What is a conduction shape factor? How is it related to the thermal resistance?

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Question: Explain the relationship between conduction shape factor and thermal resistance. Answer: The conduction shape factor and thermal resistance are related through an equation derived from Fourier's law of heat conduction. The thermal resistance (R) is inversely proportional to the product of the conduction shape factor (S) and the material's thermal conductivity (k). As the conduction shape factor increases, the thermal resistance decreases, making the material conduct heat more easily. Conversely, as the conduction shape factor decreases, the thermal resistance increases, making the material a more effective heat insulator.

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01

Definition of Conduction Shape Factor

In heat transfer, the conduction shape factor (S) is a dimensionless parameter that represents the effect of the geometry of a conducting body on the rate of heat conduction. It depends on the material's shape and is used to simplify complex heat conduction problems. It is especially useful when dealing with non-uniform heat flow through conducting materials.
02

Definition of Thermal Resistance

Thermal resistance (R) is a property of a material that quantifies its resistance to heat flow. It is usually denoted by the letter 'R' and is calculated using the formula: R = \frac{ΔT}{Q} Where ΔT is the temperature difference between the two sides of the material and Q is the rate of heat flow through the material (measured in watts).
03

Functional Relationship between Conduction Shape Factor and Thermal Resistance

The conduction shape factor and thermal resistance are related through the following equation, which is derived using Fourier's law of heat conduction: Q = S * k * ΔT Where: - Q is the rate of heat flow through the object (measured in watts) - S is the conduction shape factor (dimensionless) - k is the thermal conductivity of the material (measured in watts per meter-kelvin, W/m.K) - ΔT is the temperature difference between the two sides of the object (measured in kelvin, K) This equation shows that the rate of heat transfer through an object is directly proportional to the conduction shape factor and the thermal conductivity of the material. The conduction shape factor accounts for the effects of the object's geometry on heat transfer, while the thermal conductivity represents the material's intrinsic ability to conduct heat. To find the thermal resistance (R), we can rearrange the equation as follows: R = \frac{ΔT}{Q} = \frac{1}{S * k} This equation shows that the thermal resistance is inversely proportional to the product of the conduction shape factor and the material's thermal conductivity. Thus, as the conduction shape factor increases, the thermal resistance decreases, and the material will conduct heat more easily. Conversely, as the conduction shape factor decreases, the thermal resistance increases, and the material becomes a more effective heat insulator.

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Most popular questions from this chapter

Two finned surfaces with long fins are identical, except that the convection heat transfer coefficient for the first finned surface is twice that of the second one. What statement below is accurate for the efficiency and effectiveness of the first finned surface relative to the second one? (a) Higher efficiency and higher effectiveness (b) Higher efficiency but lower effectiveness (c) Lower efficiency but higher effectiveness (d) Lower efficiency and lower effectiveness (e) Equal efficiency and equal effectiveness

A \(2.2\)-mm-diameter and 10-m-long electric wire is tightly wrapped with a \(1-m m\)-thick plastic cover whose thermal conductivity is \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Electrical measurements indicate that a current of \(13 \mathrm{~A}\) passes through the wire and there is a voltage drop of \(8 \mathrm{~V}\) along the wire. If the insulated wire is exposed to a medium at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Two flow passages with different cross-sectional shapes, one circular another square, are each centered in a square solid bar of the same dimension and thermal conductivity. Both configurations have the same length, \(T_{1}\), and \(T_{2}\). Determine which configuration has the higher rate of heat transfer through the square solid bar for \((a) a=1.2 b\) and \((b) a=2 b\).

Chilled water enters a thin-shelled 5-cm-diameter, 150-mlong pipe at \(7^{\circ} \mathrm{C}\) at a rate of \(0.98 \mathrm{~kg} / \mathrm{s}\) and leaves at \(8^{\circ} \mathrm{C}\). The pipe is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the pipe is to be insulated with glass wool insulation \((k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in order to decrease the temperature rise of water to \(0.25^{\circ} \mathrm{C}\), determine the required thickness of the insulation.

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