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Consider heat conduction through a wall of thickness \(L\) and area \(A\). Under what conditions will the temperature distributions in the wall be a straight line?

Short Answer

Expert verified
Answer: The temperature distribution in a wall will be a straight line when the wall is of uniform composition with constant thermal conductivity (k), the heat transfer rate (Q) is constant along the wall's thickness, and the area (A) of the wall remains constant throughout its thickness.

Step by step solution

01

Setting Up Fourier's Law for Heat Conduction

Fourier's law for one-dimensional heat conduction through a solid wall states that the heat transfer rate (Q) is proportional to the temperature gradient and the surface area of the wall. It is given by the formula: \(Q = -kA\frac{dT}{dx}\) Where: - \(Q\) is the heat transfer rate through the wall; - \(k\) is the thermal conductivity of the wall material (constant); - \(A\) is the area of the wall; - \(\frac{dT}{dx}\) is the temperature gradient in the wall (change in temperature per unit length in the x-direction).
02

Analyzing the Condition for a Linear Temperature Distribution

In order for the temperature distribution to be a straight line, \(\frac{dT}{dx}\) must be constant. Since this quantity represents the temperature gradient, it means that the change in temperature must be the same along the entire thickness of the wall. This would happen when: 1. The thermal conductivity \(k\) is constant throughout the wall, meaning that the wall is of uniform composition; 2. The heat transfer rate \(Q\) remains constant along the length of the wall; 3. The area \(A\) of the wall remains constant along the wall's thickness.
03

Interpretation of Results

For a straight-line temperature distribution, the following conditions must be met: - The wall must be of uniform composition and constant thermal conductivity \(k\); - The heat transfer rate \(Q\) must be constant along the wall's thickness; - The area \(A\) of the wall must remain constant throughout its thickness. In conclusion, if these conditions are satisfied, the temperature distribution in the wall will be a straight line.

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Most popular questions from this chapter

Hot- and cold-water pipes \(12 \mathrm{~m}\) long run parallel to each other in a thick concrete layer. The diameters of both pipes are \(6 \mathrm{~cm}\), and the distance between the centerlines of the pipes is \(40 \mathrm{~cm}\). The surface temperatures of the hot and cold pipes are \(60^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively. Taking the thermal conductivity of the concrete to be \(k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the rate of heat transfer between the pipes.

Two 4-m-long and \(0.4-\mathrm{cm}\)-thick cast iron $(k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( steam pipes of outer diameter \)10 \mathrm{~cm}$ are connected to each other through two 1 -cm-thick flanges of outer diameter 18 \(\mathrm{cm}\). The steam flows inside the pipe at an average temperature of \(200^{\circ} \mathrm{C}\) with a heat transfer coefficient of $180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The outer surface of the pipe is exposed to an ambient at \(12^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Disregarding the flanges, determine the average outer surface temperature of the pipe. \((b)\) Using this temperature for the base of the flange and treating the flanges as the fins, determine the fin efficiency and the rate of heat transfer from the flanges. (c) What length of pipe is the flange section equivalent to for heat transfer purposes?

A \(20-\mathrm{cm}\)-diameter hot sphere at \(120^{\circ} \mathrm{C}\) is buried in the ground with a thermal conductivity of $1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The distance between the center of the sphere and the ground surface is \(0.8 \mathrm{~m}\), and the ground surface temperature is \(15^{\circ} \mathrm{C}\). The rate of heat loss from the sphere is (a) \(169 \mathrm{~W}\) (b) \(20 \mathrm{~W}\) (c) \(217 \mathrm{~W}\) (d) \(312 \mathrm{~W}\) (e) \(1.8 \mathrm{~W}\)

A \(3-\mathrm{cm}\)-long, \(2-\mathrm{mm} \times 2-\mathrm{mm}\) rectangular cross-section aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the fin efficiency is 65 percent, the effectiveness of this single fin is (a) 39 (b) 30 (c) 24 (d) 18 (e) 7

A stainless steel plate is connected to a copper plate by long ASTM B98 copper-silicon bolts of \(9.5 \mathrm{~mm}\) in diameter. The portion of the bolts exposed to convection heat transfer with air is \(5 \mathrm{~cm}\) long. The air temperature for convection is at \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of $5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. The thermal conductivity of the bolt is known to be \)36 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The copper plate has a uniform temperature of \(70^{\circ} \mathrm{C}\). If each bolt is estimated to have a rate of heat loss to convection of \(5 \mathrm{~W}\), determine the temperature \(T_{b}\) at the upper surface of the stainless steel plate. According to the ASME Code for Process Piping (ASME B31.3-2014, Table A-2M), the maximum use temperature for the ASTM B 98 copper-silicon bolt is \(149^{\circ} \mathrm{C}\). Does the use of the ASTM B 98 bolts in this condition comply with the ASME code?

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