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Chapter 3: Problem 169

Determine the winter \(R\)-value and the \(U\)-factor of a masonry cavity wall that is built around 4-in-thick concrete blocks made of lightweight aggregate. The outside is finished with 4 -in face brick with \(\frac{1}{2}\)-in cement mortar between the bricks and concrete blocks. The inside finish consists of \(\frac{1}{2}\)-in gypsum wallboard separated from the concrete block by \(\frac{3}{4}\)-in-thick (1-in by 3 -in nominal) vertical furring whose center- to-center distance is 16 in. Neither side of the \(\frac{3}{4}\)-in-thick air space between the concrete block and the gypsum board is coated with any reflective film. When determining the \(R\)-value of the air space, the temperature difference across it can be taken to be \(30^{\circ} \mathrm{F}\) with a mean air temperature of \(50^{\circ} \mathrm{F}\). The air space constitutes 80 percent of the heat transmission area, while the vertical furring and similar structures constitute 20 percent.

Short Answer

Expert verified
Answer: To find the winter R-value and the U-factor of the masonry cavity wall, follow the steps in the provided solution. Step 1 involves calculating the R-value for each layer, Step 2 focuses on the air space and vertical furring, Step 3 calculates the equivalent R-value for the entire wall, and Step 4 provides the U-factor. Once you've completed these calculations, you'll have determined the winter R-value and U-factor for the masonry cavity wall.

Step by step solution

01

Calculate the R-value for each layer

The R-value is resistance to heat flow and is given by the thickness of the material divided by its thermal conductivity (\(k\) value). For each layer, find the R-value using the formula: \(R = \frac{thickness}{k}\) We use the standard thermal conductivity (\(k\) value) for each material: - Face brick: \(k = 5.0\) - Cement mortar: \(k = 5.9\) - Concrete blocks (lightweight aggregate): \(k = 3.0\) - Air space: We are given the temperature difference across it (\(30^{\circ} \mathrm{F}\)) and the mean air temperature (\(50^{\circ} \mathrm{F}\)). With these values, we can find the approximate \(R\)-value using an R-value table for an air space. Alternatively, if the air space were absent, these values would not affect the calculation. - Gypsum wallboard: \(k = 0.50\)
02

Calculate the R-value for the air space and vertical furring

Find the R-value for the air space and vertical furring as previously mentioned. From the R-value table, the air space R-value is found to be \(R_{air space} = 1.23\). For the vertical furring, calculate the R-value based on the given percentage (20%) as follows: \(R_{furring} = 0.2 * R_{gypsum~wallboard}\)
03

Calculate the equivalent R-value for the wall

Find the equivalent R-value for the whole wall by summing the R-values of the individual layers (brick, mortar, concrete blocks, air space, and gypsum wallboard) and the vertical furring: \(R_{total} = R_{brick} + R_{mortar} + R_{concrete~blocks} + R_{air space} + R_{gypsum~wallboard} + R_{furring}\) Remember to use the formula for R-value from Step 1 for each material.
04

Calculate the U-factor

Finally, calculate the U-factor (which is the inverse of the R-value) for the masonry cavity wall: \(U = \frac{1}{R_{total}}\) You have successfully determined the winter R-value and the U-factor of the masonry cavity 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 process by which heat moves from a warmer area to a cooler one. In buildings, this concept is key to understanding how efficiently a wall can insulate interior spaces from external temperatures. Heat can be transferred through three main methods:
  • Conduction: Transfer of heat through direct contact.
  • Convection: Transfer via fluid movement, such as air or water.
  • Radiation: Transfer through electromagnetic waves.
In the context of a masonry cavity wall, understanding heat transfer helps in assessing how different materials contribute to thermal resistance. Each layer in the wall has a unique role in either facilitating or resisting the flow of heat, thus affecting the overall energy efficiency of a building.
Masonry Cavity Wall
A masonry cavity wall is a typical structure used in construction for effective insulation. It comprises two layers, or "wythes," which often include an outer layer of brick and an inner layer of concrete blocks. The space between these layers is the cavity, which can be filled with insulating materials or left as an air gap. The cavity serves to:
  • Reduce the transfer of moisture.
  • Improve thermal resistance by breaking the heat transfer path.
  • Enhance the structural stability of the wall.
This design helps to regulate a building's temperature by minimizing heat loss in winter and heat gain in summer. In our exercise, the wall includes face brick, concrete blocks, and an air space, all contributing to the wall's overall thermal resistance.
Thermal Conductivity
Thermal conductivity is the measure of how well a material conducts heat. It is represented by the symbol "k." Materials with high thermal conductivity transfer heat easily, while those with low conductivity are more resistant to heat flow. In the exercise:
  • Face brick has a relatively high conductivity (\(k = 5.0\)).
  • Cement mortar further aids in conducting heat (\(k = 5.9\)).
  • Concrete blocks, being lightweight, offer some resistance (\(k = 3.0\)).
  • Gypsum wallboard is a good insulator (\(k = 0.50\)).
Calculating the R-value of each material involves dividing the thickness by its thermal conductivity. This helps determine how well each layer resists heat flow, providing different resistance levels to the final wall construction.
U-factor Calculation
The U-factor, or thermal transmittance, quantifies how well a building element conducts heat. It is the inverse of the R-value and is expressed as \(-\( U = \frac{1}{R_{total}}\)\). The U-factor provides a means to evaluate energy efficiency. A lower U-factor indicates better insulation properties, meaning the wall will retain heat more effectively in winter.
  • Sum the R-values of all wall components.
  • Determine the total R-value of the wall.
  • Calculate the U-factor using its formula.
For a masonry cavity wall, calculating the U-factor paints a clear picture of how well the wall insulates, ensuring the insulation needs are met according to climatic conditions and energy efficiency goals.

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Most popular questions from this chapter

The unit thermal resistances ( \(R\)-values) of both 40-mm and 90-mm vertical air spaces are given in Table 3-9 to be \(0.22 \mathrm{~m}^{2} \cdot \mathrm{C} / \mathrm{W}\), which implies that more than doubling the thickness of air space in a wall has no effect on heat transfer through the wall. Do you think this is a typing error? Explain.

A \(5-\mathrm{mm}\)-diameter spherical ball at \(50^{\circ} \mathrm{C}\) is covered by a \(1-\mathrm{mm}\)-thick plastic insulation \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The ball is exposed to a medium at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine if the plastic insulation on the ball will help or hurt heat transfer from the ball.

Consider a flat ceiling that is built around \(38-\mathrm{mm} \times\) \(90-\mathrm{mm}\) wood studs with a center-to-center distance of \(400 \mathrm{~mm}\). The lower part of the ceiling is finished with 13-mm gypsum wallboard, while the upper part consists of a wood subfloor \(\left(R=0.166 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), a \(13-\mathrm{mm}\) plywood, a layer of felt \(\left(R=0.011 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), and linoleum \(\left(R=0.009 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\). Both sides of the ceiling are exposed to still air. The air space constitutes 82 percent of the heat transmission area, while the studs and headers constitute 18 percent. Determine the winter \(R\)-value and the \(U\)-factor of the ceiling assuming the 90 -mm-wide air space between the studs ( \(a\) ) does not have any reflective surface, (b) has a reflective surface with \(\varepsilon=0.05\) on one side, and ( ) has reflective surfaces with \(\varepsilon=0.05\) on both sides. Assume a mean temperature of \(10^{\circ} \mathrm{C}\) and a temperature difference of \(5.6^{\circ} \mathrm{C}\) for the air space.

A 3-cm-long, 2-mm \(\times 2-\mathrm{mm}\) rectangular crosssection aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the fin efficiency is 65 percent, the effectiveness of this single fin is (a) 39 (b) 30 (c) 24 (d) \(18 \quad(e) 7\)

Steam at \(450^{\circ} \mathrm{F}\) is flowing through a steel pipe \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) whose inner and outer diameters are \(3.5\) in and \(4.0\) in, respectively, in an environment at \(55^{\circ} \mathrm{F}\). The pipe is insulated with 2 -in-thick fiberglass insulation \((k=\) \(\left.0.020 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\). If the heat transfer coefficients on the inside and the outside of the pipe are 30 and \(5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), respectively, determine the rate of heat loss from the steam per foot length of the pipe. What is the error involved in neglecting the thermal resistance of the steel pipe in calculations?
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