Question

1. In order to determine the size of a home-heating furnace, it is necessary to estimate the heat loss during the coldest day in winter. Provide the rates of heat loss per unit surface area for the following surfaces commonly encountered in house construction. You may assume an inside air temperature of 25°C, an outside temperature of -5°C, and heat transfer coefficients of 20 Wm-2K-1 and 5 Wm-2K-1 between air and the outside and inside surfaces, respectively. The thermal conductivities of glass and air are 1.4 and 0.026 Wm-1K-1, respectively.

a. A 3 mm-thick single-pane glass window.

b. A double-pane glass window consisting of two 3 mm-thick panels separated by a 6 mm-thick layer of stagnant air.

c. A composite wall consisting of a 10-cm-thick brick exterior (k = 1.0 Wm-1K-1), a 10-cm-thick layer of loosely packed glass fiber insulation (k = 0.043 Wm-1K-1), and an inside gypsum plaster wall (k = 0.17 Wm-1K-1) that is 1 cm thick.

d. Workmen installing a wall of the design described in part (c) are asking for a premium to install a vapor barrier on the plaster wall to prevent moisture from diffusing out of the house and condensing in the glass fiber insulation. If the dew point of the moist air in the house were 10°C would you pay the premium? Why?

Answer 1

a) A_surface/R_th = 1/h_in + L_glass/k_glass + 1/h_out

= 1/5 + 0.003/1.4 + 1/20 = 0.25214 m²K/W

Q/A= Heat flux = ΔT/( A_surface/R_th) = 30K / 0.25214 m²K/W = 118.98 W/m²

^b) A_surface/R_th = A_surface/R_th obtained in a +  L_glass/k_glass + L_air/k_air = 0.25214 + (0.003/1.4) + (0.006/0.026)

= 0.48505 m²K/W

Q/A= Heat flux = ΔT/( A_surface/R_th) = 30K / 0.48505 m²K/W = 61.849 W/m²

^c) A_surface/R_th = 1/h_in + Lbrick/k_brick + L_fiber/k_fiber + L_plaster/k_plaster + 1/h_out

= 1/5 + 0.1/1 + 0.1/0.043 + 0.01/0.17 + 1/20 = 12.6344 m²K/W

Q/A= Heat flux = ΔT/( A_surface/R_th) = 30K / 12.6344 m²K/W = 2.374 W/m²

d) Resistance across inside air and brick = (1/5 + 0.1/1) = 10.2

Thus, temperature drop across this= R_{till brick}/R_total x Total temp drop = 10.2/12.6344 x 30 =24.2195 C

Thus, temp at the fibre = 25 C - 24.2195 C = 0.7804 C < Dew point

Thus, moist air will condense at the fibre interface (provided it diffuses completely)

Hence, the premium is required