voltage drop

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101010

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I have a voltage drop question. I have a job where the panel is about 200 feet from the first fixture. Then there are 10 2x4 fixtures on the ckt. (3 bulb 120v) these fixtures are spread out about another 80 feet. It has been a long time since I had to figure out this calculation and not sure where to begin. I would think 8's to first fixture then # 10's. Any help would be appreciated.
 
apparently

apparently

if they are flourescent fixtures they are more tolerant of reduced voltage.
 
101010 said:
I have a voltage drop question. I have a job where the panel is about 200 feet from the first fixture. Then there are 10 2x4 fixtures on the ckt. (3 bulb 120v) these fixtures are spread out about another 80 feet. It has been a long time since I had to figure out this calculation and not sure where to begin. I would think 8's to first fixture then # 10's. Any help would be appreciated.
Click to expand...

Wire size I believe will have to remain the same..but with out knowing the load and total length of run..I would guess off the top of my head that #10's should be good enough because at 10 amps at 250ft should be 6 volts or 5%..but if run is longer than #8 would be required or if load is higher..
 
Voltage Drop for Multiple Load Lengths

Voltage Drop for Multiple Load Lengths

101010 said:
I have a voltage drop question. I have a job where the panel is about 200 feet from the first fixture. Then there are 10 2x4 fixtures on the ckt. (3 bulb 120v) these fixtures are spread out about another 80 feet. It has been a long time since I had to figure out this calculation and not sure where to begin. I would think 8's to first fixture then # 10's. Any help would be appreciated.
Click to expand...

Making a few assumptions as an example. First I figure the LCL (Load-Center-Length) to obtain a single distance to use for a calculation using the total circuit load. As an example if your first fixture is at 200' and the balance of (9) fixtures are spaced 8ft apart you have distances ranging from 200' out to 272ft. You take each individual load and multiply it by the distance from the panel. I assumed a load of .80 amps per fixture. So 200 x .8 = 160 Ampere/Ft and 272 x .80= 217.6 Ampere/ft. You do this calculation for each fixture. Then get the sum of the ampere/ft for all loads which in my example would be 1888 ampere/ft. Take the total load of .8 x 10 fix = 8 amps and divide it into the total ampere/ft and you get 236ft. The 236ft is the distance I would use for my calculation. Figuring 8amps at 236ft I get a 2.95 volt drop using #8 copper.
 
080516-1237 EST

101010:

A very rough calculation for simplification.

#10 is about 1 ohm/1000 ft at room temperature, #8 is 0.63, and #6 is 0.4 .

Assume your maximum length and this produces a worst case. The one way distance is thus assumed at 280 ft. Also assume total load is 10 A and at the far end.

#10 voltage drop is 2*0.28*10 = 5.6 V or 4.7% drop at 120 V.

If code will allow make the neutral #8 or #6, then
for #8 neutral and #10 hot the voltage drop is 0.28*10 + 0.28*0.63*10 = 4.6 V or 3.8% at 120 v.
for #6 neutral and #10 hot the voltage drop is 0.28*10 + 0.28*0.4*10 = 3.92 V or 3.3% at 120 v.

Now consider a 240 V to 120 V transformer at the 240 ft point. Use #12 wire with a loop resistance of 2*0.28*1.6 = 0.9 ohms. The current in this line is 5 A so the drop is 4.5 V or 1.9% at 240 V and at the midpoint of the fixtures.

Is there any cost advantage or are there other problems?

Check my math for any mistakes..

.
 
1) The first step is to figure out what sort of voltage drop you can tolerate. The NEC has a recommendation (not a requirement) for 3%; but really what matters is how the load will respond to voltage drop. Will it operate correctly? Will it be less (or more!) efficient? Etc.

2) Next you calculate the voltage drop using the various methods described by others in this thread.

3) You adjust the design to compensate for the voltage drop.

IMHO you should strongly consider using a multi-wire branch circuit for this application. This has the effect of turning your 120V load into a 240V (or 208V three phase) load, both reducing the load current and at the same time increasing the allowable voltage drop. If you keep the same %voltage drop, the numeric voltage drop goes up since the load voltage goes up.

-Jon
 
I was trying to follow the format you guys are doing for a job site I am working on but I am either doing it way wrong or I need larger wiring.

I am working with 208V 1Ph (2hots and 1ground) and I have a circuit with the following loads and distances.

Service to first pole is a distance of 225ft, this pole contains a 440W.
Second pole is 75 feet away from the first and contains 880W.
Third pole is 120 feet away from the second and contains 440W.
Fourth pole is 80 feet away from third and contains 440W.
Fifth pole is 100 feet away from the fourth and contains 880W.

So the total circuit length if I would do worst case is 600 feet with a total load of 3080W which is about 15A. To go with the recommendation of 3% voltage drop, and using worst case I would have to use #4 wiring for this one circuit.

Now if I wanted to get really in depth how could I calculate from the actual going pole to pole and not just at a single location at the end? It seems to me I would calculate the full load of the circuit on the first pole then decrease the load as I go through the circuit but increase the distance, but then again would that mean I am ignoring the first poles load. Is the only way to find a more exact number taking the avg load per foot like Grant had mentioned?

Hope i didn't break any rules and hijacking this thread please let me know if I should have started a new one (Kinda new here)

Thanks for the help,
Brad
Electrical EIT
 
Str1ve13 said:
I was trying to follow the format you guys are doing for a job site I am working on but I am either doing it way wrong or I need larger wiring.

I am working with 208V 1Ph (2hots and 1ground) and I have a circuit with the following loads and distances.

Service to first pole is a distance of 225ft, this pole contains a 440W.
Second pole is 75 feet away from the first and contains 880W.
Third pole is 120 feet away from the second and contains 440W.
Fourth pole is 80 feet away from third and contains 440W.
Fifth pole is 100 feet away from the fourth and contains 880W.

So the total circuit length if I would do worst case is 600 feet with a total load of 3080W which is about 15A. To go with the recommendation of 3% voltage drop, and using worst case I would have to use #4 wiring for this one circuit.

Now if I wanted to get really in depth how could I calculate from the actual going pole to pole and not just at a single location at the end? It seems to me I would calculate the full load of the circuit on the first pole then decrease the load as I go through the circuit but increase the distance, but then again would that mean I am ignoring the first poles load. Is the only way to find a more exact number taking the avg load per foot like Grant had mentioned?

Hope i didn't break any rules and hijacking this thread please let me know if I should have started a new one (Kinda new here)

Thanks for the help,
Brad
Electrical EIT
Click to expand...

I would convert the application to a three phase distribution, adding one more wire and one CB at the panel.

You heading in the right direction with the calcs though, just don't forget that you are not starting with full voltage at point #2. So you add each segment's voltage drop based on 208 and the sum should not exceed 3%.
 
Ah yes that's what i was forgetting, about the voltage not being 208 at the next poles. Thanks and never thought about running it 3 phase until I started reading some of the posts. I am only a couple years into the field. I read the light specs and see 120V, 208V, 277V, 480V and think its all single phase and has to be wired with 2 hots and a ground. I guess what you are saying is my thinking of this is not true. So if I run it 3 phase, will the lights alternate between the phases? First pole would get phases A-B, second B-C, third A-C, and so on?

Thanks
 
you must have three phase available first..so you are going to have like 1320 watts, 880 watts, 880 watts would this be correct..in single phase if you are running 2 hots and a ground you have like 1320 watts and 1760 watts so you started off wrong..try it again and see what you come up with..now remember the code states (08) that you must open all ungrounded conductors during servicing or a fault..so can this application afford total darkness if you have a ballast fail and it trips the breaker..
 
The site has 3 runs of lights so the odds of all 3 failing at the same time seem very rare to me so the site should never be in total darkness.

I do have 3phase service.
So you are saying that for the two hots I should be calculating the voltage drop across each conductor (meaning each will have half of the entire load to balance the circuit i guess). Therefore for the scenerio I mentioned before in (worst case) 2#6CU THHN (gives me a 4V vd across the 1320W loaded conductor and a 5V vd across the 1760W loaded conductor at 600ft)? Is this what is being said about using the 2 hots and a ground method and if so do I then add the 4V and 5V so overall I get a 9V vd across the full 208V circuit? Which this is greater then 3% but is that the correct method?

Thanks
 
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Str1ve13 said:
Ah yes that's what i was forgetting, about the voltage not being 208 at the next poles. Thanks and never thought about running it 3 phase until I started reading some of the posts. I am only a couple years into the field. I read the light specs and see 120V, 208V, 277V, 480V and think its all single phase and has to be wired with 2 hots and a ground. I guess what you are saying is my thinking of this is not true. So if I run it 3 phase, will the lights alternate between the phases? First pole would get phases A-B, second B-C, third A-C, and so on?

Thanks
Click to expand...

Yes. Balance the load between the three phases so you have equal, or close to equal number of fixtures between each three phase combinations. seems like you're going to have 2/2/3, if 440W represents one fixture. Use the ballast namepolate amperage data for your calcualtions and don't forget the power factor. You run a 3 conductor cable on a messenger, that is also your ground. Put a junction box on each pole to tap off the needed conductors/phases.
 
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