voltage drop calculation

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gmarano

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Can anyone tell me where to look in the code book so i can calculate a voltage drop I have to make a 300 ft run to device in the field and i want to know where the calculation of the voltage drop will be in the code to size the condutor correctly to correct the drop.
and yes i will be using a calculator.
 
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I am not sure what it is your are asking.

The NEC doesn't really address voltage drop, except in fine print notes that aren't enforceable as code. (With the exception of fire pumps.)

Table 8 in Chapter 9 gives you conductor properties and Table 9 of Chapter 9 gives you A/C resistance.

Is this what you are looking for?

Chris
 
raider1 said:
I am not sure what it is your are asking.

The NEC doesn't really address voltage drop, except in fine print notes that aren't enforceable as code. (With the exception of fire pumps.)

Table 8 in Chapter 9 gives you conductor properties and Table 9 of Chapter 9 gives you A/C resistance.

Is this what you are looking for?

Chris
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I have to make a 300 ft run to a device in the field and since it is a long run and their will be a drop i want to know where to find the calculation for the drop in the code.
 
gmarano said:
I have to make a 300 ft run to a device in the field and since it is a long run and their will be a drop i want to know where to find the calculation for the drop in the code.
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It is not in the code. The code does not tell us how to do a voltage drop calculation. The NEC is not a design manual.
 
For the exact formula* look in the American Electricians Handbook, Ugly's, Electricians Dictionary...ECT.....or do the simple thing use an on line calculator as noted above.

*Some of the formulas give you an estimated K (constant, typically 12.5 for copper) others show you how to calculate K. I have never calculated exact K for all the wire sizes but off the top of my head I think it was a bit higher for AWG 12-8.
 
you won't find the formula in the NEC, to do a basic calc. go to ch 9 tbl 9, match up your correct pipe and wire type, then using the ohm to Neutral per 1000/ft (the bottom of the two numbers), divide the length by 1000, and multiply by the footage, you can then multiply by the amperage. Don't forget to count the return of the run for the length. If you know the load of what your feeding you can use:

2xKxIxL / cmil

2 = number of wires, K = 12.9 (Cu), or 21.2 (Al), I = amperes, L = length and cmil = circular mil of the conductor (get this from ch 9 tbl 8).

In your case, assuming copper, #12, and a 10 amp load:
2 x 12.9 x 10 x 300 / 6530 = 11.85 vd, which if you were running 480 (480 x 3% = 14.4 allowable VD) you would be ok, otherwise not so.

Or you can use an online calculator.
 
Voltage Drop in the NEC

Voltage Drop in the NEC

OK, I have another question about voltage drop. From where did the Vd=(2xLxKxI)/CM equation come? I see it as the recomended method for voltage drop calculations. (Replace 2 with sqrt(3) for three phase). But this equation is nowhere in the NEC and doesn't take into account power factor.

The NEC has Chapter 9 Table 9. Note 2 gives the NEC method of calculating voltage drop (this probably sounds like heresy). Voltage drop is simply Ohm's law. The effective Z is given as Rcos(theta) + Xsin(theta). R and X are given in Table 9. This voltage drop calculation takes into account power factor.

Eric
 
Some voltage drop tables I put together for a job where the customer was concerned with the design and installation in regards to possible equipment operation. Based upon the first table electrical contractors should be thankful the NEC does not limit VD.

VD.jpg


MOREVD.jpg
 
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eric stromberg said:
OK, I have another question about voltage drop. From where did the Vd=(2xLxKxI)/CM equation come? I see it as the recommended method for voltage drop calculations. (Replace 2 with sqrt(3) for three phase). But this equation is nowhere in the NEC and doesn't take into account power factor.
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Vd=(2xLxKxI)/CM is basically Ohms Law VD = I x R.
K = Ohms-Cmil/ft which ranges from 10 to 12.5 depending on the temperature.
If you rearrange the formula VD = [(2 x L x K)/CM] x (I), [(2 x L x K)/CM]= R . Result being I x R.

The NEC has Chapter 9 Table 9. Note 2 gives the NEC method of calculating voltage drop (this probably sounds like heresy). Voltage drop is simply Ohm's law. The effective Z is given as Rcos(theta) + Xsin(theta). R and X are given in Table 9. This voltage drop calculation takes into account power factor.
Eric
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This is the formula that I use and is quoted in the red IEEE Book.
 
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Red Book Voltage Drop

Red Book Voltage Drop

bob said:
Vd=(2xLxKxI)/CM is basically Ohms Law VD = I x R.
K = Ohms-Cmil/ft which ranges from 10 to 12.5 depending on the temperature.
If you rearrange the formula VD = [(2 x L x K)/CM] x (I), [(2 x L x K)/CM]= R . Result being I x R.


This is the formula that I use and is quoted in the red IEEE Book.
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OK, so it comes from the Red book. I suppose i need to take that one off the shelf and look through it.

I was wondering where this came from, because i keep seeing it in Exam prep courses and material written for the NEC. I just thought it a little odd that the NEC has one way of calculating voltage drop and everyone (it seems) uses a different method than what is in the notes.

Thanks for the clarification
 
eric stromberg said:
OK, so it comes from the Red book. I suppose i need to take that one off the shelf and look through it.

I was wondering where this came from, because i keep seeing it in Exam prep courses and material written for the NEC. I just thought it a little odd that the NEC has one way of calculating voltage drop and everyone (it seems) uses a different method than what is in the notes.

Thanks for the clarification
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Eric
In most cases, the VD = IxR will get you an answer that is close enough to
determine the correct size wire. I see most people use the K = 12.5 which is the value at 65C. At 20C K = 10.5.
If you are in a conditioned area, the lower value could save you a few bucks by using a smaller conductor. If you
are in an unconditioned warehouse it may not get to 65C but using k =12.5 gives you a safety margin.
 
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bob said:
Eric
In most cases, the VD = IxR will get you an answer that is close enough to
determine the correct size wire. I see most people use the K = 12.5 which is the value at 65C. At 20C K = 10.5.
If you are in a conditioned area, the lower value could save you a few bucks by using a smaller conductor. If you
are in an unconditioned warehouse it may not get to 65C but using k =12.5 gives you a safety margin.
Click to expand...

And i've always used Vd = [Rcos(theta) + Xsin(theta)] with the values of R and X coming from Table 9, based on the conductor size and type of raceway. Of course, cos(theta) is equal to the power factor and sin(theta) must be back calulated from sin(arccos(PF))
This method takes into account both the raceway (ferromagnetic or not) and the power factor. Next time i perform some voltage drop calculations, i think i'll use the CM formula and the chapter 9/table 9/ note 2 formula and see how they compare.

I remember an exam question from the Professional Engineering Exam that i took that concerned a voltage drop during motor start up when the power factor was 30%. I'll do these calculations also (with both formula) and compare the results.
 
During motor startup, you sometimes also need to take into account the transformer KVa for the drop realized while the transformer is being taxed. The so called "flicker factor".
 
mdshunk said:
During motor startup, you sometimes also need to take into account the transformer KVa for the drop realized while the transformer is being taxed. The so called "flicker factor".
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Correct. I often find myself in a compromise between the transformer rating and the size of the wire to the motor. The larger the transformer, the smaller the condutor to the motor can be. When the transformer is too small, the motor conductors have to be large in order to mitigate the voltage drop.
Also, when the transformer is too small, the loads have to be started sequentially. With too many loads on the transformer, some motors won't start.
 
eric stromberg said:
Correct. I often find myself in a compromise between the transformer rating and the size of the wire to the motor. The larger the transformer, the smaller the condutor to the motor can be. When the transformer is too small, the motor conductors have to be large in order to mitigate the voltage drop.
Also, when the transformer is too small, the loads have to be started sequentially. With too many loads on the transformer, some motors won't start.
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What's the VD formula when you're taking into account the transformer rating? I think that some guys have more of a visceral feeling rather than a mathematical formula. I use some of these: http://hi-line-engineering.com/downloads.asp
 
eric stromberg said:
I use the transformer curves that show the voltage drop within the transformer as per the load
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From the manufacturer's catalog, eh. Handy how they do that. That's one of the few times where knowing the motor LRA comes in handy.
 
mdshunk said:
From the manufacturer's catalog, eh. Handy how they do that. That's one of the few times where knowing the motor LRA comes in handy.
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Hey, I'll take whatever i can get. Sometimes this electrical work is like putting together a jigsaw puzzle. Every piece is welcome. (Ok, OK, now let's not go to town with that last statement).
 
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