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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.

Short Answer

Expert verified
The expression for the thermal conductivity \(k(x)\) under the given conditions is: \(k(x) = \frac{6000}{(1-x)(600-900x^2)}\)

Step by step solution

01

Write down the heat conduction equation

For one-dimensional steady-state heat conduction with no internal heat generation, the heat conduction equation is: \(q = -kA(x)\frac{dT(x)}{dx}\)
02

Solve for k(x)

Divide both sides by -A(x) to isolate k(x): \(k(x) = -\frac{q}{A(x)\frac{dT(x)}{dx}}\)
03

Differentiate T(x) with respect to x

Differentiate T(x) with respect to x: \(\frac{dT(x)}{dx} = \frac{d}{dx}(300(1-2x-x^3))\) We have: \(\frac{dT(x)}{dx} = -600 + 900x^2\)
04

Substitute the given values of A(x), T(x), and q

Now, substitute the given values of A(x), T(x), and q into the expression for k(x): \(k(x) = -\frac{6000 \, \text{W}}{(1-x)(-600+900x^2)}\)
05

Simplify the expression for k(x)

Finally, simplify the expression for k(x): \(k(x) = \frac{6000}{(1-x)(600-900x^2)}\) This is the expression for the thermal conductivity k(x) under the given conditions.

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Key Concepts

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

Heat Conduction
Heat conduction is the process where thermal energy is transferred from areas of high temperature to areas of low temperature within a material or between materials in direct contact. Imagine a hot coffee mug; heat travels from the hot coffee, through the mug, and into your hands. This movement of heat is what we refer to as conduction.

In our example exercise, we're focusing on heat conduction along a single-dimensional path. This means heat moves in one direction, much like a train moving along its tracks. The equation used for this process, \[q = -kA(x)\frac{dT(x)}{dx}\]illustrates how the rate of energy transfer (q) relies on three primary factors:
  • Thermal conductivity \(k\), which indicates how well a material can conduct heat.
  • The cross-sectional area \(A(x)\) through which heat is moving.
  • The temperature gradient \(\frac{dT(x)}{dx}\), or the rate at which temperature changes along the material.
Understanding these components can help in designing materials with the desired level of heat conduction, crucial in fields ranging from electronics to building insulation.
Steady-State
Steady-state is a term used to describe a system in thermal equilibrium. This means that, despite the ongoing transfer of heat, the temperatures within the system are not changing over time. In other words, input and output of energy are balanced, resulting in a stable, unchanging temperature distribution.

In many engineering problems, assuming a steady-state simplifies the analysis because we don't have to deal with time-dependent changes. In our exercise, the heat conduction is steady-state as the temperature distribution described by \(T(x) = 300 (1-2x-x^{3})\) does not have any time-related variables.
When dealing with steady-state conditions, engineers can focus on spatial variations rather than worrying about how these temperatures evolve with time, making calculations and designs more straightforward.
One-Dimensional Conduction
One-dimensional conduction refers to the movement of heat in a single direction, as opposed to multi-dimensional cases where heat diffuses outward in multiple directions.

Think about a metal rod heated at one end. In this case, heat travels linearly along the rod, making it a one-dimensional conduction problem. This simplifies the mathematical treatment compared to, say, a sphere or a block where heat might move in all directions.
In our exercise, the function for area \(A(x) = (1-x)\) highlights this simplicity—only changes along this single dimension (the x-axis) are considered. This focus is crucial to developing accurate models for heat conduction in scenarios where components have a linear heating or cooling design.
Furthermore, such assumptions reduce computational complexity and make it easier for engineers to predict thermal behavior.
Differential Equations
Differential equations are mathematical tools essential for solving problems involving rates of change, such as heat conduction. They describe how a quantity changes over another, which is particularly useful when dealing with processes like heat flow.

In our specific problem, the differential equation part is represented as \(\frac{dT(x)}{dx}\), which is the temperature gradient—the rate of change of temperature concerning distance. Solving this involves taking derivatives, like we did for the function \(T(x) = 300(1-2x-x^3)\), resulting in \(-600 + 900x^2\).
Being familiar with differential equations allows one to model and predict complex thermal systems accurately. This not only helps in understanding current setups but also in innovating new applications where controlled heat flow is critical, such as in sustainability technology or advanced manufacturing.

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Most popular questions from this chapter

A cylindrical rod of stainless steel is insulated on its exterior surface except for the ends. The steady-state temperature distribution is \(T(x)=a-b x / L\), where \(a=305 \mathrm{~K}\) and \(b=10 \mathrm{~K}\). The diameter and length of the rod are \(D=20 \mathrm{~mm}\) and \(L=100 \mathrm{~mm}\), respectively. Determine the heat flux along the rod, \(q_{x}^{\prime \prime}\). Hint: The mass of the rod is \(M=0.248 \mathrm{~kg}\).

A plane wall that is insulated on one side \((x=0)\) is initially at a uniform temperature \(T_{i}\), when its exposed surface at \(x=L\) is suddenly raised to a temperature \(T_{s}\). (a) Verify that the following equation satisfies the heat equation and boundary conditions: $$ \frac{T(x, t)-T_{s}}{T_{i}-T_{s}}=C_{1} \exp \left(-\frac{\pi^{2}}{4} \frac{\alpha t}{L^{2}}\right) \cos \left(\frac{\pi}{2} \frac{x}{L}\right) $$ where \(C_{1}\) is a constant and \(\alpha\) is the thermal diffusivity. (b) Obtain expressions for the heat flux at \(x=0\) and \(x=L\). (c) Sketch the temperature distribution \(T(x)\) at \(t=0\), at \(t \rightarrow \infty\), and at an intermediate time. Sketch the variation with time of the heat flux at \(x=L, q_{L}^{\prime \prime}(t)\). (d) What effect does \(\alpha\) have on the thermal response of the material to a change in surface temperature?

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 \(0 \leq r \leq r_{2}\).

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 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?

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