212
SELECTION PROCEDURE (With 580F240 Example)I DETERMINE COOLING AND HEATING REQUIRE-
MENTS AT DESIGN CONDITIONS.
Given:
Required Cooling Capacity. . . . . . . . . . . . . . 230,000 Btuh
Sensible Heat Capacity (SHC). . . . . . . . . . . 170,000 Btuh
Required Heating Capacity. . . . . . . . . . . . . .215,000 Btuh
Condenser Entering Air Temp. . . . . . . . . . .95 F (Summer)
Evaporator Entering Air Temp. . . . . . . . . . . . . . . 80 F edb,
67 F ewb
Evaporator Air Quantity . . . . . . . . . . . . . . . . . . . 8,000 cfm
External Static Pressure. . . . . . . . . . . . . . . . . . . 0.6 in. wg
Electrical Characteristics (V-Ph-Hz) . . . . . . . . . . 460-3-60
Vertical discharge unit with optional EconoMi$er IV
required.
edb — Entering dry-bulb
ewb — Entering wet-bulb
II SELECT UNIT BASED ON REQUIRED COOLING
CAPACITY.
Enter Cooling Capacities table for 580F240 (page 183) at
condenser entering temperature 95 F, evaporator air
entering at 8,000 cfm and 67 F wb. The unit will provide a
total cooling capacity of 249,000 Btuh and a sensible heat-
ing capacity (SHC) of 188,000 Btuh. For air entering evap-
orator at temperatures other than 80 F edb, calculate SHC
correction as required.
NOTE: Unit ratings are gross capacities and do not
include the effect of evaporator-fan motor heat. To calcu-
late net capacities, see Step V.
III SELECT HEATING CAPACITY OF UNIT TO PROVIDE
DESIGN CONDITION REQUIREMENTS.
In the ARI Heating Capacities and Efficiencies table note
that the 580F240360 will provide an output capacity of
292,000 Btuh, which is adequate for the given application.
IV DETERMINE FAN SPEED AND POWER REQUIRE-
MENTS AT DESIGN CONDITIONS.
Before entering the Fan Performance tables, calculate the
total static pressure required based on unit components.
From the given and the Accessory/FIOP Static Pressure
table on page 191 find:
External static pressure 0.60 in. wg
EconoMi$er IV static pressure 0.10 in. wg
Total static pressure 0.70 in. wg
Enter the Fan Performance table 580F240360 at
8,000 cfm and 0.70 in. wg external static pressure. By
interpolation, find that the rpm is 1243 and the watts
are 5368.
V DETERMINE NET COOLING CAPACITY.
Cooling capacities are gross capacities and do not include
indoor (evaporator) fan motor (IFM) heat. Use the watts
input power to the motor calculated in Section IV above.
IFM Watts = 5368
Determine net cooling capacity using the following
formula:
Net capacity = Gross capacity – IFM heat
= 249,000 Btuh – 5368 Watts
= 249,000 Btuh – 18,316 Btuh
= 230,684 Btuh
Net sensible capacity = 188,000 Btuh – 18,316 Btuh
= 169,684 Btuh
The calculations show that a 580F240360 unit with the
standard motor and standard low-medium static drive is
the correct selection for the given conditions.
(3.412 Btuh )
Watts