Selection Procedure - Carrier WEATHERMASTER 50HJQ004-016 Product Data

Selection procedure (with 50HJQ005 example)
I Determine cooling and heating requirements
at design conditions:
Given:
Required Cooling Capacity (TC) . . . . .38,000 Btuh
Sensible Heat Capacity (SHC). . . . . . .24,000 Btuh
Required Heating Capacity. . . . . . . . .35,000 Btuh
Outdoor Entering-Air Temperature . . . . . . . . .95 F
Outdoor-Air Winter Design Temperature . . . . . 0° F
Indoor-Air Winter Design Temperature . . . . . .70 F
Indoor Entering-Air
Temperature . . . 80 F edb (entering air, dry bulb),
Indoor-Air Quantity . . . . . . . . . . . . . . . . 1600 cfm
External Static Pressure . . . . . . . . . . . . 0.45 in. wg
Electrical Characteristics (V-Ph-Hz) . . . . . 230-3-60
II 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, cal-
culate 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 cal-
culate net capacities, see Step V.
III Select electric heat.
Enter the Instantaneous and Integrated Heating Rat-
ings 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 value since deductions for 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) addi-
tional electric heat is required.
Determine additional electric heat capacity in kW.
12,600 Btuh
= 3.7 kW of heat required.
3413 Btuh/kW
92
67 F ewb (entering air, wet bulb)
Enter the Electric Heating Capacities table for
50HJQ005 at 208/230, 3 phase. The 6.5-kW
heater at 240 v most closely satisfies the heating
required. To calculate kW at 230 v, use the Multipli-
cation 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).
IV Determine fan speed and power require-
ments at design conditions.
Before entering Fan Performance tables, calculate
the total static pressure required based on unit com-
ponents. 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.
V 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
Integrated heating capacity is maximum (instanta-
neous) capacity less the effect of frost on the out-
door coil and the heat required to defrost it.
Therefore, net capacity is equal to 42,810 Btuh, the
total heating capacity determined in Step III.
.45 in. wg
.07 in. wg
.09 in. wg
.61 in. wg
Btuh/Watts)
= 32,935 Btuh