Question

Consider a large 3-cm-thick stainless steel plate $(k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ in which heat is generated uniformly at a rate of \(5 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). Both sides of the plate are exposed to an environment at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Explain where in the plate the highest and the lowest temperatures will occur, and determine their values.

Short answer

Answer: The highest temperature within the 3-cm-thick stainless steel plate is approximately \(71.63^{\circ} \mathrm{C}\), and the lowest temperature is approximately \(52.81^{\circ} \mathrm{C}\).

Step-by-step solution

Define the given variables

Define the variables available in the given problem: - Thickness of the stainless steel plate, \(L = 3\times 10^{-2} \mathrm{~m}\) - Thermal conductivity of stainless steel, \(k = 15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Heat generated within the plate, \(q = 5 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\) - Ambient temperature, \(T_{\infty} = 30^{\circ} \mathrm{C}\) - Heat transfer coefficient, \(h = 60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Apply Fourier's Law of heat conduction

Considering one-dimensional steady-state heat conduction with uniform heat generation, the general heat conduction equation is: $$-\frac{d}{dx}(k\frac{dT}{dx})+q=0$$ Integrating the above equation twice with respect to \(x\) yields: $$\frac{dT}{dx}=-\frac{qx^2}{2k}+Cx$$ where C is an integration constant.

Apply the boundary conditions

Apply the boundary conditions at both sides of the plate: 1) Left boundary condition: At \(x=0\), the rate of heat transfer from the convection is equal to the heat flux due to conduction: $$-k(\frac{dT}{dx})\Big|_{x=0}=h(T_{\infty}-T|x=0)$$ Substituting \(\frac{dT}{dx}\) from the previous step: $$-\frac{q(0)^2}{2k}+C(0)=\frac{h}{k}(T_{\infty}-T|x=0)$$ Solving for C: $$C=\frac{h}{k}(T_{\infty}-T|x=0)$$ 2) Right boundary condition: At \(x=L\), the rate of heat transfer from the convection is equal to the heat flux due to conduction: $$-k(\frac{dT}{dx})\Big|_{x=L}=h(T_{\infty}-T|x=L)$$ Substituting \(\frac{dT}{dx}\) from the previous step and replacing C: $$-\frac{q(L)^2}{2k}+\frac{h}{k}(T_{\infty}-T|x=0)(L)=\frac{h}{k}(T_{\infty}-T|x=L)$$ Rearranging the equation, we get: $$T|x=L=\frac{(\frac{hL}{k})(T_{\infty}-T|x=0)}{1-\frac{qh^2L^3}{6k^2}}+T_{\infty}$$

Find the highest and lowest temperatures

The temperature distribution is parabolic within the plate, hence the highest temperature will be at the center of the plate (\(x=0\)), and the lowest temperature will be at the edges (\(x=L\)). 1) Highest temperature: At \(x=0\), \(T_{max} = T_{0}\), $$T_{max}=T_{0}=T_{\infty}+\frac{qL^2}{2hk}$$ 2) Lowest temperature: At \(x=L\), \(T_{min} = T_{L}\), use the expression obtained in Step 3, $$T_{min}=T_{L}=\frac{(\frac{hL}{k})(T_{\infty}-T|x=0)}{1-\frac{qh^2L^3}{6k^2}}+T_{\infty}$$ Now, substitute the known values to calculate \(T_{max}\) and \(T_{min}\): \(T_{max} = 30+\frac{(5\times 10^{5})(3\times 10^{-2})^2}{2(60)(15.1)} \approx 71.63^{\circ} \mathrm{C}\) To find \(T_{min}\), we first find \(T|x=0\) as follows: $$T|x=0=T_{\infty}+\frac{qL^2}{4h}=30+\frac{(5\times 10^{5})(3\times 10^{-2})^2}{4(60)}\approx 57.97^{\circ} \mathrm{C}$$ Now, plug \(T|x=0\) value into the equation for \(T_{min}\): \(T_{min} = \frac{(\frac{(60)(3\times10^{-2})}{15.1})(30-57.97)}{1-\frac{(5\times10^5)(60)^2(3\times10^{-2})^3}{6(15.1)^2}}+30 \approx 52.81^{\circ} \mathrm{C}\)

Conclusion

The highest temperature within the 3-cm-thick stainless steel plate occurs at the center of the plate with a value of approximately \(71.63^{\circ} \mathrm{C}\). The lowest temperature occurs at both edges of the plate with a value of approximately \(52.81^{\circ} \mathrm{C}\).