Question

The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between sheet metal panels. Consider a wall made from fiberglass insulation of thermal conductivity ki = 0.046 W/m K and thickness Li = 50 mm and steel panels, each of thermal conductivity kp = 60 W/m K and thickness Lp = 3 mm. If the wall separates refrigerated air at TX, i = 4 degree C from ambient air at T infinity, 0 = 25 degree C, what is the heat gain per unit surface area? Coefficients associated with natural convection at the inner and outer surfaces may be approximated as hi = h0 = 5 W/m2 K.

Answer 1

q" = 14.122 W/m²

Step-by-step explanation:

We are given;

Thermal conductivity of Fibre glass; k_i = 0.046 W/m.K

thickness of Fibre glass; L_i = 50 mm = 0.05 m

Thermal conductivity of steel panels; k_p = 60 W/m.K

Thickness of steel panels; L_p = 3 mm = 0.003 m

refrigerated air temperature; T∞_i = 4° C

Ambient air temperature; T∞_o = 25° C

h_i = h_0 = 5 W/m².K

Formula to calculate the heat gain per unit surface area in this case is;

q" = (T∞_o - T∞_I)/((1/h_o) + (L_p/k_p) + (L_i/k_i) + (L_p/k_p) + (1/h_i)

Where; ((1/h_o) + (L_p/k_p) + (L_i/k_i) + (L_p/k_p) + (1/h_i) is thermal resistance (R_th)

Plugging in the relevant values;

q" = (25 - 4)/((1/5) + (0.003/60) + (0.05/0.046) + (0.003/60) + (1/5))

q" = 14.122 W/m²