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The unit thermal resistances ( \(R\)-values) of both \(40-\mathrm{mm}\) and \(90-\mathrm{mm}\) vertical airspaces 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 airspace in a wall has no effect on heat transfer through the wall. Do you think this is a typing error? Explain. 3-171C What is a radiant barrier? What kinds of materials are suitable for use as radiant barriers? Is it worthwhile to use radiant barriers in the attics of homes?

Short Answer

Expert verified
Explain the factors affecting insulation and the role of radiant barriers in homes. Answer: No, there is not necessarily a typing error regarding the fact that doubling the thickness of airspace does not affect heat transfer. Insulation is affected by factors like conductivity, composition, orientation, temperature, etc., and not solely by thickness. Radiant barriers are reflective surfaces that reduce heat transfer by radiation, and using them in the attic of homes can help reduce energy consumption for heating and cooling, resulting in lower energy bills.

Step by step solution

01

Analyze the given information about unit thermal resistances

Unit thermal resistance (R-value) represents how effective a material is at preventing heat transfer. The higher the R-value, the better the insulation. The given information states that the R-values of both 40mm and 90mm vertical airspaces are 0.22 \(m^2 \cdot C / W\). This suggests that even if the thickness is more than doubled, there is no change in the insulation property of the airspace. #Step 2: Identify if there is a typing error in the text#
02

Identify if there is a typing error in the text

The statement that the insulation property remains the same despite increasing the thickness by more than double may look suspicious. However, this is not necessarily a typing error. The reason is that insulation is affected not only by the thickness of the material but also by factors like conductivity, composition, orientation, temperature, etc. So, if the other factors remain constant, doubling the thickness will not necessarily double the R-value. #Step 3: Define a radiant barrier#
03

Define a radiant barrier

A radiant barrier is a reflective surface that reduces the heat transfer by radiation. It works by reflecting radiant heat instead of suppressing conduction or convection heat transfer like conventional insulation. #Step 4: Discuss suitable materials for radiant barriers#
04

Discuss suitable materials for radiant barriers

Generally, materials that have a low emissivity and high reflectivity are suitable for radiant barriers. Common materials used include aluminum, copper, and other metal foils, which reflect thermal radiation effectively. #Step 5: Discuss the worthiness of using radiant barriers in attics of homes#
05

Discuss the worthiness of using radiant barriers in attics of homes

It is worthwhile to use radiant barriers in the attics of homes because it can help to reduce heat gain in the summer and retain heat during winter. This helps to keep the attic cooler in the summer and warmer in the winter, resulting in reduced energy consumption for heating and cooling, and consequently lower energy bills.

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Most popular questions from this chapter

A pipe is insulated to reduce the heat loss from it. However, measurements indicate that the rate of heat loss has increased instead of decreasing. Can the measurements be right?

Two very long, slender rods of the same diameter and length are given. One rod (Rod 1) is made of aluminum and has a thermal conductivity $k_{1}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$, but the thermal conductivity of Rod \(2, k_{2}\), is not known. To determine the thermal conductivity of Rod 2 , both rods at one end are thermally attached to a metal surface which is maintained at a constant temperature \(T_{b}\). Both rods are losing heat by convection, with a convection heat transfer coefficient \(h\) into the ambient air at \(T_{\infty}\). The surface temperature of each rod is measured at various distances from the hot base surface. The measurements reveal that the temperature of the aluminum rod (Rod 1) at \(x_{1}=40 \mathrm{~cm}\) from the base is the same as that of the rod of unknown thermal conductivity (Rod 2) at \(x_{2}=20 \mathrm{~cm}\) from the base. Determine the thermal conductivity \(k_{2}\) of the second rod \((\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\).

A \(2.5\)-m-high, 4-m-wide, and 20 -cm-thick wall of a house has a thermal resistance of \(0.025^{\circ} \mathrm{C} / \mathrm{W}\). The thermal conductivity of the wall is (a) \(0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (b) \(1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (c) \(3.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(5.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(8.0 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

Heat is lost at a rate of \(275 \mathrm{~W}\) per \(\mathrm{m}^{2}\) area of a 15 -cm-thick wall with a thermal conductivity of $k=1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The temperature drop across the wall is (a) \(37.5^{\circ} \mathrm{C}\) (b) \(27.5^{\circ} \mathrm{C}\) (c) \(16.0^{\circ} \mathrm{C}\) (d) \(8.0^{\circ} \mathrm{C}\) (e) \(4.0^{\circ} \mathrm{C}\)

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.

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