Selection Procedure
This section outlines a
Summer Design:
•Summer outdoor design conditions - 95 DB/76 WB ambient temperature
•Summer room design conditions
•Total cooling load - 980 MBH (81.6 tons)
•Sensible cooling load - 735 MBH (61.25 tons)
•Outdoor air ventilation load - 154.0 MBH (12.8 tons)
•Return air temperature - 78 DB/65 WB
Winter Design:
•Winter outdoor design condition is 0°F.
•Total return air temperature is 70°F.
•Total heating load - 720 MBH
•Winter outdoor air ventilation load - 288.6 MBH
•Total winter heating load - 1008.6 MBH
Air Delivery Data:
• | Supply fan CFM - 36000 CFM |
• | Supply duct static pressure - 1.86 |
| 2.2 in wg |
•Economizer
•Modulating 100 percent exhaust
Cooling Capacity Selection:
Step 1 - Coil and Fan
Selection
A summation of the peak cooling load and the outside air ventilation load shows: 980 MBH + 154.0 MBH = 1134.0 MBH required unit capacity.
The supply fan air flow requirement is 36,000 cfm.
From Table 10, p. 39, a 4 row W coil with 144 fpf (fins per foot) and no turbulators at 80 DB/67 WB and 36000 supply air cfm has a total cooling capacity of 1336 MBH and sensible cooling capacity of 969 MBH. With chilled water coil capacity data at 80 DB/67 WB only, TOPSS is required for an accurate selection at other conditions. TOPSS is also required to select the correct water control valve for proper flow control, in this case a 2 ½ "or 3" valve.
Table 3, p. 34 - General Data shows that air handler "C" can provide 36000 total supply CFM.
Thus air handler "C" with a 4 row 144 fpf W coil having no turbulators at 45°F entering water and a 10°F rise with a 2 ½" valve is selected. The coil water flow rate is 266 GPM and water side pressure drop is 13.7 ft of water.
Step 3 - Determine Supply Fan Motor Heat Gain
Having selected air handler casing "C" with a 4 row 144 fpf W coil and no turbulators, 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:
Supply Duct Static Pressure | 2.2" |
Chilled Water Coil Table 33, p. 72 | 0.64" |
Return Duct Negative Static | 0.30" |
Pressure |
|
Heat Exchanger Table 34, p. 72 | 0.03" |
Throwaway Filter Table 35, p. 73 | 0.26" |
Return Damper Table 34, p. 72 | 0.34" |
Economizer Damper(i) Table 34, | 0.57" |
p. 72 |
|
Unit Total Static Pressure | 4.0" |
(i)Add either the economizer damper value or return damper value, depending on which static pressure is greater. (Do not use both.)
Using total of 36000 CFM and total static pressure of 4.0 inches, enter Table 17, p. 48. The table shows 40.4
BHP with 1097 rpm required for the 36" supply fan.
From Figure 17, p. 30 supply fan motor heat gain = 109.0 MBH, or
109.0MBH x 1000 ÷ ( 36000 CFM x
1.085 ) = 2.8°F supply fan motor heat
• Minimum outdoor air ventilation | |
| - 3600 CFM |
• | Exhaust fan CFM - 36000 CFM |
• | Return air duct negative static |
| pressure - 0.3 in wg |
Electrical Characteristics:
•Voltage/cycle/phase - 460/60/3
Unit Accessories:
•Gas fired heat exchanger - High Heat
•Downflow supply and upflow return
•High Efficiency Throwaway filters
Step 2 - Cooling Coil Entering Conditions
Mixed air dry bulb temperature determination:
Using the minimum percent of OA (3600 CFM ÷ 36000 CFM = 10 percent), determine the mixture dry bulb to the cooling coil.
RADB + % OA (OADB - RADB) = 78 + (0.10) (95 - 78) = 78 + 1.5 = 79.5°F
Approximate wet bulb mixture temperature:
RAWB + % OA (OAWB - RAWB) = 65 + (0.10) (76 - 65) = 65 + 1.1 = 66.1°F
Step 4 - Determine Total Required Cooling Capacity
Required capacity = Total peak load + OA load + supply air fan motor heat
Required capacity = 980.0 + 154.0 + 109.0 = 1243.0 MBH
Step 5 - Determine Unit Capacity
The coil entering air conditions of
79.5DB/66.1 WB are close to the capacity data table at 80 DB/67 WB used for the original selection. The unit capacity with the 4 row 144 fpf W coil with no turbulators at 45°F entering water a 10°F rise, 36000 cfm supply air flow and 10% outside air
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