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 \sigma _{e} and also low k. Nonoxide ceramics (i.e., nitrates such as aluminum nitrite) have low \sigma _{e} , 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., Q_{k,H-1} = \dot{S}_{ e,J}.
(a) Draw the thermal circuit diagram.
(b) Determine the temperature T_{H}.
(c) Determine the temperature drop T_{H} − T_{1}.
Question 4.12: Electrically heated range-top elements glow when they are in...
(a) The thermal circuit diagram of the heat flow is shown in Figure (b). Here we are given the heat flow rate Q_{H-1}. From Figure (b), we have
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}(b) Here we first determine E_{b,1} = \sigma _{SB}T^{ 4}_{ 1} , by solving the above for E_{b,1}, i.e.,
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}].The radiation resistances are
(R_{r,\epsilon })_{1}=\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} \right) _{1} ,A_{r,1}=\pi D_{1}L(R_{r,F })_{1-3}=\frac{1}{A_{r,1}F _{1-3}} , F_{1-3}=1
(R_{r,\epsilon })_{3}=\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} \right) _{3} ,A_{r,3}\gg A_{r,1}
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}} \right) =\frac{1}{A_{r.1}}\left(\frac{1}{\epsilon _{r,1}}-1+1 \right) =\frac{1}{A_{r,1}\epsilon _{r,1}}
Then we have
\sigma _{SB}T_{1}^{4}=E_{b,3}+\dot{S}_{e,J}R_{r,\sum }Solving for T_{1}, we have
T_{1}=\left(\frac{E_{b,3}+\dot{S}_{e,J}R_{e,\sum }}{\sigma _{SB}} \right) ^{1/4}Next, we use Q_{H,1}+Q_{H,2}=\dot{S}_{e,J} and 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}} again, but this time we solve for T_{H}, i.e.,
T_{H} = T_{1} + \dot{S}_{ e,J}R_{k,H-1},
where R_{k,H-1} is for a cylindrical shell and is given in Table as
R_{k,H-1}=\frac{ln(D_{1}/D_{H})}{2\pi Lk_{1}} Table
Now substituting for T_{1} and R_{k,H-1}, in the above expression for T_{H}, we have
T_{H}=\left[\frac{E_{b,3}+\dot{S}_{e,J}\frac{1}{A_{r,1}\epsilon _{r,1}} }{\sigma _{SB}} \right] ^{1/4}+\dot{S}_{e,J}\frac{ln(D_{1}/D_{H})}{2\pi Lk_{1}}(c) The difference in temperature, T_{H} − T_{1}, is found from the first equation written above, i.e.,
T_{H} − T_{1} = \dot{S}_{ e,J}R_{k,H-1}Now using the numerical values, we have
T_{H}=\left\{\frac{5.670\times 10^{-8}(W/m^{2}-K^{4})\times (293.15)^{4}(K)^{4}}{5.670\times 10^{-8}(W/m^{2}-K^{4})} +\frac{5\times 10^{2}(W)\left[\frac{1}{\pi \times 0.01(m)\times 0.3(m)\times 0.9} \right] }{5.670\times 10^{-8}(W/m^{2}-K^{4})} \right\} ^{1/4}+5\times 10^{2}(W) \frac{ln(0.01/0.002)}{2\pi \times 0.3(m)\times 1.5(W/m-k)}
=\left[\frac{4.187\times 10^{2}+5\times 10^{2}\times 117.9(1/m^{2})}{5.670\times 10^{-8}(W/m^{2}-K^{4})} \right] ^{1/4}(K)
+5\times 10^{2}(W)\times 0.5691(K/W)
T_{H}=\left(\frac{4.187\times 10^{2}+5.895\times 10^{4}}{5.670\times 10^{-8}} \right) ^{1/4}(K)+(5\times 10^{2}\times 0.5691)(K)
=1.011\times 10^{3}(K)+2.846\times 10^{2}(K)=1,296K
Solving for the temperature difference, we have
T_{H} − T_{1} = 5 × 10^{2} (W) × 0.5691(K/W) = 284.6 KT_{1} = 1,011 K
