Answered

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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

Sagot :

batolisis

Answer:

14.12 w /m^2

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

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