Bryant Selection Procedure with 549B048 Example, II Select Unit Based on Required Cooling

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-

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:

External static pressure	.45 in. wg
Durablade economizer	.05 in. wg
Electric heat	.09 in. wg
Total static pressure	.59 in. wg

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:

ature of 95 F, indoor air entering at 1600 cfm and 67 F

Watts =

746 x Bhp

549B036-120

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.

14,600 Btuh

= 4.3 kW of heat required.

3413 Btuh/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).

Motor Efficiency

746 x 1.15

Watts =

.75

Watts = 1144

VDETERMINE 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

=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

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.

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