Sizing MCA per UL 1995 - Fan Powered HVAC units

Status
Not open for further replies.

Eddie702

Licensed Electrician
Location
Western Massachusetts
Occupation
Electrician
This is a little on subject. But I'm trying to figure out why the heater resistance load gets the 125% towards MCA? I know UL 1995 says so, but wondering why, I get motor can have an inrush but resistance can't, and with VFD's even the motor wont see it on the . I'm just curious if any one knows.
I don't usually have to deal with heaters in sizing as units my company builds have compressors so much larger and don't operate at same time.

example unit: 480V/3
Blower Motor: 103A (on VFD)
Compressors: 53A + 34A + 34A + 34A + 34A
Condenser Fans: 8.8A
Control breaker: 5A
Heaters: 30A + 30A
MCA: 332A
MOP: 400A

Now have a customer demanding a huge 246A electric heater (don't know why so big I lost that battle showing 350% larger then will ever need)
MCA becomes 442A with the 125% of those large heaters and MOP 450A.


Any HVAC equipment in my opinion gets 125% because it can be a continuous load
 

Eddie702

Licensed Electrician
Location
Western Massachusetts
Occupation
Electrician
Hi everyone,

Long time observer, first time poster. I'm in the middle of a discussion with an HVAC rep regarding the MCA and MOCP values for their units. During the design we were not given the MCA and MOCP values so I had to calculate them myself. Now that the submittals are coming in, I'm finding that they have indicated slightly larger MCA and MOCP values than what we had calculated. I was hoping I could have some insight from the group here. What are your thoughts?

I've provided an example for us to work through. For the sake of time let's just discuss the MCA calculation.

We have a 480V, 1-PH fan powered box with 9.5kW electric resistance heat and a 5.5A, 277V, 1-PH fan motor. The electric heater is 19.8A at 480V, 1-PH.

Per UL 1995 MCA is derived from 125% of the largest motor rated current plus 125% of the electric heater rated current plus 100% of all other loads.

My calculation for MCA:
MCA = ((1.25*5.5A*277V) + (1.25*9,500VA)) / 480V
MCA = ((1,904VA) + (11,875VA)) / 480V
MCA = (13,779VA) / 480V
MCA = 28.7A at 480V, 1-PH

Manufacturer's calculation for MCA:
MCA = (1.25*5.5A) + (1.25*19.8A)
MCA = (6.9A) + (24.8A)
MCA = 31.6A at 480V, 1-PH

I know the discrepancy is that they are adding 277V and 480V currents together, which I claim is incorrect. I've emailed them and they came back with a response posted below. I will concede that the way UL 1995 is written, MCA is calculated by using the rated current values and therefore their calculation is done correctly according to UL 1995. However, I am not in 100% agreement with the rep's interpretation. I don't believe their explanation and supporting graphic is correct.

Manufacturer's response.
"The calculations we use for MCA, taken from UL 1995, below, are appropriate for our units.

'37.14 c) For a combination load, a load consisting of one or more motors, electric heaters, and any
other loads, not involving a hermetic refrigerant motor compressor, [MCA is] the sum of the rated
currents multiplied by 125 per cent.

(Clause 37.14(c))
MCA = 1.25 ´ (LOAD1 + LOAD2 + LOAD4) + (1.25* ´ LOAD3)'

While we can use kW or kVA to calculate the average current as the customer explained, this is only accurate on the balanced three phase system. The issue is that our unit is an unbalanced system and we must size the wires – and the breaker – for the worst case loading due to the unbalance. Since the UL code is mainly concerned with safety, they consider only the addition of the currents onto the worst case line. See the example below.

In the drawing below, Line 2 sees only the heater currents.
Line 1 sees the sum of both heater and motor currents. This will be a higher current than that on Line 2, and so will be the worst case.

I hope this helps explain why UL uses the sum of currents rather than the average in the sizing calculations."

View attachment 2388827

Thanks for everyone's time.
If the boxes are not made yet (all that stuff is built to order) maybe the MFG. could build them with a 480/277 volt 4 wire branch circuit feed you would then have a MCA of 18.4 would run on #12s
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Now have a customer demanding a huge 246A electric heater (don't know why so big I lost that battle showing 350% larger then will ever need)
MCA becomes 442A with the 125% of those large heaters and MOP 450A.

I believe UL 1995 allows for 100% of the heater load (instead of 125%) to be used for heaters that large.
 

TheOath

Member
Location
California
Occupation
Engineer
I believe UL 1995 allows for 100% of the heater load (instead of 125%) to be used for heaters that large.
yes, I've read exception if field wiring terminal larger than 50 kW..... but what constitutes "field wiring" as OEM heater is installed internal to unit in factory.
 
Status
Not open for further replies.
Top