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Can the thermal resistance concept be used for a solid cylinder or sphere in steady operation? Explain.

Short Answer

Expert verified
Yes, the thermal resistance concept can be applied to solid cylinders and spheres in steady operation, provided that the specific geometries of these shapes are taken into account. For solid cylinders and spheres, heat transfer occurs mainly through conduction and can be modeled using the appropriate equations for cylindrical and spherical coordinates. By correctly adapting the formulas to radial coordinates, we can accurately describe the heat transfer and temperature distribution within these shapes, and utilize the concept of thermal resistance for various applications and analyses.

Step by step solution

01

Understand the thermal resistance concept

Thermal resistance is a concept used in heat transfer to quantify the resistance to heat flow across a material or medium. It's defined as the temperature difference between two points divided by the heat transfer rate per unit area that goes through those points. Mathematically, for systems having directionally proportional temperature distribution to length, we can represent it as: Thermal Resistance (R) = \( \frac{ΔT}{q} \) Where ΔT is the temperature difference between two points, and q is the heat transfer rate per unit area.
02

Discuss conduction in solid cylinders and spheres

In a solid cylinder or sphere, heat transfer occurs mainly through conduction. The temperature distribution across the thickness of the cylinder or sphere depends on the geometry and the material's thermal conductivity (k). For simple shapes like infinite planar walls, the heat transfer along the thickness is one-dimensional. But for cylindrical and spherical objects, the heat transfer is radially oriented, causing the temperature distribution to be position-dependent and involving multiple directions.
03

Explain the applicability of the thermal resistance concept to solid cylinders and spheres

The thermal resistance concept can be applied to solid cylinders and spheres in steady operation, as long as we properly account for the specific geometry of these shapes. In these cases, the radial heat transfer can be modeled using the appropriate equations for cylindrical and spherical coordinates. For a solid cylinder, the thermal resistance can be calculated as: \( R_{cylindrical} = \frac{ ln(\frac{r_2}{r_1})}{2πLk} \) Where \(r_1\) and \(r_2\) are the inner and outer radii of the cylinder, L is the length of the cylinder, and k is the thermal conductivity of the material. For a solid sphere, the thermal resistance can be calculated as: \( R_{spherical} = \frac{r_2 - r_1}{4πr_1r_2k} \) Where \(r_1\) and \(r_2\) are the inner and outer radii of the sphere, and k is the thermal conductivity of the material.
04

Conclusion

In conclusion, the thermal resistance concept can be applied to solid cylinders and spheres in steady operation, as long as their specific geometries are taken into account. By adapting the formulas to radial coordinates, we can accurately describe the heat transfer and temperature distribution within these shapes and use the concept of thermal resistance for various applications and analyses.

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Key Concepts

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

Heat Transfer
Heat transfer is the process by which thermal energy moves from one place to another. This transfer often occurs due to a temperature difference, leading to energy moving from a hot section to a cooler one. There are three primary methods of heat transfer: conduction, convection, and radiation.
Conduction refers to heat transfer through a material without the material itself moving. For instance, when a metal rod is heated at one end, heat travels to the cooler end. This process can be visualized as the transfer of kinetic energy from one molecule to the next.
Convection involves heat transfer through the movement of fluids, such as air or water, while radiation involves heat transfer through electromagnetic waves. Understanding these basic mechanisms is crucial for applying concepts such as thermal resistance to objects like cylinders and spheres.
Conduction
Conduction is a significant method of heat transfer, especially in solid objects. It occurs when heat flows through a solid material. This can be visualized as the transfer of vibrational energy between closely packed molecules.
In an intricate shape like a cylinder or a sphere, conduction becomes more complex due to geometric factors. The heat transfer path isn't straight but radial, meaning it extends outward in all directions.
In engineering applications, understanding conduction is vital as it allows us to design materials and structures that can efficiently manage heat flow. This includes knowing properties such as thermal conductivity, which is a measure of a material's ability to conduct heat. The higher the thermal conductivity, the more efficiently heat is transferred through that material.
Cylindrical and Spherical Geometry
When considering heat transfer in cylindrical and spherical geometries, understanding the shape is key. Unlike flat surfaces, cylinders and spheres challenge us to account for a radial heat flow path.
In a cylindrical object, heat flows radially from the inner surface to the outer surface. This is because the curvature means that heat conducts through a curved, rather than a flat, surface. The length of the path along which heat travels influences how heat is distributed.
Spherical objects, however, involve heat transfer across a spherical shell. The radial symmetry here means that heat travels outward evenly from the center, leading to calculations that adjust for this symmetry.
The radial nature of heat transfer in these geometries necessitates special formulas to accurately calculate thermal resistance, which consider the curvature and dimensions of each object.
Steady State Operation
In steady state operation, the system's properties remain constant over time. For heat transfer problems, it means that the temperature distribution does not change as time progresses.
This is a critical assumption for analyzing thermal resistance because it implies that the heat entering a system equals the heat exiting it, leaving the temperatures at points in the material stable.
Understanding steady state operation simplifies the mathematical treatment of heat problems, allowing us to focus on spatial variations without worrying about time-dependent changes.
In practical terms, steady state analysis allows engineers to design systems for efficient heat management, ensuring that systems remain reliable and function optimally under specified conditions.

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Most popular questions from this chapter

Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend attaching fins both inside and outside the tubes?

Obtain a relation for the fin efficiency for a fin of constant cross-sectional area \(A_{c}\), perimeter \(p\), length \(L\), and thermal conductivity \(k\) exposed to convection to a medium at \(T_{\infty}\) with a heat transfer coefficient \(h\). Assume the fins are sufficiently long so that the temperature of the fin at the tip is nearly \(T_{\infty}\). Take the temperature of the fin at the base to be \(T_{b}\) and neglect heat transfer from the fin tips. Simplify the relation for \((a)\) a circular fin of diameter \(D\) and \((b)\) rectangular fins of thickness \(t\).

A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating \(0.04 \mathrm{~W}\). The board is impregnated with copper fillings and has an effective thermal conductivity of \(30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(40^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the temperatures on the two sides of the circuit board. (b) Now a \(0.2\)-cm-thick, 12-cm-high, and 18-cmlong aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with 864 2-cm-long aluminum pin fins of diameter \(0.25 \mathrm{~cm}\) is attached to the back side of the circuit board with a \(0.02\)-cm-thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

What is a radiant barrier? What kind of materials are suitable for use as radiant barriers? Is it worthwhile to use radiant barriers in the attics of homes?

Heat is generated steadily in a 3-cm-diameter spherical ball. The ball is exposed to ambient air at \(26^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(7.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The ball is to be covered with a material of thermal conductivity \(0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thickness of the covering material that will maximize heat generation within the ball while maintaining ball surface temperature constant is (a) \(0.5 \mathrm{~cm}\) (b) \(1.0 \mathrm{~cm}\) (c) \(1.5 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(2.5 \mathrm{~cm}\)

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