4_12_16 - PROBLEM 4.14 KNOWN: Heat generation in a buried...

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PROBLEM 4.14 KNOWN: Heat generation in a buried spherical container. FIND: (a) Outer surface temperature of the container, (b) Representative isotherms and heat flow lines. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Soil is a homogeneous medium with constant properties. PROPERTIES: Table A-3 , Soil (300K): k = 0.52 W/m K. ANALYSIS: (a) From an energy balance on the container, g qE = & and from the first entry in Table 4.1, ( ) 12 2D q k TT. l D/4z π = Hence, ( ) q 1 D/4z 500W 1 2m/40m T T 20 C+ 92. 7C W k 2 D 2 2m 0.52 mK ππ −− = + == oo < (b) The isotherms may be viewed as spherical surfaces whose center moves downward with increasing radius. The surface of the soil is an isotherm for which the center is at z = .
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PROBLEM 4.15 KNOWN: 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 convection and pipe wall conduction resistances. PROPERTIES: Table A-3 , Soil (300K): k = 0.52 W/m K; Table A-3 , Cellular glass (365K): k = 0.069 W/m K. ANALYSIS: The heat rate can be expressed as 12 tot TT q R = where the thermal resistance is R tot = R ins + R soil . From Eq. 3.28, ( ) ( ) 21 nD / D n 0.7m/0.5m 0.776m K/W R. 2 Lk 2 L 0.069 W/m KL ππ = == ×⋅ ll From Eq. 4.27 and Table 4.1, ( ) ( ) ( ) -1 -1 2 soil soil soil cosh 2z/D cosh 3/0.7 1 0.653 R m K/W. SK 2 0.52 W/mK LL = = = = Hence, ( ) ( ) 12 00C W q 8 4L 1 mK m 0.776 0.653 LW = = × + o q q/L 84 W/m. ʹ < COMMENTS: (1) Contributions of the soil and insulation to the total resistance are approximately the same. The heat loss may be reduced by burying the pipe deeper or adding more insulation. (2) The convection resistance associated with the oil flow through the pipe may be significant, in which case the foregoing result would overestimate the heat loss. A calculation of this resistance may be based on results presented in Chapter 8. (3) Since z > 3D/2, the shape factor for the soil can also be evaluated from S = 2 π L/ n l (4z/D) of Table 4.1, and an equivalent result is obtained.
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PROBLEM 4.16 KNOWN: Operating conditions of a buried superconducting cable.
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This note was uploaded on 01/24/2012 for the course MECHE ALL taught by Professor Any during the Spring '11 term at Carnegie Mellon.

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4_12_16 - PROBLEM 4.14 KNOWN: Heat generation in a buried...

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