3. (25 pts) Consider a flat plate solar collector for water heating installed on a roof. The collector has a single layer of glass (glazing) above the absorber to retain heat. Glass cover Air space Absorber plate Insulation The geometry and operating conditions for the solar collector are summarized as follows The face area of the collector assembly is H= 1 m W = 1 m. The assembly is inclined 0 = 30 from the horizontal. Tips: The air gap between the absorber plate and the glass cover is L = 15 mm thick. The absorber plate is at a uniform temperature T = 60C and has emissivity &a=0.9. The glass cover plate is thin, and has an emissivity of &G = 0.5. The ambient air is blowing over the surface of the absorber at U = 2 m s with T% = 10C. a. (12 pts) Assume, for now, the glass cover is at uniform temperature TG = 30C. Calculate the natural convection and radiation heat transfer rates from the absorber plate to the glass cover. Review textbook section 9.8.1 for procedures and correlations to estimate natural convection in inclined rectangular enclosures See notes from class 32 for calculating radiation heat transfer between parallel plates b. (13 pts) Solve for the steady state temperature of the glass cover plate and the total heat loss rate through the front face of the solar collector. To do this, perform the following steps: Calculate the external forced convection and radiation heat transfer rate from the glass cover plate to the ambient (assuming TG = 30C) Set up a steady state energy balance for the glass plate: Qconv,AG + Qrad,AG = Qconv,G + Qrad,G0 Adjust TG until the above energy balance equation is satisfied. Make sure to update the radiation heat transfer rate and natural convection heat coefficient at each iteration. o If using EES, you can just "Update Guesses" (CTRL-G), add the energy balance equation, and comment out the guess for Ts.