Selection Procedure

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

1

Summer design conditions — 95 DB/76 WB, 95 F entering air to condenser.

2

Summer room design conditions — 76 DB/66 WB.

3

Total peak cooling load — 321 MBh (27.75 tons).

4

Total peak supply cfm — 12,000 cfm.

5

External static pressure — 1.0 inches.

6

Return air temperatures — 80 DB/66 WB.

7

Return air cfm — 4250 cfm.

8

Outside air ventilation cfm and load — 1200 cfm and 18.23 MBh (1.52 tons).

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: 27.75 tons + 1.52 tons = 29.27 required unit capacity. From Table 18-2, 30-ton unit capacity at 80 DB/67 WB, 95 F entering the condenser and 12,000 total peak supply cfm, is 30.0 tons. Thus, a nominal 30-ton unit is selected.

Step 2 — Having selected a nominal 30-ton 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       1.20inches

Heat exchanger

(Table 24-1)                            .14inches

High efficiency filter 2"

(Table24-1)                             .09inches

Economizer

(Table24-1)                             .076inches

Unit total static pressure    1.50 inches

Using total cfm of 12,000 and total static pressure of 1.50 inches, enter Table 22-1. Table 22-1 shows 7.27 bhp with 652 rpm.

Step 3 — Determine evaporator coil entering air conditions. Mixed air dry bulb temperature determination.

Using the minimum percent of OA (1,200 cfm ÷ 12,000 cfm = 10 percent), determine the mixture dry bulb to the evaporator. RADB + %OA (OADB -RADB) = 80 + (0.10) (95 - 80) = 80 + 1.5 = 81.5F

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 7.27 bhp = 20.6 MBh.

Capacity = 321 + 18.23 + 20.6 = 359.8 MBh (30 tons)

Step 5 — Determine unit capacity: From Table 18-2 unit capacity at 81.5 DB. 67 WB entering the evaporator, 12000 supply air cfm, 95 F entering the condenser is 361 MBh (30.1 tons) 279 sensible MBh.

Step 6 — Determine leaving air temperature:

Unit sensible heat capacity, corrected for supply air fan motor heat 279 - 20.6 = 258.4 MBh.

Supply air dry bulb temperature difference = 258.4 MBh ÷ (1.085 x 12,000 cfm) = 19.8 F.

Supply air dry bulb: 81.5 - 19.8 = 61.7.

Unit enthalpy difference = 361 ÷ (4.5 x 12,000) = 6.7

Btu/lb leaving enthalpy = h (ent WB) = 31.62

Leaving enthalpy = 31.62 Btu/lb -6.7 Btu/lb = 24.9 Btu/lb.

From Table 17-1, the leaving air wet bulb temperature corresponding to an enthalpy of 24.9 Btu/lb = 57.5. Leaving air temperatures = 61.7 DB/57.5 WB