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74
SELECTION PROCEDURE (WITH 549B048 EXAMPLE)
I DETERMINE COOLING AND HEATING REQUIRE-
MENTS 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 temper-
ature of 95 F, indoor air entering at 1600 cfm and 67 F
ewb. The 549B048 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, 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 capac-
ities, 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
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) additional electric heat is
required.
Determine additional electric heat capacity in kW.
Enter the Electric Heating Capacities table for 549B048 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 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 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 components.
From the given and the Pressure Drop tables, find:
Enter the Fan Performance table for 549B048 vertical dis-
charge. At 1600 cfm and 230-v, the standard motor will
deliver 1.2 in. wg static pressure and 1.15 brake horse-
power (bhp). This will adequately handle job requirements.
NOTE:
Convert bhp to Watts using the formula below and
the motor efficiency found in the Indoor-Fan Motor Data
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 (instantaneous)
capacity less the effect of frost on the outdoor 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
50TFQ004-012
549B036-120