greichling
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- Location
- Cincinnati, Ohio
NEC 430.24 & 430.122 Application to the feeder of a packaging machine control panel.
I have seen this code applied 3 different ways with regard to several motor control panels that have a mixture of Adjustable / Variable Speed Drives and Starters. So I have several questions.
These packaging / material handling machines are cyclic in nature. Motors turn on and off for different intervals based on the program and physical limitations of the equipment.
430.24 Exception 3 allows discounting motors when circuitry is interlocked but in practice hardwired circuity is rarely used. The PLC program controls the cycle.
1) Is the program cycle sufficient to satisfy exception 3. See Example below - One of the 2 HP Motors is on a Conveyor that is raised and lowered by an elevator. It will only run when the Elevator (5 HP motor on VFD) is stopped in the Up or Down position. Logic is in the PLC to prevent the conveyor from running unless the elevator is stopped at the up or down position. Could the 2 HP motor Smallest motor be discounted in the wire size and disconnect size calculations?
2) The measured peak load of this equipment is 65% of the calculated values with the continuous load being less than 40% of the calculated value. Is this type of measured data sufficient for to satisfy 430.26? Do you have any other recommendations?
3) What is the correct method for calculating the Feeder to the a control panel containing Variable Frequency Drive controllers (VFD)? I have seen this calculated in (3) different ways as shown below. The Largest motor is 7.5 Hp but the 5 HP VFD has an input of 13 amps. Which one is correct based on the intent of the code? Or is there yet another interpretation?
Given a 460 VAC three phase machine - 1 motor per drive. With the following Control panel
QNTY Motor / Controller
(7) 1 HP VFDs - Drive rated input 3.2 Amps, Motor = 2.1 Amps from NEC table.
(1) 5 Hp VFD Drive rated input 13 Amps, Motor = 7.6 Amps from NEC table.
(1) 7.5 Hp Motor on a starter 11 Amps from Table
(2) 2 Hp Motor on a Starter 3.4 Amps from NEC
Method #1: (Find the largest FLA (Input of drive from Manufacturer doc or Motor FLC from NEC Table) use 125% of this value in the calc per 430.24 here the 5 HP drive Input is used with the 125% as it is larger than the 7.5 Hp motor in the calc. Then add the remaining VFD input FLA and Motor FLCs from the NEC Table.)
(7 * 3.2) + (13 * 1.25) + 11 + (2 * 3.4)
22.4 + 16.25 + 11 + 6.8
= 56.45
Method #2 (This method interprets 430.122 as modifying 430.24 and that all VFDS are added in at 125%. Since the largest motor is not on a VFD it is also added at 125% per 430.24 )
((7 * 3.2)+13) * 1.25 + (11 * 1.25) + (2 * 3.4)
(35.4 * 1.25) + (11 *1.25) + 3.4
44.25 + 13.75 + 6.8
= 64.8
Method #3 (This interpretation is that does not modify 430.24 as 430.122 does not address the several motor scenario that 430.24 deals with.)
(7*2.1) + 7.6 + (1.25*11) + (2 * 3.4)
14.7 +7.6 + 13.75 + 6.8
= 42.85
I have seen this code applied 3 different ways with regard to several motor control panels that have a mixture of Adjustable / Variable Speed Drives and Starters. So I have several questions.
These packaging / material handling machines are cyclic in nature. Motors turn on and off for different intervals based on the program and physical limitations of the equipment.
430.24 Exception 3 allows discounting motors when circuitry is interlocked but in practice hardwired circuity is rarely used. The PLC program controls the cycle.
1) Is the program cycle sufficient to satisfy exception 3. See Example below - One of the 2 HP Motors is on a Conveyor that is raised and lowered by an elevator. It will only run when the Elevator (5 HP motor on VFD) is stopped in the Up or Down position. Logic is in the PLC to prevent the conveyor from running unless the elevator is stopped at the up or down position. Could the 2 HP motor Smallest motor be discounted in the wire size and disconnect size calculations?
2) The measured peak load of this equipment is 65% of the calculated values with the continuous load being less than 40% of the calculated value. Is this type of measured data sufficient for to satisfy 430.26? Do you have any other recommendations?
3) What is the correct method for calculating the Feeder to the a control panel containing Variable Frequency Drive controllers (VFD)? I have seen this calculated in (3) different ways as shown below. The Largest motor is 7.5 Hp but the 5 HP VFD has an input of 13 amps. Which one is correct based on the intent of the code? Or is there yet another interpretation?
Given a 460 VAC three phase machine - 1 motor per drive. With the following Control panel
QNTY Motor / Controller
(7) 1 HP VFDs - Drive rated input 3.2 Amps, Motor = 2.1 Amps from NEC table.
(1) 5 Hp VFD Drive rated input 13 Amps, Motor = 7.6 Amps from NEC table.
(1) 7.5 Hp Motor on a starter 11 Amps from Table
(2) 2 Hp Motor on a Starter 3.4 Amps from NEC
Method #1: (Find the largest FLA (Input of drive from Manufacturer doc or Motor FLC from NEC Table) use 125% of this value in the calc per 430.24 here the 5 HP drive Input is used with the 125% as it is larger than the 7.5 Hp motor in the calc. Then add the remaining VFD input FLA and Motor FLCs from the NEC Table.)
(7 * 3.2) + (13 * 1.25) + 11 + (2 * 3.4)
22.4 + 16.25 + 11 + 6.8
= 56.45
Method #2 (This method interprets 430.122 as modifying 430.24 and that all VFDS are added in at 125%. Since the largest motor is not on a VFD it is also added at 125% per 430.24 )
((7 * 3.2)+13) * 1.25 + (11 * 1.25) + (2 * 3.4)
(35.4 * 1.25) + (11 *1.25) + 3.4
44.25 + 13.75 + 6.8
= 64.8
Method #3 (This interpretation is that does not modify 430.24 as 430.122 does not address the several motor scenario that 430.24 deals with.)
(7*2.1) + 7.6 + (1.25*11) + (2 * 3.4)
14.7 +7.6 + 13.75 + 6.8
= 42.85