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Total Load Calculation for RTU based on 220.50

Merry Christmas
Location
PA
Occupation
Intern in Highschool
Let's say I have 3 RTU units at 208/3P
RTU #1: 48A MCA, 60A MOCP
RTU #2: 30A MCA, 40 MOCP
RTU #3: 22A MCA, 30 MOCP

Based on 220.50, I need to take 125% of the largest motor.
But do I use the 48 MCA or do I need to find the FLA.
But doesn't the MCA already calculate 125% of the largest load when sizing the MCA from the manufacturer?

Also, after finding the largest load, what load do I use when I calculate RTU #2 and RTU #3 for the load calculation at 100%?
Do I use the 30A and 22A of the MCA or FLA?

Also if there is refrigeration (Walk in) equipment, do I use 100% load if it's not the largest load?
Or can I consider the Walk-in as a Kitchen Equipment Load and use Table 220.56?
 

suemarkp

Senior Member
Location
Kent, WA
Occupation
Retired Engineer
MCA has the 125% factor for the largest motor. The easiest thing to do is add the MCAs and call it a day. That will provide some margin, as it is larger than it needs to be.

Otherwise, use the largest MCA and add the components (RLA, FLA) of the other units. This isnt always straightforward though, as there could be control circuits or heaters in those units. If you can get the pieces with 25% added to the largest motor to add up to the MCA, then you can be confident you have all the components covered.
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
Let's say I have 3 RTU units at 208/3P
RTU #1: 48A MCA, 60A MOCP
RTU #2: 30A MCA, 40 MOCP
RTU #3: 22A MCA, 30 MOCP

Based on 220.50, I need to take 125% of the largest motor.
But do I use the 48 MCA or do I need to find the FLA.
You need more data from each unit's nameplate, namely the largest motor FLA. So to make up some numbers, you might find after checking the nameplates:

RTU #1: 48A MCA, 60A MOCP, 10A largest FLA
RTU #2: 30A MCA, 40 MOCP, 8A largest FLA
RTU #3: 22A MCA, 30 MOCP, 6A largest FLA

Then when MCA was calculated for each unit, the formula was MCA = sum of loads + 25% extra of the largest FLA. Which means you can back calculate "sum of loads" in each unit as MCA - 25% of largest FLA.

That would make your feeder MCA = 48A MCA (the MCA with the largest FLA) + (30A MCA - 25% * 8A) + (22A MCA - 25% * 6A) = 96.5A. And it would make your feeder MOCP = 60A MOCP (the MOCP with the largest FLA) + (30A MCA - 25% * 8A) + (22A MCA - 25% * 6A) = 108.5A. I don't see any allowance in 430.62(A) to further round up the MOCP, so you'd be limited to 100A.

Of course, you'll have to check the nameplates to find the actual largest FLAs and redo the calculations.

Cheers, Wayne
 
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