voltage drop based on 240 or 120

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hh76

Member
240v 1ph circuit 16.7amps

using 8awg I get a drop of 2.5v

Is that 2% based on 120v from each conductor to ground, or 1% of 240v.
 

Jljohnson

Senior Member
Location
Colorado
That would be 1% of 240v (approximately). Since you didn't give us the length of your circuit, we can not check your calculation but then you didn't ask us to do that anyway. I will go out on a limb and say that your circuit is 95' from the panel.
 

hh76

Member
100ft run.

rounding my numbers in forum full of engineers, sorry.

percentage of 240v is what I have always used, but I've got an engineer sizing wire based on the voltage for the individual conductor.
 

Rockyd

Senior Member
Location
Nevada
Occupation
Retired after 40 years as an electrician.
Permited voltage drop for a 240 volt circuit is 7.2 volts.

Not to be the CODE police but voltage drop is a recommendation per 210.19(1) ex. 2 FPN per 90.5(C).
 

hh76

Member
in this case the design perameters call for a 1.5% drop or less.

can anyone point to something in writing that could help to argue my case?
 

hardworkingstiff

Senior Member
Location
Wilmington, NC
in this case the design perameters call for a 1.5% drop or less.

can anyone point to something in writing that could help to argue my case?

What are you trying to argue, that the VD is for 240-volt and not 120-volt?

If this is a straight 240-volt circuit, I think that fact would be the supporting fact. If it's a 120/240-volt circuit, then with a 1.5% limitation and someone was using it at 120-volts then you would exceed the 1.5% limitation.

Your OP did not indicate a 120/240-volt circuit, so I'd say you meet the design parameters.
 

hh76

Member
I understand that 2% of 240 is the same as 1% of 120.

When I run calculations I get a 2.5v drop. When somebody needs that in terms of % drop, do I use 240v, or 120v?

It is a 120/240v circuit.
 

kwired

Electron manager
Location
NE Nebraska
If it is a 120/240 circuit the neutral should be carrying unbalanced current from the other two lines and voltage drop is probably minimal compared to the drop in the presumably heavier load between the ungrounded conductors. The more balanced the neutral load the less voltage drop on the neutral.
 

mivey

Senior Member
When somebody needs that in terms of % drop, do I use 240v, or 120v?
It will be based on the voltage on the load and/or net load. If the 16.7 amps made up of a bunch of 120 volt loads, you will need to do a 120 volt calc for the worst unbalanced case.

If it is 16.7 amps of 240 volt loads, the drop is a % of 240 volts.
 

hardworkingstiff

Senior Member
Location
Wilmington, NC
I understand that 2% of 240 is the same as 1% of 120.

When I run calculations I get a 2.5v drop. When somebody needs that in terms of % drop, do I use 240v, or 120v?

It is a 120/240v circuit.

Now it gets a little dicey. If you have loads on both phases, the VD drop acts like a 240v circuit and the % is split between the 2 loads. The VD drop will split between the 2 loads. If you have 16.7 amps on each phase, then the VD will be 2.5v on the overall circuit and each phase will see 1.25v drop (it acts like a series circuit since the neutral is carrying 0-amps).

If you had 16.7-amps on A phase and 8.35-amps on B phase then the VD for 8.35-amps is split between the 2 phases and the VD drop for the additional 8.35-amps for A phase will be seen by A phase only. VD for 8.35-amps is 1.25v /2 = .625VD on A and .625VD on B. Now we need to add the 2nd 1.25VD (for the rest of A's load) to A only (since B's total load has been accounted for) for a total VD on A of 1.875-amps.

Now if B was turned off and A left on, A's VD would go up to 2.5V since B is not loaded.

(Sorry I couldn't give you a more clear explanation, I never went to school for this and don't know how it's taught).
 

pnelspec

Member
Location
Australia
A voltage drop % is usually referred as "per unit" (PU). Which means, related to the nominal value. Now, if your load "uses" 240V and you measure a voltage drop of 2.5V out of 240V, like 237.5V instead of 240V, so your voltage drop is about 1%.
But, if the 2.5V you measure across the wire only, you obviously don?t know what's the nominal is, since the load can be connected as 120 or 240 using the same wire. So, in that case you better do not use % term with reference to that drop.

Pol
 

mivey

Senior Member
you obviously don?t know what's the nominal is, since the load can be connected as 120 or 240 using the same wire. So, in that case you better do not use % term with reference to that drop.
You should because that will determine the current value and which impedance it flows though.

You can calculate in per unit current as well. For distribution systems, one method I use is per unit currents, voltages, and impedances. I only convert back to volts and amps at the end if I need it.
 

Smart $

Esteemed Member
Location
Ohio
... but I've got an engineer sizing wire based on the voltage for the individual conductor.

in this case the design perameters call for a 1.5% drop or less.

A 1.5% voltage drop based on the individual conductors is the same as saying an allowable 3% VD for a total length of a 2-wire circuit, but disallows any "improvements" as a result of balanced 1? 3-wire circuit (i.e. 120/240), or a balanced 3? 4-wire circuit (e.g. 208/120 or 480/277).

In the case of a 240V 1? 2-wire circuit, your circuit total VD is 50% per conductor.
 
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charlie b

Moderator
Staff member
Location
Seattle, WA
Occupation
Electrical Engineer
Let me try to explain it this way. If you want to look at the individual wires, and if the total VD is 2.5 volts, then the VD along each wire is 1.25 volts. You can say that the voltage from the source to the load is 120 volts, and that the VD along that path is 1.25, and divide those two to discover that 1.25 is about 1% of 120 volts. You can do the same for the other wire, and see that it also has a VD of 1% of 120 volts. Or you can look at the entire circuit, take note that the total VD is 2.5 volts, and see that that value is about 1% of 240 volts. The answer is 1% in either case. Your engineer has the wrong notions here.
 
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