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Under what conditions can a plane wall be treated as a semi-infinite medium?

Short Answer

Expert verified
Answer: A plane wall can be treated as a semi-infinite medium if its thickness is much greater than the penetration depth, given by the relation: L >> √(4αt), where L is the wall's thickness, α is the material's thermal diffusivity, and t is the time of interest. This condition ensures that heat conduction occurs for a large distance in one direction compared to the other two directions, simplifying the analysis of heat transfer within the plane wall.

Step by step solution

01

Define the plane wall dimensions and properties

Consider a plane wall with thickness L and thermal conductivity k. Heat is conducted through the wall, and we need to determine the conditions that will allow us to treat this wall as a semi-infinite medium.
02

Define semi-infinite medium condition

A plane wall can be treated as a semi-infinite medium if the heat flow occurs for a large distance in one direction compared to the other two directions. In other words, the wall thickness should be much greater than the distance traveled by heat during the time frame of interest.
03

Calculate the penetration depth

To determine if the plane wall can be treated as a semi-infinite medium, we need to compare its thickness to the penetration depth, which is the distance heat travels in the material during a certain time. The penetration depth (δ) is calculated using the following formula: \[δ = \sqrt{4αt}\] where α is the material's thermal diffusivity and t is the time of interest.
04

Analyze the relation between wall thickness and penetration depth

A plane wall can be treated as a semi-infinite medium if its thickness (L) is much greater than the penetration depth (δ). The general condition for this is: \[L >> \sqrt{4αt}\]
05

Conclusion

A plane wall can be treated as a semi-infinite medium under the condition that its thickness is much greater than the penetration depth, which is given by the relation: \[L >> \sqrt{4αt}\] By ensuring this condition is met, we assume that heat conduction occurs for a large distance in one direction compared to the other two directions, allowing us to simplify our calculations and analysis of heat transfer within the plane wall.

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Key Concepts

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

Plane Wall Heat Transfer
When studying how heat moves through a solid object, such as a wall, one common scenario to analyze is the case of plane wall heat transfer. This situation involves a flat, two-dimensional surface with heat flowing in one primary direction, typically from one side of the wall to the other. Understanding this process is vital in fields such as insulation design, building energy efficiency, and material science.

To simplify the analysis, we make assumptions about how the heat is transferring through the material. In reality, heat can move in three dimensions, but for a wall that is relatively thin compared to its length and height, we can assume that the main direction of heat transfer is through its thickness. That simplification leads us to model the wall as a semi-infinite medium, a hypothetical concept where one dimension is significantly larger than the others, effectively extending to infinity. This assumption allows us to disregard the edge effects of the wall and focus on the heat transfer through its thickness.
Thermal Conductivity
To get to the heart of plane wall heat transfer, we need to understand thermal conductivity, a material property denoted by the symbol k. Thermal conductivity reflects how easily heat can pass through a material, and it is a fundamental concept to grasp when solving heat transfer problems.

Materials with high thermal conductivity, like metals, are excellent heat conductors, meaning they can transfer heat quickly from one side to the other. Conversely, materials with low thermal conductivity, such as wood or insulation foam, are poor heat conductors and serve as good insulators. k depends not only on the material's composition but also on factors like temperature and moisture content.

In the context of treating a plane wall as a semi-infinite medium, the thermal conductivity helps determine how rapidly the temperature within the wall will change due to heat flow. This property, combined with the wall's thickness and the heat transfer time frame, influences whether we can accurately model the wall as semi-infinite.
Penetration Depth
A critical concept in determining if it is appropriate to treat a plane wall as a semi-infinite medium is penetration depth. Penetration depth, denoted by \(\delta\), represents the distance into the wall that heat will penetrate during a certain period. Intuitively, it tells us how far heat 'travels' into the medium over time, making it an essential factor in thermal analysis.

The penetration depth calculation relies on the thermal diffusivity of the material, represented by \(\alpha\), which combines the thermal conductivity, density, and specific heat capacity of the material. Using the formula \(\delta = \sqrt{4\alpha t}\), where \(t\) is the time of interest, we can quantify the penetration depth.

For the assumption of semi-infinite medium to be valid, the thickness of the plane wall, \(L\), must be significantly larger than \(\delta\). This guarantees that the time it would take heat to affect the entire thickness of the wall is much longer than the time period we are interested in, ensuring that heat only affects a 'semi-infinite' portion of the wall, which simplifies our mathematical models and analysis.

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Most popular questions from this chapter

In a production facility, large plates made of stainless steel \(\left(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) of \(40 \mathrm{~cm}\) thickness are taken out of an oven at a uniform temperature of \(750^{\circ} \mathrm{C}\). The plates are placed in a water bath that is kept at a constant temperature of \(20^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(600 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The time it takes for the surface temperature of the plates to drop to \(100^{\circ} \mathrm{C}\) is (a) \(0.28 \mathrm{~h}\) (b) \(0.99 \mathrm{~h}\) (c) \(2.05 \mathrm{~h}\) (d) \(3.55 \mathrm{~h}\) (e) \(5.33 \mathrm{~h}\)

A large heated steel block \(\left(\rho=7832 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\right.\) \(434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=63.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=18.8 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) is allowed to cool in a room at \(25^{\circ} \mathrm{C}\). The steel block has an initial temperature of \(450^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that the steel block can be treated as a quarter-infinite medium, determine the temperature at the edge of the steel block after 10 minutes of cooling.

Carbon steel balls ( \(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}, k=64 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\left.c_{p}=434 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(150^{\circ} \mathrm{C}\) are quenched in an oil bath at \(20^{\circ} \mathrm{C}\) for a period of 3 minutes. If the balls have a diameter of \(5 \mathrm{~cm}\) and the convection heat transfer coefficient is \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The center temperature of the balls after quenching will be (Hint: Check the Biot number). (a) \(27.4^{\circ} \mathrm{C}\) (b) \(143^{\circ} \mathrm{C}\) (c) \(12.7^{\circ} \mathrm{C}\) (d) \(48.2^{\circ} \mathrm{C}\) (e) \(76.9^{\circ} \mathrm{C}\)

Layers of 6-in-thick meat slabs \(\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\) initially at a uniform temperature of \(50^{\circ} \mathrm{F}\) are cooled by refrigerated air at \(23^{\circ} \mathrm{F}\) to a temperature of \(36^{\circ} \mathrm{F}\) at their center in \(12 \mathrm{~h}\). Estimate the average heat transfer coefficient during this cooling process. Solve this problem using the Heisler charts. Answer: \(1.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\)

Consider a hot semi-infinite solid at an initial temperature of \(T_{i}\) that is exposed to convection to a cooler medium at a constant temperature of \(T_{\infty}\), with a heat transfer coefficient of \(h\). Explain how you can determine the total amount of heat transfer from the solid up to a specified time \(t_{o}\).

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