PROBLEM 4.10KNOWN:Temperature, diameter and burial depth of an insulated pipe. FIND:Heat loss per unit length of pipe. SCHEMATIC:ASSUMPTIONS:(1) Steady-state conditions, (2) One-dimensional conduction through insulation, two-dimensional through soil, (3) Constant properties, (4) Negligible oil onvection and pipe wall conduction resistances. cPROPERTIES:Table A-3, Soil (300K): k = 0.52 W/m⋅K; Table A-3, Cellular glass (365K): = 0.069 W/m⋅K. kANALYSIS:The heat rate can be expressed as 12totTTqR−=where the thermal resistance is Rtot= Rins+ Rsoil. From Equation 3.28, ()( )21insinsn D / Dn 0.7m/0.5m0.776m K/WR.2 Lk2 L 0.069 W/m KLππ⋅===×⋅AAFrom Equation 4.21 and Table 4.1, -1-12soilsoilsoilcosh2z/Dcosh3/ 0.710.653RmSk20.52 W/m K LL==⋅K/W.Hence, 120 0CW
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This note was uploaded on 12/07/2010 for the course MAE Heat Trans taught by Professor Lee,j.s. during the Spring '10 term at Seoul National.