Voltage Drop Visited again.

[kid_stevens]

New Mexico's NMAC 14.10.4 Electrical code uses 2LRI/1000 with I at 80%.

I find 2LRI/P on Mikes Excel spread sheet. For tests purposes here in NM If it does not hint at NM you use the NEC and national info not NMs. No constant is shown on our tests they are as simple as; You have 8 ga THW going 450 feet at 40 amps to a motor at 120V, What is the VD. They are looking for correct math not correct electrical answer so the Formula only matters.

Length = L
Parallel = P
Resistance = R per foot
Current = I

So I get using the NMAC version 2*450*.35*32 (40*.8)/1000 = 10.08 VD

Where the Spread sheet gives me 40*.35*450*2/1= 12600 VD

If I use the per foot value for the resistance I get

2*450*.000778*32/1000 = .02 VD

Spread Sheet says 40*.000778*450*2 = 28.01 VD
And actually using the Spread sheet at 40Amps is 28.01 V

Using the spread sheet with 32Amps (40*.8) is 22.41 V

Which way is right. I never got this answer correct on the test in two times. another monkey in the mess is Telecomm teaches power Techs 2*K*L*I/CM for VD but that is DC only. And we had to carry a constant in our heads for K.

So anyone got a good answer?

 

[bob]

New Mexico's NMAC 14.10.4 Electrical code uses 2LRI/1000 with I at 80%.

I find 2LRI/P on Mikes Excel spread sheet. For tests purposes here in NM If it does not hint at NM you use the NEC and national info not NMs. No constant is shown on our tests they are as simple as; You have 8 ga THW going 450 feet at 40 amps to a motor at 120V, What is the VD. They are looking for correct math not correct electrical answer so the Formula only matters.

Length = L
Parallel = P
Resistance = R per foot
Current = I

So I get using the NMAC version 2*450*.35*32 (40*.8)/1000 = 10.08 VD?

Your problem here is R = 0.78 and not .35. I would use VD = IxR for an good approximation. 2*450*0.78*40/1000 = 28 volts.

Spread Sheet says 40*.000778*450*2 = 28.01 VD
And actually using the Spread sheet at 40Amps is 28.01 V

Using the spread sheet with 32Amps (40*.8) is 22.41 V

Which way is right. I never got this answer correct on the test in two times. another monkey in the mess is Telecomm teaches power Techs 2*K*L*I/CM for VD but that is DC only. And we had to carry a constant in our heads for K.

Using 2*K*L*I/CM is a DC caculation but the error is small for an AC result.

 

[haskindm]

There are many ways to figure voltage drop. The two ways I teach in my class are what you refer to as the Telecomm method and using the table eight value for resistance per 1000 feet.
You are correct the Telecomm method is based on DC resistance, but so is the value from table 8. I believe that this will yield a result that is "close enough" even though we usually use AC current. This is especially true in the real world where many electricians either do not understand voltage drop or choose to ignore it.
Using table eight in the NEC voltage drop would be calculated as:
2*L*R/1000*A = VD
L=one-way length of conductor
A=amperage of circuit.

You may also use table eight to compute the exact K value for the "Telecomm Method"
Take the ohm/1000 ft figure multiply it by the circle mil and the divide by 1000. You will notice that the figure for 1000 kcmil uncoated copper will give you a k-factor of exactly 12.9. # 12 will give you an exact k-factor of 12.9294 so for practical purposes 12.9 is close enough.

 

[zazmat]

I found this link very useful in doing VD calculations

http://www.electrician2.com/calculators/wireocpd_ver_1.html

 

[Smart $]

New Mexico's NMAC 14.10.4 Electrical code uses 2LRI/1000 with I at 80%.

I have to assume here that "I" at 100% is circuit ampacity and at 80% is max continuous load permitted on that circuit. This is calculation is not for voltage drop at the load but rather maximum voltage drop for a continuous load circuit (assuming again the OCPD is not 100% duty rated?which is typical).

Basic voltage drop considers only the load current, for which circuit ampacity has no bearing. The example you give, as quoted below, would use the current passing through the circuit (40) as the "I" value in your calculation.

For tests purposes here in NM If it does not hint at NM you use the NEC and national info not NMs.

So why did you try to calculate it in your example, which "does not hint at NM"?!!

No constant is shown on our tests they are as simple as; You have 8 ga THW going 450 feet at 40 amps to a motor at 120V, What is the VD. They are looking for correct math not correct electrical answer so the Formula only matters.

Length = L
Parallel = P
Resistance = R per foot
Current = I

Where is it written that your Resistance (R) value is in an "ohms per foot" unit?

Here?

I find 2LRI/P on Mikes Excel spread sheet.

Traditional formulas use an R value in "ohms per 1000Ft." and in doing so also have a divisor of 1000 on the same side of the equal sign to make a ratio of L to 1000Ft. For instance, your example has an L value of 450Ft. The 450:1000 ratio times an "ohms per 1000Ft" value will yield a result in "ohms". As bob pointed out VD (voltage drop) = I (amperes) ? R (resistance in ohms)
Hint: If you write out the units for all your values, after numerator/denominator cancellations of same units (e.g. 36 arnsts/12 arnsts = 3, a unitless value), the remaining units will always take the form: volts = amperes ? ohms...

2 * 450 feet * 0.778 ohms ? 1000 feet ? 40 amperes = ? volts

2 * 450 * 0.778 ohms ? 1000 ? 40 amperes = 28 volts

or

2 * 450 feet * 0.000778 ohms ? 1 foot ? 40 amperes = ? volts

2 * 450 * 0.000778 ohms ? 40 amperes = 28 volts

 

[kid_stevens]

But in NM the NMAC has the formula for 2LRI(I*.80 no matter what)/1000 When I ran that none of answers matched. So I tried 2KLI/CM and it did not get close so the purpose of this post was to find out the other ways of doing VD on the test. And trying them all since we are allowed to annotate our books before the test but not write in them during the test. So now I have written all the different ways on the back cover page.

I did make a mistake in my example because (.778/1000)* L(450) produces .35 resistance so 2*.35*32/1000 = .02 I had used 450 twice in my example.
Voltage = Amperage * Resistance so 40*.35(.778/1000*(450)) = 14.00

Here we are called EE98 for Master Contractors and JE98 for Master Journeymen. Master in my meaning, means the entire electrical realm is ours as they have ER1 and JR1 for residential. ERs don't have to do Voltage drops in the test. EEs take the same residential test.

I hate not getting 100%.

monday i retake the commercial test and with all these formulas I shall find he correct one.

 

[jdsmith]

Hint: If you write out the units for all your values, after numerator/denominator cancellations of same units (e.g. 36 arnsts/12 arnsts = 3, a unitless value), the remaining units will always take the form: volts = amperes ? ohms

This is by far the easiest way to make sure your calculations are correct. Chances are several of the formulas you have tried would work and give a very close answer as long as all of the units agree. If you have R in your formula, when you plug the numbers in substitute (X ohms/1000 ft). Write it in the equation as a fraction. When you plug the numbers in the calculator to get an answer, also divide, multiply, and cancel the units. It's fairly difficult to get an answer with the units you expect that is the wrong value.

Jeremy

 

[Smart $]

But in NM the NMAC has the formula for 2LRI(I*.80 no matter what)/1000

And you also said if there was no hint of NM in the question, you are to "use the NEC and national info not NMs." If this truly be the case, you would not use the NMAC method, period. Your example did not hint of NM, so do not calculate that way. What would happen if you were given a question the same as your example with different values and wire size, and one of the multiple choice answers matched using the NMAC method... would that be the correct choice?

As far as I at 80% (or times .80) makes no difference to me, being I'm located in Ohio. Regardless, I feel you are not providing all the info, or you are not fully knowledgable on the issue. Yes, I could be wrong... but there has to be some reason for doing it that way and it is most certainly not plainly evident as to what that reason is. I gave you my best assumption above. Go with it, stay in the dark, or prove me wrong... your choice.

I did make a mistake in my example because (.778/1000)* L(450) produces .35 resistance so 2*.35*32/1000 = .02 I had used 450 twice in my example.
Voltage = Amperage * Resistance so 40*.35(.778/1000*(450)) = 14.00

Glad to see you caught your mistake!

monday i retake the commercial test and with all these formulas I shall find he correct one.

Truth be told, there is only one "formula"... but several "variations". The current IEEE formula is another...

 

[kid_stevens]

I found which formula PSI exams is using in New Mexico. The Correct answer always came out with Evd=I*R(divided by 1000*Length)

 

[ramsy]

Truth be told, there is only one "formula"... but several "variations". The current IEEE formula is another...

Thats the one on my desktop spreadsheet, but the only portable field gadget I own for this is my Hewlett Packard calculator (HP48). I wish HP made a cell phone so I could work in the field with one gadget.