Greenhouse Ventilation and Cooling
J. Raymond Kessler, Jr.
Fan and pad cooling systems must be properly designed to obtain maximum evaporative efficiency during periods of intense hot weather. This requires a non-turbulent potential flow through the greenhouse to avoid mixing the cooled air at plant level with warmer air in the top of the greenhouse.
Exhaust Fans The fan and pad cooling system consists of large volume exhaust fans and a correctly sized continuous wet pad system, both properly located with respect to the greenhouse layout. The fans exhaust the air and develop a slight vacuum or negative pressure throughout the entire house because it is substantially air tight.
This slight vacuum draws air in through the cooling pad system and causes cooled air to move smoothly through the growing region of the crops absorbing heat. The warmed air is then expelled by the exhaust fans in the opposite wall. This system produces a potential type air flow that moves a cool layer of non-turbulent air through the plants for best cooling efficiency.
The pad system requires a sufficient flow area to accommodate the large air volume needed to remove the intense solar heat. It is composed of wettable, fibrous material, in the form of self supporting special fluted cellulose cooling cells. It is kept wet by water recirculating through it. The pad system also distributes the air uniformly and by virtue of its resistance, restricts the turbulence from the outdoor air, delivering a smooth, laminar flow of cool air into the house.
Rate of Air Exchange
Because the solar heat comes into a greenhouse on a ground surface area basis, the air flow rates for ventilation are always determined on a cubic feet per minute (cfm) for each square foot of floor area. The basic air flow rate of 8 cfm per square foot has been determined to be sufficient for moderately shaded greenhouses having a maximum interior light intensity of about 5000 foot candles. However, in warm climates and houses with tall gutters (>12 feet), 11-14 cfm per square foot is advisable. This basic air flow rate is then adjusted for elevations over 1,000 feet above sea level, the expected interior light intensity, the allowable greenhouse temperature increase, and the distance from the pad to the fan.
Exchange Rate Adjustments
Elevation: The heat removal capacity of air depends on its weight and not on its volume. Because air is less dense at higher altitudes the elevation of the greenhouse must be considered in design calculations. At higher elevations a greater volume of air is needed to provide the equivalent weight of air required at normal elevations. Corrections for elevations (FElev) greater than 1000 feet are in table 1.
Light Intensity: The interior light intensity, which depends on the location of the greenhouse and the amount of shading, determines the amount of heat input into the greenhouse. The interior light intensity, measured in foot candles (FC), corrections (FLight) are in table 2.
Temperature Increase: The greenhouse temperature increase from pad to fan is a design factor. It is inversely proportional to the air flow rate and can be adjusted to any value desired. Usually a 7°F rise in temperature is tolerated. If it is important to hold a more constant temperature across the greenhouse, it will be necessary to raise the velocity of air movement through the greenhouse. Corrections for pad-to-fan temperature increase (FTemp) are in table 3.
This completes the adjustment and design factors necessary for a heat balance. Combining all these factors determines the house adjustment factor (FHouse) as follows:
Pad-to-fan Distance: The pad and fans should be located on opposite walls. The preferred pad-to-fan distance ranges from 100 to 200 feet. This distance is an important design consideration. Distances greater than 200 feet can result in unacceptable temperature increases across the house. For very long houses (>200 feet), consider installing pads in each end and roof-mounted fans at the midpoint. For short pad-to-fan distances (<100 feet) the cross sectional air flow velocity within the house becomes too low and the house feels clammy or stuffy even though the air flow rate is technically correct. This can be compensated for by increasing the size of the fans which increases the cost of the system. The correction for several distances less than 100 feet (FVel) is in table 4.
Total Air Flow Required
The correct factor FVel is ignored for pad-to-fan distances of 100 feet or greater. For pad-to-fan distances less than 100 feet, calculate BOTH FHouse and FVel and use the LARGER of the two to complete the total air flow requirement where,
TOTAL CFM = L × W × 8 cfm/ft2 × FHouse (unless FVel is > FHouse)
Next select the size and number of fans that collectively equal or exceed the rate of air movement required and should be rate to do so at a static water pressure of 0.1 inches. If slant-wall fans are used, the fans should be rate to do so at a static water pressure of 0.5 inches. The static pressure rating takes into account the resistance encountered by drawing air through the pad and the fan itself. Fans should not be spaced more than 25 feet apart and should be evenly spaced.
The size of the pad system is determined by adding the total cfm for each exhaust fan selected and dividing the cfm that can be moved through one square foot of pad per minute. Cross-fluted cellulose pads, 4 inches thick can move 250 cfm/ft2 and cross-fluted cellulose pads, 6 inches thick can move 400 cfm/ft2 (6-inch pad flutes are designed differently than 4-inch pads). This area is then divided by the length of the wall on which the pads will be mounted to determine the actual pad height (not including hardware).
Water must be delivered to the top of a 4-inch thick pad at the rate of 0.5 gpm per linear foot of pad. For pad lengths of 30 to 50 feet, a 1¼-inch water-distribution pipe is required, while for lengths of 50 to 60 feet, a 1½-inch pipe is needed. Sixty feet is the longest recommended pipe length. A 120-foot pad length could be serviced from a water supply at the midpoint supplying two 60-foot distribution pipes. At every 3 inches, 1/8-inch holes should be made in the pipe.
The flow rate for a 6-inch thick pad is 0. 75 gpm per linear foot of pad. A 1¼-inch distribution pipe is used for pads 30 feet and shorter, while a 1½-inch pipe is used for 30- to 50-foot pad lengths. The longest pipe length recommended is 50 ft. Again, 1/8-inch holes are spaced 3 inches apart in these distribution pipes.
Sump Tank Volume
The sump tank volume should be at least 0.75 gal/ft2 of 4-inch thick pad and 1.0 gal/ft2 of 6-inch thick pad. These sump volumes are designed to operate at half the depth of the tank and will provide room to accommodate water returning from the pad when the system is turned off.
Example: Design a pad-and-fan system for a detached, glass-covered greenhouse that is 100 feet long and 50 feet wide at an elevation of 3000 feet above sea level. The pads and fans will be mounted on opposite walls over the 100 foot distance (50 foot width). The greenhouse has a moderate covering of shade cloth with a maximum interior light intensity of 5,000 FC. A 7°F rise in temperature can be tolerated. Assume a design air flow rate of 8 cfm/ft2 and 4-inch cross-fluted cellulose pads.
1. Determine FElev, FLight, and FTemp from tables 1, 2, and 3, respectively. Calculate FHouse.
2. Look up the FVel for a pad-to-fan distance of 100 feet in table 4. Because the pad-to-fan distance is 100 feet or greater, FVel = 1.0, this factor can be ignored.
3. Calculate the TOTAL CFM required for the greenhouse.
TOTAL CFM = 100 × 50 × 8 cfm/ft2 × 1.12 = 44,800 cfm
4. Determine the number of fans required. Because fans should not be over 25 feet apart and will be mounted on a 50 foot wall: 50 feet / 25 feet = 2 fans.
5. Determine the minimum size for each fan by dividing the TOTAL CFM by the number of fans found in step 4: 44,800 cfm / 2 fans = 22,400 cfm per fan.
6. Next determine the pad area. Divide the capacity of all fans by the capacity of the 4-inch pads per square foot: (22,400 cfm × 2 fans) / 250 cfm = 179.2 square feet.
7. Divide this value by the length of the wall to get the required pad height: 179.2 / 50 = 3.6 feet
8. The pump capacity for a 4-inch thick pad is 0.5 gpm per linear foot of pad and 1¼-inch water-distribution pipe. The pad length is 50 feet: pump capacity = 0.5 gpm × 50 ft = 25 gpm
9. The sump tank is 0.75 gal/ft2 of 4-inch thick pad. Multiply this value time the total square feet of pad area found in #6: 0.75 gal/ft2 × 179.9 ft2 = 134.4 gallons
Table 1. Factor used to correct rate of air removal for elevation above sea level.
Table 2. Factor used to correct rate of air removal for interior light level.
Table 3. Factor used to correct rate of air removal for pad-to-fan temperature rise.
Table 4. Factor used to correct rate of air removal for pad-to-fan distance.