another case of multiple motors on a single circuit

[habattack]

I appreciate any help in interpreting 430 for multiple motors that run simultaneously. I have:
1. Motor #1 - 8.5HP on 460v/3p, with FLA of 18.6A and LRA of 125A
2. Motors #2 and #3 - connected on L1 and L2 of the 3p conductor, with FLA = 1.9A each

I want to verify size of the common OCPD. I read at a disconnect vendor site that I would need to "add together one equivalent full-load current and one equivalent locked-rotor current". This would put the disconnect at over 125A.
1. I don't see the "adding LRC" requirement in the code (430.52). Is this requirement correct?
2. Also, should motors #2 and #3 be handled differently?

Thanks in advance for any guidance or input.

 

[petersonra]

This would be a feeder circuit. There are rules in article 430 for sizing motor feeder conductors and OCPD.

 

[wwhitney]

I'll take a stab at it, as I'm learning a bit about 430. Hopefully someone more experienced will confirm or correct.

- Table 430.250 gives FLC for 3 phase 460V motors as 11A for 7.5 HP and 14A for 10 HP. Interpolating for 8.5HP gives 12A, but as the specified FLA is higher, we'll used that. The table doesn't include any columns for 277V supply, and the HP is not given, so we'll just go with the 1.9A FLA for the two smaller ones.

- 430.24 tells you that the conductors supplying the 3 motors need to have an ampacity of 125% of the largest motor FLC and the sum of the other motor FLCs. So that's 125% * 18.6 + 1.9 + 1.9 = 27A.

- If this is a branch circuit and meets all the conditions in 430.53(C), then 430.53(C)(4) says the branch circuit can be protected by an inverse-time circuit breaker not greater than 250% of the largest motor FLC and the sum of the other motor FLCs. So 250% * 18.6 + 1.9 + 1.9 = 50A

- If this is a feeder, then 430.62(A) gives the same computation, so 50A OCPD is the maximum when the conductors are sized via 430.24.

Cheers, Wayne

 

[habattack]

Wayne,

Thanks for the detailed explanation. I agree with your approach. I spent some time in the code today and it turns out that I also should have provided a little more detail on the application. In fact, the motors are part of a refrigeration condensing unit. Accordingly, the unit is actually subject to article 440.

• 440.8 = the unit meets the requirement of a single machine and so can have one disconnect as per 430.112
• 440.32 = conductor must have ampacity of 125% of rated-load current
• 440.52(A) = OCPD can be 140% max of rated-load current [para. (1)], and fuse of not more than 125% of rated load current [para. (3)].

The overall unit’s rated-load current should be on the equipment nameplate, which is not yet purchased. So I think we can add the three FLAs = 18.6A + 1.9A + 1.9A = 22.4A to be equivalent to the overall unit’s rated load current. Therefore:

• Conductor = 125% * 22.4A = 28A -> AWG 10
• max fuse size = 125% * 22.4A = 28A -> 25A.
• max OCPD = 140% * 22.4A = 31.4A -> 30A

Hopefully I got it right. Thoughts?

 

[wwhitney]

The overall unit’s rated-load current should be on the equipment nameplate, which is not yet purchased. So I think we can add the three FLAs = 18.6A + 1.9A + 1.9A = 22.4A to be equivalent to the overall unit’s rated load current.

I'm not very proficient with 440, but I have no confidence in the above conclusion. Seems to me you need to get the equipment nameplate data before doing the circuit design.

Cheers, Wayne