120 vs 240 when calculating 3% efficiency

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hitehm

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Hello All - I was taking a quick online quiz and for the question below I got the answer wrong b/c I used 240V instead of 120V which gave me a 3% Vdrop of 7.2V @ 240V instead of 3.6V @ 120V. The question says the source power is 120/240 but the loads described to engineer the circuit are all 240 and the current used in the formula is produced "at 240". It's also odd the explanation says 3V is the maximum when it's actually 3.6V and this is also the only indication the author used 120V since it says in explanation "3V maximum"

So am I correct to use 240 or is 120 correct and if so why? It's actually not the only answer I questioned out of 20 sample questions and before I purchase the entire book I want to make sure it's accurate info.

You need to install a feeder to a subpanel in a newly constructed remote workshop at a single family dwelling. The source power is 120/240V single phase. The load in the workshop is engineered to be 85 amperes continuous at 240V, and a non-continuous 240V load of 35 amperes. The wire length will be 245' from the main service termination point to the workshop termination point. Using the Informational notes in Article 215 and the following formula, size the THWN copper feeder wires to provide reasonable efficiency of operation by preventing excessive voltage drop (keeping the wires as small as possible). The formula is: VD = 2 x L x R x I ÷ 1000. VD = voltage drop, L = length, R = resistance in /1000'.

answer choices: a: 1/0 b: 4/0 c: 350kcmil d: 500kcmil (350kcmil is shown as the correct answer, see below)

explanation given:
215.2(A)(1)(a,b) directs the load to be calculated as follows: 35A @ 100%, 85A @ 125%. 35 x 106.25 = 141.25A. Inf. note 2 shows the need to maintain the voltage drop to less than 3%. The wire resistance is found in Ch. 9 Table 8, and the wire ampacity is from Table 310.15(B)(16). Using the resistance from #350 kcmil: 2 x 245' x 0.0382 x 141.25A = 2643.9175 ÷ 1000 = 2.643V, just under the 3V maximum.
 
hitehm said:
Hello All - I was taking a quick online quiz and for the question below I got the answer wrong b/c I used 240V instead of 120V which gave me a 3% Vdrop of 7.2V @ 240V instead of 3.6V @ 120V.
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To respond to just this part, to deliver a given power load requires twice the current at 120v as compared to 240v. Sinve voltage drop is dependent on current and not voltage, the drop at 120v should be twice that at 240v.

Now, if they're comparing delivering 120v vs 240v to a load designed for 240v only, then of course the current will be half as much at 120v, and the delivered power will be one-fourth that as it would if supplied at 240v.
 
The book answer is wrong, IMHO.
You are given a voltage drop formula which explicitly contains a factor of two times the run distance. That is appropriate for calculating the voltage drop seen by a 240V load where you have to include both the L1 and the L2 wires in the circuit. So to get a percentage voltage drop to satisfy the NEC informational note you need to divide that absolute voltage by 240. No ambiguity there. And 3% of 240V is indeed 7.2V.
But if you assume a balanced load and situation and look at two independent (but matched) 120V loads instead of the 240V load, there is no current in the neutral and so the voltage drop seen is calculated using the 245' length without a factor of 2. That means that each 120V load sees a voltage drop (in the L1-N or L2-N voltage) of half that. A VD of 3% (of 120) will give you 3.6V.
So the VD at 240 is 7.2 and the VD at 120 is 3.6. If you look only at L1 and L2 you have no idea whether the load is at 240V or is two balanced loads at 120V. I would say that the answer given for the problem is flat out wrong. Your instinct and analysis are correct, and if that is representative of the writing and editing that went into the whole book, I am not encouraged.
Note that if you are instead given the situation where the load is completely unbalanced (85A on the L1 side only), then the wire size needs to be bigger, since the VD will be composed of a drop along L1 and an equal drop along N. That will give you the book answer, but it is not what is requested by the problem!
The casual rounding of 3.6V to 3V while doing the rest of the calculation to tenths of a volt is also inexcusable for a textbook.
Finally, your title of the thread is also a bit confusing since a 3%f voltage drop does not correspond to a 3% efficiency. 3% efficiency would be only 3% of the power getting to the load. :) For a resistive load (which is not necessarily a reasonable assumption) a 3% voltage drop will reduce both the current and the voltage at the load by 3%, leading to a 6% reduction in load power.
 
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LarryFine said:
You're saying voltage drop limits, not actual load-driven voltage drops, right?
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Right. I see that I was not precise in my statements, since there is also some confusion as to whether it is more appropriate to use actual load current or continuous duty adjusted load current for VD calculations. The later is mandated for wire ampacity calculations, but arguably not for voltage drop calculations.
 
I agree that the book answer is wrong. With their formula they should use a 7.2V threshold.

Also voltage drop depends on the actual current flowing, so there is no reason to add 25% for continuous loads.

-Jon
 
GoldDigger said:
The book answer is wrong, IMHO.
You are given a voltage drop formula which explicitly contains a factor of two times the run distance. That is appropriate for calculating the voltage drop seen by a 240V load where you have to include both the L1 and the L2 wires in the circuit. So to get a percentage voltage drop to satisfy the NEC informational note you need to divide that absolute voltage by 240. No ambiguity there. And 3% of 240V is indeed 7.2V.
But if you assume a balanced load and situation and look at two independent (but matched) 120V loads instead of the 240V load, there is no current in the neutral and so the voltage drop seen is calculated using the 245' length without a factor of 2. That means that each 120V load sees a voltage drop (in the L1-N or L2-N voltage) of half that. A VD of 3% (of 120) will give you 3.6V.
So the VD at 240 is 7.2 and the VD at 120 is 3.6. If you look only at L1 and L2 you have no idea whether the load is at 240V or is two balanced loads at 120V. I would say that the answer given for the problem is flat out wrong. Your instinct and analysis are correct, and if that is representative of the writing and editing that went into the whole book, I am not encouraged.
Note that if you are instead given the situation where the load is completely unbalanced (85A on the L1 side only), then the wire size needs to be bigger, since the VD will be composed of a drop along L1 and an equal drop along N. That will give you the book answer, but it is not what is requested by the problem!
The casual rounding of 3.6V to 3V while doing the rest of the calculation to tenths of a volt is also inexcusable for a textbook.
Finally, your title of the thread is also a bit confusing since a 3%f voltage drop does not correspond to a 3% efficiency. 3% efficiency would be only 3% of the power getting to the load. :) For a resistive load (which is not necessarily a reasonable assumption) a 3% voltage drop will reduce both the current and the voltage at the load by 3%, leading to a 6% reduction in load power.
Click to expand...
Thanks for the detailed reply. I doubt the author was referencing balanced 120 volt load across both phases to neutral. He only mentions the 240 volt line-to-line loads. I agree, my title isn't entirely clear. I should've titled it something like "when calculating a max VD of 3% for an efficient circuit" or something similar.
 
winnie said:
I agree that the book answer is wrong. With their formula they should use a 7.2V threshold.

Also voltage drop depends on the actual current flowing, so there is no reason to add 25% for continuous loads.

-Jon
Click to expand...
I agree about your statement on the actual current flow but I think the reason the added 25% is needed to be in the vdrop calculation is because the resistance of the wire and any heat buildup in the wire can also increase the resistance and in turn, effect the current flow. The added 25% for continuous loads adjusts how we select the cable ampacity and therefore cable size. So I guess in short, when engineering the circuit for a 3% wire VD its best to assume the current flowing will be 125% of the continuous load current and that forced increase in wire size will in a way, balance out any increased voltage drop from the continuous load.
 
My understanding is that the added 25% for continuous loads in order to protect the OCPD from overheating.

Normal breakers are designed to operate at 80% of their trip rating on a continuous basis. So a 100A breaker can only serve a 80A continuous load.

However the _wire_ must be protected by that 100A breaker. So in addition to having a breaker rated at 125% of the continuous load, you need a conductor rated at 125% of the continuous load.

Note, however that if you have a '100% rated _breaker_' you are permitted to use conductors at 100% of their continuous rating.

Nothing about the voltage drop in the cables.

It is absolutely true that running a conductor at its 100% rating will mean that the conductor runs warmer, and therefore will have higher resistance and thus greater voltage drop. I just don't think this 'temperature coefficient of voltage drop' makes it into the code requirements.

-Jon
 
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