To remain at 0 K heat absorbed by the medium must be removed in the amount of. dq dx = q(l) q(0) dx. Q = [1 2E 3 (τ L )] σ(t T 4 2 ).

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1 294 RADIATIVE HEAT TRANSFER 13.7 Two infinite, isothermal plates at temperatures T 1 and T 2 are separated by a cold, gray medium of optical thickness τ L = L (no scattering). (a) Calculate the radiative heat flux at the bottom plate, the top plate, and the net radiative energy going into the gray medium, assuming that both plates are black. (b) Repeat (a), but assume that both plates have the same temperature T, and that both plates are gray with equal emittance ɛ (diffuse emission and reflection). For a nonscattering cold medium S (τ) = I b (τ) = 0, and equation (13.55) simplifies to q(τ) = 2J 1 E 3 (τ) 2J 2 E 3 (τ L τ). (a) If the walls are black, J = σt 4 and q(τ) = 2σT 4 1 E 3(τ) 2σT 4 2 E 3(τ L τ), or q(0) = σt 4 1 2σT 4 2 E 3(τ L ), q(l) = [ σt 4 2 2σT 4 1 E 3(τ L ) ]. To remain at 0 K heat absorbed by the medium must be removed in the amount of L 0 dx = L 0 dx = q(l) q(0) dx = [1 2E 3 (τ L )] σ(t T 4 2 ). This amount is negative since energy must be removed to keep the medium cold. (b) If the medium is gray, and both plates have identical conditions, we must replace σt1 4 = σt 2 4 by J w, and q(0) = q(l) = J w [1 2E 3 (τ L )]. The radiosity is determined from equation (13.44) as J w = σtw 4 ɛ 1 q(0) = σtw 4 1 [1 2E 3 (τ L )] J w, ɛ 1 J w = σt 4 w 1 + (1/ɛ 1) [1 2E 3 (τ L )]. q(0) = q(l) = 1 2 = [1 2E 3 (τ L )] σt 4 w 1 + (1/ɛ 1) [1 2E 3 (τ L )].

2 312 RADIATIVE HEAT TRANSFER Consider a space enclosed by infinite, diffuse-gray, parallel plates 1 m apart filled with a gray, nonscattering medium ( = 5 m 1 ). The surfaces are isothermal (both at T w = 500 K with emittance = 0.6), and there is uniform and constant heat generation within the medium per unit volume, = 10 6 W/m 3. Conduction and convection are negligible such that q =. Determine the radiative heat flux to the walls as well as the maximum temperature within the medium, using the diffusion approximation with jump boundary conditions. For radiative equilibrium we have and, using symmetry q = dz =, Ψ = or q / = τ 1 2 τ L. = dτ, Using the diffusion approximation, equation (14.13) leads to q = 4π 3 di b dτ, or Ψ = 4 dφ b 3 dτ, Φ b = E b J w /. ( dφ b dτ = 3 4 Ψ = 3 τ 1 ) 4 2 τ L, Φ b = C τ(τ L τ). The boundary condition is, from equation (14.25), or E b (0) J w = 2 de b 3 dτ (0) 1 d 2 E b 2 dτ 2, Φ b (0) = 2 3 Φ b (0) 1 2 Φ b (0), C = τ L 1 ( 2 3 ) = τ L 4. Φ b (τ) = τ L [1 + τ(τ L τ)]. The radiosity is eliminated from Φ b using equation (13.44), i.e., J w = E bw 1 q(0), Φ = E b E bw / = Φ b + J w E bw / ( 1 = 1 2 = Φ b 1 Ψ(0) ) [1 + τ(τ L τ)].

3 CHAPTER The maximum temperature occurs at the center, τ = τ L /2: ( 1 Φ max = 1 ) τ2 L. 8 4 Substituting the given values: τ L = 5 m 1 1 m = 5, 5 Φ max = ( ) = 5.635, 8 4 E bmax = E bw + Φ max Tmax 4 = Tw 4 + Φ max σ = 10 6 W/m W/m 3 K = K 4 q wall = Ψ(0) T max = 2113 K = W/m 3 5 m 1 = W/m 2.

4 314 RADIATIVE HEAT TRANSFER Do Problem using the Schuster-Schwarzschild approximation. For radiative equilibrium we have and, using symmetry q = dz =, Ψ(τ) = or dτ =, q / = τ 1 2 τ L. For the 2-flux approximation we have from equations (14.35) through (14.37) dτ = 4E b G =, 4E b = ( dg = 4q = 4 τ 1 ) dτ 2 τ L. G = and, using the boundary condition at τ = 0, G(0) + 2q(0) = 4J w = E b = [C + 2τ(τ L τ)], C = τ L + 4J w / + G, [ ( C )] 2 τ L. [ τ L ] 2 τ(τ L τ) + J w, Φ b = E b J w / = τ L τ(τ L τ). Eliminating the wall radiosity from Φ b using equation (13.44) gives, J w = E bw 1 q(0) Φ = E b E bw / = Φ b + J w E bw / = ( = Φ b 1 Ψ(0) ) τ(τ L τ). The maximum temperature occurs at the center τ = τ L /2: ( 1 Φ max = 1 ) τ2 L 8.

5 CHAPTER Substituting the given values: τ L = 5 m 1 1 m = 5, 5 Φ max = = 6.292, E bmax = E bw + Φ max, Tmax 4 = Tw 4 + Φ max σ = 10 6 W/m W/m 3 K = K 4, q wall = Ψ(0) T max = 2172 K, = W/m 3 5 m 1 = W/m 2.

6 316 RADIATIVE HEAT TRANSFER Do Problem using the Milne-Eddington approximation. For radiative equilibrium we have and, using symmetry q = dz =, Ψ(τ) = or dτ =, q / = τ 1 2 τ L. For the differential approximation equations (14.43) through (14.45) are applicable, and dτ = 4E b G =, 4E b = ( dg = 3q = 3 τ 1 ) dτ 2 τ L. G = and, using the boundary condition at τ = 0, G(0) + 2q(0) = 4J w = E b = [ C + 3 ] 2 τ(τ L τ), C = τ L + 4J w / + G, [ ( C )] 2 τ L,. [ τ L ] 8 τ(τ L τ) + J w, Φ b = E b J w / = τ L τ(τ L τ). Eliminating the wall radiosity from Φ b using equation (13.44) gives J w = E bw 1 q(0), Φ = E b E bw / = Φ b + J w E bw / = ( = Φ b 1 Ψ(0) ) τ(τ L τ). The maximum temperature occurs at the center τ = τ L /2: Φ max = 1 ( ) τ2 L.

7 CHAPTER Substituting the given values: τ L = 5 m 1 1 m = 5, Φ max = = 5.510, 32 E bmax = E bw + Φ max, Tmax 4 = Tw 4 + Φ max σ = 10 6 W/m W/m 3 K = K 4, q wall = Ψ(0) T max = 2101 K, = W/m 3 5 m 1 = W/m 2.

8 CHAPTER Consider two parallel black plates both at 1000 K, which are 2 m apart. The medium between the plates emits and absorbs (but does not scatter) with an absorption coefficient of = cm 1 (gray medium). Heat is generated by the medium according to the formula = CσT 4, C = cm 1, where T is the local temperature of the medium between the plates. Assuming that radiation is the only important mode of heat transfer, determine the heat flux to the plates. For the given problem τ L = cm cm = , and we may expect the P 1 - approximation to give accurate results. The governing equations for a one-dimensional slab are from equations (15.34), (15.36) and (15.46), and applying the condition of radiative equilibrium with internal heat generation, dz = (4σT 4 G) = = CσT 4, dg dz = 3q, z = 0 : 2q = 4σT 4 w G, z = L : 2q = 4σT 4 w G. Solving the first equation for G leads to G = 4σT 4 C ( σt 4 = 4 C ) σt 4, dg (4 dz = C ) d dz (σt 4 ), and d 2 q dz 2 = C d ( ) 3q dz (σt 4 ) = C = 3C 4 C/ 4 C 2 q. With τ = z and 3C/(4 C) = /( ) = this becomes subject to d 2 q q = 0, dτ2 ( τ = 0 : 2q = 4σTw 4 4 C ) σt 4 = 4σTw 4 ( 4 ) C 1 dτ (0), and a similar boundary condition at τ = τ L ; however, the second boundary condition may be replaced by q(τ L /2) = 0, using symmetry. q(τ) = C 1 sin 0.1 ( τ 1 2 τ L) + C2 cos 0.1 ( τ 1 2 τ L), where the argument τ 1 2 τ L has been chosen to exploit symmetry, leading to C 2 = 0. From the first boundary condition: ( ) 4 τ = 0 : 2C 1 sin 0.05τ L = 4σTw 4 C 1 0.1C 1 cos 0.05τ L C 1 = = 4σT 4 w 0.1(4/C 1) cos 0.05τ L 2 sin 0.05τ L W/m 2 0.1( / ) cos( ) 2 sin( ) = W m 2

9 340 RADIATIVE HEAT TRANSFER q(0) = C 1 sin( ) = 4539 W/m 2.

10 354 RADIATIVE HEAT TRANSFER 16.3 Consider parallel, black plates, spaced 1 m apart, at constant temperatures T 1 and T 2. Due to pressure variations, the (gray) absorption coefficient is equal to = x; 0 = 0.01 cm 1 ; 1 = cm 2, where x is measured from plate 1. The medium does not scatter radiation. Determine, for radiative equilibrium, the nondimensional heat flux Ψ = q/σ(t 4 1 T 4 2 ) by the exact method, and the S 2- approximation. Defining dτ = dz leaves the governing equations in the same form as for a constant, regardless of the method employed, with τ L = L 0 (z) dz = (a) Exact: From Table 13.1 Ψ = L 0 ( z) dz = 0 L L 2 = = 2. (b) S 2 -approximation: From Example 16.1, with J i = σt 4 i and A 1 = 0, Ψ = 2µ 1 + τ L /2µ, where µ = for the nonsymmetric, and µ = 1/ 3 for the symmetric S 2 -approximation. Ψ nonsymmetric = Ψ symmetric = 1 = τ L 3 = , 1 3/2 + 3 /4 = Note that neither S 2 -approximation is as accurate at the P 1 -approximation for which (Example 14.5) Ψ P1 = τ L = (However, the nonsymmetric S 2 - approximation is more accurate for small τ L ).