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 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 \(20^{\circ} \mathrm{C}\).

Short answer

Answer: The outer surface temperature of the container is 100.4°C, and the tank can supply 3.45 × 10⁻³ kg/s of water at 100°C with the cold water entering at 20°C.

Step-by-step solution

Expressing the differential equation and boundary conditions

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}

Solving the differential 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}

Evaluating the outer surface temperature

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}

Determining the amount of water supplied

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 \(20^{\circ} \mathrm{C}\).

Key concepts

Differential Equation

Understanding how heat transfers through materials can be explained using differential equations. A differential equation is a mathematical expression that relates a function with its derivatives. In the context of heat conduction, we're interested in how temperature varies across a material, governed by the rate of heat flow.
For a spherical container, the differential equation that describes steady one-dimensional heat conduction is:\[\frac{1}{r^{2}}\frac{d}{dr}\left(r^{2}\frac{dT}{dr}\right)=0\]Here:

• \(r\) is the radial distance (radius),
• \(T\) is the temperature,
• \(\frac{dT}{dr}\) is the rate of change of temperature with respect to \(r\).

The equation implies that, under steady state conditions and without any internal heat generation, the temperature distribution is solely due to thermal conduction. Integrating this equation helps to derive the temperature distribution across the material.

Boundary Conditions

Boundary conditions are essential to finding unique solutions to differential equations. They provide specific information about the system at given points, allowing us to solve for unknown constants.
In our problem, we have two boundary conditions:

• The temperature at the inner surface of the spherical container is given as \(120^{\circ} \mathrm{C}\): \( T(r_1) = 120^{\circ} \mathrm{C}\).
• The heat flow at the outer surface is connected to the heater's power output. This is expressed as: \[k\frac{dT}{dr}\bigg|_{r=r_{2}}=-\frac{0.9 \times 800}{4 \pi r_{2}^{2}}\]

Here, \(k\) is the thermal conductivity, and the term on the right accounts for the effective power output due to heat loss through insulation. Applying these conditions allows us to solve the differential equation for the temperature distribution in the container.

Temperature Distribution

Temperature distribution describes how temperature varies across a material. In our scenario, the temperature in the container changes between its inner and outer surfaces.
By solving the differential equation with given boundary conditions, we determine that the temperature distribution in the container is:\[T(r)=\frac{-3397.07}{r}+205.18^{\circ} \mathrm{C}\]This function tells us how temperature depends on the radius \(r\) of the container. It’s derived by considering both structural properties and external heat input. This equation allows for calculating the temperature at any point within the container, including at the outer surface to check its practicality or safety.

Specific Heat Capacity

Specific heat capacity is an important concept in heat transfer problems because it defines how much heat energy a material can hold per unit mass per degree change in temperature.
For water, the specific heat capacity is \(4.186 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). This means water can absorb or release about \(4.186 \mathrm{~kJ}\) for every kilogram of its mass for each degree temperature change.
In the case of our spherical container, knowing the specific heat capacity helps determine how much water at \(100^{\circ} \mathrm{C}\) can be supplied continuously. Using:\[Q_{H}=\dot{m}c_{p}(T_{out}-T_{in})\]We relate the heater's power to the mass flow rate \( \dot{m} \), where:

• \(Q_H\) is the heat power from the heater,
• \(c_p\) is the specific heat capacity,
• \(T_{out}\) and \(T_{in}\) are the outlet and inlet temperatures, respectively.

Calculating \( \dot{m} \) tells us how much heated water can be directly supplied, showcasing practical applications of conducting heat and specific heat capacity together.