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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

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Question # 1 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 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 m Refrigerated air Ambient air Tooi = 4°C h; = 5 W/m2-K To,o ho = 5 W/m2-K = 25°C T,i To,o Panel (2) k, = 60 W/m-K Insulation q" 1/h, Lp/Kp L/ki Lp/kp 1/ho k; = 0.046 W/m-K
Transcribed Image Text:Question # 1 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 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 m Refrigerated air Ambient air Tooi = 4°C h; = 5 W/m2-K To,o ho = 5 W/m2-K = 25°C T,i To,o Panel (2) k, = 60 W/m-K Insulation q" 1/h, Lp/Kp L/ki Lp/kp 1/ho k; = 0.046 W/m-K
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