Voltage Drop

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tkirk911

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What is the actual formula for voltage drop. I have seen many and a little confused.
1. vd=2xLxRxLxI/100 (Hand book says R from table 8 but that is for dc)
2. vd=2xKxDxI/ cm (K-should it be 12.9 or should we use exact k (RxCM/100')
3. vd=2xLxRxLxI/100 (where R= table 9 "alternating current" dependent on conduit type.

I understand to change 2 to 1.732 for 3 phase. Just the formula is confusing me.

Assume a 208 volt single phase 60 hz circuit.

Any help would be greatly appreciated.
 
This is what I've learned from school:
v_d.jpg


NEC art. 215, page 98 has formulas for DC, 1 phase, 3 phase

As I understand, in real life, cos and sine are not easy to define when you need a quick calculation, so those other factors such as K and Cm are derived, and just another form of the formula above.. and how the formulars are formed, I won't go into details

I have developed spreadsheet uses those formulas to comply with code, and pretty much works, well at least my boss approved and clients have had no objection.. If you look up in the Ugly's Book, very close as well..

just a thought...

*edit for correction*
 
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VD= 2*xKxDxI/CM

*: 2 is for single phase, for 3 phase change 2 to 1.732 (sq. root of 3)

K=RxCM/1000 (not 100)

D= distance 1 way in circuit

I obviously is current

CM is circular mils
 
tkirk911 said:
...Just the formula is confusing me. ...
Click to expand...
Me too. Not one of those formulas make any sense to me. Well, #2 is okay, but not my preferred.

My method is to look at this as an Ohm's Law problem:

Vd = 2 x R/1000 x L x I (single phase or DC)
where:
R is ohms/1000' from table 8 or table 9
L is the one-way distance in feet
I is the load current

So, the one-way resistance is R/1000 x L. The round trip resistance is 2 x R/1000 x L

The Voltage drop is: Vd = I x R(total) or, for this case: Vd = 2 x R/1000 x L x I

Pretty much following barbeer, except I tend to just look up the R/1000' as opposed to calculate k factors and look up conductor area.

You pretty well have the rest of it correct.

carl
 
tkirk911 said:
What is the actual formula for voltage drop. I have seen many and a little confused.
Click to expand...
There's just a lot of different ways to estimate voltage drop. It's like guessing somebody's weight; with there being unknowns (what is the ambient temperature at each section of the wire on it's route?) there are many ways to come to an answer. :)

I've listed my preferred formula (like Carl's) here, with some links to other discussions. I'd say in general, Carl and I are in the minority about a preferred formula.
 
tkirk911 said:
What is the actual formula for voltage drop. I have seen many and a little confused....

Any help would be greatly appreciated.
Click to expand...
Visit this web page: http://www.dolphins-software.com/voltageDrop.htm

[Edit to add...] While the referenced page shows how to compensate for ambient temperature, it does not show—nor does the formula contain—a means of adjusting resistance change created by the amount of current, or lack thereof, on the conductor.
 
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bhoang said:
v_d.jpg

As I understand, in real life, cos and sine are not easy to define when you need a quick calculation...
Click to expand...
Perhaps the following will help...
cos θ = Power Factor (PF)
sin θ = √(1 ? (cos θ)?) = √(1 ? PF?)​
Not too difficult for less-than-1 power factors if you're using a [scientific] calculator :grin:
 
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