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Chapter 3: Problem 69

A 12-m-long and 5-m-high wall is constructed of two layers of \(1-\mathrm{cm}\)-thick sheetrock \((k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) spaced \(16 \mathrm{~cm}\) by wood studs \((k=0.11 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose cross section is \(16 \mathrm{~cm} \times 5 \mathrm{~cm}\). The studs are placed vertically \(60 \mathrm{~cm}\) apart, and the space between them is filled with fiberglass insulation \((k=0.034 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The house is maintained at \(20^{\circ} \mathrm{C}\) and the ambient temperature outside is \(-9^{\circ} \mathrm{C}\). Taking the heat transfer coefficients at the inner and outer surfaces of the house to be \(8.3\) and \(34 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine \((a)\) the thermal resistance of the wall considering a representative section of it and (b) the rate of heat transfer through the wall.

Short Answer

Expert verified
Based on the provided information, calculate the overall thermal resistance of a representative section of the wall and the rate of heat transfer through the wall. Consider the sheetrock, wood studs, fiberglass insulation, and inner and outer surface heat transfer coefficients.

Step by step solution

01

1. Identify the relevant parameters and materials

Here's what we know: - Sheetrock thickness: 1cm (0.01 m); k=0.17 W/m.K - Wood studs: 16 cm x 5 cm cross-section; k=0.11 W/m.K; placed 60 cm (0.6 m) apart - Fiberglass insulation: k=0.034 W/m.K - Wall dimensions: 12 m x 5 m - Indoor temperature: 20掳C - Outdoor temperature: -9掳C - Inner surface heat transfer coefficient: 8.3 W/m虏.K - Outer surface heat transfer coefficient: 34 W/m虏.K
02

2. Calculate the thermal resistance of each component

We will separate the calculations for the sheetrock, wood studs, and fiberglass insulation: (1) Sheetrock resistance: The thermal resistance is given by: $R_S = \frac{t_s}{k_s A_s} \\ R_s = \frac{0.01\,\text{m}}{0.17\,\text{W/m.K} \cdot * (12\,\text{m}\cdot 5\,\text{m})}$ (2) Wood studs resistance: The distance between studs is 0.6 m. If we consider a representative section of the wall between two studs 1 m high, the wood stud cross-sectional area is 5 cm x 1 m = 0.05 m虏. Then, the thermal resistance is: $R_{WS} = \frac{t_{WS}}{k_{WS} A_{WS}} \\ R_{WS} = \frac{0.16\,\text{m}}{0.11\,\text{W/m.K} \cdot * 0.05\,\text{m}^{2}}$ (3) Fiberglass insulation resistance: The geometric resistance per unit length of the fiberglass insulation (since it's the same height as the wall) will be: $R_{FI} = \frac{t_{FI}}{k_{FI} A_{FI}} \\ R_{FI} = \frac{0.16\,\text{m}}{0.034\,\text{W/m.K} \cdot * (0.6\,\text{m} - 0.05\,\text{m})}$ (4) Inner and outer surface resistance: The inner surface resistance is given by: $R_{in} = \frac{1}{h_{in}A_{in}} \\ R_{in} = \frac{1}{8.3\,\frac{\text{W}}{\text{m}^{2}\cdot \text{K}} \cdot (12\,\text{m}\cdot 5\,\text{m})}$ The outer surface resistance is given by: $R_{out} = \frac{1}{h_{out}A_{out}} \\ R_{out} = \frac{1}{34\,\frac{\text{W}}{\text{m}^{2}\cdot \text{K}} \cdot(12\,\text{m}\cdot 5\,\text{m})}$
03

3. Calculate the overall thermal resistance of a representative section of the wall

Considering that every 0.6 m, there is a wood stud, and the space between studs is filled with insulation, we can calculate the overall thermal resistance of a representative section (1 m high): \(R = R_{in} + \frac{0.05\,\text{m}}{0.6\,\text{m}} R_{WS} + \frac{0.55\,\text{m}}{0.6\,\text{m}} R_{FI} + R_S + R_{out}\)
04

4. Determine the rate of heat transfer through the wall

The rate of heat transfer through the wall can be found using the equation: \(q = \frac{\Delta T}{R}\), where \(\Delta T = T_{in} - T_{out} = 20掳C - (-9掳C)\) So, \(q = \frac{(20掳C - (-9掳C))}{R}\)

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Key Concepts

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

Heat Transfer
Heat transfer is the process by which thermal energy moves from a region of higher temperature to a region of lower temperature. This is a fundamental concept in thermodynamics and plays a crucial role in various applications, including building design and insulation. Understanding how heat moves through different materials helps in designing structures that are energy efficient.

There are three primary modes of heat transfer: conduction, convection, and radiation. Conduction occurs within a material or between materials in direct contact. Convection involves the movement of fluid (like air or water) transferring heat, and radiation involves heat transfer through electromagnetic waves without the need for a medium.

In the context of building walls, like those mentioned in the exercise, conduction is the primary mechanism of heat transfer. The materials used, such as sheetrock and fiberglass insulation, play significant roles in determining how much heat is transferred between the interior and exterior environments. Calculating the rate of heat transfer, as shown in the exercise, helps in assessing the wall's efficiency and thermal performance.
Building Insulation
Building insulation is a material or method used to reduce the rate of heat transfer in walls, roofs, and other parts of a building. Effective insulation is crucial for energy efficiency, as it helps maintain indoor temperatures and reduces the energy required for heating and cooling.

In our exercise, fiberglass insulation is used between the wooden studs. Fiberglass is a popular insulation material due to its low thermal conductivity, meaning it provides a high resistance to heat flow. This reduces the amount of heat that escapes the house in the winter or enters in the summer, leading to lower energy bills.

There are various types of insulation materials, each with different properties:
  • Fiberglass: Known for its affordability and good thermal resistance.
  • Foam board: Provides high insulation value with a thin layer.
  • Spray foam: Can fill gaps and provide excellent insulation.
Choosing the right insulation depends on factors like climate, budget, and the specific requirements of the building.
Conduction
Conduction is a type of heat transfer that occurs when thermal energy moves through a solid material from a high-temperature area to a low-temperature area. It is one of the simplest forms of heat transfer and is characterized by the flow of heat through a medium due to temperature difference.

In the exercise, conduction is the main type of heat transfer occurring in the wall. Each component of the wall, from sheetrock to wood studs and fiberglass insulation, has its own thermal conductivity, which determines how effectively it conducts heat. The resistance to heat flow through these materials is calculated to understand their cumulative effect on thermal performance.

For conduction, the formula used is:\[ R = \frac{t}{k \, A} \]where \( t \) is the thickness, \( k \) is the thermal conductivity, and \( A \) is the area. This formula helps in determining the thermal resistance, which is key to understanding how well a structure can insulate and resist heat flow.
Thermal Conductivity
Thermal conductivity is a material property that indicates how well a material can conduct heat. It is denoted by the symbol \( k \) and is measured in watts per meter per degree Kelvin (\( \text{W/m} \cdot \text{K} \)). A higher thermal conductivity means a material is a good conductor of heat, while a lower value means it is a better insulator.

In the exercise, different materials such as sheetrock, wood, and fiberglass each have unique thermal conductivity values. These values are crucial for determining the overall thermal resistance of the wall. Materials with low thermal conductivity, like fiberglass insulation (\( k = 0.034 \, \text{W/m} \cdot \text{K} \)), are used to insulate, whereas wood studs have a slightly higher thermal conductivity of \( k = 0.11 \, \text{W/m} \cdot \text{K} \), indicating that they conduct heat more than fiberglass but less than many other materials.

Understanding thermal conductivity allows engineers and builders to select the best combination of materials for energy-efficient buildings by minimizing unwanted heat flow while maintaining structural integrity.

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Most popular questions from this chapter

Steam at \(450^{\circ} \mathrm{F}\) is flowing through a steel pipe \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) whose inner and outer diameters are \(3.5\) in and \(4.0\) in, respectively, in an environment at \(55^{\circ} \mathrm{F}\). The pipe is insulated with 2 -in-thick fiberglass insulation \((k=\) \(\left.0.020 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\). If the heat transfer coefficients on the inside and the outside of the pipe are 30 and \(5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), respectively, determine the rate of heat loss from the steam per foot length of the pipe. What is the error involved in neglecting the thermal resistance of the steel pipe in calculations?

In a combined heat and power (CHP) generation process, by-product heat is used for domestic or industrial heating purposes. Hot steam is carried from a CHP generation plant by a tube with diameter of \(127 \mathrm{~mm}\) centered at a square crosssection solid bar made of concrete with thermal conductivity of \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The surface temperature of the tube is constant at \(120^{\circ} \mathrm{C}\), while the square concrete bar is exposed to air with temperature of \(-5^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the temperature difference between the outer surface of the square concrete bar and the ambient air is to be maintained at \(5^{\circ} \mathrm{C}\), determine the width of the square concrete bar and the rate of heat loss per meter length.

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 be very long, the rate of heat transfer from the fin is (a) \(2.2 \mathrm{~W}\) (b) \(3 \mathrm{~W}\) (c) \(3.7 \mathrm{~W}\) (d) \(4 \mathrm{~W}\) (e) \(4.7 \mathrm{~W}\)

Consider a \(1.5\)-m-high and 2 -m-wide triple pane window. The thickness of each glass layer \((k=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\), and the thickness of each air space \((k=0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(1 \mathrm{~cm}\). If the inner and outer surface temperatures of the window are \(10^{\circ} \mathrm{C}\) and \(0^{\circ} \mathrm{C}\), respectively, the rate of heat loss through the window is (a) \(75 \mathrm{~W}\) (b) \(12 \mathrm{~W}\) (c) \(46 \mathrm{~W}\) (d) \(25 \mathrm{~W}\) (e) \(37 \mathrm{~W}\)

A 3-m-diameter spherical tank containing some radioactive material is buried in the ground \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The distance between the top surface of the tank and the ground surface is \(4 \mathrm{~m}\). If the surface temperatures of the tank and the ground are \(140^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer from the tank.
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