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Chapter 2: Problem 16

Steady-state, one-dimensional conduction occurs in a rod of constant thermal conductivity \(k\) and variable crosssectional area \(A_{x}(x)=A_{o} e^{a x}\), where \(A_{o}\) and \(a\) are constants. The lateral surface of the rod is well insulated. (a) Write an expression for the conduction heat rate, \(q_{x}(x)\). Use this expression to determine the temperature distribution \(T(x)\) and qualitatively sketch the distribution for \(T(0)>T(L)\). (b) Now consider conditions for which thermal energy is generated in the rod at a volumetric rate \(\dot{q}=\dot{q}_{s} \exp (-a x)\), where \(\dot{q}_{o}\) is a constant. Obtain an expression for \(q_{x}(x)\) when the left face \((x=0)\) is well insulated.

Short Answer

Expert verified
We found that the conduction heat rate, \(q_{x}(x)\), can be expressed as: \[q_{x}(x) = -k A_{o} e^{ax} \frac{dT(x)}{dx}\] Applying the boundary conditions, we found that the temperature distribution, \(T(x)\), can be described by: \[T(x) = -\frac{1}{k A_{o}} \int q_{x}(x) e^{-ax} dx + C\] For the case with thermal energy generation, the heat rate, \(q_{x}(x)\), is given by: \[q_{x}(x) = \frac{k A_{o} e^{ax}}{A_{o}} \left.\frac{dT(x)}{dx} \right|_{x=0} + \dot{q}_{s}\]

Step by step solution

01

We start by recalling the one-dimensional conduction heat rate equation, which can be described as: \[q_{x}(x) = -k A_{x}(x) \frac{dT(x)}{dx}\] This equation states that the conduction heat rate depends on the thermal conductivity, the cross-sectional area, and the temperature gradient. #Step 2: Set up and solve the temperature distribution expression#

We insert the given in the heat rate equation which is \(A_{x}(x) = A_{o} e^{ax}\): \[q_{x}(x) = -k A_{o} e^{ax} \frac{dT(x)}{dx}\] Now, to find the temperature distribution, we need to integrate this expression with respect to \(x\). Separate variables and rearrange to get: \[\frac{dT(x)}{dx} = -\frac{q_{x}(x)}{k A_{o} e^{ax}}\] Now, integrate both sides with respect to \(x\): \[T(x) = -\frac{1}{k A_{o}} \int q_{x}(x) e^{-ax} dx + C\] where \(C\) is the integration constant. #Step 3: Determine boundary conditions and apply them#
02

Since we aim to get the qualitative sketch rather than the actual temperature values, we can assume the boundary conditions as \(T(0) = T_{1}\) and \(T(L) = T_{2}\). It is mentioned that \(T(0) > T(L)\), i.e., \(T_{1} > T_{2}\), meaning the rod is hotter on the left side and colder on the right side. Therefore, the conduction heat rate, \(q_{x}(x)\), will be flowing from the left (hotter) to the right (colder). Hence, \(q_{x}(x)\) is constant since the lateral surface is well insulated, and the temperature distribution is described by the expression obtained in step 2. #Step 4: (b) - Apply the conditions for thermal energy generation in the rod#

For the second part of the problem, we are given that the thermal energy is generated at a volumetric rate of \(\dot{q} = \dot{q}_{s} e^{-ax}\). We can expand the heat diffusion equation by adding the thermal generation term: \[q_{x}(x) = -k A_{x}(x) \frac{dT(x)}{dx} + A_{x}(x) \dot{q}\] Insert the given values of \(A_{x}(x)\) and \(\dot{q}\) into this equation: \[q_{x}(x) = -k A_{o} e^{ax}\frac{dT(x)}{dx} + A_{o} e^{ax} \dot{q}_{s} e^{-ax}\] Now, it is mentioned that the left face \((x = 0)\) is well insulated. This means that no heat can flow through the left face, and \(q_{x}(0) = 0\). Let's apply this condition to our equation: \[-k A_{o} \left. \frac{dT(x)}{dx} \right|_{x=0} + A_{o} \dot{q}_{s} = 0\] Simplify the equation, and we have the expression for the conduction heat rate, \(q_{x}(x)\): \[q_{x}(x) = \frac{k A_{o} e^{ax}}{A_{o}} \left.\frac{dT(x)}{dx} \right|_{x=0} + \dot{q}_{s}\]

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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 of materials that describes their ability to conduct heat. It is denoted by the symbol \(k\) in equations. This property plays a significant role in heat conduction processes, as it dictates how easily heat can flow through a material. Materials with high thermal conductivity, like metals, transfer heat quickly, while those with low thermal conductivity, like wood, are better insulators.

In the context of the rod described in the problem, thermal conductivity determines how fast heat moves along its length. The equation for heat conduction, \(q_{x}(x) = -k A_{x}(x) \frac{dT(x)}{dx}\), shows the dependence on \(k\). This tells us that, for a given temperature gradient \(\frac{dT}{dx}\), higher thermal conductivity means a higher rate of heat transfer.
  • High \(k\): Efficient heat transfer.
  • Low \(k\): Better for insulation needs.
Understanding this property helps us predict how different materials will behave in thermal applications.
Temperature Distribution
Temperature distribution refers to how temperature varies over a certain region or object. It is often described mathematically as a function of position, such as \(T(x)\) along a rod. In the given problem, the temperature distribution is determined by integrating the expression for heat conduction.

To derive \(T(x)\), the expression \(\frac{dT(x)}{dx} = -\frac{q_{x}(x)}{k A_{o} e^{ax}}\) must be integrated, resulting in a temperature profile that shows how temperature changes from one end of the rod to the other. The boundary condition \(T(0) > T(L)\) indicates the temperature is highest at the start \(x = 0\) and decreases towards \(x = L\).

This distribution can qualitatively be understood by how efficiently the rod conducts heat, maintaining a steady decline in temperature along its length. The constant lateral insulation ensures that heat transfer is only one-dimensional, simplifying the analysis and what happens to heat through the rod.
Volumetric Heat Generation
Volumetric heat generation refers to the production of heat within a material. This process is denoted in the exercise by the volumetric rate \(\dot{q} = \dot{q}_{s} e^{-ax}\), indicating that heat is being generated and distributed within the rod at a varying rate depending on \(x\).

In scenarios where heat is not only conducted but also generated within the object, the heat conduction equation is modified. The equation \(q_{x}(x) = -k A_{x}(x) \frac{dT(x)}{dx} + A_{x}(x) \dot{q}\) accounts for both conduction and this internal generation.
  • Heat is generated internally, adding to the complexity of heat transfer analysis.
  • The rate of generation varies with position, affecting the temperature profile.
When applying this concept to the rod, the insulation condition at \(x = 0\) (no heat flow through the left face) helps balance the energy generated internally, ensuring proper energy conservation and leading to the specific understanding of \(q_{x}(x)\). The presence of a heat source inside the material alters how quickly temperatures change, indicating areas of higher heat generation and complex patterns of heat flow.

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Most popular questions from this chapter

Consider a one-dimensional plane wall with constant properties and uniform internal generation \(\dot{q}\). The left face is insulated, and the right face is held at a uniform temperature. (a) Using the appropriate form of the heat equation, derive an expression for the \(x\)-dependence of the steady-state heat flux \(q^{\prime \prime}(x)\). (b) Using a finite volume spanning the range \(0 \leq\) \(x \leq \xi\), derive an expression for \(q^{\prime \prime}(\xi)\) and compare the expression to your result for part (a).

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 solid, truncated cone serves as a support for a system that maintains the top (truncated) face of the cone at a temperature \(T_{1}\), while the base of the cone is at a temperature \(T_{2}

Passage of an electric current through a long conducting rod of radius \(r_{i}\) and thermal conductivity \(k_{r}\) results in uniform volumetric heating at a rate of \(\dot{q}\). The conducting rod is wrapped in an electrically nonconducting cladding material of outer radius \(r_{o}\) and thermal conductivity \(k_{c}\), 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 appropriate boundary conditions for the solution of these equations.

An apparatus for measuring thermal conductivity employs an electrical heater sandwiched between two identical samples of diameter \(30 \mathrm{~mm}\) and length \(60 \mathrm{~mm}\), which are pressed between plates maintained at a uniform temperature \(T_{o}=77^{\circ} \mathrm{C}\) by a circulating fluid. A conducting grease is placed between all the surfaces to ensure good thermal contact. Differential thermocouples are imbedded in the samples with a spacing of \(15 \mathrm{~mm}\). The lateral sides of the samples are insulated to ensure onedimensional heat transfer through the samples. (a) With two samples of SS 316 in the apparatus, the heater draws \(0.353 \mathrm{~A}\) at \(100 \mathrm{~V}\), and the differential thermocouples indicate \(\Delta T_{1}=\Delta T_{2}=25.0^{\circ} \mathrm{C}\). What is the thermal conductivity of the stainless steel sample material? What is the average temperature of the samples? Compare your result with the thermal conductivity value reported for this material in Table A.1. (b) By mistake, an Armco iron sample is placed in the lower position of the apparatus with one of the SS316 samples from part (a) in the upper portion. For this situation, the heater draws \(0.601 \mathrm{~A}\) at \(100 \mathrm{~V}\), and the differential thermocouples indicate \(\Delta T_{1}=\Delta T_{2}=\) \(15.0^{\circ} \mathrm{C}\). What are the thermal conductivity and average temperature of the Armco iron sample? (c) What is the advantage in constructing the apparatus with two identical samples sandwiching the heater rather than with a single heater-sample combination? When would heat leakage out of the lateral surfaces of the samples become significant? Under what conditions would you expect \(\Delta T_{1} \neq \Delta T_{2} ?\)
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