Voltage drop calculations basics

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arnettda

Senior Member
I have a garage 450 feet away from the electrical service. I am figuring voltage drop. This is a storage garage with a few lights and outlets. Not totally sure what will be used in it. A vacuum cleaner? Circular saw. Maybe one day a electric heater. I understand the formula to figure it out but what values do I use. If I do a load calc on the building do I use 120 or 240 volts. Do I do both and see which one gives me a large conductor size.
Should I just assume one 20 amp 120 volt load? I get confused when there is so much unknowen. Thanks
 

kwired

Electron manager
Location
NE Nebraska
How much load do you want to be able to run, and how much voltage drop do you want to be able to tolerate is first thing you need to ask yourself.

A single 20 amp 120 volt load will see more voltage drop then if you had an identical 20 amp load on the opposing line (a multiwire circuit with neutral balanced - is effectively still 20 amps but at 240 volts)

So figure max load at 120 volts as well as max load at 240 volts that you expect to have, and throw in some extra for down the road if you wish.

Unless someone is doing serious work in there there probably isn't much more then one or two major load power tools in use at one time.

If they do welding voltage will fluctuate easily as one starts and stops their welding arc and will be noticeable by fluctuations in lighting levels.
 

arnettda

Senior Member
How much load do you want to be able to run, and how much voltage drop do you want to be able to tolerate is first thing you need to ask yourself.

A single 20 amp 120 volt load will see more voltage drop then if you had an identical 20 amp load on the opposing line (a multiwire circuit with neutral balanced - is effectively still 20 amps but at 240 volts)

So figure max load at 120 volts as well as max load at 240 volts that you expect to have, and throw in some extra for down the road if you wish.

Unless someone is doing serious work in there there probably isn't much more then one or two major load power tools in use at one time.

If they do welding voltage will fluctuate easily as one starts and stops their welding arc and will be noticeable by fluctuations in lighting levels.


It is one persons garage so the likely hood of him using two tools at once is very small so I could limit the current value at say 20 amps. And throw in some more for lighting and other small appliances. radio and fridge.
If

Would you limit the voltage drop to 3%?
 

arnettda

Senior Member
How much load do you want to be able to run, and how much voltage drop do you want to be able to tolerate is first thing you need to ask yourself.

A single 20 amp 120 volt load will see more voltage drop then if you had an identical 20 amp load on the opposing line (a multiwire circuit with neutral balanced - is effectively still 20 amps but at 240 volts)

So figure max load at 120 volts as well as max load at 240 volts that you expect to have, and throw in some extra for down the road if you wish.

Unless someone is doing serious work in there there probably isn't much more then one or two major load power tools in use at one time.

If they do welding voltage will fluctuate easily as one starts and stops their welding arc and will be noticeable by fluctuations in lighting levels.

If I planed for balanced loads I could use 240 volts for my V rating but the likely hood of them being balanced is small. I would be better using 120 volts?
 

kwired

Electron manager
Location
NE Nebraska
If I planed for balanced loads I could use 240 volts for my V rating but the likely hood of them being balanced is small. I would be better using 120 volts?

I would calculate for the worst case acceptable @ 120 volts at whatever % voltage drop you want to allow.

Around these parts utility supply voltages tend to be in the 123/246 range for typical 120/240 service - some parts of the country they may typically be only 115/230 ish on average - take that into consideration as well. If you don't like to see lights dim when you start a power tool, then you need to not only consider the full load rating of that tool but how much surge current it draws when starting, that will be the worst voltage drop situation.
 

arnettda

Senior Member
I would calculate for the worst case acceptable @ 120 volts at whatever % voltage drop you want to allow.

Around these parts utility supply voltages tend to be in the 123/246 range for typical 120/240 service - some parts of the country they may typically be only 115/230 ish on average - take that into consideration as well. If you don't like to see lights dim when you start a power tool, then you need to not only consider the full load rating of that tool but how much surge current it draws when starting, that will be the worst voltage drop situation.

Is the 3% value in the code a actual code or just a Informational note? That 3% would be from 120 volt rating so if my utility voltage was higher I could essentially have a larger voltage drop? If the 3% from the code is just a recommendation what could you use that would still let things work?
 

kwired

Electron manager
Location
NE Nebraska
Is the 3% value in the code a actual code or just a Informational note? That 3% would be from 120 volt rating so if my utility voltage was higher I could essentially have a larger voltage drop? If the 3% from the code is just a recommendation what could you use that would still let things work?
That is just an informational note and a suggestion. If your equipment can tolerate 20% voltage drop then nothing wrong with that. May not be so desired to have that much drop on a general purpose feeder or a lighting circuit - but for an individual circuit feeding a resistive load - may not really hurt anything to have that much drop.
 

Carultch

Senior Member
Location
Massachusetts
Is the 3% value in the code a actual code or just a Informational note? That 3% would be from 120 volt rating so if my utility voltage was higher I could essentially have a larger voltage drop? If the 3% from the code is just a recommendation what could you use that would still let things work?

When you are taking an exam, voltage drop is per the NEC recommendations, and voltage always starts out at the nominal value, unless otherwise indicated.

When you are using the concept in practice, this rule is a lot less rigid. The voltage you get at the source of your circuit is a moving target. Measure when all your neighbors are running a lot of loads at once...and you might have 117 Volts. Wait until the middle of the night when everyone's loads are off, and at the same service, you might have 122 Volts.

The 3%, 3%, 5% rule works pretty well, with a good safety factor for most variation in the initial voltage, and most tolerances of typical equipment that matters. Calculate it assuming you begin with nominal voltage at your source, and it is probably good enough.
 

Carultch

Senior Member
Location
Massachusetts
I have a garage 450 feet away from the electrical service. I am figuring voltage drop. This is a storage garage with a few lights and outlets. Not totally sure what will be used in it. A vacuum cleaner? Circular saw. Maybe one day a electric heater. I understand the formula to figure it out but what values do I use. If I do a load calc on the building do I use 120 or 240 volts. Do I do both and see which one gives me a large conductor size.
Should I just assume one 20 amp 120 volt load? I get confused when there is so much unknowen. Thanks

If the loads are balanced, the neutral will only carry a minimal amount of current. The feeder will act as if it is a single load across the two live lines, at 240V, equal in KVA to the sum total of the loads. In this case, you'd use the 240V as the basis for your calculation. Usually, this is good enough for most purposes, because with a good panelboard layout, it is unlikely that one line will operate at full amps while the other line draws no current. If only one line is loaded, it will usually be much less than the full capacity.

Panelboards should always be laid out with as much effort as possible, to balance the loads across all lines or phases. This helps you get the best utilization out of every amp brought to the panelboard, and helps you maintain decent balanced voltages throughout the system.

If you do not have a 240V feeder connected to both lines, then you use the 120V as your basis. Because your circuit would only be connected line-to-neutral.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
150813-2351 EDT

I have the following suggestion:

Assume there will be something like a DeWalt radial arm saw with a 12" blade. That is a large inertia load.

From experience with a saw such as this wired for 120 V and at the end of about 120 ft of #12 copper wire the voltage drop was about 30 V or more during startup. Way too much drop and startup was several seconds.

Wire the garage for 240 V, and have large loads wired for 240 V.

Use #4 copper which will be about 0.25 ohms for 900 ft of wire. At a starting current of 50 A the voltage drop during startup will be about 12 V or 5% of 240 V. Startup should be a fraction of a second.

If you used the same wire, but at 120 V, the voltage drop would be in the 24 V or greater range. Now the drop is around 20%.

.

.
 

ActionDave

Chief Moderator
Staff member
Location
Durango, CO, 10 h 20 min from the winged horses.
Occupation
Licensed Electrician
When you are taking an exam, voltage drop is per the NEC recommendations, and voltage always starts out at the nominal value, unless otherwise indicated.

When you are using the concept in practice, this rule is a lot less rigid. The voltage you get at the source of your circuit is a moving target. Measure when all your neighbors are running a lot of loads at once...and you might have 117 Volts. Wait until the middle of the night when everyone's loads are off, and at the same service, you might have 122 Volts.

The 3%, 3%, 5% rule works pretty well, with a good safety factor for most variation in the initial voltage, and most tolerances of typical equipment that matters. Calculate it assuming you begin with nominal voltage at your source, and it is probably good enough.
I see things the way you do, except I think 10% is well within tolerance for real world.
 

ActionDave

Chief Moderator
Staff member
Location
Durango, CO, 10 h 20 min from the winged horses.
Occupation
Licensed Electrician
I have a garage 450 feet away from the electrical service. I am figuring voltage drop. This is a storage garage with a few lights and outlets. Not totally sure what will be used in it. A vacuum cleaner? Circular saw. Maybe one day a electric heater. I understand the formula to figure it out but what values do I use. If I do a load calc on the building do I use 120 or 240 volts. Do I do both and see which one gives me a large conductor size.
Should I just assume one 20 amp 120 volt load? I get confused when there is so much unknowen. Thanks
For me, I would put the scratch pad, pencil, and calculator away and run 2,2,4,6 mobile home feeder cable, tag it to a 90A breaker and be happy that I did a good job and gave the customer the best bang for the buck considering.....

1) Take electric heat and A/C out of the equation and you use more power in the morning making toast and coffee while the girls house use the hair dryers and curling irons in the morning than you do the rest of the day.

2) I ran everything electric I could in my house one Saturday afternoon just to see how much load I could pull, electric clothes dryer, double electric oven, microwave, stereo, hair dryers in two bathrooms, every light in the house,....... put my ampmeter on the main and I topped out at around 78A.

3) Back when I was framing houses we would run three to five skill saws, a table saw, and an air compressor off a temp panel with two 20A breakers.

4) Most anything that every day people use can tolerate more that 10% voltage drop.
 

Fitzdrew516

Senior Member
Location
Cincinnati, OH
I have a garage 450 feet away from the electrical service. I am figuring voltage drop. This is a storage garage with a few lights and outlets. Not totally sure what will be used in it. A vacuum cleaner? Circular saw. Maybe one day a electric heater. I understand the formula to figure it out but what values do I use. If I do a load calc on the building do I use 120 or 240 volts. Do I do both and see which one gives me a large conductor size.
Should I just assume one 20 amp 120 volt load? I get confused when there is so much unknowen. Thanks

Do your load calcs in KVA that way it stays consistent. I'm assuming you're running a subpanel out there. I always do my voltage drop on what it CAN be loaded to, not what it currently is. This way you can assure power quality throughout the life of the panel. So if you run a 240V/1P 60A panel you'd have to use #1/0 to get under the standard 3% voltage drop (copper conductors) if you're accounting for the whole 60A. This also makes your life easier because you don't have to calculate your loads to determine your voltage drop; just use the 60A number.

It's worth noting that the 2%, 3%, 5% rule is not an NEC requirement as mentioned before, HOWEVER, energy codes often require you to abide by this rule. Whether or not you have to depends on location, your AHJ, and if you are in fact permitting :D
 

Fitzdrew516

Senior Member
Location
Cincinnati, OH
I have a couple excel calculators I made (3ph & 1ph) that I tried to attach, but they are .xlsx format so it wouldn't let me attach. I tried to save them down to .xls, but it got rid of some of the stored formulas in them. Boo :slaphead:
 

Fitzdrew516

Senior Member
Location
Cincinnati, OH
Rename the files to .xls and let the users download them and rename them to .xlsx.

See attached. As GoldDigger said you'll have to rename the extension to xlsx.
 

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  • 1ph Voltage Drop (COPPER CONDUCTORS).xls
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  • 3ph Voltage Drop (COPPER CONDUCTORS).xls
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  • 1ph Voltage Drop (ALUMINUM CONDUCTORS).xls
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  • 3ph Voltage Drop (ALUMINUM CONDUCTORS).xls
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