Bryant 558F SELECTION PROCEDURE With 551B048 Example, V Determine Net Capacities, 551B036-150

SELECTION PROCEDURE (With 551B048 Example)

551B036-150

I DETERMINE COOLING AND HEATING REQUIRE- MENTS AT DESIGN CONDITIONS.

Given:

Required Cooling Capacity (TC). . . . . . . . . . . 44,000 Btuh Sensible Heat Capacity (SHC) . . . . . . . . . . . . 32,000 Btuh Required Heating Capacity. . . . . . . . . . . . . . . 50,000 Btuh Condenser Entering-Air Temperature . . . . . . . . . . . . .95 F Evaporator Entering-Air Temperature . . . . . . . . . 82 F edb,

67 F ewb Evaporator Air Quantity . . . . . . . . . . . . . . . . . . . 1600 cfm External Static Pressure (ESP) . . . . . . . . . . . . 0.75 in. wg Electrical Characteristics (V-Ph-Hz) . . . . . . . . . . 230-3-60

IISELECT UNIT BASED ON REQUIRED COOLING CAPACITY.

Enter the Cooling Capacities table on page 79 at con- denser entering temperature of 95 F, evaporator-air enter- ing at 1600 cfm and 80 F db and 67 F wb. The 551B048 unit will provide cooling capacity of 48,900 Btuh and a sensible heat capacity of 34,700 Btuh. For evaporator-air temperature other than 80 F edb, calculate sensible heat capacity correction, as required, using the formula found in the notes following the Cooling Capacities tables.

For this example:

Correction factor = 1.1 x (1 – .21) x (82 – 80)

= 1.738

Multiply the correction factor by 1600 cfm (a total of 2781). From the Gross Cooling Capacities tables find that the sensible heat capacity at 80 F is 34.7 MBtuh (equivalent to 34,700 Btuh). Add 34,700 and 2781 to get the corrected sensible heat capacity of 37.48 MBtuh (37,481 Btuh).

NOTE: Unit ratings are gross capacities and do not include the effect of evaporator-fan motor heat. To calcu- late net capacities, see Step V.

IIISELECT ELECTRIC HEAT.

Heating load required is 50,000 Btuh.

50,000

= 14.6 kW of heat required

3.413 Btuh/kW

Enter the electric Heating Capacities table on page 97 for 551B048 at 208/230 (3 phase). The accessory 16.0 kW heater at 240 v most closely satisfies the heating required. To calculate kW at 230 v, use the Multiplication Factors table on page 97.

16.0 x .92 = 14.7 kW

16.0 x .92 x 3413 = 50,240 Btuh gross capacity

IV DETERMINE FAN SPEED AND POWER REQUIRE- MENTS AT DESIGN CONDITIONS.

Before entering the Fan Performance tables, calculate the total static pressure required based on unit components. From the given and the Pressure Drop tables on page 94, find:

External static pressure	.75 in. wg
Electric heat	.09 in. wg
Total static pressure	.84 in. wg

Enter the Fan Performance table for 551B048 vertical dis- charge unit on page 85. At 1600 cfm, the standard motor will deliver 1.20 in. wg static pressure and 1.15 Bhp. This will adequately handle the job requirements.

VDETERMINE NET CAPACITIES.

Capacities are gross and do not include the effect of indoor (evaporator) fan motor (IFM) heat. To determine input power to the motor, convert bhp to watts using the Evaporator-Fan Motor Efficiency table on page 94.

Bhp x 746 Watts/Bhp

IFM Watts =

Motor Efficiency

1.15 x 746

=

.75

=1144

Determine net cooling capacity as follows:

Net capacity = gross capacity – IFM heat

= 48,900 Btuh – (1144 Watts

x 3.413 WattsBtuh )

= 48,900 Btuh – 3904 Btuh

=44,996 Btuh

Net sensible capacity = 37,481 Btuh – 3904 Btuh

=33,577 Btuh

Determine net heating capacity as follows:

Net capacity = gross capacity + IFM heat

=50,240 + 3904

=54,144 Btuh

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