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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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Most popular questions from this chapter

A 3-m-diameter spherical tank containing some radioactive material is buried in the ground \((k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The distance between the top surface of the tank and the ground surface is \(4 \mathrm{~m}\). If the surface temperatures of the tank and the ground are \(140^{\circ} \mathrm{C}\) and \(15^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer from the tank.

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.

A \(2.2\)-mm-diameter and 10-m-long electric wire is tightly wrapped with a \(1-m m\)-thick plastic cover whose thermal conductivity is \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Electrical measurements indicate that a current of \(13 \mathrm{~A}\) passes through the wire and there is a voltage drop of \(8 \mathrm{~V}\) along the wire. If the insulated wire is exposed to a medium at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

A hot plane surface at \(100^{\circ} \mathrm{C}\) is exposed to air at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat loss from the surface is to be reduced by half by covering it with sufficient insulation with a thermal conductivity of \(0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming the heat transfer coefficient to remain constant, the required thickness of insulation is (a) \(0.1 \mathrm{~cm}\) (b) \(0.5 \mathrm{~cm}\) (c) \(1.0 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(5 \mathrm{~cm}\)

A turbine blade made of a metal alloy \((k=\) \(17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) has a length of \(5.3 \mathrm{~cm}\), a perimeter of \(11 \mathrm{~cm}\), and a cross-sectional area of \(5.13 \mathrm{~cm}^{2}\). The turbine blade is exposed to hot gas from the combustion chamber at \(973^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(538 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The base of the turbine blade maintains a constant temperature of \(450^{\circ} \mathrm{C}\) and the tip is adiabatic. Determine the heat transfer rate to the turbine blade and temperature at the tip.
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