Question

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 \(500-\mathrm{W}\) electric strip heater and then insulated. The temperature of the inner surface of the container is observed to be nearly \(100^{\circ} \mathrm{C}\) 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 \(20^{\circ} \mathrm{C}\).

Short answer

Question: Find the outer surface temperature of the container and the amount of water at 100°C that the container can continuously supply. Answer: The outer surface temperature of the container, T_2, can be found using the equation: \(T_2 = T_1 - \int_{r_1}^{r_2}\frac{dT}{dr} dr\) where T_1 is the temperature at the inner surface, r_1 is the inner radius, and r_2 is the outer radius of the container. The amount of water at 100°C that the container can continuously supply can be found using the mass flow rate: \(m = \frac{Q}{c\Delta T}\) where Q is the heat transfer rate, c is the specific heat capacity of water, and ΔT is the temperature difference between hot and cold water.

Step-by-step solution

Formulate the differential equation and boundary conditions.

We will use the heat conduction equation for a spherical coordinate system, which is given by: \(\frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{dT}{dr}\right) = 0\) The boundary conditions are at the inner and outer surfaces of the container, denoted by \(T_{1}\) at \(r_1\) and \(T_{2}\) at \(r_2\).

Solve the differential equation.

To solve the heat conduction equation, we first integrate with respect to \(r\), then divide by \(r^2\) and integrate once more: \(\frac{d}{dr}\left(r^{2}\frac{dT}{dr}\right) = C_1\) \(\frac{d}{dr}\left(r^{2}\frac{dT}{dr}\right) = C_1r^{2}\) Next, we integrate: \(r^{2}\frac{dT}{dr} = \frac{1}{3}C_1r^{3} + C_2\)

Apply boundary conditions.

Now we apply the boundary conditions with the given temperatures at the inner and outer radii: 1. At \(r=r_1\), \(T=T_1\): \(r_1^2\frac{dT}{dr}|_{r=r_1} = \frac{1}{3}C_1r_1^3 + C_2\) 2. At \(r=r_2\), \(T=T_2\): \(r_2^2\frac{dT}{dr}|_{r=r_2} = \frac{1}{3}C_1r_2^3 + C_2\) Now, we can solve for constants \(C_1\) and \(C_2\): \(C_1 = 3 \frac{r_1^2\frac{dT}{dr}|_{r=r_1} - r_2^2\frac{dT}{dr}|_{r=r_2}}{r_1^3 - r_2^3}\) \(C_2 = r_1^2\frac{dT}{dr}|_{r=r_1} - \frac{1}{3}C_1r_1^3\)

Determine outer surface temperature of the container.

In order to calculate the outer surface, we need to find the rate of heat transfer through the container wall using heater provided: Total heat generated by the heater is \(500W\). 10 percent will be lost through the insulation, so only 90 percent is used for heating the water: \(Q = 0.9\cdot500=450 \mathrm{~W}\) Heat transfer rate through the container can be calculated using the formula: \(Q = -kA\frac{dT}{dr}|_{r=r_2}\) where \(A\) is the area of the outer surface: \(A = 4\pi r_2^2\) Now, we can rearrange the equation for \(\frac{dT}{dr}\) at the outer surface and substitute the given values of \(k\) and \(A\) to find \(\frac{dT}{dr}|_{r=r_2}\): \(\frac{dT}{dr}|_{r=r_2} = - \frac{Q}{kA}\) From Step 3, we can plug this value into the expression for \(C_1\) and \(C_2\) and insert them into the temperature equation for the outer surface: \(T_2 = T_1 - \int_{r_1}^{r_2}\frac{dT}{dr} dr\)

Determine how much water at \(100^{\circ} \mathrm{C}\) the tank can supply steadily if the cold water enters at \(20^{\circ} \mathrm{C}\).

To calculate the amount of water supplied, we can use the relationship: \(Q = mc\Delta T\) where \(Q\) is the heat transfer rate, \(m\) is the mass flow rate, \(c\) is the specific heat capacity of water, and \(\Delta T\) is the temperature difference between hot and cold water. We can rearrange the formula for the mass flow rate: \(m = \frac{Q}{c\Delta T}\) Now, we can substitute the given values for \(Q\), \(c\), \(T_1\), and \(T_{cold}\): \(m = \frac{450 \mathrm{~W}}{4.186 \mathrm{~kJ/kg \cdot K} \cdot (100^{\circ}\mathrm{C}-20^{\circ}\mathrm{C})}\) Finally, we can calculate the mass flow rate of the hot water that can be supplied steadily by the container.