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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
Factors Used In Unit Cooling
Selection:
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 |
