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Why is the thermal conductivity of superinsulation orders of magnitude lower than the thermal conductivity of ordinary insulation?

Short Answer

Expert verified
Answer: The thermal conductivity of superinsulation is significantly lower than that of ordinary insulation materials because it is specifically designed to minimize all three heat transfer mechanisms: solid conduction, gas conduction, and radiation. Superinsulation has an extremely low-density structure, gas-filled pores with low-conductivity gases, and micron-sized pores with multiple reflective layers to effectively reduce heat transfer, making it a superior insulator compared to ordinary insulation materials.

Step by step solution

01

Understand the terms

Thermal conductivity is a property of a material that measures its ability to conduct heat. A lower thermal conductivity means that the material is a better insulator, as it prevents heat from being transferred easily. Superinsulation is an insulation material with exceptionally low thermal conductivity, which is orders of magnitude lower than that of ordinary insulation materials.
02

Identify the factors affecting thermal conductivity

Thermal conductivity of insulation materials depends on three main factors: 1. Solid conduction: The transfer of heat through the solid structure of the material. 2. Gas conduction: The transfer of heat through the gas trapped within the material's pores or cavities. 3. Radiation: The transfer of heat as electromagnetic waves, mainly in the infrared range.
03

Analyze the structure and properties of superinsulation

Superinsulation is designed to minimize all three heat transfer mechanisms mentioned above. The key features of superinsulation include: 1. Extremely low-density structure: Reduces solid conduction by minimizing contact between the solid material particles. 2. Gas-filled pores: These pores are filled with low-conductivity gases, such as air, argon, or krypton, reducing the overall gas conduction. 3. Micron-sized pores and multiple layers: The small pore size and the presence of multiple reflective layers help to minimize the radiation heat transfer mechanism.
04

Compare the thermal conductivity of superinsulation with ordinary insulation

Ordinary insulation materials may have structures with higher density, larger pores, or fewer layers than superinsulation, which will cause them to have a higher thermal conductivity. Consequently, they are less effective at preventing heat transfer. The design and properties of superinsulation materials serve to minimize all three heat transfer mechanisms, resulting in a significantly lower thermal conductivity compared to ordinary insulation materials. Hence, the thermal conductivity of superinsulation is orders of magnitude lower than that of ordinary insulation.

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Most popular questions from this chapter

A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by radiation. Both the air in the room and the surrounding surfaces are at \(20^{\circ} \mathrm{C}\). The exposed surface of the person is \(1.5 \mathrm{~m}^{2}\) and has an average temperature of \(32^{\circ} \mathrm{C}\) and an emissivity of \(0.90\). If the rates of heat transfer from the person by convection and by radiation are equal, the combined heat transfer coefficient is (a) \(0.008 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\) (b) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(5.5 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\) (d) \(8.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(10.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Consider a 3-m \(\times 3-\mathrm{m} \times 3-\mathrm{m}\) cubical furnace whose top and side surfaces closely approximate black surfaces at a temperature of \(1200 \mathrm{~K}\). The base surface has an emissivity of \(\varepsilon=0.7\), and is maintained at \(800 \mathrm{~K}\). Determine the net rate of radiation heat transfer to the base surface from the top and side surfaces.

The inner and outer surfaces of a \(4-\mathrm{m} \times 7-\mathrm{m}\) brick wall of thickness \(30 \mathrm{~cm}\) and thermal conductivity $0.69 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( are maintained at temperatures of \)20^{\circ} \mathrm{C}\( and \)5^{\circ} \mathrm{C}$, respectively. Determine the rate of heat transfer through the wall in W.

A 1000-W iron is left on an ironing board with its base exposed to the air at \(20^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the base surface and the surrounding air is $35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. If the base has an emissivity of \)0.6$ and a surface area of \(0.02 \mathrm{~m}^{2}\), determine the temperature of the base of the iron. Answer: \(674^{\circ} \mathrm{C}\)

What is stratification? Is it more likely to occur at places with low ceilings or places with high ceilings? How does it cause thermal discomfort for a room's occupants? How can stratification be prevented?

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