voltage drop based on 240 or 120

[weressl]

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.

But that is NOT what you said in the original post. You said it is a 240V cisruit. One presume that it is a 2 pole breaker. Or are you mixing 120 and 240V loads on the same circuit?

 

[Smart $]

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.

Well, nice try, but that's not exactly what I meant :roll:

Using 240 volt 1? 2-wire circuit with a 16.7A load as in the OP, with a "total" VD of 2.5VD using #8, that is a total of just over 1%VD. However, if the design criteria specs VD per individual conductor, the above scenario would be 1.25VD per conductor, or just over 0.5%VD@240V. You don't calculate each conductor at 120V.

It's essentially the same calculation as total VD but instead of using the 2? multiplier for total VD, you eliminate the multiplier altogether (or use 1?)... giving you the VD from source to load for either of the two conductors.

As for the engineer, I just think he's being lazy and passing off his liability... he needs to spec the wire he wants and be done with it.

 

[mivey]

Well, nice try, but that's not exactly what I meant :roll:

Using 240 volt 1? 2-wire circuit with a 16.7A load as in the OP, with a "total" VD of 2.5VD using #8, that is a total of just over 1%VD. However, if the design criteria specs VD per individual conductor, the above scenario would be 1.25VD per conductor, or just over 0.5%VD@240V. You don't calculate each conductor at 120V.

It's essentially the same calculation as total VD but instead of using the 2? multiplier for total VD, you eliminate the multiplier altogether (or use 1?)... giving you the VD from source to load for either of the two conductors.

As for the engineer, I just think he's being lazy and passing off his liability... he needs to spec the wire he wants and be done with it.

So for a 120 volt load, I can have 1.8 volts on the ungrounded conductor for 1.8/120=1.5%. I will also have 1.8 volts on the neutral for 1.8/0 = ∞%.

That's not going to work out too good on a per-conductor basis. :grin:

 

[weressl]

Well, nice try, but that's not exactly what I meant :roll:

Using 240 volt 1? 2-wire circuit with a 16.7A load as in the OP, with a "total" VD of 2.5VD using #8, that is a total of just over 1%VD. However, if the design criteria specs VD per individual conductor, the above scenario would be 1.25VD per conductor, or just over 0.5%VD@240V. You don't calculate each conductor at 120V.

It's essentially the same calculation as total VD but instead of using the 2? multiplier for total VD, you eliminate the multiplier altogether (or use 1?)... giving you the VD from source to load for either of the two conductors.

As for the engineer, I just think he's being lazy and passing off his liability... he needs to spec the wire he wants and be done with it.

Voltage drop is calculated per circuit NOT PER INDIVIDUAL CONDUCTORS as it can not occur on only one of the conductors. Since you need - at least - two wire for ANY circuit it is obvious that that one half of the voltage drop occurs on one and the other half of the voltage drop occurs on the other wire. Since it is a series circuit the voltage drops are additive.

Voltage drop is current dependent - and power factor, but most here just ignore it - and the resultant voltage drop is then expressed as a percentage of the supply voltage. The supply voltage in this case if 240v.

 

[Smart $]

Voltage drop is calculated per circuit NOT PER INDIVIDUAL CONDUCTORS as it can not occur on only one of the conductors. Since you need - at least - two wire for ANY circuit it is obvious that that one half of the voltage drop occurs on one and the other half of the voltage drop occurs on the other wire. Since it is a series circuit the voltage drops are additive.

Voltage drop is current dependent - and power factor, but most here just ignore it - and the resultant voltage drop is then expressed as a percentage of the supply voltage. The supply voltage in this case if 240v.

Well youra preachin' to the choir there, buddy!

I'm just trying to explain why the engineer—referenced by the OP—is spec'ing 1.5%VD max. per conductor (which amounts to 3% total allowed on a two-wire circuit). Not all circuits are two-wire, and some calculate up to 3% allowed on one conductor in a balanced mwbc due to no neutral current, and subsequently no VD on the neutral (same as calculating a 3? 4-wire mwbc as a 3? load and using the 1.732 multiplier at L-L voltage). Spec'ing 1.5%VD max per conductor forces a "you can't cheat on the calculation" rule on them.

 

[Smart $]

So for a 120 volt load, I can have 1.8 volts on the ungrounded conductor for 1.8/120=1.5%. I will also have 1.8 volts on the neutral for 1.8/0 = ∞%.

That's not going to work out too good on a per-conductor basis. :grin:

Ohhh... didn't see your post hiding above Laszlo's. :roll:

Well, for one, if there is 0V on the neutral from one end of it to the other, then there is no voltage drop

For two, there technically isn't 120V from one end to the other on the ungrounded conductor either.

So is it that you don't understand what I mean, or are you just trying to give me a hard time

Anyway, it is as Lazlo explained, but additionally as I expanded on in my reply. If you really don't understand what I'm attempting to say, I will continue as best I can until you do...

 

[mivey]

So is it that you don't understand what I mean, or are you just trying to give me a hard time

Just giving you a hard time. :grin:

 

[weressl]

Well youra preachin' to the choir there, buddy!

I'm just trying to explain why the engineer?referenced by the OP?is spec'ing 1.5%VD max. per conductor (which amounts to 3% total allowed on a two-wire circuit). Not all circuits are two-wire, and some calculate up to 3% allowed on one conductor in a balanced mwbc due to no neutral current, and subsequently no VD on the neutral (same as calculating a 3? 4-wire mwbc as a 3? load and using the 1.732 multiplier at L-L voltage). Spec'ing 1.5%VD max per conductor forces a "you can't cheat on the calculation" rule on them.

That is silly. As long as the total voltage drop does not exceed 3% to the load supplied it is immaterial what the 'combined' voltage loss is in the neutral.

(BTW, I said; two conductors - AT LEAST - are needed to complete a circuit.)

 

[Smart $]

That is silly.

Perhaps... perhaps not.

As long as the total voltage drop does not exceed 3% to the load supplied it is immaterial what the 'combined' voltage loss is in the neutral.

But it is material. Given a balanced mwbc operating consistently at full load has a lower VD than the same mwbc with only one subcircuit pulling full load. I know it. You know it. And guess what? Many contractors know it. On a hard bid project, you give a contractor the means to save and/or make a few bucks, he's going to do it.

(BTW, I said; two conductors - AT LEAST - are needed to complete a circuit.)

...and I said nothing contrary. If you feel I was implying such, you are mistaken.

 

[LarryFine]

For calculating like this, I'd look at loads on a 120/240v circuit of feeder the same as a 240v one. Any current on the neutral is current not on one of the lines, which is also why we need not count it as a CCC.

 

[Smart $]

For calculating like this, I'd look at loads on a 120/240v circuit of feeder the same as a 240v one. Any current on the neutral is current not on one of the lines, which is also why we need not count it as a CCC.

So what are you saying, Larry?

First, we are talking voltage drop, though we haven't exactly determined whether the OP case is a feeder or a branch circuit.

Though you made no clear statement regarding the voltage drop issue, it sounds as though you calculate VD as though all balanced L-N loads are L-L loads, thereby halving the VD for L-N loads... on a feeder. IMO, that's reasonable when supplying a fair many branch circuits likely to be on and off at various times. That's because it is quite unlikely all of only one phase of L-N loads will be on while all other phase L-N loads are off.

But the likelihood changes when it gets down to the mwbc level. On a 3? 4W mwbc, there's a fairly good chance one subcircuit will be on while the other two are off. But then there are those siituations where the entire load is consistently on or off concurrently... depends on the scenario.