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. – Jon Custer May 13 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? – Oliver Lines May 13 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. – Jon Custer May 13 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 May 13 at 13:52
• @uhoh I'm happy with how you edited it, thanks. – Oliver Lines May 14 at 14:44

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