Can someone please explain this to me?

[jrohe]

I am still studying for my WV Master Electricians Exam. I have been going along pretty good but I just hit a brick wall. Here is the question from my practice test.
A 30 amp 230 volt load is located 100 feet from the source. What is the minimum size branch circuit conductor required to operate within the limits for voltage drop that is recommended be the NEC?
a. # 10 conductor
b. # 8 conductor
c. # 6 conductor
d. #4 conductor

The book says the correct answer is B. I have no idea how to come to this conclusion. Can anyone help me?

I suspect they are expecting you to use the method that uses DC-constant resistance for determining voltage drop to calculate the minimum wire area, in cmils, required.

For single phase circuits (assumed because that information is missing in the test in the question), the equation is:

VD = 2 x K x I x D / CM

Rearranging the equation to solve for CM results in the equation:

CM = 2 x K x I x D / VD

Making assumptions for copper conductors and VD% cannot exceed 3 percent because the information is missing in the test question:

K = 12.9 ohms
I = 30 amps
D = 100 feet
VD = 230 volts x 0.3 = 6.9 volts

Therefore:

CM = 2 x 12.9 x 30 x 100 / 6.9 = 11217.4 cmils.

Looking at Chapter 9, Table 8, the smallest conductor with this area is #8 AWG conductor at 16510 cmils. A #10 conductor has an area of 10380 cmils, which is too small, at least according to this calculation method.

 

[steve66]

I get "B" or #8 if the PF = 1.

#10 gives a 3.1% drop. #8 gives 2%.

This is using X and R from NEC table #9 to calculate %Z. I get 1.2 ohms/1000 feet for Z for #10 wire, and 0.69 for #8 wire. (The conduit type only has a minor effect on the numbers).

Then I use the formula:

Vdrop Line to Line = 2x distance x amps x Z/1000.

Then %drop = 100 x VdropL-L / 230.

Actually, I have this all in an excel sheet.

If you use a PF of .85 (maybe by using the values of Z in table #9) you get #10 wire.

 

[jimo144]

Lots to remember. Break the Question down into parts.

Lots to remember. Break the Question down into parts.

kefox81,

I read your question and the replies and moved on. You had received good replies and how to get them. After a while I realized that I might be able to offer something. You have lots to remember and 'The Test' often adds anxiety and what if you forget something?

My suggestion is to first realize that the question is a Two Part Question. Part one is the Amp Capacity of the Conductor. The second part is the 100 Ft... or every 100 Ft. (What if you face a 200 Ft distance question?)

If you break the question down that way, it will be easier to remember:
30 Amps = #10 Conductor.
Each 100 Ft, next size larger Conductor, #8.
Answer, B would be correct for this question.

That might be easier to remember.

JimO

 

[480sparky]

Well since the question said 100', that's what I put in.

You've got 200 feet of conductor...... two times one hundred.

 

[480sparky]

Well since the question said 100', that's what I put in.

Once the current goes out 100 feet, it's gotta come back that 100 feet.

 

[Daniel malack]

Never heard that before. It's drop at the appliance not baking at source.

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[Daniel malack]

Never heard that before. It's drop at the appliance not baking at source.

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Not back at source. Spell check got me.

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[steve66]

Not back at source. Spell check got me.

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There is voltage drop along both conductors. But it really just depends what the equation or calculator you are using is set up for.

The equation I posted has a factor of 2x included. So it uses the one way length, and calculates twice the voltage drop.

I haven't used the Mike Holt calculator, so I can't say which it expects.

Like equation or calculator, you just have to make sure you enter the right numbers.

 

[Daniel malack]

What would be the concern for voltage drop on way back? It is irellavent at that point.

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[ggunn]

What would be the concern for voltage drop on way back? It is irellavent at that point.

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No, it's not. Picture it as three resistors in series with the first and third as the conductors and the middle one as the load. The sum of the voltage drops across all three loads equals the supply voltage, therefore the voltage drop across the load is the supply voltage minus the other two.

 

[480sparky]

Never heard that before. It's drop at the appliance not baking at source.

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The circuit length is 200 feet. Two 100' conductors. Both conductors will create voltage drop, not just one of them.

You have the resistance of 100 feet of conductor, the resistance of the load, and the resistance of the other 100 feet of conductor.

 

[Daniel malack]

The question was , A 30 amp 230v load. 1 single load connected 100ft you only consider 100 foot of conductor. Not 200

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[Daniel malack]

No, it's not. Picture it as three resistors in series with the first and third as the conductors and the middle one as the load. The sum of the voltage drops across all three loads equals the supply voltage, therefore the voltage drop across the load is the supply voltage minus the other two.

210.19 conductors- minimum ampacity and size. FUN No 4: Conductors for branch circuits as defined in Article 100, sized to prevent a voltage drop exceeding 3 % at the farthest outlet of power etc.. etc..

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[wwhitney]

The question was , A 30 amp 230v load. 1 single load connected 100ft you only consider 100 foot of conductor. Not 200

That's not correct. Here's one way to look at it:

Let's pick one side of the supply as our 0V reference, and suppose the supply voltage is exactly 240V, so the other side of the supply is at 240V. Suppose that with the 30 amp load, the voltage drop on one of the two 100 foot conductors going to the load is 5V.

Then the voltages at the two sides of the load will be 235V and 5V. That is, from a DC analogy, if the current starts off at one side of the supply at 240V above reference, it takes 5V to push the current through the first 100 feet of conductor. It takes 230V to push the current through the load, i.e. the load sees only 230V. And then it takes another 5V to push the current back to the other side of the supply.

So for voltage drop purposes, the relevant length is 200 feet, not 100 feet.

Cheers, Wayne

 

[Daniel malack]

That's not correct. Here's one way to look at it:

Let's pick one side of the supply as our 0V reference, and suppose the supply voltage is exactly 240V, so the other side of the supply is at 240V. Suppose that with the 30 amp load, the voltage drop on one of the two 100 foot conductors going to the load is 5V.

Then the voltages at the two sides of the load will be 235V and 5V. That is, from a DC analogy, if the current starts off at one side of the supply at 240V above reference, it takes 5V to push the current through the first 100 feet of conductor. It takes 230V to push the current through the load, i.e. the load sees only 230V. And then it takes another 5V to push the current back to the other side of the supply.

So for voltage drop purposes, the relevant length is 200 feet, not 100 feet.

Cheers, Wayne

210.19 conductors- minimum ampacity and size. FUN No 4: Conductors for branch circuits as defined in Article 100, sized to prevent a voltage drop exceeding 3 % at the farthest outlet of power etc.. etc..
Let's use the code book

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[steve66]

The question was , A 30 amp 230v load. 1 single load connected 100ft you only consider 100 foot of conductor. Not 200

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You do also have to consider the voltage drop on the neutral or return wire. Otherwise, we could make every single neutral wire really small. However, I believe most calculators already have the 2x included in the formula, so you are normally correct to use a one way distance.

The circuit length is 200 feet. Two 100' conductors. Both conductors will create voltage drop, not just one of them.

You have the resistance of 100 feet of conductor, the resistance of the load, and the resistance of the other 100 feet of conductor.

It looks to me like the southwire calculator includes the 2x factor in their formula. So they expect someone to enter the one-way distance, and 100' would be correct for this example.

However, it also looks like the Southwire calculator assumes 3 phase and a 90% power factor. I don't believe either of these are correct for this problem.

I feel that it was a poorly worded question. Using slightly different versions of the voltage drop formula or slightly different wire parameters makes enough of a difference that either #8 or #10 could be correctly calculated.

 

[480sparky]

The question was , A 30 amp 230v load. 1 single load connected 100ft you only consider 100 foot of conductor. Not 200

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So how do you get 230 volts with just one conductor? The OP's question states the load is 100 feet from the source. Not '100 feet of conductor'.

.............A 30 amp 230 volt load is located 100 feet from the source. ............

So you must have 200 feet of conductor.

 

[tw1156]

240/480V single Phase is also possible

240/480V single Phase is also possible

240/480V single phase transformers are a thing and we used them for municipal lighting on thoroughfares for their design. The initial problem doesn't state 120/240V single, 240/480V single, 240V 3 Phase, etc...there's a bunch of different ways to look at it.

 

[GoldDigger]

Once you have sorted out what number of feet to use in the manual calculation, you also need to know that various online voltage drop calculators will ask you whether the circuit is three phase, single phase, or DC.
You may ask why they want to know that, and the answer is that some (not necessarily all) calculators multiply the distance you enter by two for single phase AC and for DC but take the raw number you entered when you select three phase, on the assumption that you are talking about a balanced circuit and so there is no neutral voltage drop to consider.

 

[ggunn]

210.19 conductors- minimum ampacity and size. FUN No 4: Conductors for branch circuits as defined in Article 100, sized to prevent a voltage drop exceeding 3 % at the farthest outlet of power etc.. etc..

And your point is...? If you use an on line calculator for single phase voltage drop it is very likely that the X2 is already figured in. If you calculate it from first principles, i.e., starting from V=IR, then you MUST calculate the voltage drop in both current carrying conductors and add them together. Doubling the distance accomplishes this.

For example, if your conductors have a 1 ohm resistance each, your load is 8 ohms, and you apply a 10 volt source to the circuit, the current is throughout the circuit is 1A, right? Your 1 ohm resistors get 1V apiece and your load gets 8V. Your voltage drop is 10V - 8V = 2V, and (2V/10V)(100%) = 20%.