Product data

93
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),
67 F ewb (entering air, wet 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 47,400 Btuh and
a sensible heat capacity of 32,800 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 20,400 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, 14,600 Btuh (35,000 – 20,400) addi-
tional electric heat is required.
Determine additional electric heat capacity in kW.
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 40,810 Btuh (20,400
+ 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:
Enter the Fan Performance table for 50HJQ005
vertical discharge. At 1600 cfm and 230-v, the
standard motor will deliver 1.2 in. wg static pressure
and 1.15 brake horsepower (bhp). This will ade-
quately handle job requirements.
NOTE: Convert bhp to Watts using the formula
found in the note following the Indoor-Fan Motor
Efficiency table.
For this example:
Watts = 1144
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:
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 40,810 Btuh, the
total heating capacity determined in Step III.
14,600 Btuh
= 4.3 kW of heat required.
3413 Btuh/kW
External static pressure .45 in. wg
Durablade economizer .05 in. wg
Electric heat .09 in. wg
Total static pressure .59 in. wg
Watts =
746 x Bhp
Motor Efficiency
Watts =
746 x 1.15
.75
Net capacity = Total capacity – IFM heat
= 47,400 Btuh – (1144 Watts x 3.413
Btuh/Watts)
= 47,400 Btuh – 3904 Btuh
= 43,496
Net sensible capacity = 32,800 Btuh – 3904 Btuh
= 28,896 Btuh
Selection procedure (with 50HJQ005 example)
50TFQ004-01250HJQ004-012