In a rapid thermal processing, solid objects are heated or cooled for a very short time such that the effect of heating or cooling does not penetrate far into the substrate and/or high temperatures are sustained for a short period to avoid material-property degradation (e.g., segregation of dopants

Question

In a rapid thermal processing, solid objects are heated or cooled for a very short time such that the effect of heating or cooling does not penetrate far into the substrate and/or high temperatures are sustained for a short period to avoid
material-property degradation (e.g., segregation of dopants in semiconductors). In a continuous process (as compared to a batch process) thermal processing oven, a large electrically heated plate irradiates upon the integrated-circuit chips
facing it as they move through the oven. This is depicted in Figure (a). Assume that the conveyer-chips and the heater make up a two-surface enclosure. Then the planar chip surface has a view factor to the heater that is unity. The chip surface has areas where the emissivity is large, i.e.,[latex] \epsilon _{r,1} = 1[/latex], and areas where the surface emissivity is small, [latex]\epsilon _{r,1} = 0.2[/latex]. Assume that the emissivity nonuniformity would still allow for a two-surface treatment.
(a) Draw the thermal circuit diagram.
(b) For [latex]T_{1} = 250^{\circ }C[/latex] and [latex]T_{2} = 900^{\circ }C[/latex], determine the surface radiative heat flux [latex]q_{r,1-2}[/latex] to the high and low emissivity portions of the surface.

Accepted Answer

(a) The thermal circuit model is shown in Figure (b).

(b) Since [latex]A_{r,2} \gg A_{r,1}[/latex] and [latex]F_{1-2} = 1[/latex], we use [latex]Q_{r,1-2}=A_{r,1}\epsilon _{r,1}[E_{b,1}(T_{1})-E_{b,2}(T_{2})] for F_{1-2}=1 with \frac{A_{r,1}}{A_{r,2}} \simeq 0 or \epsilon _{r,2}=1[/latex] and for [latex]q_{r,1-2} = Q_{r,1-2}/A_{r,1}[/latex], we have

[latex] q_{r,1-2}=\epsilon _{r,1}[E_{b,1}(T_{1})-E_{b,2}(T_{2})][/latex]

[latex]=\epsilon _{r,1}(\sigma _{SB}T_{1}^{4}-\sigma _{SB}T_{2}^{4})[/latex]

[latex]=\epsilon _{r,1}\sigma _{SB}(T_{1}^{4}-T_{2}^{4}) [/latex]

We now use the numerical values, and for areas with [latex]\epsilon _{r,1} = 1[/latex] we have

[latex] q_{r,1-2}=1 imes 5.670 imes 10^{-8}(W/m-K^{4})[(523.15)^{4}(K)^{4}-(1,173.15)^{4}(K)^{4}] [/latex]

[latex]q_{r,1-2}=-1.032 imes 10^{5}W/m^{2} for \epsilon _{r,1}=1 [/latex]

For [latex] \epsilon _{r,1}=0.2 [/latex] we have

[latex] q_{r,1-2}=0.2 imes 5.670 imes 10^{-8}(W/m-K^{4})[(523.15)^{4}(K)^{4}-(1,173.15)^{4}(K)^{4}][/latex]

[latex]=-2.063 imes 10^{4}W/m^{2} [/latex]

Question 4.5: In a rapid thermal processing, solid objects are heated or c...

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