Example: Heat Transfer - radiation

Understanding the situation

In our first example we considered the flow of heat by conduction. Convection is a little harder to model and calculate, especially for moving organisms -- and it depends on the local situation of air or water flow. But radiation is fairly clean to calculate since the Stefan-Boltzmann equation (see Heat transfer) gives an upper limit for the electromagnetic radiation energy that a warm body emits.

Presenting a sample problem

Suppose you are designing clothing for a polar scientist. You have material that has an R value* of 0.6 K-m²/W. If you make clothing out of this stuff do you have to worry about heat loss due to electromagnetic radiation? Or if you have handled the conduction loss will that be good enough? You want the clothing to be effective at -20 ^oC.

Solving this problem

Even though it doesn't say so explicitly, this sounds like an estimation problem. We don't know how big the explorer is or what his temperature is. So let's treat it like one.

To solve this, we need to estimate the rates of conductive heat loss compared to the radiative heat loss. The rate of loss per unit area due to conduction expressed in terms of the R value was given in the page Example: Heat Transfer - conduction. The result was

$$\Phi = \frac{A \Delta T}{R}$$

To calculate the full heat loss due to conduction we need the area, $A$, the temperature difference, $\Delta T$, and $R$. $R$ is given in the problem; we can estimate the body temperature of the scientist to be 30 ^oC (assuming he is healthy); we only have to estimate his area. Before we do that, let's look at the heat loss due to radiation.

[Be careful! In this example $R$ does NOT stand for the radius, but for the "R value" — the thermal heat flow parameter.]

The energy loss flux per unit area and the total flux for an area $A$ due to radiation are given by 

$$J = \sigma T^4$$

$$\Phi = JA = A\sigma T^4$$

We have to compare the two total fluxes. We know $\sigma$ (5.67 x 10^-8 W/m²K⁴) and the absolute temperature of a person is about 37 ^oK or  $T$ = 310 ^oK. We only don't know the area so that's all we have to estimate.

But wait! We can estimate his surface area pretty easily — say by approximating him by a cylinder. But the problem doesn't ask us for an absolute value of the thermal energy emitted (though that might be an interesting question). It just asks which is more important. Since both fluxes are proportional to the same A, we only have to compare $ΔT/R$ and $σT^4$. Let's do it. (Note the size of a Celsius and Kelvin degree are the same so in doing $\Delta T$ we can report it in Kelvin degrees.)

$$\frac{\Delta T}{R} = \frac{50 \mathrm{\;K}}{0.6 \mathrm{\;K m^2/W}} = \frac{50 \mathrm{\;K}}{(3/5) \mathrm{\;K m^2/W}} = \frac{250}{3} \frac{\mathrm{W}}{\mathrm{m^2}} \approx 80 \frac{\mathrm{W}}{\mathrm{m^2}} $$

$$\sigma T^4 = \bigg(5.67 \times 10^{-8} \frac{\mathrm{W}}{\mathrm{m^2 K^4}}\bigg)(3.1 \times 10^2 \mathrm{\;K})^4 \approx 524 \frac{\mathrm{W}}{\mathrm{m^2}}$$

Whoa! A bit surprising! The radiation loss is almost 6 times bigger. We're not quite done, though. We have to remember that our object is not just emitting radiation into its environment, it's also absorbing radiation from its environment. But since the temperature of the environment is less, the amount absorbed will be less than the amount emitted. It's the difference that we need to calculate.

Since the temperature of the environment is -20 ^oC or about 253 ^oK, we get that the absorbed radiation will be

$$\sigma T^4 = \bigg(5.67 \times 10^{-8} \frac{\mathrm{W}}{\mathrm{m^2 K^4}}\bigg)(2.5 \times 10^2 \mathrm{\;K})^4 \approx 222 \frac{\mathrm{W}}{\mathrm{m^2}}$$

The difference is still about 300 W/m², almost 4 times greater than the heat lost to conduction.

That says you definitely have to worry about the radiation properties of your clothing. You don't want your scientist to be a perfect radiator in a polar environment. 

* See the Wikipedia article on Clothing Insulation. Note that the clothing industry uses two units for the R value. The "clo" = 0.155 K-m²/W and the "tog" = 0.1 K-m²/W. 

Joe Redish 11/24/14