A circular disk (ignore its thickness) of D = 1 m diameter is placed over a barbecue pit maintained at T₁ = T2 = 600 K. The upper surface of the disk is exposed to quiescent ambient air and surroundings at 300 K. The bottom surface and top surface of the disk has emissivities of εd.1 = 1.0 and d2 = 0.9, respectively. The bottom and cylindrical side surface of the barbecue pit have emissivities of 81 = 0.6 and ɛ2 = 1.0, respectively. Assume the disk and barbecue pit surfaces are diffuse and gray. A) Determine the net heat transfer rate to the cover, qnet,d, when Ta= 500 K. B) Plot qnet,d as a function of the disk temperature for 300 K < Ta< 600 K, using Excel or MATLAB. What is the steady-state temperature of the disk? g Quiescent Air Tsur T300 K Disk- -Ed.2 = 0.9 Ed,1 = 1.0 -T₂, ₂ = 1 T₁, ₁ = 0.6; L= 0.5 m D=1m

2_5190874259647370... A cylindrical shape is with diameter of 1.2m and length of 1.2m. The top surface has an emissivity of 0.7 and temperature 500 K. The bottom surface has an emissivity of o.5 and temperature 650 K. The side surface has an emissivity of 0.4. Determine the shape factors for all surfaces. E 0.0 L=650 K II >

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A composite wall is comprised of two large plates separated by sheets of refractory insulation. In the installation process, the sheets of thickness L = 50 mm and thermal conductivity k = 0.05 W/mK are separated at 1-m intervals by gaps of width w = 10 mm. The hot and cold plates have temperatures and emissivities of T1 = 400 deg C, emissivity1 = 0.85 and T2 = 35 deg C, emissivity2 = 0.5, respectively. Assume that the plates and insulation are diffuse-gray surfaces. %3D Determine the heat loss by radiation through the gap per unit length of the composite wall (normal to the page). Recognizing that the gaps are located on a 1-m spacing, determine what fraction of the total heat loss through the composite wall is due to transfer by radiation through the insulation gap. Hot side Gap w = 10 mm A. 47 W/m, 9.2% T1 = 400°C B. 47 W/m, 10.2% L = 50 mm C. 37 W/m, 10.2% D. 37 W/m, 9.2% T2 = 35°C Cold side 1 m Insulation, k = 0.05 W/m-K