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.

The one-dimensional heat conduction equation with heat generation is given by: \(\frac{d}{dx}\left(k \frac{dT}{dx}\right) + \dot{q}_{gen} =0 \) Where \(x\) is the direction perpendicular to the plate, \(k\) is the thermal conductivity of the material, \(T\) is the temperature as a function of \(x\), and \(\dot{q}_{gen}\) is the heat generation rate per unit volume.

Since both sides of the plate are exposed to the same environment, we can assume the temperature distribution is symmetric across the plate. Due to this symmetry, we only need to analyze half of the plate (from \(x=0\) to \(x=L/2\), where \(L\) is the plate's thickness). The boundary conditions are: 1. At the center of the thickness, there is symmetry, so \(\frac{dT}{dx}(0) = 0\). 2. At \(x=L/2\), we have a convective boundary condition, and can use Newton's law of cooling: \(-k \frac{dT}{dx}(L/2) = h[T(L/2) - T_\infty]\), where \(h\) is the heat transfer coefficient, and \(T_\infty\) is the ambient temperature.

Integrate the heat equation once, considering the first boundary condition: \(\int_{0}^x \frac{d}{dx'}(k \frac{dT}{dx'})dx' = -\int_{0}^x \dot{q}_{gen} dx'\) \(k \frac{dT}{dx} = -\dot{q}_{gen}x + C_1 \) Using the first boundary condition, we find \(C_1 = 0\): \(k \frac{dT}{dx}(0) = 0 \Rightarrow C_1 = 0\) Then the temperature distribution is given by: \(\frac{dT}{dx} = -\frac{\dot{q}_{gen}}{k}x \) Integrating this expression, we obtain the temperature distribution as a function of \(x\): \(T(x) = -\frac{\dot{q}_{gen}}{2k}x^2 + C_2\)

To determine the integration constant \(C_2\), we use the second boundary condition: \(-k \frac{dT}{dx}(\frac{L}{2}) = h[T(\frac{L}{2}) - T_\infty]\) Substitute the temperature expression: \(-k\left(-\frac{\dot{q}_{gen}}{k}\frac{L}{2}\right) = h\left[-\frac{\dot{q}_{gen}}{2k}\left(\frac{L}{2}\right)^2 + C_2 - T_\infty\right]\) Solve for \(C_2\): \(C_2 = T_\infty - \frac{\dot{q}_{gen}L^2}{16k} + \frac{\dot{q}_{gen}L}{8h}\) Now, we have the complete temperature distribution within the plate: \(T(x) = -\frac{\dot{q}_{gen}}{2k}x^2 + T_\infty - \frac{\dot{q}_{gen}L^2}{16k} + \frac{\dot{q}_{gen}L}{8h}\)

The highest temperature occurs at the center of the plate, which is at \(x=0\): \(T_{max} = T(0) = T_\infty - \frac{\dot{q}_{gen}L^2}{16k} + \frac{\dot{q}_{gen}L}{8h}\) The lowest temperature occurs at the surface of the plate, which is at \(x=L/2\): \(T_{min} = T(\frac{L}{2}) = T_\infty\) Plug in the given values and calculate the highest and lowest temperatures: \(T_{max} = 30 - \frac{5\times10^5 \times (0.03)^2}{16\times 15.1} + \frac{5\times 10^5\times 0.03}{8\times 60} \approx 50.981 ^{\circ}C\) \(T_{min} = 30^{\circ}C\) Therefore, the highest temperature occurs at the center of the plate and is approximately \(50.981^{\circ}C\), while the lowest temperature occurs at the surface of the plate and is \(30^{\circ}C\).