MCA and MOCP

[EE_Intern_2]

I have two questions.

Question 1. I have seen two different sets of formulas for calculating MCA and MOCP for HVAC equipment (compressor motor 16Amps RLA, condenser fan motor 3.0 A RLA, and supply & exhaust fan motor 2.8A RLA). Which set of equations below is the correct set?

MCA = 1.25 * largest motor load + all other motor loads
MCA = 1.75 * largest motor load + all other motor loads

MOCP = 1.75 * largest motor load + all other motor loads
MOCP = 2.25 * largest motor load + all other motor loads


Question 2.
I can have a 26KW electric heater that draws 76A added as an option to the above HVAC unit.

How can my heater Overcurrent protection device for the heater be larger than the Maximum over current protection for the HVAC motors when they are the same unit?

are the heater and compressor motors on separate breakers?

Thanks

 

[gndrod]

I have two questions.

Question 1. I have seen two different sets of formulas for calculating MCA and MOCP for HVAC equipment (compressor motor 16Amps RLA, condenser fan motor 3.0 A RLA, and supply & exhaust fan motor 2.8A RLA). Which set of equations below is the correct set?

MCA = 1.25 * largest motor load + all other motor loads
MCA = 1.75 * largest motor load + all other motor loads

MOCP = 1.75 * largest motor load + all other motor loads
MOCP = 2.25 * largest motor load + all other motor loads


Question 2.
I can have a 26KW electric heater that draws 76A added as an option to the above HVAC unit.

How can my heater Overcurrent protection device for the heater be larger than the Maximum over current protection for the HVAC motors when they are the same unit?

are the heater and compressor motors on separate breakers?

Thanks

The motor and heater elements are separate disconnects at the unit from a main feeder disco. The heater elements may have multiple slots that the AHJ may require for the total unit loading capacity.

 

[augie47]

I have two questions.

Question 1. I have seen two different sets of formulas for calculating MCA and MOCP for HVAC equipment (compressor motor 16Amps RLA, condenser fan motor 3.0 A RLA, and supply & exhaust fan motor 2.8A RLA). Which set of equations below is the correct set?

Both the MCA & MOCP are normally provided by the manufacturer per 440.4(B).

MCA = 1.25 * largest motor load + all other motor loads
MCA = 1.75 * largest motor load + all other motor loads

MOCP = 1.75 * largest motor load + all other motor loads
MOCP = 2.25 * largest motor load + all other motor loads

If you do need to figure them:
MCA is 125% + largest (440.33)
MCPOP is 1.75 unless that is insufficient then 2.25% + other loads. (440.22)

Question 2.
I can have a 26KW electric heater that draws 76A added as an option to the above HVAC unit.

How can my heater Overcurrent protection device for the heater be larger than the Maximum over current protection for the HVAC motors when they are the same unit?

are the heater and compressor motors on separate breakers?

Thanks

Not always. The mfg nameplate will normally give info on combined supply and seperate supply. Internal OCP may be a part of the listing,
Keep in mind 424.22 limits the OCP to 60 amps.

 

[steve66]

If the unit has a single feed, the heater current *1.25 would be added in with the motor currents to find the MCA.

Gus: Yes, the heater loads need to be divided into circuits 60A or less, but on units that large, that is often accomplished by additional fuses in the HVAC units control panel. The HVAC unit still might have a MCA much larger than the 60A.

Steve

 

[EE_Intern_2]

Thank you for the replies.

The HVAC system I'm looking at has @ 208V:

one - 16.0 Amp RLA compressor motor
one- 3.0 Amp Condenser Fan Motor
one -2.8 Amp Supply and Exhaust Fan Motor

The optional 26.3 KW Electric Heater @ 208V draws 73.0 Amps

what is the MCA and MOCP for this unit?

 

[Mgraw]

Best guess. 100.6 MCA 110 MOCP Assuming 208v 3ph and it is not a heatpump.

 

[steve66]

Best guess. 100.6 MCA 110 MOCP Assuming 208v 3ph and it is not a heatpump.

I'm not 100% sure, but I think the heater gets an extra 25% for being a continuous load. THat would make it about 114 MCA.

If the heater does get another 25%, I'm not sure if you still have to add an extra 25% for a smaller motor.

Steve

 

[skeshesh]

Steve I think you have to consider both.

OP is on the right track and doing these calcs is a handy skill to have, but, as a practical note, another effetive approach is to push the manufacturer to supply the MCA and MOCP. I've been in contact with the vendor reps that our mechanical department deals with regularly and they're proven pretty resourceful in tracking down all the info I need.

 

[EE_Intern_2]

The HVAC system I'm looking at has @ 208V:

one - 16.0 Amp RLA compressor motor
one- 3.0 Amp Condenser Fan Motor
one -2.8 Amp Supply and Exhaust Fan Motor

The optional 26.3 KW Electric Heater @ 208V draws 73.0 Amps


why are we adding the compressor motor load and the condenser motor load to the MCA equations? the heat and the A/C are not going to be running at the same time. I am assuming that the supply and exhaust fan will run in both heating and cooling. Shouldn't the MCA calculations only take into account the larger motor loads either for heating or for cooling? e.g. (1.25*73A)+2.8A=94.5A


Mgraw how did you arrive at your numbers?

 

[skeshesh]

If you calculating by hand per NEC the sequence of operation doesnt help you much and you have to add all loads connected to the main disconnect whether or not they'll be operating at the same time. This is why it's helpful to contact the vendor because they can take the operating sequence into consideration and provide a more reasonable MCA on the nameplate, which you can use directly to size your conductors and OCPD.
The 1.25 mentioned in earlier posts applies to motors not the heater which is a resistive element, so your largest motor load is still the 16A. The next comment by Steve was that a 1.25 factor could be applied to it as it may be a continuous load (operating over 3hrs).

 

[steve66]

The HVAC system I'm looking at has @ 208V:

one - 16.0 Amp RLA compressor motor
one- 3.0 Amp Condenser Fan Motor
one -2.8 Amp Supply and Exhaust Fan Motor

The optional 26.3 KW Electric Heater @ 208V draws 73.0 Amps


why are we adding the compressor motor load and the condenser motor load to the MCA equations? the heat and the A/C are not going to be running at the same time. I am assuming that the supply and exhaust fan will run in both heating and cooling. Shouldn't the MCA calculations only take into account the larger motor loads either for heating or for cooling? e.g. (1.25*73A)+2.8A=94.5A


Mgraw how did you arrive at your numbers?

Sorry, I was asleep at the wheel.

Sometimes there is electric reheat. The air is cooled more than necessary, and then warmed back up to a comfortable temperature. This might be done to eliminate humidity, or to balance temperature in different zones.

But if what you have is cooling only, or heating only, you are right.

You probably should check the manufacturers info. to make sure it works the way you think, and to double check your calculation for MCA. At the very least, field verify the label before you connect the power.

 

[Mgraw]

Mgraw how did you arrive at your numbers?

I would like to tell you it is from my vast knowledge but alas I can not. I cheated. I looked up a unit with the same numbers and used its MCA and MOCP. I will say using motor fla and heater amps x 1.25 doesn't match manufacturers numbers. From what I have read this is because UL 1955 requires manufacturers to use "worst case" numbers when calculating MCA.

 

[Chamuit]

The HVAC system I'm looking at has @ 208V:

one - 16.0 Amp RLA compressor motor
one- 3.0 Amp Condenser Fan Motor
one -2.8 Amp Supply and Exhaust Fan Motor

The optional 26.3 KW Electric Heater @ 208V draws 73.0 Amps


why are we adding the compressor motor load and the condenser motor load to the MCA equations? the heat and the A/C are not going to be running at the same time. I am assuming that the supply and exhaust fan will run in both heating and cooling. Shouldn't the MCA calculations only take into account the larger motor loads either for heating or for cooling? e.g. (1.25*73A)+2.8A=94.5A


Mgraw how did you arrive at your numbers?

I would find a nameplate.

For cooling (16.0+3.0+2.8)*1.25 = 27.25MCA
For heat (73.0+3.0+2.8)*1.25 = 98.5MCA

The CFM and S&EFM are concurrent loads with either the heating or cooling. They are all part of one assembly (one motor load) and I do not see them as seperate motors.

 

[Mgraw]

Why would the CFM be used in the heating cycle?