Model 1 All Greenhouse Heat Transfer Through The Walls
Contents
The Equations
The heat transfer through a flat wall or floor due to thermal conductivity is given by:
<math>Q_{w} = A_{w} (J/t) \Delta T=J {A_{w} (T_{o} - T_{i}) \over t}={(T_{o} - T_{i}) \over R_{w}} </math>
Where:
Aw is the area of the greenhouse walls
To is the temperature on the outside wall
TI is the temperature of the inside wall
J is the thermal conductivity of the wall
t is the wall thickness
Rw is the thermal resistance of the wall
The heat transfer through a half cylinder arch due to thermal conductivity is:
<math>Q_{arch} = A_{arch} {J \over t_{mean}} \Delta T = {\pi J z \over ln[r /(r - t)]} (T_{o} - T_{i}) = {(T_{o} - T_{i}) \over R_{arch}}</math>
Where:
Aarch is the area of the overhead arch
tmean is the mean wall thickness
z is the cylinder length
r is the cylinder radius
Rarch is the thermal resistance of the arch
An approximately half cylinder greenhouse should be relatively easy to construct, so we can assume one of that shape for an example.
If we neglect radiative transfer and convection effects, then the total heat transfer through the greenhouse walls and floor then becomes:
<math>\Sigma Q = Q_{w} + Q_{arch} = {\Delta T \over R_{total}} </math>
where:
Rtotal is the total thermal resistance of the greenhouse
Qw is the heat transfer through the end walls and floor
<math> Q_{w} = Q_{ends} + Q_{floor} </math>
which is not complicated if you simply assume that the walls and floor both have the same thermal resistance.
If we assume that all daytime heating of the greenhouse heating is solar, then:
<math> \Sigma Q = Q_{collector} = \epsilon A_{collector} C_{solar}</math>
where:
Qcollector is the heat transferred to the greenhouse by its solar thermal collector (which need not actually fit within the greenhouse itself)
Acollector is the collector area
Csolar is the insolation
<math>\epsilon</math> is the collector efficiency
Daytime Cooling Example
Start by assuming <math>\Delta</math>Tfloor = <math>\Delta</math>Tw (which is possible for an elevated greenhouse with airflow under the floor).
Let
Ti = 20oC
To = -20oC
z = 8m
r = 4m
Afloor = 64 m2
Aends = 50.3 m2
Aarch = 100.5 m2
Acollector = Afloor
Csolar = 440 W/m2
<math>\epsilon</math> = 10%
The collector in this case produces a heat transfer rate of 2820 W into the greenhouse, which at equilibrium is equal to the heat transfer out, <math>\Sigma</math>Q. This can be used to compute the total thermal resistance required to maintain Ti inside the greenhouse.
<math>R_{total} = {\Delta T \over \Sigma Q}</math> = 0.0142 K/W
which, taken over the entire surface area of the greenhouse yields the inverse of the equivalent heat transfer coefficient:
<math>R_{total} (A_{w} + A_{floor}) = (0.0142 K/W) \cdot (64 m^2 + 50.3 m^2 + 100.5 m^2) = 3.05 Km^2/W</math>
which is equivalent to an average R14.5 insulation (in english system units) on all surfaces, including the windows. That is the level of insulation required to maintain equilibrium inside an elevated half cylinder greenhouse of this size with uniform wall thickness and solar heating provided only by a collector the size of the greenhouse. If provided solely by material insulation, that’s barely practical for a sun-illuminated greenhouse.
Each of these assumed variables can be altered for more efficient operation, however. For example, assuming a soil sheltered greenhouse makes for more complicated computation but creates a more stable temperature distribution across the floor and walls and provides additional insulation across the bottom. The ends can have additional insulation to compensate for the poor insulation of the windows. The solar collector can be larger, too. And while the total resistance does need to be this high, additional resistance due to poor convection in the martian atmosphere will provide a considerable portion of that. The thermal resistance of the Martian Atmosphere is not negligible, contrary to the assumptions used to derive the equations here.
Nighttime Cooling Example
Day-to-night temperature swings are quite dramatic on Mars due to the thin Martian atmosphere, and the outside temperature will be considerably lower at night.
Let
Ti = 20oC
To = -60oC
z = 8m
r = 4m
Afloor = 64 m2
Aends = 50.3 m2
Aarch = 50.3 m2
V = 201 m3
Qcollector = 0 W
Rtotal = 0.0142 K/W
In this case, with no heat source provided, the greenhouse cools at a rate of:
<math>\Sigma Q = {\Delta T \over R_{total}}</math> = 5630 W
This implies that, with the assumed configuration, the greenhouse will radiate more heat at night than it absorbs during the day.
If the specific heat of the air inside the greenhouse is cp=1007 J/kgK at 20oC, and its density is 1kg/m3, then at this heat transfer rate the greenhouse air could cool from Ti to freezing in only 13 minutes, even with R14 insulation.
Conclusions from this Model
Some source of nighttime heat is required to sustain growing conditions in the greenhouse. Added thermal mass (heat capacitance), such as rocks or tanks of water, can provide that additional heat. Constructing the greenhouse as a full cylinder rather than a half cylinder would make half its volume available for thermal mass, but will require additional solar collector area during the day to heat it sufficiently to supply heat overnight.
It would also be useful to consider adding exterior insulation at night, such as foam batting that can be drawn down over the greenhouse windows.
If the total thermal resistance can be doubled, the introduction of 10T of water to the greenhouse thermal mass (cp=4180 J/kgK for water at 20oC) would extend the frost-over time from 13 minutes to 2.5 days. A cylindrical greenhouse with water tanks stored below the solarium could easily have room for 100T of water in a greenhouse of the floor size assumed in the examples here.
Rapid nighttime cooling of the greenhouse can be addressed with a nighttime cover and/or greater thermal mass. Some source of heat beyond what the solarium alone can absorb will be necessary, but additional thermal collectors (external to the solarium) and a heat exchanger system can be added to contribute to the thermal mass of the greenhouse.
With proper design, a Martian greenhouse can be heated primarily by solar radiation.
Limitations of This Model
This model is accurate only so far as it treats the greenhouse as a homogenous control volume. It does not deal well with heat losses though the ground, convection effects or losses through black body radiation. The Martian atmosphere may not be able dissipate heat quickly enough to keep the outside wall the same temperature as the Martian air. This is a useful model for selecting a heating system, but not necessarily for design of the greenhouse itself.
More thermodynamic models of the Martian greenhouse can be found on the page:
Thermodynamics of the greenhouse
See Also:
