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Chapter 2: Problem 25

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 \(M=0.248 \mathrm{~kg}\).

Short Answer

Expert verified
The heat flux along the cylindrical stainless steel rod is \(q_{x}^{\prime \prime} = 150 \pi \frac{W}{m^2}\).

Step by step solution

01

Calculate the cross-sectional area of the rod

We know the diameter of the rod, D = 20 mm. We convert it to meters: \(D = 0.02 \ m\). To calculate the cross-sectional area, we can use the formula for the area of a circle, because the rod is cylindrical: \(A = \pi (D/2)^2\). \[A = \pi (0.02/2)^2 = \pi (0.01)^2 = \pi \times 10^{-4} \ m^2\]
02

Derive the temperature gradient

The given temperature distribution function is \(T(x) = a - b \frac{x}{L}\), where \(a = 305 \ K\), and \(b = 10 \ K\). To find the temperature gradient, we need to differentiate the function with respect to x: \[\frac{dT(x)}{dx} = \frac{d}{dx} \left( a - b \frac{x}{L} \right) = -b \frac{1}{L}\]
03

Apply Fourier's law of heat conduction

Fourier's law of heat conduction is given by: \[ q'_x = -kA\frac{dT(x)}{dx} \] Here, \(k\) is the thermal conductivity of stainless steel, which is approximately equal to \(15 \ \frac{W}{m \cdot K}\). Now substituting the values we found for the area and the temperature gradient: \[q'_x = -15 \times (\pi \times 10^{-4}) \times (-10)\frac{1}{0.1} = 150 \pi \frac{W}{m^2}\] The heat flux along the rod is \(q_{x}^{\prime \prime} = 150 \pi \frac{W}{m^2}\).

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Key Concepts

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

Fourier's Law of Heat Conduction
Fourier's Law of Heat Conduction is a fundamental principle that describes how heat energy is transferred as a result of temperature differences. According to this law, the rate of heat transfer through a material is directly proportional to the negative of the temperature gradient and the cross-sectional area through which heat is flowing, and is inversely proportional to the thickness of the material.

In simpler terms, this means that heat will flow from areas of higher temperature to areas of lower temperature, and the rate of this flow depends on how much hotter one side is compared to the other (the temperature gradient), as well as the size of the 'door' through which heat is escaping (the cross-sectional area). A steep temperature gradient and a larger area result in a higher rate of heat transfer. An important application of this law is in calculating the heat transfer in objects like the cylindrical rod in the given exercise.

Mathematically, Fourier's Law is expressed as \[ q' = -kA \frac{dT}{dx} \]where \( q' \) is the heat flux, \( k \) is the thermal conductivity of the material, \( A \) is the cross-sectional area, and \( \frac{dT}{dx} \) is the temperature gradient. The negative sign indicates that heat flows in the direction of decreasing temperature.
Steady-State Temperature Distribution
The term 'steady-state temperature distribution' refers to a scenario where the temperature of a material does not change over time at any point, which means the amount of heat entering a point within the material is equal to the amount of heat leaving it. In the context of our cylindrical rod example, the temperature profile along the length of the rod is steady, which allows us to use a fixed function, such as \( T(x) = a - b \frac{x}{L} \), to describe the temperature at any position \( x \).

Understanding steady-state conditions is critical because it simplifies the process of analyzing heat transfer problems. By ensuring that the thermal conditions remain constant over time, we eliminate the complexity involved with transient heat conduction, where temperatures change with time. This is why steady-state analysis is commonly used in engineering applications to predict system performance under normal operating conditions.

Furthermore, in steady-state, the heat flux can be calculated for any point along the length of the rod without concern for changes over time, making it perfect for educational examples and real-world applications alike.
Thermal Conductivity
Thermal conductivity, denoted as \( k \), is a material property that indicates a material's ability to conduct heat. It is often measured in units of \( W/m \cdot K \) (watts per meter-kelvin). A high thermal conductivity means that the material is a good conductor of heat, like most metals, while a low thermal conductivity indicates that the material is a good insulator, like rubber or wood.

In our cylindrical rod exercise, the thermal conductivity of stainless steel plays a pivotal role in determining how quickly heat will flow through the rod. The higher the thermal conductivity, the more efficient the material is at transferring heat from the hot end to the cool end.

Thus, in real-life applications, selecting the proper material with the right thermal conductivity is essential for managing heat transfer in devices ranging from electronic components to architectural materials. For instance, in heat exchangers, high thermal conductivity is desirable to efficiently transfer heat between substances, whereas in building insulation, materials with low thermal conductivity are preferred to prevent unwanted heat flow.

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Most popular questions from this chapter

A plane wall of thickness \(L=0.1 \mathrm{~m}\) experiences uniform volumetric heating at a rate \(\dot{q}\). One surface of the wall \((x=0)\) is insulated, and the other surface is exposed to \(\mathrm{a}\) fluid at \(T_{\infty}=20^{\circ} \mathrm{C}\), with convection heat transfer characterized by \(h=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Initially, the temperature distribution in the wall is \(T(x, 0)=a+b x^{2}\), where \(a=300^{\circ} \mathrm{C}, b=-1.0 \times 10^{40} \mathrm{C} / \mathrm{m}^{2}\), and \(x\) is in meters. Suddenly, the volumetric heat generation is deactivated ( \(\dot{q}=0\) for \(t \geq 0\) ), while convection heat transfer continues to occur at \(x=L\). The properties of the wall are \(\rho=7000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=90 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Determine the magnitude of the volumetric energy generation rate \(\dot{q}\) associated with the initial condition \((t<0)\). (b) On \(T-x\) coordinates, sketch the temperature distribution for the following conditions: initial condition \((t<0)\), steady-state condition \((t \rightarrow \infty)\), and two intermediate conditions. (c) On \(q_{x}^{\prime \prime}-t\) coordinates, sketch the variation with time of the heat flux at the boundary exposed to the convection process, \(q_{x}^{\prime \prime}(L, t)\). Calculate the corresponding value of the heat flux at \(t=0, q_{x}^{\prime \prime}(L, 0)\). (d) Calculate the amount of energy removed from the wall per unit area \(\left(\mathrm{J} / \mathrm{m}^{2}\right)\) by the fluid stream as the wall cools from its initial to steady-state condition.

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.

Consider a small but known volume of metal that has a large thermal conductivity. (a) Since the thermal conductivity is large, spatial temperature gradients that develop within the metal in response to mild heating are small. Neglecting spatial temperature gradients, derive a differential equation that could be solved for the temperature of the metal versus time \(T(t)\) if the metal is subjected to a fixed surface heat rate \(q\) supplied by an electric heater. (b) A student proposes to identify the unknown metal by comparing measured and predicted thermal responses. Once a match is made, relevant thermophysical properties might be determined, and, in turn, the metal may be identified by comparison to published property data. Will this approach work? Consider aluminum, gold, and silver as the candidate metals.

The steady-state temperature distribution in a semitransparent material of thermal conductivity \(k\) and thickness \(L\) exposed to laser irradiation is of the form $$ T(x)=-\frac{A}{k a^{2}} e^{-a x}+B x+C $$ (a) Obtain expressions for the conduction heat fluxes at the front and rear surfaces. (b) Derive an expression for \(\dot{q}(x)\). (c) Derive an expression for the rate at which radiation is absorbed in the entire material, per unit surface area. Express your result in terms of the known constants for the temperature distribution, the thermal conductivity of the material, and its thickness. where \(A, a, B\), and \(C\) are known constants. For this situation, radiation absorption in the material is manifested by a distributed heat generation term, \(\dot{q}(x)\).

A plane wall has constant properties, no internal heat generation, and is initially at a uniform temperature \(T_{i \cdot}\) Suddenly, the surface at \(x=L\) is heated by a fluid at \(T_{\infty}\) having a convection coefficient \(h\). At the same instant, the electrical heater is energized, providing a constant heat flux \(q_{o}^{\prime \prime}\) at \(x=0\). (a) On \(T-x\) 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. (b) On \(q_{x}^{\prime \prime}-x\) coordinates, sketch the heat flux corresponding to the four temperature distributions of part (a). (c) On \(q_{x}^{n}-t\) coordinates, sketch the heat flux at the locations \(x=0\) and \(x=L\). That is, show qualitatively how \(q_{x}^{\prime \prime}(0, t)\) and \(q_{x}^{\prime \prime}(L, t)\) vary with time. (d) Derive an expression for the steady-state temperature at the heater surface, \(T(0, \infty)\), in terms of \(q_{o}^{\prime \prime}\), \(T_{\infty}, k, h\), and \(L\).
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