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

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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 ?
 
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.
 
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?
 
Horizon213 said:
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?
Click to expand...

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.)
 
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)
 
david luchini said:
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.)
Click to expand...

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?
 
lielec11 said:
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?
Click to expand...

I don't see the need to use the 1.25 except for one (the largest) motor.
 
I agree with Gus, only need to use the 1.25 once.

augie47 said:
I don't see the need to use the 1.25 except for one (the largest) motor.
Click to expand...


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).


lielec11 said:
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?
Click to expand...

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


 
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
 
david luchini said:
I agree with Gus, only need to use the 1.25 once.




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).




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


You are correct... no extra 5. :roll:
 
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?
 
Horizon213 said:
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?
Click to expand...

((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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"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.
 
david luchini said:
((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)
Click to expand...



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?
 
augie47 said:
"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.
Click to expand...

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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