Trane TWE050A Selection Procedure, Cooling Capacity, Heating Capacity, Air Delivery Selection

Selection Procedure

Cooling Capacity

Step 1 — Calculate the building’s total and sensible cooling loads at design conditions.

Step 2 — Size the equipment usingTable PD-1. Match the cooling loads at design conditions.

Example:The following are the building cooling requirements

a

Electrical Characteristics: 380-415/50/3 b

Summer Design Conditions: Entering

Evaporator Coil: 80 DB/67WB (27 DB/19WB°C)

Outdoor Ambient: 95°F (35° C) c

Total Cooling Load: 75 MBh (22kW) d

Sensible Cooling Load: 53 MBh (15.5 kW)

e

Airflow: 2500 cfm (4247 m3/h)

External Static Pressure: 0.77 in. wg (193 Pa)

Table PD-1 shows that aTWA075A matched with aTWE075A has a gross cooling capacity of 85.1 MBh (24.9 kW) and 60.8 MBh (17.8 kW) sensible capacity at 95 DB (35°C) ambient and 2500 cfm (4247 m3/h) and 80 DB/67WB (27 DB/19 WB) air entering the evaporator.

To find the net cooling capacities, fan motor heat must be subtracted. Determine the total unit static pressure:

External Static	0.77 in (193 Pa)
Standard Filter	0.10 in (25 Pa)
Supplementary Electric Heat
	0.23 in (57 Pa)
Total Static Pressure	1.10 in (275 Pa)

Note:The Evaporator Fan Performance Table has included the effect of a 1 in. (250 Pa) filter already.Therefore, the actualTotal Static Pressure is 1.10 - 0.10 = 1.00 in. (275 - 25 = 250 Pa).

With 2500 cfm (4247 m3/h) and 1.00 inches (250 Pa),Table PD-15 shows a 1.07 bhp (0.8 kW).

Note:The formula below the table can be used to calculate Fan Motor Heat, Constant x Motor Power =

Fan Motor Heat
3.5 x bhp	=	MBh
3.5 x 1.07	=	3.75 MBh

1.375 x (kW) = kW

1.375 x 0.8 = 1.1 kW

NetTotal Cooling Capacity

= 85.1 MBh – 3.75 = 81.35 MBh

=24.9 kW - 1.1 = 23.8 kW Net Sensible Cooling Capacity

= 60.8 MBh – 3.75 = 57.05 MBh = 17.8 MBh - 1.1 = 16.7 kW

Heating Capacity

Step 1 — Calculate the building heating load using theTrane calculation form or any other standard accepted method.

Step 2 — Size the equipment usingTable PD-9 to match the heating loads at design conditions.The following are building heating requirements:

a

Total Heating Load: 110 MBh (32.2 kW) b

Outdoor Ambient (Winter): 17°F (-8.3°C) DB

c

Indoor ReturnTemperature: 70°F (21.1°C) DB

d

Airflow: 2500 cfm (4247 m3/h)

Table PD-9 indicates the mechanical heating portion of the heat pump will provide 49.1 MBh (14.4 kW) for the winter design conditions.

Step 3 — Because 49.1 MBh (14.4 kW) is less than the building’s required heating capacity, a supplementary heater must be selected. 110 - 49.1 = 60.9 MBh (32.2 -

14.4= 17.8 kW) minimum heater capacity.

FromTable PD-25, the 24.22 kW heater has a capacity of 82,670 Btu. FromTable ED-5, the 24.22 kW heater at 400V indicates the heater model number is BAYHTRL435A.This heater will be adequate to cover the residual heat capacity needed for the application.

Air Delivery Selection

External static pressure drop through the air distribution system has been calculated to be 0.77 inches (192.5 Pa) of water gauge. FromTable PD-24 static pressure drop through the electric heater is 0.12 inches (30 Pa) of water (0.77 + 0.12

=.89 in.) (192.5 + 30 = 222.5 Pa). Enter Table PD-15 forTWE090A4 at 2500 cfm (4247 m3/h) and .90 in. (225 Pa) static pressure.The standard motor at 790 rpm will give the desired airflow.

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