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A \(2.5 \mathrm{~m}\)-high, 4-m-wide, and 20 -cm-thick wall of a house has a thermal resistance of \(0.0125^{\circ} \mathrm{C} / \mathrm{W}\). The thermal conductivity of the wall is (a) \(0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (b) \(1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (c) \(1.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(32 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

Short Answer

Expert verified
a) 0.8 W/(m.K) b) 1.2 W/(m.K) c) 1.6 W/(m.K) d) 2.0 W/(m.K) Answer: c) 1.6 W/(m.K)

Step by step solution

01

Identify the given values

We are given: - Height of the wall (h) : \(2.5 \mathrm{~m}\) - Width of the wall (w) : \(4.0 \mathrm{~m}\) - Thickness of the wall (d) : \(0.20 \mathrm{~m}\) - Thermal resistance (R) : \(0.0125 \mathrm{~C}/\mathrm{W} \)
02

Calculate the area of the wall

To find the area of the wall (A), multiply the height by the width: \(A = h \times w = 2.5 \mathrm{~m} \times 4.0 \mathrm{~m} = 10 \mathrm{~m^2}\)
03

Rearrange the thermal resistance formula to find thermal conductivity

To find the thermal conductivity (k), we can rearrange the formula for thermal resistance (R) as follows: \(k = \frac{d}{RA}\)
04

Plug the values into the formula and solve for thermal conductivity

Now, we can plug the values of d, R, and A into the formula and solve for k: \(k = \frac{0.20 \mathrm{~m}}{0.0125 \mathrm{~C}/\mathrm{W} \times 10 \mathrm{~m^2}} = 1.6 \mathrm{~W}/(\mathrm{m} \cdot \mathrm{K})\) Hence, the correct option is (c) \(1.6 \mathrm{~W}/(\mathrm{m} \cdot \mathrm{K})\).

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Key Concepts

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

Thermal Resistance
Understanding thermal resistance is crucial when studying heat transfer in materials. It quantifies how well a material resists the flow of heat. Much like electrical resistance in circuits, thermal resistance offers a way to measure how insulation or other barriers slow down the energy transfer between two points. It involves variables such as material thickness, surface area, and temperature difference. In practical scenarios, such as in the given exercise where a wall's thermal resistance is known, one could determine the wall's effectiveness in insulating a space against temperature changes. The lower the thermal resistance, the better the material conducts heat, resulting in more efficient heat transfer.
Heat Transfer
Heat transfer is a fundamental concept in thermal engineering, involving the movement of thermal energy from one place to another due to temperature differences. There are three primary modes of heat transfer: conduction, convection, and radiation. In the context of the exercise, we are primarily dealing with conduction through the wall. Conduction is the transfer of heat through a material without the actual movement of the substance; this process is significantly influenced by the material's thermal conductivity. The understanding of heat transfer is essential for designing energy-efficient buildings, maintaining comfortable indoor environments, and numerous engineering applications.
Material Properties
The properties of materials, such as thermal conductivity, density, and specific heat, play a pivotal role in how they behave under thermal stress. Thermal conductivity, the property highlighted in the exercise, indicates a material's ability to conduct heat. It fundamentally affects how quickly heat can pass through a material when exposed to a temperature gradient. The selection of building materials for walls, insulation, and other construction elements is heavily dependent on these thermal properties. Engineers and architects must understand these properties to make informed decisions that affect the safety, comfort, and energy efficiency of buildings.
Thermal Engineering Education
Thermal engineering education involves the study of energy conversion and heat transfer which are central to designing systems such as heating, cooling, and refrigeration units. The curriculum bridges practical problems, like the one found in this exercise, with theoretical concepts, aiming to develop an intuitive understanding alongside rigorous calculus-based learning. Effective education in this field typically emphasizes real-world applications, such as building efficient thermal barriers or creating temperature control mechanisms, to prepare students for the challenges they will face in their professional roles as engineers or designers.

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Most popular questions from this chapter

A plane wall surface at \(200^{\circ} \mathrm{C}\) is to be cooled with aluminum pin fins of parabolic profile with blunt tips. Each fin has a length of \(25 \mathrm{~mm}\) and a base diameter of \(4 \mathrm{~mm}\). The fins are exposed to an ambient air condition of \(25^{\circ} \mathrm{C}\) and the heat transfer coefficient is \(45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the thermal conductivity of the fins is \(230 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the heat transfer rate from a single fin and the increase in the rate of heat transfer per \(\mathrm{m}^{2}\) surface area as a result of attaching fins. Assume there are 100 fins per \(\mathrm{m}^{2}\) surface area.

One wall of a refrigerated warehouse is \(10.0\)-m-high and \(5.0\)-m-wide. The wall is made of three layers: \(1.0\)-cm-thick aluminum \((k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), 8.0\)-cm-thick fibreglass \((k=\) \(0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and \(3.0-\mathrm{cm}\) thick gypsum board \((k=\) \(0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The warehouse inside and outside temperatures are \(-10^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, and the average value of both inside and outside heat transfer coefficients is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Calculate the rate of heat transfer across the warehouse wall in steady operation. (b) Suppose that 400 metal bolts \((k=43 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), each \(2.0 \mathrm{~cm}\) in diameter and \(12.0 \mathrm{~cm}\) long, are used to fasten (i.e., hold together) the three wall layers. Calculate the rate of heat transfer for the "bolted" wall. (c) What is the percent change in the rate of heat transfer across the wall due to metal bolts?

A total of 10 rectangular aluminum fins \((k=\) \(203 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) are placed on the outside flat surface of an electronic device. Each fin is \(100 \mathrm{~mm}\) wide, \(20 \mathrm{~mm}\) high and \(4 \mathrm{~mm}\) thick. The fins are located parallel to each other at a center- tocenter distance of \(8 \mathrm{~mm}\). The temperature at the outside surface of the electronic device is \(60^{\circ} \mathrm{C}\). The air is at \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine \((a)\) the rate of heat loss from the electronic device to the surrounding air and \((b)\) the fin effectiveness.

Steam exiting the turbine of a steam power plant at \(100^{\circ} \mathrm{F}\) is to be condensed in a large condenser by cooling water flowing through copper pipes \(\left(k=223 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) of inner diameter \(0.4\) in and outer diameter \(0.6\) in at anerage temperature of \(70^{\circ} \mathrm{F}\). The heat of vaporization of water at \(100^{\circ} \mathrm{F}\) is \(1037 \mathrm{Btu} / \mathrm{lbm}\). The heat transfer coefficients are \(1500 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the steam side and \(35 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the water side. Determine the length of the tube required to condense steam at a rate of \(120 \mathrm{lbm} / \mathrm{h}\). Answer: \(1150 \mathrm{ft}\)

A 1-cm-diameter, 30-cm-long fin made of aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface at \(80^{\circ} \mathrm{C}\). The surface is exposed to ambient air at \(22^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the fin can be assumed to bery long, its efficiency is (a) \(0.60\) (b) \(0.67\) (c) \(0.72\) (d) \(0.77\) (e) \(0.88\)

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