Calculating radiative heat transfer between air and walls, what to use for the emissivity of air?

For example, say I have heated some air, I push it through a tunnel with cold black walls. Will the air radiate heat to the walls or will the walls radiate heat to the air?

I understand that the wall would both radiate and absorb radiation, as it has an emissivity and area and a temperature. But I am not sure how to approximate the emissivity of the air.

For example, if I wanted to calculate the radiation from air as follows

$$Q_{rad} = \sigma \epsilon AT^4$$

where $$\epsilon$$ is emissivity, $$\sigma$$ is Stefan-Boltzmann constant, $$A$$ is radiation area and $$T$$ is temperature, what would I use as the emissivity for air?

• Well, they both radiate, and given time will come into thermal equilibrium. Although, in your example, conduction and convection will likely be better means of heat transfer. Commented May 13, 2019 at 13:31
• Yeah, i understand that the wall would radiate as it has an emissivity and area and a temperature. But i wasn't sure how to approximate the emissivity of air, for example if i wanted to calculate the radiation from air as follows > Q_rad = EstefA*T^4, where E is emissivity, stef is stefan boltzmann constant, a is radiation area and T is temperature. What would i use as the emissivity?
– olly
Commented May 13, 2019 at 13:36
• For the purposes of planetary energy balances, the Earth's atmosphere, considered as one layer, has an effective emissivity of ~0.8. Commented May 13, 2019 at 13:52
• @JonCuster in order to assign an emissivity in the thermal infrared band you need a layer of air so thick (and humid) that it can absorb and re-emit photons a few times, i.e. opaque for at least some wavelengths associated with $k_BT$. For this question the length scale is short and so you probably can not use those assumptions, so the concept of emissivity will probably not apply.
– uhoh
Commented May 13, 2019 at 13:52
• @uhoh I'm happy with how you edited it, thanks.
– olly
Commented May 14, 2019 at 14:44

Basically: you shouldn't use an emissivity in this situation because the most appropriate model is $$Q_\text{rad} = 0$$.