# Model 4 A Wind Cooled Greenhouse

In this model, the greenhouse wall is modelled as a set of parallel thermal resistances. Convective heat transfer and conductive heat transfer are considered. Radiative heat transfer is neglected.

## The Equations

The greenhouse can be modeled as being composed of a composite wall separating two thermal reservoirs of the same temperature. In this model, thermal resistances are arranged in parallel, so that the total thermal resistance is a mean resistance. In this model, the greenhouse is separated into three sections: a floor, an arch wall, and the combined area of its ends. Thermal effects at the interfaces are neglected.

Since a half cylinder greenhouse is relatively easy to build, we will assume this is the shape of our greenhouse.

Heat transfer through the floor is primarily due to conduction. The thermal resistance of a flat floor is:

$R_{floor} = R_{fi} + R_{f} + R_{F} = {t_{f} \over k_{f} A_{floor}} + {t_{F} \over k_{F} A_{floor}}$

where:
Rfloor is the thermal resistance of the floor
A floor is the floor area
Rfi is the thermal resistance of the internal convection layer at the floor
Rf is the thermal resistance of the flooring material
RF is the thermal resistance of the foundation
tf is the thickness of the wall
kf is the thermal conductivity of the wall
tF is the thickness of the foundation
kF is the thermal conductivity of the foundation

The walls and ends are membranes separating two gas flows: the air inside the greenhouse and the air outside it. Three thermal resistances in series must be estimated for each: the resistance of the interior convection boundary layer, the resistance of the outside convection boundary layer, and the thermal resistance of the greenhouse wall.

The hydraulic radius of the greenhouse arch is:

$D_{h} = {4 A_{h} \over P_{h}} = {2 \pi r \over \pi + 2}$

where:
Dh is the hydraulic radius of the arch
r is the radius of the arch
Ah is the cross-sectional area of the flow
Ph is the perimeter of the cross-section in contact with the arch

The Reynolds number of each flow (inside and out) is:

$Re = {\rho v D_{h} \over \mu}$

where:
Re is the Reynolds number of the flow
$\rho$ is the density of the gas
v is the flow velocity
$\mu$ is the dynamic viscosity of the gas

Assuming air flows from one end of the greenhouse to the other, the Nusselt number of the interior flow is:

$Nu_{i} = 0.023 Re^{0.8} Pr^{1/3}$

where:
Nu is the Nusselt number of the flow
Pr is the Prandtl number of the gas

if the flow is turbulent, or Nui = 4.66 if the flow is laminar. (Laminar flow typically has a Reynolds number of 2000 or less.)

The Nusselt number for an exterior flow parallel to the greenhouse is:

$Nu_{o} = 0.463 Re^{1/2} Pr^{1/3}$

From these Nusselt numbers, the heat transfer coefficient of the interior and exterior flows can be estimated.

$h_{i} = {Nu_{i} k_{i} \over D_{h}}$

$h_{o} = {Nu_{o} k_{o} \over D_{h}}$

where:
hi is the heat transfer coefficient of the inside surface
ki is the thermal conductivity of the gas inside the greenhouse
ho is the heat transfer coefficient of the outside surface
ko is the thermal conductivity of the gas outside the greenhouse

The corresponding thermal resistances can also be estimated.

$R_{i} = {1 \over h_{i} A_{i}}$

$R_{o} = {1 \over h_{o} A_{o}}$

where:
Ri is the thermal resistance of the inside convection layer
Ro is the thermal resistance of the outside convection layer

The thermal resistance of a half-cylindrical wall is:

$R_{w} = {1 \over \pi z k_{w}} ln{{r \over(r - t_{w})}}$

where:
Rw is the thermal resistance of the wall material
tw is the thickness of the wall
kw is the thermal conductivity of the wall
z is the length of the greenhouse

This means that the total thermal resistance of the arch is:

$R_{arch} = R_{i}+R_{w}+R_{o}$

We can also introduce a definition of Rfi in terms of Ri:

$R_{fi} = R_{i} {A_{floor} \over A_{i}}$

The total thermal resistance of the end walls can be found similarly, except that it’s easier to assume the heat transfer coefficients of the end walls are the same as those of the arch. This allows combining the expression for the arch and ends.

$R_{endwall} = {t_{w} \over \pi r^2 k_{w}}$

$R_{cover} = ( R_{i} + R_{o} ){A_{i} \over A_{cover}} + { R_{w} R_{endwall} \over R_{w} + R_{ends}}$

where:
Rendwall is the combined thermal resistance of the endwalls
Rcover is the thermal resistance of the greenhouse’s entire top cover (arch and ends)
Acover is the area of the greenhouse’s entire top cover

The assumptions made for the end walls are not exact, but make the math easier without introducing unacceptable errors.

The total thermal resistance of three parallel thermal resistances is:

$R_{total} = {R_{floor} R_{cover} \over R_{floor}+R_{cover}}$

The total heat flow rate can be computed from this:

$\Sigma Q = {\Delta T \over R_{total}}$

where:
$\Sigma$Q is the total heat transfer rate
$\Delta$T is the temperature difference between the inside and outside

At the temperatures involved for a Martian greenhouse, this heat transfer is primarily from the greenhouse to the Martian air. During the day, heat can be absorbed from sunlight to offset this heat transfer rate out of the greenhouse.

$Q_{solar} = \epsilon C_{solar} A_{collector}$

where:
Qsolar is the solar heat absorbed by the greenhouse’s thermal collectors (not necessarily contained within the greenhouse itself)
Csolar is the insolation of the collector
$\epsilon$ is the efficiency of the collector

## Daytime Example

Assume the greenhouse walls are thin, uninsulated polyethylene sheeting.

Let the dimensions of the greenhouse be:
r = 4 m
z = 8 m
tw = 0.001 m
tf = tw
tF = 1 m

This implies that:
Dh = 4.89 m
Afloor = 64 m2
Aends = 50.3 m2
Aarch = 100.5 m2
Acover = 153.8 m2

Let the external and internal conditions be:
Ti = 20 oC
To = -30 oC
$\rho_{o}$ = 0.015 kg/m3
$\rho_{i}$ = 1.0 kg/m3
ki = 0.02544 W/mK
ki = kfi
ko = 0.01295 W/mK
kw = 0.33 W/mK
kw = kf
kF = 1.0 W/mK
Pri = 0.710
Pro = 0.780
$\mu_{i}$ = 1.795 x 10-5 Pa-s
$\mu_{o}$ = 1.95 x 10-5 Pa-s
vi = 0.2 m/s
vo = 10 m/s

The Reynolds numbers for the exterior and interior flows are then:

Rei = 47430
Reo = 37620

Even at the low internal flow rate assumed, the internal circulation of the greenhouse is not predominantly laminar in character. Using the appropriate equations for the Nusselt number of each flow, this implies that the interior and exterior heat transfer coefficients are:

hi = 0.656 W/m2K
ho = 0.219 W/m2K

This implies an insulation value of at least R28 (in English units) for the top cover, just by virtue of the poor convective transfer available. That figure does not include the thermal resistance of the greenhouse walls.

The thermal resistance of the walls can be computed from the thermal resistances of the arch and endwalls.

Rendwalls = 0.003 W/K
Rw = 0.0001 W/K

Rcover = 0.040 W/K

Which is equivalent to R28.2 insulation. The thermal resistances of the wall materials are nearly negligible in comparison to that of the convection layers.

Similarly, the thermal resistance of the flooring material can be ignored as well if it is of the same thickness. However, the resistance of the foundation cannot.

$R_{floor} = R_{F} + R_{fi} = 0.039 W/K$

which, given Afloor, is equivalent to R11 insulation.

The total thermal resistance of the greenhouse with its foundation on the ground is:

Rtotal = 0.020 K/W

Which is equivalent to R20 overall insulation. It is arguable in this case that the greenhouse would benefit from an elevated foundation allowing airflow beneath, which would raise the thermal resistance of the floor to nearly that of the cover, R28.

Under the given conditions, the total heat transfer rate out of the greenhouse would be:

$\Sigma$Q = 2500 W

A solar collector can be used to offset this heat loss. Assume the following for a hypothetical solar thermal collector:

Acollector = Afloor
Csolar = 440 W/m2
$\epsilon$ = 10%

This implies that a solar thermal collector the size of the greenhouse floor (which could be contained within the greenhouse) can provide up to 2820 W of heat to the greenhouse, 300W more than the heat loss rate. A collector of this size would heat the greenhouse faster than it could shed heat, causing its internal temperature to rise unacceptably.

The greenhouse’s thermal collector area would either need to be less than that of the greenhouse floor, or thermal mass would need to be provided to absorb that excess heat. Use of an elevated greenhouse with an R28 insulation value would only exacerbate this problem, reducing the total heat loss rate to only 1800 W without necessarily affecting the solar heating.

## Nighttime Example

Temperatures drop precipitously on Mars at night due to the thin atmosphere.

For the greenhouse described above, let:
To = -50 oC
Ti = 20 oC
Rtotal = 0.020 K/W
Qsolar = 0 W

In this case, the heat flow rate out of the greenhouse would accelerate to 3500 W. At this rate of heat loss, if the internal gas had a specific heat of 1007 J/kgK, then its temperature could drop from Ti to freezing in only 20 minutes with no other source of heat. The addition of 10T of water, with a specific heat of 4180 J/kgK, would provide sufficient heat to bring the greenhouse through the night if a heat exchanger were used to transfer heat between the thermal mass and greenhouse. A cylindrical greenhouse with a half cylinder solarium and the remaining volume devoted to tankage could easily accommodate at least 100T of water. However, a greenhouse of this size could not contain a solar collector large enough to absorb sufficient heat during the day for this amount of thermal mass.

The greenhouse thermal collector must extend an additional 130% beyond the solarium (more than double the greenhouse area) for solar thermal heating to be adequate.

Additional reduction of the heat loss rate can be had by increasing the insulation of the greenhouse. The endwalls, for example, can be soil sheltered, or the greenhouse can be elevated to improve the floor insulation. A removable cover such as foam batting can be placed over the windows at night to further reduce heat loss.

## Conclusions of this Model

The thermal insulation value of the thin Martian air is remarkable – likely better than that of 10 cm Styrofoam. It is sufficient to allow solar heating alone to sustain a greenhouse on Mars if a thermal mass sufficient to keep the greenhouse warm overnight (at least 10T water or equivalent) is included, and a heat exchanger provided to keep the heat transfer constant.

It is also advantageous to have an external airspace outside both the walls and floor of the greenhouse. Additional insulation of the end walls is also useful, as is additional batting which can be drawn closed at night. The solar thermal collector used by the greenhouse must extend outside the solarium in order to collect enough heat over the course of the day, but this can be done.

A larger greenhouse should also be considered, as it would have a larger floor area for collecting solar heat and could contain more thermal mass.

## Limitations of this Model

The layers of assumptions involved limit accuracy to +/-10% at best. Also, use of Nusselt numbers to derive heat transfer coefficients is particularly inaccurate under variable conditions, and change in the gas flow rates or flow geometries can affect the result significantly. Actual heat flow rates could easily vary from calculated values by a factor of 2.

This model also neglects radiative heat transfer, which while not as great as the heat transfer rate due to convection and conduction, is still significant.

This model’s qualitative outcome is reliable, and it can be used to predict the most effective distribution of insulation on the greenhouse. However, any use of this model for greenhouse design should preferably allow a considerable safety factor.

Discussion of other models for greenhouse thermodynamics can be found under Thermodynamics of the greenhouse.