the square root of three which comes from the three phase. I'll get back to you on where the 2 comes from.
Doug

 

[n8586m]

SPYDERLK

OK, you are correct, it is the hot going out and the neutral coming back. On the three phase it is the the square root of the three hot wires only.
Doug

 

[ray66v]

It really boils down to how much drop is acceptable. Is the person willing to accept a 5% drop or a 6 or more percent drop? NEC suggests a 5% or less drop, and that is a good rule of thumb. I have found that in some cases a drop larger than 5% can produce acceptable results.

 

[westbrooklawn]

So does that mean in your example the 300 should have been 600? (double the length of the distance due to "out and back" wires? That makes a substantial difference in the results.

 

[SamWalton]

I attempted to attached a searchable NEC 2002 electrical code book, but it exceeds the file size. Unless you are running your feeders excessive distances or powering sensitive equipment the voltage drop shouldn't be a problem. It would certainly help to get a little back round on what you are doing, like the distance of your feeders and the devices you are delivering power too.

 

[n8586m]

westbrooklawn

No, the "D" is the one way distance, the "2" in the equation takes care of the "in and out" distance. Note only does the NEC not require a consideration for voltage drop, they don't require a change in wire size because of it. That is somewhat of a surprise as the NEC normally over engineers everything.
Doug