Carrier 50TFQ008-012 Product Data page 12

Selection procedure
(with 50TFQ008 example) — English
I. Determine cooling and heating requirements
at design conditions.
Given:
Required Cooling Capacity (TC) . . . . .78,000 Btuh
Sensible Heat Capacity (SHC) . . . . . .51,000 Btuh
Required Heating Capacity . . . . . . . .72,000 Btuh
Outdoor Entering-Air Temperature . . . . . . . . .95 F
Outdoor-Air Winter Design Temperature . . . . .10 F
Indoor-Air Winter Design Temperature . . . . . .70 F
Indoor Entering-Air Temperature . . . . . . 80 F, edb
Indoor-Air Quantity . . . . . . . . . . . . . . . . 2600 cfm
External Static Pressure . . . . . . . . . . . . . 70 in. wg
Electrical Characteristics (V-Ph-Hz) . . . . . 380-3-50
II. Select unit based on required cooling capacity.
Enter Cooling Capacities table at outdoor entering
temperature of 95 F, indoor air entering at
2600 cfm and 67 F ewb. The 50TFQ008 unit will
provide a total cooling capacity of 83,400 Btuh and
a sensible heat capacity of 56,200 Btuh.
For indoor-air temperature other than 80 F edb, cal-
culate 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 cal-
culate net capacities, see Step V.
III. Select electric heat.
Enter the Instantaneous and Integrated Heating Rat-
ings table at 2600 cfm. At 70 F return indoor air
and 10 F air entering outdoor coil, the integrated
heating capacity is 32,200 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 72,000 Btuh.
Therefore, 39,800 Btuh (72,000 – 32,200) addi-
tional electric heat is required.
Determine additional electric heat capacity in kW.
39,800 Btuh
3413 Btuh/kW
Enter the Electric Heating Capacities table for
50TFQ008 at 400-3-50. The 11.5 kW heater at
400 v most closely satisfies the heating required. To
calculate kW at 380 v, use the Multiplication Factors
table.
11.5 kW x .902 = 10.4 kW
11.5 kW x .902 x 3413 = 35,403 Btuh
12
67 F, ewb
= 11.7 kW of heat required.
Total unit heating capacity is 67,603 Btuh (32,200
+ 35,403).
IV. Determine fan speed and power require-
ments at design conditions.
Before entering Fan Performance tables, calculate
the total static pressure required based on unit com-
ponents. From the given and the Pressure Drop
tables on page 28, find:
External static pressure
Economizer
Electric heat
Total static pressure
Enter the Fan Performance table for 50TFQ008 ver-
tical discharge. Find fan R/s and BkW at 0.75 in. wg
and 2600 cfm. Note that the fan speed is 747 rpm
and power required is 1.40 Bhp (interpolation not
shown). The standard 1.5 hp motor is satisfactory.
NOTE: Convert bhp to Watts using the formula
found in the note following the Indoor-Fan Motor
Efficiency table found on page 28.
For this example:
746 x Bhp
Watts =
Motor Efficiency
746 x 1.40
Watts =
.80
Watts = 1306
V. 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 = Gross capacity – IFM heat
= 83,400 Btuh – (1306 watts
x 3.413 Btuh/Watts)
= 83,400 Btuh – 4457 Btuh
= 78,943 Btuh
Net sensible capacity = 56,200 Btuh – 4457 Btuh
Determine net heating capacity as follows:
Net capacity = Gross capacity + IFM heat
= 67,603 Btuh + 4457 Btuh
= 72,060 Btuh
.70 in. wg
.02 in. wg
.03 in. wg
.75 in. wg
= 51,743 Btuh