A spherical metal ball of radius \(r_{o}\) is heated in an oven to a temperature of \(T_{i}\) throughout and is then taken out of the oven and allowed to cool in ambient air at \(T_{\infty}\) by convection and radiation. The emissivity of the outer surface of the cylinder is \(\varepsilon\), and the temperature of the surrounding surfaces is \(T_{\text {surr }}\). The average convection heat transfer coefficient is estimated to be \(h\). Assuming variable thermal conductivity and transient one-dimensional heat transfer, express the mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem. Do not solve.

The general heat diffusion equation for a variable thermal conductivity in spherical coordinates is given by: \(\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \alpha \frac{\partial T}{\partial r}\right)=\frac{\partial T}{\partial t}\) where \(\alpha\) is the thermal conductivity, \(T\) is the temperature, \(r\) is the radius, and \(t\) is the time.

For one-dimensional heat transfer (only radial direction), the equation provided in Step 1 simplifies to: \(\frac{\partial}{\partial r}\left(r^2 \alpha \frac{\partial T}{\partial r}\right) = r^2 \frac{\partial T}{\partial t}\) Now, we have the simplified heat diffusion equation for our problem. Next, we need to find the boundary and initial conditions for this problem.

The initial condition is that the sphere is uniformly heated to a temperature of \(T_i\) throughout. Therefore, at the initial time \(t=0\), the temperature at any point within the sphere is given by: \(T(r, 0) = T_i\) for \(0\le r\le r_o\)

We need to determine boundary conditions at the surface of the sphere \((r=r_o)\) and at the center \((r=0)\). Boundary condition at the center \((r=0)\): To maintain smoothness of the solution, the temperature gradient at the center should be zero. Therefore, \(\frac{\partial T}{\partial r}(0, t) = 0\) Boundary condition at the surface \((r=r_o)\): At the surface, convection and radiation take place. The heat flux due to convection and radiation can be given by: \(q_{conv} = h(T(r_o, t) - T_\infty)\) \(q_{rad} = \varepsilon \sigma (T(r_o, t)^4 - T_{\text{surr}}^4)\) The total heat flux at the surface is the sum of these two, and it is also equal to the heat conducted through the sphere. Hence, \(k\frac{\partial T}{\partial r}(r_o, t) = q_{conv} + q_{rad}\) Substituting the values for \(q_{conv}\) and \(q_{rad}\), \(k\frac{\partial T}{\partial r}(r_o, t) = h(T(r_o, t) - T_\infty)+\varepsilon \sigma (T(r_o, t)^4 - T_{\text{surr}}^4)\) Now, we have the boundary conditions for both ends. To summarize, 1. The differential equation governing the cooling process of the spherical ball: \(\frac{\partial}{\partial r}\left(r^2 \alpha \frac{\partial T}{\partial r}\right) = r^2 \frac{\partial T}{\partial t}\) 2. The initial condition: \(T(r, 0) = T_i\) for \(0\le r\le r_o\) 3. Boundary conditions: \(\frac{\partial T}{\partial r}(0, t) = 0\) \(k\frac{\partial T}{\partial r}(r_o, t) = h(T(r_o, t) - T_\infty)+\varepsilon \sigma (T(r_o, t)^4 - T_{\text{surr}}^4)\)