Can someone please explain this to me?

[kevin]

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.

 

[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

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.

 

[jumper]

perhaps
but he is correct

Agree with both.

 

[ActionDave]

Agree with both.

Don't push me.

 

[Ingenieur]

Agree with both.

:lol:

Without being too snarky not sure how he could disagree lol

 

[Davidson Electrical]

8 awg is correct!

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

 

[kwired]

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.

 

[benedetti]

Branch Circuit VD is 3%

Branch Circuit VD is 3%

VD= 2KIL/CM

(2 X 12.9 X 30 X 100) / 16510 = 4.68 Volts

230 X .03 = 6.9 volts allowed

B is correct

#10 would allow 7.45 volts

 

[ggunn]

For balanced loads another way to look at is is Vd = IR in a single conductor one way regardless of service type (single or three phase) and then multiply by the line to phase voltage ratio (2 or sqrt3).

 

[oldsparky52]

For balanced loads another way to look at is is Vd = IR in a single conductor one way regardless of service type (single or three phase) and then multiply by the line to phase voltage ratio (2 or sqrt3).

That's all I do for sizing a wire to a known load. Any more complicated and an EE does it.

 

[Open Neutral]

Not an issue here, but note that telco twisted pair cable may well spec loop resistance per k-ft; so there it's

K-ft*Ω/k-ft

not 2x that length. It pays to read the table carefully.

 

[Jmoney]

Vd=(2kil)/acmil.... same as acmil=(2kil)/vd... 230×.03=6.9=vd, 2= 1phase, k=12.9, i=30a, l=100', thus acmil=[(2)(12.9)(30)(100)]/6.9.... acmil=11,217... the acmil got#10 wire (per chapter 9, take 8) 10,380. #10 not big enough so you go up to the next size. #8 good up to 16,510.

 

[WAMA]

This is why the answer is 8 AWG.

This is why the answer is 8 AWG.

E (voltage drop) = 2 X K X I X D /cmils

K=12.9 ohms for copper
I=30A
D=100 Feet
Voltage drop is 3% of the nominal voltage which is 240 volts you always have to use nominal voltage.
.03 X 240 volts = 7.2 volts.

now solve for cmils = 2 x k x I x D / E (voltage drop)
substitute the values now:

cmils = 2 X 12.9 x 30 X 100 / 7.2
cmils = 77400/7.2
cmils= 10,750

go to table 8 in chapter 9 second column Area
go down to the number closest to 10,750 it is 16510,,,and that is 8 AWG.

I hope this answers your question.

 

[Carultch]

E (voltage drop) = 2 X K X I X D /cmils

K=12.9 ohms for copper
I=30A
D=100 Feet
Voltage drop is 3% of the nominal voltage which is 240 volts you always have to use nominal voltage.
.03 X 240 volts = 7.2 volts.

now solve for cmils = 2 x k x I x D / E (voltage drop)
substitute the values now:

cmils = 2 X 12.9 x 30 X 100 / 7.2
cmils = 77400/7.2
cmils= 10,750

go to table 8 in chapter 9 second column Area
go down to the number closest to 10,750 it is 16510,,,and that is 8 AWG.

I hope this answers your question.

Suppose a question is written like this question, where the words focus on voltage drop. However, suppose the size that the voltage drop equation will calculate, is not code-compliant with the load amps specified. Which answer would be correct?

The wire size that will limit voltage drop to the NEC recommendations (or problem statement specification)?
Or the wire size that is code-compliant, and is larger than what is needed to curtail voltage drop?

 

[ggunn]

Suppose a question is written like this question, where the words focus on voltage drop. However, suppose the size that the voltage drop equation will calculate, is not code-compliant with the load amps specified. Which answer would be correct?

The wire size that will limit voltage drop to the NEC recommendations (or problem statement specification)?
Or the wire size that is code-compliant, and is larger than what is needed to curtail voltage drop?

For short runs the minimum size conductor for voltage drop will often be smaller than the minimum size for ampacity. Ampacity rules.

 

[Carultch]

For short runs the minimum size conductor for voltage drop will often be smaller than the minimum size for ampacity. Ampacity rules.

Suppose you are taking an exam (as opposed to doing the calculations practically for a design), and ampacity rules the example in question. Does this mean that a question can be written like this one, with voltage drop as a red herring, while the correct answer is based on the ampacity?

 

[ggunn]

Suppose you are taking an exam (as opposed to doing the calculations practically for a design), and ampacity rules the example in question. Does this mean that a question can be written like this one, with voltage drop as a red herring, while the correct answer is based on the ampacity?

Well, sure. In taking any exam you have to read the questions carefully and make sure you answer what they are asking for. On the NABCEP exam I took, some of the questions were half a page long and full of descriptions and measurements, most of which were irrelevant to what was being asked. In an exam like that, the skill being tested is not whether you can perform the calculation without making a mistake, but rather whether you can sift through the data and pull out the salient points that determine the correct answer. Kinda like the real world.

 

[Unbridled]

I am still studying for my WV Master Electricians Exam. I have been going along pretty good but I just hit a brick wall. Here is the question from my practice test.
A 30 amp 230 volt load is located 100 feet from the source. What is the minimum size branch circuit conductor required to operate within the limits for voltage drop that is recommended be the NEC?
a. # 10 conductor
b. # 8 conductor
c. # 6 conductor
d. #4 conductor

The book says the correct answer is B. I have no idea how to come to this conclusion. Can anyone help me?

I use this formula when looking for conductor size in Single Phase Vd questions.

CM=2KIL/Vd

CM=2*12.9*30*100 / (230*.03)
CM=77,400/6.9
CM=11217
Using Table 8 in Chapter 9 = #8 (16500CM) #10 is only 10380 CM

The Length is already double by the use of "2" in the formula.

 

[crispysonofa]

Don't mean to Hi-jack the thread but what is the constant K derived from?

 

[Smart $]

Don't mean to Hi-jack the thread but what is the constant K derived from?

Actually it is not a constant though many call it such.

It's a value based on the conductivity of a material. Units are Siemens, and the reciprocal of resistivity. A conductor of uniform cross section has a resistance proportional to its resistivity and length and inversely proportional to its cross-sectional area.

Go to Chapter 9 Table 8. Pick any conductor and multiply its cmil area by its resistance per 1,000 ft and divide by 1,000.