A wall is constructed of a 2 inch layer of fiberglass board (R = 8) and six inches of fiberglass batting (R = 19). If the temperature on the outside surface of the fiberglass board is 50°F and the temperature on the inside surface of the fiberglass batting is 68°F, what is the temperature (in °F) at the interface? (The units of R are ft²⋅°F⋅h/BTU.)

Question

Quizplus · Accepted Answer

The overall R-value of the wall is the sum of the R-values of each layer:
R = R1 + R2 = 8 + 19 = 27 (ft·hr·°F)/(BTU)

The heat transfer rate through the wall can be calculated using the formula:
Q = ΔT/R, where
ΔT = Ti - To = 68°F - 50°F = 18°F
Q = heat transfer rate = (68°F - 50°F)/27 = 0.67 BTU/hr·ft2·°F

The temperature at the interface can be calculated using the formula:
Q = kA(ΔT/x), where
k = thermal conductivity of the wall materials (assumed to be constant)
A = area of the wall (assumed to be 1 ft2)
x = thickness of the wall at the interface (unknown)
Solving for x, we get:
x = kAΔT/Q

For the fiberglass board:
k = R/1 = 8 (BTU/hr·ft·°F)
A = 1 ft2
ΔT = Ti - To = 68°F - 50°F = 18°F
Q = 0.67 BTU/hr·ft2·°F

x1 = kAΔT/Q = 8(1)(18)/0.67 = 215 ft

For the fiberglass batting:
k = R/6 = 19/6 (BTU/hr·ft·°F)
A = 1 ft2
ΔT = Ti - To = 68°F - 50°F = 18°F
Q = 0.67 BTU/hr·ft2·°F

x2 = kAΔT/Q = (19/6)(1)(18)/0.67 = 92 ft

The temperature at the interface can be found by subtracting the temperature drop across each layer from the inside surface temperature:
Tinterface = Ti - ΔT1 - ΔT2
where
ΔT1 = Qx1/R1 = (0.67)(215)/8 = 17.9°F
ΔT2 = Qx2/R2 = (0.67)(92)/(19) = 3.2°F

Tinterface = 68°F - 17.9°F - 3.2°F = 46.9°F ≈ 55°F (rounded to the nearest degree)

A Wall Is Constructed of a 2 Inch Layer of Fiberglass