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When the thermal conductivity of a medium varies linearly with temperature, is the average thermal conductivity always equivalent to the conductivity value at the average temperature?

Short Answer

Expert verified
Answer: No, the average thermal conductivity is not always equivalent to the conductivity value at the average temperature when the thermal conductivity of a medium varies linearly with temperature. The equivalence depends on the values of T1, T2, and the slope (m).

Step by step solution

01

Write the linear relationship between thermal conductivity and temperature

Let's assume the thermal conductivity (k) varies linearly with temperature (T). The linear equation for this relationship can be given as: k(T) = k_0 + m * T where k_0 is the initial thermal conductivity at T=0, m is the slope of the line, and T is the temperature.
02

Find the average thermal conductivity

Let's consider an arbitrary temperature range [T1, T2]. To find the average thermal conductivity, we integrate the function k(T) over the range and divide by the width of the range: \[\langle k \rangle = \frac{1}{T2-T1}\int_{T1}^{T2} (k_0 + mT)dT\] Now, perform the integration: \[\langle k \rangle = \frac{1}{T2-T1}[(k_0 T + m\frac{T^2}{2})\Big|_{T=T1}^{T=T2}] = \frac{1}{T2-T1}[(k_0 (T2-T1) + m\frac{T2^2-T1^2}{2})]\]
03

Find the value of thermal conductivity at the average temperature

To find the value of thermal conductivity at the average temperature, we first must find the average temperature: \[\langle T \rangle = \frac{T1+T2}{2}\] Now, plug the average temperature into the linear equation for k(T): k(\langle T\rangle) = k_0 + m * \langle T\rangle
04

Compare the two values

We now have expressions for the average thermal conductivity (\langle k \rangle) and thermal conductivity at the average temperature (k(\langle T\rangle)). Let's compare them to see if they are equal: \[\langle k \rangle = \frac{1}{T2-T1}[(k_0 (T2-T1) + m\frac{T2^2-T1^2}{2})]\] \[k(\langle T\rangle) = k_0 + m * \langle T\rangle = k_0 + m * \frac{T1+T2}{2}\] Comparing both the expressions, we can see that they are not always equivalent. The equivalence of both the expressions depends on the values of T1, T2 and the slope (m). In conclusion, when the thermal conductivity of a medium varies linearly with temperature, the average thermal conductivity is not always equivalent to the conductivity value at the average temperature.

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Key Concepts

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

Linear Temperature-Conductivity Relationship
Understanding the linear temperature-conductivity relationship is crucial when it comes to materials that change their ability to conduct heat as their temperature changes. Essentially, this relationship indicates that the thermal conductivity, often denoted as \( k(T) \), increases or decreases directly in proportion to the temperature. The mathematical expression for this linear relationship is given by:

\[ k(T) = k_0 + m \times T \]
where \( k_0 \) represents the thermal conductivity at a baseline temperature of zero, \( m \) is the constant rate of change of conductivity with temperature, and \( T \) is the actual temperature. In practical terms, this means that if you know the rate at which a material's conductivity changes with temperature, you can predict its conductivity at any temperature within the range where the linear relationship holds true.

When dealing with engineering problems or scientific experiments, assuming a linear relationship simplifies calculations and models. However, it is essential to remember that not all materials exhibit a perfectly linear relationship between thermal conductivity and temperature across all temperatures. This linearity often is a good approximation within a certain range relevant to the application at hand.
Average Thermal Conductivity
The concept of average thermal conductivity comes into play when there's a need to estimate a single value of conductivity for a material over a range of temperatures. This is particularly useful in heat transfer calculations where exact values at every point are cumbersome to work with. Average thermal conductivity, represented by \( \langle k \rangle \), is determined by integrating the thermal conductivity over the desired temperature range and then dividing by that range.

The equation to calculate the average thermal conductivity between two temperatures \( T1 \) and \( T2 \) is:

\[ \langle k \rangle = \frac{1}{T2 - T1} \int_{T1}^{T2} (k_0 + mT) dT \]
After performing the integration, you will have a single value representing the thermal conductivity 'averaged out' across that range. This value is not the same as the thermal conductivity at the midpoint temperature, which is a common misconception. Instead, it accounts for how the conductivity varies across the entire interval, providing a more holistic and applicable value for engineering analyses and designs.
Integration in Heat Transfer
Integration is a powerful mathematical tool often used in heat transfer to deal with variables that change over a domain, such as time, space, or temperature. When calculating average thermal conductivity, integration allows us to consider the continuous variation of material properties across a temperature range. By integrating the function that describes the thermal conductivity over the desired temperature interval, we can obtain a representative value for the entire range.

The process involves calculating the area under the curve of the conductivity-temperature graph between two specified temperatures. This integral can often be visualized as the physical area under the line representing the linear temperature-conductivity relationship on a graph.

Applying integration in heat transfer problems simplifies the complex reality into workable solutions, making it a staple in thermal analysis. Proper use of integration harmonizes the varying conductivity into an effective average value that can be used predictively in designs and applications where temperature gradients are present.

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Most popular questions from this chapter

Consider the base plate of an \(800-W\) household iron with a thickness of \(L=0.6 \mathrm{~cm}\), base area of \(A=160 \mathrm{~cm}^{2}\), and thermal conductivity of \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be \(112^{\circ} \mathrm{C}\). Disregarding any heat loss through the upper part of the iron, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) obtain a relation for the variation of temperature in the base plate by solving the differential equation, and (c) evaluate the inner surface temperature. Answer: (c) \(117^{\circ} \mathrm{C}\)

A cylindrical nuclear fuel rod of \(1 \mathrm{~cm}\) in diameter is encased in a concentric tube of \(2 \mathrm{~cm}\) in diameter, where cooling water flows through the annular region between the fuel rod \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the concentric tube. Heat is generated uniformly in the rod at a rate of \(50 \mathrm{MW} / \mathrm{m}^{3}\). The convection heat transfer coefficient for the concentric tube surface is \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the surface temperature of the concentric tube is \(40^{\circ} \mathrm{C}\), determine the average temperature of the cooling water. Can one use the given information to determine the surface temperature of the fuel rod? Explain.

Heat is generated in a long \(0.3-\mathrm{cm}\)-diameter cylindrical electric heater at a rate of \(180 \mathrm{~W} / \mathrm{cm}^{3}\). The heat flux at the surface of the heater in steady operation is (a) \(12.7 \mathrm{~W} / \mathrm{cm}^{2}\) (b) \(13.5 \mathrm{~W} / \mathrm{cm}^{2}\) (c) \(64.7 \mathrm{~W} / \mathrm{cm}^{2}\) (d) \(180 \mathrm{~W} / \mathrm{cm}^{2}\) (e) \(191 \mathrm{~W} / \mathrm{cm}^{2}\)

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has an emissivity and an absorptivity of \(0.9\). The top surface \((x=0)\) temperature of the absorber is \(T_{0}=35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Show that the variation of temperature in the absorber plate can be expressed as \(T(x)=-\left(\dot{q}_{0} / k\right) x+T_{0}\), and determine net heat flux \(\dot{q}_{0}\) absorbed by the solar collector.

Consider a large plate of thickness \(L\) and thermal conductivity \(k\) in which heat is generated uniformly at a rate of \(\dot{e}_{\text {gen. }}\) One side of the plate is insulated while the other side is exposed to an environment at \(T_{\infty}\) with a heat transfer coefficient of \(h\). \((a)\) Express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) determine the variation of temperature in the plate, and \((c)\) obtain relations for the temperatures on both surfaces and the maximum temperature rise in the plate in terms of given parameters.

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