1. Member
Join Date
Feb 2003
Location
Post Falls, ID
Posts
74
Originally Posted by jrohe
I suspect they are expecting you to use the method that uses DC-constant resistance for determining voltage drop to calculate the minimum wire area, in cmils, required.

For single phase circuits (assumed because that information is missing in the test in the question), the equation is:

VD = 2 x K x I x D / CM

Rearranging the equation to solve for CM results in the equation:

CM = 2 x K x I x D / VD

Making assumptions for copper conductors and VD% cannot exceed 3 percent because the information is missing in the test question:

K = 12.9 ohms
I = 30 amps
D = 100 feet
VD = 230 volts x 0.3 = 6.9 volts

Therefore:

CM = 2 x 12.9 x 30 x 100 / 6.9 = 11217.4 cmils.

Looking at Chapter 9, Table 8, the smallest conductor with this area is #8 AWG conductor at 16510 cmils. A #10 conductor has an area of 10380 cmils, which is too small, at least according to this calculation method.

I agree completely with this reply by Jason; he is spot-on correct.

2. Moderator
Join Date
Jun 2003
Posts
6,601
Originally Posted by steve66
Actually, I think the Southwire calculator assumes 3 phase. That makes a bigger difference than the power factor. I don't seen anywhere on the Southwire calculator where I can choose 1 phase.

But back to the power factor - it makes some difference in the calculated impedance of the wire. If you know R and X for a wire, then Z depends on R and X and the power factor.

Z = X*PF+R*Sin(Acos(PF))

(I think that's right - I'm trying to read it from an excel formula.)

Just as an example, my Square D calculator lists L-L voltage drop per amp per 100' in magnetic copper wire: (trying to make a table, which isn't working)

....................... ..... #8 Wire #10 wire
1 phase: 90% PF. . 0 .15 ....... . 0 .23
1 phase: 80% PF ...0.13 ........ 0 .20
3 phase: 90% PF .. 0 .13 ........ 0 .20
3 phase: 80% PF... 0 .11 ........ 0 .17
Originally Posted by oldsparky52
So, ... as PF goes farther away from 1, the resistance in the circuit becomes less? So the worse the PF the less resistance?
Not quite. First of all, we are talking about the wire impedance, not just resistance.

And a low PF makes the voltage drop a little less for small wire where the wires resistance is fairly large compared to the wires reactance.

But for larger wires, where reactance is larger, the voltage drop is less for higher power factors, with 2/0 wire actually having about the same drop for 95% PF and 80% PF.

3. Originally Posted by Ingenieur
perhaps
but he is correct
Agree with both.

4. Originally Posted by jumper
Agree with both.
Don't push me.

5. Originally Posted by jumper
Agree with both.

Without being too snarky not sure how he could disagree lol

6. Junior Member
Join Date
Aug 2017
Location
Orange tx
Posts
5

## 8 awg is correct!

Mike uses the formula 2xKxIxD/Vd to find the circular mils of the wire.
nec recommends no more the 3% so..
vd=6.9 voltage drop (3%of 230)
D=100' distance
I=30a current
K=12.9 for copper wire.

cmils= 2x12.9x30x100/6.9 = 11,217cmils

Chapter 9 table 8 back of book
10awg = 10380 =too small to limit vd to 3%
8awg = 16510= correct

7. Originally Posted by Daniel malack
I just looked on several breakers GE 40°, Sq D home line 40°, Sq D QO 40°, Siemens 40°, Cutler Hammer Ch 60-75° Cutler Hammer bolt in 60-75° most are 40°. I got no reason to lie to ya.

Sent from my SM-T350 using Tapatalk
40 deg rating is not terminal temp rating it is the ambient temp that the trip rating was calibrated for.

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