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SELECTION PROCEDURE (With 559F180 Example)
I Determine cooling and heating requirements at design
conditions.
Given:
Cooling Capacity Required (TC) .........177,000 Btuh
Sensible Heat Capacity (SHC) ..........127,000 Btuh
Condenser Entering-Air
Temperature..................95FEdb/78 F Ewb
Indoor-Air Temperature ...........80FEdb/67 F Ewb
Evaporator Air Quantity ..................6,000 cfm
External Static Pressure .................0.66 in. wg
Heating Capacity ......................76,000 Btuh
Power Supply (V-Ph-Hz) ..................230-3-60
Edb — Entering dry bulb
Ewb — Entering wet bulb
II Select unit based on required cooling capacity.
Enter Gross Cooling Capacities table on page 22 for
559F180 at condenser entering dry bulb temperature
95 F, air entering evaporator at 6000 cfm, 80 F edb and
67 F ewb. The 559F180 unit will provide a total cooling ca-
pacity of 188,000 Btuh, a sensible heating capacity of
136,000 Btuh and a total unit kW of 17.8 kW. For
evaporator-air temperatures other than 80 F edb, calculate
sensible heat capacity correction, as required, using the
formula in the notes following the Cooling Capacities
tables.
NOTE: Unit ratings are gross capacities and do not include
the effect of evaporator-fan motor heat. To calculate net
capacities, see Step V.
III Select electric heat.
Heating load required is 76,000 Btuh.
76,000 Btuh
= kW of heat required
3,412 Btu/kW
= 22.3 kW
Enter the Electric Resistance Heater Data table for the
559F180 on page 64 at 230-3-60. The 34 kW electric
heater most closely satisfies the heat required.
IV Determine fan speed and motor horsepower require-
ments at design conditions.
Enter Accessory/FIOP Static Pressure Drop table on
page 47 at selected unit size and heater kW.
Find that at given air quantity (6000 cfm), pressure loss is
.09 in. wg.
Before entering the Air Delivery tables, calculate the total
static pressure required based on unit components. From
the given and the Accessory/FIOP Pressure table
(page 47) find:
External static pressure 0.66 in. wg
34 kW Heater static pressure 0.09 in. wg
Total static pressure 0.75 in. wg
Enter Air Delivery table for unit 559F180 on page 42. In-
terpolate to see that at 6000 cfm and 0.75 in. wg external
static pressure, the fan speed is 1184 rpm, the watts are
2631, and the bhp is 2.96. The standard motor with a field-
supplied drive is suitable.
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 from the Air Deliv-
ery table (calculated in Section IV above).
IFM watts = 2631
Determine net cooling capacity using the following
formula:
Net capacity = Gross capacity − IFM heat
Btuh
= 188,000 Btuh − 2631 watts (3.412 )
Watts
= 188,000 Btuh − 8977 Btuh
= 179,023 Btuh
Net sensible capacity = 136,000 Btuh − 8977 Btuh
= 127,023 Btuh
As a result of the given conditions, the 559F180 unit is the
correct selection.
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