A Sheet on Selective Surfaces (Updated: Oct. 4, 2011, BBB) Stefan's Law: Stefan's Law tells us about the power radiated by a body with area, A, and temperature, T. P/A = T4 where = 5.67 x 10-8 Wm-2K-4 and is the emissivity.

BbetteremitterABlack Body = 1The emissivity of the body, is a dimensionless number between 0 and 1 that tells how effectively this body radiates energy. A table of emissivity values from Kraushaar and Ristenen (1994) give a feeling for this variable. MaterialEmissivity (T ~ 300 0K)

Aluminum

Polished0.04

Rough plate0.06

Oxidized0.15

Cast iron0.50

Sheet steel0.7

Wood, black lacquer, white enamel, plaster, roofing paper0.9

Porcelain, marble, brick, glass, rubber, water0.94

Gold (at 100 0C)0.02

Good absorbers are good emitters: We know from Stefans Law above, that an object at temperature, T, emits energy in the form of electromagnetic radiation. How, then, do we select an object that has emissivity that is, one that emits energy really well? A perfect absorber is often called a black body. It absorbs all the radiation that hits it. Somewhat counter-intuitively, a good absorber is also a good emitter. That is, a really good absorber like the surface of a black road is also a really good emitter with an emissivity of close to 1. It is not hard to show that an ideal absorber of a particular wavelength of radiation is also the best possible emitter at that wavelength. That is, no object at temperature T can emit more radiation at wavelength than a black body.Proof: Take a look at the diagram above. Start, for example, with two objects A and B that are close to each other and at the same temperature. Suppose that A is an ideal black body and B is not. While A absorbs all radiation that hits it, B does not. It reflects some. The question is whether B can emit more radiation than A. Since A absorbs all of the radiation emitted by B it would get hotter than B if B really could emit more radiation than A. But, this situation is a direct violation of the Second Law of Thermodynamics. We are not allowed to start with two bodies at the same temperature and find that one heats while the other cools! This means that a black body, the perfect absorber is not only the best absorber but also the best emitter. In summary, excellent absorbers are also excellent emitters. That is:Example 1: Radiators. In the old days homes often used water or steam radiators to heat rooms. From a practical point of view, you really wouldn't want to make your radiator out of shiny aluminum. The low emissivity of aluminum means that it is both a poor emitter and a poor radiator and would give out less heat than one made of cast iron, for example. Example 2: Home Insulation. Aluminum foil on insulating panels has very small emissivity, It is therefore a very poor emitter of infrared radiation, a desirable feature. Example 2: Hot black roads. On a clear summer day a black asphalt road in the sun gets hot as it absorbs radiation from the sun. Most of this radiation has short wavelength as it comes from the sun's surface with a temperature of some 5800 K. To the extent that the hot black road is a "black body", it absorbs all the sun's incident radiation. It's emissivity is nearly 1 for this incident short wavelength light. The warm road also emits infrared radiation and continues to heat up until the power emitted, Pout = AT4 , balances the power absorbed from the sun, Pin = I0A. Here, I0 is the sun's intensity at the hot black road, typically 1000 W/m2. The hot black road's emissivity, , is also nearly 1 for these longer infrared wavelengths. With Pin = Pout we solve for T and find that a hot black road has a temperature of 364 K (91 0C), hot enough to fry an egg, but not hot enough to boil water! Example 3: You standing in a bathroom. With no clothes, taking your area to be ~2 m2, and your skin temperature to be ~300 K, with an emissivity of 1.0, you would radiate a power of Pout = AT4 = 919 watts, clearly an unsustainable value. In empty space you would indeed radiate heat away at this value. However, suppose you are in a bathroom with walls at 20 0C (293 K), ones with their own emissivity of 1. Then, the net heat radiated by you is given by Pout = A(Tyou4 - Twall4) = 83 watts, a much more reasonable number. The general expression for power exchanged between two parallel surfaces with emissivities and temperatures {1, T1} and {2, T2} is P = A(T14 - T24)/ [1/1 + 1/2 - 1]. Ref: Kraushaar & Ristinen, Energy and Problems of a Technological Society, p 156. With this equation, you can calculate the power exchange between two surfaces with different emissivities and temperatures. Building materials with aluminum foil come to mind. Selective Surfaces: Interestingly, the emissivity values of many materials change with wavelength of the radiation being emitted. For example, silicon is an excellent emitter of visible light, but is essentially transparent to infrared radiation. We find below that good emitters are also good absorbers. That is:

InfraredIt would be great fun to find a way to create a surface that could get hotter than a hot black road in the sun, maybe even one that could boil water. To do this we need either to absorb more of the sun's radiation coming in or emit less. We assumed that our hot black road was a perfect absorber of short wavelength light (short wavelength = 1) and a perfect emitter of infrared (long wavelength = 1). If we make the emissivities less, then we reduce both the absorbed radiation from the sun and the radiated radiation from the road. Are we stuck? The trick lies in creating a surface that has short wavelength > long wavelength . This does not violate the second law. To create our selective surface, we start with a layer of stainless steel and add a thin layer of gold and, on top of that, a thin layer of silicon. The silicon layer looks black to visible light and has short wavelength ~ 1. Since silicon is essentially transparent to infrared light, our selective surface behaves as a gold surface for infrared. Gold has an emissivity of only 0.10 for infrared wavelengths. This combination, then, is an excellent absorber of short wavelength light from the sun and a poor emitter of infrared light. Repeating our calculation, we find that this selective surface can rise to a temperature of 648 K (375 0C)!

thin layer of goldstainless steel basepowered silicon

References: Kraushaar, J.J. & Ristinen, R.A., Energy and Problems of a Technical Society, 2nd Edition, pp. 155-158, Wiley, New York. Romer, R., Energy An Introduction to Physics, pp. 380, 534-538, Freeman, San Francisco, 1976. ********************************* End Selective Surfaces *************************