Greenhouse HeatingGreenhouse Heating

Exploring Heat Transfer Occurrences in a Common Greenhouse

Ryan Vorwaller

Dustin Peterson

ME 340

12/1/2010

BackgroundBackgroundThe use of

greenhouses is an effective way to gather heat energy from the sun through radiation.

Contrary to popular belief, the sun does not directly heat the air inside a greenhouse. Rather, radiant heat from the sun passes through the greenhouse glass, heating up the ground and plantation. The ground and plantation then emit the heat and the enclosed greenhouse simply retains that heat.

ProblemProblemUsing specific dimensions, conditions, materials,

and the sun as the only heat source, how low can the temperature outside of an empty greenhouse drop while the greenhouse maintains an inside temperature of 80˚F assuming the sun is directly over the greenhouse?

Note:While they are effective

at containing heat energy, greenhouses often have heaters inside to aid in maintaining a constant high temperature.

Given ValuesGiven Values

The greenhouse is made of 3.2mm thick glass with a transmissivity of 0.79. The average convective coefficient of the air outside is 10 W/m^2*K. The soil’s absorptivity is 0.94 and there is no plantation.

CalculationsCalculations Greenhouse Dimensions - 3 meter wide, 5 meter long, 3 meter high rectangle Gsun = 1353 W/m^2 Transmittance of 3.2 mm thick glass = 0.79 k (glass) = 1.4   Q out = Q in Energy Balance – Qcond (across glass) = Qconvection (glass surface to outside air)  = Qrad (from sun)   Q rad = Area of enclosed ground*transmittance*Gsun*Soil Absorptivity           = 15*0.79*1353*0.94           = 15071 W   Assume inside surface is equal to inside temperature 80 F = 26.7 C Qcond = Area of exposed glass*k*(26.7 - Temp of Outer Surface) / glass thickness             = 63*1.4*(26.7-Ts)/0.0032             = Q rad             = 15071 W   Therefore Temperature of outer surface = 26.12 C   Assume h (air) = 10 W/m^2*K Qconv = Area of exposed glass*h*(Temp of outer surface - Temp of air)            = 63*10*(26.12-Temp of air)            = Qcond            = 15071 W

Temperature of outside air = 2.2˚C

= 36˚F

Check ResultsCheck Results

An online greenhouse calculator suggests that given a heat loss factor, our conditions would actually require some form of added heating. However, this could simply indicate the ideal nature of our assumptions as discussed on the next slide.

* See References

ConclusionConclusionOur results could be quite different from an

actual greenhouse. Many assumptions were made, including the fact that the sun is always shining directly over the greenhouse and that no heat is lost when a door is opened. Nevertheless, given these ideal circumstances, our calculations show that the outside temperature could drop to 36˚F and the inside of the greenhouse could still remain at 80˚F. Further work could be done to refine these results by including additional heat loss factors and refining our assumptions.

ReferencesReferences

Values and Constants from:Incorpera, DeWitt, Berdman, and

Lavine, Fundamentals of Heat and Mass Transfer,

6th Edition, Wiley and Sons.

Other References:https://www.greenhousecatalog.com/

greenhou se_calculator.php