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15
I Determine cooling and heating loads 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 db . . . . . . .95 F
Outdoor-Air Winter Design Temperature . . . . . 0° F
Indoor-Air Winter Design Temperature . . . . . .70 F
Indoor-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 50TFQ005 unit will
provide a total cooling capacity of 49,900 Btuh and
a sensible heat capacity of 33,900 Btuh.
For indoor-air temperature other than 80 F edb, cal-
culate sensible heat capacity correction, as required,
using the formula found in Note 2 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 18,000 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, 17,000 Btuh (35,000 – 18,000) addi-
tional electric heat is required.
Determine additional electric heat capacity in kW.
Enter the Electric Heating Capacities table (page 51)
for 50TFQ005 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 38,410 Btuh (18,000
+ 20,410).
IV Determine fan speed and power requirements
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 50TFQ005 ver-
tical discharge. At 1600 cfm and 230-v high speed,
the standard motor will deliver 0.76 in. wg static
pressure, 723 watts, and 0.64 brake horsepower
(bhp). This 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
= 49,900 Btuh – (723 Watts x 3.413
Btuh/Watts)
= 49,900 Btuh – 2468 Btuh
= 47,432
Net sensible capacity = 33,900 Btuh – 2468 Btuh
= 31,432 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 38,410 Btuh, the
total heating capacity determined in Step III.
17,000 Btuh
= 5.0 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
Selection procedure (with 50TFQ005 example)