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Space Blankets

U.S. Army Pfc. Robbin M. Chambers, a driver for 2nd Platoon, Alpha Company, 2nd Battalion, 162nd Infantry, 41st Infantry Brigade Combat Team, Oregon Army National Guard, opens a space blanket during a training exercise Oct. 24, 2014 in Kabul, Afghanistan.

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Thermal Radiation

Space blankets, (a.k.a survival blankets) such as the one seen in the previous image, are very thin and have a thermal conductivity roughly 5x greater than air^[2], therefore they are a poor a preventing heat loss by conduction. However, they do significantly reduce heat transfer by thermal radiation, which is the spontaneous emission of electromagnetic radiation by objects with temperature above absolute zero (so everything). Electromagnetic radiation sounds like a big deal, but its really just the descriptive scientific way to say light waves. Depending on the temperature of the object, the emitted light may not be visible to us, but it’s there nonetheless. For example, your body emits thermal radiation that we cannot see, but thermal imaging cameras can detect such light and allow us to “see” objects that are at a different temperature from their environment even when no external light source is present.

Thermograph of a persons face produced by a thermal imaging camera. The camera represents the intensity and color of detected thermal radiation that humans can’t see with variations in visible color to create a false-color image. The actual light emitted by the person and detected by the thermal imaging camera has a color that we cannot see. Image Credit: “Self Portrait with Thermal Imager” by Nadya Peek via Wikimedia Commons

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Space blankets reflect the electromagnetic radiation emitted by your body back to you, rather than letting it escape, thereby reducing the rate at which your body loses thermal energy to the environment. Thermal radiation transfer is the reason why clear nights feel colder than cloudy ones and why you frost forms on top of your car, but not on the ground beneath it. In order to explain these observed phenomenon and quantify heat loss from your body, we need to take a deeper look at thermal radiation.

Stephan-Boltzmann Law

Some materials are more efficient at converting thermal energy to light than others and this material property, known as the emissivity ( $epsilon$ ), affects the thermal radiation power ( $P_{out}$ ). The radiated power also depends on the object’s surface area ( $A$ ) and temperature in Kelvin ( $T_o$ ). The Stephan-Boltzmann Law relates the radiated power to all of these variables and the Stephan-Bolztmann constant ( $\sigma = 5.67 \times 10^{-8} \,\bold{W/(m K^4)}$ ):

(1) $\begin{equation*} P_{out} = \sigma \epsilon AT_{o}^4 \end{equation*}$

Materials which are good at converting thermal energy into light will also be good at the reverse process of absorbing light and converting it to thermal energy in the material. As a result, the rate of heat transfer into an object by radiation ( $\text{[math]}$ " title="Rendered by QuickLaTeX.com" height="13" width="75" style="vertical-align: -2px;">) can also be modeled using the Stephan-Bolztmann Law with the same emissivity value, only the incoming radiation power is determined by the temperature of the environment ( $T_{env}$ ) rather than the object.

(2) $\begin{equation*} P_{in} = \sigma \epsilon AT_{env}^4 \end{equation*}$

Exercises

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Net Thermal Radiation Rate

Subtracting the emitted radiation power from the absorbed radiation power we can determine the net radiation power to the object:

(3) $\begin{equation*} P_{net} = \sigma \epsilon A(T_{env}^4-T_{o}^4) \end{equation*}$

Notice that when the object is warmer than its environment, $P_{net}$ will be negative because more radiation will be leaving the object than is absorbed.

Everyday Example: Space Blankets

Let’s evaluate the effectiveness of adding a space blanket during the wilderness survival example from the previous chapter. The situation was a 25 °F (-3.9 °C) day with a 10 mph wind and you are thin clothes that don’t stop the wind very well. Layering the space blanket on top should cut the wind, so right off the bat you save most of the 1100 W of heat loss rate due to forced convection (wind chill effect) that we calculated in the last chapter. The blanket will reduce conduction somewhat by trapping a layer of air, but natural convection within that layer will move heat to the blanket where it will be conducted across, so you still experience much of the 160 W conductive heat loss we calculated previously.

A space blanket would effectively eliminate the thermal radiation heat loss by reflecting your emitted radiation back to you. Even though you will still transfer thermal energy to the inside of the blanket by conduction and natural convection from the inside, the blanket will do a poor job of radiating that energy away to outside because it has a relatively low emissivity.

Let’s figure out your heat loss rate without the space blanket in order to see what heat loss it actually saves you. To make the calculation easier, let’s assume a there is a layer of low clouds or heavy forest vegetation so that very little of the cold upper atmosphere is visible. In that case, the overall environmental temperature is just the -3.9 °C air temperature. We know body temperature 37 °C , but before we can calculate the net heat loss due rate to thermal radiation we must convert our temperatures to Kelvin:

$\begin{equation*} T_{env} = -3.9\,\bold{^\{circ}C} + 273.15 = 269.25\,\bold{K} T_{o} = 37\,\bold{^\{circ}C} + 273.15 = 310.15\,\bold{K} \end{equation*}$

Now we can work to apply the Stephan-Boltzmann Law for net radiation power:

$\begin{equation*} P_{net} = \sigma \epsilon A(T_{env}^4-T_{o}^4) \end{equation*}$

Using the methods in Chapter 17 we estimate the surface area of the upper body to be $\approx 1\,\bold{m^2}$ . The typical emissivity of human skin is 0.985^[4]

$\begin{equation*} P_{net} = (5.67 \times 10^{-8} \,\bold{W/(m K^4)}) (0.985) (1\,\bold{m^2})(269.25^4-310.15^4) \approx 200\,\bold{W} \end{equation*}$

We find that the rate of radiative heat loss would be approximately 200 W without a space blanket. Therefore the space blanket saves you 200 W of radiative heat loss and 1100 W of convective heat loss, leaving only the 160 W of conductive heat loss. We see that a space blanket can significantly reduce heat loss rate is some situations. Considering this benefit compared to its small weight and volume, a space blanket seems like a reasonable addition to a survival kit. However, a space blanket will not serve as a substitute for appropriate clothing. A typical person has a resting thermal power of roughly 100 W, therefore the person in our example would still have a 60 W thermal power deficit. Over time resulting energy loss would lower the body temperature until hypothermia triggered a shivering response, which could boost the thermal power by up to 2.5 times, or up to 250 W.^[5] This strategy would only work short term, until the person was too tired to shiver. Alternatively, if the person in this example had gotten wet while wearing cotton then the resulting rate of heat loss by conduction would be roughly 1100 W (calculated in the Cotton Kills chapter) and shivering would not be able to make up for the thermal power deficit, even in the short term. Even shivering would not significantly delay a dangerously low body temperature in the wet cotton situation.

Reinforcement Exercises: Space Walk

Astronaut Sunita L. Williams, Expedition 14 flight engineer, participates in the mission’s third planned session of extravehicular activity (EVA) as construction resumes on the International Space Station. Astronaut Robert L. Curbeam, (out of frame), STS-116 mission specialist, also participated in the 7-hour, 31-minute spacewalk. Image Credit: “Sunita Williams astronaut spacewalk” by NASA, via Wikimedia Commons

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An interactive or media element has been excluded from this version of the text. You can view it online here:
https://openoregon.pressbooks.pub/bodyphysics/?p=2788

We have a complication to mention: many materials have different emissivity at different frequency,which is the property of light that we perceive as colors. If the fraction electromagnetic radiation reflected by an object is the same at all visible frequencies, the object is gray; if the fraction depends on the frequency, the object has some other color. For instance, a red or reddish object reflects red light more strongly than other visible frequencies and because it absorbs less red, it radiates less red when hot. Therefore its emissivity would be lower at frequencies we see as red. Differential reflection and absorption of frequencies outside the visible range have no effect on what we see, but they may have physically important effects with regard to thermal radiation. Skin is a very good absorber and emitter of infrared radiation, having an emissivity of 0.97 in the infrared. This high infrared emissivity is why we can so easily feel infrared radiation from a campfire on warming our skin, but also why our bodies readily lose thermal energy by infrared radiation. OpenStax University Physics, University Physics Volume 2. OpenStax CNX. Nov 12, 2018 http://cnx.org/contents/7a0f9770-1c44-4acd-9920-1cd9a99f2a1e@14.10[/footnote]

Thermal Radiation Spectra

TBH...just interesting health-y books

Notes

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