15
SelectionProcedure
Selection of Trane commercial air
conditioners is divided into five basic
areas:
1
Cooling capacity
2
Heating capacity
3
Air delivery
4
Unit electrical requirements
5
Unit designation
Factors Used In Unit Cooling Selection:
1
Summer design conditions  95 DB/
76 WB (35/24.4°C), 95°F (35°C) entering
air to condenser.
2
Summer room design conditions 
76 DB/66 WB (24.4/18.9°C).
3
Total peak cooling load  270 MBh
(79 kW) (22.5 tons).
4
Total peak supply cfm  10,000 cfm
(4720 l/s).
5
External static pressure  1.0 inches wc
(249 Pa).
6
Return air temperatures  80 DB/66°F
WB (26.7/18.9°C).
7
Return air cfm  3540 cfm (1671 l/s).
8
Outside air ventilation cfm and load 
1000 cfm and 15.19 MBh (1.27 tons or
4.45 kW) 472 l/s.
9
Unit accessories include:
a
Aluminized heat exchanger  high heat
module.
b
2 Hi-efficiency throwaway filters.
c
Exhaust fan.
d
Economizer cycle.
Step 1  A summation of the peak
cooling load and the outside air
ventilation load shows: 22.5 tons + 1.27
tons = 23.77 (79 kW + 4.45 kW = 83.45)
required unit capacity. From Table 19-1,
25 ton (89 kW) unit capacity at 80 DB/67
WB (27/19°C), 95°F entering the
condenser and 10,000 total peak supply
cfm (4720 l/s), is YC/TC/TE*305.
Step 2  Having selected the correct
unit, the supply fan and exhaust fan
motor bhp must be determined.
Supply Air Fan:
Determine unit static pressure at
design supply cfm:
External static pressure 1.25 inches
(310 Pa)
Heat exchanger
(Table 28-1) .12 inches
(30 Pa)
High efficiency filter 2 (25 mm)
(Table 28-1) .07 inches
(17 Pa)
Economizer
(Table 28-1) .07 inches
(17 Pa)
Unit total static pressure 1.50 inches
(374 Pa)
Using total cfm of 10,000 (4720 l/s) and
total static pressure of 1.50 inches
(38 mm), enter Table 24-1. Table 24-1
shows 5.35 bhp (4 kW) with 616 rpm.
Step 3  Determine evaporator coil
entering air conditions. Mixed air dry
bulb temperature determination.
Using the minimum percent of OA
(1,000 cfm ÷ 10,000 cfm = 10 percent),
determine the mixture dry bulb to the
evaporator. RADB + % OA
(OADB - RADB) = 80 + (0.10) (95 - 80) =
80 + 1.5 = 81.5°F [26.7 + 1.5 = 28°C).
Approximate wet bulb mixture
temperature:
RAWB + OA (OAWB - RAWB) =
66 + (0.10) (76-66) = 68 + 1 = 67°F.
A psychrometric chart can be used to
more accurately determine the mixture
temperature to the evaporator coil.
Step 4  Determine total required unit
cooling capacity:
Required capacity = total peak load +
O.A. load + supply air fan motor heat.
From Figure 16-1, the supply air fan
motor heat for 5.85 bhp = 15 MBh.
Capacity = 270 + 15 + 15 =
300 MBh (89 kW)
Step 5  Determine unit capacity:
From Table 19-2 unit capacity at 81.5
DB/67 WB entering the evaporator,
10,000 supply air cfm, 95°F (35°C)
entering the condenser about 305.6
MBh (89.5 kW) with 241 MBh (70.6 kW)
sensible.
Step 6  Determine leaving air
temperature:
Unit sensible heat capacity, corrected
for supply air fan motor heat 241 - 15 =
226 MBh (66.2 kW).
Supply air dry bulb temperature
difference = 226 MBh ÷ (1.085 x
10,000 cfm) = 20.8°F (-6.2°C)
Supply air dry bulb: 81.5-20.8 = 60.7
(15.9°C)
Unit enthalpy difference = 305.6 ÷
(4.5 x 10,000) = 6.79
Btu/lb leaving enthalpy = h (ent WB)
= 31.62
Leaving enthalpy = 31.62 Btu/lb -
6.79 Btu/lb = 24.83 Btu/lb.
From Table 17-1, the leaving air wet
bulb temperature corresponding to an
enthalpy of 24.8 Btu/lb = 57.5.
Leaving air temperatures = 61.7 DB/
57.5 WB (15.9/13.9°C).