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Chapter 2: Problem 23

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 \(T\left({ }^{\circ} \mathrm{C}\right)=a+b \mathrm{x}^{2}\). 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 \(q\) 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
(a) Heat generation rate: 200,000 W/m鲁. (b) Heat fluxes: 0 W/m虏 at x=0 and 10,000 W/m虏 at x=0.05 m. They are related by the balance and distribution of the generated heat.

Step by step solution

01

Determine the Heat Generation Rate

In a steady-state condition with uniform heat generation, the heat equation in one-dimension is given by \( \frac{d^2T}{dx^2} = -\frac{q}{k} \), where \( q \) is the heat generation rate, and \( k \) is the thermal conductivity. By differentiating the given temperature distribution function \( T(x) = a + b x^2 \), we can find the second derivative: \( \frac{dT}{dx} = 2bx \) and \( \frac{d^2T}{dx^2} = 2b \). Substitute \( 2b \) into the heat equation: \(-\frac{q}{k} = 2b\).Now solve for \( q \):\[ q = -2b \times k = -2(-2000) \times 50 = 200,000 \text{ W/m}^3 \]
02

Calculate Heat Flux at Wall Faces

To calculate the heat flux, use Fourier's law, \( q'' = -k \frac{dT}{dx} \). First, find \( \frac{dT}{dx} = 2bx \). Evaluate this expression at the wall faces:At \( x = 0 \):\[ \frac{dT}{dx} = 2b \times 0 = 0 \]\[ q''_0 = -k \times 0 = 0 \text{ W/m}^2 \]At \( x = 0.05 \):\[ \frac{dT}{dx} = 2b \times 0.05 = 2(-2000) \times 0.05 = -200 \]\[ q''_{0.05} = -k \times (-200) = 50 \times 200 = 10,000 \text{ W/m}^2 \]
03

Relate Heat Flux to Generation Rate

The difference in heat flux between the two faces (0 and 0.05 meters) represents the net heat generated within that section of the wall. The calculated heat generation rate \( 200,000 \text{ W/m}^3 \), after considering the total volume (\( A \cdot \Delta x \), where \( A \) is the cross-sectional area) for thickness \( \Delta x = 0.05 \) meters, matches the total heat transfer across the wall per square meter. The net flux is implicitly balanced by the internal generation throughout the material.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a fundamental property that measures a material's ability to conduct heat. It is denoted by the symbol \( k \). This property is crucial in determining how heat is transferred through materials. Conductivity values vary significantly across different materials. For instance, metals typically have high thermal conductivity, which means they can easily conduct heat, while materials like wood or rubber have much lower conductivity.
In the exercise, the thermal conductivity \( k \) of the wall material is given as \( 50 \text{ W/m} \cdot \text{K} \). This value tells us that 50 watts of thermal energy per meter of thickness per degree Kelvin of temperature difference will flow through the wall. A higher thermal conductivity implies that heat can be transferred more efficiently through the material.
Steady-State Temperature Distribution
In the context of heat transfer, a steady-state temperature distribution means that the temperature at any given point in a material does not change over time. There is a balance between heat entering and leaving any section of the material, and so the temperature remains constant.
In the above problem, the equation for temperature distribution is given as \( T(x) = a + bx^2 \), where \( a \) and \( b \) are constants. This parabolic distribution is typical when uniform heat generation occurs within a material. For our specific example, the steady temperature is described as
  • \( a = 200^{\circ} \text{C} \)
  • \( b = -2000^{\circ} \text{C} / \text{m}^2 \)
The term \( bx^2 \) shows that the temperature changes with the square of the position \( x \) within the wall.
Heat Generation Rate
The heat generation rate, represented by \( q \), measures how much heat is generated per unit volume within a material. This is especially important for materials that produce heat internally, such as electronic components or reactors.
Using the second derivative of the temperature distribution equation, we find the heat generation rate by solving \[-\frac{q}{k} = \frac{d^2T}{dx^2}\]. Substituting the values from our equation gives us \[q = -2b \times k = -2(-2000) \times 50 = 200,000 \text{ W/m}^3\]. This result indicates that 200,000 watts of heat is generated per cubic meter of the wall, providing a greater insight into the heat dynamics occurring within the material.
Fourier's Law
Fourier's Law is a cornerstone of heat transfer and defines the relationship between heat flux and temperature gradient within a material. It states that the heat flux \( q'' \) is proportional to the negative gradient of the temperature, \[q'' = -k \frac{dT}{dx}\]
This law is instrumental in understanding how heat flows through materials. In our exercise, we applied Fourier's Law to determine the heat flux at the faces of the wall. It was calculated that:
  • At \( x = 0 \), the heat flux \( q'' \) was zero.
  • At \( x = 0.05 \) meters, the flux was 10,000 \( \text{ W/m}^2 \).
This means the heat is being transferred away from the hotter inside regions towards the cooler regions outside, as predicted by Fourier's Law. The net heat flux is balanced by the internal heat generation, ensuring a consistent temperature distribution.

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Most popular questions from this chapter

A method for determining the thermal conductivity \(k\) and the specific heat \(c_{p}\) of a material is illustrated in the sketch. Initially the two identical samples of diameter \(D=60 \mathrm{~mm}\) and thickness \(I\), \(=10 \mathrm{~mm}\) and the thin heater are at a uniform temperature of \(T_{i}=23.00^{\circ} \mathrm{C}\). while surrounded by an insulating powder. Suddenly the heater is energized to provide a uniform heat flux \(q_{e}^{\prime \prime}\) on each of the sample interfaces, and the heat fiux is maintained constant for a period of time, \(\Delta t_{t r}\) A short time after sudden heating is initiated, the temperature at this interface \(T_{\omega}\) is related to the heat flux as $$ T_{d}(t)-T_{i}=2 e^{\prime \prime}\left(\frac{t}{\pi \alpha c k}\right)^{i / 2} $$ For a particular test run, the electrical heater dissipates 15.0 W for a period of \(\Delta t_{e}=120 \mathrm{~s}\) and the temperature at the interface is \(T,(30 \mathrm{~s})=24.57^{\circ} \mathrm{C}\) after \(30 \mathrm{~s}\) of heating. A long time after the heiter is deenergized, \(t>\Delta t_{\text {e. }}\) the samples reach the uniform temperature of \(T_{d}(\propto)=\) \(33.50^{\circ} \mathrm{C}\). The density of the sample materials, determined by measurement of volume and mass, is \(\rho=\) \(3965 \mathrm{~kg} / \mathrm{m}^{3}\), Determine the specific heat and thermal conductivity of the test material. By looking at values of the thermophysical properties in Table A.I or A.2, identify the test sample material.

A plane layer of coal of thickness \(L=1 \mathrm{~m}\) experiences uniform volumetric generation at a rate of \(\hat{q}=20 \mathrm{~W} / \mathrm{m}^{2}\) due to slow exidation of the coal particles. Averuged over a daily period, the top surface of the layer transfers heat by convection to ambient air for which \(h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\mathrm{m}}=25^{\circ} \mathrm{C}\), while receiving solar irrudiation in the amoent \(G_{5}=400 \mathrm{~W} / \mathrm{m}^{2}\). Irnatiation from the amosphere may be neglected. The solar absorptivity and emissivity of the surface are each \(\alpha_{S}=\varepsilon=0.95\). (a) Write the steady-sate form of the heat diffusion equation for the layer of coul. Verify that this equation is satisfied by a temperiture distribution of the form $$ T(x)=T_{x}+\frac{\dot{q} L^{2}}{2 k}\left(1-\frac{x^{2}}{L^{2}}\right) $$ From this distribution, what can you say about conditions at the bottom surface \((x=0)\) ? Sketch the temperature distribution and label key features. (b) Obtain an expression for the rate of heat transfer by conduction per unit area at \(x=L\). Applying an energy balance to a control surface about the top surface of the layer, obtain an expression for \(T_{r}\) Evaluate \(T_{4}\) and \(T(0)\) for the prescribed conditions. (c) Daily average values of \(G_{5}\) and \(h\) depend on a number of factors such as time of year, cloud cover, and wind conditions. For \(h=5 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\), compute and plot \(T_{\text {t and }} T(0)\) as a function of \(G_{5}\) for \(50 \leq G_{s} \leq 500 \mathrm{~W} / \mathrm{m}^{2}\). For \(G_{5}=400 \mathrm{~W} / \mathrm{m}^{2}\), compute and plot \(T_{\text {, and }} T(0)\) as a function of \(h\) for \(5 \leq h \leq 50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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

A TV advertisement by a well-known insulation manufacturer states: it isn't the thickness of the insulating material that counts, it's the R-value. The ad shows that to ottain an R-value of 19, you need \(18 \mathrm{ft}\) of rock, 15 in. of wood. or just 6 in. of the manufacturer's insulation. Is this advertisement techinically reasonable? If you are like most TV viewers, you don't know the R-value is defined as \(L / k\). where \(L\). \(\mathrm{in}\).) is the thickness of the insulation and \(k\) (Btu * in \(/ \mathrm{hr}-\mathrm{ft}^{2}+{ }^{\mathrm{F}}\) ) is the thermal conductivity of the muterial.

Parsage of an electric current through a long conducting rod of radius \(r_{i}\) and themal conductivify \(k_{r}\) results in uniform volumetric heating at a nute of \(\phi\). The conducting rod is wrapped in an clectrically nonconducting cladding material of outer radius \(r_{e}\) and thermal conductivity \(k_{0}\) 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 appropriste boundary conditions for the solution of these equations.
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