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Chapter 2: Problem 63

A spherical particle of radius \(r_{1}\) experiences uniform thermal generation at a rate of \(\dot{q}\). The particle is encapsulated by a spherical shell of outside radius \(r_{2}\) that is cooled by ambient air. The thermal conductivities of the particle and shell are \(k_{1}\) and \(k_{2}\), respectively, where \(k_{1}=2 k_{2}\). (a) By applying the conservation of energy principle to spherical control volume \(A\), which is placed at an arbitrary location within the sphere, determine a relationship between the temperature gradient \(d T / d r\) and the local radius \(r\), for \(0 \leq r \leq r_{1}\). (b) By applying the conservation of energy principle to spherical control volume \(\mathrm{B}\), which is placed at an arbitrary location within the spherical shell, determine a relationship between the temperature gradient \(d T / d r\) and the local radius \(r\), for \(r_{1} \leq r \leq r_{2}\). (c) On \(T-r\) coordinates, sketch the temperature distribution over the range \(0 \leq r \leq r_{2}\).

Short Answer

Expert verified
In summary, the temperature gradient as a function of local radius r for the spherical particle (0 鈮� r 鈮� r鈧�) is given by \( \frac{dT}{dr} = -\frac{3\dot{q} \cdot r}{k_1} \) and for the shell (r鈧� 鈮� r 鈮� r鈧�) is given by \( \frac{dT}{dr} = \frac{C_1}{r^2} \). The temperature distribution in the T-r coordinate system is represented by a smooth curve with a negative slope for the particle and a positive slope for the shell, with the slope being continuous and smooth at r = r鈧�.

Step by step solution

01

Determine the temperature gradient for spherical control volume A

For the particle 0 鈮� r 鈮� r鈧�, we apply the conservation of energy principle to an arbitrary spherical control volume A. The rate of heat generation inside control volume A is given by: \( \dot{Q}_{gen} = \dot{q} \times V \) Where, V is the volume of control volume A and is equal to (4/3)蟺r鲁. Now, the rate of heat conduction through the control surface (sphere with radius r) is given by: \( \dot{Q}_{cond} = -k_1 A \frac{dT}{dr} \) Where, A is the surface area of the control surface and is equal to 4蟺r虏. Applying the conservation of energy principle, we get: \( \dot{Q}_{gen} = \dot{Q}_{cond} \) Substituting the expressions for volume and surface area, we get: \( \dot{q} \times \frac{4}{3}\pi r^3 = -k_1 \times 4\pi r^2 \frac{dT}{dr} \) Now, reorganize the equation to express the temperature gradient as a function of r for 0 鈮� r 鈮� r鈧�: \( \frac{dT}{dr} = -\frac{3\dot{q} \cdot r}{k_1} \)
02

Determine the temperature gradient for spherical control volume B

For shell r鈧� 鈮� r 鈮� r鈧�, we apply the conservation of energy principle to an arbitrary spherical control volume B. The rate of heat conduction through the control surface (sphere with radius r) can be expressed as: \( \dot{Q}_{cond} = -k_2 \frac{dT}{dr} \times 4\pi r^2 \) Applying the conservation of energy principle, we get: \( \dot{Q}_{cond\in} = \dot{Q}_{cond\out} \) Now, we can write the temperature gradient for the shell using the conservation of energy principle as a function of r for r鈧� 鈮� r 鈮� r鈧�: \( \frac{dT}{dr} = \frac{C_1}{r^2} \) Where C鈧� is an integration constant.
03

Sketch the temperature distribution for 0 鈮� r 鈮� r鈧�

Since the temperature gradient is continuous and has a unique value at every r, we can sketch a smooth curve representing the temperature distribution function in the T-r coordinate system. For 0 鈮� r 鈮� r鈧�, the curve will follow the equation: \( \frac{dT}{dr} = -\frac{3\dot{q} \cdot r}{k_1} \) For r鈧� 鈮� r 鈮� r鈧�, the curve will follow the equation: \( \frac{dT}{dr} = \frac{C_1}{r^2} \) The curve will have a negative slope for 0 鈮� r 鈮� r鈧� and will represent the particle. The curve will have a positive slope for r鈧� 鈮� r 鈮� r鈧� and will represent the shell. The slope at r = r鈧� will be continuous and smooth.

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is pivotal in understanding heat transfer problems. It states that energy cannot be created or destroyed; it can only be transferred or converted from one form to another. In the context of this problem, this principle is used to analyze the heat flow within a spherical particle and the surrounding shell.

For the spherical particle, energy generated within the volume must equal the energy conducted through the surface. Thus, the math involves balancing heat generation and conduction at every point inside the particle. Given that heat generation inside the particle is uniform, the volume (\( V = \frac{4}{3}\pi r^3 \)) plays a key role in determining the gradient in temperature as you move from the center towards the surface.
  • Heat generation: \( \dot{Q}_{gen} = \dot{q} \times V \)
  • Heat conduction: \( \dot{Q}_{cond} = -k_1 A \frac{dT}{dr} \)
  • Surface area: \( A = 4\pi r^2 \)
For the shell, though no generation term exists, the energy flowing in at the inside surface must match that flowing out at the outer surface. Again, energy balance gives us the relation required to describe the temperature gradient in the shell, compensating for an absence of internal generation.

This energy balance for both particle and shell regions gives us essential boundary conditions for temperature distribution.
Thermal Conductivity
Thermal conductivity (\( k \)) indicates the ability of a material to conduct heat. In this problem, the particle and the shell have different thermal conductivities (\( k_1 \) and \( k_2 \)), with \( k_1 = 2k_2 \). This difference implies that the particle conducts heat twice as efficiently as the shell.
  • For the particle between \( 0 \leq r \leq r_1 \), the heat conduction is dependent on \( k_1 \).
  • For the shell between \( r_1 \leq r \leq r_2 \), it relies on \( k_2 \).
This property is crucial as it affects the rate at which temperature changes within each section. More specifically:

- High thermal conductivity in the particle means that even with uniform heat generation, changes in temperature are efficiently distributed. With a direct influence on the derived gradient equation, the negative value for the particle's temperature gradient signifies heat moving outward efficiently.
- Conversely, the shell's lower conductivity slows the outward heat flow, impacting the temperature gradient as energy "percolates" through the material.

Understanding these different thermal conductances helps in predicting how quickly or slowly the system will react to internal or external temperature changes.
Temperature Distribution
The temperature distribution is the graphical representation of how temperature varies with radius (\( r \)) within the particle and shell structure. Knowing the temperature distribution is crucial for predicting how effective the system is in dissipating heat.
  • For the region \( 0 \leq r \leq r_1 \), the temperature decreases from the center outward, following the equation \( \frac{dT}{dr} = -\frac{3\dot{q} \cdot r}{k_1} \).
  • In the shell, \( r_1 \leq r \leq r_2 \), temperature decreases more slowly, following \( \frac{dT}{dr} = \frac{C_1}{r^2} \).
The transition in temperature slope at \( r = r_1 \) is smooth and continuous due to the energy balance ensuring consistency in heat flow.

This change in slopes illustrates:- An efficient heat spread within the particle, attributed to its higher thermal conductivity.- A more gradual temperature change within the shell, indicating slower heat transfer.

Producing this distribution requires applying insights from conservation of energy and thermal conductivity principles to solve the differential equations governing the system's thermal behavior. The differential equations provide a mathematical framework to describe how various factors like heat generation rate and material properties influence the temperature profile.

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Most popular questions from this chapter

Consider a plane wall \(100 \mathrm{~mm}\) thick and of thermal conductivity \(100 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Steady-state conditions are known to exist with \(T_{1}=400 \mathrm{~K}\) and \(T_{2}=600 \mathrm{~K}\). Determine the heat flux \(q_{x}^{\prime \prime}\) and the temperature gradient \(d T / d x\) for the coordinate systems shown.

The steady-state temperature distribution in a semitransparent material of thermal conductivity \(k\) and thickness \(L\) exposed to laser irradiation is of the form $$ T(x)=-\frac{A}{k a^{2}} e^{-a x}+B x+C $$ (a) Obtain expressions for the conduction heat fluxes at the front and rear surfaces. (b) Derive an expression for \(\dot{q}(x)\). (c) Derive an expression for the rate at which radiation is absorbed in the entire material, per unit surface area. Express your result in terms of the known constants for the temperature distribution, the thermal conductivity of the material, and its thickness. where \(A, a, B\), and \(C\) are known constants. For this situation, radiation absorption in the material is manifested by a distributed heat generation term, \(\dot{q}(x)\).

A spherical shell of inner and outer radii \(r_{i}\) and \(r_{o}\), respectively, contains heat-dissipating components, and at a particular instant the temperature distribution in the shell is known to be of the form $$ T(r)=\frac{C_{1}}{r}+C_{2} $$ Are conditions steady-state or transient? How do the heat flux and heat rate vary with radius?

Passage of an electric current through a long conducting rod of radius \(r_{i}\) and thermal conductivity \(k_{r}\) results in uniform volumetric heating at a rate of \(\dot{q}\). The conducting rod is wrapped in an electrically nonconducting cladding material of outer radius \(r_{o}\) and thermal conductivity \(k_{c}\), and convection cooling is provided by an adjoining fluid. For steady-state conditions, write appropriate forms of the heat equations for the rod and cladding. Express appropriate boundary conditions for the solution of these equations.

To determine the effect of the temperature dependence of the thermal conductivity on the temperature distribution in a solid, consider a material for which this dependence may be represented as $$ k=k_{o}+a T $$ where \(k_{o}\) is a positive constant and \(a\) is a coefficient that may be positive or negative. Sketch the steady-state temperature distribution associated with heat transfer in a plane wall for three cases corresponding to \(a>0\), \(a=0\), and \(a<0\).
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