A plane layer of radiating and isotropically scattering medium with constant thermal conductivity k, constant radiative properties, optical thickness τD, and albedo ω is between infinite parallel

Question

plane layer of radiating and isotropically scattering medium with constant thermal conductivity k, constant radiative properties, optical thickness [latex]τ_D[/latex], and albedo ω is between infinite parallel diffuse-gray walls of emissivity ϵw at temperatures [latex]T_{w1} [/latex] and [latex]T_{w2} [/latex]. The medium has a uniform internal heat source [latex]\dot{q}[/latex]. Derive the relations needed to obtain the heat flux to each bounding wall, and the temperature distribution in the medium using the [latex]P_1[/latex] approximation.

Accepted Answer

With conduction, radiation, and internal heat generation, the energy equation (Equation 10.1)

[latex] ho c_p\frac{DT}{Dt}=\beta T\frac{DP}{Dt}+ abla . (k abla T-q_r)+\dot{q} +\phi _d[/latex] (10.1)

is [latex] abla .(k abla T-q_r)+\dot{q}=0[/latex] or, using a summation form for the radiation and conduction fluxes,

[latex]\sum\limits_{i=1}^{3}{\frac{\partial q_{r,i}}{\partial x_i} } +\sum\limits_{i=1}^{3}{\frac{\partial q_{e,i}}{\partial x_i}}=\dot{q} [/latex] (13.55)

For the present one-dimensional problem, since [latex]q_r = I^{(1)} [/latex], where [latex] I^{(1)} [/latex] is the first moment of the intensity, the energy equation can be put in the dimensionless form

[latex]\frac{d \widetilde{I} ^{(1)}}{d au _1} =4N\frac{d^2\vartheta }{d au ^2_1} +\frac{\dot{S} }{ au _D} [/latex] (13.56)

where [latex]N =k (κ + σ_s )/4σ T^3_{w1},\dot{S} =\dot{q} D/\sigma T^4_{w1}, au _1=(k+\sigma _s)x_1, au _D=(k+\sigma _s)D,\widetilde{I} ^{(1)}=I^{(1)}/\sigma T^4_{w1}, [/latex] and [latex]ϑ = T/ T_{w1}.[/latex] This is the defining equation for the derivative of [latex] I^{(1)} [/latex]. In the pure radiation solution of Example 13.3 without internal heat sources, the derivative of the first moment in Equation 13.43

[latex]\frac{dI^{(1)}}{d au _1} =(1-\omega )(4\pi I_b-I^{(0)}) [/latex] (13.43)

was set equal to zero, because for these conditions [latex]\widetilde{I}^{(1)}=q_r/\sigma T^4_{w1} [/latex] is constant and [latex]{d\widetilde{I}^{(1)}}/{d au _1}=0 [/latex].This is not the case here with conduction and an internal heat source included. The presence of a second derivative of dimensionless temperature requires two boundary conditions for solving the energy equation: [latex]\vartheta (τ = 0) = 1 [/latex] and [latex]\vartheta (τ_1 = au _D) = T_{w2}/T_{w1} [/latex].

To proceed with the solution, two coupled second-order differential equations are derived for[latex]\widetilde{I}^{(0)} [/latex] and[latex]\widetilde{I}^{(1)} [/latex] (the nondimensional intensity integrated over all solid angles, and the radiative flux). The first is obtained by equating Equation 13.56 and the first-moment differential equation (Equation 13.43) in the [latex]P_N[/latex] method. For the present one-dimensional problem, this yields

[latex]\frac{4N}{1-\omega } \frac{d^2\vartheta }{d au ^2_1}+\frac{\dot{S} }{ au _D(1-\omega )}-4\vartheta ^4=-\widetilde{I} ^{(0)} [/latex] (13.57)

where [latex]\widetilde{I} ^{(0)}=I^{(0)} /\sigma T^4_{w1} [/latex]. The second equation is found by substituting the closure Equation 13.38 into the second-moment differential Equation 13.34 to obtain the relation between [latex]\widetilde{I} ^{(0)} [/latex] and [latex]\widetilde{I} ^{(1)} [/latex]

[latex]I^{(ij)} =\frac{1}{3}\delta _{ij}I^{(0)} [/latex] (13.38)

[latex]\sum\limits_{i=1}^{3}{\frac{\partial I^{(ij)}}{\partial au _i} } =-I^{(j)} [/latex] (3 equations: j=1,2,3) (13.34)

[latex]\frac{d\widetilde{I} ^{(11)}}{d au _1} =\frac{1}{3}\frac{d\widetilde{I}^{(0)} }{d au _1} =-\widetilde{I}^{(1)} [/latex] (13.58)

Now, Equation 13.58 is differentiated with respect to [latex]τ_1[/latex], and the result is substituted into Equation 13.56 to yield

[latex]\frac{d^2 \widetilde{I} ^{(0)}}{d au^2 _1} +12N\frac{d^2\vartheta }{d au ^2_1} +\frac{3\dot{S} }{ au _D}=0 [/latex] (13.59)

Equations 13.57 and 13.59 can be solved simultaneously for [latex] \widetilde{I} ^{(0)} [/latex] and ϑ.They can be combined into a single fourth-order equation in ϑ by differentiating Equation 13.57 twice with respect to[latex] τ_1[/latex] and the result is substituted into Equation 13.59 to eliminate the second derivative of [latex] \widetilde{I} ^{(0)} [/latex]. The resulting fourth-order equation, or the two second-order equations, require two boundary conditions in addition to the known boundary surface temperatures. These are generated from the conditions in Equations 13.40 and 13.41 using Equation 13.50 to eliminate J, which results in

[latex]\int_{\Omega =0}^{2\pi }{I_0(\Omega )} Y_l^m(\Omega ) d\Omega =\int_{0}^{2\pi }{\left[\epsilon (\Omega )I_{b,w}+\frac{1}{\pi }\int_{\Omega _i=0}^{2\pi }{ ho (\Omega ,\Omega _i)I(\Omega _i)l_id\Omega _i} ight]d\Omega } [/latex] (13.40)

[latex]\int_{\Omega =0}^{2\pi }{I_0(\Omega )} Y_l^m(\Omega ) d\Omega =\int_{\Omega =0}^{2\pi }{I_0(\Omega ) l_i^{2n-1}d\Omega } = \int_{\Omega =0}^{2\pi }{I_0(\Omega )l_id\Omega }=j [/latex] (13.41)

[latex]J( au _1=0)=\sigma T^4_{w1}-\frac{\left(1-\epsilon _w ight) }{\epsilon _w}q_r [/latex] (13.50)

[latex]\frac{1}{4} \widetilde{I}^{(0)} ( au _1=0)=1-\left\lgroup\frac{1}{\epsilon _w}-\frac{1}{2} ight group \widetilde{I}^{(1)} ( au _1=0)[/latex]
[latex]\frac{1}{4} \widetilde{I}^{(0)} ( au _1= au _D)=\vartheta ^4_{w2}+\left\lgroup\frac{1}{\epsilon _w}-\frac{1}{2} ight group \widetilde{I}^{(1)} ( au _1= au _D)[/latex] (13.60)

or

[latex]\frac{1}{4} \widetilde{I}^{(0)}_i =\vartheta ^4_{wi}\pm \left\lgroup\frac{1}{\epsilon _w}-\frac{1}{2} ight group \widetilde{I}^{(1)}_i [/latex] (13.61)

where the i subscript denotes walls 1 or 2 and the positive sign applies at wall i = 2. Inserting Equation 13.58 to eliminate [latex]\widetilde{I}^{(1)}[/latex] results in the final boundary relations for [latex]\widetilde{I}^{(0)}[/latex]

[latex]\pm \left\lgroup\frac{d\widetilde{I}^{(0)} }{d au _1} ight group_i=-\frac{3}{4\left\lgroup\frac{1}{\epsilon _w} -\frac{1}{2} ight group }[\widetilde{I}_i^{(0)} -4\vartheta ^4_{wi}] [/latex] (13.62)

These can be directly applied as the boundary conditions for Equation 13.59. The problem is completely specified with two second-order differential equations, that is, Equations 13.57 and 13.59 and the four boundary conditions, the specified boundary temperatures for Equation 13.57, and the conditions (i = 1, 2) in Equation 13.62 for Equation 13.59. An iterative numerical solution can be used to obtain ϑ(τ ) and [latex]\widetilde{I}^{(0)}_i ( au _1)[/latex]. Then [latex]\widetilde{I}^{(1)}[/latex] at the boundaries (i = 1, 2) is found from Equation 13.61, which gives the desired radiative fluxes at the boundaries. The total energy transfer also requires the amount of heat conduction. Since the temperature distribution has been determined, this can be found by evaluating −kdT /dx at the boundaries.

Question 13.4: A plane layer of radiating and isotropically scattering medi...

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