Service calculation load for 8 Air handlers . 480 volt 3ph. Name plate pic. attached

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augie47

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Do we assume no heat load, or do we note a violation of 110.3(B) and 424.28 for not marking the nameplate ?
 

Horizon213

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Location
Atlanta, GA
That is true, they didn't mark the name plate, but this unit has 5KW heat strips. Customer may want to use one of the larger options of heat strips listed on the name plate. Does this help? I would like to know load demand for the worse case scenario.
 

Horizon213

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Atlanta, GA
I calculated it by multiplying the MCA for the 15kw heat strips X 8 for the number of units. Then 25% of the largest rated motor, which in this case I used the compressor motor. Is this correct?
 

david luchini

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I calculated it by multiplying the MCA for the 15kw heat strips X 8 for the number of units. Then 25% of the largest rated motor, which in this case I used the compressor motor. Is this correct?

I think I would take the full load amps of each unit (6+0.5+3+18 = 27.5) times 8 units, and the add 25% of the largest motor.

27.5 * 8 + 1.5 = 221.5A

(I think the 208/230V listed for the indoor fan is a typo.)
 

augie47

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This may have some interesting discussion.
Since you mentioned "service calculation", I would think our guide would be 220.14 which it seems would have you use the ampere rating of the load.
220.14(c) would have you add 25% to the largest motor load.
I don't see clarity as to which number to use for the heat load. IMO, the actual FLA would be the one used for service calculations.
My numbers would be same as david's in post #6 (he posted as I was typing)
 

lielec11

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Location
Charlotte, NC
I think I would take the full load amps of each unit (6+0.5+3+18 = 27.5) times 8 units, and the add 25% of the largest motor.

27.5 * 8 + 1.5 = 221.5A

(I think the 208/230V listed for the indoor fan is a typo.)

I agree although my math is a bit different:

6 x 1.25 (for largest motor) + 3 + 5 +.5 +18 (heater) = 30.5

30.5 x 8 = 244A ~ 250A service.

Thoughts?
 

augie47

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State Electrical Inspector (Retired)
I agree although my math is a bit different:

6 x 1.25 (for largest motor) + 3 + 5 +.5 +18 (heater) = 30.5

30.5 x 8 = 244A ~ 250A service.

Thoughts?

I don't see the need to use the 1.25 except for one (the largest) motor.
 

david luchini

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I agree with Gus, only need to use the 1.25 once.

I don't see the need to use the 1.25 except for one (the largest) motor.


I'm not sure how you arrived at your answer, though...6 x 1.25 (for largest motor) + 3 + 5 +.5 +18 (heater) = 34 (not 30.5).


I agree although my math is a bit different:

6 x 1.25 (for largest motor) + 3 + 5 +.5 +18 (heater) = 30.5

30.5 x 8 = 244A ~ 250A service.

Thoughts?

Also, I'm not sure where the "5" is coming from in your calculation.


 

lielec11

Senior Member
Location
Charlotte, NC
So then my numbers would changed to...

(6 x 1.25) + 3 + 5 +.5 + 18 = 34A (math was wrong first time)

+

7 x (6 + 3 + 5 +.5 +18) = 227.5A+ 34A = 261A
 

Horizon213

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Location
Atlanta, GA
Thanks guys! I calculated MCA which would include all the motors and heaters 33.5 x 8 x 480 x 1.73 = 222,547 va

Then for the largest motor I calculated using the compressor 6 x 1.25 x 480 x 1.73 = 6228

Add the two together 222,547 + 6228 = 228,775 va

228,775 / 480 x 1.73 = 275 amp.

What do you guys think?
 

david luchini

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Thanks guys! I calculated MCA which would include all the motors and heaters 33.5 x 8 x 480 x 1.73 = 222,547 va

Then for the largest motor I calculated using the compressor 6 x 1.25 x 480 x 1.73 = 6228

Add the two together 222,547 + 6228 = 228,775 va

228,775 / 480 x 1.73 = 275 amp.

What do you guys think?

((6+0.5+3+18) * 8) + (6 *0.25) = 221.5A is your load.

You don't need to use the MCA of each unit, just the FLA. (27.5*8*480*1.732=182,905VA)

And for the largest motor, you would want to add 25%, you added 125%. (6*0.25*480*1.732=1247VA)

(182,905VA + 1247VA = 184,152 VA = 221.5A @ 480V-3ph)
 
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augie47

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Tennessee
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State Electrical Inspector (Retired)
"what do you guys think ?"
??
david showed what he thought twice and I agreed with him.
I see nothing in service calculations that requires you to use MCA.

If you like 275.5 use it.
Do you need to use that number.. not in our opinion.

but

The Code is minimum required.
Your number is great unless someone is bidding against you and uses the required number and you loose the job.
 

Horizon213

Member
Location
Atlanta, GA
((6+0.5+3+18) * 8) + (6 *0.25) = 221.5A is your load.

You don't need to use the MCA of each unit, just the FLA. (27.5*8*480*1.732=182,905VA)

And for the largest motor, you would want to add 25%, you added 125%. (6*0.25*480*1.732=1247VA)

(182,905VA + 1247VA = 184,152 VA = 221.5A @ 480V-3ph)



Thank you very much David. You are sure about this?

What is the extra amperage for in the MCA to get to 33.5 from 27.5?
 

Horizon213

Member
Location
Atlanta, GA
"what do you guys think ?"
??
david showed what he thought twice and I agreed with him.
I see nothing in service calculations that requires you to use MCA.

If you like 275.5 use it.
Do you need to use that number.. not in our opinion.

but

The Code is minimum required.
Your number is great unless someone is bidding against you and uses the required number and you loose the job.

You are right Augie about competitor bidding, but I am more concerned about having enough in the service to handle the load on the building and just wanted some other opinions. You guys are great. Thanks for your time.
 
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