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A pipe is insulated to reduce the heat loss from it. However, measurements indicate that the rate of heat loss has increased instead of decreasing. Can the measurements be right?

Short Answer

Expert verified
Explain the factors to consider in this situation and how they might affect the results. Answer: Yes, the measurements can be correct if any of the factors affecting heat loss has changed significantly. These factors include the thermal conductivity of the insulating material, the thickness of the insulation layer, the surface area of the pipe, and the temperature difference between the pipe and surroundings. Changes in these factors can contribute to an increase in heat loss despite the added insulation, and further investigation would be needed to identify and resolve the issue.

Step by step solution

01

Understand heat conduction and insulation

Heat conduction through a material is given by Fourier's Law: $$q = k * A * \frac{\Delta T}{d}$$ where \(q\) is the heat flow rate, \(k\) is the thermal conductivity of the material, \(A\) is the surface area, \(\Delta T\) is the temperature difference between the pipe and surroundings, and \(d\) is the thickness of the insulating layer. Insulation works by reducing the heat flow rate due to its lower thermal conductivity compared to the pipe's material.
02

Analyze possible causes of increased heat loss

There are a few possible factors that might cause the increased rate of heat loss despite the insulation: 1. The insulating material has a higher thermal conductivity than intended or expected, causing more heat to be conducted through. 2. The insulation layer is too thin, reducing its effectiveness in preventing heat loss. 3. An increase in the surface area of the pipe due to additional insulation, allowing more heat transfer between the pipe and surroundings. 4. The temperature difference between the pipe and surroundings has significantly increased after adding the insulation, possibly due to external factors or changes in operating conditions.
03

Evaluate the factors

To determine if the measurements can be right, we need to consider how each factor might contribute to the increased heat loss: 1. Check the specifications of the insulating material used for its thermal conductivity and compare it with the value needed to reduce heat loss from the pipe. If the value is significantly higher, this could be a contributing factor. 2. Measure the thickness of the insulating layer and compare it with the required thickness for effective heat loss reduction. If it's too thin, this may be the cause of the increased heat loss. 3. Calculate the surface area of the pipe before and after adding the insulation. If there's a significant increase in the surface area, it might lead to more heat being lost to the surroundings. 4. Compare the temperature difference between the pipe and surroundings before and after adding the insulation. If there's a significant increase, it can lead to a higher heat loss rate even with the insulation.
04

End Note

If any of these factors has changed significantly enough to cause the increase in heat loss, then the measurements can indeed be correct. The insulation might not be working as intended, and further investigation would be required to identify and solve the issue.

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Key Concepts

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

Fourier's Law
Fourier's Law governs how heat flows through materials, making it a cornerstone of understanding heat loss, especially regarding insulation. This law states that the heat transfer rate through a solid substance is proportional to the temperature difference across the substance and inversely proportional to its thickness. In simple terms, if one side of a material is hotter than the other, heat will flow from the hot side to the cooler side.

The formula according to Fourier's Law is: \[q = k \times A \times \frac{\Delta T}{d}\],where \(q\) is the heat flow rate, \(k\) stands for thermal conductivity, \(A\) is the cross-sectional area through which heat is flowing, \(\Delta T\) represents the temperature difference across the material, and \(d\) denotes the thickness of the material.

Facilitating education in thermal dynamics, this principle helps students predict and analyze the performance of insulating materials in real-world applications, such as the insulation wrapped around a pipe.
Thermal Conductivity
Thermal conductivity is a material-specific value that measures its ability to conduct heat. When we say a material has high thermal conductivity, like metal, it means heat passes through it easily. Conversely, a good insulator has low thermal conductivity, impeding heat flow, akin to a thermal barrier.

In practice, choosing the right insulating material with appropriate thermal conductivity, which is usually a low value, is crucial for reducing energy loss in homes, industries, and various engineering applications. Problems in homework often require students to calculate heat transfer through materials, necessitating a clear grasp of this concept.

Understanding the nuance, like why certain materials feel cooler to the touch even at the same temperature, can be attributed to their differing thermal conductivities, is an essential part of the student's learning journey.
Insulating Materials
Insulating materials are the unsung heroes in our battle to control heat loss. They are selected based on their low thermal conductivity, providing resistance to heat flow. Insulators include a range of materials such as fiberglass, foam, and even air, which is used in double-paned windows.

Classroom exercises often involve comparing various insulators to determine which minimizes heat loss effectively, incorporating real-life decision making. For a student delving into the world of thermal insulation, figuring out that thicker or more insulating layers make a notable difference can be a significant leap forward in understanding heat dynamics.

Exploring the properties and appropriate applications of different insulating materials allows students to appreciate their role in energy conservation and temperature regulation across numerous industries.
Heat Transfer Rate
The heat transfer rate is the speed at which heat energy moves through a material. It's a vital parameter in many engineering problems, especially when we're trying to keep energy consumption down or maintain a specific temperature. In essence, the rate at which a pipe loses heat to its surroundings can be a telling sign of the effectiveness of its insulation.

Homework problems often touch upon this rate, requiring students to understand how changes in material, thickness, or surface area may influence it. For instance, they learn that increasing the thickness of insulation or decreasing its thermal conductivity can reduce the heat transfer rate, leading to more energy-efficient systems.

Having a solid comprehension of what influences heat transfer rate equips learners with skills to analyze possible improvements in practical situations like home insulation, HVAC systems, and industrial process design.

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Most popular questions from this chapter

The walls of a food storage facility are made of a 2 -cm-thick layer of wood \((k=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in contact with a 5 -cm- thick layer of polyurethane foam \((k=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the temperature of the surface of the wood is \(-10^{\circ} \mathrm{C}\) and the temperature of the surface of the polyurethane foam is \(20^{\circ} \mathrm{C}\), the temperature of the surface where the two layers are in contact is (a) \(-7^{\circ} \mathrm{C}\) (b) \(-2^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(8^{\circ} \mathrm{C}\) (e) \(11^{\circ} \mathrm{C}\)

A plane brick wall \((k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(10 \mathrm{~cm}\) thick. The thermal resistance of this wall per unit of wall area is (a) \(0.143 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (b) \(0.250 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (c) \(0.327 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (d) \(0.448 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (e) \(0.524 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

The \(700 \mathrm{~m}^{2}\) ceiling of a building has a thermal resistance of \(0.52 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The rate at which heat is lost through this ceiling on a cold winter day when the ambient temperature is \(-10^{\circ} \mathrm{C}\) and the interior is at \(20^{\circ} \mathrm{C}\) is (a) \(23.1 \mathrm{~kW} \quad\) (b) \(40.4 \mathrm{~kW}\) (c) \(55.6 \mathrm{~kW}\) (d) \(68.1 \mathrm{~kW}\) (e) \(88.6 \mathrm{~kW}\)

Steam in a heating system flows through tubes whose outer diameter is \(5 \mathrm{~cm}\) and whose walls are maintained at a temperature of \(180^{\circ} \mathrm{C}\). Circular aluminum alloy 2024-T6 fins \((k=186 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of outer diameter \(6 \mathrm{~cm}\) and constant thickness \(1 \mathrm{~mm}\) are attached to the tube. The space between the fins is \(3 \mathrm{~mm}\), and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins.

Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend attaching fins both inside and outside the tubes?

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