Voltage drop calculations

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  #1  
Old 10-13-08, 10:59 PM
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Voltage drop calculations

Greets all. I have a question as to which table is more accurate.. Table 8 or table 9 of Ch. 9... 99.999% sure it's table 9 but my instructor is insisting everyone use the DC table for AC residential which makes no sense to me. Thoughts on that one?

I was self taught to use Ed = I x R with Table 9

My instructor is insisting on Ed = K x I x L x 2 / CMA with table 8

A 100' 120v 16amp load on 12/2 my way gives me 6.4 volts

Same scenario the instructor way gives 5.88 volts.

Quite a diff if you ask me. I have a feeling the higher guage wire you use, the more skewed Table 8 will become. Am I right on that?

I don't like K being a canned CU=12 and AL=20. I'm assuming this is where the discrepancy creeps in, no?

Thanks all.
 
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Old 10-13-08, 11:29 PM
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O.k... I think K is CU=12.9 and AL=21.2

According to Mike Holt anyway...

That puts Table 8 & CMA much closer @ 6.32 volts...

Is this correct?

Thanks again.
 
  #3  
Old 10-14-08, 12:29 AM
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Eh... Nevermind... Figured it out on my own.

K = Resistance x CMA / 1000

which puts you at 12.6 thru 12.8 for #18 to 4/0...

Sorry to eat up space here.
 
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Old 10-15-08, 11:29 AM
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O.k... I could really use some help here. First off, this is not a homework question. I came up with a fictional homerun and then used 4 different calculations to arrive at 4 different answers.

Which one is right?!@#*?!

Please use your own methods for finding voltage drop otherwise this is a pointless exercise. eg- ignore my previous posts. The more people who answer, the better this will go.

Bob runs a dedicated 14/2 home run, which is 120 feet long, and uses a single pole 15 amp breaker in the panel. The expected amperage draw is 12 amps. What will the voltage drop be on that circuit?

A) 8.41 v

B) 9.04 v

C) 8.84 v

D) 8.93 v

E) None of the above (explain why)

Thanks guys for participating. I believe I'm being taught a skewed wrong way to do voltage drop calculations and I can't likely correct my instructor without getting an attitude. So, I'm turning to you all for the right answer. I'll still have to use the wrong way on tests but I don't want to do that in the field.
 
  #5  
Old 10-15-08, 02:34 PM
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B) 9 volts. Bob needs to run #10 AWG.
 
  #6  
Old 10-15-08, 07:09 PM
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Please read this article. It's a great explanation, but it won't get an A on your homework assignment.

Truth is, a 9-volt drop on a 120-volt circuit is outside the NEC's recommendations, but not its requirements.
 
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Old 10-15-08, 07:44 PM
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So was your vote B as well?

Thanks for the link. Knew it already except for the bottom reverse equation stuff. My whole point was to see what methods people use to arrive at an answer, not to learn more about voltage drop. Links are always good tho. Bookmarked.

B, C and D are all technically right and they differ so little that it doesn't really matter. I'm in the opinion that D is the most correct answer (table 9), then progressively down to answer A (which I think is wrong).

I could live with using table 8 but our curriculum is stating the direct current constant is 12/20, not 12.9/21.2.... Doesn't even state K = R x CMA / 1000, which is marginally better than using a canned 12.9....

Anyway... Thanks for playing thinman. You can pick up your door prize on the way out... One person isn't much of a test base but I guess it'll have to do.
 
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Old 10-15-08, 11:07 PM
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Madpenguin.,

I know this is not really fair but I did used the electrician calucator and I ran few diffrent angles and came up with 8.6 volt drop.

With manual figues { hand and brain work Beer 4U2 }

It came up 8.90 volt drop

Merci,Marc
 
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Old 10-15-08, 11:46 PM
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Mad
Remember that the tables are considering an operating temperature of the wire of 75C. Usually or the majority of the time the wire is not at 75C. but a lower temperature. A 12 AWG copper wire with 75 degree insulation has an ampacity of 25 amperes in Table 310.16. Using 310.16 this means that with 3 current carrying conductors in a raceway and a ambient temperature of 30 degrees C. the #12 will reach 75 degrees C, when 25 amperes is applied. I haven't done the math but if your instructor is using k=12 for copper I would assume he is considering a 16 amp load (I believe K is around 12 at 16 amps) which would be the maximum load for a 12 awg copper wire considering a 20 amp circuit breaker supplying a continuous load. When ever you use the k method you must be aware that K changes with the temperature of the wire in relation to the amperage flow. At at 16 amps using 12 awg copper.... K is no longer 12.9. and 75C its more like 12 and 47 C.
 
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Old 10-16-08, 01:36 AM
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There he is....

Originally Posted by Bruto View Post
using k=12 for copper I would assume he is considering a 16 amp load (I believe K is around 12 at 16 amps) which would be the maximum load for a 12 awg copper wire considering a 20 amp circuit breaker supplying a continuous load.
I don't understand. This wasn't just on a vd calc concerning a 16 amp continuous load.... We are supposed to use k=12 for everything. This comes from the EWR (electrical wiring residential), chapter 4, which is part of our curriculum.

K = approximate resistance in ohms per circular-mil foot at 167F (75C).
For uncoated copper wire, use 12 ohms*
For aluminum wire, use 20 ohms*

* Derived values from Tables 8 and 9, Chapter 9, NEC.
Originally Posted by Bruto View Post
When ever you use the k method you must be aware that K changes with the temperature of the wire in relation to the amperage flow.
Ah.... Good to know. This book makes no mention of that. It flat out states that K=12 period for all residential voltage drop calculations. But also says "It is sufficiently accurate for voltage drop calculations necessary for residential wiring".

Up until now, I've been using VD=IxR from the values in Table 9. This is the best way, no? Seems to take more into consideration... Haven't the faintest idea what effective Z and Power Factor are tho... "Notes 2" is total greek to me.

The rest of your post I'll have to mull over after I've slept.

Thanks Bruto.
 
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Old 10-16-08, 09:39 AM
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A k of 12 is probably more realistic for residential wiring since wire temperatures would not often reach 75C. The 75C column of 310.16 is generally the max termination rating of most equipment. and the corresponding amperage would relate to a K of 12.9 temperatures lower than that would have smaller resistances and K factors. For wires smaller than 1/0 awg the difference between ac and dc resistances is negligable using the same k. The DC resistance of table 8 must be adjusted for larger wire above 1/0 by calculating a different K that adjusts for eddy current effects and is often called the Q factor. that factor is obtained by dividing the ac ohms to neutral impedance in chapter 9 table 9 by the dc resistance in table 8.
 
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Old 10-16-08, 10:54 AM
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Thanks Bruto for the patience and explanation. It's most appreciated. I am a first year, after all. No doubt, It'll come eventually.

Thanks.
 
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Old 10-16-08, 11:33 AM
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Your welcome. Voltage drop calculators use 12.9 for copper and 21.2 for aluminum this simply insures that a large enough wire will be chosen. Advanced voltage drop calculators will have you input ambients, the expected operating amperage, continuous or non-continuous load but these calculators would most likely never be used in residential wiring design. These calculators consider the change in k for the voltage drop.
 
 

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