A heated enclosure is the circular tube in Figure 5.15, open at both ends and insulated on the outside surface [Usiskin and Siegel (1960)]. For a uniform heat addition (such as by electrical heating in the tube wall) to the inside surface of the tube wall and a surrounding environment at 0 K,

Question

A heated enclosure is the circular tube in Figure 5.15, open at both ends and insulated on the outside surface [Usiskin and Siegel (1960)]. For a uniform heat addition (such as by electrical heating in the tube wall) to the inside surface of the tube wall and a surrounding environment at 0 K, what is the temperature distribution along the tube? If the surroundings are at [latex]T_e,[/latex] how does this influence the temperature distribution?

Accepted Answer

Since the open ends of the tube are nonreflecting, they can be assumed to act as black disks at the surrounding temperature 0 K. Then with [latex]ϵ_1 = ϵ_3 = 1,[/latex] 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)

gives [latex]J_3= σT_1^4 = σT_3^4 = 0.[/latex] Consequently, the summation in Equation 5.64b provides

[latex]J_k(r_k) -\sum\limits_{j=1}^{N}{\int_{A_j}^{}{J_j(r_j)dF_{dk-dj}(r_j,r_k)}=q_k(r_k) } [/latex] [latex] m+1≤k≤N[/latex] (5.64b)

only radiation from surface 2 to itself. Since the tube is axisymmetric, the two differential areas [latex]dA_k[/latex] and [latex]dA^*_k[/latex] can be rings located at x and y. Then Equation 5.64b yields

[latex]J_2(\xi ) -\int_{\eta =0}^{\eta =1}{J_2(\eta )dF_{d\xi -d\eta }(\left|\eta -\xi \right| )} =q_2 [/latex] (5.67a)

where [latex]ξ = x/D, η = y/D, I= L/D,[/latex] and [latex] dF_{d\xi -d\eta }(\left|\eta -\xi \right| )[/latex] is the configuration factor for two rings a distance [latex] |η – ξ|[/latex] apart and is given by factor 15 in Appendix C as

[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)

Absolute-value signs are used because the configuration factor depends only on the magnitude of the distance between the rings. When [latex]|η – ξ| = 0, dF = dη,[/latex] and this is the configuration factor from a differential ring to itself. Equation 5.67a can be divided by the constant [latex]q_2[/latex] and the dimensionless quantity [latex]J_2(ξ)/q_2[/latex] found by numerical or approximate solution methods for linear integral equations as discussed in Section 5.4.2. The [latex]J_2(ξ)/q_2[/latex] distribution is in Figure 5.16b for a tube four diameters in length. From Equation 5.65b, the [latex]T_2^4(ξ)[/latex] along the tube is given by

[latex]\sigma T_2^4(\xi )=\frac{1-\epsilon _2}{\epsilon _2} q_2+J_2(\xi )[/latex]

Since [latex]q_2[/latex] is constant, the [latex]T_2^4(ξ)[/latex] has the same shape as [latex]J_2(ξ).[/latex] The wall temperature is high in the central region of the tube and low near the ends where energy is radiated more readily to the low-temperature environment. Now consider the environment at [latex] T_e ≠ 0.[/latex] The open ends of the tube can be regarded as perfectly absorbing disks at [latex]T_e,[/latex] and the integral Equation 5.64b yields

[latex]J_2(\xi ) -\int_{\eta =0}^{I}{J_2(\eta )dF_{d\xi -d\eta }(\left|\eta -\xi \right| )-\sigma T_e^4F_{d\xi -1}(\xi )-\sigma T_e^4F_{d\xi -3}(I-\xi )} =q_2 [/latex]

where [latex]F_{dξ–1}(ξ)[/latex] is the configuration factor from a ring element at [latex] ξ [/latex] to disk 1 at [latex] ξ = 0,[/latex],[latex]F_{d\xi -1}(\xi )=[(\xi ^2+\frac{1}{2} )/(\xi ^2+1)^{1/2}]-\xi [/latex] . Since the integral equation is linear in [latex]J_2(ξ)[/latex] let a trial solution be the sum of two parts where, for each part, either [latex]T_e = 0 [/latex] or [latex]q_2 = 0[/latex]:

[latex]J_2(\xi ) =J_2(\xi )\left|_{T_e=0}+J_2(\xi )\right|_{q_2=0} [/latex]

Substitute the trial solution into the integral equation to get

[latex]J_2(\xi )\left|_{T_e=0}+J_2(\xi )\right|_{q_2=0}-\int_{\eta =0}^{I}{J_2(\eta )\left|_{T_e=0}dF_{d\xi -d\eta }(\left|\eta -\xi \right| )-\int_{\eta =0}^{I}{J_2(\eta )} \right|_{q_2=0}}dF_{d\xi -d\eta }(\left|\eta -\xi \right| )-\sigma T_e^4F_{d\xi -1}(\xi ) -\sigma T_e^4F_{d\xi -3}(I-\xi )=q_2[/latex]

For [latex] T_e = 0,[/latex] Equation 5.67a applies; subtract this equation to give

[latex]J_2(\xi )\left|_{q_2=0}-\int_{\eta =0}^{I}{J_2}(\eta ) \right|_{q_2=0} dF_{d\xi -d\eta }(\left|\eta -\xi \right| )-\sigma T_e^4F_{d\xi -1}(\xi )-\sigma T_e^4F_{d\xi -3}(I-\xi )=0[/latex]

The solution is [latex]J_2\mid _{q_2=0}=\sigma T_e^4;[/latex] this would be expected physically for an unheated surface in a uniformtemperature environment. The tube temperature distribution is found from Equation 5.65b as

[latex]\sigma T_e^4(\xi ) =\frac{1-\epsilon _2}{\epsilon _2}q_2+J_2(\xi )\left|_{T_e=0}+J_2(\xi )\right|_{q_2=0} [/latex]
[latex]=\frac{1-\epsilon _2}{\epsilon _2}q_2+J_2(\xi )\mid _{T_e=0}+\sigma T_e^4 [/latex]

where [latex]J_2(\xi )\mid _{T_e=0} [/latex] was found in the first part of this example. The environment has thus added [latex]σT_e^4[/latex] to the solution for [latex]σT_2^4 (ξ)[/latex] found previously for [latex]T_e = 0.[/latex]

Question 5.19: A heated enclosure is the circular tube in Figure 5.15, open...

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