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 Subject: RADIATION HEAT TRANSFER Category: Science > Technology Asked by: jasonwjs-ga List Price: \$50.00 Posted: 24 Sep 2002 01:52 PDT Expires: 24 Oct 2002 01:52 PDT Question ID: 68346
 ```1.A specific flat surface orbiting the earth is coated with black nickel. Assume the surface is a diffuse emitter with the plate is insulated on one surface and the other surface is exposed to solar radiation.The solar irration is 1350w/m^2.Find equilibrium Temperature?``` ```Hi jasonwjs, this is by no means an easy question you are asking! First I will have to make a few assumptions for the calculation to be possible at all: - Temperature in deep space usually only reaches 3°C above absolute zero. However, you are asking about an object in earth orbit. If I assume that it is actually orbiting the earth, (planetary and) solar radiation would heat up an object that perfectly absorbs and emits to 280° K (7° C) (or about 5° C if planetary radiation and shielding effects are taken into account) - which is what I will assume as your environmental temperature. ( http://www.faqs.org/faqs/astronomy/faq/part4/section-14.html ) - I will assume that your black nickel has an emissivity of 0.06 and an absorbtivity of 0.93 of the incoming radiation ( http://www.efdlant.navfac.navy.mil/criteria/documents/MH/1003_13A.PDF ) - I will also assume that one surface is hit by the radiation at a 90° angle. To calculate an equilibrium temperature we will have to use the Stefan-Boltzmann Law: Radiation Power ( * absporptivity ) = emissivity * Stefan's Constant * Radiation Area (Temperature of Radiator^4 - Temperature of surroundings^4) In your case we assumed emissivity as 0.06, absorptivity 0.93 , Stefan's Constant is 5.6703*10^-8 watt/((m^2)(K^4)), Radiation Area is assumed to be 1 m^2 for this calculation and Temperature of surroundings is 280°K. To achieve equilibrium we also have to assume that the emission of energy is equal to the solar radiation: 1350 watt/m^2. This brings us to the following equation: 1350 W/m^2 * 0,93 = 0.06 * 5.6703*10^-8 W/((m^2)(K^4)) * 1 m^2 * ( T^4 - 6146560000 K^4 ) If we bring T to the left side and all the other parts to the right, the equation looks like this: 1350 W/m^2 * 0.93 = 0.06 * 5.6703*10^-8 W*m^2/K^4 * T^4 - 0.06 * 6146560000 * 5.6703*10^-8 W*m^2 T^4 = ((0.93 * 1350 + 0.66 * 6146560000 * 5.6703*10^-8)/0.06 * 5.6703*10^-8) K^4 T^4 = (0.93 * 1350/(0.06 * 5.6703*10^-8) + 6146560000) K^4 T =~ 783° K However, be aware that if this was a homework question or similar, that you teacher might have assumed that the temperature in space is 2.7° K due to the famous background radiation only, without taking planetary and solar radiation into account for the calculation of surrounding temperature (it is very difficult to say what the surrounding temperature in a certain area of space really is). If you insert approximately 3° K into the above equation the result would be T =~ 779° The resulting temperatures seem to be very high - that is due to the black nickel coating being a nearly ideal absorber of solar radiation while being a much poorer emitter. I have also found an article describing the equilibrium temperature of ballistic missiles ( http://www.ucsusa.org/security/CM_apA-E.pdf ) where the scientists took the temperature of the surrounding out of the equation and instead integrated albedo effects into the equation. However, it is not so easy to use their equation because it requires a spherical rotating body. If you assume that your body is orbiting with one of its thin sides towards earth and in some distance to it, you could assume that the radiation from the planet is fairly small and can be ignored. In such a case you would just have to assume the surrounding temperature as zero and do the calculation with that in mind: T^4 = 1350 * 0.93/(0.06 * 5.6703*10^-8) K^4 T =~ 779 K Please request a clarification if you have any questions about the answer! Additional Resources Stefan-Boltzmann Law http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c3 Is Space Hot or Cold? http://spider.ipac.caltech.edu/staff/waw/mad/mad5.html Search Terms equilibrium temperature space diffuse ://www.google.com/search?sourceid=navclient&q=equilibrium+temperature+space+diffuse temperature of space ://www.google.com/search?sourceid=navclient&q=temperature+of+space absorption radiation "black nickel" ://www.google.com/search?sourceid=navclient&q=absorption+radiation+%22black+nickel%22 emissivity nickel ://www.google.com/search?sourceid=navclient&q=emissivity+nickel``` ```The answer which has been posted is appropriate to the question but ignores the bigger picture. In assuming an area of 1m2, voyager-ga has actually established an equation with different units on either side, the LHS is in W/m2 and the RHS in W. The bigger picture is to add to the LHS, the area of the body which is absorbing energy, and change the area on the RHS to the area which is emitting energy. In this example, they would be one and the same, hence use of 1m2 gives the right numerical answer, but for most real life cases, the two areas are different, with the emitting area larger than the absorbing area.``` 