Voltage Drop equation?

[JasonCo]

I have a homework question, I've already worked it out. I just want to make sure I got the right answer.

Scenario: For a load, the conductors consist of two 12 AWG THHN/THWN solid coated copper conductors connected to a 2-pole breaker in a 120/208-volt, 3phase, 4-wire panelboard. The circuit serves a load that draws 9.8 amps. The distance between the panel and the load is 150 feet.

Question: Refer to the scenario, Find the voltage dropped in the conductors. The voltage dropped in the circuit conductors is __ %?

A: 4.92
B: 3.0
C: 2.2
D: 4.5
E. 2.84

Okay, I'm reviewing for my test and looking back at old quizes I took in class. For WHATEVER REASON I circled E as my answer. But re-doing the question now, I'm getting A as my answer. Anyone able to double check my work?

Equation is 2KIL/CMA.
I did 2 x 13.1253 x 9.8 x 150 / 6530 = 5.9094 voltage dropped across the conductors. But question is asking the %. So I did 5.9094 / 120 = 4.92% as the answer.

Not sure if I'm doing the calculations correctly. If I need to double 150 feet. Or if I need to divide by 208 instead (which I doubt it because its 2-pole). But anyways, just wanted someone to hopefully proof me! Thanks a bunch for your help!!

 

[GoldDigger]

You do need to divide by 208 since that is the load voltage you are dropping from.
You can actually make it even more complicated with the vectors involved, but the answer still comes out the same.
The percent voltage drop at the load is not the sum of the percent voltage drops on the two lines, which makes it a little weird if you try to do it that way.

 

[JasonCo]

its a 2-pole breaker so that means its running under single phase correct?

Oh........ okay nvm I understand now. I would use 208 because it is 2-pole BUT NOT use 1.732 in the equation because its single phase and not a 3-pole. Yeah had a bit of a brain fart. I understand fully now. Thanks a bunch man, appreciate that!

 

[topgone]

I have a homework question, I've already worked it out. I just want to make sure I got the right answer.

Scenario: For a load, the conductors consist of two 12 AWG THHN/THWN solid coated copper conductors connected to a 2-pole breaker in a 120/208-volt, 3phase, 4-wire panelboard. The circuit serves a load that draws 9.8 amps. The distance between the panel and the load is 150 feet.

Question: Refer to the scenario, Find the voltage dropped in the conductors. The voltage dropped in the circuit conductors is __ %?

A: 4.92
B: 3.0
C: 2.2
D: 4.5
E. 2.84

Okay, I'm reviewing for my test and looking back at old quizes I took in class. For WHATEVER REASON I circled E as my answer. But re-doing the question now, I'm getting A as my answer. Anyone able to double check my work?

Equation is 2KIL/CMA.
I did 2 x 13.1253 x 9.8 x 150 / 6530 = 5.9094 voltage dropped across the conductors. But question is asking the %. So I did 5.9094 / 120 = 4.92% as the answer.

Not sure if I'm doing the calculations correctly. If I need to double 150 feet. Or if I need to divide by 208 instead (which I doubt it because its 2-pole). But anyways, just wanted someone to hopefully proof me! Thanks a bunch for your help!!

Where did you get K=13.1253?

 

[JasonCo]

Where did you get K=13.1253?

We know from the scenario, it says that the conductors being used are 12 AWG THHN/THWN solid coated copper conductors. With this information you can find K. To do this, all you need to do is go to Chapter 9, table 8 in the code book towards the back of the code book. Find the part of the table that is for wire size 12. Then look for the key word that says "solid", it'll either say solid or stranded. Sense it is solid, we will base our information off of the section under Quantity 1 (not 7), on the table. Now with the key word "copper coated" we can go to that part of the table and find the number 2.01 under kFT. Once we have that, we need to convert it into feet. We know that kilo = 1000 per foot so 2.01 kFT converted into FT = .00201 ft. Now that we have this information, lets focus our energy towards the formula for finding K. Which is K = CM x FT(ohms). We will multiply .00201 (ohms_ft) by the circular mill area on the same table. Go over to the part that says Circular mils, and you will see for number 12 wire, the circular mils is 6530.

Now that we have 6530 and .00201. plug it into the formula I just said, which is K = CM x FT(ohms).
So 6530 x .00201 = 13.1253. This is how you find K when the question doesn't just give it to you.

 

[Julius Right]

Why at first to multiply and then to divide by cmils?
It is not very important here, however I should use Z from the Table 9 [even it is built for 3 conductors, since the ac resistance is the same and the reactance also -in equilateral configuration].
and a pf of 0.85 -usually-it is close to .

 

[JasonCo]

Why at first to multiply and then to divide by cmils?
It is not very important here, however I should use Z from the Table 9 [even it is built for 3 conductors, since the ac resistance is the same and the reactance also -in equilateral configuration].
and a pf of 0.85 -usually-it is close to .

Not sure if I'm follow what you are saying exactly. The reason I multiplied all that and then divided by the cmils is because the formula asks you to do so. Formula for single phase loads is (2KIL)/Cmils

2 = 2
K = 13.1253 (Using formula for finding K and using Chapter 9, table 8)
I = 9.8 amps
L = 150 feet
Cmils = 6530 (Shown in Chapter 9, table 8)

So (2 x 13.1253 x 9.8 x 150) / 6530 = 5.9094 volts dropped.
Then to find %, 5.9094 / 208 = 2.84%

I don't think you can do this any other way, at least theoretically for test taking purposes.

 

[RichB]

I used the generic 12.9=k and got the same answer

 

[Smart $]

FWIW, the scenario states solid coated copper, but I've never known THHN/THWN to contain solid coated copper conductors. :happysad:

 

[K8MHZ]

If you know the ohms per 1000 feet, why bother with the K? Maybe I am missing something, but it looks like figuring K in is a bunch of extra work if the ohms per 1000 feet is also listed.

 

[Smart $]

If you know the ohms per 1000 feet, why bother with the K? Maybe I am missing something, but it looks like figuring K in is a bunch of extra work if the ohms per 1000 feet is also listed.

I agree... but many seem to be adamant on using the (2)KIL/CMIL formula rather than (2)RID.

2×2.01ohms/1000ft×9.8A×150ft=5.9V

 

[K8MHZ]

2×2.01ohms/1000ft×9.8A×150ft=5.9V

Yep, I got 5.91. Divide that by 208 to get .00284. Times 100% = 2.84%.

 

[K8MHZ]

I agree... but many seem to be adamant on using the (2)KIL/CMIL formula rather than (2)RID.

I wonder how many of them got my 55 gallon drum question right?

 

[Julius Right]

The impedance Z from the NEC Table 9 of Chapter 9 Tables is calculated using formula:
Z=R*cos(fi)+X*sin(fi) an approximation used on IEEE 141/1993
ch.3.11 Calculation of voltage drops-for instance. Then voltage drop VD in absolute value |VD|:
|VD|=|Z|*|I| [Volt]
:weeping:

 

[K8MHZ]

Occam's Razor anyone?

I haven't heard that mentioned here in a while, and I kind of couldn't resist bringing it up.

 

[K8MHZ]

The impedance Z from the NEC Table 9 of Chapter 9 Tables is calculated using formula:
Z=R*cos(fi)+X*sin(fi) an approximation used on IEEE 141/1993
ch.3.11 Calculation of voltage drops-for instance. Then voltage drop VD in absolute value |VD|:
|VD|=|Z|*|I| [Volt]
:weeping:

In my book, Table 9 is for three phase, three single conductors in conduit. The OP's question was about a 2 wire load (one phase) so I don't think Table 9 applies.

Also, Chapter 9 has no values for coated wires.

 

[Smart $]

Occam's Razor anyone?...

Per wikipedia...

Occam's razor (also written as Ockham's razor and in Latin lex parsimoniae) is a problem-solving principle devised by William of Ockham (c. 1287–1347). It states that among competing hypotheses, the one with the fewest assumptions should be selected.

 

[K8MHZ]

Per wikipedia...

From the same article:

The widespread layman's formulation that "the simplest explanation is usually the correct one" appears to have been derived from Occam's razor.

 

[junkhound]

Good education here on how different folks look at simple problems differently.

Original question does not give the temperature, so 2.2 vs 2.84 correct if a more probable 20C assumed vs 75C.


Or is the NEC so sacred that the chap 9 is one size fits all ?

 

[junkhound]

FWIW, the scenario states solid coated copper, but I've never known THHN/THWN to contain solid coated copper conductors. :happysad:

+1, never have seen tinned or plated solid THHN either. Maybe some old R wire.
Stranded Nickel plated copper yes in 260C aerospace rated wire, or silver plated for 200C TFE wire, but not in THHN.

The original homework question very poorly set up, aka 'unobtainium', never seen in real life.