Heat Chap03-120

Heat Chap03-120 - Chapter 3 Steady Heat Conduction Heat...

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Chapter 3 Steady Heat Conduction Heat Transfer In Common Configurations 3-120C Under steady conditions, the rate of heat transfer between two surfaces is expressed as ( ) Q Sk T T = - 1 2 where S is the conduction shape factor. It is related to the thermal resistance by S=1/ (kR) . 3-121C It provides an easy way of calculating the steady rate of heat transfer between two isothermal surfaces in common configurations. 3-122 The hot water pipe of a district heating system is buried in the soil. The rate of heat loss from the pipe is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the soil is constant. Properties The thermal conductivity of the soil is given to be k = 0.9 W/m °C. Analysis Since z>1.5D, the shape factor for this configuration is given in Table 3-5 to be m 07 . 34 )] m 08 . 0 /( ) m 8 . 0 ( 4 ln[ ) m 20 ( 2 ) / 4 ln( 2 = = = π D z L S Then the steady rate of heat transfer from the pipe becomes W 1686 = ° - = - = C ) 5 60 )( C W/m. 9 . 0 )( m 07 . 34 ( ) ( o 2 1 T T Sk Q 3-84 60 ° C L = 20 m D = 8 5 ° C 80 cm
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Chapter 3 Steady Heat Conduction 3-123 "!PROBLEM 3-123" "GIVEN" L=20 "[m]" D=0.08 "[m]" "z=0.80 [m], parameter to be varied" T_1=60 "[C]" T_2=5 "[C]" k=0.9 "[W/m-C]" "ANALYSIS" S=(2*pi*L)/ln(4*z/D) Q_dot=S*k*(T_1-T_2) z [m] Q [W] 0.2 2701 0.38 2113 0.56 1867 0.74 1723 0.92 1625 1.1 1552 1.28 1496 1.46 1450 1.64 1412 1.82 1379 2 1351 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 1200 1400 1600 1800 2000 2200 2400 2600 2800 z [m] Q [W] 3-85
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Chapter 3 Steady Heat Conduction 3-124 Hot and cold water pipes run parallel to each other in a thick concrete layer. The rate of heat transfer between the pipes is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat transfer is two-dimensional (no change in the axial direction). 3 Thermal conductivity of the concrete is constant. Properties The thermal conductivity of concrete is given to be k = 0.75 W/m °C. Analysis The shape factor for this configuration is given in Table 3-5 to be m 078 . 9 ) m 05 . 0 )( m 05 . 0 ( 2 ) m 05 . 0 ( ) m 05 . 0 ( ) m 4 . 0 ( 4 cosh ) m 8 ( 2 2 4 cosh 2 2 2 2 1 2 1 2 2 2 1 2 1 = - - π = - - π = - - D D D D z L S Then the steady rate of heat transfer between the pipes becomes W 306 = ° - ° = - = C ) 15 60 )( C W/m. 75 . 0 )( m 078 . 9 ( ) ( 2 1 T T Sk Q 3-86 T 2 = 15 ° C D = 5 cm T 1 = 60 ° C L = 8 m z = 40 cm
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Chapter 3 Steady Heat Conduction 3-125 "!PROBLEM 3-125" "GIVEN" L=8 "[m]" D_1=0.05 "[m]" D_2=D_1 "z=0.40 [m], parameter to be varied" T_1=60 "[C]" T_2=15 "[C]" k=0.75 "[W/m-C]" "ANALYSIS" S=(2*pi*L)/(arccosh((4*z^2-D_1^2-D_2^2)/(2*D_1*D_2))) Q_dot=S*k*(T_1-T_2) z [m] Q [W] 0.1 644.1 0.2 411.1 0.3 342.3 0.4 306.4 0.5 283.4 0.6 267 0.7 254.7 0.8 244.8 0.9 236.8 1 230 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 200 250 300 350 400 450 500 550 600 650 z [m] Q [W] 3-87
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Chapter 3
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Heat Chap03-120 - Chapter 3 Steady Heat Conduction Heat...

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