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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Most popular questions from this chapter

A hot surface at \(80^{\circ} \mathrm{C}\) in air at \(20^{\circ} \mathrm{C}\) is to be cooled by attaching 10 -cm-long and 1 -cm-diameter cylindrical fins. The combined heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer from the fin tip is negligible. If the fin efficiency is \(0.75\), the rate of heat loss from 100 fins is (a) \(325 \mathrm{~W}\) (b) \(707 \mathrm{~W}\) (c) \(566 \mathrm{~W}\) (d) \(424 \mathrm{~W}\) (e) \(754 \mathrm{~W}\)

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum \((k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a \(10-\mathrm{cm}\)-thick layer of insulation \((k=0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The rate at which the liquid oxygen gains heat is (a) \(141 \mathrm{~W}\) (b) \(176 \mathrm{~W}\) (c) \(181 \mathrm{~W}\) (d) \(201 \mathrm{~W}\) (e) \(221 \mathrm{~W}\)

Two finned surfaces are identical, except that the convection heat transfer coefficient of one of them is twice that of the other. For which finned surface is the \((a)\) fin effectiveness and \((b)\) fin efficiency higher? Explain.

The fins attached to a surface are determined to have an effectiveness of \(0.9\). Do you think the rate of heat transfer from the surface has increased or decreased as a result of the addition of these fins?

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially that of the surrounding air, i.e. \(20^{\circ} \mathrm{C}\). Its width is \(5.0 \mathrm{~cm}\); thickness is \(1.0 \mathrm{~mm}\); thermal conductivity is \(200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\); and base temperature is \(40^{\circ} \mathrm{C}\). The heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the fin temperature at a distance of \(5.0 \mathrm{~cm}\) from the base and the rate of heat loss from the entire fin.

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