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Sizing MCA per UL 1995 - Fan Powered HVAC units

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    Sizing MCA per UL 1995 - Fan Powered HVAC units

    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."

    Click image for larger version

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    Thanks for everyone's time.

    #2
    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.

    Comment


      #3
      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.

      Comment


        #4
        Originally posted by Ohno Raccoon View Post
        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.

        Comment


          #5
          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.
          __________________________________________________ ____________________________
          Many people are shocked when they discover I am not a good electrician...

          I'm in California, ergo I am still stuck on the 2014 NEC... We'll get around to the 2017 code in around 2021.

          Comment


            #6
            Originally posted by texie View Post
            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.


            Originally posted by Jraef View Post
            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.

            Click image for larger version

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            Comment


              #7
              Originally posted by Ohno Raccoon View Post
              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.
              Master Electrician
              Electrical Contractor
              Richmond, VA

              Comment


                #8
                Originally posted by texie View Post
                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.
                In his example, it's a #10 vs #10 conductor...same size required in both calculations.

                Comment


                  #9
                  Originally posted by david luchini View Post
                  In his example, it's a #10 vs #10 conductor...same size required in both calculations.
                  Actually, it's going to go from a #10 up to a #8. I didn't mention it in the original post but for this instance the breaker is going from 30A up to 35A.

                  Comment


                    #10
                    Originally posted by Ohno Raccoon View Post
                    Actually, it's going to go from a #10 up to a #8. I didn't mention it in the original post but for this instance the breaker is going from 30A up to 35A.
                    #10 should be sufficient for an MCA of 31.6A.

                    Comment


                      #11
                      Originally posted by david luchini View Post
                      In his example, it's a #10 vs #10 conductor...same size required in both calculations.
                      Yep, you are correct. Don't know where I got #12.

                      Comment


                        #12
                        Originally posted by david luchini View Post
                        #10 should be sufficient for an MCA of 31.6A.
                        Yes, but the breaker will need to be 35A. Wouldn't NEC 240.4(D)(7) require the conductors be #8?

                        Comment


                          #13
                          Originally posted by Ohno Raccoon View Post
                          Yes, but the breaker will need to be 35A. Wouldn't NEC 240.4(D)(7) require the conductors be #8?
                          See 240.4(G) - Specific conductor applications.

                          Comment


                            #14
                            Originally posted by david luchini View Post
                            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

                            Comment


                              #15
                              Originally posted by Ohno Raccoon View Post

                              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
                              The heater current is 19.8A, not 17.1A.

                              Comment

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