Answer to Question #178302 in Mechanical Engineering for luke

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 k=0.046W/m-K and thickness 50mm and steel panels, each of thermal conductivity k=60W/m-K and thickness 3mm. If the wall separates refrigerated air at 4C from ambient air at 25C, what is the heat gain per surface area? Convection heat transfer at both surfaces may be approximated as 5W/m2-K (answer in W/m2)

Answer 1

The walls of the refrigerator are exposed to refrigerated air on one side and ambient air on the other side. The mode of heat transfer between the refrigerator wall and air are,

• Conduction through the walls of the refrigerator.

• Convection between refrigerated air and layer of fluid adjacent to the refrigerator walls.

• Convection between ambient air and layer of fluid adjacent to the refrigerator walls.

The thermal resistance of the composite wall by considering convection and conduction can be calculated by using:

$R_{th}=\frac{1}{h_o}+\frac{L_p}{k_p}+\frac{L_i}{k_i}+\frac{L_p}{k_p}+\frac{1}{h_i}$

$R_{th}=\frac{1}{5}+\frac{3 \times 0.001}{60}+\frac{50 \times0.001}{0.046}+\frac{3 \times0.001}{60}+\frac{1}{5}=1.487 m^2K/W$

$Q=\frac{T_0-T_i}{R_{th}}=\frac{25-4}{1.487}=14.12 W/m^2$