ELECTRICAL DATA
UNIT |
|
| VOLTAGE | COMPRESSOR | COND FAN | INDOOR FAN | POWER SUPPLY | ||||
SIZE | RANGE | MOTOR | MOTOR | ||||||||
|
|
|
| ||||||||
582A |
|
| Min | Max | RLA | LRA | FLA | FLA | MCA | MOCP* | |
018 |
| 187 | 253 | 9.0 | 45.0 | 0.8 | 1.8 | 13.9 | 20 | ||
024 |
| 187 | 253 | 12.8 | 61.0 | 0.8 | 2.0 | 18.8 | 30 | ||
030 |
| 187 | 253 | 14.4 | 73.0 | 0.8 | 2.0 | 20.8 | 30 | ||
| 187 | 253 | 8.3 | 68.0 | 0.8 | 2.0 | 13.2 | 20 | |||
|
| ||||||||||
|
| 187 | 253 | 15.1 | 81.0 | 1.6 | 3.6 | 24.1 | 35 | ||
036 |
| 187 | 253 | 10.9 | 78.0 | 1.6 | 3.6 | 18.8 | 25 | ||
|
| 414 | 506 | 5.8 | 40.0 | 0.9 | 1.9 | 10.1 | 15 | ||
|
| 187 | 253 | 18.6 | 105.0 | 1.6 | 3.8 | 28.9 | 45 | ||
042 |
| 187 | 253 | 10.7 | 85.0 | 1.6 | 3.8 | 18.8 | 25 | ||
|
| 414 | 506 | 5.3 | 42.0 | 0.9 | 2.0 | 9.5 | 15 | ||
|
| 197 | 253 | 25.3 | 131.0 | 1.6 | 3.8 | 37.0 | 60 | ||
048 |
| 187 | 253 | 14.6 | 108.0 | 1.6 | 3.8 | 23.7 | 35 | ||
|
| 414 | 506 | 7.3 | 47.5 | 0.9 | 2.0 | 12.0 | 15 | ||
|
| 187 | 253 | 28.9 | 147.0 | 1.6 | 6.2 | 43.9 | 60 | ||
060 |
| 187 | 253 | 18.6 | 125.0 | 1.6 | 6.2 | 31.1 | 45 | ||
|
| 414 | 506 | 8.5 | 66.5 | 0.9 | 3.2 | 14.7 | 20 | ||
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| LEGEND |
|
FLA | — | Full Load Amps |
|
LRA | — | Locked Rotor Amps | ® |
MCA | — | Minimum Circuit Amps |
|
MOCP — Maximum Overcurrent Protection |
| ||
RLA | — | Rated Load Amps |
|
NOTES:
1.In compliance with NEC (National Electrical Code) requirements for multimotor and combination load equipment (refer to NEC Articles 430 and 440), the overcurrent protective device for the unit shall be Power Supply fuse . The CGA (Canadian Gas Association) units may be fuse or circuit breaker.
2.Minimum wire size is based on 60 C copper wire. If other than 60 C wire is used, or if length exceeds wire length in table, determine size from NEC.
3.Unbalanced
Never operate a motor where a phase imbalance in supply volt- age is greater than 2%. Use the following formula to determine the percentage of voltage imbalance.
%Voltage imbalance
=100 x max voltage deviation from average voltage average voltage
EXAMPLE: Supply voltage is
AB = 452 v BC = 464 v AC = 455 v
Average Voltage = 452 + 464 + 455 3
= 1371
3 = 457
Determine maximum deviation from average voltage. (AB) 457 452 = 5 v
(BC) 464 457 = 7 v (AC) 457 455 = 2 v
Maximum deviation is 7 v.
Determine percent of voltage imbalance.
% Voltage Imbalance = 100 x |
| 7 | |
457 | |||
| |||
= 1.53%
This amount of phase imbalance is satisfactory as it is below the maximum allowable 2%.
IMPORTANT: If the supply voltage phase imbalance is more than 2%, contact your local electric utility company immediately.
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