Evaporative Cooling
Energy Efficient Buildings
Evaporative Cooling
Examples of evaporative cooling:
Boiling water
Evaporation from body
Evaporation from lakes and oceans
Roof ponds
Roof misting
Swimming pools
Cooling towers
Focus today is cooling air for space conditioning.
Residential size evaporative coolers look like:
[pic]
Air: gives up sensible heat to evaporate water, i.e. cools
picks up sensible heat, i.e. humidity
Overall: process is adiabatic
Common name: "swamp cooler"
If Arizona @ T = 90 F and RH = 25%, evaporative cooling reduces temperature at constant enthalpy. Minimum air temperature occurs when air is 100% saturated. On psychrometric chart, process looks like:
[pic]
Define effectiveness:
[pic]
Most coolers about 80% effective. In Arizona example:
[pic]
Also, apparent why evaporative coolers used only in dry climates. Mississippi example: Twb1 = T2min = 82 F and RH = 100%, too hot and humid for "cooling". Use design wet-bulb temperature as guide to climate.
[pic]
Sizing Stand-Alone Evaporative Coolers
[pic]
1) Solve for Tsup at design conditions.
[pic]
If Tsup < Tset, then evaporative cooler can provide cooling.
2) Determine msup from energy balance on the house. Consider sensible heat load only since evaporative cooler is incapable of removing moisture.
[pic]
Evaporative-Cooling Assisted Systems
Unfortunately, cooling capacity declines when cooling loads increase.
[pic]
Thus, direct evaporative cooling often combined with mechanical or indirect evaporative cooling system.
Example
[pic]
Problem Statement:
Can direct/indirect evaporative cooling system in Boulder Colorado maintain indoor air conditions of T4 = 72 F at RH4 = 60% or less with 100% outside air, when outside air is at design conditions of T0 = 91 F, Twb,0 = 59 F? If so, find the volume flow rate of supply air and water consumption of the evaporative cooler.
Direct / Indirect Evaporative Cooling System:
[pic]
[pic]
Solution:
Tc is function of Twb and Th. At given conditions, Tc = 73 F. Assume T2 = 78 F.
For i3 = 24.7 Btu/lba and i4 = 26.8 Btu/lba, RH4 = 43%.
Energy balance on space gives:
ma(i3 - i4) + Qspace = 0
ma = Qspace / (i4 - i3) = 60,000 Btu/hr / (26.8 - 24.7) Btu/lba = 28,571 lba/hr
Va = ma v0 = 28,571 lba/hr x 16.8 ft3/lba / 60 min/hr = 8,000 cfm
But could cool air further and reduce Va. If ε = 80%, then
[pic]
At T3' = 58.8 F, i3' = 24.7 Btu/lba, i4' = 29.6 Btu/lba and RH4' = 55%.
Energy balance on space gives:
ma = Qspace / (i4' - i3') = 60,000 Btu/hr / (29.6 - 24.7) Btu/lba = 12,245 lba/hr
Va = ma v0 = 12,245 lba/hr x 16.8 ft3/lba / 60 min/hr = 3,429 cfm
Mass balance on water through evaporative cooler gives:
mw = ma (W3' - W2) = 12,245 lba/hr (0.0096 - 0.0056) lbw/lba
mw = 49.0 lb/hr = 5.9 gallons/hr
Assuming the total water + sewer charge for water is $5.00 /ccf, the cost of water would be about:
Cw = 5.9 gallons/hr $5.00 /ccf / (7.48 gal/ft3 100 ft3/cf) = $0.039 /hr
For comparison, the cost of electricity for a 0.6 kW/ton chiller if electricity cost $0.10 /kWh would be about:
Ce = 60,000 Btu/hr 0.6 kW/ton / (12,000 Btu/ton-hr) x $0.10 /kWh = $0.300 /hr
In this example, the cost of electrical cooling is about 7 times greater than the cost of evaporative cooling.
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