Voltage Drop

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bphgravity

Senior Member
Location
Florida
Re: Voltage Drop

Before someone comes on here and gives you the what for, try to put a little more effort into your questions. First, look to see if these questions have ever been asked before. Second, try to find these answers for yourself, and if you still don't understand, or if you have ideas or suggestions, post you question then. Alot of MA exam questions are being repeated here over and over.
 

peter

Senior Member
Location
San Diego
Re: Voltage Drop

Mike lists two methods--
Ohm's Law Method - Single phase only
VD = I X R

Formula Method --
VD [single phase] = 2 X (K X Q) X I X D
_____________________

CM

In which K = voltage constant which is 12.9 ohm for copper and 21.2 ohm for aluminum.
Q = is an alternating current adjustment factor which you can forget about [he says].

For three phase, multiply the upper part of the equation by the square root of 3 [=1.732].

Ask and ye shall receive.

~Peter
 

tpwd

Member
Location
Texas
Re: Voltage Drop

on a 120/240 volt single-phase system.
VD=2xLxRxI divided by 1000

VD= voltage drop [based on conductors temperature of 75-degree]
L=one-way length of circuit
R= conductor resistance in ohms per thousand feet [from chapter 9, table 8][I use uncoated copper]
I= load current [amps]
you are allowed up to 3% drop on branch circuits or 5% drop on feeder and branch circuit
;)
 

rudy.v

Member
Location
Texas
Re: Voltage Drop

I was wondering could the following calculation be used to find voltage drop
E=26.4xDxI/cmils
E= voltage drop
26.4: being a constant factor for copper
D:distance (one way)
I: current
cmils: of given wire size
 

peter

Senior Member
Location
San Diego
Re: Voltage Drop

I don't know where the 26.4 comes from?
I think the 2 gets replaced by the square root of 3 [=1.732] for three phase.
 

dereckbc

Moderator
Staff member
Location
Plano, TX
Re: Voltage Drop

Rudy, your constant is too high. It is fine to multiply the constant by two and use the one-way distance to simplify the equation. K for copper is in the range of 10.8 to 11.2 depending on the source. I use 11.1 to be safe, so using your example 22.2 would equal K
 

bphgravity

Senior Member
Location
Florida
Re: Voltage Drop

"K" is the resistance of a circular mil foot at a respective temperature. The common and most accepted value is 12.9 for copper and 21.2 for AL. Table 8 of the NEC is based at a 75 degree and exact K can be found by taking the R at 1000 feet X cm / 1000. I have seen other resources use values as low as 10.8 based on a temperature of 60 degrees. Just FYI.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Re: Voltage Drop

All of the formulas discussed above provide approximate values for voltage drop. The approximation is considered ?good enough? for the purposes of ?code compliance.? Actually, that?s not hard, since there is no NEC requirement, only a vague hint, relating to voltage drop.

Using the value of 26.4, as compared to 25.8 (i.e., 2 times a K value of 12.9), would give also an ?approximate value,? but it will be higher (by about 2%) than that use by most of the industry. It would be conservative (i.e., would be safe), but it might lead you to base your bid on a cable that is one size larger than your competitor will be using.

But now to mathematics. To say that ?K is 12.9? is to use the ?Approximate K Method? of voltage drop. The ?Exact K Method? requires you to calculate a value for K, based on information in Table 8. The formula is: K = (R * CM) / 1000, where R is the tabulated resistance (in ohms per 1000 ft) and CM is the area in circular mils. I created a quick spreadsheet using both Uncoated Copper and Coated Copper, and looking at stranded and un-stranded (up to size 8) conductors from sizes 18 to 500 MCM. I calculated ?Exact K? values ranging from 12.6 to 12.9 (Uncoated Copper) and from 13.0 to 13.7 (Coated Copper). I found an ?Exact K? of 13.2 (corresponding to the 26.4 of Rudy V?s formula) on Coated 4/0 and Coated 400 MCM.
 
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