11.12 In order to increase the thermal resistance of a typical exterior frame wall, such as the one shown in Example 11.11, it is customary to use 2 x 6 studs instead of 2 x 4 studs to allow for placement of more insulation within the wall cavity. A typical exterior (2 x 6) frame wall of a house consists of the materials shown in the accompanying figure. Assume an inside room temperature of 68 °F and an outside air temperature of 20 °F with an exposed area of 150 ft². Determine the heat loss through this wall. The figure shows a section of an exterior frame wall. The frame wall consists of an exterior sliding wood attached to sheathing. The exterior sliding wood is coated with an outside film. On the other end of the wall is a gypsum wall board coated with an inside film. The insulation batt is attached between the exterior and interior ends of the wall. The outside and inside films are marked 1 and 6 respectively. The sliding wood is marked 2 and the sheathing is marked 3. The insulation batt is marked 4. The gypsum wallboard is marked 5.
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The thermal resistance (R) is calculated by dividing the thickness of the material (d) by its thermal conductivity (k). However, the problem does not provide the thickness or thermal conductivity of the materials used in the wall. Therefore, we cannot calculate Show more…
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EXAMPLE 11.9 Energy Transfer Through a Concrete Wall Goal Apply the equation of heat conduction. Problem Find the energy transferred in 1.00 h by conduction through a concrete wall 2.0 m high, 3.65 m long, and 0.20 m thick if one side of the wall is held at 20°C and the other side is at 5°C. Strategy 𝒫 = kA(Th - Tc)/L gives the rate of energy transfer by conduction in joules per second. Multiply by the time and substitute given values to find the total thermal energy transferred. SOLUTION Multiply the energy transfer equation by Δt to find an expression for the total energy Q transferred through the wall. Q = 𝒫Δt = kA(Th - Tc /L)Δt Substitute the numerical values to obtain Q, consulting the table for k. Q = (1.3 J/s · m °C)(7.3 m²)(15°C/0.20 m)(3600 s) = 2.6 × 10¹ J LEARN MORE Remarks Early houses were insulated with thick masonry walls, which restrict energy loss by conduction because k is relatively low. The large thickness L also decreases energy loss by conduction, as shown by energy transfer equation. There are much better insulating materials, however, and layering is also helpful. Despite the low thermal conductivity of masonry, the amount of energy lost is still rather large, enough to raise the temperature of 600 kg of water by more than 1°C. Question True or False: Materials having high thermal conductivities provide better insulation than materials having low thermal conductivities. False. A good insulator does not transfer energy by heat readily through it, and has low thermal conductivity. True. The rate of heat flow in a material is proportional to its thermal conductivity. True. Thermal conductivity measures a material's usefulness as heat insulation. False. A good insulator transfers energy by heat effectively, and has low thermal conductivity. PRACTICE IT Use the worked example above to help you solve this problem. Find the energy transferred in 1 h by conduction through a concrete wall 2.3 m high, 3.20 m long, and 0.20 m thick if one side of the wall is held at 20°C and the other side is at 5°C. 1391040 Your response differs from the correct answer by more than 10%. Double check your calculations. EXERCISE HINTS: GETTING STARTED | I'M STUCK! A wooden shelter has walls constructed of wooden planks 1.00 cm thick. If the exterior temperature is -20.0°C and the interior is 5.00°C, find the rate of energy loss through a wall that has dimensions 1.95 m by 2.30 m. 2466 Do not follow the example too closely. What question is asked in the example? What question is asked in this exercise? How are these related?
Keerti J.
A furnace wall consists of 120 mm thick refractory brick and 120 mm thick insulating fire brick separated by an air gap. The outer wall of the furnace, after the insulating fire brick, is covered with 12 mm thick plaster. The inner surface of the furnace wall (refractory brick layer) is at 1090°C, and the room temperature is 20°C. The heat transfer coefficient from the outside wall surface to the air in the room is 18 W/m^2°C, and the resistance to heat flow of the air gap is 0.16 K/W. If the thermal conductivities of the refractory brick, insulating fire brick, and plaster are 1.6 W/mK, 0.3 W/mK, and 0.14 W/mK, respectively. Sketch the thermal resistance network. Calculate: 1. Rate at which heat is lost per m^2 of the wall surface 2. Each interface temperature 3. Temperature of the outside surface of the wall.
Shaiju T.
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