Question 10.4: Obtain relations for the energy transfer in an enclosure of ...

Equation 10.72,

\sum\limits_{j=1}^{N}{}\left\lgroup \frac{\delta_{kj}}{\epsilon _{\lambda,k }}-F_{k-j}\frac{1-\epsilon _{ \lambda ,k}}{\epsilon _{ \lambda ,k}}\overline{t}_{\lambda ,k-j} \right\rgroup q_{\lambda,j } =\sum\limits_{j=1}^{N}{[(\delta_{kj} -F_{k-j}\overline{t}_{\lambda ,k-j} )E_{\lambda b,j}-F_{k-j}\overline{\alpha}_{\lambda ,k-j} E_{ \lambda b,g}]}                       (10.72)

applied to a two-surface enclosure, gives for k = 1 and 2 (note that F_{1–1} =F_{2–2} = 0):

\frac{1}{\epsilon _{\lambda ,1}} q_{\lambda ,1}-F_{1-2}\frac{1-\epsilon _{\lambda ,2}}{\epsilon _{\lambda ,2}} \overline{t} _{\lambda ,1-2}q_{\lambda ,2}=E_{\lambda b,1}-F_{1-2} \overline{t} _{\lambda ,1-2}E_{\lambda b,2}-F_{1-2}\overline{\alpha}_{\lambda ,1-2} E_{\lambda b,g}                                (10.75a)

-F_{2-1}\frac{1-\epsilon _{\lambda ,1}}{\epsilon _{\lambda ,1}} \overline{t} _{\lambda ,2-1}q_{\lambda ,1}+\frac{1}{\epsilon _{\lambda ,2}} q_{\lambda ,2}=-F_{2-1} \overline{t} _{\lambda ,2-1}E_{\lambda b,1}-F_{2-1}\overline{\alpha}_{\lambda ,2-1} E_{\lambda b,g}+E_{\lambda b,g}                        (10.75b)

For infinite parallel plates, F_{1–2 }= F_{2–1} = 1 , and from Equations 10.69 and 10.70 \overline{t} _{\lambda ,2-1}=\overline{t} _{\lambda ,1-2} and \overline{\alpha } _{\lambda ,2-1}=\overline{\alpha} _{\lambda ,1-2} For simplicity, the numerical subscripts on \overline{t} and \overline{\alpha } are omitted. Then Equations 10.75a and 10.75b become

\frac{1}{\epsilon _{\lambda ,1}} q_{\lambda ,1}-\frac{1-\epsilon _{\lambda ,2}}{\epsilon _{\lambda ,2}} \overline{t} _{\lambda }q_{\lambda ,2}=E_{\lambda b,1}-\overline{t} _{\lambda }E_{\lambda b,2}-\overline{\alpha}_{\lambda } E_{\lambda b,g}          (10.76a)

-\frac{1-\epsilon _{\lambda ,1}}{\epsilon _{\lambda ,1}} \overline{t} _{\lambda }q_{\lambda ,1}+\frac{1}{\epsilon _{\lambda ,2}}q_{\lambda ,2} =\overline{t} _{\lambda }E_{\lambda b,1}+E_{\lambda b,2}-\overline{\alpha}_{\lambda } E_{\lambda b,g}              (10.76b)

Equations 10.76a and 10.76b are solved for q_{λ,1} and q_{λ,2}. After using the relation \overline{\alpha }_λ = 1− \overline{t } , this yields

q_{\lambda ,1}=\frac{1}{1-(1-\epsilon _{\lambda ,1})(1-\epsilon _{\lambda ,2})\overline{t}_{\lambda }^{2} }\left\{\epsilon _{\lambda ,1}\epsilon _{\lambda ,2}\overline{t}_\lambda (E_{\lambda b,1}-E_{\lambda b,2})+\epsilon _{\lambda ,1}(1-\overline{t}_\lambda )\times \left[1+(1-\epsilon _{\lambda ,2})\overline{t}_\lambda \right](E_{\lambda b,1}-E_{\lambda b,g}) \right\}                         (10.77a)

q_{\lambda ,2}=\frac{1}{1-(1-\epsilon _{\lambda ,1})(1-\epsilon _{\lambda ,2})\overline{t}_{\lambda }^{2} }\left\{\epsilon _{\lambda ,1}\epsilon _{\lambda ,2}\overline{t}_\lambda (E_{\lambda b,2}-E_{\lambda b,1})+\epsilon _{\lambda ,2}(1-\overline{t}_\lambda )\times \left[1+(1-\epsilon _{\lambda ,1})\overline{t}_\lambda \right](E_{\lambda b,2}-E_{\lambda b,g}) \right\}             (10.77b)

The total energy fluxes added to surfaces 1 and 2 are

q_1=\int_{\lambda =0}^{\infty }{q_{\lambda ,1}}d\lambda and q_2=\int_{\lambda =0}^{\infty }{q_{\lambda ,2}}d\lambda               (10.78)

The total energy added to the gas to maintain its temperature T_g is equal to the net energy leaving the parallel plates. Hence, per unit area of the plates,

q_g = −(q_1 + q_2 )          (10.79)

When the medium between the plates does not absorb or emit radiation, then \overline{t}_λ = 1 and Equations 10.77a and 10.77b reduce to Equation 5.5.

Q_{1 \xrightleftharpoons[]{} 2}=Q_{1\longrightarrow 2}-Q_{2\longrightarrow 1}=\sigma T_1^4A_1F_{1-2}-\sigma T_2^4A_2F_{2-1}            (5.5)

With an absorbing radiating gas, the numerical integration of Equations 10.77a and 10.77b over all λ to obtain q_1 and q_2 is difficult because of the very irregular variations of the gas absorption coefficient with λ; some methods for detailed integrations were discussed in Chapter 9.