Electrically heated range-top elements glow when they are in high-power settings. The elements have electrical resistance wires that are electrically insulated from the element covers (i.e., casing) by a dielectric filler (e.g., oxide or monoxide ceramic). This ceramic should have a very low

Question

Electrically heated range-top elements glow when they are in high-power settings. The elements have electrical resistance wires that are electrically insulated from the element covers (i.e., casing) by a dielectric filler (e.g., oxide or monoxide ceramic). This ceramic should have a very low electrical conductivity and yet be able to conduct the heat from the heating element to the casing. Oxide ceramics have a low [latex]\sigma _{e} [/latex] and also low k. Nonoxide ceramics (i.e., nitrates such as aluminum nitrite) have low [latex]\sigma _{e} [/latex], but high k. The melting temperature of the resistance wire and the dielectric are also important. The heating element is a coiled wire in a cylindrical form, as shown in Figure (a). All the Joule heating leaves through surface 1, i.e., [latex]Q_{k,H-1} = \dot{S}_{ e,J}[/latex].
(a) Draw the thermal circuit diagram.
(b) Determine the temperature [latex]T_{H}[/latex].
(c) Determine the temperature drop [latex]T_{H} − T_{1}[/latex].

[latex] \epsilon _{r,1} = 0.9 , F_{1−3} = 1 , T_{3} = 20^{\circ }C , D_{H} = 2 mm , D_{1} = 10 mm , L = 30 cm , A_{r,3} \gg A_{r,1} , k_{1} = 1.5 W/m-K, and \dot{S}_{ e,J} = 500 W [/latex]

Accepted Answer

(a) The thermal circuit diagram of the heat flow is shown in Figure (b). Here we are given the heat flow rate [latex]Q_{H-1}[/latex]. From Figure (b), we have

[latex] Q_{H-1}=\frac{T_{H}-T_{1}}{R_{k,H-1}} =Q_{r,1-3}=\frac{E_{b,1}-E_{b,3}}{(R_{r,\epsilon })_{1}+(R_{r,F})_{1-3}+(R_{r,\epsilon })_{3}} =\dot{S}_{e,J} [/latex]

(b) Here we first determine [latex]E_{b,1} = \sigma _{SB}T^{ 4}_{ 1} [/latex], by solving the above for [latex]E_{b,1}[/latex], i.e.,

[latex] E_{b,1} = \sigma_{SB}T^{ 4}_{1} = E_{b,3} + \dot{S}_{ e,J}[(R_{r,\epsilon })_{1} + (R_{r,F} )_{1-3} + (R_{r,\epsilon })_{3}]. [/latex]

The radiation resistances are

[latex](R_{r,\epsilon })_{1}=\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} ight) _{1} ,A_{r,1}=\pi D_{1}L[/latex]

[latex] (R_{r,F })_{1-3}=\frac{1}{A_{r,1}F _{1-3}} , F_{1-3}=1 [/latex]

[latex] (R_{r,\epsilon })_{3}=\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} ight) _{3} ,A_{r,3}\gg A_{r,1} [/latex]

[latex] R_{r,\sum }=\frac{1}{A_{r,1}}\left(\frac{1-\epsilon _{r,1}}{\epsilon _{r,1}}+1+\frac{A_{r,1}}{A_{r,3}}\frac{1-\epsilon _{r,3}}{\epsilon _{r,3}} ight) =\frac{1}{A_{r.1}}\left(\frac{1}{\epsilon _{r,1}}-1+1 ight) =\frac{1}{A_{r,1}\epsilon _{r,1}} [/latex]

Then we have

[latex] \sigma _{SB}T_{1}^{4}=E_{b,3}+\dot{S}_{e,J}R_{r,\sum } [/latex]

Solving for [latex]T_{1}[/latex], we have

[latex]T_{1}=\left(\frac{E_{b,3}+\dot{S}_{e,J}R_{e,\sum }}{\sigma _{SB}} ight) ^{1/4} [/latex]

Next, we use [latex]Q_{H,1}+Q_{H,2}=\dot{S}_{e,J}[/latex] and [latex]Q_{H,1}=Q_{k,H-1}=Q_{r,1-3}=\frac{T_{H}-T_{1}}{R_{k,H-1}} =\frac{E_{b,1}-E_{b,3}}{R_{r,\sum }}=\frac{E_{b,1}-E_{b,3}}{(R_{r,\epsilon })_{1}+(R_{r,F })_{1-3}+(R_{r,\epsilon })_{3}} [/latex] again, but this time we solve for [latex]T_{H}[/latex], i.e.,

[latex]T_{H} = T_{1} + \dot{S}_{ e,J}R_{k,H-1}[/latex],

where [latex]R_{k,H-1}[/latex] is for a cylindrical shell and is given in Table as

[latex]R_{k,H-1}=\frac{ln(D_{1}/D_{H})}{2\pi Lk_{1}} [/latex] Table

Now substituting for [latex]T_{1}[/latex] and [latex]R_{k,H-1}[/latex], in the above expression for [latex]T_{H}[/latex], we have

[latex] T_{H}=\left[\frac{E_{b,3}+\dot{S}_{e,J}\frac{1}{A_{r,1}\epsilon _{r,1}} }{\sigma _{SB}} ight] ^{1/4}+\dot{S}_{e,J}\frac{ln(D_{1}/D_{H})}{2\pi Lk_{1}} [/latex]

(c) The difference in temperature, [latex]T_{H} − T_{1}[/latex], is found from the first equation written above, i.e.,

[latex]T_{H} − T_{1} = \dot{S}_{ e,J}R_{k,H-1} [/latex]

Now using the numerical values, we have

[latex] T_{H}=\left\{\frac{5.670 imes 10^{-8}(W/m^{2}-K^{4}) imes (293.15)^{4}(K)^{4}}{5.670 imes 10^{-8}(W/m^{2}-K^{4})} +\frac{5 imes 10^{2}(W)\left[\frac{1}{\pi imes 0.01(m) imes 0.3(m) imes 0.9} ight] }{5.670 imes 10^{-8}(W/m^{2}-K^{4})} ight\} ^{1/4} [/latex]

[latex] +5 imes 10^{2}(W) \frac{ln(0.01/0.002)}{2\pi imes 0.3(m) imes 1.5(W/m-k)} [/latex]

[latex] =\left[\frac{4.187 imes 10^{2}+5 imes 10^{2} imes 117.9(1/m^{2})}{5.670 imes 10^{-8}(W/m^{2}-K^{4})} ight] ^{1/4}(K) [/latex]

[latex] +5 imes 10^{2}(W) imes 0.5691(K/W) [/latex]

[latex] T_{H}=\left(\frac{4.187 imes 10^{2}+5.895 imes 10^{4}}{5.670 imes 10^{-8}} ight) ^{1/4}(K)+(5 imes 10^{2} imes 0.5691)(K) [/latex]

[latex] =1.011 imes 10^{3}(K)+2.846 imes 10^{2}(K)=1,296K [/latex]

Solving for the temperature difference, we have

[latex]T_{H} − T_{1} = 5 × 10^{2} (W) × 0.5691(K/W) = 284.6 K [/latex]

[latex] T_{1} = 1,011 K [/latex]

Question 4.12: Electrically heated range-top elements glow when they are in...

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