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The thermal conductivity of a sheet of rigid, extruded insulation is reported to be \(k=0.029 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The measured temperature difference across a 20 -mm-thick sheet of the material is \(T_{1}-T_{2}=10^{F} \mathrm{C}\). (a) What is the heat flux through a \(2 \mathrm{~m} \times 2 \mathrm{~m}\) shect of the insulation? (b) What is the rate of heat transfer through the sheet of insulation?

Short Answer

Expert verified
(a) Heat flux is 14.5 W/m². (b) Heat transfer rate is 58 W.

Step by step solution

01

Understand the Problem

We need to find both the heat flux through the insulation and the total rate of heat transfer across it. The thermal conductivity, the thickness of the insulation, the temperature difference, and the dimensions of the sheet are given.
02

Calculate Heat Flux

The formula for heat flux is given by:\[q'' = \frac{k \, (T_1 - T_2)}{L}\]where:- \(k = 0.029 \, \text{W/m} \cdot \text{K}\) is the thermal conductivity,- \(T_1 - T_2 = 10\, ^\circ \text{C}\), the temperature difference,- \(L = 0.020 \, \text{m}\) is the thickness of the sheet.Substitute the values:\[q'' = \frac{0.029 \, \text{W/m} \cdot \text{K} \times 10}{0.020} = 14.5 \, \text{W/m}^2\]
03

Calculate Total Heat Transfer Rate

The total heat transfer rate, \( \dot{Q} \), through the sheet is calculated using:\[\dot{Q} = q'' \times A \]where:- \(q'' = 14.5 \, \text{W/m}^2\) (calculated from Step 2),- \(A = 2 \, \text{m} \times 2 \, \text{m} = 4 \, \text{m}^2\).Substitute the values:\[\dot{Q} = 14.5 \, \text{W/m}^2 \times 4 \, \text{m}^2 = 58 \, \text{W}\]

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Key Concepts

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

Heat Flux
Heat flux refers to the rate at which heat energy passes through a given surface. It's often measured in watts per square meter (W/m²). Imagine it as the amount of heat energy flowing through each square meter of a material every second.
In the context of our problem with the insulation material, the heat flux is determined using the formula:\[ q'' = \frac{k \cdot (T_1 - T_2)}{L} \]Here, "k" is the thermal conductivity, which indicates how well the material can conduct heat. The temperature difference, \( T_1 - T_2 \), drives the heat to flow through the sheet. "L" represents the thickness of the insulation. Given the higher the thermal conductivity and the temperature difference, or the thinner the material, the greater the heat flux. This is why in our exercise, with a thermal conductivity \( k = 0.029 \, \text{W/m} \cdot \text{K} \), temperature difference \( 10 \, ^\circ \text{C} \), and thickness \( L = 0.020 \, \text{m} \), the formula calculates the heat flux as \( q'' = 14.5 \, W/m^2 \). This means each square meter of that insulation allows 14.5 watts to pass through due to the given conditions.
Rate of Heat Transfer
The rate of heat transfer is the total amount of heat being transferred through an entire object or system, rather than just per unit area. It tells us how much heat is being transferred in total, which is especially useful for larger spaces or materials.
To calculate it, we multiply the heat flux, found previously, by the total area through which the heat is transferring:\[ \dot{Q} = q'' \times A \]In our task, the area "A" is a sheet measuring \(2 \, \text{m} \times 2 \, \text{m} = 4 \, \text{m}^2\). Substituting in the heat flux \( q'' = 14.5 \, \text{W/m}^2 \), we find the rate of heat transfer is \( \dot{Q} = 58 \, \text{W} \). This calculation tells us that the entire sheet lets through 58 watts of heat energy at any moment. It's vital to understand this concept, as larger areas generally mean more heat transferring across them. When planning insulation or heating systems, knowing both the heat flux and the rate of heat transfer helps in designing for energy efficiency.
Thermal Insulation
Thermal insulation is a crucial concept in controlling how much heat moves between different environments. It essentially acts as a barrier, reducing the rate of heat flow, thereby helping to maintain a desired temperature.
To understand how insulation works, we consider its thermal conductivity. A lower "k" value means that the material is a better insulator, since it allows less heat to pass through. The exercise uses a material with \( k = 0.029 \, \text{W/m} \cdot \text{K} \), indicating decent insulating properties.Several factors influence the effectiveness of insulation:- **Material Thickness:** Thicker materials usually provide better insulation, as they create a larger barrier for the heat to overcome.- **Temperature Difference:** The greater the difference in temperature across the insulation, the more potential there is for heat transfer. Insulation works harder to maintain a balance when this difference is large.- **Surface Area:** More surface area means more exposure to heat transfer, even if the insulation itself is effective.Understanding these aspects helps in choosing the right insulation material for your needs, whether for a cozy home, a fridge, or industrial applications. Effective insulation is central to energy efficiency, driving down heating and cooling costs and reducing environmental impact.

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Most popular questions from this chapter

The inner and outer surface temperatures of a glass window \(5 \mathrm{~mm}\) thick are 15 and \(5^{\circ} \mathrm{C}\). What is the heat loss through a window that is \(1 \mathrm{~m}\) by \(3 \mathrm{~m}\) on a side? The thermal conductivity of glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at \(20^{\circ} \mathrm{C}\) throughout the year, while the walls of the room are nominally at \(27^{\circ} \mathrm{C}\) and \(14^{\circ} \mathrm{C}\) in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of \(32^{\circ} \mathrm{C}\) throughout the year and to have an emissivity of \(0.90\). The coefficient associated with heat transfer by natural convection between the person and the room air is approximately \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

The clectrical-substitation radiometer shown schematically determines the optical (radiant) power of a beam by measuring the electrical power required to heat the receiver to the same temperature. With a beam, such as a luser of optical power \(P_{\text {epv }}\) incident on the receiver, its temperature, \(T_{p}\) increases above that of the chamber walls held al a uniform temperature, \(T_{\text {we }}=77 \mathrm{~K}\), With the optical beam blocked, the heater on the hackside of the receiver is encrgized and the electrical power, \(P_{\text {eine: }}\) required to reach the same value of \(T_{\text {, is measured. The }}\) purpose of your analysis is to determine the relationship tetween the clectrical and optical power, considering heat transfer processes experienced by the receiver. Consider a radiomeler with a \(15-\mathrm{mm}\)-diameter receiver having a blackened surface with an emissivity of \(0.95\) and an absorptivity of \(0.98\) for the optical beam. When operating in the optical mode, conduction heat losses from the backside of the receiver are negligible. In the electricul mode, the loss amounts to \(5 \%\) of the electrical power. What is the optical power of a beam when the indicated electrical power is \(20.64 \mathrm{~mW}\) ? What is the corresponding recciver temperature?

In the thermal processing of semiconductor materials. annealing is accomplished by heating a silicon wafer according to a temiperature-titue recipe and then maintaining a fixed elevated temperature for a prescribed period of time. For the process tool arrangement shown as follow5, the wafer is in an evacuated chamber whose walls are maintained at \(27 \mathrm{C}\) and wiahin which heating lamps maintain a radiant flux \(q\), at its upper surface. The wafer is \(0.78 \mathrm{~mm}\) thick, has a thermul conductivity of 30 W/m \(\cdot \mathrm{K}\), and an emiskivity that cquals its absorptivity to the radiant flux \(\left(\varepsilon=\alpha_{i}=0.65\right)\). For \(q_{t}^{n}=3.0 \times 10^{5}\) \(\mathrm{W} / \mathrm{m}^{2}\), the temperalure on its lower surface is measured by a ridiation themometer and found to have a value of \(T_{\alpha l}=997^{\circ} \mathrm{C}\). To avoid warping the wafer and inducing slip planes in the crystal structure, the temperature difference across the thickness of the wafer must be less than \(2^{\circ} \mathrm{C}\). Is this cundition being met?

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