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Chapter 3: Problem 40

Consider a house whose walls are \(12 \mathrm{ft}\) high and \(40 \mathrm{ft}\) long. Two of the walls of the house have no windows, while each of the other two walls has four windows made of \(0.25\)-in-thick glass \(\left(k=0.45 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right), 3 \mathrm{ft} \times 5 \mathrm{ft}\) in size. The walls are certified to have an \(R\)-value of 19 (i.e., an \(\mathrm{L} / k\) value of \(19 \mathrm{~h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\) ). Disregarding any direct radiation gain or loss through the windows and taking the heat transfer coefficients at the inner and outer surfaces of the house to be 2 and \(4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), respectively, determine the ratio of the heat transfer through the walls with and without windows.

Short Answer

Expert verified
The ratio of heat transfer through the walls with and without windows is approximately 1.18.

Step by step solution

01

Calculate the area of the house walls with and without windows

First, let's calculate the total area of the walls and the windows: Area of the walls without windows = Wall height 脳 Wall length = \(12 \mathrm{ft} 脳 40 \mathrm{ft} = 480 \mathrm{ft^2}\) Area of the windows = Number of windows 脳 Window height 脳 Window width = \(4 脳 3 \mathrm{ft} 脳 5 \mathrm{ft} = 60 \mathrm{ft^2}\) Area of the walls with windows = Total area of walls - Total area of windows = \(480 \mathrm{ft^2} - 60 \mathrm{ft^2} = 420 \mathrm{ft^2}\)
02

Calculate the R-value of the windows

Now, let's calculate the R-value of windows using the given thickness of windows and the thermal conductivity value k: R_value of windows (R_window) = Glass thickness / k = \((0.25 / 12\:\mathrm{ft}) / 0.45 \:\mathrm{Btu/h\cdot ft\cdot { }^{\circ} F} \approx 1.48 \:\mathrm{h\cdot ft^2 \cdot { }^{\circ} F / Btu}\)
03

Calculate the series resistance for walls with and without windows

Next, calculate the series resistance value for both walls with windows and without windows. R_wall = given R-value of the wall = 19 \( \mathrm{~h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu} \) R_inner = 1 / (heat transfer coefficient at the inner surface) = 1 / 2 \(\mathrm{h\cdot ft^2\cdot { }^{\circ} F / Btu}\) R_outer = 1 / (heat transfer coefficient at the outer surface) = 1 / 4 \(\mathrm{h\cdot ft^2\cdot { }^{\circ} F / Btu}\) Total Series Resistance for walls without windows = R_inner + R_wall + R_outer Total Series Resistance for walls without windows = 1/2 + 19 + 1/4 = \(19.75 \:\mathrm{h\cdot ft^2 \cdot { }^{\circ} F / Btu}\)
04

Calculate the parallel resistance for walls with windows

Knowing that there is a window, we need to calculate parallel resistance value for walls with windows. R_parallel = [(Area of windows 脳 R_window) + (Area of walls with windows 脳 R_wall)] / Total Area of the walls R_parallel = [\((60 脳 1.48) + (420 脳 19) / 480\:\mathrm{h\cdot ft^2\cdot { }^{\circ} F / Btu}\approx 16.02\)
05

Calculate heat transfer ratio for walls with and without windows

Now we will find the total heat transfer ratio for walls with and without windows. Total Resistance for walls with windows = R_inner + R_parallel + R_outer Total Resistance for walls with windows = 1/2 + 16.02 + 1/4 = \(16.77\:\mathrm{h\cdot ft^2\cdot { }^{\circ} F / Btu}\) Heat Transfer Ratio = Resistance without windows / Resistance with windows = \(19.75 / 16.77 \approx 1.18\). So, the ratio of heat transfer through the walls with and without windows is approximately 1.18.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It is represented by the symbol k and is an intrinsic property of the material, meaning it does not change with the amount or shape of the material. The units for thermal conductivity in the Imperial system are 叠迟耻/(丑路蹿迟路掳贵). The higher the thermal conductivity of a material, the more easily heat passes through it. For instance, in the context of the exercise, the glass windows have a thermal conductivity (k) of 0.45 叠迟耻/(丑路蹿迟路掳贵), indicating that glass is not a very good insulator compared to materials with lower thermal conductivity values.

In practical terms, selecting materials for the walls of a building with low thermal conductivity can minimize the amount of heat that escapes during winter or enters during summer, thus improving energy efficiency. Conversely, glass, having higher thermal conductivity than insulating wall materials, allows more heat transfer, thereby affecting the overall insulation of the house.
R-value
The R-value is a metric used to indicate the thermal resistance of a material or assembly of materials (such as a wall with insulation). The R-value quantifies how well a two-dimensional barrier, such as a layer of insulation, resists the conductive flow of heat. It is usually expressed in units of 丑路蹿迟虏路掳贵/叠迟耻. The R-value is the inverse of the heat transfer coefficient, U-value, and is calculated by dividing the thickness of the material (L) by its thermal conductivity (k).

In the solved exercise, the walls are described to have an R-value of 19, signifying a high level of thermal resistance which implies they are effective at insulating the home. The calculation of the windows' R-value in the exercise demonstrates the process of determining the thermal resistance of different components of a structure, which in turn allows us to understand and compare how each component contributes to the overall thermal performance.
Thermal Resistance
Thermal resistance is a concept closely related to the R-value and is expressed with the same units: 丑路蹿迟虏路掳贵/叠迟耻. It represents the ability of a physical object to resist the flow of heat. In a building scenario, the total thermal resistance for a wall or a window is influenced not only by the materials used but also by the boundary air films on either side. These boundary layers have their own thermal resistance values, which must be added to the material's R-value to determine the total thermal resistance to heat flow.

The exercise involves calculating the total series thermal resistance for the house's walls with and without windows. The exercise improved the understanding of how components with differing thermal resistances combine to affect the overall heat transfer through the structure. By analyzing the sum of the individual R-values and the configuration of these components (in series or parallel), you can predict the overall insulating effectiveness of the combined structure. This analysis is crucial for evaluating the energy efficiency of buildings and for making informed decisions when selecting materials for construction or renovation projects.

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Most popular questions from this chapter

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.

Consider a tube for transporting steam that is not centered properly in a cylindrical insulation material \((k=\) \(0.73 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The tube diameter is \(D_{1}=20 \mathrm{~cm}\) and the insulation diameter is \(D_{2}=40 \mathrm{~cm}\). The distance between the center of the tube and the center of the insulation is \(z=5 \mathrm{~mm}\). If the surface of the tube maintains a temperature of \(100^{\circ} \mathrm{C}\) and the outer surface temperature of the insulation is constant at \(30^{\circ} \mathrm{C}\), determine the rate of heat transfer per unit length of the tube through the insulation.

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.

A mixture of chemicals is flowing in a pipe \(\left(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\right.\), and \(L=10 \mathrm{~m}\) ). During the transport, the mixture undergoes an exothermic reaction having an average temperature of \(135^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent any incident of thermal burn, the pipe needs to be insulated. However, due to the vicinity of the pipe, there is only enough room to fit a \(2.5\)-cm-thick layer of insulation over the pipe. The pipe is situated in a plant, where the average ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the insulation for the pipe such that the thermal conductivity of the insulation is sufficient to maintain the outside surface temperature at \(45^{\circ} \mathrm{C}\) or lower.

In the United States, building insulation is specified by the \(R\)-value (thermal resistance in \(\mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu units). A homeowner decides to save on the cost of heating the home by adding additional insulation in the attic. If the total \(R\)-value is increased from 15 to 25 , the homeowner can expect the heat loss through the ceiling to be reduced by (a) \(25 \%\) (b) \(40 \%\) (c) \(50 \%\) (d) \(60 \%\) (e) \(75 \%\)
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