Solar Loads on a greenhouse project.

Related Items

Air Speed
Plant Transpiration


As stated in the preamble on the Fequently Asked Question sheet there are many items that will effect the total solar loading on your greenhouse which result in the selection and sizing of the cooling/ventilation systems.

The primary tools/devises/methods used to offset the solar loads are primarily:

The items that effect the solar loading on your structure are:

 

In the Previous vent write-ups we have been asked how we arrived at the 15,000,000 BTU/hour total solar load.

 

The project information from the example is as follows:

"Say a grower is planning on building a 10 bay gutter connect project that is located near Fort Wayne, Indiana. The project is comprised of 30' wide gutter houses each 240 feet long for a total area of 72000 sq. ft . and wishes to use only roofing vents for cooling.

Environmental Data for the site is 92 deg. F DB/73 deg. F wet bulb summer outdoor design-4 deg. F winter outdoor design, the project is located at 41 deg. F north latitude at a elevation of 793 feet above sea level."

At our office we use a computer model that allows hourly calculations but here by hand we will demonstrate the method.

Again using ASHRAE standards, this is the rational.

At 40 deg. N latitude solar load imposed on the greenhouse structure will be predominately roof load. At 40 deg. north latitude the greatest solar loading will occur on June 21 at 12:00 PM. Based on Simpson's rule with time interval equal to 10 minutes. ( ASHRAE - fundamentals/fenestration ) the load is 267 BTU's per sq. ft. (826 watts/M2). This is used for our max. load calculations.

Note: We assumed no shading factor ( SC ) other than the cladding product which is 0.91 and based on a clear day.

Chart One


Area (sq. Ft. )
SHGF
SC/trans factor
Load-Btus/hr

East Wall

2880
41
0.91
107,452

West Wall

2880
41
0.91
107,452

North Wall

4500
38
0.91
155,610

South Wall

4500
95
0.91
389,025

Roof

72000
267
0.91
17,493,840

Slope factor on roof

28800
(38+95)/2
.091
1,723,680

Total Solar Heat gain

18,253,377

Note: The above chart is assuming it to be a perfectly clear day, free of clouds and haze, and in the real world this seldom happens so we like to apply a haze/cloud factor, as illustrated below.

June 21: 12:00 PM, 40 deg. N latitude

Chart 2

Corrected

Area (sq. Ft. )
SHGF
SC/trans factor
Load-Btus/hr

East Wall

2880
41
0.77
90,921

West Wall

2880
41
0.77
90,921

North Wall

4500
38
0.77
131,670

South Wall

4500
95
0.77
32,9175

Roof

72000
267
0.77
14,802,480

Adjusted Solar Heat Gain

15,445,167

Note: The above chart would closer reflect an average typical day.

 

Now to throw a real kicker.

Let's move this entire greenhouse to Logan Utah.

According to weather data ( again as published in ASHRAE - fundamentals ) Logan has a 1% summer design of 93 deg. F DB design which is not that different than Fort Wayne at 92 deg. F DB. Both projects are located at about 41 deg. N latitude.

So would it be safe to say the loads will be the same !!!!!!!!!

If you did you would be incorrect.


The greenhouse in Logan Utah would be faced with a higher solar load just by virtue of the increased elevation from sea level. ....... Logan is 4,775 feet about sea level. Whereas, Fort Wayne is 791 feet. ...... Solar loading increases 0.07% every 1000 feet in elevation. So the solar loading at Logan would be that much greater.

Rule of Thumb


Typically, ( again by rule of thumb ), plant transpiration ( we're assuming the house is full of crop ), will look after 1/2 of the solar gain. So the ventilation and cooling systems will need to look after 7,700,000 BTU's of heat. For those that wish this expressed in refrigeration terms this is a 641 Tons mechanical cooling load.


Just to back track a bit. Lets assume the grower wished to use fan cooling. What would the total airflow rate be if they wished to maintain a spacial temperature of no more than 10 deg. F from ambient.

Using Q = H / 60 X C * D ( ti-to)

where H = 7,700,000 BTUs/hr
C= specific heat of air = 0.245 BTU/Lb X F
D = density of air 0.075 lbs/ft 3
Ti - To = 10 def. F

which yields an air flow rate of 698,412 CFM of air flow.

Check using .... the Rule of thumb air flow rate of 1 AC/min based on 8 foot height.

= 72,000 sq. ft. X 8 feet = 576,000 CFM

By now you should realize that the "old rule of thumb can leave you short" in cases. 698,000 CFM verses 576,000 CFM. ( The old rule of thumb may work well in a greenhouse with soil or gravel floors - where an additional evaporative cooling base is created. However, in modern houses with covered floors and drip irrigation, this evaporative cooling base is not available. ). Basically ... if you short yourself on cooling and ventilation, the crop will suffer stress simply by the increased in transpiration rates. " They ... the plants are spending all their energy just to keep cool , not growing or producing fruit etc. "


Daily Load Profile Curve

For those interested the chart below is a sample of a the cooling load profile curve for the roof only on the above project. The values calculated is solar loading in BTU's/hr. Only the months of Jan to Aug have been listed.

Chart 3 - Roof Only ( Jan to Aug )

Solar Time
Jan
Feb
Mar
Apr
May
June
July
Aug
5
Nil
Nil
Nil
Nil
Nil
Nil
Nil
Nil
6
Nil
Nil
Nil
609840
1718640
2217600
1774080
665280
7
Nil
Nil
1441440
3381840
4823280
5377680
4878720
3437280
8
776160
2383920
4712400
6819120
8094240
8482320
8038800
6763680
9
3049200
5433120
7927920
9812880
10810800
11143440
10755360
9646560
10
5322240
7927920
10311840
12030480
12972960
13194720
12806640
11864160
11
6874560
9480240
11808720
13471920
14248080
14414400
14081760
13250160
12
7373520
9979200
12363120
13970880
14691600
14802480
14525280
13693680
13
6874560
9480240
11808720
13471920
14248080
14414400
14081760
13250160
14
5322240
7927920
10311840
12030480
12972960
13194720
12806640
11864160
15
3049200
5433120
7927920
9812880
10810800
11143440
10755360
9646560
16
776160
2383920
4712400
6819120
8094240
8482320
8038800
6763680
17
Nil
Nil
1441440
3381840
4823280
5377680
4878720
3437280
18
Nil
Nil
Nil
609840
1718640
2217600
1774080
665280
19
Nil
Nil
Nil
Nil
Nil
Nil
Nil
Nil
20
Nil
Nil
Nil
Nil
Nil
Nil
Nil
Nil
21
Nil
Nil
Nil
Nil
Nil
Nil
Nil
Nil
22
Nil
Nil
Nil
Nil
Nil
Nil
Nil
Nil
23
Nil
Nil
Nil
Nil
Nil
Nil
Nil
Nil
24
Nil
Nil
Nil
Nil
Nil
Nil
Nil
Nil

Jan
Feb
Mar
Apr
May
June
July
Aug
Daily totals
39417840
60429600
84767760
106223040
120027600
124462800
119196000
104947920

LATITUDE

What would happen if we moved this project further north say up to 48 deg. north latitude.

June 21: 12:00 PM, 48 deg. N latitude

Chart 4

Corrected

Area (sq. Ft. )
SHGF
SC/trans factor
Load-Btus/hr

East Wall

2880
40
0.77
88,704

West Wall

2880
40
0.77
88,704

North Wall

4500
37
0.77
128,205

South Wall

4500
134
0.77
464,310

Roof

72000
252
0.77
13,970,880

Adjusted Solar Heat Gain

14,740,803

Note: The above chart would closer reflect a typical day. Compare this with Chart 2.
Even though we have moved this structure further north, the solar loading really didn't fall that much. The roof loading sure dropped, but take a look at how the south wall gain increased. Interesting.

 

Air Speed

Air speed influences many factors in plant growth. These include transpiration and evaporation, leaf temperature and carbon dioxide. In general air speeds of 20 to 50 feet per minute across the leaf surfaces facilitate the uptake of available carbon dioxide, promotes transpiration/evaporation and reduces leaf temperature. At high air speeds 100-200 feet per minute plant growth is inhibited.

Here's a way to check the air speed.

Say a greenhouse structure is comprised of 21' wide X 10' feet high under the gutter gutter houses. Each house is ventilated with an exhaust fan rated at 19,000 CFM air flow. The exhaust fans are located at mid point of the height under the gutter.

The air flow rate would be 19,000 / (21 X 10) = 90 feet per minute, assuming a bench type crop.

If it was a tall vine like crop such as tomatoes or cucumbers, the face area could be as high as 180 feet per minute.
In the case of the tomato house, it would be unwise to add more air flow for cooling, if cooling was problem since all that would occur is greater mechanical induces stress on the crop. here the grower may wise to consider evaporative cooling of high pressure fog cooling to reduce the heat gain of the structure or by adding some form of shading which would also limit the solar gains.


Plant Transpiration

In the project sample above, assume the grower is wishing to raise tomatoes. The plant density that they are targeting for a one plant per four sq. feet. Therefore, in the 72,000 sq. feet of greenhouse structure, they should try to acheive 18,000 plant spaces. Generally it is accepted to provide up to 1.0 USGal (4 litres) per plant per day.
Now lets assumed the old 50% rule of thumb ( ie 50% of the daily solar load is offset by plant transpiration. If you look at chart 4 for the month of June, the total ( average/projected ) dailey solar load is
124,462,800 BTU's. 50% of this load is assumed to be controlled by plant transpiration. So 50% of 124,462,800 is 62,231,400 BTU's.

To offset this load would require the evaporation of 59,268 lbs of water ( 7,115 USGals or 26,933 litres ) or 0.39 USgal/plant ( 1.4 lires/plant ). So almost 1/2 of the water uptake by the plants on this day is just just for cooling. The question to ask is ...... gees not much growing or production is left avaiable for the plants ..... ????. USing the 50% rule.

 

 

 


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