Jimmy7
Senior Member
- Location
- Boston, MA
- Occupation
- Electrician
HVAC for Load Calc
- Thread starter Jimmy7
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Art 220.14(C) refers you to 440.6.
With that in mind, for calculations I would take your FLA (Total) X Voltage.
The examples in Annex D seem to support that method.
That said, the nameplate gives RLA and not FLA so there may be some discussion concerning using the RLA.
With that in mind, for calculations I would take your FLA (Total) X Voltage.
The examples in Annex D seem to support that method.
That said, the nameplate gives RLA and not FLA so there may be some discussion concerning using the RLA.
Rick 0920
Senior Member
- Location
- Jacksonville, FL
- Occupation
- Electrical Instructor
Jimmy, you have to combine the loads from both the compressor and the fan motor and multiply by the voltage specified. (230v)
24.4 amps (RLA = running load amps) + 1.4A (FLA = full load amps) = 25.8A. The minimum circuit amps of 31.9 is for your combined load x 125% to size your conductors. 24.4 + 1.4 = 25.8A x 230v = 5,934va If you have electric heat, you can omit the smaller load from your calculation for noncoincident loads. See NEC Art. 220.60. If you're working under the 2020 NEC, even if you omit the AC from the calculation, you may have to count it as your largest motor later in your calculation.
24.4 amps (RLA = running load amps) + 1.4A (FLA = full load amps) = 25.8A. The minimum circuit amps of 31.9 is for your combined load x 125% to size your conductors. 24.4 + 1.4 = 25.8A x 230v = 5,934va If you have electric heat, you can omit the smaller load from your calculation for noncoincident loads. See NEC Art. 220.60. If you're working under the 2020 NEC, even if you omit the AC from the calculation, you may have to count it as your largest motor later in your calculation.
It is a bit of a conundrum as you are dealing with FLA, RLA and MCA and applying them to both load calculations and sizing the conductors. With that nameplate, sizing the branch circuit is the easy part, 31.9 amps. Calculating the load, not so much.. do you assume RLA = FLA ??
wwhitney
Senior Member
- Location
- Berkeley, CA
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- Retired
(2017) 220.14(C) says "Loads for motor outlets shall be calculated in accordance with the requirements in 430.22, 430.24, and 440.6."
430.22 tells you that for a single motor, use a current of 125% times the FLA.
430.24 tells you for for a group of loads, the 125% factor applies only to the largest motor and any continuous non-motor loads.
440.6 doesn't actually tell you how to do a computation. It just tells you to use the rated load current in the computations. In fact 430.7 goes on to implicitly tell you to use 430.24 for the ampacity computation.
So as far as I can see, 220.14(C) is still directing you use to 125% of the largest motor for Article 440 equipment. I.e. just use the MCA for the load calc.
[If 220.14(C) said "430.22, 430.24, or 440.6" then there might be an argument that you can choose to read 440.6 and ignore 430.24. But since it says "and" the requirements of both apply.]
Cheers, Wayne
430.22 tells you that for a single motor, use a current of 125% times the FLA.
430.24 tells you for for a group of loads, the 125% factor applies only to the largest motor and any continuous non-motor loads.
440.6 doesn't actually tell you how to do a computation. It just tells you to use the rated load current in the computations. In fact 430.7 goes on to implicitly tell you to use 430.24 for the ampacity computation.
So as far as I can see, 220.14(C) is still directing you use to 125% of the largest motor for Article 440 equipment. I.e. just use the MCA for the load calc.
[If 220.14(C) said "430.22, 430.24, or 440.6" then there might be an argument that you can choose to read 440.6 and ignore 430.24. But since it says "and" the requirements of both apply.]
Cheers, Wayne
Frank DuVal
Senior Member
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- Electrical Contractor, Electrical Engineer
Yes, running load amps is full load amps.
ptonsparky
Tom
- Occupation
- EC - retired
I just divide the MCA by 1.25 & use that.
I don't know if anyone questioned my calculations in the 40 years.
I don't know if anyone questioned my calculations in the 40 years.
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
Should not be dividing by 1.25. If your calculating a load for a feeder that includes only one piece of HVAC equipment, and no other motors, then just use MCA. If there are multiple pieces of HVAC equipment/motors, then you need the RLA/FLA of all the HVAC/motors, so you can add them up and just add 25% of the largest one.
This earlier thread seems to cover it:
HVAC Design Load
what is the proper method when doing load calculations for HVAC in a commercial application. i have a plan reviewer that wants to see the justification for the loads. it is not as easy as just looking at the nameplate on the unit, where you have MOP and MCA. i typically use 440.22(B)1. i find...
forums.mikeholt.com
In particular, the reference to 220.3 and Table 220.3.
Cheers, Wayne
ptonsparky
Tom
- Occupation
- EC - retired
I figure the FLA of the unit by dividing the MCA by 1.25. I don't need to know the RLA of each compressor or # of fans and their amperage.
Yes, use MCA for a single unit.
Yes, use MCA for a single unit.
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
Well, dividing by 1.25 will give you a non-conservative approximation. As only the largest motor gets multiplied by 1.25 when calculating the MCA, not all of them. When the largest motor dominates the total load, then the error is small.
So for an exact answer, you need to at least know the FLA/RLA of the largest load. And then just subtract 25% of that number from MCA.
Cheers, Wayne
lordofthisworld
Senior Member
- Location
- Massachusetts
With HVAC equipment, the manufacture has already calculated the conductor size to be based on the the total of all of the motor loads in the combination load equipment times 125%. It is not necessary to do these calculations again. You just have to verify that the branch circuit conductors supplying the equipment have an ampacity equal to or greater than the minimum circuit ampacity marked on the nameplate.
True for sizing the branch circuit, but the question here is regarding calculation of the entire dwelling load in order to size the service.
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