/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 A composite slab is prepared by ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A composite slab is prepared by pasting two plates of thicknesses \(L_{1}\) and \(L_{2}\) and thermal conductivities \(K_{1}\) and \(K_{2} .\) The slabs have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

Short Answer

Expert verified
The equivalent conductivity is \(K_{eq} = \frac{L_1 + L_2}{\frac{L_1}{K_1} + \frac{L_2}{K_2}}\).

Step by step solution

01

Understanding the Problem

We have two plates with thicknesses \(L_1\) and \(L_2\) and thermal conductivities \(K_1\) and \(K_2\) that form a composite slab. We are to find the equivalent thermal conductivity for the entire slab.
02

Identify Series Configuration

The composite slab is created by stacking the two plates one behind the other, indicating a series configuration. This means the thermal conduction occurs through one plate and then the other.
03

Formula for Equivalent Thermal Conductivity

In a series arrangement, the equivalent thermal conductivity \(K_{eq}\) can be found using the formula: \[ \frac{L}{K_{eq}} = \frac{L_1}{K_1} + \frac{L_2}{K_2} \] where \(L = L_1 + L_2\) is the total thickness.
04

Substituting and Solving for \(K_{eq}\)

Simplify the formula by substituting \(L = L_1 + L_2\):\[ \frac{L_1 + L_2}{K_{eq}} = \frac{L_1}{K_1} + \frac{L_2}{K_2} \]Multiply both sides by \(K_{eq}\):\[ K_{eq} = \frac{L_1 + L_2}{\frac{L_1}{K_1} + \frac{L_2}{K_2}} \]
05

Final Formula

Thus, the equivalent thermal conductivity of the composite slab is: \[ K_{eq} = \frac{L_1 + L_2}{\frac{L_1}{K_1} + \frac{L_2}{K_2}} \] This accounts for both the thicknesses and conductivities of the individual slabs.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Composite Slab
A composite slab is formed by joining two or more layers of different materials together. In this case, we are dealing with two plates, each with distinct properties, joined to create a single slab. Each plate has its own thickness and thermal conductivity, which influence how heat is transferred through the overall structure.

The composite slab allows us to evaluate the combined effect of the differing layers on heat transfer. Engineers often use composite slabs in applications where optimizing heat conduction is critical while maintaining structural integrity. By combining materials, we can create slabs with specific thermal behaviors desired for particular applications.
Series Configuration
In the context of a composite slab, series configuration refers to the arrangement of layers such that heat flows sequentially through each layer. This means that the heat needs to pass through one layer completely before entering the next.

Imagine stacking two slices of bread one on top of the other. Heat travels from one slice to the other in a linear path. When layers are placed in a series configuration, each layer acts as a barrier to heat, one after the next. This arrangement is significant because it dictates how the total thermal resistance is calculated, influencing how heat travels across the slab.
Equivalent Conductivity
Equivalent conductivity is a measure that represents how heat is transferred through a composite material as if it were a single uniform layer. Given the complexity of multiple materials working together, equivalent conductivity gives us a single value that simplifies analysis.

For engineers and scientists, calculating equivalent conductivity helps in comparing the composite's heat transfer capability to that of a single material slab. It is especially useful in thermal management, where precise control over temperature regulation is necessary. Achieving an accurate equivalent conductivity can ensure materials are chosen and arranged optimally for the intended purpose.
Thermal Conductivities in Series
When computing thermal conductivities for materials arranged in series, we rely on a specific formula. The total heat resistance is the sum of the individual resistances; however, thermal resistance varies inversely with thermal conductivity.

In mathematical terms, dividing the thickness by thermal conductivity gives each layer's thermal resistance. For two layers of thicknesses \(L_1\) and \(L_2\) and conductivities \(K_1\) and \(K_2\), the formula becomes
  • \( \frac{L}{K_{eq}} = \frac{L_1}{K_1} + \frac{L_2}{K_2} \)
Here, \(L\) represents the total thickness \(L_1 + L_2\). Solving this gives the equivalent conductivity of the composite slab. Understanding this allows precise predictions of how laminated materials will conduct heat, which is critical in numerous industrial and engineering applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

On a winter day when the atmospheric temperature drops to \(-10^{\circ} \mathrm{C}\), ice forms on the surface of a lake. (a) Calculate the rate of increase of thickness of the ice when \(10 \mathrm{~cm}\) of ice is already formed. (b) Calculate the total time taken in forming \(10 \mathrm{~cm}\) of ice. Assume that the temperature of the entire water reaches \(0^{\circ} \mathrm{C}\) before the ice starts forming. Density of water \(=1000 \mathrm{~kg} \mathrm{~m}^{-3}\), latent heat of fusion of ice \(=3.36 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}\) and thermal conductivity of ice \(=1.7 \mathrm{~W} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}\). Neglect the expansion of water on freezing.

A spherical tungsten piece of radius \(1 \cdot 0 \mathrm{~cm}\) is suspended in an evacuated chamber maintained at \(300 \mathrm{~K}\). The piece is maintained at \(1000 \mathrm{~K}\) by heating it electrically. Find the rate at which the electrical energy must be supplied. The emissivity of tungsten is \(0: 30\) and the Stefan constant \(\sigma\) is \(6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}\)

A metal ball of mass \(1 \mathrm{~kg}\) is heated by means of a \(20 \mathrm{~W}\) heater in a room at \(20^{\circ} \mathrm{C}\). The temperature of the ball becomes steady at \(50^{\circ} \mathrm{C}\). (a) Find the rate of loss of heat to the surrounding when the ball is at \(50^{\circ} \mathrm{C}\). (b) Assuming Newton's law of cooling, calculate the rate of loss of heat to the surrounding when the ball is at \(30^{\circ} \mathrm{C}\). (c) Assume that the temperature of the ball rises uniformly from \(20^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\) in 5 minutes. Find the total loss of heat to the surrounding during this period. (d) Calculate the specific heat capacity of the metal.

Water at \(50^{\circ} \mathrm{C}\) is filled in a closed cylindrical vessel of height \(10 \mathrm{~cm}\) and cross sectional area \(10 \mathrm{~cm}^{2}\). The walls of the vessel are adiabatic but the flat parts are made of \(1-\mathrm{mm} \quad\) thick \(\quad\) aluminium \(\quad\left(K=200 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1}{ }^{-1} \mathrm{C}^{-1}\right)\). Assume that the outside temperature is \(20^{\circ} \mathrm{C}\). The density of water is \(1000 \mathrm{~kg} \mathrm{~m}^{-5}\), and the specific heat capacity of water \(=4200 \mathrm{~J} \mathrm{k} \stackrel{-\mathrm{l}}{\mathrm{g}} \mathrm{C}^{-1}\). Estimate the time taken for the temperature to fall by \(1 \cdot 0^{\circ} \mathrm{C}\). Make any simplifying assumptions you need but specify them.

A cubical block of mass \(1 \cdot 0 \mathrm{~kg}\) and edge \(5 \cdot 0 \mathrm{~cm}\) is heated to \(227^{\circ} \mathrm{C}\). It is kept in an evacuated chamber maintained at \(27^{\circ} \mathrm{C}\). Assuming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decrease. Specific heat capacity of the material of the block is \(400 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.