Sizing a Subpanel

[wwhitney]

It should be 440.33 Motor-Compressor(s) With or Without Additional Motor Loads.

Again, that is in Part IV "Branch Circuit Conductors" so it does not apply to feeder conductors.

Cheers, Wayne

 

[TX+ MASTER#4544]

Again, that is in Part IV "Branch Circuit Conductors" so it does not apply to feeder conductors.

Cheers, Wayne

2023 NEC
Part IV Circuit Conductors.
440.32 Single Motor-Compressor. .....the greater of the following: (1) (2)
Most motors are considered to have a continuous duty application.
You have a choice of two methods for selecting the greater of them all.
So if you will have a feeder for the motors then you will select only one of them.

440.33 Motor-Compressor(s) With or without Additional Motor Loads.
(1) Chose one that is greater.....of all motor compressor(s).
(2)The FLC of all other motors (if any)
(3) 25 % of the highest full load motor current in the group

This is required in order to select the feeder conductors.
If you don't have branch circuit conductors why is a feeder necessary?
To size the feeder you need to know the B/Cs ampacity to calculate the feeder size and eventually size the feeder OCPD.
I did not do that in my OP. It will take more time to select type of OCPD for each motor.

Did this explain it better?
TX+MASTER#4544

 

[TX+ MASTER#4544]

2023 NEC

Hey, thanks, just an 'ol master electrician and your mentor.

A feeder for a hermetic refrigerant motor compressor is found in Part IV of 440.43 ...."not less than the sum of each of the following"......
Item (1) use the rated load or B/C current...whichever is greater ....of motor-compressor(s). You will select the larger of the two, ie., either the rated load current or the B/C selection, whichever is greater.

See 440.32 (1) 125% of the rated load current
See 440.32 (2) 125 % of the B/C selection

Item (2) 440.33 ...sum of full load current of all motors. None in this example.

Item (3) .25 % of highest rated motor current (amps) in the group.

I will pause here for a moment; if these refrigerant motors are three phase then you might sketch them out on paper and balance them on each phase, or each line #1 or line #2 ( group) which is what is listed in item (3).

I think they are all single phase 120/240 or 240 single phase so therefore if they are all single phase (2 wire) then they are naturally balanced with a motor on line 1 and one motor on line 2 the (group).

I calculated this to be 38 amps. What type of wiring method is this installation?
If it's NM (romex) select from the 60 degree column.T.310.16 and chose a size 8 AWG with ground (EGC).
If its conduit and type THHN, select from 75 degree column and chose 2 size 8 AWG plus an EGC if required.
The EGC can be selected from T.250.122 but cannot be determined until OCPD is calculated. But never larger than the circuit conductors.
The feeder OCPD cannot be rounded up to the next size. See 440.22 (B) (1) and 430.62 (A)

I have blabbered too long on this subject so I will let someone else add to it.

Thanks for reading.
Comments accepted.
TX+MASTER#4544

Again, the 440.43 is incorrect. No such section. Should be 440.33
TX+MASTER#4544

 

[david]

The mca has 125% built in so you don't use it twice.

18.9 amps /1.25 = 15

15 x 1.25 = 18.9

9 amp / 1.25 = 7 amps

18.9 + 7 = 26 amps.

You already said the 26 amps was good for the feeder supply conductors
440.22 as you already mentioned in post #10
Is this how you read 440.22 (175 % and 225%)?
and 430.62?

18.9 × 1.75= 33 amps max
2.25=42 amps max
9× 1.75=16 amps max
2.25= 20 amps max

Feeder protection can not round up.

Max feeder overcurrent protection. Your supposed to use the 175 % unless the motors won't start than the 225 % but I know you already knew that


Based on 18.9 Max of 33 amps + 6 fla amps smaller compressor = 39 amps for feeder
35 amps standard size not exceeding max

Based on 18.9 amps max of 42 amps + 6 amps smaller compressor =48 amps for feeder
45 amps standard size not exceeding max

 

[david]

What he is telling you article 440 does not adress feeders.
Feeder conductor
ampacity is addressed in Article 430

The overcurrent protection for the feeder 430.62 sends you to 440.22 (A) 175/225 % of the running load current /or the branch circuit selection current, the larger of the two.

The OP only submitted the branch circuit selection current.

 

[david]

What he is telling you article 440 does not adress feeders.
Feeder conductor
ampacity is addressed in Article 430

The overcurrent protection for the feeder 430.62 sends you to 440.22 (A) 175/225 % of the running load current /or the branch circuit selection current, the larger of the two.

The OP only submitted the branch circuit selection current.

Rated load current late and I'm tired
Min circuit ampacity/ branch circuit selection current

 

[david]

Again, that is in Part IV "Branch Circuit Conductors" so it does not apply to feeder conductors.

Cheers, Wayne

I hate to say this McGraw-Hill's National Electrical Code 2014 Handbook
Frederic P. Hartwell, Joseph F. Mc Partland, Brian J. McPartland

Say on page 1100 that he is right. Referring to the feeder conductors

In part "Breakers in the subfeeder panel are rated using 430.62 and the conductors by 440.33

I know it is a commentary and you may not be interested in there comments

 

[david]

The mca has 125% built in so you don't use it twice.

18.9 amps /1.25 = 15

15 x 1.25 = 18.9

9 amp / 1.25 = 7 amps

18.9 + 7 = 26 amps.

Recognizing there is a lot of dispute in determing the feeder min ampacity

Assuming the min Circuit ampacities are the same as the branch circuit selection current, and we were not given the rated load currents

According to McGraw Hill's commentary you go to 440.33 to size the feeder

18.9 +9 + 4.725 (25% of the largest) = 32.625 for a feeder min ampacity

 

[Dennis Alwon]

This thread has gotten so convoluted that I have no idea what the correct method is for sizing a feeder for a/c units as the op asked. My understanding for motors is you take the largest ampacity motor and then multiply by 250%(based on Table 430.52) to get the largest ocpd for that motor. Then you would add the ampacity of each motor.

 

[Dennis Alwon]

Section 430.62 reference 440.22 which allows us to use 175% or 225%. I am assuming 175% for the largest ocpd plus the sum of the other motors?????

 

[wwhitney]

McGraw-Hill's National Electrical Code 2014 Handbook
. . .
In part "Breakers in the subfeeder panel are rated using 430.62 and the conductors by 440.33

Clearly wrong as 440.33 is in "Part IV. Branch-Circuit Conductors" so it does not apply to feeder conductors. Feeder conductors and feeder OCPD are governed by the rules in Article 430.

Cheers, Wayne

 

[david]

I hate to say this McGraw-Hill's National Electrical Code 2014 Handbook
Frederic P. Hartwell, Joseph F. Mc Partland, Brian J. McPartland

Say on page 1100 that he is right. Referring to the feeder conductors

In part "Breakers in the subfeeder panel are rated using 430.62 and the conductors by 440.33

I know it is a commentary and you may not be interested in there comments

 

[wwhitney]

The mini split has an MCA of 9A on a 20A breaker.

If this unit is subject to Article 440, I don't see how those number can comply with both Part IV Branch Circuit Conductors and Part III Branch Circuit SCGF, as 440.4(B) requires them to.

Part IV basically says the says MCA = 125% * compressor RLA + 100% * other loads. While 440.22 says MOCP shall not exceed (no rounding up to the next standard size like in Article 430) 225% * compressor RLA + 100% * other loads.

The difference between those two computations is 100% * compressor RLA. So the difference between MOCP and MCA is at most 100% * compressor RLA. Which from the nameplate data implies the compressor RLA is at least 11A. But that contradicts the MCA being 9A.

So now I am confused. My answers heretofore were based on the mistaken assumption that the formulas in Part III of Article 440 matches those in Article 430. There's a difference between 250%, round up (Article 430 OCPD selection) and 225%, round down (440.22).

Cheers, Wayne

 

[wwhitney]

[An image containing the following text]

Compressor CB: 175% x 26 = 45.5 amps (45-amp CB)
Fan CBs: 250% * 6.0 = 15 amps (15-amp CB)
Subfeeder conductors: (125% * 26) + 6.0 + 6.0 = 44.5A (6 AWG)
Subfeeder CB: (45 A comp. CB) + 6.0 + 6.0 = 57A (60A CB)

Figure 440-9. Circuit breakers may be used for multimotor A/C assemblies (Sec. 440.22)

The computation for the "subfeeder CB" in the above is not correct. The caption implies that the author is applying 440.22, but that section has no provision for rounding up to the next largest standard OCPD size. 57A would be the maximum size given the computation.

Moreover, 440.22 doesn't apply to the subfeeder CB sizing, as it is Part III Branch Circuit SCGF. Instead, 430.62 applies. However, it gives the same computation, and it again does not have a provision for rounding up to the next largest standard OCPD size.

Cheers, Wayne

 

[david]

The computation for the "subfeeder CB" in the above is not correct. The caption implies that the author is applying 440.22, but that section has no provision for rounding up to the next largest standard OCPD size. 57A would be the maximum size given the computation.

Moreover, 440.22 doesn't apply to the subfeeder CB sizing, as it is Part III Branch Circuit SCGF. Instead, 430.62 applies. However, it gives the same computation, and it again does not have a provision for rounding up to the next largest standard OCPD size.

Cheers, Wayne

 

[wwhitney]

[some more text from a book including]
. . . Breakers in subfeeder panel are rated using 430.62 . . .

Right, but the figure is wrong, as 430.62 does not allow rounding up from 57A to a 60A breaker.

Edit: but the figure is useful to illustrate my issue with 430.63. Change one of the 6A FLC fans to a 6A non-motor load. Now 430.63 applies instead of 430.62. And 430.63 says "not less than" rather than the "not greater than" of 430.62, but otherwise the computation is the same. So 60A becomes the minimum size OCPD allowed, and there is no maximum size OCPD specified. Seems like throwing in a 100A breaker would be comply with 430.63.

Cheers, Wayne

 

[david]

Right, but the figure is wrong, as 430.62 does not allow rounding up from 57A to a 60A breaker.

Edit: but the figure is useful to illustrate my issue with 430.63. Change one of the 6A FLC fans to a 6A non-motor load. Now 430.63 applies instead of 430.62. And 430.63 says "not less than" rather than the "not greater than" of 430.62, but otherwise the computation is the same. So 60A becomes the minimum size OCPD allowed, and there is no maximum size OCPD specified. Seems like throwing in a 100A breaker would be comply with 430.63.

Cheers, Wayne

So back to the OP question
18.9 × 1.75= 33 amps
30 amp breaker
9 amps x 1.75 = 15.75
15 amp breaker

Feeder 30 amp breaker + 9 amps = 39 amps
No rounding up
The feeder breaker 35 amps max correct?

 

[david]

Feeder if it was based on440.33
18.9 +9 + (25% of the largest fl amps) 3.78
31.68 amps
32 amps correct?

 

[wwhitney]

So back to the OP question
18.9 × 1.75= 33 amps
30 amp breaker
9 amps x 1.75 = 15.75
15 amp breaker

Feeder 30 amp breaker + 9 amps = 39 amps
No rounding up
The feeder breaker 35 amps max correct?

None of that is correct. We can't do those computations without knowing the nameplate data from the other two units. They will typically have a compressor RLA which is the largest motor, along with other loads, both motors and possibly non-motors. We need that largest motor RLA to proceed with the calculations.

So say the 18.9 MCA unit has a compressor RLA of 12.9A, to make up a plausible number. That means the other loads are 18.9 - 125% * 12.9 = 2.8A. [This works with an MOCP of 30A if the manufacturer used the 225% allowance of 440.22, 225% * 12.9 + 2.8 = 31.8, round down to 30A MOCP.] As commented previously, for the other unit, a 9A MCA doesn't seem compatible with a 20A MOCP, so I'll ignore that 20A MOCP figure. And to make up another number, say it has a compressor RLA of 6A. That means the other loads are 9 - 125% * 6 = 1.5A.

For the feeder OCPD, we need to apply 430.62 (assuming only motor loads), as 440.22 only covered branch circuits. 430.62 says to figure out the branch MOCP for the largest motor or compressor alone. That's the 12.9 RLA compressor and using the 225% allowance in 440.22 (since apparently the manufacturer did, that's the only way that unit can get a 30A MOCP), that gives 12.9 * 225% = 29.0A. The largest allowable OCPD size is that MOCP plus all the other loads, so 29A + 2.8A + 6A + 1.5A = 39.3A. If we are restricted to standard sizes, that means 35A.

But that's just one example for the assumed RLAs of 12.9A and 6A. We need the nameplate data to get the true answer. Depending on the RLA values, the largest allowable standard size under 430.62(A) may be 30A, 35A, or 40A.

Also, if the #10 Cu wire is a 75C rated method with 75C rated terminations, its ampacity is 35A. So 430.62(B) would then allow a 35A OCPD, even if the 430.62(A) computation comes out to 30A maximum OCPD.

Cheers, Wayne

 

[david]

Feeder if it was based on440.33
18.9 +9 + (25% of the largest fl amps) 3.78
31.68 amps
32 amps correct?

Now I'm trying to compute the feeder size based on 430.62 (430.24)
430.24 (430.6 (A)
430.6(A) [ table 430.248]

I need the HP rating off of a article 440 compressor

The data plate is going to give the locked Rotor current but table 430.25 (A)

Says to use only with 430.110
How do I get the Hp rating to use table 430.248?

Edit story I didn't see post 59 I need to read that first