Sizing MCA per UL 1995 - Fan Powered HVAC units

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Ohno Raccoon

Member
Location
Fairfax, VA
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."

enviro.jpg

Thanks for everyone's time.
 

texie

Senior Member
Location
Fort Collins, Colorado
Occupation
Electrician, Contractor, Inspector
I agree with the manfacturer's calc. The standard is "current" not balanced VA or watts. This same issue arises with motor circuit feeders sometimes when you have, say, 1, single phase and 1, 3 phase motor on the feeder.
 

Ohno Raccoon

Member
Location
Fairfax, VA
Thanks for the input. Yeah, based on the phrasing of UL 1995, I tend to agree.

I guess the support tech's reasoning behind the approach didn't sound right to me.
 

texie

Senior Member
Location
Fort Collins, Colorado
Occupation
Electrician, Contractor, Inspector
Thanks for the input. Yeah, based on the phrasing of UL 1995, I tend to agree.

I guess the support tech's reasoning behind the approach didn't sound right to me.

In your example the impact is kind of a big deal (#10 VS #12 conductor), so I see why you are having heart burn over this. Sorry.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
You are incorrectly assuming that the single phase loads are 277V, they are not (from what I can make out on that drawing). 277V would be single phase, Line to NEUTRAL. What I see is single phase, Line to LINE, which would be 480V.
 

Ohno Raccoon

Member
Location
Fairfax, VA
In your example the impact is kind of a big deal (#10 VS #12 conductor), so I see why you are having heart burn over this. Sorry.
Yeah, the real problem is that this will affect multiple unit types across the whole project including conductors, MOCPs and disconnects. I'll admit I should have pressed harder to get this info during the design.


You are incorrectly assuming that the single phase loads are 277V, they are not (from what I can make out on that drawing). 277V would be single phase, Line to NEUTRAL. What I see is single phase, Line to LINE, which would be 480V.
No, I did not make an assumption. The manufacturer listed the fan as 277V, 1-PH and the electric heater at 480V, 1-PH. Sorry, the image did not post at its original size, it's probably too difficult to read in the original post. I'll try it again.

enviro.jpg
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
No, I did not make an assumption. The manufacturer listed the fan as 277V, 1-PH and the electric heater at 480V, 1-PH.
That's exactly what this diagram shows. Line 2 supplies the heaters only. This is just like a typical residential clothes dryer.

One thing not seen here, but may be part of the control circuitry, is anything to prevent simultaneous operation.
 

Ohno Raccoon

Member
Location
Fairfax, VA
See 240.4(G) - Specific conductor applications.
Got it, thank you.

Does anyone have any input regarding the calculation of MCA to cover the "worst case line", L1? Based on the explanation from the manufacturer's rep and the provided diagram, L1 serves both the heaters and the fan; L2 serves just the heaters. I admit that my initial calculation inaccurately divides the fan load evenly across both lines.

But I figured the load for L1 would be:
(1523VA + 4750VA) = 6273VA
L1-N = 22.6A at 277V
MCA for L1-N = 1.25*22.6A = 28.3A at 277V

And the load for L2 would be just 4750VA:
L2-N = 17.1A at 277V
MCA for L2-N = 1.25*17.1A = 21.4A at 277V
 

Ohno Raccoon

Member
Location
Fairfax, VA
The heater current is 19.8A, not 17.1A.
Right, the heater current is 19.8A at 480V, 1PH (L1-L2). I was suggesting the heater current between L2-N is 17.1A.

What about the current on L1? I'm still hung up with their explanation that the current on L1 is simply 19.8A at 480V, 1PH plus 5.5A at 277V, 1PH. Is that really an accurate calculation for getting the line current on L1?
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Right, the heater current is 19.8A at 480V, 1PH (L1-L2). I was suggesting the heater current between L2-N is 17.1A.

It's not 17.1A. There is no heater current between L2-N (or L1-N).


What about the current on L1? I'm still hung up with their explanation that the current on L1 is simply 19.8A at 480V, 1PH plus 5.5A at 277V, 1PH. Is that really an accurate calculation for getting the line current on L1?

If you wanted a precise number, you would need to know the power factor of the motor. But for simplicity's sake, the current on L1 is the heater current plus the motor current.
 

kwired

Electron manager
Location
NE Nebraska
You have a single phase multiwire circuit - doesn't matter if it is 277/480 or 120/240, one of the ungrounded conductors has more load connected to it than the other, at the very least that one conductor needs higher ampacity than the other. Figuring out the net KVA and then dividing by line to line volts does not reflect actual load on each conductor here.
 

Ohno Raccoon

Member
Location
Fairfax, VA
It's not 17.1A. There is no heater current between L2-N (or L1-N).




If you wanted a precise number, you would need to know the power factor of the motor. But for simplicity's sake, the current on L1 is the heater current plus the motor current.
Thanks for the reminder. Yes, when you put it that way it makes sense. I was having a mental block adding current ratings across different operating voltages.

You have a single phase multiwire circuit - doesn't matter if it is 277/480 or 120/240, one of the ungrounded conductors has more load connected to it than the other, at the very least that one conductor needs higher ampacity than the other. Figuring out the net KVA and then dividing by line to line volts does not reflect actual load on each conductor here.
Yes, I understand and admitted that taking the net KVA and dividing by line to line volts was not accurate.

Thanks for the input everyone.
 

TheOath

Member
Location
California
Occupation
Engineer
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.
 
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