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 and thickness Li 50 mm and steel panels, each of thermal conductivity kp 60 and thickness Lp 3 mm. If the wall separates refrigerated air at T,i 4 C from ambient air at T,o 25 C, what is the heat transfer rate per unit surface area

Answer 1

14.12 w /m^2

Step-by-step explanation:

Determine the heat transfer rate per unit surface area

First step : determine the thermal resistance of the composite wall considering all present forms of heat transfer

Rth = 1 / h₀ + Lp / Kp + Li / Ki + Lp / Kp + 1 / hi --------- ( 1 )

where : h₀ and hi = 5 W/m^2 , Li = 50 mm , Ki = 0.046 W/m.k, Lp = 3 mm,

Kp = 60 W/m.k

Insert values into equation 1 above

Rth = 1.487 m^2. K/W

Next : determine the heat gain per unit surface area ( heat transfer rate )

Q = ( To - Ti ) / Rth

= 21° C / 1.487

= 14.12 W/m^2