16
Table 4 – Electrical Data, Standard Motors
UNIT
V---PH---Hz
{
VOLTAGE
LIMITS
FAN MOTOR
POWER SUPPLY
Hp (kW)
FLA
Minimum
Circuit Amps
MOCP
524J*25A
524J*25H
208/230---3---60
187---253
5.0 (3.73)
14.7/13.6
18.4/17.0
30/25
460---3---60
414---506
5.0 (3.73)
6.8
8.5
15
575---3---60
518---632
5.0 (3.73)
5.1
6.4
15
524J*28A
208/230---3---60
187---253
7.5 (5.59)
21.4/19.4
26.9/24.3
45/40
460---3---60
414---506
7.5 (5.59)
9.7
12.1
20
575---3---60
518---632
7.5 (5.59)
7.8
9.8
15
524J*30A
208/230---3---60
187---253
10.0 (7.46)
28.0/25.2
35.0/31.5
60/50
460---3---60
414---506
10.0 (7.46)
12.6
15.8
25
575---3---60
518---632
10.0 (7.46)
10.3
12.9
20
See Legend and Notes below.
Table 5 – Electrical Data, Alternate Motors
UNIT
V---PH---Hz*
VOLTAGE
LIMITS
FAN MOTOR
POWER SUPPLY
Hp (kW)
FLA
Minimum
Circuit Amps
MOCP
524J*25A
524J*25H
208/230---3---60
187---253
7.5 (5.59)
21.4/19.4
26.9/24.3
45/40
460---3---60
414---506
7.5 (5.59)
9.7
12.1
20
575---3---60
518---632
7.5 (5.59)
7.8
9.8
15
524J*28A
524J*30A
208/230---3---60
187---253
10.0 (7.46)
28.0/25.2
35.0/31.5
60/50
460---3---60
414---506
10.0 (7.46)
12.6
15.8
25
575---3---60
518---632
10.0 (7.46)
10.3
12.9
20
See Legend and Notes below.
Legend and Notes for Tables 4 and 5
LEGEND:
FLA
--- Full Load Amps
MOCP --- Maximum Overcurrent Protection
{
Motors are designed for satisfactory operation within 10% of
normal voltage shown. Voltages should not exceed the limits
shown in the Voltage Limits column.
NOTES
:
1. Minimum circuit amps (MCA) and MOCP values are calcu-
lated in accordance with The NEC. Article 440.
2. Motor FLA values are established in accordance with Under-
writers’ Laboratories (UL). Standard 1995.
3.
Unbalanced 3-Phase Supply Voltage
Never operate a motor where a phase imbalance in supply
voltage is greater than 2%. Use the formula in the example
(see column to the right) to determine the percentage of volt-
age imbalance.
4. Installation with Accessory Electric Heaters
Size the Field Power Wiring between the heater TB1 and the
524J indoor fan motor per NEC Article 430---28 (1) or (2)
(depends on length of conduit between heater enclosure and
524J power entry location). Install wires in field---installed
conduit.
Example: Supply voltage is 230-3-60
% Voltage Imbalance
= 100 x
max voltage deviation from average voltage
average voltage
AB = 393 v
BC = 403 v
AC = 396 v
Average Voltage =
(393 + 403 + 396)
=
1192
3
3
=
397
Determine maximum deviation from average voltage.
(AB) 397 – 393 = 4 v
(BC) 403 – 397 = 6 v
(AC) 397 – 396 = 1 v
Maximum deviation is 4 v.
Determine percent of voltage imbalance.
% Voltage Imbalance
= 100 x
6
397
= 1.5%
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.
524J
Содержание 524J Series
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