Question

*Problem 13.056 An electric furnace consisting of two heater sections, top and bottom, is used to heat treat a coating that is applied to both surfaces of a thin metal plate inserted midway between the heaters. Heater - Plate Sidewalls L L L LLLLLLLLL ||||||||| Heater The heaters and the plate are 2 m x 2 m on a side, and each heater is separated from the plate by a distance of L = 0.5 m. Each heater is well insulated on its back side has a temperature of 880 K, and has an emissivity of 0.65 at its exposed surface. The plate and the sidewalls have emissivities of 0.6 and 0.3, respectively. Additionally, the sidewalls have a temperature of 400 K. For the prescribed conditions, obtain the required electrical power, in kW, and the plate temperature, in K. 1 kW Yele c = Tplate =

Answer 1

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Eron Solution / L = 0.5m no The Heater and the plater are amx am le A = Az = 4 m2 A2 = (0.5x2) x4 = 4 m2 The plate Emissvities (E3) = 0.6 The sidewall Emissivites (E.)= 013 heater sill temperature (T) =880k Sielewell temperaturu (TJ) = 400K El = 0.9 1- El SAL Jimmy 1-E2 Agez Altin Jo M в Л МИ — Ebi Jz = Eb3 - Symmetry of swem About plate only one-half Need be consider plate is adiabatic

I from figure 7 = x = 4 F13 = 0.62 F 2 = 1 - F,3 = 0,38 F32 = F2 = 0.38 Hence, A, F12 = 4X (0.38) = 1152m² A fiz = 4 X (0.62) = 2.48m2 Az F23 = 48(0.38) = 1.52m² (1-Ey) - 0.7 = 01583m-2 & A2 - EL - - 0.0278m 2 0.6 A, EI Esi = TTY Ebi = 5.67X10-8 X (880)4 Ebi = 34002.727 W/m2

Eb2 = TTY = 5,67710-8X (400)" Eby 1451.52 W/m2 Calculate The Net Radiation Transfor from Single Heater is 9, Ebi Eba I-EL + - A, F12 + A + AIE, I +I-E2 A 3 F32) Az Ez TAIF substituling valus, 34002, 927 - 1451.52 + 0.583 0.0278 t - 1.52 + 1 tt 2.48 1.52 32551. 207 0.99823

19, = 326099kw Thus, The furnance es 32.6089kw. lower Requirement Cele = 221 = 32.6089 q = Eb, - J, (l-Ei . AIE 1 - Eo, -2, (1- & ) [AEI = 34002-727 – 32608.9 x 010278 Jo = 33549,463 W/m2 Jp = Ebr 22 ( 1 & 2 = 145/152 - (- 32608.9 ) X 0.583 J2 = 10957.014 W/m2

J, - J3 Jz - Ja Ai fia A3F32 J- Ja Jg - Ja - Aulash A3 832 AF13 - 1152 2148 = 0.613 1.613 J3 = J, 70 1613 J2 1.613 J = 33549. 4637 0.613 8109571014 Js = 24.963.492 W/m2 J3 = Ebz T = E)

2 4963.492 5 67 8 10 8 3 7₂ = 814.574K Thus, I The Plate Temperature is 8141574 K