Hey guys,

Please check my math?

I need to push 120 VAC, 300 feet. Calculating for Voltage Drop of > 3%. I came up with a run of 300ft. #8AWG.

The load is not important for this question, it's purely a Voltage drop question.

How about a current? 20 A, 15 A, 0 A.

At zero amps any gage will do.

Cu or Aluminum?

How about a current? 20 A, 15 A, 0 A.

At zero amps any gage will do.

Cu or Aluminum?
LOL! It's good to see a sense of humor!

I agree!

Don,
Post your current. (I love calculation questions--well sometimes! )

The load is not important for this question, it's purely a Voltage drop question.Sorry Don, the guys are right.
The load is not important, it's essential.
Without a load there is NO voltage drop.

Hey guys,

Please check my math?

I need to push 120 VAC, 300 feet. Calculating for Voltage Drop of > 3%. I came up with a run of 300ft. #8AWG.

The load is not important for this question, it's purely a Voltage drop question.

Don,

is there anyway you can show us how you would work the problem?

I have some questions for you (Assuming UF cable):
1. How would you calculate the distance of a 300 foot run. Do you multiply it by 2= 600 feet:cool:
2. What if you had a multi-conductor (two hots and a neutral). Does this make a difference in how you would calculate the distance of the run.:cool:

Note: Lets make this clear. I am not asking the question because I do not know. I do know!

Washington1:

Then for #2, I think you need to add, "for a 240/120 volt circuit", otherwise only you can solve it. We could be working in Burma with 230/400 mains.

Details, details, why would you want them? Brain dead moment. I guess I just assumed you all would read my mind which is apparently a "Flat line" at the moment.

The load is 20 amps, the medium is copper the distance is 300 ft.

The formula was Ed = KxIxLx2/CMA becomes: Ed = 144000/16510 = 2.6 VAC.

I get #3 copper 2.9V drop 2.4%. #4 copper is a 3.1% drop.

This would be a 20 AMP continuous load.

Got it, thanks, Ron. I typed the wrong CMA in, I typed faster than I read!

I get 2.7 VAC with a CMA of 52,630.

OK Don. Similar problem:
120/240 volt service, (2 hots, neutral, ground), Load is 20 A at 240V, 300 ft away, copper.

Washington1:

Then for #2, I think you need to add, "for a 240/120 volt circuit", otherwise only you can solve it. We could be working in Burma with 230/400 mains.
True!



I guess I tuned-in too late!

Ron,

I came up with a Ed of 5.2 using a #4 AWG.

Here's the formula I used: Ed=(12)(20)(300)(3)/41742 = 5.2 (Max Allowed = 7.2)

Don:

The answer is exactly the same as you got before. No additional calculations were needed.
I guess you can call it a trick question. The answer might have been different if it was a 240 volt circuit, but it wasn't. It was a 240/120 volt circuit. The lower voltage prevails. Gotcha.

What is 240/120:)

It's L1, L2, N and Gnd. Between L1 and L2 = 240, Between L1 and N = 120, Between L2 and N = 120. So, 240/120 or 240 V 4-wire.

It's L1, L2, N and Gnd. Between L1 and L2 = 240, Between L1 and N = 120, Between L2 and N = 120. So, 240/120 or 240 V 4-wire.
:eek:

Just joking with you keep!

Trying to patch-up froming giving you a bad mark! Wish I could erase it!:rolleyes:

Ok, 240/120=2

Ok, 240/120=2
And all this time I thought it was 240/120=4-2

Whoa, the calculation has to change because the number of conductors is three not two. At least according to the formula I'm testing.

The formula given is Ed=K*I*L*3/CMA. Becomes Ed=(12)(20)(300)(3)/41742
Ed=5.2

Do you ignore the 240 VAC [ L(1) to L(2)] and only count the L(1) to N?

Why?

I can't answer your first question, but I can the second.

What if the circuit was loaded as 20 amps of 120 on L1 and 0 amps on L2?

You now have a 120 V, 20 Amp circuit and thus you have to calculate the voltage drop on 120 V.

e.g.. If it were 1% your after; 1% of 120 is 1.2 and 1% of 240 is 2.4, so that shows that the smaller voltage prevails.

You need to educate me on the 3 conductor thing and how would NM-B or and cable in conduit be any different. If the 240 was equally loaded then there would be no neutral current, so your really only using 2 wires or sharing the current within 3 wires.

Now if it were 3 phase, I'd say there are 3 conductors, but with 240 single phase, it doesn't make sense to count the grounded conductor (Neutral). But code,they could say count it. In any case, with a 2 wire or 4 wire 240 system the sum of the power losses will always be the same. e.g. L1,L2 = 20A, N=0; L1=20,N=20,L2=0; or L2=20, L1-0, N=20 or even L1=15; L2=5 and N=15. There is no extra heat no matter what the sharing is, but it has to handle the worst case of the very unbalanced L1 L2.

Ron,

A grounded conductor is still a conductor as was pointed out by Stan. You have to count the three conductors? Wouldn't you considered the 120 to be a derivative of the 240?

The 120 VAC represents a line to Neutral, the 240 is the Line to Line connection.

What is your formula for Ed?

Don,

Formula method:
2KId/CM
2=Two lenghts of wire to the load
K=12.9 for copper
I=Amperage
D=Distance
CM=Circular Mil

You can also use Ohm's law to get VD.
Yet, in this case you would need the conductors resistance.

The formula would be:
I(R)=VD

Voltage drop is only recommended by the NEC, but is not required. What's recommended:
-Max combined volatge drop for both feeder and branch circuit should not exceed 5%
-Max on the feeder or branch circuit should not exceed 3%

So this means. The conductor volatge (feeder only) drop on a 120V source would be: 120(.03)=3.6V then 120-3.6=116v recommended for the feeder.
If 240, then: 240(.03)=7.2V then 240-7.2=233V recommended

I use Lectri-Calc for voltage drop.

Much as I like math, the math to figure VD it too tedious for me. :D

A lot of simplifications come from 2 basic equations:

R = pL/A; Resistance = Resistivity*Length*(Cross sectional area) and
V= IR

p=16.8 nΩ-m; yep nano-Ohms-meter

Don:

This should give you an intuitive idea as to why you would not consider the conductors as 3. I made the math EASY by saying that each conductor has 1 ohm of resistance.
Note that when a 240 volt load is connected you are dissipating 800 W and when a single 120 V load is connected your also dissipating 800 W. And when you allocate between all three conductors only 350 watts are dissipated.

So, you can hopefully see that you must use 3% of 120 V because of case 2 which is the worst 120 V load case. In case 1, you would take 3% of 240, Case 2 and 3, 3% of 120.
In case #1, assume the neutral wire isn't run.

.... Amp...0hms......Power (I^2*R)
L1... 20... 1... 400
L2... 20... 1... 400
N... 0... 1... 0
Ptotal......................800 W Case 1

L1... 20... 1... 400
L2... 0... 1... 0
N... 20... 1... 400
Ptotal......................800 W Case 2

L1... 15... 1... 225
L2... 5... 1... 25
N... 10... 1... 100
Ptotal .......................350 W Case 3

What this says is that with a 20 A protection device on a dual breaker, you cannot exceed 800 W.

Remember used 1 ohm to make the math very EASY.