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

Circular fins of uniform cross section, with diameter of \(10 \mathrm{~mm}\) and length of \(50 \mathrm{~mm}\), are attached to a wall with surface temperature of \(350^{\circ} \mathrm{C}\). The fins are made of material with thermal conductivity of \(240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and they are exposed to an ambient air condition of \(25^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heat transfer rate and plot the temperature variation of a single fin for the following boundary conditions: (a) Infinitely long fin (b) Adiabatic fin tip (c) Fin with tip temperature of \(250^{\circ} \mathrm{C}\) (d) Convection from the fin tip

Exposure to high concentration of gaseous ammonia can cause lung damage. To prevent gaseous ammonia from leaking out, ammonia is transported in its liquid state through a pipe \(\left(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}\right.\), \(D_{o}=4 \mathrm{~cm}\), and \(L=10 \mathrm{~m}\) ). Since liquid ammonia has a normal boiling point of \(-33.3^{\circ} \mathrm{C}\), the pipe needs to be properly insulated to prevent the surrounding heat from causing the ammonia to boil. The pipe is situated in a laboratory, where the average ambient air temperature is \(20^{\circ} \mathrm{C}\). The convection heat transfer coefficients of the liquid hydrogen and the ambient air are \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine the insulation thickness for the pipe using a material with \(k=\) \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to keep the liquid ammonia flowing at an average temperature of \(-35^{\circ} \mathrm{C}\), while maintaining the insulated pipe outer surface temperature at \(10^{\circ} \mathrm{C}\).

Steam at \(235^{\circ} \mathrm{C}\) is flowing inside a steel pipe \((k=\) \(61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(10 \mathrm{~cm}\) and \(12 \mathrm{~cm}\), respectively, in an environment at \(20^{\circ} \mathrm{C}\). The heat transfer coefficients inside and outside the pipe are \(105 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine ( \(a\) ) the thickness of the insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) needed to reduce the heat loss by 95 percent and \((b)\) the thickness of the insulation needed to reduce the exposed surface temperature of insulated pipe to \(40^{\circ} \mathrm{C}\) for safety reasons.

What is the value of conduction shape factors in engineering?

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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