Radiation: the third of the loss nobody counted

The pipe also glows

• The bare pipe's 110 W/m counted only convection. A 60 °C surface also radiates, and leaving that out underestimates the loss by about a third.
• Every surface emits $\varepsilon\sigma T^4$, with emissivity $\varepsilon$ and absolute temperature $T$, and $\sigma = 5.67 \times 10^{-8}$ W/(m²·K⁴).

A network for radiation

• The view factor $F_{ij}$ is the fraction of radiation leaving $i$ that reaches $j$. For grey surfaces, each surface adds $(1 - \varepsilon)/(\varepsilon A)$ and each pair adds $1/(A_iF_{ij})$; the radiosity $J$ is the node between them.
• For large parallel plates: $q = \sigma(T_1^4 - T_2^4)/(1/\varepsilon_1 + 1/\varepsilon_2 - 1)$.

The bare pipe, honestly

A radiation shield

• An oven wall at 600 K faces a casing at 300 K, both $\varepsilon = 0.8$: $q = \mathbf{4593}$ W/m².
• A polished shield ($\varepsilon = 0.05$) between them adds $2/0.05 - 1 = 39$ to the sum: $q = \mathbf{170}$ W/m², 3.7% of before.

White in sunlight, black in infrared

• Emissivity depends on wavelength. White paint absorbs only about 20% of sunlight, which peaks near 0.5 µm, but emits about 90% at the 10 µm of room-temperature infrared.
• So a white roof stays cool in sun, and a polished metal roof (low emissivity) stays warm at night. Choose a surface for the wavelength it meets.

Worked example: the thermocouple that reads low

• A thermocouple in a duct of 200 °C gas, with walls at 100 °C, $h = 100$ and $\varepsilon = 0.8$, balances convection in against radiation out: $h(T_g - T) = \varepsilon\sigma(T^4 - T_w^4)$.
• Solving gives $T = \mathbf{188}$ °C: it reads 12 K low, with no fault in the sensor. A polished radiation shield around the tip cuts the error.

Three radiation slips

Recap