Consider emission from a long cylindrical hole drilled into a material at uniform temperature $T_w$
(Figure 5.17a). The hole is long, so the portion of its internal surface at its bottom end can be neglected in the radiative energy balances. The outside environment is at $T_e.$

Question

Consider emission from a long cylindrical hole drilled into a material at uniform temperature [latex]T_w[/latex]
(Figure 5.17a). The hole is long, so the portion of its internal surface at its bottom end can be neglected in the radiative energy balances. The outside environment is at [latex]T_e.[/latex]

Accepted Answer

As in the previous discussion, the solution is obtained by using the reduced temperature [latex]T_r = (T_w^4 − T_e^4 )^{1/4}.[/latex] If a position is viewed at x on the cylindrical side wall, the energy leaving is J(x). An apparent emissivity is defined as [latex]\epsilon_a(x) = J(x)/σT_r^4[/latex] . The analysis determines how [latex]\epsilon_a(x)[/latex] is related to the surface emissivity [latex]\epsilon,[/latex] where [latex]\epsilon[/latex] is constant over the cylindrical side of the hole. The integral equation governing the radiation exchange within a hole was first derived by Buckley (1927, 1928) and later by Eckert (1935); both obtained approximate analytical solutions. Results were evaluated numerically by Sparrow and Albers (1960). Using the reduced temperature, the opening of the hole is approximated by a perfectly absorbing disk at zero reduced temperature. Then, from Equation 5.65b

[latex]\sigma T_k^4(r_k)=\frac{1-\epsilon _k}{\epsilon _k}q_k(r_k)+J_k(r_k) [/latex] [latex] m+1≤k≤N [/latex] (5.65b)

(because [latex]\epsilon=1 [/latex] and [latex]T_r = 0[/latex] for the opening area), J = 0 from the opening into the cavity. Hence, the governing equation for the enclosure is Equation 5.64a,

[latex]J_k(r_k) -(1-\epsilon _k)\sum\limits_{j=1}^{N}{\int_{A_j}^{}{J_j(r_j)dF_{dk-dj}(r_j,r_k)}=\epsilon _k\sigma T_k^4(r_k) } [/latex] [latex] 1≤k≤m[/latex] (5.64a)

written for the cylindrical side wall and including only radiation from the cylindrical wall to itself. As in Example 5.19, the configuration factor is for a ring of differential length on the cylindrical enclosure exchanging radiation with a ring at a different axial location. Equation 5.64a then yields along the surface of the hole

[latex]J(\xi )-(1-\epsilon )\int_{\eta =0}^{\infty }{J(\eta )}dF_{d\xi -d\eta } (\left|\eta -\xi \right| )=\epsilon \sigma T^4[/latex] (5.68)

where [latex] ξ = x/D, η = y/D,[/latex] and [latex]dF_{d\xi -d\eta } (\left|\eta -\xi \right| )[/latex] is in Equation 5.67b.

[latex]dF_{d\xi -d\eta }(\left|\eta -\xi \right| )=\left\{1-\frac{|η – ξ|^3+\frac{3}{2}|η – ξ| }{[(η – ξ)^2+1]^{3/2}} \right\}d\eta [/latex] (5.67b)

After division by the constant [latex]σT_r^4,[/latex] the apparent emissivity [latex] ϵ_a(ξ)[/latex] is governed by the integral equation

[latex]\epsilon _a(\xi )-(1-\epsilon )\int_{\eta =0}^{\infty }{\epsilon _a}(\eta )dF_{d\xi -d\eta }(\left|\eta -\xi \right| )=\epsilon [/latex] (5.69)

The solution of Equation 5.69 was carried out for various surface emissivities [latex]\epsilon [/latex],and the results are in Figure 5.17b. The radiation leaving the surface approaches that of a blackbody as the wall position is at greater depths into the hole. At the mouth of the hole [latex]\epsilon_a≈\sqrt{\epsilon } [/latex] , as shown in Buckley (1927, 1928).

Question 5.20: Consider emission from a long cylindrical hole drilled into ...

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