The ceiling of a tract house is constructed of wooden studs with fiberglass insulation between them. On the interior of the ceiling is plaster and on the exterior is a thin layer of sheet metal. A cross section of the ceiling with dimensions is shown below. (a) The R-factor describes the thermal resistance of insulation and is defined by: R-factor = L/keff = ?T/(q/A) Calculate the R-factor for this type of ceiling and compare the value of this R-factor with that for a similar thickness of fiberglass. Why are the two different? (b) Estimate the rate of heat transfer per square meter through the ceiling if the interior temperature is 22°C and the exterior temperature is –5°C.

Question

GIVEN



FIND



ASSUMPTIONS

- Steady state heat transfer

- One dimensional conduction through the ceiling

- Thermal resistance of the sheet metal is negligible

SKETCH



PROPERTIES AND CONSTANTS

Pine or fir wood studs (kw) = 0.15 W/(m K) at 20°C

Fiberglass (kfg) = 0.035 W/(m K) at 20°C

Plaster (kp) = 0.814 W/(m K) at 20°C

Answer 1

The thermal circuit for the ceiling with studs is shown below



where Rp = thermal resistance of the plaster

Rw = thermal resistance of the wood

Rfg = thermal resistance of the fiberglass

Each of these resistances can be evaluated



To convert these all to a wall area basis the fraction of the total wall area taken by the wood studs and

the fiberglass must be calculated



Therefore the resistances of the studs and the fiberglass based on the wall area are



The R-Factor of the wall is related to the total thermal resistance of the wall by



For 120 mm. of fiberglass alone, the R-factor is



The R-factor of the ceiling is only 68% that of the same thickness of fiberglass. This is mainly due to

the fact that the wood studs act as a ‘thermal short’ conducting heat through the ceiling more quickly

than the surrounding fiberglass.

(b) The rate of heat transfer through the ceiling is