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

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

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

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

and the moment Equations 13.33 and 13.34 used for the P_1 approximation become

\sum\limits_{i=1}^{3}{\frac{\partial I^{(i)}}{\partial \tau _i} } =(1-\omega )(4\pi I_b-I^{(0)})                     (13.33)

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

\frac{dI^{(1)}}{d\tau _1}  =(1-\omega )(4\pi I_b-I^{(0)})                     (13.43)

\frac{d I^{(11)}}{d\tau _1} =-I^{(1)};     \frac{d I^{(12)}}{d\tau _1} =0;  \frac{d I^{(13)}}{d\tau _1} =0;                          (13.44)

The closure condition for P_1, Equation 13.38,

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

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

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

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

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

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}              (12.30)

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

\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}                        (12.46)

\sigma T^4(\tau _1)=-\frac{3q_r}{4}\tau _1+C             (12.47)

The boundary conditions are applied to relate the temperature distribution to the known temperatures and obtain the integration constant C. Measuring \tau_1     from the wall at T_{w1}, Equation 13.40 at this boundary becomes [using Equation 13.41 and Y_1^0 = (3/4π)^{1/2} cosθ]

\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 }{\rho (\Omega ,\Omega _i)I(\Omega _i)l_id\Omega _i}  \right]d\Omega  }                         (13.40)

\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                (13.41)

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

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

I(\tau _1=0,\theta_i ) =\frac{1}{4\pi }[I^{(0)}+3I^{(1)}\cos \theta_i ]= \frac{1}{4\pi }[4\sigma T^4(\tau _1=0)+3q_r\cos \theta_i ]

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

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

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

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

J(\tau _1=0)=\sigma T^4_{w1}-\frac{\left(1-\epsilon _w\right) }{\epsilon _w}q_r                  (13.50)

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

\sigma T^4(\tau _1=0)=\sigma T^4_{w1}-\left\lgroup\frac{1}{\epsilon _w}-\frac{1}{2}  \right\rgroup q_r          (13.51)

A similar analysis applied at wall 2 gives

\sigma T^4(\tau _1=\tau _D)=\sigma T^4_{w2}+\left\lgroup\frac{1}{\epsilon _w}-\frac{1}{2}  \right\rgroup q_r                         (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 \sigma T^4(\tau _1=0)=C , resulting in

\sigma T^4(\tau _1)=\sigma T^4_{w1}-\left\lgroup\frac{1}{\epsilon _w}-\frac{1}{2}  \right\rgroup q_r – \frac{3q_r}{4} \tau _1      (13.53)

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

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

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