In a food processing facility, a spherical container of inner radius \(r_{1}=40 \mathrm{~cm}\), outer radius \(r_{2}=41 \mathrm{~cm}\), and thermal conductivity \(k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is used to store hot water and to keep it at \(100^{\circ} \mathrm{C}\) at all times. To accomplish this, the outer surface of the container is wrapped with a 800 -W electric strip heater and then insulated. The temperature of the inner surface of the container is observed to be nearly \(120^{\circ} \mathrm{C}\) at all times. Assuming 10 percent of the heat generated in the heater is lost through the insulation, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the container, \((b)\) obtain a relation for the variation of temperature in the container material by solving the differential equation, and \((c)\) evaluate the outer surface temperature of the container. Also determine how much water at \(100^{\circ} \mathrm{C}\) this tank can supply steadily if the cold water enters at \(20^{\circ} \mathrm{C}\).

The differential equation for steady one-dimensional heat conduction in spherical coordinates is: \begin{equation} \frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{dT}{dr}\right)=0 \end{equation} The boundary conditions are given by the temperature of the inner surface (\(120^{\circ} \mathrm{C}\)) and the thermal conductivity (\(1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)). These can be written as: \begin{equation} T(r_{1})=120^{\circ} \mathrm{C} \end{equation} \begin{equation} k\frac{dT}{dr}\bigg|_{r=r_{2}}=-\frac{0.9 \times 800}{4 \pi r_{2}^{2}} \end{equation}

To solve the differential equation, we begin by integrating with respect to \(r\): \begin{equation} \int\frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{dT}{dr}\right)dr=0 \end{equation} This yields: \begin{equation} r^{2}\frac{dT}{dr}=C_{1} \end{equation} Integrating once more with respect to \(r\) and solving for \(T(r)\): \begin{equation} T(r)=\frac{C_{1}}{r}+C_{2} \end{equation} Now, we can apply the boundary conditions to find the constants \(C_{1}\) and \(C_{2}\). Using the boundary condition from equation (2): \begin{equation} 120=\frac{C_{1}}{40 \times 10^{-2}}+C_{2} \end{equation} Using the boundary condition from equation (3): \begin{equation} C_{1}=\frac{0.9 \times 800}{-1.5 \times 4 \pi (41 \times 10^{-2})^{2}} \end{equation} Solving equations (6) and (7) for \(C_{1}\) and \(C_{2}\), we get: \begin{equation} C_{1}=-3397.07\mathrm{~W} \cdot \mathrm{m} \cdot \mathrm{K} \end{equation} \begin{equation} C_{2}=205.18^{\circ} \mathrm{C} \end{equation} Thus, the temperature distribution in the container material is: \begin{equation} T(r)=\frac{-3397.07}{r}+205.18^{\circ} \mathrm{C} \end{equation}

We can find the outer surface temperature by substituting \(r_{2}=41\mathrm{~cm}\) into the temperature distribution equation: \begin{equation} T(41 \times 10^{-2})=\frac{-3397.07}{41 \times 10^{-2}}+205.18^{\circ} \mathrm{C} \end{equation} Evaluating this expression, we find the outer surface temperature to be: \begin{equation} T(r_{2})=100.4^{\circ} \mathrm{C} \end{equation}

Let's determine the amount of water at \(100^{\circ} \mathrm{C}\) that can be supplied steadily by the tank when the temperature of the cold water entering is \(20^{\circ} \mathrm{C}\). The heat power from the strip heater is given by: \begin{equation} Q_{H}=(0.9)(800\mathrm{~W})=720\mathrm{~W} \end{equation} To find the mass flow rate of water (\(\dot{m}\)), we can use the formula: \begin{equation} Q_{H}=\dot{m}c_{p}(T_{out}-T_{in}) \end{equation} where \(c_{p}\) is the specific heat capacity of water, \(4.186 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), \(T_{out}=100^{\circ} \mathrm{C}\), and \(T_{in}=20^{\circ} \mathrm{C}\). Solving for \(\dot{m}\), we get: \begin{equation} \dot{m}=\frac{720\mathrm{~W}}{4.186 \times 10^{3} \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}(100-20)^{\circ}\mathrm{C}} \end{equation} Evaluating this expression, we find the mass flow rate of water to be: \begin{equation} \dot{m}=3.45 \times 10^{-3} \frac{\mathrm{kg}}{\mathrm{s}} \end{equation} Thus, the tank can supply \(3.45 \times 10^{-3}\,\mathrm{kg/s}\) of water at \(100^{\circ} \mathrm{C}\) when the cold water entering has a temperature of \(20^{\circ} \mathrm{C}\).