Question 11.3: Consider radiative cooling from within a translucent materia...
Consider radiative cooling from within a translucent material in the limit when heat conduction is negligible compared with radiation. A hot rectangular bar of emitting, absorbing, and scattering material with nonreflecting boundaries such as in Figure 11.10 is initially at uniform temperature T_i. It is suddenly placed in a very low-temperature environment. The medium is gray with constant radiative properties, scatters isotropically, has density ρ and specific heat c. Derive the energy relation to obtain the transient temperature distribution.
Using Equation 11.78
∇ ⋅ q_r=4\pi k\left\lgroup\frac{\sigma T^4}{\pi } -\overline{I} \right\rgroup =4 πβ(I-\overline{I} )=4\pi \beta\frac{k}{\sigma _s}\left\lgroup\frac{\sigma T^4}{\pi }-\widehat{I} \right\rgroup (11.78)
for ∇ ⋅q_r , the energy equation for T(x, y, t) is
\rho c\frac{\partial T(x,y,t)}{\partial t}=-4\beta \frac{k}{\sigma _s}[\sigma T^4(x,y,t)-\pi \widehat{I} (x,y,t)] ;T(x,y,t)=T_i (11.88)
This is solved simultaneously with the integral equation found by substituting I from Equation 11.87 into Equation 11.79.
4\overline{I}(x,y)=(b-y)\int_{x_0=0}^{d}{I(x_0,b)\frac{S_2[\beta \rho _0(x_0,b)]}{\rho _0^2(x_0,b)}dx_0 +(d-x)\int_{y_0=0}^{b}{I(d,y_0)\frac{S_2[\beta \rho _0(d,y_0)]}{\rho _0^2(d,y_0)}dy_0 +y\int_{x_0=0}^{d}{I(x_0,0)}\frac{S_2[\beta \rho _0(x_0,0)]}{\rho _0^2(x_0,0)}dx_0 +x\int_{y_0=0}^{b}{I(0,y_0)} \frac{S_2[\beta \rho _0(0,y_0)]}{\rho _0^2(0,y_0)}dy_0}+\beta \int_{x^*=0}^{d}{\int_{y^*=0}^{b}{\widehat{I}(x^*,y^*)\frac{S_1(\beta \rho ^*)}{\rho ^*} dx^*dy^* } } } (11.87)
\widehat{I}=\frac{1}{\beta } \left\lgroup k\frac{\sigma T^4}{\pi }+\sigma _s\overline{I} \right\rgroup =(1-\omega ) \frac{\sigma T^4}{\pi }+\omega \overline{I} (11.79)
The incoming intensities at the boundaries are zero as the surroundings are at low temperature and the boundaries are nonreflecting. Then
\widehat{I}(x,y,t)=\frac{k}{\beta }\frac{\sigma T^4(x,y,t)}{\pi }+\frac{\sigma _s}{4}\int_{x^*=0}^{d}{\int_{y^*=0}^{b}{\widehat{I}(x^*,y^*,t) \frac{S_1(\beta \rho^*)}{\rho ^*} }dx^*dy^* } (11.89)
where \rho^*=[(x-x^*)^2+(y-y^*)^2]^{1/2}. Equations 11.88 and 11.89 are placed in dimensionless form and solved numerically. Starting at t = 0, the T(x,y,0) in Equation 11.89 is set equal to T_i. Then the integral equation is solved numerically for \widehat{I}(x,y,0).The T(x,y,0) and \widehat{I}(x,y,0) are substituted into the right side of Equation 11.88, and the resulting temperature derivative is used to extrapolate forward in time to obtain T(x, y, Δt). With this temperature distribution, the \widehat{I}(x,y,Δt) distribution is found by iteration of Equation 11.89, and the process is continued to move forward in time. The numerical method is discussed in Chapter 13.