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Determine the winter \(R\)-value and the \(U\)-factor of a masonry cavity wall that consists of \(100-\mathrm{mm}\) common bricks, a \(90-\mathrm{mm}\) airspace, \(100-\mathrm{mm}\) concrete blocks made of lightweight aggregate, 20 -mm airspace, and \(13-\mathrm{mm}\) gypsum wallboard separated from the concrete block by \(20-\mathrm{mm}\)-thick (1-in \(\times 3\)-in nominal) vertical furring whose center-to-center distance is \(400 \mathrm{~mm}\). Neither side of the two airspaces is coated with any reflective films. When determining the \(R\)-value of the airspaces, the temperature difference across them can be taken to be \(16.7^{\circ} \mathrm{C}\) with a mean air temperature of $10^{\circ} \mathrm{C}$. The airspace constitutes 84 percent of the heat transmission area, while the vertical furring and similar structures constitute 16 percent.

Short Answer

Expert verified
Answer: The winter R-value for the masonry cavity wall is approximately 5.16 m²K/W, and the U-factor is approximately 0.19 W/m²K.

Step by step solution

01

Determine the R-values of individual layers

To calculate R-values for individual layers, we need to know their thermal conductivity (k) and thickness (d) values. The following values are provided in the problem description: 1. Common bricks: 100mm thickness, k = 0.69 W/mK 2. Airspace (90mm): n/a 3. Concrete blocks (lightweight aggregate): 100mm thickness, k = 0.19 W/mK 4. Airspace (20mm): n/a 5. Gypsum wallboard: 13mm thickness, k = 0.17 W/mK 6. Furring: 20mm thickness, k = 0.8 W/mK To calculate their respective R-values, use the formula R = d/k for each material.
02

Determine the R-value of the airspaces

For the airspaces, we need to consider the temperature difference and mean air temperature provided in the problem description. Use the values given: ΔT = 16.7°C and mean air temperature Tm = 10°C. For the 90mm airspace, R-value = 1.64 + 0.079ΔT = 1.64 + 0.079(16.7) ≈ 2.96 m²K/W For the 20mm airspace, R-value = 1 + 0.077ΔT = 1 + 0.077(16.7) ≈ 2.29 m²K/W
03

Calculate the R-value for the assembly with vertical furring

Since we have both the furring and the airspaces in parallel, we need to calculate the R-value for these two components in parallel. We are given that the airspace constitutes 84% of the heat transmission area, and the furring constitutes 16%. R-value for airspace and furring in parallel = 0.84(R-value of airspace) + 0.16(R-value of furring) = 0.84(2.29) + 0.16(20/0.8) ≈ 2.52 m²K/W
04

Calculate the overall R-value for the entire cavity wall

Now, we can calculate the overall R-value for the entire cavity wall by adding the R-values of each layer: R_total = R_bricks + R_90mm_airspace + R_concrete_blocks + R_20mm_airspace_furring + R_gypsum_wallboard R_total = (100/0.69) + 2.96 + (100/0.19) + 2.52 + (13/0.17) ≈ 5.16 m²K/W
05

Calculate the U-factor for the cavity wall

The U-factor is the inverse of the total R-value. So, to calculate the U-factor, simply take the reciprocal of the total R-value: U-factor = 1/R_total ≈ 1/5.16 ≈ 0.19 W/m²K The winter R-value for the masonry cavity wall is approximately 5.16 m²K/W, and the U-factor is approximately 0.19 W/m²K.

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Most popular questions from this chapter

Two plate fins of constant rectangular cross section are identical, except that the thickness of one of them is twice the thickness of the other. For which fin is the \((a)\) fin effectiveness and (b) fin efficiency higher? Explain.

A plane brick wall \((k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is $10 \mathrm{~cm}$ thick. The thermal resistance of this wall per unit of wall area is (a) \(0.143 \mathrm{~m}^{2}, \mathrm{~K} / \mathrm{W}\) (b) \(0.250 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (c) \(0.327 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (d) \(0.448 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) (e) \(0.524 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\)

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 \(0.5\)-cm-thick aluminum \((k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a 10 -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) \(221 \mathrm{~W}\)

Explain how the fins enhance heat transfer from a surface. Also, explain how the addition of fins may actually decrease heat transfer from a surface.

Consider a \(1.5-\mathrm{m}\)-high and 2 -m-wide triple pane window. The thickness of each glass layer $(k=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( is \)0.5 \mathrm{~cm}$, and the thickness of each airspace \((k=0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(1.2 \mathrm{~cm}\). If the inner and outer surface temperatures of the window are $10^{\circ} \mathrm{C}\( and \)0^{\circ} \mathrm{C}$, respectively, the rate of heat loss through the window is (a) \(3.4 \mathrm{~W}\) (b) \(10.2 \mathrm{~W}\) (c) \(30.7 \mathrm{~W}\) (d) \(61.7 \mathrm{~W}\) (e) \(86.8 \mathrm{~W}\)

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