(III) Suppose the insulating qualities of the wall of a house come mainly from a 4.0-in. layer of brick and an R-19 layer of insulation, as shown in Fig. 14–19. What is the total rate of heat loss through T, such a wall, if its total area is 195 ft? and the temperature Brick Insulation (R¡) (R2) T1 Heat flow difference across it is 35 F°? FIGURE 14–19 Problem 45. Two layers insulating a wall.

There is a 1.20-cm-thick stagnant air pocket. A) What thickness of cork would have the same R-factor as the stagnant air pocket? The thermal conductivity of air is 0.0230 W/m·K and of cork is 0.0460 W/m·K.in cm  B) What thickness of tin would be required for the same R-factor as a 1.20-cm-thick stagnant air pocket? The thermal conductivity of air is 0.0230 W/m·K and of tin is 66.8 W/m·K . in m i asked how to do this but got the wrong soloution

1) A wall section composed from the outside to inside, of concrete brick 100mm thick, an air space of 50mm, two layers of fiberglass insulation 38mm thick, concrete block 150mm thick and air space of 19mm and a gypsum board of 13mm. The exterior conditions are -20C and RH=90% while the interior conditions are 23C and RH=40%. The surface temperature from the outside to inside are shown in the figure. The permeances are: Still Air: µ = 175 ng/pa.s.m Fiber glass: M=2560 ng/pa.s.m? Concrete block: M= 200 ng/pa.s.m? Gypsum board: M= 2000 ng/pa.s.m? Concrete brick: µ = 4.55 ng/pa.s.m Determine if there is a risk of condensation, and if there is, what is the condensation rate? What would happen if a vapor barrier (M=0.2 ng/pa.s.m²) was installed on the warm side of the insulation (Surface 4). 21с 19C 16 С 12C -10 -15C -18C -19C Gyp Air Concrete Fiber Fiber Air Concrete Outdoor Indoor sum block Glass Glass brick 230 19 50 mm -20C 13 150 mm 38 mm | 38 mm 100 mm RH=90% RH=40% mm mm Surf. 1 Surf.…

Calculate the heat loss through a 100-ft² wall with an inside temperature of 65°F and an outside temperature of 35°F. Assume the exterior wall is composed of 2- in. of material having a 'k' factor of 0.80, and 2-in. of insulation having a conductance of 0.16. RTotal = 8.75 & Q = 342-Btu/hr RTotal = 9.2 & Q = 399-Btu/hr RTotal = 8.75 & Q = 399-Btu/hr RTotal = 9.2 & Q = 342-Btu/hr Hide hint for Question 3 Utilize the (RTotal = 1/C + x1/k1) equation.