I am trying to model the warming of a greenhouse for a differential equations project because I'm interested in math and gardening. I was hoping to work it all out myself from some basic thermodynamics equations but I've spent over 10 hours trying to figure it out to no avail.

I'm hoping someone knows of a way to calculate how quickly the ground will warm due to sunlight. I am having a really hard time converting surface illumination from the sun (or any source) to the surface temperature. If I can figure this out the rest of my project should be easy. Skip to the end if you want to avoid the explanation of what I know so far.

I've seen the heat conductance formula $\frac{Q}{t}=\frac{-kA}{L}\Delta T$, where k is thermal conductivity, A is area, and L is the thickness of the material doing the conducting but this requires I know the $\Delta T$.

I've seen the horizontal surface convection formula for air, roughly $\frac{Q}{t}=hA\Delta T$, where h is the Heat Transfer Coefficient.

And the Stefan-Boltzmann equation applied to surface temperature, $\frac{Q}{t}=\epsilon \sigma AT^{4}$, where $\epsilon$ is the emmisivity, A is surface area doing the emitting and receiving, $\sigma$ is the Stephen-Boltzmann Constant $5.67 \times 10^{-8}$, and T is temperature in Kelvin.

Newton's Law of Cooling $T(t)= T_{ambient}+Ce^{-kt}$

And the general heat transfer equation of $Q=m\sigma \Delta T$, where $\sigma$ is the heat transfer coefficient, m is mass, T is Kelvin, and Q is total energy transferred (not $\frac{J}{s}$).

I hope you can see how I'm having trouble calculating how quickly the air in the greenhouse will heat up because I don't know how quickly the ground will heat up.

The thinking of my model is that greenhouses are warmer than ambient because they eliminate the heat loss due to convection (due to air heating and moving away) and instead convert that to losses by conduction by the walls, which if the walls are slightly insulated is a slower process. But I can't figure out how quickly the air heats (and thus can't figure out the temperature of the walls of the greenhouse) without knowing how quickly the ground heats.

I know that after going some distance down into the earth the temp is constant, and I know that temperature, so I've toyed around with conductance, but my answers end up being ridiculous, so I know I'm doing something wrong here.

So in conclusion, how would one calculate the heating of the ground due to solar radiation of a known source? Assume I live at the equator when the sun is directly overhead and am only concerned with the T(t) of over day. I know that's a lot, and any tips help. Thanks for reading!


1 Answer 1


But I can't figure out how quickly the air heats (and thus can't figure out the temperature of the walls of the greenhouse) without knowing how quickly the ground heats.

This is going to step up the model complexity, as the ground is essentially a semi-infinite medium (also known as a half-space) with respect to building-scale heat transfer above it, and the nature of the infinite extent precludes using Fourier's law of conduction over a finite thickness.

For simplicity, let's idealize the problem as one-dimensional; that is, we'll ignore edge effects. (This idealization grows less appropriate for smaller ground areas, as lateral conduction through the ground ultimately predominates over vertical forcing. In that case of a small footprint, it might be best to set the ground temperature simply as the average ground surface temperature outside.)

The standard approach to solving 1D semi-infinite media problems, as described by Incropera & DeWitt, for example, is to define a similarity variable $\eta\equiv\frac{x}{(4\alpha t)^{1/2}}$, with depth $x$, ground thermal diffusivity $\alpha$, and time $t$. Note that

$$\frac{\partial T}{\partial x}=\frac{dT}{d\eta}\frac{\partial\eta}{\partial x}=\frac{1}{(4\alpha t)^{1/2}}\frac{dT}{d\eta};$$

$$\frac{\partial^2T}{\partial x^2}=\frac{d}{d\eta}\left(\frac{\partial T}{\partial x}\right)\frac{d\eta}{dx}=\frac{1}{4\alpha t}\frac{d^2T}{d\eta^2};$$

$$\frac{\partial T}{\partial t}=\frac{dT}{d\eta}\frac{\partial\eta}{\partial t}=-\frac{x}{2t(4\alpha t)^{1/2}}\frac{dT}{d\eta}.$$

We're going to use these relations in conjunction with the 1D conduction heat equation

$$\frac{\partial^2 T}{\partial x^2}=\alpha\frac{\partial T}{\partial t}$$ to give the ordinary differential equation

$$\frac{d^2 T}{d\eta}=-2\eta\frac{dT}{d\eta}.$$

Integrating twice, we obtain

$$T(\eta)=C_1\int_0^\eta \exp(-u^2)\,du+C_2=C_3\,\text{erf}(\eta)+C_2,$$

where $u$ is a dummy variable, $\text{erf}()$ represents the error function, and $C_1$, $C_2$, and $C_3$ are constants based on the boundary and initial conditions.

Any textbook or handbook that addresses this topic will give solutions for a constant surface heat flux, surface convection with a constant convection coefficient, etc. Your case involves both radiative and convective effects at the surface, and so I'd expect the solution to be fairly complex, even with the (possibly questionable) assumptions already made.

  • $\begingroup$ Thank you! I have to digest that but I'll get back to you on how it helps! $\endgroup$ Mar 13, 2023 at 5:30

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