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Chapter 2: Problem 40

The steady-state temperature distribution in a onedimensional wall of thermal conductivity \(k\) and thickness \(L\) is of the form \(T=a x^{3}+b x^{2}+c x+d .\) Derive expressions for the heat generation rate per unit volume in the wall and the heat fluxes at the two wall faces \((x=0, L)\).

Short Answer

Expert verified
The heat generation rate per unit volume in the wall is given by: \(q(x) = -k(6ax + 2b)\) And the heat fluxes at the two wall faces are: \(q_{0} = -kc\) \(q_{L} = -k(3aL^2 + 2bL + c)\)

Step by step solution

01

Calculate the first and second derivatives of the temperature function

First, we need to calculate the first and second derivatives of the given temperature function with respect to x: \(T(x) = ax^3 + bx^2 + cx + d\) First derivative: \(\frac{dT}{dx} = 3ax^2 + 2bx + c\) Second derivative: \(\frac{d^2T}{dx^2} = 6ax + 2b\)
02

Calculate the heat generation rate per unit volume

Now, we can use the heat conduction equation to find the heat generation rate per unit volume (\(q\)): \(\frac{d^2 T}{dx^2} = - \frac{q}{k}\) Substitute the second derivative of T into the equation: \(6ax + 2b = - \frac{q}{k}\) We need to find 'q', so multiply both sides by -k: \(q = -k(6ax + 2b)\) Thus, the heat generation rate per unit volume in the wall is: \(q(x) = -k(6ax + 2b)\)
03

Calculate the heat fluxes at the two wall faces

We can now use the first derivative and the heat flux equation to find the heat fluxes at the two wall faces (x = 0, L): \(q_{x} = -k\frac{dT}{dx}\) Substitute the first derivative of T into the equation: \(q_{x} = -k(3ax^2 + 2bx + c)\) At \(x = 0\), the heat flux is: \(q_{0} = -k(3a(0)^2 + 2b(0) + c)\) Therefore, \(q_{0} = -kc\) At \(x = L\), the heat flux is: \(q_{L} = -k(3aL^2 + 2bL + c)\) In summary, the heat generation rate per unit volume in the wall is: \(q(x) = -k(6ax + 2b)\) And the heat fluxes at the two wall faces are given by: \(q_{0} = -kc\) \(q_{L} = -k(3aL^2 + 2bL + c)\)

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Key Concepts

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

Thermal Conductivity
Understanding thermal conductivity is crucial when studying heat transfer in materials. Thermal conductivity, represented by the symbol \(k\), is a measure of a material's ability to conduct heat. It specifies the amount of heat that passes through a unit area of the material with a given temperature gradient per unit time. The higher the value of \(k\), the more efficient the material is at transferring heat. For instance, metals like copper have high thermal conductivity, which is why they feel colder on touch compared to wood, which has a low \(k\) value. In the textbook exercise, the coefficient \(k\) appears in the equation \(q(x) = -k(6ax + 2b)\), depicting the relationship between the material's thermal conductivity and the rate of heat being generated per unit volume within the wall.
Temperature Distribution
Temperature distribution refers to how temperature varies within an object. It鈥檚 often described by a mathematical function, such as \(T=ax^3+bx^2+cx+d\) in the provided exercise, which shows how temperature changes along the one-dimensional wall. Uniform temperature distribution implies that every part of the object is at the same temperature, which is rare in real-world scenarios due to various factors such as heat sources and sinks. Non-uniform temperature distribution, as illustrated in the exercise, is characterized by the temperature changing as a function of position. This is essential for predicting how heat will flow through the material, which is based on the principle that heat moves from regions of higher temperature to regions of lower temperature. Understanding temperature distribution helps in controlling heat within systems, such as in building insulation or electronics cooling.
Heat Flux
Heat flux is the rate of heat energy transfer per unit surface area. It's a vector quantity, which means it has both a magnitude and a direction. In the context of the textbook problem, the heat flux \(q_x\) can be calculated using the formula \(q_x = -k\frac{dT}{dx}\). The negative sign indicates that heat flows from higher to lower temperatures, adhering to the second law of thermodynamics. Heat flux can vary within a material if the temperature distribution is non-uniform. For example, at the wall faces \(x = 0\) and \(x = L\), the heat fluxes, \(q_0\) and \(q_L\), are determined not only by the thermal conductivity but also by how steeply the temperature gradient changes at these specific points. Analyzing heat flux is imperative for designing thermal management systems to ensure materials and structures can handle the anticipated heat loads without failure.

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Most popular questions from this chapter

A cylinder of radius \(r_{o}\), length \(L\), and thermal conductivity \(k\) is immersed in a fluid of convection coefficient \(h\) and unknown temperature \(T_{\infty}\). At a certain instant the temperature distribution in the cylinder is \(T(r)=a+b r^{2}\), where \(a\) and \(b\) are constants. Obtain expressions for the heat transfer rate at \(r_{o}\) and the fluid temperature.

A plane wall has constant properties, no internal heat generation, and is initially at a uniform temperature \(T_{i \cdot}\) Suddenly, the surface at \(x=L\) is heated by a fluid at \(T_{\infty}\) having a convection coefficient \(h\). At the same instant, the electrical heater is energized, providing a constant heat flux \(q_{o}^{\prime \prime}\) at \(x=0\). (a) On \(T-x\) coordinates, sketch the temperature distributions for the following conditions: initial condition \((t \leq 0)\), steady-state condition \((t \rightarrow \infty)\), and for two intermediate times. (b) On \(q_{x}^{\prime \prime}-x\) coordinates, sketch the heat flux corresponding to the four temperature distributions of part (a). (c) On \(q_{x}^{n}-t\) coordinates, sketch the heat flux at the locations \(x=0\) and \(x=L\). That is, show qualitatively how \(q_{x}^{\prime \prime}(0, t)\) and \(q_{x}^{\prime \prime}(L, t)\) vary with time. (d) Derive an expression for the steady-state temperature at the heater surface, \(T(0, \infty)\), in terms of \(q_{o}^{\prime \prime}\), \(T_{\infty}, k, h\), and \(L\).

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 \(a<0\).

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?

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, \(\dot{q}(x)\).
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