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Chapter 2: Problem 30

The steady-state temperature distribution in a onedimensional wall of thermal conductivity \(50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(50 \mathrm{~mm}\) is observed to be \(T\left({ }^{\circ} \mathrm{C}\right)=a+b x^{2}\), where \(a=200^{\circ} \mathrm{C}, b=-2000^{\circ} \mathrm{C} / \mathrm{m}^{2}\), and \(x\) is in meters. (a) What is the heat generation rate \(\dot{q}\) in the wall? (b) Determine the heat fluxes at the two wall faces. In what manner are these heat fluxes related to the heat generation rate?

Short Answer

Expert verified
The heat generation rate in the wall is \(200000 \mathrm{W} / \mathrm{m}^3\). The heat flux at the left wall face is \(0 \mathrm{W} / \mathrm{m}^2\), and the heat flux at the right wall face is \(10000 \mathrm{W} / \mathrm{m}^2\). The heat fluxes show that all the heat generated within the wall is being transferred to the right wall face, with no net heat flow at the left face.

Step by step solution

01

Write down the temperature distribution function

: We are provided with the steady-state temperature distribution function in the wall as: \(T(x) = a + b x^2\), where \(a = 200^{\circ} \mathrm{C}\), \(b = - 2000^{\circ} \mathrm{C} / \mathrm{m}^2\), and \(x\) in meters.
02

Use the heat equation for a one-dimensional medium

: For a one-dimensional medium, the heat equation at steady state is given by: \(\frac{d}{dx}\left(k\frac{dT}{dx}\right) + \dot{q} = 0\), where \(k\) is the thermal conductivity and \(\dot{q}\) is the heat generation rate.
03

Compute the first and second derivatives of the temperature distribution

: To use the heat equation, we first need to find the first derivative of the given temperature distribution function: \(\frac{dT}{dx} = \frac{d}{dx}(a + b x^2) = 2bx\) Now, compute the second derivative: \(\frac{d^2T}{dx^2} = \frac{d^2}{dx^2}(a + b x^2) = 2b\)
04

Solve for the heat generation rate

: Now, we can plug the second derivative and the thermal conductivity into the heat equation to obtain the heat generation rate: \(\frac{d}{dx}\left(k\frac{dT}{dx}\right) + \dot{q} = 0\) \(\frac{d}{dx}\left(50\cdot 2bx\right) + \dot{q} = 0\) Substitute the value of \(b\): \(\frac{d}{dx}\left(50\cdot 2(-2000)x\right) + \dot{q} = 0\) Now, take the derivative and solve for the heat generation rate, \(\dot{q}\): \(-200000 + \dot{q} = 0\) \(\dot{q} = 200000 \mathrm{W} / \mathrm{m}^3\) So, the heat generation rate in the wall is \(200000 \mathrm{W} / \mathrm{m}^3\).
05

Obtain the heat fluxes at the two wall faces using Fourier's law

: Fourier's law of heat conduction states that: \(q_x = -k \frac{dT}{dx}\) For the heat flux at the left face (x=0): \(q_{x=0} = -50 \cdot 2b \cdot 0 = 0\) For the heat flux at the right face (x=50 mm = 0.05 m): \(q_{x=0.05} = -50 \cdot 2(-2000) \cdot 0.05 = 10000~\mathrm{W} / \mathrm{m}^2\)
06

Analyze the relation between heat fluxes and heat generation rate

: We have calculated the heat generation rate, \(\dot{q}\), which characterizes the heat generated uniformly within the material, and we have determined the heat fluxes at the two wall faces. We can observe that although there is heat generation within the wall, there is no net heat flow at the left face. The net heat flow at the right face matches the total heat generated in the wall, indicating that all the generated heat is being transferred to the right wall face as heat flux.

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Key Concepts

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

Steady-State Temperature Distribution
Understanding the steady-state temperature distribution in materials is crucial in various fields, including engineering and environmental science. This concept refers to a condition where the temperature in a material does not change with time despite the possible presence of heat sources, sinks, or flow. In a mathematic description, like in our exercise example, the temperature distribution function is expressed as a function of spatial coordinates and can be represented by an equation such as

\( T(x) = a + bx^2 \),

where \( a \) and \( b \) are constants, and \( x \) represents the position within the material. At steady state, the temperature's rate of change within any part of the system is zero. This simplifies analyzing heat transfer problems as it implies a balance between heat generated within the system and the heat lost to the surroundings.
Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It is denoted by the symbol \( k \) and is usually expressed in units of \( W/(m \cdot K) \). Higher thermal conductivity indicates that the material can transfer heat more efficiently. This property plays a pivotal role in determining the rate of heat transfer through materials. For instance, metals typically have high thermal conductivity and are therefore used in applications that require efficient heat dissipation, such as heatsinks in computers. Understanding thermal conductivity is essential when analyzing problems involving heat transfer, as it directly impacts the thermal behavior of the system, such as in the heat flux calculation in our exercise.
Heat Generation Rate
The heat generation rate, represented by \( \dot{q} \), is a term that quantifies the amount of heat produced within a unit volume of material per unit time. Its unit is \( W/m^3 \). In our exercise, the heat generation rate is the result of some process or reaction happening within the wall, which in steady-state conditions directly relates to the heat leaving the system. Situations involving heat generation include electrical resistive heating in conductors, chemical reactions, or radioactive decay. This concept is vital when we want to understand and manage the temperatures in devices or materials that generate heat internally.
Fourier's Law of Heat Conduction
Fourier's law of heat conduction is a foundational principle defining the relationship between the heat flux and the temperature gradient within a material. Mathematically, it's expressed as:

\( q_x = -k \frac{dT}{dx} \),

where \( q_x \) is the heat flux in the direction of x, \( k \) is the thermal conductivity, and \( \frac{dT}{dx} \) is the temperature gradient in the material. The negative sign shows that heat flows from high to low temperature. In our exercise, we apply Fourier's law to determine the heat fluxes at the wall faces. This law underpins much of the analysis in heat transfer problems, allowing for the evaluation of how heat is distributed or dissipated across a material.

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Most popular questions from this chapter

A pan is used to boil water by placing it on a stove, from which heat is transferred at a fixed rate \(q_{\sigma}\). There are two stages to the process. In Stage 1, the water is taken from its initial (room) temperature \(T_{i}\) to the boiling point, as heat is transferred from the pan by natural convection. During this stage, a constant value of the convection coefficient \(h\) may be assumed, while the bulk temperature of the water increases with time, \(T_{\infty}=T_{\infty}(t)\). In Stage 2, the water has come to a boil, and its temperature remains at a fixed value, \(T_{\infty}=T_{b}\), as heating continues. Consider a pan bottom of thickness \(L\) and diameter \(D\), with a coordinate system corresponding to \(x=0\) and \(x=L\) for the surfaces in contact with the stove and water, respectively. (a) Write the form of the heat equation and the boundary/ initial conditions that determine the variation of temperature with position and time, \(T(x, t)\), in the pan bottom during Stage 1. Express your result in terms of the parameters \(q_{o}, D, L, h\), and \(T_{\infty}\), as well as appropriate properties of the pan material. (b) During Stage 2, the surface of the pan in contact with the water is at a fixed temperature, \(T(L, t)=\) \(T_{L}>T_{b}\). Write the form of the heat equation and boundary conditions that determine the temperature distribution \(T(x)\) in the pan bottom. Express your result in terms of the parameters \(q_{o}, D, L\), and \(T_{L}\), as well as appropriate properties of the pan material.

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 plane wall of thickness \(2 L=40 \mathrm{~mm}\) and thermal conductivity \(k=5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) experiences uniform volumetric heat generation at a rate \(\dot{q}\), while convection heat transfer occurs at both of its surfaces \((x=-L,+L)\), each of which is exposed to a fluid of temperature \(T_{\infty}=20^{\circ} \mathrm{C}\). Under steady-state conditions, the temperature distribution in the wall is of the form \(T(x)=a+b x+c x^{2}\) where \(a=82.0^{\circ} \mathrm{C}, b=-210^{\circ} \mathrm{C} / \mathrm{m}, c=-2 \times 10^{4 \circ} \mathrm{C} / \mathrm{m}^{2}\), and \(x\) is in meters. The origin of the \(x\)-coordinate is at the midplane of the wall. (a) Sketch the temperature distribution and identify significant physical features. (b) What is the volumetric rate of heat generation \(\dot{q}\) in the wall? (c) Determine the surface heat fluxes, \(q_{x}^{\prime \prime}(-L)\) and \(q_{x}^{\prime \prime}(+L)\). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the surfaces at \(x=-L\) and \(x=+L\) ? (e) Obtain an expression for the heat flux distribution \(q_{x}^{\prime \prime}(x)\). Is the heat flux zero at any location? Explain any significant features of the distribution. (f) If the source of the heat generation is suddenly deactivated \((\dot{q}=0)\), what is the rate of change of energy stored in the wall at this instant? (g) What temperature will the wall eventually reach with \(\dot{q}=0\) ? How much energy must be removed by the fluid per unit area of the wall \(\left(\mathrm{J} / \mathrm{m}^{2}\right)\) to reach this state? The density and specific heat of the wall material are \(2600 \mathrm{~kg} / \mathrm{m}^{3}\) and \(800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

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.

A steam pipe is wrapped with insulation of inner and outer radii \(r_{i}\) and \(r_{o}\), respectively. At a particular instant the temperature distribution in the insulation is known to be of the form $$ T(r)=C_{1} \ln \left(\frac{r}{r_{o}}\right)+C_{2} $$ Are conditions steady-state or transient? How do the heat flux and heat rate vary with radius?
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