Trane RT-PRC010-EN manual Selection Procedure

Selection Procedure

This section outlines a step-by-step procedure that may be used to select a Trane single-zone air conditioner.The sample selection is based on the following conditions:

•Summer outdoor design conditions — 95 DB/76 WB ambient temperature

•Summer room design conditions —78 DB/65 WB

•Total cooling load — 430 MBh (35.8 tons)

•Sensible cooling load — 345 MBh (28.8 tons)

•Outdoor air ventilation load — 66.9 MBh

•Return air temperature — 80 DB/65 WB

Winter Design:

•Winter outdoor design conditions —0 F

•Return air temperature — 70 F

•Total heating load — 475 MBh

•Winter outdoor air ventilation load — 133 MBh

Air Delivery Data:

•Supply fan cfm — 17,500 cfm

•External static pressure — 1.2 in wg

•Minimum outdoor air ventilation — 1,750 cfm

•Exhaust fan cfm — 12,000 cfm

•Return air duct negative static pressure

— 0.65 in wg

Electrical Characteristics:

•Voltage/cycle/phase — 460/60/3 Unit Accessories:

•Gas fired heat exchanger — high heat module

•Throwaway filters

•Economizer

•Modulating 100 percent exhaust/ return fan

COOLING CAPACITY SELECTION

Step 1 — Nominal Unit Size Selection

A summation of the peak cooling load and the outside air ventilation load shows: 430 MBh + 66.9 MBh = 496.9 MBh required unit capacity. From

Table PD-9, a 50 ton unit capacity with standard capacity evaporator coil at 80 DB/65 WB, 95 F outdoor air temperature and 17,500 total supply cfm is 551 MBh total and 422 MBh sensible. Thus, a nominal 50 ton unit with standard capacity evaporator coil

is selected.

Step 2 — Evaporator Coil Entering Conditions

Mixed air dry bulb temperature determination:

Using the minimum percent of OA (1,750 cfm ÷ 17,500 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

Approximate wet bulb mixture temperature:

RAWB + % OA (OAWB - RAWB) = 65 + (0.10) (76 - 65) = 65 + 1.1 = 66.1 F

Step 3 — Determine Supply Fan Motor Heat Gain

Having selected a nominal 50 ton unit, the supply fan bhp can be calculated.The supply fan motor heat gain must be considered in final determination of unit capacity.

Supply Air Fan

Determine unit total static pressure at design supply cfm:

External Static Pressure	1.2 inches
Evaporator Coil	0.25 inches
(Table PD-43)
Return Duct Negative	0.65 inches
Static Pressure
Heat Exchanger	0.31 inches
(Table PD-43)
Throwaway Filter	0.10 inches
(Table PD-43)

Economizer w/Exhaust Fan 0.12 inches

(Table PD-43)
Trane Roof Curb	0.13 inches
(Table PD-43)
UnitTotal Static Pressure	2.76 inches

Using total of 17,500 cfm and total

static pressure of 2.76 inches, enterTable PD-36.Table PD-36 shows 15.3 bhp with 924 rpm.

From Chart SP-1 supply fan motor heat gain = 46.0 MBh.

Step 4 — DetermineTotal Required Cooling Capacity

Required capacity =Total peak load + OA load + supply air fan motor heat.

Required capacity = 430 + 66.9 + 46.0 = 543 MBh (45.2 tons)

Step 5 — Determine Unit Capacity

FromTable PD-9, unit capacity at 81.5 DB/

66.1WB entering the evaporator, 17,500 supply air cfm, 95 F outdoor ambient, is 561 MBh (45.8 tons) with 426 MBh sensible.

Step 6 — Determine Leaving Air Temperature

Unit sensible heat capacity corrected for supply air fan motor heat = 426 MBh - 46 MBh = 380 MBh.

Supply air dry bulb temperature difference =

Sensible Btu	=
1.085 x Supply cfm

380 MBh ÷ (1.085 x 17,500 cfm) = 20.0 F

Supply air dry bulb = 81.5 DB - 20.0 = 61.5 F

Unit enthalpy difference =

Total Btu = 4.5 x Supply cfm

561 MBh ÷ (4.5 x 17,500 cfm) =

7.12 Btu/lb

Leaving enthalpy = h(ent WB) - h(diff). From Table 21-1 h(ent WB) = 30.9 Btu/lb

Leaving enthalpy = 30.9 Btu/lb - 7.12 Btu/lb = 23.78 Btu/lb

Supply air wet bulb = 55.9

Leaving air temperature = 61.5

DB/55.9 WB

16 RT-PRC010-EN