Voltage Drop

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jotw

Member
Location
Texas
Hello all,

I am working on an underground pull for a 400A service. The total distance is just over 600'. I have worked voltage drop calcs myself and compared to online calculators and either they are wrong or I am (probably the latter). I appreciate any input. Also, (and maybe this is where I'm getting confused), do I double the cmils in the calc if I'm parralleling a set of conductors? If not, how do you account for a paralleled set? And, I want to keep this under 3%. Anyway, here goes,

400' to first j box (the ditch isn't straight so I'm putting in pull boxes)
VD=2x21.2x1.028x400x400/750kcmil

VD= 9.29 (3.8%)

I know that is over 3%, but, I just want to be sure my math is right. If I parallel that 750 do I just divide the VD in half? The 1.028 is the Q factor (for conductors over 2/0) which is a new one for me.

I'll work on the remainder of the UG once I feel more comfortable about the calculation concerning voltage drop. :confused:
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
400' to first j box (the ditch isn't straight so I'm putting in pull boxes)
VD=2x21.2x1.028x400x400/750kcmil
Could you elaborate a little? Those numbers work out to 9298.603, and I assume there's a divide by 1000 you didn't include, but if you'd at least show units with the numbers I could maybe be of more help. Are you using Table 8 from the NEC2008? I don't see a 0.0171 (the resistance of 750 kcmil Cu wire in ohms/1000') in your numbers anywhere.

If you double up the wire run you would indeed halve the resistance and therefore the voltage drop.
 
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charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
I don't have time right now to look through your math. But I put the numbers into a spreadsheet I have, and my results were half of yours. Perhaps someone else can help. By the way, is this a 3 phase, 240 volt line-to-line system? :-?
 

kingpb

Senior Member
Location
SE USA as far as you can go
Occupation
Engineer, Registered
the Resistance (impedance) for the conductors goes down as you parallel them.

i.e. 2 sets the impedance is 1/2, 3 sets is 1/3, 4 sets it's 1/4 etc.

Thats how you can account for parallel conductors.

If it's 480V 3 phase, 1 set of 750KCMIL should keep you just under 3%. If after it's installed and is slightly over 3%, use the transformer taps to bump you back to where you want to be.
 
Last edited:

jotw

Member
Location
Texas
Sorry about that. This is single phase, and I'm looking at using aluminum not copper. And, according to Mike Holt's VD calc, the Q factor for ohms/kft for 750 would be 1.028. I've never used a "Q" factor before, but then again, I've not had to do much voltage drop math either. thanks
 

jotw

Member
Location
Texas
I've looked at that one. I would rather learn to fish than have one given to me...or something like that. Know what I mean?
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Hello all,

I am working on an underground pull for a 400A service. The total distance is just over 600'. I have worked voltage drop calcs myself and compared to online calculators and either they are wrong or I am (probably the latter). I appreciate any input. Also, (and maybe this is where I'm getting confused), do I double the cmils in the calc if I'm parralleling a set of conductors? If not, how do you account for a paralleled set? And, I want to keep this under 3%. Anyway, here goes,

400' to first j box (the ditch isn't straight so I'm putting in pull boxes)
VD=2x21.2x1.028x400x400/750kcmil

VD= 9.29 (3.8%)

I know that is over 3%, but, I just want to be sure my math is right. If I parallel that 750 do I just divide the VD in half? The 1.028 is the Q factor (for conductors over 2/0) which is a new one for me.

I'll work on the remainder of the UG once I feel more comfortable about the calculation concerning voltage drop. :confused:

A single run of 750 kcmil uncoated copper wire 400 feet long that is carrying 400A will have a voltage drop of 2.74 volts, I believe.

V = IR = (400A)(.0171 ohms per 1000')(400'/1000') = 2.736V.

Does that help?
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
VD=2x21.2x1.028x400x400/750kcmil

You are using the correct equation, but it really will help you understand the equation (and help ask questions about the equation) if you write down what each number means, and the units involved.

Code:
VD (in volts line to line) =
     2 (constant for number of phases, use 1.732 for three phase systems)
* 21.2 (constant for Aluminum, in units of ohm circular mil /foot)
* 1.028 ('Q' factor which compensates for 'skin effect')
* 400 (length in feet)
* 400 (amps)
/750 (kcmil)
/1000 (conversion for kcmil to circular mil)

A piece of Aluminum wire, 1 foot long with a cross section of 1 circular mil, has a resistance of 21.2 ohms. Resistance is proportional to length, so you multiply by the length. Resistance is inversely proportional to cross section, so you divide by the cross section. 21.2 ohm cirmil/foot * 400 feet / 750000 cirmil gives a one way resistance of 0.0113 ohms. You scale for the 'Q' factor, and you multiply by 2 because you are going there and back. You apply Ohms law: Voltage = Current * Resistance. Bingo you have voltage drop.

Some details that you can improve: the resistance value of a conductor changes with temperature. I believe that 21.2 applies to Aluminum at 75C; at 20C the number is 16.7. The 75C number is often used because when a conductor is used at its ampacity it will self heat...but when conductors are oversized for voltage drop then the conductor temperature will be similar to ambient.

Another detail: selecting the proper amp value as a basis for calculation. A 400A service will rarely actually be used at 400A, so most calculations you wouldn't really need to use the 400A number. On the other hand, if you have large motor loads (say big HVAC units) with big startup currents, then you might need to run your calculation with a number bigger than 400A in order to protect other loads from momentary voltage drops when the big motors start. Figuring out the correct amp value to use is an important design consideration.

-Jon

ncreases with length, and goes down as cross section increa
 

jotw

Member
Location
Texas
I guess I'm not understanding why my math, your math (ggunn), and the online calculator I was looking at all come up with different answers. ??? In terms of cost, I guess a single run of 750 cu. would be cheaper than two runs of alum. paralleled, but, then again, I haven't checked to be sure.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
A single run of 750 kcmil uncoated copper wire 400 feet long that is carrying 400A will have a voltage drop of 2.74 volts, I believe.

V = IR = (400A)(.0171 ohms per 1000')(400'/1000') = 2.736V.

Does that help?
I missed where you said it was going to be aluminum.

V = IR = (400A)(0.0282 ohms per 1000')(400'/1000') = 4.512V
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
You are welcome. Now do you see how your equation matches ggun's? He is giving the one way voltage drop, yours is the two way voltage drop. He is working from a table which gives ohms per 1000 feet of wire, you are using the resistivity of the aluminum and the size of the wire to calculate ohms per foot.

-Jon
 

jumper

Senior Member
You are welcome. Now do you see how your equation matches ggun's? He is giving the one way voltage drop, yours is the two way voltage drop. He is working from a table which gives ohms per 1000 feet of wire, you are using the resistivity of the aluminum and the size of the wire to calculate ohms per foot.

-Jon

Both of you two made sense to me. Thanks also.
 

jotw

Member
Location
Texas
I didn't go the whole 600' because I have a bend @ 400' where I'm installing a pull box (hand hole, whatever). I didn't want to waste any more time wondering if I was doing the math right. I figured that I could change wire size for the remaining 200', too. I understand everyone's calculations, but that still doesn't explain why it seems that everyone has a different VD calc. resulting in different results. Should I keep my wire size the same for the entire length? There are 12 ways to skin a cat, and umpteen thousand more ways to run an UG. I guess I'm looking for the most efficient way. Thanks for the posts.
 

bob

Senior Member
Location
Alabama
I guess I'm not understanding why my math, your math (ggunn), and the online calculator I was looking at all come up with different answers. ??? In terms of cost, I guess a single run of 750 cu. would be cheaper than two runs of alum. paralleled, but, then again, I haven't checked to be sure.

Everyone is using 400 ft. You posted 600 ft. Which is it? Also it does not appear that you are reducing the impedance for parallel conductors.
 

jotw

Member
Location
Texas
the total was 600'. I stopped @ 400' for assistance from this forum. I even used 400' because that is where the first j box will be located.
 
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