HVAC Design Load

gmayeux

Member
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 these motor values on the data sheet of the unit, and also include heat strip if applicable, and plug that KW number into the total design load, i am using 220.18(A) for motor loads and 220.51 for heat load, as a basis for design, but do you or should you consider the AC compressor/motors a continuous load ? and therefore add the 1.25,. i cannot find a specific code reference in article 220 or 440 to state requirements
 
The MCA on the unit already includes a 125%, you do not need to add it again.

The simplest way for multiple units is to add up the MCA's, however you will end up over counting since you end up counting that 25% for largest motor on every unit instead of once. Only problem is it's a bit of a hassle to dissect all the loads on all the units.
 
What about using 80% for all but the largest unit?
I think the potential problem with that is if the unit has multiple compressors then you would end up undercounting? Say it has 3 compressors @ 20A each (ignore the small stuff like controls and fan motors). That MCA should be 25+20+20=65. Now say you took 80% of that to try to avoid the over count, you be counting that unit has 52A.

What I have usually done is just add up all the MCA's, then when I have the total service (or a feeder) load I just see where I am at and if it behooves me to be a bit lower due to equipment or conductor sizes, Ill take a closer look.....maybe a few receps servicing dedicated equipment will also get changed to general use receps :D:angel:
 

gmayeux

Member
Several replies have suggested using MCA's. but please correct me if I'm wrong based on 440.4(B) the MCA is used for sizing the conductors, minimum supply circuit conductor capacity,
 

JoeStillman

Senior Member
Location
West Chester, PA
It's hard to get the info you need from the Mechanical Department. They find a sheet with electrical data and think they're done when they find you the MCA and MOCP. But we know that's not enough info. I often google the model number myself and bypass the air-heads.

When you are aggregating several HVAC-type loads (typically condensing units) you need compressor RLA (rated load amps) and fan HP or FLA. You will be over-designed if you consider the MCA to be the load. Over-design is never going to hurt anyone. But it can drive up the cost.
 

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Electrical Contractor
What are you trying to calculate? The branch circuit size? the panel load?the service load? etc....

Branch Circuit load for an a/c is calculated per Part IV of art 440 but I am not sure that is what you are asking
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
David
How does Table 220.3 "Referencing 440" get you to using the MCA for load calculation ?
Is there a specific line in 440 that states this ?
220.3 references Art 440, section IV, which tells you the minimum ampacity (mca) required for branch circuit conductors.

In other words, the branch circuit load is calculated by the required branch circuit minimum ampacity.
 

topgone

Senior Member
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 these motor values on the data sheet of the unit, and also include heat strip if applicable, and plug that KW number into the total design load, i am using 220.18(A) for motor loads and 220.51 for heat load, as a basis for design, but do you or should you consider the AC compressor/motors a continuous load ? and therefore add the 1.25,. i cannot find a specific code reference in article 220 or 440 to state requirements
What others said, MCA (minimum circuit amps) will be used by you in choosing your wire size while MOP(maximum overcurrent protection) will tell you what size of protection device is.

Your problem is not knowing the FLA (or RLA) of your units which will be needed in computing the main circuit protection device! The MCA of the units gives you a clue on what your RLA is. Equate MCA (of your supply conductor) to the unit's maximum continuous current (MCC). UL says you can get the RLA (rated load amps) of the unit by dividing the MCC by 1.56 (units with currents 9.1A up to 20A or by 1.40 if current is above 20A, else use 1.70 ( also see 430.32 (A) (2).

Example: MCA or MCC= 40A. Therefore, your RLA = 40/1.4 = 28.6A. Do this to the rest of the HVAC units you intend to be supplied with. Then you can compute for the group's MCA (125 of greatest unit RLA + sum of the rest) and your MOP (2.25 x biggest unit + the other loads).
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Your problem is not knowing the FLA (or RLA) of your units which will be needed in computing the main circuit protection device! The MCA of the units gives you a clue on what your RLA is. Equate MCA (of your supply conductor) to the unit's maximum continuous current (MCC). UL says you can get the RLA (rated load amps) of the unit by dividing the MCC by 1.56 (units with currents 9.1A up to 20A or by 1.40 if current is above 20A, else use 1.70 ( also see 430.32 (A) (2).
This is not right. I have a cutsheet of a unit with motors of 11A, 9.7A, 6.2A, 2.1A, 2.1A, 1.6A, 1A and 0.29A.

The cutsheet lists the unit FLA as 33.99A (the sum of the currents above) and the unit MCA as 36.7 (the FLA plus 25% of the largest motor.)

Taking the MCA divided by 1.4 is 26.21A which is 7.78A lower than the actual unit FLA. That would lead to the service being undersized.
 
This is not right. I have a cutsheet of a unit with motors of 11A, 9.7A, 6.2A, 2.1A, 2.1A, 1.6A, 1A and 0.29A.

The cutsheet lists the unit FLA as 33.99A (the sum of the currents above) and the unit MCA as 36.7 (the FLA plus 25% of the largest motor.)

Taking the MCA divided by 1.4 is 26.21A which is 7.78A lower than the actual unit FLA. That would lead to the service being undersized.
Although we cant use it for calculation purposes (perhaps as a "preview" for using 220.87) Topgone's divide by 1.4 figure is very close to what I find the actual load of HVAC units to be. 1/1.4 is 71%, I usually find 66% of MCA to be almost always right on for the actual load.
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Although we cant use it for calculation purposes (perhaps as a "preview" for using 220.87) Topgone's divide by 1.4 figure is very close to what I find the actual load of HVAC units to be. 1/1.4 is 71%, I usually find 66% of MCA to be almost always right on for the actual load.
33.99/36.7 is 92.6%.

That's nowhere near 66% or 71%.
 

gmayeux

Member
if the MCA is used to determine the actual load of the HVAC, in this case the MCA 38 amps, a 40 amp breaker will not suffice, unless i am missing something here.
 
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