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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 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.

At a given instant of time, the temperature distribution within an infinite homogeneous body is given by the function $$ T(x, y, z)=x^{2}-2 y^{2}+z^{2}-x y+2 y z $$ Assuming constant properties and no internal heat generation, determine the regions where the temperature changes with time.

Typically, air is heated in a hair dryer by blowing it across a coiled wire through which an electric current is passed. Thermal energy is generated by electric resistance heating within the wire and is transferred by convection from the surface of the wire to the air. Consider conditions for which the wire is initially at room temperature, \(T_{i}\), and resistance heating is concurrently initiated with airflow at \(t=0\). (a) For a wire radius \(r_{o}\), an air temperature \(T_{\infty}\), and a convection coefficient \(h\), write the form of the heat equation and the boundary/initial conditions that govern the transient thermal response, \(T(r, t)\), of the wire. (b) If the length and radius of the wire are \(500 \mathrm{~mm}\) and \(1 \mathrm{~mm}\), respectively, what is the volumetric rate of thermal energy generation for a power consumption of \(P_{\text {elec }}=500 \mathrm{~W}\) ? What is the convection heat flux under steady-state conditions? (c) On \(T-r\) 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. (d) On \(q_{r}^{\prime \prime}-t\) coordinates, sketch the variation of the heat flux with time for locations at \(r=0\) and \(r=r_{o^{*}}\).

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 \(q_{\phi}\).

Two-dimensional, steady-state conduction occurs in a hollow cylindrical solid of thermal conductivity \(k=16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), outer radius \(r_{o}=1 \mathrm{~m}\) and overall length \(2 z_{o}=5 \mathrm{~m}\), where the origin of the coordinate system is located at the midpoint of the center line. The inner surface of the cylinder is insulated, and the temperature distribution within the cylinder has the form \(T(r, z)=a+b r^{2}+c \ln r+d z^{2}\), where \(a=\) \(-20^{\circ} \mathrm{C}, \quad b=150^{\circ} \mathrm{C} / \mathrm{m}^{2}, c=-12^{\circ} \mathrm{C}, d=-300^{\circ} \mathrm{C} / \mathrm{m}^{2}\) and \(r\) and \(z\) are in meters. (a) Determine the inner radius \(r_{i}\) of the cylinder. (b) Obtain an expression for the volumetric rate of heat generation, \(\dot{q}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\). (c) Determine the axial distribution of the heat flux at the outer surface, \(q_{r}^{\prime \prime}\left(r_{o}, z\right)\). What is the heat rate at the outer surface? Is it into or out of the cylinder? (d) Determine the radial distribution of the heat flux at the end faces of the cylinder, \(q_{r}^{\prime \prime}\left(r,+z_{o}\right)\) and \(q_{r}^{\prime \prime}\left(r,-z_{o}\right)\). What are the corresponding heat rates? Are they into or out of the cylinder? (e) Verify that your results are consistent with an overall energy balance on the cylinder.

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