In a previous post I used statistical mechanics to show how you’d calculate the thermal conductivity of any gas and showed why the insulating power of the best normal insulating materials was usually identical to ambient air. That analysis only considered the motion of molecules and not of photons (black-body radiation) and thus under-predicted heat transfer in most circumstances. Though black body radiation is often ignored in chemical engineering calculations, it is often the major heat transfer mechanism, even at modest temperatures.

One can show from quantum mechanics that the radiative heat transfer between two surfaces of temperature T and To is proportional to the difference of the fourth power of the two temperatures in absolute (Kelvin) scale.

Heat transfer rate = P = A ε σ( T^4 – To^4).

Here, A is the area of the surfaces, σ is the Stefan–Boltzmann constant, ε is the surface emissivity, a number that is 1 for most non-metals and .3 for stainless steel.  For A measured in m², σ = 5.67×10⁻⁸ W m⁻² K⁻⁴.

Unlike with conduction, heat transfer does not depend on the distances between the surfaces but only on the temperature and the infra-red (IR) reflectivity. This is different from normal reflectivity as seen in the below infra-red photo of a lightly dressed person standing in a normal room. The fellow has a black plastic bag on his arm, but you can hardly see it here, as it hardly affects heat loss. His clothes, don’t do much either, but his hair and eyeglasses are reasonably effective blocks to radiative heat loss.

As an illustrative example, lets calculate the radiative and conductive heat transfer heat transfer rates of the person in the picture, assuming he has  2 m² of surface area, an emissivity of 1, and a body and clothes temperature of about 86°F; that is, his skin/clothes temperature is 30°C or 303K in absolute. If this person stands in a room at 71.6°F, 295K, the radiative heat loss is calculated from the equation above: 2 *1* 5.67×10⁻⁸ * (8.43×10⁹ -7.57×10⁹) = 97.5 W. This is 23.36 cal/second or 84.1 Cal/hr or 2020 Cal/day; this is nearly the expected basal calorie use of a person this size.

The conductive heat loss is typically much smaller. As discussed previously in my analysis of curtains, the rate is inversely proportional to the heat transfer distance and proportional to the temperature difference. For the fellow in the picture, assuming he’s standing in relatively stagnant air, the heat boundary layer thickness will be about 2 cm (0.02m). Multiplying the thermal conductivity of air, 0.024 W/mK, by the surface area and the temperature difference and dividing by the boundary layer thickness, we find a Wattage of heat loss of 2*.024*(30-22)/.02 = 19.2 W. This is 16.56 Cal/hr, or 397 Cal/day: about 20% of the radiative heat loss, suggesting that some 5/6 of a sedentary person’s heat transfer may be from black body radiation.

We can expect that black-body radiation dominates conduction when looking at heat-shedding losses from hot chemical equipment because this equipment is typically much warmer than a human body. We’ve found, with our hydrogen purifiers for example, that it is critically important to choose a thermal insulation that is opaque or reflective to black body radiation. We use an infra-red opaque ceramic wrapped with aluminum foil to provide more insulation to a hot pipe than many inches of ceramic could. Aluminum has a far lower emissivity than the nonreflective surfaces of ceramic, and gold has an even lower emissivity at most temperatures.

Many popular insulation materials are not black-body opaque, and most hot surfaces are not reflectively coated. Because of this, you can find that the heat loss rate goes up as you add too much insulation. After a point, the extra insulation increases the surface area for radiation while barely reducing the surface temperature; it starts to act like a heat fin. While the space-shuttle tiles are fairly mediocre in terms of conduction, they are excellent in terms of black-body radiation.

There are applications where you want to increase heat transfer without having to resort to direct contact with corrosive chemicals or heat-transfer fluids. Often black body radiation can be used. As an example, heat transfers quite well from a cartridge heater or band heater to a piece of equipment even if they do not fit particularly tightly, especially if the outer surfaces are coated with black oxide. Black body radiation works well with stainless steel and most liquids, but most gases are nearly transparent to black body radiation. For heat transfer to most gases, it’s usually necessary to make use of turbulence or better yet, chaos.

Robert Buxbaum