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Chapter 13: Problem 48

Consider the attic of a home located in a hot climate. The floor of the attic is characterized by a width of \(L_{1}=10 \mathrm{~m}\) while the roof makes an angle of \(\theta=30^{\circ}\) from the horizontal direction, as shown in the schematic. The homeowner wishes to reduce the heat load to the home by adhering bright aluminum foil \(\left(\varepsilon_{f}=0.07\right)\) onto the surfaces of the attic space. Prior to installation of the foil, the surfaces are of emissivity \(s_{e}=0.85\). (a) Consider installation on the bottom of the attic roof only. Determine the ratio of the radiation heat transfer after to before the installation of the foil. (b) Determine the ratio of the radiation heat transfer after to before installation if the foil is installed only on the top of the attic floor. (c) Determine the ratio of the radiation heat transfer if the foil is installed on both the roof bottom and the floor top.

Short Answer

Expert verified
The radiation heat transfer ratios for each scenario are as follows: a) Installation on the bottom of the attic roof only: \(\frac{Q_{after}}{Q_{before}} = \frac{0.07}{0.85}\) b) Installation on the top of the attic floor only: \(\frac{Q_{after}}{Q_{before}} = \frac{0.07}{0.85}\) c) Installation on both the roof bottom and the floor top: \(\frac{Q_{after}}{Q_{before}} = \frac{0.07}{0.85}\)

Step by step solution

01

a) Installation on the bottom of the attic roof only

Step 1: Calculate the original heat transfer before the installation. Since \(\varepsilon_{before} = 0.85\), the initial heat transfer is: \[Q_{before} \propto 0.85\] Step 2: Calculate the heat transfer after the installation of foil. Since \(\varepsilon_{after} = 0.07\) (emissivity of the aluminum foil) for the roof only, the heat transfer after the installation is: \[Q_{after} \propto 0.07\] Step 3 : Compute the ratio of the radiation heat transfer after to before the installation of the foil. Since we are not given the temperature, we can skip the \(\sigma T^4\) part and just consider the ratio of the emissivities. Therefore, the ratio of the radiation heat transfer after to before is: \[\frac{Q_{after}}{Q_{before}} = \frac{0.07}{0.85}\]
02

b) Installation on the top of the attic floor only

Step 1: Calculate the original heat transfer before the installation. Since \(\varepsilon_{before} = 0.85\), the initial heat transfer is: \[Q_{before} \propto 0.85\] Step 2: Calculate the heat transfer after the installation of foil. Now, the aluminum foil covers only the attic floor, so the emissivity after the installation is \(\varepsilon_{after} = 0.07\) for the floor only. We have: \[Q_{after} \propto 0.07\] Step 3 : Compute the ratio of the radiation heat transfer after to before the installation of the foil. Again, the ratio of the radiation heat transfer after to before is: \[\frac{Q_{after}}{Q_{before}} = \frac{0.07}{0.85}\]
03

c) Installation on both the roof bottom and the floor top

Step 1: Calculate the original heat transfer before the installation. Since \(\varepsilon_{before} = 0.85\), the initial heat transfer is: \[Q_{before} \propto 0.85\] Step 2: Calculate the heat transfer after the installation of foil. For this case, let's suppose both surfaces, roof and floor, are covered with the foil. Therefore, the emissivity after the installation is \(\varepsilon_{after} = 0.07\) for both the roof and the floor. We have: \[Q_{after} \propto 0.07\] Step 3 : Compute the ratio of the radiation heat transfer after to before the installation of the foil. Finally, the ratio of the radiation heat transfer after to before is: \[\frac{Q_{after}}{Q_{before}} = \frac{0.07}{0.85}\]

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Key Concepts

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

Emissivity
In the context of radiation heat transfer, emissivity is a crucial concept. It refers to the efficiency with which a surface emits thermal radiation. Emissivity values range from 0 to 1. A high emissivity implies that a surface is very efficient at radiating energy, while a low emissivity suggests poor emission.
Understanding emissivity is vital when predicting how much heat a material will transfer via radiation. In this particular problem, the attic undergoes a transformation by the addition of aluminum foil with an emissivity of 0.07. Compared to the initial surfaces with an emissivity of 0.85, this represents a significant reduction in the ability to radiate heat.
- High emissivity = more heat radiation (e.g., the original attic surface). - Low emissivity = less heat radiation (e.g., surface with aluminum foil).
Reducing emissivity through material modification, like adding aluminum foil, directly decreases the radiation heat transfer. This is beneficial in hot climates, where it is desirable to keep heat out of living spaces to maintain a cooler interior climate.
Thermal Insulation
Thermal insulation is a way to prevent heat from entering or leaving a building, significantly impacting energy efficiency.
In hot climates, such as in the exercise, thermal insulation is critical to help maintain comfortable indoor temperatures and reduce cooling costs. By covering attic surfaces with aluminum foil, the goal is to reduce heat transfer by minimizing radiation losses. This is achieved by lowering the emissivity of the surfaces, which in turn enhances the insulating properties.
A few benefits of effective thermal insulation include:
  • Reduced energy consumption necessary for heating or cooling.
  • Enhanced indoor comfort by moderating temperature variations.
  • Potential for reduced carbon footprint due to lower energy demands.
Thermal insulation extends beyond just reflective coatings. It often involves materials like fiberglass, foam, and reflective barriers like aluminum foil, which work together to restrict heat flow into or out of a space, ensuring that desired temperatures are maintained more efficiently.
Heat Load Reduction
Reducing the heat load in a building means decreasing the amount of heat that needs to be managed, either by cooling or heating systems. Accomplishing this in hot climates is essential for maintaining lower energy bills and ensuring comfortable living conditions throughout the year.
In the scenario described, heat load reduction is achieved through the application of aluminum foil with low emissivity. As discussed, the reflective nature of aluminum foil helps mitigate the radiant heat entering the attic spaces, thus decreasing the overall heat load on the building.
Strategies to reduce heat load include:
  • Using reflective materials to diminish incoming radiant heat.
  • Incorporating thermally insulating materials to prevent conduction.
  • Improving air sealing to prevent unwanted air leakage which could carry heat.
Together, these strategies help in maintaining comfort while simultaneously reducing the workload on cooling systems, thereby conserving energy and cutting costs. The application of low emissivity materials is particularly effective in radiant heat transfer scenarios, as shown by the exercise.

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Most popular questions from this chapter

Most architects know that the ceiling of an ice-skating rink must have a high reflectivity. Otherwise, condensation may occur on the ceiling, and water may drip onto the ice, causing bumps on the skating surface. Condensation will occur on the ceiling when its surface temperature drops below the dew point of the rink air. Your assignment is to perform an analysis to determine the effect of the ceiling emissivity on the ceiling temperature, and hence the propensity for condensation. The rink has a diameter of \(D=50 \mathrm{~m}\) and a height of \(L=10 \mathrm{~m}\), and the temperatures of the ice and walls are \(-5^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively. The rink air temperature is \(15^{\circ} \mathrm{C}\), and a convection coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) characterizes conditions on the ceiling surface. The thickness and thermal conductivity of the ceiling insulation are \(0.3 \mathrm{~m}\) and \(0.035 \mathrm{~W} / \mathrm{m}-\mathrm{K}\), respectively, and the temperature of the outdoor air is \(-5^{\circ} \mathrm{C}\). Assume that the ceiling is a diffuse-gray surface and that the Walls and ice may be approximated as blackbodies. (a) Consider a flat ceiling having an emissivity of \(0.05\) (highly reflective panels) or \(0.94\) (painted panels). Perform an energy balance on the ceiling to calculate the corresponding values of the ceiling temperature. If the relative humidity of the rink air is \(70 \%\), will condensation occur for either or both of the emissivities? (b) For each of the emissivities, calculate and plot the ceiling temperature as a function of the insulation thickness for \(0.1 \leq r \leq 1 \mathrm{~m}\). Identify conditions for which condensation will occur on the ceiling.

The cylindrical peephole in a furnace wall of thickness \(L=250 \mathrm{~mm}\) has a diameter of \(D=125 \mathrm{~mm}\). The furnace interior has a temperature of \(1300 \mathrm{~K}\), and the surroundings outside the furnace have a temperature of \(300 \mathrm{~K}\). Determine the heat loss by radiation through the peephole.

A flat-plate solar collector, consisting of an absorber plate and single cover plate, is inclined at an angle of \(\tau=60^{\circ}\) relative to the horizontal. Consider conditions for which the incident solar radiation is collimated at an angle of \(60^{\circ}\) relative to the horizontal and the solar flux is \(900 \mathrm{~W} / \mathrm{m}^{2}\). The cover plate is perfectly transparent to solar radiation \((\lambda \leq 3\) \(\mu m)\) and is opaque to radiation of larger wavelengths. The cover and absorber plates are diffuse surfaces having the spectral absorptivities shown. The length and width of the absorber and cover plates are much larger than the plate spacing \(L\). What is the rate at which solar radiation is absorbed per unit area of the absorber plate? With the absorber plate well insulated from below and absorber and cover plate temperarures \(T_{a}\) and \(T_{c}\) of \(70^{\circ} \mathrm{C}\) and \(27^{\circ} \mathrm{C}\), respectively, what is the heat loss per unit area of the absorber plate?

Two concentric spheres of diameter \(D_{1}=0.8 \mathrm{~m}\) and \(D_{2}=1.2 \mathrm{~m}\) are separated by an air space and have surface temperatures of \(T_{1}=400 \mathrm{~K}\) and \(T_{2}=300 \mathrm{~K}\). (a) If the surfaces are black, what is the net rate of radiation exchange between the spheres? (b) What is the net rate of radiation exchange between the surfaces if they are diffuse and gray with \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.05\) ? (c) What is the net rate of radiation exchange if \(D_{2}\) is increased to \(20 \mathrm{~m}\), with \(\varepsilon_{2}=0.05, \varepsilon_{1}=0.5\), and \(D_{1}=0.8 \mathrm{~m}\) ? What error would be introduced by assuming blackbody behavior for the outer surface \(\left(\varepsilon_{2}=1\right)\), with all other conditions remaining the same? (d) For \(D_{2}=1.2 \mathrm{~m}\) and emissivities of \(s_{1}=0.1,0.5\), and \(1.0\), compute and plot the net rate of radiation exchange as a function of \(\varepsilon_{2}\) for \(0.05 \leq \varepsilon_{2} \leq 1.0\).

Liquid oxygen is stored in a thin-walled, spherical container \(0.8 \mathrm{~m}\) in diameter, which is enclosed within a second thin-walled, spherical container \(1.2 \mathrm{~m}\) in diameter. The opaque, diffuse, gray container surfaces have an emissivity of \(0.05\) and are separated by an evacuated space. If the outer surface is at \(280 \mathrm{~K}\) and the inner surface is at \(95 \mathrm{~K}\), what is the mass rate of oxygen lost due to evaporation? (The latent heat of vaporization of oxygen is \(2.13 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).)
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