Starting with an energy balance on a spherical shell volume element, derive the one-dimensional transient heat conduction equation for a sphere with constant thermal conductivity and no heat generation.

Short Answer

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Answer: The one-dimensional transient heat conduction equation for a sphere with constant thermal conductivity and no heat generation is given by: \(\frac{1}{r^2}\frac{\partial (r^2 \rho c_p T)}{\partial t} = -k \frac{d^2T}{d r^2}\), where \(r\) is the radial coordinate, \(T\) is the temperature, \(\rho\) is the density of the material, \(c_p\) is the specific heat capacity, and \(k\) is the thermal conductivity.

Step by step solution

01

Energy Balance for a Spherical Shell Volume Element

Consider a small spherical shell of thickness \(\delta r\) and radius \(r\). The volume of the shell is given by, \(V = 4 \pi r^2 \delta r\). Let the heat transferred into the shell be \(q_{in}\), and the heat transferred out of the shell be \(q_{out}\). The difference between these two will lead to a change in the internal energy of the shell. Therefore, the energy balance can be written as: \(\frac{\partial E}{\partial t} = q_{in} - q_{out}\).
02

Assumption of Constant Thermal Conductivity and No Heat Generation

We will assume that the thermal conductivity of the sphere, \(k\), is constant and there is no heat generation within the sphere. This means that there are no additional heat sources or sinks within the sphere, and heat only flows due to temperature gradients.
03

Deriving the One-Dimensional Transient Heat Conduction Equation for a Sphere

According to Fourier's law of heat conduction, the heat flux in the radial direction is given by \(q_r = -k \frac{dT}{d r}\), where \(T\) is the temperature and \(r\) is the radial coordinate. The heat entering the shell is \(q_{in} = -k \frac{dT}{dr} \Big|_{r}4 \pi r^2\) and the heat exiting the shell is \(q_{out} = -k \frac{dT}{dr} \Big|_{r+\delta r}4 \pi (r + \delta r)^2\). Now we can rewrite the energy balance equation: \begin{align*} \frac{\partial E}{\partial t} &= q_{in} - q_{out} \\ \frac{\partial (V \rho c_p T)}{\partial t} &= -k \frac{dT}{dr} \Big|_{r}4 \pi r^2 - (-k \frac{dT}{dr} \Big|_{r+\delta r}4 \pi (r + \delta r)^2) \end{align*} Here, \(\rho\) is the density of the material and \(c_p\) is the specific heat capacity. Dividing both sides by \(V= 4\pi r^2\delta r\), we get: \begin{align*} \frac{1}{r^2}\frac{\partial (r^2 \rho c_p T)}{\partial t} &= -\frac{k}{\delta r}\left( \frac{dT}{dr} \Big|_{r+\delta r} - \frac{dT}{dr} \Big|_{r} \right) \\ \end{align*} Now we take the limit as \(\delta r \to 0\): \begin{align*} \frac{1}{r^2}\frac{\partial (r^2 \rho c_p T)}{\partial t} &= -k \frac{d^2T}{d r^2} \end{align*} This is the one-dimensional transient heat conduction equation for a sphere with constant thermal conductivity and no heat generation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fourier's Law of Heat Conduction
Understanding how heat flows within materials is crucial for numerous applications, from designing efficient engines to keeping our homes warm. Fourier's Law of Heat Conduction is a fundamental principle that describes this heat flow, stating that the rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area through which heat is flowing. In simpler terms, it explains that heat moves from regions of higher temperature to regions of lower temperature, and it does so more rapidly across larger areas and steeper temperature differences.

For a spherical object, the law is applied in the radial direction because heat flows in or out in the direction of the radius. This relationship can be represented mathematically as \(q_r = -k \frac{dT}{dr}\), where \(q_r\) is the radial heat flux, \(k\) is the thermal conductivity, \(r\) is the radius, and \(T\) is the temperature. The negative sign signifies that heat flows against the increasing radius direction, or towards lower temperatures. Fourier's law is a cornerstone of thermal physics and is the starting point for deriving more complex heat transfer equations, such as the transient heat conduction equation for a sphere.
Energy Balance
The concept of energy balance is central to thermodynamics and heat transfer. It simply states that the change in energy of a system over time is equal to the energy entering the system minus the energy leaving the system. In the context of a spherical volume element, you can imagine a thin spherical shell where we track the heat entering and leaving.

An equation representing energy balance for heat conduction through a spherical shell might look like this: \(\frac{\partial E}{\partial t} = q_{in} - q_{out}\), where \(E\) is the internal energy, \(t\) is time, and \(q_{in}\) and \(q_{out}\) are the heat transfer rates into and out of the shell, respectively. Using this principle, we can derive more detailed equations that describe how temperature changes within the sphere over time. This approach is a fundamental method in thermal analysis for predicting and understanding temperature profiles in various materials and shapes.
Thermal Conductivity
Thermal conductivity, represented by the symbol \(k\), is a material property that quantifies its ability to conduct heat. It appears in Fourier's Law and is a key factor in determining how quickly heat will move through a material. Materials with high thermal conductivity, like many metals, are efficient at transferring heat and are therefore used in applications where heat needs to be spread or dissipated quickly. Conversely, materials with low thermal conductivity, such as insulating foam, are used to prevent heat flow.

In the context of our spherical example, we assume that the thermal conductivity is constant, which simplifies the analysis. In reality, thermal conductivity can vary with temperature and even with direction in some materials. This property has a direct impact on the temperature distribution and rate of heat transfer within an object. Understanding thermal conductivity helps engineers create more efficient systems, whether they're trying to keep something hot or cool it down.

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Most popular questions from this chapter

Consider a solid cylindrical rod whose ends are maintained at constant but different temperatures while the side surface is perfectly insulated. There is no heat generation. It is claimed that the temperature along the axis of the rod varies linearly during steady heat conduction. Do you agree with this claim? Why?

Heat is generated uniformly at a rate of \(4.2 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\) in a spherical ball \((k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of diameter \(24 \mathrm{~cm}\). The ball is exposed to iced-water at \(0^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperatures at the center and the surface of the ball.

A large steel plate having a thickness of \(L=4\) in, thermal conductivity of \(k=7.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), and an emissivity of \(\varepsilon=0.7\) is lying on the ground. The exposed surface of the plate at \(x=L\) is known to exchange heat by convection with the ambient air at \(T_{\infty}=90^{\circ} \mathrm{F}\) with an average heat transfer coefficient of \(h=12 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) as well as by radiation with the open sky with an equivalent sky temperature of \(T_{\text {sky }}=480 \mathrm{R}\). Also, the temperature of the upper surface of the plate is measured to be \(80^{\circ} \mathrm{F}\). Assuming steady onedimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the plate, \((b)\) obtain a relation for the variation of temperature in the plate by solving the differential equation, and \((c)\) determine the value of the lower surface temperature of the plate at \(x=0\).

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

Consider a steam pipe of length \(L\), inner radius \(r_{1}\), outer radius \(r_{2}\), and constant thermal conductivity \(k\). Steam flows inside the pipe at an average temperature of \(T_{i}\) with a convection heat transfer coefficient of \(h_{i}\). The outer surface of the pipe is exposed to convection to the surrounding air at a temperature of \(T_{0}\) with a heat transfer coefficient of \(h_{o^{*}}\) Assuming steady one-dimensional heat conduction through the pipe, \((a)\) express the differential equation and the boundary conditions for heat conduction through the pipe material, \((b)\) obtain a relation for the variation of temperature in the pipe material by solving the differential equation, and (c) obtain a relation for the temperature of the outer surface of the pipe.

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