Selection Procedure (With 50Tfq005 Example); Determine Cooling And Heating Loads At Design Conditions; Select Unit Based On Required Cooling Capacity; Determine Net Capacities - Carrier 50HJQ004-016 Product Data

SELECTION PROCEDURE (WITH 50TFQ005 EXAMPLE)

A. Determine cooling and heating loads at design conditions.
Given:
Required Cooling Capacity (TC)
Sensible Heat Capacity (SHC)
Required Heating Capacity
Outdoor Entering-Air Temperature db 95_F
Outdoor-Air Winter Design Temperature
Indoor-Air Winter Design Temperature 70_F
Indoor-Air Temperature
80_F edb (entering air, dry bulb)
67_F ewb (entering air, wet bulb)
Indoor-Air Quantity
External Static Pressure
Electrical Characteristics (V-Ph-Hz)
B. Select unit based on required cooling capacity.
Enter Cooling Capacities table at outdoor entering
temperature of 95_F, indoor air entering at 1600 cfm and
67_F ewb. The 50TFQ005 unit will provide a total cooling
capacity of 49,900 Btuh and a sensible heat capacity of
33,900 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 capacities, see
Step E.
C. 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 18,000 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,
17,000 Btuh (35,000 – 18,000) additional electric heat is
required.
Determine additional electric heat capacity in kW.
17,000 Btuh = 5.0 kW of heat required.
3413 Btuh/kW
38,000 Btuh
24,000 Btuh
35,000 Btuh
0°F
1600 cfm
0.45 in. wg
230-3-60
23
Enter the Electric Heating Capacities table for 50TFQ005 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 38,410 Btuh (18,000 + 20,410).
D. Determine fan speed and power requirements 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
EconoMi$er IV
Electric heat
Total static pressure
Enter the Fan Performance table for 50TFQ005 vertical
discharge. At 1600 cfm and 230-v high speed, the standard
motor will deliver 0.76 in. wg static pressure, 723 watts, and
0.64 brake horsepower (bhp). This will adequately handle job
requirements.
E. Determine 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
= 49,900 Btuh – (723 Watts x
= 49,900 Btuh – 2468 Btuh
= 47,432
Net sensible capacity = 33,900 Btuh – 2468 Btuh
= 31,432 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
38,410 Btuh, the total heating capacity determined in Step C.
.45 in. wg
.07 in. wg
.09 in. wg
.61 in. wg
3.413Btuh/Watts)