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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 \(a<0\).

Short Answer

Expert verified
In conclusion, for the given relationship between thermal conductivity 'k' and temperature 'T' as \( k = k_{0} + aT \), the steady-state temperature distribution in a plane wall is: - Linear for \(a = 0\) - Convex for \(a > 0\) - Concave for \(a < 0\)

Step by step solution

01

Fourier's Law of Heat Conduction

To begin, we'll use Fourier's law of heat conduction to relate the heat flux 'q' to the thermal conductivity 'k' and temperature gradient. Fourier's law is given by: \( q = -k \frac{dT}{dx} \) Given that the thermal conductivity 'k' has a temperature dependence, we can substitute the expression. So, \( q = -(k_{0} + aT) \frac{dT}{dx} \)
02

Analyzing the Case with a = 0

First, let's analyze the case when 'a = 0'. In this case, the thermal conductivity 'k' is constant and not dependent on temperature. The equation becomes: \( q = -k_{0} \frac{dT}{dx} \) In this case, the temperature distribution in the plane wall will be linear. This is due to the fact that the thermal conductivity is constant and therefore doesn't depend on temperature. As a result, the temperature gradient, and consequently the heat flux, will also remain constant.
03

Analyzing the Case with a > 0

Now, let's analyze the case when 'a > 0'. In this case, the thermal conductivity increases with an increase in temperature. The equation remains: \( q = -(k_{0} + aT) \frac{dT}{dx} \) In this case, as the temperature increases, the thermal conductivity also increases, making it easier for heat to be conducted. As a result, the heat flux will increase and the temperature gradient will become more pronounced. This results in a convex temperature profile across the plane wall.
04

Analyzing the Case with a < 0

Finally, let's analyze the case when 'a < 0'. In this case, the thermal conductivity decreases with an increase in temperature. The equation remains: \( q = -(k_{0} + aT) \frac{dT}{dx} \) In this case, as the temperature increases, the thermal conductivity decreases, making it harder for heat to be conducted with increased temperature. As a result, the heat flux will decrease and the temperature gradient will become less pronounced. This results in a concave temperature profile across the plane wall. In conclusion, for the given relationship between thermal conductivity 'k' and temperature 'T', the steady-state temperature distribution is: - Linear for \(a = 0\) - Convex for \(a > 0\) - Concave for \(a < 0\)

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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
Fourier's Law of Heat Conduction is a fundamental principle in the study of heat transfer within solids. It describes how heat energy moves from regions of higher temperature to lower temperature.

Mathematically, Fourier's Law is expressed as: \[ q = -k \frac{dT}{dx} \] where \(q\) is the heat flux, \(k\) is the thermal conductivity of the material, \(dT\) represents the temperature differential, and \(dx\) denotes the change in distance along the direction of heat transfer.

The negative sign indicates that heat flows in the direction of decreasing temperature. In the context of the exercise, a temperature-dependent thermal conductivity is considered, reflected by the equation \(k = k_{o} + aT\). When \(a\) equals zero, thermal conductivity is constant, leading to a straightforward linear temperature distribution, which is a direct consequence of Fourier's Law.
Thermal Conductivity
Thermal conductivity, represented by the symbol \(k\), is a measure of a material's ability to conduct heat. It is an inherent property that varies among different substances and can depend on factors such as temperature, as shown in the given exercise.

The equation \(k = k_{o} + aT\) highlights how thermal conductivity can vary with temperature. In this instance:\(k_{o}\) is a baseline thermal conductivity, while \(a\) is a coefficient that adjusts conductivity based on the material's temperature. This coefficient can be positive or negative, leading to different effects on how heat spreads through the material. Different values of \(a\) result in varying temperature distribution profiles, which is crucial in the design and analysis of systems requiring thermal management.
Steady-State Heat Transfer
Steady-state heat transfer refers to the condition when the temperature distribution in a material does not change over time, meaning that the heat entering a system equals the heat leaving it.

In the steady-state, the heat flux, temperature gradients, and thermal conductivity achieve an equilibrium where their values remain constant over time. The physical interpretation of the steady-state condition is that, after an initial transitory phase, the system has settled into a pattern of heat flow that does not vary. The exercise demonstrates this by evaluating how different thermal conductivity relationships with temperature, characterized by the coefficient \(a\), affect the steady-state temperature distribution across a plane wall, whether it is linear, convex, or concave. This understanding is essential for engineers and scientists as they design materials and structures to handle thermal loads effectively.

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Most popular questions from this chapter

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 \((x=0, L)\).

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?

The cylindrical system illustrated has negligible variation of temperature in the \(r\) - and \(z\)-directions. Assume that \(\Delta r=r_{o}-r_{i}\) is small compared to \(r_{i}\), and denote the length in the z-direction, normal to the page, as \(L\). (a) Beginning with a properly defined control volume and considering energy generation and storage effects, derive the differential equation that prescribes the variation in temperature with the angular coordinate \(\phi\). Compare your result with Equation 2.26. (b) For steady-state conditions with no internal heat generation and constant properties, determine the temperature distribution \(T(\phi)\) in terms of the constants \(T_{1}, T_{2}, r_{i}\), and \(r_{\sigma}\). Is this distribution linear in \(\phi\) ? (c) For the conditions of part (b) write the expression for the heat rate \(q_{\phi}\).

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.

Assume steady-state, one-dimensional heat conduction through the symmetric shape shown. Assuming that there is no internal heat generation, derive an expression for the thermal conductivity \(k(x)\) for these conditions: \(A(x)=(1-x), \quad T(x)=300\) \(\left(1-2 x-x^{3}\right)\), and \(q=6000 \mathrm{~W}\), where \(A\) is in square meters, \(T\) in kelvins, and \(x\) in meters.

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