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Chapter 3: Problem 88

Consider uniform thermal energy generation inside a one-dimensional plane wall of thickness \(L\) with one surface held at \(T_{s, 1}\) and the other surface insulated. (a) Find an expression for the conduction heat flux to the cold surface and the temperature of the hot surface \(T_{s, 2}\), expressing your results in terms of \(k, \dot{q}, L\), and \(T_{s, 1}\). (b) Compare the heat flux found in part (a) with the heat flux associated with a plane wall without energy generation whose surface temperatures are \(T_{s, 1}\) and \(T_{s, 2}\).

Short Answer

Expert verified
In a one-dimensional plane wall with uniform thermal energy generation, the temperature distribution is given by: \[T(x) = -\frac{\dot{q}}{2k} x^2 + T_{s, 1} + \frac{\dot{q}}{2k} L^2\] With the given boundary conditions, the heat flux to the cold surface (where \(x = 0\)) is zero due to the insulation, while the temperature of the hot surface \(T_{s, 2}\) is equal to \(T_{s, 1}\). Comparing the heat flux in this case to a plane wall without energy generation, it is observed that the heat flux in the presence of uniform thermal energy generation is zero at both surfaces, indicating the heat is absorbed within the wall. In contrast, for a wall without internal heat generation, the heat flux is due to the temperature difference between the surfaces and is given by: \[q = -k\frac{T_{s, 2} - T_{s, 1}}{L}\]

Step by step solution

01

(Step 1: Write down the heat diffusion equation)

For a one-dimensional plane wall with uniform thermal energy generation, the heat diffusion equation is given by: \[\frac{d^2T}{dx^2} = -\frac{\dot{q}}{k}\] Where \(T\) is the temperature, \(x\) is the position along the wall, \(\dot{q}\) is the rate of heat generation per unit volume, and \(k\) is the thermal conductivity.
02

(Step 2: Solve the heat diffusion equation)

To solve the equation, we take two integrations: 1. Integrate once: \[\int \frac{d^2T}{dx^2} dx = -\frac{\dot{q}}{k} x + C_1\] \[\frac{dT}{dx} = -\frac{\dot{q}}{k} x + C_1\] 2. Integrate again: \[\int \frac{dT}{dx} dx = -\frac{\dot{q}}{2k} x^2 + C_1 x + C_2\] \[T(x) = -\frac{\dot{q}}{2k} x^2 + C_1 x + C_2\]
03

(Step 3: Apply boundary conditions)

Now we apply the given boundary conditions to find the constants \(C_1\) and \(C_2\). 1. At \(x = 0\), the surface is insulated, which implies that \(\frac{dT}{dx} = 0\): \[0 = -\frac{\dot{q}}{k} \cdot 0 + C_1\] \[C_1 = 0\] 2. At \(x = L\), the surface temperature is \(T_{s, 1}\): \[T_{s, 1} = -\frac{\dot{q}}{2k} L^2 + C_2\] \[C_2 = T_{s, 1} + \frac{\dot{q}}{2k} L^2\] Finally, the temperature distribution is given by: \[T(x) = -\frac{\dot{q}}{2k} x^2 + T_{s, 1} + \frac{\dot{q}}{2k} L^2\]
04

(Step 4: Calculate heat flux to the cold surface and temperature of the hot surface)

The conduction heat flux can be found using Fourier's law: \(q = -k \frac{dT}{dx}\) At \(x=0\), the heat flux is zero because the surface is insulated. Now, we need to find the temperature of the hot surface \(T_{s, 2}\), which is at \(x = L\): \[T_{s, 2} = -\frac{\dot{q}}{2k} L^2 + T_{s, 1} + \frac{\dot{q}}{2k} L^2 = T_{s, 1}\]
05

(Step 5: Comparing the heat flux for both cases)

For case (a), we have a uniform thermal energy generation inside the wall, which results in the heat flux being zero at both surfaces. Thus, the heat is consumed within the wall itself. For case (b), the plane wall has no internal energy generation. The heat flux can be found using Fourier's law and the temperature difference between the two surfaces: \[q = -k\frac{T_{s, 2} - T_{s, 1}}{L}\] Comparing both cases, we find that in the presence of uniform thermal energy generation, the heat is absorbed in the wall itself, resulting in zero heat flux at the surfaces. In contrast, in the absence of internal heat generation, the heat flux is purely due to the temperature difference between the two surfaces.

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Key Concepts

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

Thermal Energy Generation
When we talk about thermal energy generation, we refer to the creation of heat within a material or system. This can stem from various sources including chemical reactions (like combustion), electrical energy (resistance heating in conductors), nuclear reactions (as observed in reactors), or even due to the absorption of electromagnetic radiation. In the case of the textbook exercise, thermal energy generation is presented as a uniform process occurring inside a plane wall.

Uniform thermal energy generation implies that heat is produced at the same rate throughout the entire volume of the wall, which is an assumption that simplifies the mathematical treatment of the problem. In real-life scenarios, heat generation can be non-uniform due to various factors, such as material inhomogeneity or non-uniform external conditions. This concept provides a crucial understanding when analyzing how heat impacts material temperature, which further affects how the material will conduct heat.
Heat Conduction
Heat conduction is the transfer of thermal energy through a material without the bulk movement of the material itself - basically, heat flowing from the hotter to the cooler parts of a body. This can be envisioned much like a row of dominoes falling one after the other; the molecular agitation caused by heat in a solid is transferred from one particle to the next.

Several factors influence how well a material conducts heat. One of these is the material's thermal conductivity, denoted by k in the exercise. Materials with high thermal conductivity, like metals, easily transfer heat, whereas those with low thermal conductivity, like wood or foam, are better insulators. In the exercise's scenario, where one side of a wall is insulated, the heat generated internally must travel to the other side, indicating heat conduction in action. Knowing the nuances of heat conduction not only helps in solving thermodynamics problems but also plays a fundamental role in designing and optimizing thermal systems like heat exchangers or insulation.
Fourier's Law
Fourier's law of heat conduction is a principle that determines the rate at which heat energy is transferred through a material. It states that the rate of heat flow through a material is proportional to the negative gradient of temperatures and the area through which the heat is flowing, mathematically expressed as q = -k * A * (dT/dx), where q is the heat flux, k is the thermal conductivity, A is the cross-sectional area, and (dT/dx) is the temperature gradient.

In the context of the exercise, Fourier's law helps calculate the heat flux to the cold surface of the wall given the internal generation of heat. As part of improving the exercise, understanding Fourier's law allows students to grasp why the heat flux is zero at the insulated surface, as there is no temperature gradient, and hence no heat flow. It also explains the comparison between a wall with and without internal heat generation, ultimately showing that Fourier's law is fundamental in predicting and understanding the behavior of thermal systems.

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Most popular questions from this chapter

In Problem 3.48, the electrical power required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C}\) depends on the thermal conductivity of the wall material \(k\), the thermal contact resistance \(R_{t, c}^{\prime}\) and the convection coefficient \(h\). Compute and plot the separate effect of changes in \(k\) \((1 \leq k \leq 200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), \quad R_{t, c}^{\prime} \quad\left(0 \leq R_{t, c}^{\prime} \leq 0.1 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\right)\), and \(h\left(10 \leq h \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) on the total heater power requirement, as well as the rate of heat transfer to the inner surface of the tube and to the fluid.

Finned passages are frequently formed between parallel plates to enhance convection heat transfer in compact heat exchanger cores. An important application is in electronic equipment cooling, where one or more air-cooled stacks are placed between heat-dissipating electrical components. Consider a single stack of rectangular fins of length \(L\) and thickness \(t\), with convection conditions corresponding to \(h\) and \(T_{\infty}\). (a) Obtain expressions for the fin heat transfer rates, \(q_{f, o}\) and \(q_{f, L}\), in terms of the base temperatures, \(T_{o}\) and \(T_{L}\). (b) In a specific application, a stack that is \(200 \mathrm{~mm}\) wide and \(100 \mathrm{~mm}\) deep contains 50 fins, each of length \(L=12 \mathrm{~mm}\). The entire stack is made from aluminum, which is everywhere \(1.0 \mathrm{~mm}\) thick. If temperature limitations associated with electrical components joined to opposite plates dictate maximum allowable plate temperatures of \(T_{o}=400 \mathrm{~K}\) and \(T_{L}=350 \mathrm{~K}\), what are the corresponding maximum power dissipations if \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=300 \mathrm{~K} ?\)

A composite wall separates combustion gases at \(2600^{\circ} \mathrm{C}\) from a liquid coolant at \(100^{\circ} \mathrm{C}\), with gas- and liquid-side convection coefficients of 50 and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The wall is composed of a \(10-\mathrm{mm}\)-thick layer of beryllium oxide on the gas side and a 20 -mm-thick slab of stainless steel (AISI 304) on the liquid side. The contact resistance between the oxide and the steel is \(0.05 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the heat loss per unit surface area of the composite? Sketch the temperature distribution from the gas to the liquid.

Radioactive wastes \(\left(k_{\mathrm{rw}}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) are stored in a spherical, stainless steel \(\left(k_{\mathrm{ss}}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) container of inner and outer radii equal to \(r_{i}=0.5 \mathrm{~m}\) and \(r_{o}=0.6 \mathrm{~m}\). Heat is generated volumetrically within the wastes at a uniform rate of \(\dot{q}=10^{5} \mathrm{~W} / \mathrm{m}^{3}\), and the outer surface of the container is exposed to a water flow for which \(h=\) \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\). (a) Evaluate the steady-state outer surface temperature, \(T_{s, o}\) (b) Evaluate the steady-state inner surface temperature, \(T_{s, i^{*}}\) (c) Obtain an expression for the temperature distribution, \(T(r)\), in the radioactive wastes. Express your result in terms of \(r_{i}, T_{s, i}, k_{\mathrm{rw}}\), and \(\dot{q}\). Evaluate the temperature at \(r=0\). (d) A proposed extension of the foregoing design involves storing waste materials having the same thermal conductivity but twice the heat generation \(\left(\dot{q}=2 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\right)\) in a stainless steel container of equivalent inner radius \(\left(r_{i}=0.5 \mathrm{~m}\right)\). Safety considerations dictate that the maximum system temperature not exceed \(475^{\circ} \mathrm{C}\) and that the container wall thickness be no less than \(t=0.04 \mathrm{~m}\) and preferably at or close to the original design \((t=0.1 \mathrm{~m})\). Assess the effect of varying the outside convection coefficient to a maximum achievable value of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (by increasing the water velocity) and the container wall thickness. Is the proposed extension feasible? If so, recommend suitable operating and design conditions for \(h\) and \(t\), respectively.

To maximize production and minimize pumping costs, crude oil is heated to reduce its viscosity during transportation from a production field. (a) Consider a pipe-in-pipe configuration consisting of concentric steel tubes with an intervening insulating material. The inner tube is used to transport warm crude oil through cold ocean water. The inner steel pipe \(\left(k_{s}=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) has an inside diameter of \(D_{i, 1}=150 \mathrm{~mm}\) and wall thickness \(t_{i}=10 \mathrm{~mm}\) while the outer steel pipe has an inside diameter of \(D_{i, 2}=250 \mathrm{~mm}\) and wall thickness \(t_{o}=t_{i}\). Determine the maximum allowable crude oil temperature to ensure the polyurethane foam insulation \(\left(k_{p}=\right.\) \(0.075 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) between the two pipes does not exceed its maximum service temperature of \(T_{p, \max }=\) \(70^{\circ} \mathrm{C}\). The ocean water is at \(T_{\infty, o}=-5^{\circ} \mathrm{C}\) and provides an external convection heat transfer coefficient of \(h_{o}=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection coefficient associated with the flowing crude oil is \(h_{i}=450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) It is proposed to enhance the performance of the pipe-in-pipe device by replacing a thin \(\left(t_{a}=5 \mathrm{~mm}\right)\) section of polyurethane located at the outside of the inner pipe with an aerogel insulation material \(\left(k_{a}=0.012 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). Determine the maximum allowable crude oil temperature to ensure maximum polyurethane temperatures are below \(T_{p, \max }=70^{\circ} \mathrm{C}\).
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