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How does the thermal resistance network associated with a single-layer plane wall differ from the one associated with a five-layer composite wall?

Short Answer

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Question: Explain the main difference between the thermal resistance networks of a single-layer wall and a five-layer composite wall. Answer: The main difference between the thermal resistance networks of a single-layer wall and a five-layer composite wall lies in their complexity. A single-layer wall has a simple network consisting of one resistor representing the wall's thermal resistance. In contrast, a five-layer composite wall has a more complex network with five resistors connected in series, each representing the thermal resistance of each layer. This difference in complexity affects the insulating properties and overall heat transfer through the wall.

Step by step solution

01

Understand thermal resistance

Thermal resistance is a property of an insulating material that indicates its resistance to the flow of heat energy. It depends on the material's thermal conductivity and thickness. A lower thermal resistance allows more heat energy to pass through, while a higher resistance results in better insulation.
02

Defining thermal resistance network

A thermal resistance network represents the flow of heat energy through a layered structure, like walls, with each component's thermal resistance represented as a resistor in the network.
03

Determine single-layer wall resistance network

For a single-layer wall, with a surface area A, thickness L, and thermal conductivity k, the wall's thermal resistance (R) can be calculated using the formula: R = \frac{L}{kA} Since there is only one layer, the thermal resistance network will consist of only one resistor representing the wall's thermal resistance.
04

Determine five-layer composite wall resistance network

For a five-layer composite wall, each layer will have its resistance, depending on the thickness, thermal conductivity, and surface area of each layer. In this case, the total thermal resistance (R_total) will be the sum of the resistances of the individual layers: R_{total} = R_1 + R_2 + R_3 + R_4 + R_5 The resistance network for a five-layer composite wall would include five resistors connected in series, each representing the thermal resistance of each layer.
05

Compare single-layer and five-layer resistance networks

The primary difference between the two networks lies in the complexity of the heat transfer process. In a single-layer wall, the thermal resistance network is straightforward, consisting of just one resistor. However, in a five-layer composite wall, the network becomes more complex, consisting of multiple resistors representing the different layers. This difference in complexity can impact the wall's insulating properties and overall heat transfer through the wall.

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Key Concepts

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

Heat Transfer
Heat transfer is the movement of thermal energy from one object or substance to another. This process occurs whenever there is a temperature difference between two regions. Heat transfer can happen in three main ways: conduction, convection, and radiation.
Conduction is the form of heat transfer that takes place via direct contact. In other words, heat moves through a solid material. Think of a metal spoon warming up when placed in hot soup. In the context of the thermal resistance network of walls, conduction is the primary means of heat transfer. When designing walls, particularly insulating ones, understanding conduction is crucial. The materials chosen and their arrangement determine how effectively they can resist the flow of heat, which is often quantified as thermal resistance.
To ensure efficient heat transfer control, each material layer's properties, like thermal conductivity and thickness, play a critical role. A material with high thermal conductivity will allow heat to pass through more readily, decreasing the thermal resistance. Conversely, low thermal conductivity ensures better insulation.
Thermal Resistance
Thermal resistance is a measure that indicates how well a material can resist the flow of heat. It's like the electrical resistance that restricts the flow of current, but in this case, it concerns thermal energy. The higher the thermal resistance, the better the insulator the material is. This means less heat will pass through it, keeping energy within the desired area.
There are key factors that determine a material's thermal resistance:
  • Thermal Conductivity (\(k\)) - This is a material property indicating how easily heat can pass through. Lower values denote better insulating properties.
  • Thickness (\(L\)) - The thicker the material, the more it can oppose the heat flow.
  • Surface Area (\(A\)) - A larger surface area can distribute the heat load and impact the overall resistance.
To calculate thermal resistance (\(R\)), the formula is used: \[ R = \frac{L}{kA} \]This formula shows that resistance increases with both thickness and material with low conductivity, optimizing a wall's insulating characteristics. Effective wall construction often involves selecting materials with suitable thermal resistance to minimize energy loss and maintain thermal comfort.
Composite Wall
A composite wall consists of several layers, each made from different materials with varying thermal properties. The concept of a composite wall becomes important when constructing buildings or any structure where energy efficiency and insulation are critical.
In a composite wall, each layer acts as a separate resistor, creating a thermal resistance network. The overall thermal performance of a composite wall depends on the combined effect of these layers. The total thermal resistance is calculated as the sum of the individual resistances of each layer:\[ R_{total} = R_1 + R_2 + R_3 + R_4 + \, R_n \]This arrangement allows for flexible design choices, optimizing for numerous factors like cost, strength, and thermal performance. By choosing appropriate materials for each layer, designers can tailor the wall to specific values of insulation and heat transfer requirements.
A five-layer composite wall is an example of a more complex resistance network compared to a single-layer wall. This complexity allows for highly efficient thermal regulation, with each layer contributing to the overall thermal resistance, improving insulation, and reducing energy costs. Understanding the individual roles of each layer can help in designing walls that provide maximum thermal comfort while being structurally sound.

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Most popular questions from this chapter

A 4-m-high and 6-m-wide wall consists of a long \(18-\mathrm{cm} \times\) 30 -cm cross section of horizontal bricks \((k=0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) separated by \(3-\mathrm{cm}\)-thick plaster layers \((k=0.22 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). There are also 2 -cm-thick plaster layers on each side of the wall, and a 2-cmthick rigid foam \((k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) on the inner side of the wall. The indoor and the outdoor temperatures are \(22^{\circ} \mathrm{C}\) and \(-4^{\circ} \mathrm{C}\), and the convection heat transfer coefficients on the inner and the outer sides are \(h_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{2}=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Assuming one-dimensional heat transfer and disregarding radiation, determine the rate of heat transfer through the wall.

Hot water at an average temperature of \(70^{\circ} \mathrm{C}\) is flowing through a \(15-\mathrm{m}\) section of a cast iron pipe \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(4 \mathrm{~cm}\) and \(4.6 \mathrm{~cm}\), respectively. The outer surface of the pipe, whose emissivity is \(0.7\), is exposed to the cold air at \(10^{\circ} \mathrm{C}\) in the basement, with a heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat transfer coefficient at the inner surface of the pipe is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the walls of the basement to be at \(10^{\circ} \mathrm{C}\) also, determine the rate of heat loss from the hot water. Also, determine the average velocity of the water in the pipe if the temperature of the water drops by \(3^{\circ} \mathrm{C}\) as it passes through the basement.

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) \(88.6 \mathrm{~kW}\)

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum \((k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a \(10-\mathrm{cm}\)-thick layer of insulation \((k=0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The rate at which the liquid oxygen gains heat is (a) \(141 \mathrm{~W}\) (b) \(176 \mathrm{~W}\) (c) \(181 \mathrm{~W}\) (d) \(201 \mathrm{~W}\) (e) \(221 \mathrm{~W}\)

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.

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