A rubber ring (i.e., rubber O-ring) with cross-sectional diameter D = 3 mm and ring radius Rr = 3 cm, as shown in Figure Ex. 3.20, is softened by heating in a microwave oven. The density of rubber is ρ = 1,000 kg/m^3 and its specific heat capacity is cp = 2,010 J/kg-K. A surface heat loss per unit

Question

A rubber ring (i.e., rubber O-ring) with cross-sectional diameter D = 3 mm and ring radius $R_{r} = 3 cm$ , as shown in Figure, is softened by heating in a microwave oven. The density of rubber is $\rho = 1,000 kg/m^{3}$ and its specific heat capacity is $c_{p} = 2,010 J/kg-K.$ A surface heat loss per unit area $q_{1} = 10 W/m^{2}$ (positive for heat flow out, i.e., along the surface normal) and a uniform temperature $T_{1}(t)$ is assumed. The RMS of the oscillating electric field $(\overline{e_{e}^{2}})^{1/2}$ is $8 × 10^{3} V/m$ and the frequency is 3 GHz.

Determine the elapsed time needed to raise the temperature of the rubber by $50^{\circ }C$

Determine the elapsed time needed to raise the temperature of the rubber by [latex]50^{\circ }C[/latex]

Accepted Answer

The transient, uniform temperature of a body of volume [latex]V_{1} [/latex], i.e., node 1, subject to constant surface heat transfer and volumetric energy conversion is given by [latex]T_{1}=T_{1}(t=0)+\frac{\dot{S}_{1}-Q_{1}}{( ho c_{p}V)_{1}}t =T_{1}(t=0)+\frac{\dot{S}_{1}-q_{1}A_{1}}{( ho c_{p}V)_{1}}t=T_{1}(t=0)+a_{1}t[/latex], i.e.,

[latex] T_{1}-T_{1}(t=0)+\frac{\dot{s}_{1}V_{1}-q_{1}A_{1}}{( ho c_{p}V)_{1}}t ,\dot{s}_{1}=2\pi f\epsilon _{ec}\epsilon _{o}\overline{e_{e}^{2}} [/latex]

[latex] A_{1}=2\pi R_{r}(\pi D)=2\pi ^{2}R_{r}D, V_{1}=\left(\pi \frac{D^{2}}{4} ight)2\pi R_{r}=\frac{\pi ^{2}D^{2}R_{r}}{2} [/latex]

Now, solving for t from above we have

[latex] t=\frac{[T_{1}-T_{1}(t=0)]( ho c_{p}V)_{1}}{\dot{s}_{1}V_{1}-q_{1}A_{1}} [/latex]

Using the numerical values, we have

[latex] \epsilon _{ec}=0.48 [/latex] from Table, using f = 3 GHz

[latex] \dot{S}_{1}=2\pi imes 3 imes 10^{9}(1/s) imes 0.48 imes 8.8542 imes 10^{-12}(A^{2}-s^{2}/N-m^{2}) imes (8 imes 10^{3})^{2}(V^{2}/m^{2})=5.124 imes 10^{6}W/m^{3} [/latex]

[latex] A_{1}=2\pi^{2} imes 3 imes 10^{-2}(m) imes 3 imes 10^{-3}(m)=1.775 imes 10^{-3}m^{2} [/latex]

[latex] V_{1}=\pi^{2} (3 imes 10^{-2})^{2}(m^{2}) imes 3 imes 10^{-2}(m)/2=1.331 imes 10^{-6}m^{2}[/latex]

Then, inserting the numerical values in the relation for t, we have

[latex] t=\frac{50^{\circ }C imes 1,100(Kg/m^{3}) imes 2,010(J/Kg-^{\circ }C) imes 1.331 imes 10^{-6}(m^{3})}{5.124 imes 10^{6}(W/m^{3}) imes 1.331 imes 10^{-6}(m^{3})-10(W/m^{2}) imes 1.775 imes 10^{-3}(m^{2})}=21.63 s [/latex]

Question 3.20: A rubber ring (i.e., rubber O-ring) with cross-sectional dia...

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