Question

PLEAS SHOW ALL WORK, Explain your steps for part b pleas!

12.50 A small, opaque, diffuse object at T, = 400 K is sus- pended in a large furnace whose interior walls are at T; = 2000 K. The walls are diffuse and gray and have an emissivity of 0.20. The spectral, hemispherical emissivity for the surface of the small object is given below. 0.7 0.5 w 1 3 a (um) (a) Determine the total emissivity and absorptivity of the surface. (b) Evaluate the reflected radiant flux and the net radiative flux to the surface.

Answer 1

Given Data: To = 4ook. TE = 2000k 0.7 E = 0.20 0.5 1 3 (Mm) a) Total Emissivity and al alsoppitivity of Susface : d, Ts = 1(400) => 400 Mmok de Tg = 3(400) 1200 Umok Now, From Hack body radiction functions faible; → for 400 m.k Floxlum) for 1200 Unik, Floum) = 0.002. Therefore, Total emissnity of the Surface § Ex (2) Exo (d, Tsjad :: El Ts) = Et (T) = { [Flossen) - Florrun) +86-F70-34m) =0:7(0.002 -1) +0.5 (1-0.002) → E(Ts) = 0.5 Surface is 0:5 • Emissivity of

And now, A, Tf = 1 (2000) -> 2000Mm.k. 12 T = 3 (2000) Good Mmok. From Hackbody radiation functions table : for 2000 Mnak, Floqun) -0.067. for Good look, F(+3.44) = 0.738. Therefore, Total absorptivity of Surface: 1 x (X) Gx di a = & Gà (A) da Fax (2) Exo (d. Tp) da U Eb (TF) = 2, [fro-Ban) - Form)] +az (o-F10-rsam) -0.7 (0-738 -0.067) +0.5(1-0.738) ..d = 0.60). 5 Absorptivity of Surface a = 0.601

6) Reflected Radiative fiux: .: Goef P. G. = (1-2) ELTA = (1-0.601) (5.67x1681d/m24) (2000k)4. = 3.62 +105 W/m² Geref Net Radiative flux: - ::g" = G- PG - EESTS = 26Tf EETS = (567x16€ well-264 ]-[0.Goi(2000K)4_0.5(4001434] q" 5.348x 105 W/m. 1.7 .*. Net radiative flux 2

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