Question

3 4. (30 pts) radiation; a cubical room 3 mx 3 mx 3 m is heated through the floor by maintaining it at a uniform temperature of 310K. Since the side walls are well insulated, the heat loss through them can be considered negligible. The heat loss takes place through the ceiling, which is maintained at 280K. All surfaces have an emissivity & =0.85. Answer each following question: Long #2 (50 min) ceiling Az = 9 м2 Tz = 280k side. Insulated Az = 36 m² - Floor. 4,-9 m2 Ti = 31ok YIK (i) (5pts) draw the equivalent thermal network (or circuit) with each resistant component (ii) (10pts) find the equivalent total resistance from the floor to the ceiling, using symbols such as A, E, F, etc. (iii) (5pts) find the view factors of F2, F12, (refer 0.07 to the figures) (iv) (10pts) determine the rate of heat loss by 008 0.02 radiation through the ceiling. T 02 03 05 10 YL01 2 345 1020 2 03 04 05 051 L XL Fig. 13-4 ZX Fig. 13-6

Answer 1

to the SolD : A cubical room is heated through floor & radrated to the other faces of the ceebes. The data for reedration heat loss from flow to other surface are gireen be low as, ceilling Az=gost 1 T2=280k E = €2=E3=0.85 side wall insulected Azzam? 3m x -----7 t floor pl ① T = Blok ! Alam The equivalent theral network leerth each resistance component. mm ummus Ommmmmmm Floor Ristel, R2 side cocells R=1-61 GIAL 1-63 Ceilling R3 = 1- ez R22 ERA FIRA EIAI Fiz Az E3A3 ② The equirealent total resistance from floor to the ceilling .: R=R1+R2 + R₂ = 1-EL tt t - EIAI FIZAI - E2 E A

3 The view factors of FI-2, 1-2 are. . Fint flzt flz=1 As, the surface o is a flat surface F1420 e from the figure gister in problems. Fig 13-4 given in the problem For X/L=1 & the Por X/L=1 graph, the shape factor is .. [F1-2= 0.2] u FI-3=1-0-2 [FI-320-8] 6 The rate of heat loss by radiation through the ceilting is, - The beat loss from the ceilling to the environment is calculated as, - Q = €2.6. Agzx T2" = 0.85 x 5.67X168 x 9x 2804 Q=2,666.101 klatts /