Question 11.4: For the transient cooling problem in Example 11.3, derive an...
For the transient cooling problem in Example 11.3, derive an expression for the instantaneous emittance of the rectangular region based on its instantaneous heat loss and mean temperature.
The emittance is obtained from the local heat fluxes leaving the region boundaries. For this example the externally incident intensities are zero, so only the volume integral in Equation 11.82 is needed. Along the upper boundary,
q_r(r) =\int_{A}^{}{I(r_0)[n.(r-r_0)]\frac{r-r_0}{\left|r-r_0\right|^4 }e^{-\beta \left|r-r_0\right| }dA+\beta \int_{v}^{}{\widehat{I}(r^*) }\frac{r-r^*}{\left|r-r^*\right|^3 }e^{-\beta \left|r-r^*\right| } dV^* } (11.82)
q_r(x,b)=2\beta \int_{x^*=0}^{d}{\int_{y^*=0}^{b}{\int_{z^*=0}^{\infty }{\widehat{I}(x^*,y^*) }\frac{(b-y^*)e^{-\beta [(x-x^*)^2+(b-y^*)^2+z^{*2}]^{1/2}}}{[(x-x^*)^2+(b-y^*)^2+z^{*2}]^{3/2}} } }dz^*dx^*dy^* (11.90)
Using the same transformation as in Equation 11.84, the integration over z* is expressed as an S_2 function,
\int_{\xi =0}^{\infty }{\frac{e^{-\beta \rho ^*(1+\xi ^{*2})^{1/2}}}{\rho ^{*2}(1+\xi ^{*2})} \rho ^*d\xi ^*}=\frac{1}{\rho ^*}\int_{t=1}^{\infty }{\frac{e^{-\beta \rho ^*t}}{t(t^2-1)^{1/2}}dt } =\frac{1}{\rho ^*} \frac{\pi}{2} S_1(\beta \rho ^*) (11.84)
q_r(x,b)=\pi \beta \int_{x^*=0}^{d}{\int_{y^*=0}^{b}{{\widehat{I}(x^*,y^*) }\frac{(b-y^*)}{\rho _1^{*2}} } }S_2(\beta \rho _1^*)dx^*dy^* (11.91)
where \rho _1^*=[(x-x^*)^2+(b-y^*)^2]^{1/2} . Similarly, along boundary x = d,
q_r(d,y)=\pi \beta \int_{x^*=0}^{d}{\int_{y^*=0}^{b}{{\widehat{I}(x^*,y^*) }\frac{(d-x^*)}{\rho _2^{*2}} } }S_2(\beta \rho _2^*)dx^*dy^* (11.92)
where \rho _2^*=[(d-x^*)^2+(y-y^*)^2]^{1/2} . The overall emittance of the rectangle will be based on its instantaneous mean temperature
T_m(t)=\frac{1}{bd}\int_{x=0}^{d}{\int_{y=0}^{b}{T(x,y,t)}dxdy } (11.93)
The total heat loss Q(t) is obtained by integrating the local q_r over their respective boundaries around the rectangle and using symmetry of the upper and lower boundaries and the two vertical sides. Then the overall emittance is
\epsilon (T)=\frac{Q(t)}{\sigma T_m^4(t)}=\frac{2}{\sigma T_m^4(t)}\left[\int_{0}^{d}{q_r}(x,b)dx+\int_{0}^{d}{{q_r}(d,y)dy} \right] (11.94)