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no total, 21 =° °\u2212=\u2212= \u221e R TT Q& (a) The rate of heat transfer after insulation is W9.9966615.015.0 ins noins =×== QQ && The total thermal resistance with the foam insulation is C) W.m/42.0( C/W 02253.0 )m C)(16.8 W/m.025.0( C/W 02253.0 4 2 4 conv,2foam321total °+°= °+°= ++++= L L RRRRRR The thickness of insulation is determined from cm 5.4m 0.054 ==\u23af\u2192\u23af °+° °\u2212=\u23af\u2192\u23af\u2212= \u221e 4 4total 21 ins C) W.m/42.0( C/W 02253.0 C)823( W9.99 L LR TT Q& The outer surface temperature of the wall is determined from C8.3°=\u23af\u2192\u23af° °\u2212=\u23af\u2192\u23af\u2212= \u221e 22 conv 22 ins C/W 00350.0 C)8( W9.99 T T R TT Q& PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-121 (b) The total thermal resistance with the fiberglass insulation is C) W.m/6048.0( C/W 02253.0 )m C)(16.8 W/m.036.0( C/W 02253.0 4 2 4 conv,2glassfiber 321total °+°=°+°= ++++= LL RRRRRR The thickness of insulation is determined from cm 7.7m 0.077 ==\u23af\u2192\u23af °+° °\u2212=\u23af\u2192\u23af\u2212= \u221e 4 4total 21 ins C W.m/6048.0( C/W 02253.0 C)823( W9.99 L LR TT Q& The outer surface temperature of the wall is determined from C8.3°=\u23af\u2192\u23af° °\u2212=\u23af\u2192\u23af\u2212= \u221e 22 conv 22 ins C/W00350.0 C)8( 9.99 T T R TT Q& Discussion The outer surface temperature is same for both cases since the rate of heat transfer does not change. 3-177 Cold conditioned air is flowing inside a duct of square cross-section. The maximum length of the duct for a specified temperature increase in the duct is to be determined. Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Steady one-dimensional heat conduction relations can be used due to small thickness of the duct wall. 5 When calculating the conduction thermal resistance of aluminum, the average of inner and outer surface areas will be used. Properties The thermal conductivity of aluminum is given to be 237 W/m\u22c5°C. The specific heat of air at the given temperature is cp = 1006 J/kg\u22c5°C (Table A-15). Analysis The inner and the outer surface areas of the duct per unit length and the individual thermal resistances are 2 22 2 11 m 0.1m) 1(m) 25.0(44 m 88.0m) 1(m) 22.0(44 === === LaA LaA Ri Ralum T\u221e1 T\u221e2 Ro [ ] C/W 09214.007692.000007.001515.0 C/W07692.0 )m 0.1(C). W/m13( 11 C/W 00007.0 m 2/)188.0(C) W/m.237( m 015.0 C/W01515.0 )m 88.0(C). W/m75( 11 oalumitotal 22 2 o 2alum 22 1 i °=++=++= °=°== °=+°== °=°== RRRR Ah R kA LR Ah R The rate of heat loss from the air inside the duct is W228 C/W09214.0 C)1233( total 12 =° °\u2212=\u2212= \u221e\u221e R TT Q& For a temperature rise of 1°C, the air inside the duct should gain heat at a rate of W805C)1)(CJ/kg. kg/s)(1006 8.0(total =°°=\u394= TcmQ p&& Then the maximum length of the duct becomes m 3.53=== W228 W805total Q Q L & & PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-122 3-178 Heat transfer through a window is considered. The percent error involved in the calculation of heat gain through the window assuming the window consist of glass only is to be determined. Assumptions 1 Heat transfer is steady. 2 Heat transfer is one-dimensional. 3 Thermal conductivities are constant. 4 Radiation is accounted for in heat transfer coefficients. Ri Rglass T\u221e1 T\u221e2 Ro Properties The thermal conductivities are given to be 0.7 W/m\u22c5°C for glass and 0.12 W/m\u22c5°C for pine wood. Ri Rwood T\u221e1 T\u221e2 Ro Analysis The surface areas of the glass and the wood and the individual thermal resistances are 2wood 2 glass m 45.0m) 2(m) 5.1(15.0 m 55.2m) 2(m) 5.1(85.0 ==== AA C/W 41433.117094.092593.031746.0 C/W 08787.003017.000168.005602.0 C/W17094.0 )m 45.0(C). W/m13( 11 C/W03017.0 )m 55.2(C). W/m13( 11 C/W 92593.0 )m 45.0(C) W/m.12.0( m 05.0 C/W 00168.0 )m 55.2(C) W/m.7.0( m 003.0 C/W31746.0 )m 45.0(C). W/m7( 11 C/W05602.0 )m 55.2(C). W/m7( 11 woodo,woodwoodi,woodtotal, glasso,glassglassi,glass total, 22 wood2 woodo, 22 glass2 glasso, 2 woodwood wood wood 2 glassglass glass glass 22 wood1 woodi, 22 glass1 glassi, °=++=++= °=++=++= °=°== °=°== °=°== °=°== °=°== °=°== RRRR RRRR Ah R Ah R Ak L R Ak L R Ah R Ah R The rate of heat gain through the glass and the wood and their total are W4.1933.111.182 W3.11 C/W41433.1 C)2440( W 1.182 C/W08787.0 C)2440( woodglasstotal woodtotal, 12 wood glasstotal, 12 glass =+=+= =° °\u2212=\u2212==° °\u2212=\u2212= \u221e\u221e\u221e\u221e QQQ R TT Q R TT Q &&& && If the window consists of glass only the heat gain through the window is 2 glass m 0.3m) 2(m) 5.1( ==A C/W 07469.002564.000143.004762.0 C/W02564.0 )m 0.3(C). W/m13( 11 C/W 00143.0 )m 0.3(C) W/m.7.0( m 003.0 C/W04762.0 )m 0.3(C). W/m7( 11 glasso,glassglassi,glass total, 22 glass2 glasso, 2 glassglass glass glass 22 glass1 glassi, °=++=++= °=°== °=°== °=°== RRRR Ah R Ak L R Ah R W2.214 C/W07469.0 C)2440( glasstotal, 12 glass =° °\u2212=\u2212= \u221e\u221e R TT Q& Then the percentage error involved in heat gain through the window assuming the window consist of glass only becomes 10.8%=×\u2212=\u2212= 100 4.193 4.1932.214Error % with wood with woodonly glass Q QQ & && PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 3-123 3-179 Steam is flowing inside a steel pipe. The thickness of the insulation needed to reduce the heat loss by 95 percent and the thickness of the insulation needed to reduce outer surface temperature to 40°C are to be determined. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is one-dimensional since there is thermal symmetry about the centerline and no variation in the axial direction. 3 Thermal conductivities are constant. 4 The thermal contact resistance at the interface is negligible. Properties The thermal conductivities are given to be k = 61 W/m\u22c5°C for steel and k = 0.038 W/m\u22c5°C for insulation. Analysis (a) Considering a unit length of the pipe, the inner and the outer surface areas of the pipe and the insulation are 2 3333 2 2 2 1 m 1416.3m) 1( m 3770.0m) 1(m) 12.0( m 3142.0m) 1(m) 10.0( DDLDA LDA LDA o i === === === \u3c0\u3c0 \u3c0\u3c0 \u3c0\u3c0 Ri T\u221e1 R2 Ro T\u221e2 R1 The individual thermal resistances are C/W 02274.0 0.23876 )12.0/ln( 03079.0 02274.0 0.23876 )12.0/ln( 00048.003031.0 C/W 22026.018947.000048.003031.0 C/W 02274.0 )m 1416.3(C). W/m14( 11 C/W 18947.0 )m 3770.0(C). W/m14( 11 C/W 0.23876 )12.0/ln( )m 1(C) W/m.038.0(2 )12.0/ln( 2 )/ln( C/W 00048.0 )m 1(C) W/m.61(2 )5/6ln( 2 )/ln( C/W 03031.0 )m 3142.0(C). W/m105( 11 3 3 3 3 insulationo,21insulation total, steelo,1insulation no total, 3 2 3 2insulationo, 22steelo, 33 2 23 insulation2 1 12 pipe1 22 °++= +++=+++= °=++=++= °=°== °=°== °=°=== °=°=== °=°== D D D D RRRRR RRRR DDAh R Ah R DD Lk rr RR Lk rr RR Ah R i i oo oo ii i \u3c0\u3c0 \u3c0\u3c0 Then the steady rate of heat loss from the steam per meter pipe length for the case of no insulation becomes W976.1 C/W 0.22026 C)20235(21 =° °\u2212=\u2212= \u221e\u221e totalR TT Q& The thickness of the insulation needed in order to save 95 percent of this heat loss can be determined from C/W 02274.0 0.23876 )12.0/ln( 03079.0 C)20235( W)1.97605.0( 3 3insulationtotal, 21 insulation °\u239f\u239f\u23a0 \u239e \u239c\u239c\u239d \u239b ++ °\u2212=×\u23af\u2192\u23af\u2212= \u221e\u221e D DR TT Q& whose solution is cm 10.78===\u23af\u2192\u23af= 2 12-33.55 2 - thicknessm 3355.0 233 DD D PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution