Question # 1The walls of a refrigerator are typically constructed by sand-wiching a layer of insulation between sheet metal panels.Consider a wall made from fiberglass insulation of thermalconductivity k; = 0.046 W/m · K and thickness L, = 50 mmand steel panels, each of thermal conductivity k,= 60W/m · K and thickness L, = 3 mm. If the wall separates re-frigerated air at Tai= 4°C_from ambient air at To= 25°C,what is the heat gain per unit surface area? Coefficients as-sociated with natural convection at the inner and outer sur-faces may be approximated as h; = h, = 5 W/m² · K.Hint:SCHEMATIC:L = 0.050 m –K* Lp = 0.003 mRefrigeratedairAmbient airTooi = 4°Ch; = 5 W/m2-KTo,oho = 5 W/m2-K= 25°CT,iTo,oPanel (2)k, = 60 W/m-KInsulationq"1/h, Lp/Kp L/ki Lp/kp 1/hok; = 0.046 W/m-K

Given: Inside temperature (T∞,i)=4°COutside temperature…

The walls of a refrigerator are typically constructed by sand- wiching a layer of insulation between sheet metal panels. Consider a wall made from fiberglass insulation of thermal conductivity k, = 0.046 W/m · K and thickness L, = 50 mm and steel panels, each of thermal conductivity k,= 60 W/m ·K and thickness L, = 3 mm. If the wall separates re- frigerated air at To,i what is the heat gain per unit surface area? Coefficients as- 4°C from ambient air at To= 25°C, sociated with natural convection at the inner and outer sur- faces may be approximated as h, = h, = 5 W/m² · K. Hint: SCHEMATIC: L = 0.050 m - e Lp = 0.003 m SRefrigerated air Ambient air Tooi = 4°C h = 5 W/m2-K To,o = 25°C ho = 5 W/m2-K To,i wowowawowo q" To,o Panel (2) kp = 60 W/m-K Insulation 1/h, Lp/kp L/k; Lp/kp 1/ho k; = 0.046 W/m-K

The walls of a refrigerator are typically constructed by sand- wiching a layer of insulation between sheet metal panels. Consider a wall made from fiberglass insulation of thermal conductivity k, = 0.046 W/m · K and thickness L, = 50 mm and steel panels, each of thermal conductivity k,= 60 W/m ·K and thickness L, = 3 mm. If the wall separates re- frigerated air at To,i what is the heat gain per unit surface area? Coefficients as- 4°C from ambient air at To= 25°C, sociated with natural convection at the inner and outer sur- faces may be approximated as h, = h, = 5 W/m² · K. Hint: SCHEMATIC: L = 0.050 m - e Lp = 0.003 m SRefrigerated air Ambient air Tooi = 4°C h = 5 W/m2-K To,o = 25°C ho = 5 W/m2-K To,i wowowawowo q" To,o Panel (2) kp = 60 W/m-K Insulation 1/h, Lp/kp L/k; Lp/kp 1/ho k; = 0.046 W/m-K

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter7: Forced Convection Inside Tubes And Ducts

Section: Chapter Questions

Problem 7.53P

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