Electrical heaters are used in contact heating (i.e., conduction heating) for various thermal processes. A common example is the fabric iron, rendered in Figures Ex. 3.17(a) and (b). A large fraction of the electrical power flowing into the iron, flows into the fabric. This is designated as Qs. The

Question

Electrical heaters are used in contact heating (i.e., conduction heating) for various thermal processes. A common example is the fabric iron, rendered in Figures (a) and (b). A large fraction of the electrical power flowing into the iron, flows into the fabric. This is designated as [latex]Q_{s}[/latex]. The contact area is [latex]A_{k} [/latex]. Then the heat flux is [latex]q_{s} = Q_{s}/A_{k} [/latex]. Consider a relatively thick fabric that is initially at [latex]T_{i} [/latex]. Here thick indicates that we are interested in elapsed times such that the effect of heating at the surface (designated by x = 0) has not yet penetrated to the other side of the fabric.
The fabric has thermal diffusivity [latex]\left\langle \alpha \right\rangle =\left\langle k\right\rangle /\left\langle\rho c_{p}\right\rangle = 10^{−7} m^{2}/s [/latex], and [latex]\left\langle k\right\rangle = 0.2 W/m-K [/latex]. The heat flux is [latex]q_{s} = −2 × 10^{4} W/m^{2} [/latex], and the initial temperature is [latex]T (t = 0) = 20^{\circ }C [/latex]. The elapsed time is t = 30 s. The surface temperature may exceed the charring (or scorching) temperature of fabric [latex]Tc = 180^{\circ }C [/latex], and then permanent damage is caused to the fabric.

(a) Determine the surface temperature [latex]T_{s}(t) = T (x = 0, t = 30 s)[/latex].
(b) Determine the temperature at location x = 3 mm from the surface, for this elapsed time, i.e., T (x = 3 mm, t = 30 s).

Accepted Answer

Because of the thick-layer assumption, the semi-infinite slab geometry can be used. The solution for the transient temperature distribution is given in Table and the magnitude of the error function is listed in Table. The transient temperature distribution is given in Table as

[latex] T (x, t) = T (t = 0) −\frac{q_{s}(4\left\langle \alpha ight angle t/\pi )^{1/2}}{\left\langle k ight angle } e^{-x^{2}/(4\left\langle\alpha ight angle t )}+\frac{q_{s}x}{\left\langle k ight angle } \left\{1-erf\left[\frac{x}{2(\left\langle\alpha ight angle t)^{1/2}} ight] ight\} [/latex]

(a) For x = 0 and t = 30 s, we have

[latex] T(x=0,t)=T_{t}(t)=T (t = 0) −\frac{2q_{s}(\left\langle \alpha ight angle t/\pi )^{1/2}}{\left\langle k ight angle } [/latex]

This shows that the surface temperature changes with [latex]q_{s}[/latex] to the first power and t to the 1/2 power.
Using the numerical values, we have

[latex] T (x = 0, t = 30 s)[/latex]

[latex]= 20(^{\circ }C) −\frac{2 imes [-(2 imes 10^{4})(W/m^{2})][10^{-7}(m^{2}/s) imes 30(s)/\pi ]^{1/2}}{0.2(W/m-^{\circ }C)}[/latex]

[latex]=20(^{\circ }C)+195.5(^{\circ }C)=215.5^{\circ }C [/latex]

This shows that the surface temperature exceeds the charring temperature at an elapsed time of t = 30 s.
(b) For x = 3 mm and at t = 30 s, we have

[latex] \frac{x^{2}}{4\left\langle \alpha ight angle t}=\frac{(3 imes 10^{-3})^{2}(m^{2})}{4 imes 10^{-7}(m^{2}/s) imes 30(s)} =0.75 [/latex]

[latex] erf[(0.75)^{1/2}]=erf(0.8660)=0.7850 [/latex] from Figure (a)(i) or for more accurate results from interpolation in Table

[latex]T (x = 3mm, t = 30 s)= 20(^{\circ }C) −\frac{2 imes [-(2 imes 10^{4})(W/m^{2})][10^{-7}(m^{2}/s) imes 30(s)/\pi ]^{1/2}}{0.2(W/m-^{\circ }C)}e^{-0.75}[/latex]

[latex]+\frac{[-2 imes 10^{4}(W/m^{2})] imes 3 imes 10^{-3}(m)}{0.2(W/m-^{\circ }C)}[1-erf(0.8660)][/latex]

[latex]= 20(^{\circ }C)+92.34(^{\circ }C)-64.49(^{\circ }C)=47.85^{\circ }C[/latex]

Question 3.17: Electrical heaters are used in contact heating (i.e., conduc...

Related Answered Questions