23
SELECTION PROCEDURE (WITH 50TFQ005 EXAMPLE)
A. Determine cooling and heating loads 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 db 95
_
F
Outdoor-Air Winter Design Temperature 0
°
F
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
1600 cfm
External Static Pressure
0.45 in. wg
Electrical Characteristics (V-Ph-Hz)
230-3-60
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
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
.45 in. wg
EconoMi$er IV
.07 in. wg
Electric heat
.09 in. wg
Total static pressure
.61 in. wg
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
3.413Btuh/Watts)
= 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.
50TFQ
Summary of Contents for APOLLO CONTROL 50HJQ004-016
Page 5: ...5 50HEQ003 006 ...
Page 19: ...19 BASE UNIT DIMENSIONS 50TFQ004 007 50 TFQ C06069 50TFQ ...
Page 20: ...20 BASE UNIT DIMENSIONS 50TFQ008 012 C06070 50TFQ ...
Page 76: ...76 BASE UNIT DIMENSIONS 50HJQ004 007 C06006 50HJQ ...
Page 77: ...77 BASE UNIT DIMENSIONS 50HJQ008 012 C06081 50HJQ ...
Page 132: ...132 TYPICAL WIRING SCHEMATICS 50HJQ 50HJQ016 460V SHOWN C06092 50HJQ ...
Page 144: ...144 BASE UNIT DIMENSIONS 50HEQ003 006 C06058 50HEQ ...