Selection Procedure (With 50Hjq005 Example) - Carrier 50HJQ004-016 Product Data

SELECTION PROCEDURE (WITH 50HJQ005 EXAMPLE)

(SAME PROCEDURE FOR 50HEQ)
A. Determine cooling and heating requirements at design
conditions:
Given:
Required Cooling Capacity (TC)
Sensible Heat Capacity (SHC)
Required Heating Capacity . . . . . . . . . . . . . . .
Outdoor Entering-Air Temperature
Outdoor-Air Winter Design Temperature
Indoor-Air Winter Design Temperature
Indoor Entering-Air
Temperature . . . . . . . . . . 80_F edb (entering air, dry bulb),
67_F ewb (entering air, wet bulb)
Indoor-Air Quantity . . . . . . . . . . . . . . . . . . . . . .
External Static Pressure . . . . . . . . . . . . . . . . . . .
Electrical Characteristics (V-Ph-Hz)
B. Select unit based on required cooling capacity.
Enter Cooling Capacities table at outdoor entering temperature of
95_F, indoor air entering at 1600 cfm and 67_F ewb. The
50HJQ005 unit will provide a total cooling capacity of 46,300
Btuh and a sensible heat capacity of 35,600 Btuh.
For indoor-air temperature other than 80_F edb, calculate sensible
heat capacity correction, as required, using the formula found in
Note 3 following the cooling capacities tables.
NOTE: Unit ratings are gross capacities and do not include the
effect of indoor-fan motor heat. To calculate net capacities, see
Step E.
C. Select electric heat.
Enter the Instantaneous and Integrated Heating Ratings table at
1600 cfm. At 70_F return indoor air and 0°F air entering outdoor
coil, the integrated heating capacity is 22,400 Btuh. (Select
integrated heating capacity
outdoor-coil frost and defrosting have already been made. No
correction is required.)
The required heating capacity is 35,000 Btuh. Therefore, 12,600
Btuh (35,000 – 22,400) additional electric heat is required.
Determine additional electric heat capacity in kW.
12,600 Btuh
= 3.7 kW of heat required
3413 Btuh/kW
Enter the Electric Heating Capacities table for 50HJQ005 at
208/230, 3 phase. The 6.5-kW heater at 240v most closely
satisfies the heating required. To calculate kW at 230v, use the
Multiplication Factors table.
6.5 kW x .92 = 5.98 kW
6.5 kW x .92 x 3413 = 20,410 Btuh
Total unit heating capacity is 42,810 Btuh (22,400 + 20,410).
38,000 Btuh
. . . . . . . . . . . . 24,000 Btuh
35,000 Btuh
. . . . . . . . . . . . . . . 95_F
. . . . . . . . . . . 0°F
. . . . . . . . . . . 70_F
1600 cfm
0.45 in. wg
. . . . . . . . . . . 230-3-60
value since deductions for
3 7 kW f h t i d
D. Determine fan speed and power requirements at design
conditions.
Before entering Fan Performance tables, calculate the total static
pressure required based on unit components. From the given and
the Pressure Drop tables, find:
External static pressure
EconoMi$er IV
Electric heat
Total static pressure
Enter the Fan Performance table for 50HJQ005 vertical discharge
at 1600 cfm and .61 in. wg static pressure. The bhp is 0.79 and
the watts are 781. The rpm is 997. The standard motor will
adequately handle job requirements.
E. Determine net capacities.
Capacities are gross and do not include the effect of indoor-fan
motor (IFM) heat.
Determine net cooling capacity as follows:
Net capacity =
Total capacity – IFM heat
=
46,300 Btuh – (781 Watts x 3.413
= 46,300 Btuh – 2665 Btuh
= 43,635
Net sensible capacity = 35,600 Btuh – 2665 Btuh
= 32,935 Btuh
Integrated heating capacity is maximum (instantaneous) capacity
less the effect of frost on the outdoor coil and the heat required to
defrost it. Therefore, net capacity is equal to 42,810 Btuh, the
total heating capacity determined in Step C.
84
.45 in. wg
.07 in. wg
.09 in. wg
.61 in. wg
Btuh/Watts)