I thought I'd calculate the greenhouse's innate heating capability from solar radiation for fun.

Let the greenhouse have dimensions of 3 m x 2 m, with an eaves height of 2m, and a roof slope of 1:2, giving a ridge height of 2.5 m. This yields a surface area for the walls up to the eaves of (2 * (3 m * 2 m)) + (2 * (2 m * 2 m)) = 20 m^2, and to the ridge of (2 * ((2 m * 0.5 m) / 2) = 1 m^2. The roof has a surface area of (2 * (3 m * sqrt(1^2 + 0.5^2)m)) ~= 6.71 m^2, for a total of ~= 27.7 m^2. Assume 95% (arbitrarily decided) is glass, so 26.3 m^2.

Short-wave transmissivity of typical single-pane glass is 88% [0] at a 0-degree angle of incidence. Plain glass has a refractive index of approximately 1.5, and air is 1.0. Using Snell's Law and Fresnel's Laws, we can calculate the transmissivity for realistic angles. The geographic center of the continental U.S. [1] will see a maximum solar elevation of 73.61 deg during the summer solstice, and a maximum of 26.76 deg during the winter solstice. Using Snell's Law, solving for the refracted angle yields arcsin(sin((90 - 73.61) deg)/1.5) ~= 10.84 deg and arcsin(sin((90 - 26.76) deg)/1.5) ~= 36.53 deg, respectively. These are for the sides of the building, and I assume that each side gets equal amounts of light at complementary angles throughout the day.

Re-doing these calculations for the roof's slope yielded 29.21 deg and 0.13 deg for refracted angles.

Using Fresnel's Laws (or a calculator for them [2]) we get 95.9% and 96.0% for light entering the roof, and 77.4% and 95.9% for light entering the walls. I'll use the average of each of these for yearly, so 95.95% for the roof, and 86.65% for the walls.

An online solar insolation calculator [3] with all losses turned off yielded an average daily maximum of 8.708 kWh/m^2/day, or 0.36283 kW/m^2, and an average daily minimum of 2.437 kWh/m^2/day, or 0.10154 kW/m^2. The average of these is thus 0.23219 kW/m^2.

The roof has a glass area of 6.38 m^2, and the walls have 19.92 m^2. Summing the product of insolation, area, and transmissivity together, we get 5.438 kW entering the greenhouse. The greenhouse has a volume of (3 m * 2 m * 2 m) + (3 m * (2 m * 0.5 m / 2)) = 13.5 m^3. Assuming dry air at 20 C, this is a delta-T/sec of (5438 W * 1 sec) / (13.5 m^3 * 1.2 kg/m^3 * 1005 J/(kg * K)) = 0.334 degC/sec.

Standard greenhouse glass seems to be 3mm thick, giving it a heat transfer coefficient (assuming no coatings) of 6.3 W/(m^2 * delta-degC). Thus, the combined surface area is losing 165.69 W * delta-degC. Assuming the outside temperature was 0 C, and the internal temperature was 20 C, that's 3312 W. The two forces on their own would reach equilibrium at 5438 W / 165.69 W * delta-degC = 32.82 delta-degC. This ignores transpiration from the plants inside, heat absorption characteristics of various things inside the greenhouse, etc.

This paper [4] has quite a bit of detail on parts I elided.

[0]: https://greenhouse.hosted.uark.edu/Unit03/Section02.html

[1]: https://geohack.toolforge.org/geohack.php?pagename=Geographi...

[2]: https://www.lasercalculator.com/fresnel-reflection-and-trans...

[3]: https://www.fabhabs.com/solar-insolation-calculator

[4]: https://faculty.ksu.edu.sa/sites/default/files/lmhdr_lthlth_...