HomeworkO H.W. 1:The walls of a refrigerator are typically constructed by sandwiching a layer ofinsulation between sheet metal panels. Consider a wall made from fiberglassinsulation of thermal conductivity k; = 0.046 W/m.K and thickness L, = 50 mmand steel panels, each of thermal conductivity k, = 60 W/m.K and thickness L,= 3 mm. If the wall separates refrigerated air at T = 4 C from ambient air at T,.= 25 C, what is the heat gain per unit surface area? Coefficients associatedwith natural convection at the inner and outer surfaces may be approximatedas h, = h, = 5 W/m?.K.%3D%3DL; = 0.050 m KK Lo = 0.003 mRefrigeratedairAmbient airToo.i = 4°Chi = 5 W/m2-KTo,o = 25°Cho = 5 W/m2-K%3DPanel (2)kp = 60 W/m-KInsulationk; = 0.046 W/m-K22Warith Alanbiyaa (Dr. ALI M)Heat TransferPage 11 of

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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 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 refrigerated air at Ť = 4 C from ambient air at T, 25 C, what is the heat gain per unit surface area? Coefficients associated with natural convection at the inner and outer surfaces may be approximated as h; = h. = 5 W/m?.K. %3D %3D Lj = 0.050 m- Lp = 0.003 m Refrigerated air Ambient air Too,i = 4°C h; = 5 W/m2-K To,o ho = 5 W/m2-K = 25°C %3D Panel (2) Insulation k; = 0.046 W/m-K q" ko = 60 W/m-K

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, = 0.046 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 refrigerated air at Ť = 4 C from ambient air at T, 25 C, what is the heat gain per unit surface area? Coefficients associated with natural convection at the inner and outer surfaces may be approximated as h; = h. = 5 W/m?.K. %3D %3D Lj = 0.050 m- Lp = 0.003 m Refrigerated air Ambient air Too,i = 4°C h; = 5 W/m2-K To,o ho = 5 W/m2-K = 25°C %3D Panel (2) Insulation k; = 0.046 W/m-K q" ko = 60 W/m-K

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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