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Consider a homogeneous spherical piece of radioactive material of radius \(r_{o}=0.04 \mathrm{~m}\) that is generating heat at a constant rate of \(\dot{e}_{\text {gen }}=5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). The heat generated is dissipated to the environment steadily. The outer surface of the sphere is maintained at a uniform temperature of \(110^{\circ} \mathrm{C}\) and the thermal conductivity of the sphere is \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the sphere, \((b)\) obtain a relation for the variation of temperature in the sphere by solving the differential equation, and \((c)\) determine the temperature at the center of the sphere.

Short Answer

Expert verified
Answer: The temperature at the center of the sphere is 310°C.

Step by step solution

01

Express the differential equation for heat conduction

Considering steady-state, one-dimensional heat transfer and heat generation, the heat conduction differential equation for a sphere can be expressed in spherical coordinates as: \(\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dT}{dr}\right)=-\dot{e}_{gen}\) Where r is the radial distance from the center of the sphere, T is the temperature, and \(\dot{e}_{gen}\) is the heat generation rate.
02

Set up the boundary conditions

Assuming one-dimensional heat conduction and steady state, we have two boundary conditions: 1. At r = 0, the temperature gradient is finite, meaning that the temperature gradient must not go to infinity at the center of the sphere. 2. At r = \(r_o\), the temperature is maintained at a constant value, so T(\(r_o\)) = 110°C.
03

Solve the differential equation

We can now solve the differential equation for the temperature profile. First, integrate the equation with respect to r: \(\int\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dT}{dr}\right)dr=-\dot{e}_{gen}\int dr\) Upon integrating, we get: \(r^2\frac{dT}{dr}=-\dot{e}_{gen}r+C_1\) Now, integrate the equation one more time with respect to r: \(\int r^2\frac{dT}{dr}dr=-\dot{e}_{gen}\int rdr+C_1\int r dr\) Upon integrating, we get: \(r^3\frac{dT}{dr}=-\frac{1}{2}\dot{e}_{gen}r^3+C_1r^2+C_2\)
04

Apply the boundary conditions

Now apply the boundary conditions to find the constants C1 and C2: Boundary condition 1: At r = 0, temperature gradient is finite, so: \(\frac{dT}{dr}(0) \neq \infty\) This ensures that C1 = 0. Boundary condition 2: At r = \(r_o\), the temperature is: T(\(r_o\)) = 110°C Plug this value into our previous equation: \(110=-\frac{1}{2}\dot{e}_{gen}r_o^2+C_2\) Now, solve for C2: \(C_2=110+\frac{1}{2}\dot{e}_{gen}r_o^2\) Plug in the given values for \(\dot{e}_{gen}\) and \(r_o\): \(C_2=110+\frac{1}{2}\times5\times10^{7}\times(0.04)^2\) \(C_2=210\)
05

Find the temperature profile

Now that we have found the constants, the temperature profile of the sphere is given by: \(T(r)=\frac{1}{2}(-\dot{e}_{gen}r^2+r_o^2\dot{e}_{gen})+C_2\)
06

Determine the temperature at the center

To find the temperature at the center of the sphere, evaluate the temperature profile at r = 0: \(T(0)=\frac{1}{2}(-\dot{e}_{gen}\times0^2+r_o^2\dot{e}_{gen})+C_2\) Plug in the given values for \(\dot{e}_{gen}\) and \(r_o\): \(T(0)=\frac{1}{2}(0+5\times10^{7}\times(0.04)^2)+210\) \(T(0)=310\) Therefore, the temperature at the center of the sphere is 310°C.

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Key Concepts

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

Steady-State Heat Transfer
In the context of heat conduction in spheres, steady-state heat transfer refers to a condition where the temperature within the sphere does not change with time. Imagine the sphere is constantly generating heat but reaches a point where the generated heat is uniformly dissipated throughout the material and across its boundary, maintaining an unvarying temperature distribution. During steady-state, the temperature gradient and rate of heat transfer remain constant, which simplifies the analysis as it eliminates the time variable from the differential equation used to describe heat conduction.
Spherical Coordinates
Understanding spherical coordinates is pivotal for solving problems involving spherical objects. Unlike Cartesian coordinates which use three perpendicular axes, spherical coordinates describe a point in 3D space with three values: the radial distance (r) from a central point, the polar angle (θ), and the azimuthal angle (φ). For heat conduction in spheres, only the radial distance, r, is typically of interest because the temperature is assumed to be radially symmetrical and doesn't depend on the angles, simplifying the differential equation to only include variations with respect to r.
Heat Generation Rate
The heat generation rate, denoted as \(\dot{e}_{gen}\), represents the power per unit volume produced by the material. In our scenario with a radioactive sphere, this rate is constant. The unit watt per cubic meter (W/m³) is an indicator of how much heat energy is being created inside the sphere due to radioactive decay. This heat generation affects the temperature distribution within the sphere and must be taken into account when deriving the differential equation for temperature.
Thermal Conductivity
The property thermal conductivity, symbolized by k, is a measure of a material's ability to conduct heat. Materials with high thermal conductivity transfer heat efficiently, like many metals, while those with low thermal conductivity, like rubber or wood, are less efficient and often used as insulators. This property plays a crucial role in how rapidly heat is distributed throughout the sphere. In the problem given, the thermal conductivity remains constant, enabling us to predict how temperature changes with radial distance.
Boundary Conditions
In differential equations, especially for heat transfer problems, it's essential to define boundary conditions. These conditions describe the behavior of the temperature at the sphere's limits—its surface and center. For this sphere, the temperature is specified at the outer surface (a Dirichlet boundary condition) and it is implied that the temperature gradient remains finite at the center of the sphere (a regularity condition). The boundary conditions are crucial because they allow us to solve the differential equation uniquely by determining the integration constants.
Differential Equation Solving
The process of differential equation solving in this context involves formulating a mathematical description that relates spatial variations in temperature within the sphere to the rate of internal heat generation and the thermal properties of the material. Integration of this equation twice, along with the application of specific boundary conditions, yields a mathematical expression for the temperature distribution within the sphere. It is a fundamental step in predicting how temperature behaves at any point in the spherical domain.
Temperature Profile
The temperature profile is the final result we seek: it is a mathematical function that describes how temperature varies as a function of radial distance from the center of the sphere. Understanding and predicting this profile is vital in applications such as nuclear reactor design, thermal insulation, and material science. The temperature profile provides insights into the effectiveness of heat dissipation, the thermal stability of the sphere under steady-state conditions, and is critical for ensuring the safe operation of systems that involve heat-generating materials.

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Most popular questions from this chapter

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 \(x=0\).

Consider a third order linear and homogeneous differential equation. How many arbitrary constants will its general solution involve?

A 6-m-long 3-kW electrical resistance wire is made of \(0.2\)-cm-diameter stainless steel \((k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The resistance wire operates in an environment at \(20^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(175 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the outer surface. Determine the surface temperature of the wire \((a)\) by using the applicable relation and \((b)\) by setting up the proper differential equation and solving it. Answers: (a) \(475^{\circ} \mathrm{C}\), (b) \(475^{\circ} \mathrm{C}\)

Is the thermal conductivity of a medium, in general, constant or does it vary with temperature?

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