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Determine the winter \(R\)-value and the \(U\)-factor of a masonry wall that consists of the following layers: \(100-\mathrm{mm}\) face bricks, 100 -mm common bricks, \(25-\mathrm{mm}\) urethane rigid foam insulation, and \(13-\mathrm{mm}\) gypsum wallboard.

Short Answer

Expert verified
Question: Calculate the winter R-value and U-factor for a masonry wall with the following layers and thicknesses: 100mm face bricks, 100mm common bricks, 25mm urethane rigid foam, and 13mm gypsum wallboard. Assume constant material conductivities. Answer: To determine the winter R-value and U-factor of the masonry wall, we must first find the R-values of each layer and then calculate the total R-value, followed by finding the U-factor. Applying the formula R = d/k, and the given thicknesses and corresponding conductivity values, we get the R-values for each layer. By adding these values, we obtain the total R-value for the wall. Finally, calculate the U-factor by taking the inverse of the total R-value (U = 1/R_total). Note that these values are approximations and may vary depending on external factors.

Step by step solution

01

Find the \(R\)-values for each layer of the wall

To find the \(R\)-values for each layer of the wall, we can use the following formula: $$R = \frac{d}{k}$$ Where \(R\) is the thermal resistance, \(d\) is the thickness of the material in meters, and \(k\) is the material's conductivity in \(W/(m\cdot K)\). We will need to find the appropriate \(k\) values for each material on our wall using a reference or source for material conductivity values. The thicknesses are given in the problem statement, but we will need to convert them from millimeters to meters. $$ d_{face\, bricks} =0.1\,\mathrm{m} \\ d_{common\, bricks} =0.1\,\mathrm{m} \\ d_{urethane\, rigid\, foam} =0.025\,\mathrm{m} \\ d_{gypsum\, wallboard} =0.013\,\mathrm{m} $$ To find the \(R\)-value for each layer, we can use the formula given above: $$R_{face\, bricks} = \frac{d_{face\, bricks}}{k_{face\, bricks}}$$ $$R_{common\, bricks} = \frac{d_{common\, bricks}}{k_{common\, bricks}}$$ $$R_{urethane\, rigid\, foam} = \frac{d_{urethane\, rigid\, foam}}{k_{urethane\, rigid\, foam}}$$ $$R_{gypsum\, wallboard} = \frac{d_{gypsum\, wallboard}}{k_{gypsum\, wallboard}}$$
02

Calculate the total \(R\)-value for the entire wall

To obtain the total thermal resistance of the wall, we can simply add the values we found for each layer: $$R_{total} = R_{face\, bricks} + R_{common\, bricks} + R_{urethane\, rigid\, foam} + R_{gypsum\, wallboard}$$
03

Determine the \(U\)-factor of the wall

The \(U\)-factor, or thermal transmittance, is the inverse of the total \(R\)-value. So, to calculate the \(U\)-factor for the masonry wall, simply take the reciprocal of the total \(R\)-value: $$U = \frac{1}{R_{total}}$$ Now you have determined both the winter \(R\)-value and the \(U\)-factor for the masonry wall. Keep in mind that these values are approximate and may vary depending on factors like material quality, temperature variations and other external conditions.

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Key Concepts

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

U-factor calculation
The U-factor, or thermal transmittance, measures how well a wall or surface can conduct heat. In simpler terms, it assesses how much heat is lost through a building component. It's the inverse of the R-value, which represents thermal resistance. A lower U-factor means better insulation, as less heat escapes, which helps in keeping buildings warm in winter.

To calculate the U-factor, you first need to know the total R-value, which is the sum of thermal resistances of all layers in a wall. Once you have the total R-value, you can easily find the U-factor using the formula:
  • \[ U = \frac{1}{R_{total}} \] Executing this calculation provides you the U-factor, reflecting how well the overall wall setup insulates against heat loss.
This concept is crucial in building construction and design as it plays a significant role in energy efficiency assessments.
Masonry wall insulation
In the context of masonry walls, insulation is all about maintaining the right balance between keeping the inside of a building warm during cold times and preventing excessive heat during warm periods. A typical masonry wall might consist of different layers, like bricks and gypsum, including a specific layer of insulation material such as urethane foam.

Each layer contributes differently to the wall's overall ability to resist heat flow. In a masonry wall like the one described in the exercise, materials are chosen for their specific properties:
  • Common and face bricks provide structural support and moderate insulation on their own.
  • Urethane rigid foam offers a significant boost in insulation because of its high R-value per thickness ratio.
  • Gypsum wallboard is typically used as interior finish and also contributes slightly to insulation.
The combination of these layers ensures that the building envelope effectively reduces heat transfer, aiding in both energy efficiency and comfort regulation inside the building.
Thermal conductivity
Thermal conductivity is a material-specific property that describes how well a substance can transfer heat. It's represented by the symbol \(k\) and is measured in watts per meter-Kelvin \(W/(m \cdot K)\). A high thermal conductivity implies that the material can conduct heat efficiently, while a low value suggests the opposite; it's a poor conductor and, therefore, a good insulator.

When analyzing a building material like those in a masonry wall, determining the thermal conductivity helps in understanding how each material will contribute to overall insulation. For instance:
  • Materials like urethane foam have low thermal conductivity, which makes them excellent insulators.
  • Traditional building materials like bricks have moderate to high thermal conductivity, meaning they allow more heat transfer.
Selecting materials with appropriate thermal conductivities can improve a wall's R-value, thereby enhancing energy efficiency. It is essential to correctly measure these properties to create a building envelope that effectively manages heat flow, ensuring occupant comfort while minimizing energy use.

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Most popular questions from this chapter

Steam at \(320^{\circ} \mathrm{C}\) flows in a stainless steel pipe \((k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) whose inner and outer diameters are \(5 \mathrm{~cm}\) and \(5.5 \mathrm{~cm}\), respectively. The pipe is covered with \(3-\mathrm{cm}\)-thick glass wool insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Heat is lost to the surroundings at \(5^{\circ} \mathrm{C}\) by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the heat transfer coefficient inside the pipe to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the steam per unit length of the pipe. Also determine the temperature drops across the pipe shell and the insulation.

Two flow passages with different cross-sectional shapes, one circular another square, are each centered in a square solid bar of the same dimension and thermal conductivity. Both configurations have the same length, \(T_{1}\), and \(T_{2}\). Determine which configuration has the higher rate of heat transfer through the square solid bar for \((a) a=1.2 b\) and \((b) a=2 b\).

A row of 3 -ft-long and 1-in-diameter used uranium fuel rods that are still radioactive are buried in the ground parallel to each other with a center-to- center distance of 8 in at a depth of \(15 \mathrm{ft}\) from the ground surface at a location where the thermal conductivity of the soil is \(0.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). If the surface temperature of the rods and the ground are \(350^{\circ} \mathrm{F}\) and \(60^{\circ} \mathrm{F}\), respectively, determine the rate of heat transfer from the fuel rods to the atmosphere through the soil.

An 8-m-internal-diameter spherical tank made of \(1.5\)-cm-thick stainless steel \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located in a room whose temperature is \(25^{\circ} \mathrm{C}\). The walls of the room are also at \(25^{\circ} \mathrm{C}\). The outer surface of the tank is black (emissivity \(\varepsilon=1\) ), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the tank are \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -h period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\).

What is a conduction shape factor? How is it related to the thermal resistance?

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