Unsure of Wire Size

[audiocableguy]

Hello, I need help in determining wire guage for a commercial install. I am not doing the installation, just double checking the Elect. Contractor. The following will need to be in a single conduit, indoors.

X4 NEMA L14-30 (2H+1N+G) 200ft from a 3 Phase panel. Voltage is 208 VAC and each Recepticle has an independant Neutral. Ground is common. Wire is 90deg. THHN. Total: 8H+4N+G (13 cond.) Average Voltage loss is acceptable.

The loads are UPS units which will be loaded to 22 amps 100 percent of the time! Neutrals are included to each for future expansion.

My confusion is the 3 Phases with Neutrals. Would the Neutrals be counted as current carring conductors? 3 Phase + N and it is counted but this this layout is very gray.

Distance loss, conduit fill, amperage and load factor has lead me to believe #6 conductors in 2.5" conduit. Wondering if these numbers are way off or ???

The short form:
(x4) 30 Amp / 208VAC / NEMA L14-30
200 Feet x13 Conductors THHN wire in conduit
100% Load Rating, Avg. Volt Loss,

Thanks for your time, any questions or comments welcome!!

 

[dereckbc]

Re: Unsure of Wire Size

Originally posted by audiocableguy:
My confusion is the 3 Phases with Neutrals. Would the Neutrals be counted as current car ring conductors? 3 Phase + N and it is counted but this this?layout is very gray.

The short form:
(x4) 30 Amp / 208VAC / NEMA L14-30
200 Feet x13 Conductors THHN wire in conduit
100% Load Rating, Avg. Volt Loss,

Since the circuits are supplying UPS, non-linear loads, I would consider the neutral a current carrying conductor.

I am confused about how many conductors you are putting in a raceway, but I assume 8H+4N+G with 30-amp breakers. Answer: 6 AWG with 10-20 current carrying conductors. If you are only putting 3H+1N in a raceway you only need # 10 AWG THHN on a 30-amp circuit. Voltage drop is not going to be an issue since you are up-sizing from a 10 AWG.

[ November 25, 2003, 10:21 AM: Message edited by: dereckbc ]

 

[iwire]

Re: Unsure of Wire Size

When you use only two phases of a 3 phase system the neutral must be counted as a current carrying conductor, Non-linear load or not.

310.15(B)(4)(b) In a 3-wire circuit consisting of two phase wires and the neutral of a 4-wire, 3-phase, wye-connected system, a common conductor carries approximately the same current as the line-to-neutral load currents of the other conductors and shall be counted when applying the provisions of 310.15(B)(2)(a).

12 current carrying conductors total for a 50% derating hit.

This will require 6 awg for each circuit as Dereck has said or you could apply 240.4(B) and use 8 awg.

 

[dereckbc]

Re: Unsure of Wire Size

Thanks Bob, I overlooked 240.4(B).

Audiocable guy, a question for you. You mentioned 22-amps continious. Is that with a full load on the UPS during recharge?

[ November 25, 2003, 09:04 PM: Message edited by: dereckbc ]

 

[iwire]

Re: Unsure of Wire Size

Thanks Dereck.

I do not know if I like the fact that 240.4(B) sticks in my mind.

The fact that it does not come to your mind says to me you do not engineer to the minimum.

 

[dana1028]

Re: Unsure of Wire Size

Derek-iwire - could you clarify something please?

I understand the simple arithmetic of derating and using the 90 degree column for derating purposes and coming up with a #8 AWG for these circuits (22A/50% = 44A/80% = 55A => #8 THHN)

There is no Section 240.4(B) involved in that computation.

Now comes the need for clarification: How does 240.4(B) apply to derating conductors? This code section applies to OCPD's; the allowance for using the next higher breaker....it says nothing about using a smaller sized wire than the calculated required conductor size.

 

[iwire]

Re: Unsure of Wire Size

Originally posted by dana1028:
Derek-iwire - could you clarify something please?

I understand the simple arithmetic of derating and using the 90 degree column for derating purposes and coming up with a #8 AWG for these circuits (22A/50% = 44A/80% = 55A => #8 THHN)

Hi Dana I am not following your method here, it looks like you based your figures on 22 amps and not the rating of the circuit.

Here is how I came up with my answers.

We need circuit conductors for a 30 amp circuit in a raceway with 12 current carrying conductors.

We know we suffer a 50% adjustment factor, 8 awg is rated 55 amps.

55 amps x .5 = 27.5 amps

So now we know 8 awg is rated 27.5 amps when run with 11 other current carrying conductors.

If it was not for 240.4(B) we would not be able to use this conductor on a 30 amp breaker for this single outlet branch circuit.

Without 240.4(B) we need 6 awg, 6 awg is rated 75 amps.

75 amps x .5 = 37.5 more than enough for a 30 amp branch circuit.

You can not do your ampacity adjustments based on the load.

If you where running just 3 current carrying conductors in a raceway for a 60 amp branch circuit feeding a 48 amp load would you use 60 amp rated conductors or 48 amp rated conductors?

[ November 26, 2003, 06:15 PM: Message edited by: iwire ]

 

[dana1028]

Re: Unsure of Wire Size

Thank you iwire, now I understand where you were coming from, I was just scratching my head there for a while.

However - "You can not do your ampacity adjustments based on the load."...I always start with the load and apply the various derating factors to determine the final conductor size.


Load = 22A... x 125% for continuous load (or divide by .80) = conductors must be rated for 27.5A (this is where 240.4B comes into play)....with 11 conductors in conduit you must derate by 50% (i.e. divide by .5) = 55A (i.e. your conductor must have a minimum rating of 55A to satisfy the load/derating constraints...which = a #8 AWG.

We got at the same answer coming from different directions is all.

 

[iwire]

Re: Unsure of Wire Size

Dana I believe you have a code violation in doing it that way.

240.4 Protection of Conductors.
Conductors, other than flexible cords, flexible cables, and fixture wires, shall be protected against overcurrent in accordance with their ampacities specified in 310.15, unless otherwise permitted or required in 240.4(A) through (G).

310.15(B) Tables. Ampacities for conductors rated 0 to 2000 volts shall be as specified in the Allowable Ampacity Table 310.16 through Table 310.19 and Ampacity Table 310.20 through 310.23 as modified by (1) through (6).

310.15(B)(2) Adjustment Factors.
(a) More Than Three Current-Carrying Conductors in a Raceway or Cable. Where the number of current-carrying conductors in a raceway or cable exceeds three, or where single conductors or multiconductor cables are stacked or bundled longer than 600 mm (24 in.) without maintaining spacing and are not installed in raceways, the allowable ampacity of each conductor shall be reduced as shown in Table 310.15(B)(2)(a).

The way all that reads to me is you find the ampacity of the conductor in the tables and reduce the ampacity by Table 310.15(B)(2)(a).

No where does it say use the load to determine the adjusted ampacity.

In the original question 8 awg in a conduit with 11 other current carrying conductors would have a rating of 27.5 amps.

Connecting that 27.5 amp conductor to a 30 amp breaker would violate 240.4

In this case as it was a single outlet you could apply 240.4(B).

If this was a multiple outlet circuit you would have to use 25 amp OCP or 6 awg.

If we where talking about voltage drop I would agree with you and use the load amps to determine the wire size.

Anyone else have an opinion on this?

Bob

[ November 27, 2003, 03:55 AM: Message edited by: iwire ]

 

[dereckbc]

Re: Unsure of Wire Size

Bob, from a design point of view, if voltage drop is the goal, I ignore 310.16 at the start. If the run is over 50 feet you should be safe using voltage drop method in most cases. So, at 200 feet, 208 VAC, 22-amp load, a #6 AWG would do the trick yielding a 3% voltage drop. All that is left to do is make sure the conductor meets de-rating requirements. I am lazy and use engineering tables.

 

[audiocableguy]

Re: Unsure of Wire Size

Sorry Dereck it has taken so long to reply to your question. I calculated 18 Amps as a constant load from the external equipment and 4 amps continuous for internal battery charging. The units are Ferrups from Powerware. You have helped me with in the past with these units in other posts.

6 AWG seams to be the right choice for the job due to the conduit fill derate and VD. Two conduits should help to reduce the hit on the derating factor.

 

[dana1028]

Re: Unsure of Wire Size

iwire - thank you for your explanation - I will amend my thinking on this.

 

[itasca_mn]

Re: Unsure of Wire Size

I have been taught to use 5 steps for sizing conductors. According to my continuing ed instructor, this is how the NEC requires us to size conductors.

Step 1: calculate the load at 125% of continuous load, and 100% of the noncontinuous load. For assemblies that are rated for 100% operation compute the load at 100% for both continuous and noncontinuous loads.

22 amps X 125% = 27.5 amps

Step 2: Using the equipment rating (either 60C or 75C) go to the applicable table in article 310 and select the conductor from the temperature column that corresponds to the equipment rating.

If we were going by the letter of the code, we could not assume that in this case the equipment rating is 75 degree C, but in practice it probably is with modern equipment. From the 75 degree C table of 310.16 we select a #10 cu conductor for 27.5 amps.

Note 1: If derating is not involved, the calculation is complete. If derating is required, go on to steps 3, and/or 4, and step 5.

Note 2: Derating is a completely new calculation; start over. Use 100% of the load, and select the conductor at its maximum operating temperature. For example, THHN is rated at 90C. You are permitted to use the 90C column for derating.

Step 3: Adjacent conductor adjustment is required when there are more than three current carrying conductors in a raceway or cable. Use table 310.15(B)(2)(a).

22 amps divided by .50 = 44 amps

Step 4: Ambient temperature correction is required when the conductor is going to be installed in a temperature in excess of 86F (30C). Use the correction factor table located at the bottom of the conductor ampacity table.

No ambient temperature correction is required in for this installation.

Step 5: Go to the applicable table in article 310 and select the conductor. Compare this conductor with the one selected in steps 1 and 2. The larger of the two is the correct conductor.

From the 90 degree column of 310.16 we select a #8 cu conductor for 44 amps. Compare this to the #10 selected in steps 1 and 2, the #8 is the correct size.

[ November 30, 2003, 09:47 PM: Message edited by: itasca_mn ]

 

[iwire]

Re: Unsure of Wire Size

Originally posted by itasca_mn:
Note 2: Derating is a completely new calculation; start over. Use 100% of the load,

I disagree, can you show me where in the code it tells you to derate based on the load and not the value listed in 310.16?

Read 310.15(B) and 310.15(B)(2) to me they are clear that whatever rating the conductor has in 310.16 will be reduced by the adjustment factors.

When you have a load with no derating required do you size the conductors based on the load or based on the OCPD that you have chosen?

 

[itasca_mn]

Re: Unsure of Wire Size

Sure iwire,

article 210.19(a)(1) General. Branch-circuit conductors shall have an ampacity not less than the maximum load to be served. Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the minimum branch-circuit conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the continuous load plus 125% of the continuous load.

Its just saying that the conductor you select must have an ampacity equal to at least 125% of the continuous plus the noncontinuous loads before any adjustment or correction factors are applied. If the conductor that you select after the adjustment and correction factors are applied is smaller, you cannot use it, if it is larger, you can. That is why the conductor selected in steps one and two are compared to the one selected in steps 3 and 4, and the larger of the two is the correct one.

Here are a couple scans taken directly out of my continuing ed workbook explaining it.

 

[itasca_mn]

Re: Unsure of Wire Size

I disagree, can you show me where in the code it tells you to derate based on the load and not the value listed in 310.16?

The results are the same, except that we are approaching it from opposite directions. Using your approach you must use an educated guess at a conductor size, and derate its ampacity. If it is not enough or it is too much, you must calculate again using a different sized conductor until you arrive at a conductor size that fits your requirements.

The 5 step method cuts several steps, and eliminates any guess work by arriving at an ampacity that the conductor must carry. The correct conductor size can be chosen from the answer from a single set of calculations. The results are the same. The ampacity of the conductors listed in the tables in article 310 are adjusted to meet the requirements in both cases. The only reason that we came up with different answers was because you applied the 125% for continuous loads in addition to adjustment and correction factors, whereas I did not.

When you have a load with no derating required do you size the conductors based on the load or based on the OCPD that you have chosen?

Except for general purpose branch circuits where exact loads are impossible to pin down, you always base conductor size on the load being served. However, other articles such as 240.4(d) that pertain to small conductors can limit the size of the OCPD, such is the case in step 2 of the practical example that I posted.

[ December 01, 2003, 10:44 AM: Message edited by: itasca_mn ]

 

[don_resqcapt19]

Re: Unsure of Wire Size

Itasca,
The required ampacity of the conductor in this application is 27.5 amps and not 22 amps per 210.19(A)(1). You must derate based on the required conductor ampacity and not the load.
Don

 

[itasca_mn]

Re: Unsure of Wire Size

Don,

27.5 is exactly the answer I came up with, not 22 amps. This is the minimum size that the conductor can be before derating. But derating can be done as outlined in article 210.19(a)(1). As long as the ampacity of the conductor selected after derating is the same or larger than 125% of the continuous plus the continuous load, that is all that is required. I really believe that if you study this technique, you will find it not only to be code compliant in every way, it is also a much more streamlined method.

Again we are merely approaching the problem from opposite directions. You are taking a value from the 90C column of 310.16 for an arbitrary (guessed) sized conductor.... say a #8. A #8 THHN is good for 55 amps. The correction factor for 10-20 current carrying conductors is .5 as per table 310.15(b)(2)(a). 55 amps X .5 = 27.5 amps. Sufficient for this load.

I on the other hand do not have to arbitrarilly choose a conductor (that could be the wrong size). By the answer to my calculation in steps 1 and 2, I know that the smallest sized conductor allowed before derating must have an ampacity of at least 27.5 amps. By using 100% of the load as allowed in 210.19(a)(1), I calculate 22 amps divided by .5 to give me 44 amps. I can then go directly to the 90C coulumn and choose a #8 THHN, which is good for 55 amps.

As long as the resultant answer gives me a conductor size after derating equal to or larger than 125% of the continuous plus the continuous load, that conductor is the correct size. The results are the same, they are just approached from opposite directions. You could do an infinite number of calculations using both techniques, and come up with the same answer every time. The difference will be that using the 5 step method, you would have your answers in less than half the time.

 

[don_resqcapt19]

Re: Unsure of Wire Size

Itasca,
My point is that your step 3 is not correct. You must adjust from the required conductor ampacity and not the actual load. Look what happens if the continuous load is 24 amps. 125% of 24 amps is 30 amps and a #10 is suitable if no adjustment factors are required. If you apply a 50% adjustment factor to 24 amps, you get 48 amps requiring a #8, but when the 50% adjustment factor is applied to the required conductor ampacity of 30 amps, you get 60 amps requiring a #6.

Don

 

[itasca_mn]

Re: Unsure of Wire Size

Don, you're missing my point also.

Article 210.19 states that the conductor must equal 125% of the continuous load plus the noncontinuous load before the application of any adjustment factors. A #8 THHN is good for 55 amps before the application of adjustment factors. Well above the 30 amps required for the 24 amp continuous load.

I will contact my continuing ed instructor to see if he can better clear this up.

[ December 01, 2003, 02:42 PM: Message edited by: itasca_mn ]