Greenhouse Cooling

1
Greenhouse Cooling

• Summer Cooling
• Winter Cooling

2
Conditions of Un-cooled GH

• 20-30º F warmer than outside
• Detrimental effects
• loss of stem strength
• reduced flower size
• delay in flowering
• bud abortion
• improper water and nutrient uptake

3
Cooling methods

• Evaporative cooling
• fan and pad (1954)
• package coolers (swamp coolers)
• Fog (1980)
• General ventilation and de-humidification

4
Benefits

• Reduce air temperature below outside temperature
  by 25º F.
• Improve CO2 uptake
• Improves nutrient and water uptake
• Hardier foliage and stems
• Provide constant air movement
• Control humidity levels

5
Terms

• Evaporative cooling
• Wet Bulb Temperature
• Dry Bulb Temperature
• Air Flow
• Psychrometric Chart
• Relative humidity

6
Summer Cooling System

• Fan and Pad System
• Rate of air exchange measured in cubic feet of
  air per minute (cfm)
• Factors affecting air exchange
• elevation
• light
• allowable temperature increase from pad to fan

7
Summer Cooling System

• The rate of air flow through a greenhouse is
  generally the cubic foot capacity of the
  greenhouse with a few corrections factors.
• Corrections factors?

8
Elevation

• As the elevation increases, fan capacities need
  to be increased.
• One pound of air at sea level occupies a given
  volume. As this pound of air moves up in
  elevation, its volume increases. The weight
  remains the same.
• Buy a bag of potato chips in San Bernardino, then
  drive up to Big Bear. What happens to the bag of
  chips?

9
Light

• As light intensity increases, the heat input from
  the sun increases requiring greater air removal
  from the greenhouse.
• An intensity of 5,000 foot candles is accepted as
  a desirable level for crops in general. Can be
  achieved with a coating of shading compound on
  the outside, or a screen material drawn from eave
  to eave.

10
Pad to fan temperature rises

• Solar energy warms air as it moves from pad to
  fan.
• Usually a 7º rise in temperature as air moves
  from pad to fan is acceptable.
• If a more constant temperature is desired, the
  velocity of the fans should be increased.

11
Calculation formula

• Standard cfm L x W x 8 x Fhouse
• Correction formula
• Fhouse Felev x Flight x Ftemp
• or
• L x W x 8 x Fvel, where Fvel 10/?D.
• D distance of air travel
• usually used when D is less than 100 ft.
• use larger of the two correction factors.

12
Example 1

• GH 70 ft. x 50 ft.
• at 500 ft elevation
• 5000 fc
• 7º F pad to fan increase allowed
• Calculate the cfm requirement for this GH.

13
Solution-Ex. 1

• Std cfm 70 x 50 x 8 28,000
• Fhouse 1.0 x 1.0 x 1.0 1.0
• Corrected cfm value 28,000 cfm / min.
• air will be drawn in the 70 ft. direction. Since
  it is less than 100 ft., the Fvel formula should
  also be used. In this case, it is equal to
  10/?70 1.2
• Therefore, 1.2 is used instead of 1.0
• Correct value 28,000 x 1.2 33,600 cfm

14
Example 2

• GH 70 x 50
• Elevation 5,000 ft.
• Light 6000 fc
• Allowable Pad to fan T 7º F
• Determine Cooling requirement

15
Solution Ex. 2

• 50 x 70 x 8 28,000 cfm
• Fhouse 1.2 x 1.2 x 1.0 1.44
• Corrected cfm 28,000 x 1.44 40,320
• F vel 28,000 x 1.2 33,600 cfm

16
Fvel 10/?D

• Use this formula when distances of air travel are
  less than 100 ft.
• Compare the Fhouse value with this, use the
  larger of the two.
• When air flow is less than 100 ft., the air feels
  like there is no velocity, therefore, the fan
  capacity should be increased.

17
Determining Fan Capacities

• Fans should not be more than 25 apart from
  center to center.
• Divide length of wall by 25
• 50 ft. ? 25 ft. 2 fans needed
• 40,320 cfm ? 2 20,160 cfm/fan

18
(No Transcript)
19
(No Transcript)
20
(No Transcript)
21
Cool Pad Description

• Material should provide an evaporative surface
  with large surface area.
• Material should be able to withstand weathering.

22
Cool Pad Materials

• Aspen Fiber
• Cross-fluted Cellulose

23
Cool Pad Efficiencies

• Aspen 150 cfm/ft2
• Cross-Fluted Cellulose 4 thick 250 cfm/ft2
• Cross- Fluted Cellulose 6 thick 350 cfm/ft2

24
Determining amount of Pad Material

• Total corrected cfm required / cooling efficiency
  per ft2 of pad.
• Example If 40,0000 cfm is required and a 4
  cool cell will be used
• 40,000/250 160 ft2 required.

25
Determining the Height of the Pad

• Identify length of wall on which pad will be
  installed.
• Divide total cfm required by wall length
• 160 sq. ft./ 50 ft. 3.2 ft. high

26
Determining Sump Capacity

• Sump volume 1.5 gal. Per foot of pad length.

27
Determining Pump Capacity

• Pump capacity 0.6 gal per min. x length of pad.
• 0.6 gal. Per min. x 50 ft. 30 gal per min.

28
(No Transcript)
29
GH Cooling

• Pad Location
• Water Treatment
• Sump Bleeding
• Salt Maintenance
• Fan Belt Tension
• Fan Motor Alignment
• Thermostat Calibration

30
GH Cooling

• Fan Baffles
• Air Baffles throughout GH
• Air Circulation Fans
• Louver Maintenance
• Shade cloth interior/exterior
• Shading Compound

31
Winter Cooling

• System Description
• Pressurizing Fan
• Plenum Chamber
• Convection Tube
• Motorized Inlet Shutter

32
Winter Cooling

• System Tied to Summer Cooling system
• Uses some of the exhaust fans either at full of
  low speeds.
• Air being exhausted in this mode should not
  exceed the incoming air volume.
• Process can be referred to as minimum
  ventilation.
• Process can also be used for dehumidification.

33
Winter Cooling Calculation
34
Winter Cooling Calculation

• Formula L x W x 2 x Fjet
• Where F jet Felev x Flight x Fwinter
• Fwinter factor taken from tables.

35
Winter Cooling Example

• GH 70 x 50
• Elevation 5,000 ft.
• Light 6000 fc
• Allowable Pad to fan T 7º F
• GH temp above outdoor temp 13 F
• Determine Winter Cooling requirement

36
Example (contd)

• From previous example, we needed 40,320 cfm for
  maximum cooling.
• Total cfm required 50 x 70 x 2 x Fjet
• Fwinter 1.15
• ?Fjet 1.2 x 1.2 x 1.15 1.66
• Corrected cfm 50 x 70 x 2 x 1.66 11,620

37
Example (contd)

• In the winter cooling mode, air will be entering
  the GH at a rate of 11,620 cfm.
• One of the exhaust fans should be running at a
  rate not to exceed 11,620 cfm.
• We called for 2 fans, each with a capacity of
  20,160 cfm.
• If one fan runs at ½ speed, the capacity will be
  less than 11,620 cfm

38
Example (contd)

• Another option is to divide 40,320 by 4 and use 4
  fans, each with a capacity of 10,080 cfm.
• During the winter cooling mode, one fan will run
  to expel hot humid air.
• Any other solutions?

39
Dehumidification

• Relative Humidity
• Psychrometric Chart (follow your charts)
• Curved lines RH
• Horizontal line at bottom 0 RH
• Upper most line 100 RH or total saturation

40
(No Transcript)
41
Example of RH

• At 100 RH, locate the 60 F dry bulb vertical
  line. At this condition, 1 lb. Of air contains
  77 grains of H2O vapor
• Follow the 60 F dry bulb line to 50 RH line. At
  this condition there are approximately 38 grains
  of water vapor. The air is now only 50 full and
  thus has a RH of 50.

42
Relative Humidity

• The amount of water vapor in the air compared to
  what it can actually hold at that temperature.

43
Example of RH

• We always experience a daily fluctuation in
  outdoor relative humidity.
• Assume a night temperature of 65 F with 95 RH.
  During the day, air arms air to 85 F, the RH
  drops to 45.
• Also, as air gets colder, the capacity of
  holding moisture is reduced.

44
Example 1a

• If the greenhouse is at 60 F and air contains 77
  grains of moisture and the temperature drops to
  40 F, air now contains only 38 grains of
  moisture. What happened to the other 39 grains
  of moisture?

45
Example 1b

• Assume a greenhouse condition of 65 F night
  temperature and 75 RH. With poor heat
  distribution and random cold air currents, a cold
  spot that is at 56 F develops. What would be the
  outcome?

46
Effect of Temperature on RH

• In a previous example, 1 of air at 60 F DB and
  100 RH contained 77 grains of moisture. If air
  is warmed 20 F to a temp. of 80 F, the RH is now
  50.
• Air at 80 F could hold 156 grains of moisture
  when saturated but actually contains only 77
  grains, therefore is only half full.

47
Effect of T on RH (contd)

• If air is warmed another 20 F, to 100 F, the RH
  will drop to 25.
• This shows that for every 20 F DB rise in T, the
  capacity of air to hold water vapor approximately
  doubles.

48
Reducing the RH in the GH

• If cold air w/ low water vapor content is brought
  into the GH and warmed, the capacity to hold
  water vapor increases.

49
Reducing RH in the GHExample2

• Assume a rainy weather condition, 40 F at 100 RH
  outside.
• Inside GH, 65 F at 90 RH. Too humid.
• Is it possible to lower the RH to a satisfactory
  level even if it is raining outside?

50
Reducing RH in GH

• Chart shows
• every of air holds 83 grains inside the GH.
• Every of air holds 37 grains outside.
• Therefore, with each of air removed by an
  exhaust fan, 83 grains are carried out (removed)
  while 37 grains enter the house with the make-up
  air.
• This results in a net removal of 46 grains of
  moisture.

51
Reducing RH in the GH

• A dehumidification system utilizing exhaust fans
  running to expel half of the air in the GH will
  lower the RH to 65 once the GH is heated back
  to 65 F.

52
Fog Cooling

• Evaporative Cooling
• Increase Humidity in Propagation
• Reduce Light Levels

53
(No Transcript)
54
(No Transcript)
55
(No Transcript)
56
Greenhouse CoolingSystem Description

• Use of series of fog nozzles
• High pressure 400-2000 PSI pumps
• Water treatment and filters
• Ridge Vent
• Thermostat/Humidistat sensing devices

57
(No Transcript)
58
(No Transcript)
59
Fog Cooling Facts

• Fog particle size 10-40 microns
• High pressure
• Surface area 10,000 x greater than same volume
  in 1 cup of water
• Droplets stay suspended in air, no condensation
• Droplets evaporate 40,000 faster than H20 in a
  cup
• Air cooled at 100 efficiency

60
Advantages

• Uses les electricity
• Even temperature control across greenhouse
• Cooler average temperatures achieved
• Good substitute for mist
• Can be used outdoors in dry climates

61
(No Transcript)