Mission Design for the CubeSat OUFTI-1

CHAPTER 9where c p is the material’s heat capacity [ KJ ], k the conductivity [ W ], ⃗q theKgK mKheat flux and σ the energy generated inside the body.For a steady monodimentional problem without heat generation we have :⎧⎪⎨⎪⎩dqdx = −k d2 Tdx = 0T (0) = T 1T (L) = T 2(9.7)Solving the problem we have the following expression of the heat flux:Q 1→2 = qA = − T 2 − T 1LkAR eq = LkA(9.8)where A is the contact surface through which the heat can pass.As expected, the heat goes from the hotter body to the colder. We also havethe expression of the equivalent conductive resistance.The same can be done for the radiation. With reference to figure 9.2, a bodyis in thermal equilibrium with the heat flux Q , the irradiance G, the reflectedAirradiance G R and the radiant energy E.Figure 9.2: Equilibrium for radiative heat exchange{JA = Q + GAJ = E + G R = ɛE B + ρG = ɛE B + (1 − ɛ) G(9.9)where J is the radiosity and represent the total outgoing flux.We introduce the view factor F A,B , which represents the proportion of all theradiation which leaves surface A and strikes surface B for the first time, withoutmultiple reflection:Q 1→2 = F 1,2 A 1 (J 1 − J 2 ) (9.10)Galli Stefania 90 University of Liège