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Old 12-03-12, 06:37 PM   #25
Heat recoverer
Join Date: May 2011
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Default Thermal lost estimate

Just for fun, I compute the standard thermal lost for a greenhouse of 13M2 (see attach drawing)... this is an example only but with enough detail so you might adjust it for your own configuration....

This is the formulas we use to estimate the need in heating for commercial greenhouses.

The formula is pretty accurate if:
a) the configuration of your greenhouse are the same has the one draw.
b) Toledo Spain -- Minimum temperature is 2C
c) Tomato lower recommended temperature 13C -- 55F
d) Pepper lower recommended temperature 15.5C -- 60F

This formula does not consider the "Insulation" of the shed "end Wall". This one is likely much more resistant to heat transfer then the 4mm polycarbonate panel. So, the real lost should be lower.

Step #1 Compute areas, volume and length of perimeter. (all from the drawing)

Sidewall = 2( 1.5m x 5m) = 15 m2
End area = ( 2 x 1.5m x 2.6m) + (0.5m X 2.6m) = 9.1 m2
Roof area = 2(1.39m x 5m) = 13.9 m2
Curtain W. = 3 m2 ---- (This is your concrete block wall)
Volume = (1.5m x 2.6m x 5m ) + (2.6m x .5m x 5m) = 22.75 m3
Perimeter = 2 x 2.6m + 2 x 5m = 15.2m

Step #2 List of U-Values

Polycarbonate 4mm: U = 3.9 W/(m2˚C)
Insulated perimeter: U = 1.39 W/(m˚C)
Your curtain wall (200mm Concrete Block U = 2.9 W/(m2˚C) )

Delta T -- Tomatoes don't like to go under 13C (55F) at night, For pepper, it is about 15.5C (60F). You live in Toledo, Spain -- Use 2C. Consequently, the DeltaT = 15.5C (pepper) - 2C = 13.5C

Step #3 Conductive heat loss for each identified items:

1. Sidewall =3.9W/(m2˚C) x 15 m2 x 13.5˚C = 790 watts
2. End area = 3.9 W/(m2˚C) x 9.1 m2 x 13.5˚C = 480 watts
3. Roof area = 3.9 W/(m2˚C) x 13.9 m2 x 13.5˚C = 732 watts
4. Perimeter = 1.39 W/(m˚C) x 15.2 m2 x 13.5˚C = 285 watts
5. Curtain wall = 2.9 W/(m2˚C) x 3 m2 x 13.5˚C = 118 watts

step #4 air infiltration heat loss,

Qa = 0.373 x 13.5˚C x 22.75m3 x (exchange hours == 2) = 229 watts

Between 1 to 4. 1 for well build polyethylene and 4 for badly build glasshouse

Step #5 Add everything

Qt = Qa + Qc = 229+ (790+480+732+285+118 ))
= 2674 W
= 9126 Btu/hr

Generally add 10%.
So, if your heating system is able to supply 3KW, this greenhouse might be operated 365 days a years with fancy vegetable such as Pepper.

It also means that if you have a night at 2C that last 10 hours, you need 30kw (102K BTU) of storage. (you likely need more since the sun is not always there.)

NOTE: Obviously, for December, January, February you may decide to growth crop that need much less heat at night -- like several green. For example, the deltaT might be reduce at 8C or even 6C for some crop.

NOTE2: since you are handling so many (almost pro) gadget, you might think to improve your greenhouse like the pro do by installing a "Thermal Curtain" that deploy automatically at night; You may save between 20 and 40% in energy requirement with this type of gadget.

SPAIN: I don't know much about your location, but I am told that in Spain, "Controlling the heat gain" at summer time is a much bigger problem then heating the greenhouse in the winter. You will need a pretty powerful fans to replace a roof vent when the outside temperature is in the 30C.

WIND: On my commercial unit, we open part of the roof on the opposite side of the dominant wind. The computer also use a special "mode" call "rain shedding" when it detect some water on the roof. Obviously, this is a little bit fancy and all controlled by computers.
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