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.
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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?
LOL! It's good to see a sense of humor!Quote:
Originally Posted by KeepItSimpleStupid
I agree!
Don,
Post your current. (I love calculation questions--well sometimes! )
Sorry Don, the guys are right.Quote:
Originally Posted by donf
The load is not important, it's essential.
Without a load there is NO voltage drop.
Don,Quote:
Originally Posted by donf
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.
True!Quote:
Originally Posted by KeepItSimpleStupid
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.
:eek:Quote:
Originally Posted by KeepItSimpleStupid
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
And all this time I thought it was 240/120=4-2Quote:
Originally Posted by KeepItSimpleStupid
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.
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