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SELECTION PROCEDURE (WITH 548F048 EXAMPLE)
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 temper-
ature of 95 F, indoor air entering at 1600 cfm and
67 F ewb. The 548F048 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, 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 V.
III 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
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) additional electric heat is
required.
Determine additional electric heat capacity in kW.
Enter the Electric Heating Capacities table (page 54) for
548F048 at 208/230, 3 phase. The 6.5-kW heater at
240 v most closely satisfies the heating required. To calcu-
late kW at 230 v, 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 38,410 Btuh (18,000 +
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 components.
From the given and the Pressure Drop tables, find:
Enter the Fan Performance table for 548F048 vertical dis-
charge. 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 (instantaneous)
capacity less the effect of frost on the outdoor 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
50TFQ004-012
548F036-120