Using the P1 approximation, derive relations for the temperature distribution and energy transfer between parallel plane walls at Tw1 and Tw2 for the limit of radiative equilibrium without internal heat sources. Each wall has the same diffuse-gray emissivity ϵw,

Question

Using the [latex]P_1[/latex] approximation, derive relations for the temperature distribution and energy transfer between parallel plane walls at [latex]T_{w1} [/latex] and [latex]T_{w2} [/latex] for the limit of radiative equilibrium without internal heat sources. Each wall has the same diffuse-gray emissivity [latex]\epsilon _w [/latex], and they are separated by an emitting, absorbing, and isotropically scattering medium with albedo ω and optical thickness [latex] au _D [/latex] based on the spacing between the walls.

Accepted Answer

Because the geometry requires the radiative energy flux to be only in the [latex] au _1 [/latex] direction, it follows that the first moments (equal to the fluxes) [latex]i^{(1)}=0 [/latex] for j = 2, 3. Then Equation 13.31 for the [latex]P_1[/latex]

[latex]I(S, heta ,\phi ) =\frac{1}{4\pi }[I^{(0)}+3I^{(1)}\cos heta +3I^{(2)}\sin heta \cos \phi +3I^{(3)}\sin heta \sin \phi ] [/latex] (13.31)

[latex]I(S, heta ,\phi ) =\frac{1}{4\pi }[I^{(0)}+3I^{(1)}\cos heta ] [/latex] (13.42)

and the moment Equations 13.33 and 13.34 used for the [latex]P_1[/latex] approximation become

[latex]\sum\limits_{i=1}^{3}{\frac{\partial I^{(i)}}{\partial au _i} } =(1-\omega )(4\pi I_b-I^{(0)}) [/latex] (13.33)

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

[latex]\frac{d I^{(11)}}{d au _1} =-I^{(1)}; [/latex] [latex] \frac{d I^{(12)}}{d au _1} =0; [/latex] [latex] \frac{d I^{(13)}}{d au _1} =0; [/latex] (13.44)

The closure condition for [latex]P_1[/latex], Equation 13.38,

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

gives [latex]I^{(11)} =(1/3)I^{(0)} [/latex] and [latex]I^{(12)} =I^{(13)}=0. [/latex] Substituting into the second-moment differential equations (13.44) gives

[latex]\frac{1}{3}\frac{dI^{(0)}}{dk_1}=-I^{(1)} [/latex] (13.45)

Now, because [latex]I^{(1)} = q_r [/latex], and [latex]q_r[/latex] is constant in this geometry for radiative equilibrium with [latex]\dot{q}=0 [/latex], it follows from Equation 13.43 that [latex]I^{(0)}=4\pi I_b=4\sigma T^4( au _1) . [/latex] Substituting this into Equation 13.45 gives

[latex]q_r=-\frac{4\sigma }{3} \frac{dT^4}{dk_1} [/latex] (13.46)

This is the same relation as for the diffusion solution Equation 12.30.

[latex]q(x) =-\frac{4\sigma }{3\beta (x)} \frac{d(T^4)}{dx}d\lambda =-\frac{16}{3\beta (x)} \sigma T^3\frac{dT}{dx} [/latex] (12.30)

However, for other geometries the [latex]P_1[/latex] solution does not generally provide the diffusion result. Integrating Equation 12.46 results in a linear [latex]T^4[/latex] distribution in the medium

[latex]\frac{\partial^2 E_b(r)}{\partial x^2}=-\frac{3\beta Q_1}{16\pi } \frac{(x^2+y^2+z^2)^{3/2}-3x^2(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^3} [/latex] (12.46)

[latex]\sigma T^4( au _1)=-\frac{3q_r}{4} au _1+C [/latex] (12.47)

The boundary conditions are applied to relate the temperature distribution to the known temperatures and obtain the integration constant C. Measuring [latex] au_1 [/latex] from the wall at [latex]T_{w1}[/latex], Equation 13.40 at this boundary becomes [using Equation 13.41 and [latex]Y_1^0 = (3/4π)^{1/2} cosθ] [/latex]

[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)=2\pi \int_{ heta =0}^{\pi /2}{I_0} ( au _1=0)\cos heta \sin heta d heta =2\epsilon _w\sigma T^4_{w,1}\int_{ heta =0}^{\pi /2}{\cos heta \sin heta d heta }+2(1-\epsilon _w)\int_{0}^{\pi /2}{\left[2\pi \int_{0}^{\pi /2}{I( au _1=0, heta _i)}\cos heta _i\sin heta _id heta _i ight] }\cos heta \sin heta d heta [/latex] (13.48)

The incident intensity [latex]I(τ_1 = 0, θ) [/latex] in Equation 13.48 is expressed by Equation 13.42, so that, from the moments that have been found,

[latex]I( au _1=0, heta_i ) =\frac{1}{4\pi }[I^{(0)}+3I^{(1)}\cos heta_i ]= \frac{1}{4\pi }[4\sigma T^4( au _1=0)+3q_r\cos heta_i ][/latex]

Substituting into the boundary condition, Equation 13.48 gives for the diffuse boundary

[latex]J( au _1=0)=\epsilon _wT^4_{w1}+\frac{(1-\epsilon _w)}{2}\int_{ heta _i=\pi }^{\pi /2}{[4\sigma T^4( au _1=0)+3q_r\cos heta _i]}\cos heta _i\sin heta _id heta _i [/latex] (13.49)

From the net radiation expression Equation 5.17 at wall 1 (Note: [latex] q_r = J – G = q_1) [/latex]

[latex]J_k=\sigma T^4_k-\frac{1-\epsilon _k}{\epsilon _k}q_k [/latex] or [latex]\sigma T^4_k=\frac{1-\epsilon _k}{\epsilon _k}q_k +J_k [/latex] (5.17a,b)

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

Integrating Equation 13.49 and using Equation 13.50 to eliminate J results in

[latex]\sigma T^4( au _1=0)=\sigma T^4_{w1}-\left\lgroup\frac{1}{\epsilon _w}-\frac{1}{2} ight group q_r [/latex] (13.51)

A similar analysis applied at wall 2 gives

[latex]\sigma T^4( au _1= au _D)=\sigma T^4_{w2}+\left\lgroup\frac{1}{\epsilon _w}-\frac{1}{2} ight group q_r [/latex] (13.52)

Equations 13.51 and 13.52 are the same as the diffusion solution Equation 12.41. Equation 13.51 is used to evaluate the constant in Equation 13.47 where [latex]\sigma T^4( au _1=0)=C [/latex], resulting in

[latex]\sigma T^4( au _1)=\sigma T^4_{w1}-\left\lgroup\frac{1}{\epsilon _w}-\frac{1}{2} ight group q_r - \frac{3q_r}{4} au _1 [/latex] (13.53)

Evaluating Equation 13.53 at [latex]τ_1 = τ_D, [/latex] substituting in Equation 13.52 to eliminate [latex]T^4(τ_1 = τ_d), [/latex] and solving the result for [latex]q_r[/latex] yields

[latex]\frac{q_r}{\sigma (T^4_{w1}-T^4_{w2})}=\frac{1}{3 au _D/4+2/\epsilon _w-1} [/latex] (13.54)

This expression for [latex]q_r[/latex] can be substituted into Equation 13.53 to obtain [latex] T^4(τ_1) [/latex] in terms of the known boundary conditions (see Table 13.2). This completes the solution.

TABLE 13.2 [latex]P_1[/latex] Approximations for Energy Transfer and Temperature Distribution for a Gray Translucent Medium in Radiative Equilibrium (Absorption, Emission, Isotropic Scattering) between Gray Surfaces; [latex]\dot{q}=0 [/latex]
Geometry	Relationsª
Infinite parallel plates	[latex]\psi =\frac{1}{(3\beta D/4)+E_1+E_2+1} [/latex] [latex]\phi (z) =\psi \left[\frac{3\beta }{4}(D-z)+E_2+\frac{1}{2} ight] [/latex]
Infinitely long concentric cylinders	[latex]\psi =\frac{1}{\frac{3}{8} \beta D_1\ln \left\lgroup\frac{D_2}{D_1} ight group+\left\lgroup E_1+\frac{1}{2} ight group+\frac{D_1}{D_2}\left\lgroup E_2+\frac{1}{2} ight group } [/latex] [latex]\phi (r)=\psi\left[-\frac{3}{8}\beta D_1\ln \left\lgroup\frac{D}{D_2} ight group +\left\lgroup E_2+\frac{1}{2} ight group \frac{D_1}{D_2} ight] [/latex]
Concentric spheres	[latex]\psi =\frac{1}{\frac{3}{8} \beta D_1 \left\lgroup 1-\frac{D_1}{D_2} ight group+\left\lgroup E_1+\frac{1}{2} ight group+\frac{D_2}{D_1}\left\lgroup E_2+\frac{1}{2} ight group } [/latex] [latex]\phi (r)=\psi\left[-\frac{3}{8}\beta D_1 \left\lgroup\frac{D1}{D_2} -\frac{D_1}{D} ight group +\left\lgroup E_2+\frac{1}{2} ight group \frac{D^2_1}{D^2_2} ight] [/latex]
ªDefinitions: [latex] E_N=(1-\epsilon _{wN})/\epsilon _{wN},\psi =Q_1/A_1\sigma (T^4_{w1}-T^4_{w2}),\phi (\xi )=[T^4(\xi )-T^4_{w2}]/(T^4_{w1}-T^4_{w2}) ,D=2r,\beta =k+\sigma [/latex] and for [latex]\phi (r) [/latex] the κ > 0.

Question 13.3: Using the P1 approximation, derive relations for the tempera...

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