Question

A small object with an opaque, diffuse surface at a temperature of 500 K is suspended in a large furnace with walls at 2000 K. Assume that the walls of the furnace provide a diffuse irradiation to the object at a blackbody temperature equal to the furnace wall temperature. The object’s surface has a spectral hemispherical emissivity and absorptivity as given below. (a) Determine the total emissivity and total absorptivity of the object’s surface. Partial Ans: ?=0.021 (b) Evaluate the net radiation heat flux at the surface. Ans: -18.1 kW/m2 (c) Sketch the spectral emissive power distribution over all wavelengths for the object (see figure on p. 799 of text, in Example 12.6), and also for a blackbody at the same temperature as the object.Ignore red dots in image

Problem 2 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 Emissity (5) or Absorptivity (0) ŏ 0.2 0.1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Wavelength, 1 (um)

Answer 1

(a) { = 0.8 (Fo»>, -f -) +0.2 (1 - Fe-x2) T Sook 7=7 ym. 2. T.: 3500 amk Fo», – 0.382671 12. - to 4m AT = 5000 unk fondą = 0.633747 E = 0.8 (0.633747-0.382671) +0.2 x *1-0.633741) E = 0.2741 -Ts=20ook TESOK Läs Granda (mbi I an Eu Xa Eba (2000k) da > a = Eb (2000K) * d = 0.8 (Fo»>,- fo2m) +0.2 (e-Fosna) d: 0.2 + 0.6 Fordą - 0.8 fondi T= 2000 K a, = 7.4m 2,7 = 14000 yok fornia 0.962898 ng 14m 12 T = 20,000 ymk For r2 = 0.985602 a= 0.2+0.6 x 0.985602 - 0.8%0.962898 x = 0.021

(b) Net radiation heat flux from the surface Cleaning sesface) 9" : €6500K) – OG 49" - =&& (500k) – 2 Eb ( 2000 k) 0.021 x 5-67 X10-8X2004 19" = 0.2741 x 5,67 X10-8 x 5004 q" - 18079.86 W/m2 19" , - 18.1 kW/m2 c) Amaxt = 2898 ymk 2898 = 5.8 4m A max = 2898 T 500 T = 500k Eba Ebx, Ex 1 Ex x=74m 22=104m. "max=5084m 2 →