Question

Near the equator, the Sun is approximately straight overhead at noon, and around 1000 W/m2 solar irradiance reaches Earth. Around this time of day, assuming ambient temperature of 40C, and a 10m x 10m x 2m-tall greenhouse with ¼”-thick glass walls that reflect away 20% of the incoming sunlight, and only thinking of conductive losses, estimate the maximum temperature reached in the interior of the greenhouse. Hint: when are losses and sources balanced?

glass conductivity is 1.05 W/(m k)

Answer 1

As : the area of one face of the (cubic) greenhouse

Jsun=1000W/(m2): the amount of solar power incident on the top surface of the greenhouse

Tatm = 313.15 K is the atmospheric temperature

Tg is the equilibrium temperature of the greenhouse (the answer to our question), once it has risen to a temperature where the power being radiated out is equal to the power being radiated in.

So at any rate, assuming:

1.) There is no energy transfer other than sunlight coming in, blackbody radiation from the atmosphere coming in, and blackbody radiation escaping from the glass of the greenhouse.
2.) There is no conductive heat loss to ground, or convective heat loss (hot air escaping out of the top)

we can start with:

Power in = Power out

Power in from sunlight + Power in from atmospheric blackbody radiation = Power radiated out from greenhouse+ conductive heat loss

Now:

Power In=

Jsun ∗   Ag         +     σ * (313.15)⁴ *(4*(10*2)+(10*10)

=1000(10*10) + (5.67*10^-8 * (313.15)⁴ *180

=198144.33 W

Power Out=

σ * (Tg)⁴ *(4*(10*2)+(10*10) + Conductive heat loss

= (5.67*10^-8 * (Tg)⁴ *180) + k/t A (Tg-313.15)

= (5.67*10^-8 * (Tg)⁴ *180) + [(1.05/0.00635) * 180 * (Tg-313.15)]

Using Pin = Pout

1020*10^-8 * Tg⁴ + 29763.78*( Tg-313.15) = 198144.33

Solving For Tg

Tg=316.37 = 43C

In case the value is incorrect lemme know would recheck the equation