In an electrically heated, open radiation oven, shown in Figure (a), electrical resistance wires are wrapped around a ceramic cylindrical shell. The Jouleheating rate is \dot{S}_{ e,J} . The outer surface of the ceramic is ideally insulated, while the inner surface radiates to a cylindrical workpiece (i.e., object to be heated). The ceramic shell and the workpiece have the same length L. At a given time, the ceramic is at a temperature T_{2} and the workpiece is at T_{1} . The surrounding, to which heat is exchanged through the open ends of the oven, is at T_{3} .
(a) Draw the thermal circuit diagram.
(b) Determine the view factor from the relations and compare with the graphical results of Figure.
(c) Assuming a steady state, determine the fraction of heat transferred to the surroundings.
\dot{S}_{ e,J} = 10^{4} W , T_{1} = 300^{\circ }C, T_{3} = 20^{\circ }C, \epsilon _{r,1} = 0.9, \epsilon _{r,2} = 0.8, D_{1} = 3 cm, D_{2} = 15 cm, and l = 60 cm
Question 4.7: In an electrically heated, open radiation oven, shown in Fig...
(a) This is a three-surface enclosure with surface 3 being the two ends with \epsilon _{r,3} = 1 (i.e., an imaginary blackbody surface is used to complete the enclosure). The heat generation \dot{S} _{e,J} leaves surface 2 by radiation, i.e., the outer surface of the ceramic is ideally insulated, Q_{2} = 0, and there is no energy stored in the ceramic (due to the steady-state assumption). The thermal circuit diagram is shown in Figure (b).
(b) There are seven view factors (because F_{2-2} \neq 0 ). Using the reciprocity rule A_{r,i}F_{i-j}=A_{r,j}F_{j-i}, these are reduced to four. Using the summation rule \sum\limits_{j=1}^{n}{F_{i-j}} =1, these are yet reduced to two. These two are F_{2-2} and F_{2-1}. The relations for these are given in Figures (d) and (e) and are
R^{\ast }=\frac{D_{2}}{D_{1}} , l^{\ast }=\frac{2l}{D_{1}}F_{2-2}=F_{2-2}(R^{\ast },l^{\ast })=1-\frac{1}{R^{\ast }} +\frac{2}{\pi R^{\ast }} tan^{-1}\frac{2(R^{\ast 2}-1)^{1/2}}{l^{\ast }}
-\frac{l^{\ast }}{2\pi R^{\ast }}\left\{\frac{(4 R^{\ast 2}+l^{\ast 2})^{1/2}}{l^{\ast }}sin^{-1}\frac{4( R^{\ast 2}-1)+(l^{\ast 2}/ R^{\ast 2})( R^{\ast 2}-2)}{l^{\ast 2}+4( R^{\ast }-1)}-sin^{-1}\frac{R^{\ast 2}-2}{R^{\ast 2}} +\frac{\pi }{2}\left[\frac{(4R^{\ast 2}+l^{\ast })^{1/2}}{l^{\ast }} -1\right] \right\} Figure (d)
a = ^{l\ast 2} + R^{\ast 2} − 1, b = l^{\ast 2} − R^{\ast 2} + 1
F_{2-1}=F_{2-1}(R^{\ast },l^{\ast })=\frac{1}{R^{\ast }} -\frac{1}{\pi R^{\ast }}\left\{cos^{-1}\frac{b}{a}-\frac{1}{2l^{\ast }}[(a+2)^{2}-(2R^{\ast })^{2}]^{1/2}cos^{-1}\frac{b}{aR^{\ast }}+b \sin^{-1}\frac{1}{R^{\ast }}-\frac{\pi a}{2} \right\} Figure (e)
where
F_{2-3} = 1 − F_{2-2} − F_{2-1} F_{1-3} = 1 − F_{1-2} summation rule
F_{1-2}=\frac{A_{r,2}}{A_{r,1}} F_{2-1} , F_{3-2}=\frac{A_{r,2}}{A_{r,3}} F_{2-3} , F_{3-1}=\frac{A_{r,1}}{A_{r,3}} F_{1-3} reciprocity rule
A_{r,2}=\pi D_{2}l, A_{r,1}=\pi D_{1}l,A_{r,3}=2\left[\frac{\pi }{4}(D_{2}^{2}-D_{1}^{2}) \right]Using the numerical values, we have
R^{\ast }=\frac{D_{2}}{D_{1}} =\frac{0.15(m)}{0.03(m)} =5 ,l^{\ast }=\frac{2l}{D_{1}} =\frac{2\times 0.60}{0.03}=40 F_{2-2} = 1 − 0.2 + 0.1273 × 0.240 − 1.273(1.031 × 1.180 − 1.16808 + 0.04834) = 0.7079a = 1, 624, b = 1,576
F_{2-1} = 0.2 − 0.06366[0.2437 − 0.0125 × (1.626 × 10^{3} × 1.376 + 1,576 × 0.201 − 2.550 × 10^{3} )]= 0.2 − 0.06366(0.2437 − 0.0125 × 4.152) = 0.1867
F_{2-3} = 1 − 0.7079 − 0.1878 = 0.1054F_{1-2} =\frac{D_{2}}{D_{1}} F_{2-1}=0.9337
F_{1-3} = 1 − 0.9337 = 0.06630
F_{3-1}=\frac{2D_{1}l}{D_{2}^{2}-D_{1}^{2}} F_{1-3}=0.1105
F_{3-2}=\frac{2D_{2}l}{D_{2}^{2}-D_{1}^{2}} F_{2-3}=0.8781
F_{3-3} = 1 − F_{3-2} − F_{3-1} = 0.01140
(c) Now, since Q_{2} = 0 and \dot{S} _{1} = \dot{S}_{3} = 0, -Q_{1}+\dot{S}_{1}=Q_{r,1}=\frac{(q_{r,o})_{1}-(q_{r,o})_{2}}{\frac{1}{A_{r,1}F_{1-2}} } +\frac{(q_{r,o})_{1}-(q_{r,o})_{3}}{\frac{1}{A_{r,1}F_{1-3}} } through -Q_{3}+\dot{S}_{3}=Q_{r,3}=\frac{E_{b,3}(T_{3})-(q_{r,o})_{3}}{\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}}\right) _{3}} ,E_{b,3}(T_{3})=\sigma _{SB}T_{3}^{4} become
Q_{r,1}=-Q_{1}=\frac{(q_{r,o})_{1}-(q_{r,o})_{2}}{\frac{1}{\pi D_{1}lF_{1-2}} } +\frac{(q_{r,o})_{1}-E_{b,3}(T_{3})}{\frac{1}{\pi D_{1}lF_{1-3}} }Q_{r,2}=\dot{S}_{e,J}=10^{4}(W)=\frac{(q_{r,o})_{2}-(q_{r,o})_{1}}{\frac{1}{\pi D_{2}lF_{2-1}} } +\frac{(q_{r,o})_{2}-E_{b,3}(T_{3})}{\frac{1}{\pi D_{2}lF_{2-3}} }
-Q_{3}=\frac{E_{b,3}(T_{3})-(q_{r,o})_{1}}{\frac{1}{\frac{\pi}{2} ( D_{2}^{2}-D_{1}^{2})F_{3-1}} } +\frac{E_{b,3}(T_{3})-(q_{r,o})_{2}}{\frac{1}{\frac{\pi}{2} ( D_{2}^{2}-D_{1}^{2})F_{3-2}} }
-Q_{1}=\frac{E_{b,1}-(q_{r,o})_{1}}{\frac{1-\epsilon _{r,1}}{\pi D_{1}l\epsilon _{r,1}} }
\dot{S}_{e,J}=\frac{E_{b,2}-(q_{r,o})_{2}}{\frac{1-\epsilon _{r,2}}{\pi D_{2}l\epsilon _{r,2}} } ,E_{b,2}=\sigma _{SB}T_{2}^{4}
Note that surface 3 is blackbody and from -Q_{3}+\dot{S}_{3}=Q_{r,3}=\frac{E_{b,3}(T_{3})-(q_{r,o})_{3}}{\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}}\right) _{3}} ,E_{b,3}(T_{3})=\sigma _{SB}T_{3}^{4} with (R_{r,\epsilon })_{3} = 0, we have
(q_{r,o})_{3} = E_{b,3} = \sigma _{SB}T_{3}^{4}and have used this above for (q_{r,o})_{3} .
We cannot simply solve for one unknown at a time by successive substitutions. The simultaneous solution requires iterations or use of a solver, for example MATLAB.
We now solve for the six unknowns (q_{r,o})_{1}, (q_{r,o})_{2}, (q_{r,o})_{3}, T_{1}, Q_{2}, and Q_{3} and the numerical results are
(q_{r,o})_{1} = 1.789 × 10^{4} W/m^{2}
(q_{r,o})_{2} = 1.327 × 10^{5} W/m^{2}
(q_{r,o})_{3} = 4.179 × 10^{2} W/m^{2}
T_{2} = 983.9 K
Q_{1} = 5.994 × 10^{3} W
Q_{3} = 4.006 × 10^{3} W.
