SSL4 Steady Heat Conduction

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This lecture covers:
Steady Heat Conduction in Plane Walls
Thermal Resistance Concept
Thermal Resistance Network
Multilayer Plane Walls
Thermal Contact Resistance
Generalized Thermal Resistance Networks
Heat Conduction in Cylinders and Spheres
Multilayered Cylinders and Spheres
Critical Radius of Insulation

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  • SSL4 Steady Heat Conduction

    1. 1. STEADY HEAT CONDUCTION Lecture 4 Keith Vaugh BEng (AERO) MEngReference text: Chapter 17 - Fundamentals of Thermal-Fluid Sciences, 3rd Edition Yunus A. Cengel, Robert H. Turner, John M. Cimbala McGraw-Hill, 2008 KEITH VAUGH
    2. 2. OBJECTIVES } KEITH VAUGH
    3. 3. OBJECTIVESUnderstand the concept of thermal resistance and itslimitations, and develop thermal resistance networksfor practical heat conduction problems } KEITH VAUGH
    4. 4. OBJECTIVESUnderstand the concept of thermal resistance and itslimitations, and develop thermal resistance networksfor practical heat conduction problemsSolve steady conduction problems that involve }multilayer rectangular, cylindrical, or sphericalgeometries KEITH VAUGH
    5. 5. OBJECTIVESUnderstand the concept of thermal resistance and itslimitations, and develop thermal resistance networksfor practical heat conduction problemsSolve steady conduction problems that involve }multilayer rectangular, cylindrical, or sphericalgeometriesDevelop an intuitive understanding of thermal contactresistance, and circumstances under which it may besignificant KEITH VAUGH
    6. 6. OBJECTIVESUnderstand the concept of thermal resistance and itslimitations, and develop thermal resistance networksfor practical heat conduction problemsSolve steady conduction problems that involve }multilayer rectangular, cylindrical, or sphericalgeometriesDevelop an intuitive understanding of thermal contactresistance, and circumstances under which it may besignificantIdentify applications in which insulation may actuallyincrease heat transfer KEITH VAUGH
    7. 7. STEADY HEAT CONDUCTION IN PLANE WALLS Heat transfer through the wall of a house can be modelled as steady and one-dimensional. The temperature of the wall in this case depends on one direction only (say the x-direction) and can be expressed as T(x). }Heat transfer through a wallis one-dimensional when thetemperature of the wallvaries in one direction only. KEITH VAUGH
    8. 8. STEADY HEAT CONDUCTION IN PLANE WALLS Heat transfer through the wall of a house can be modelled as steady and one-dimensional. The temperature of the wall in this case depends on one direction only (say the x-direction) and can be expressed as T(x). } ⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of change ⎞ ⎜ heat transfer ⎟ − ⎜ heat transfer ⎟ = ⎜ of the energy ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ into the wall⎠ ⎝ out of the wall⎠ ⎝ of the wall ⎠Heat transfer through a wallis one-dimensional when thetemperature of the wallvaries in one direction only. KEITH VAUGH
    9. 9. STEADY HEAT CONDUCTION IN PLANE WALLS Heat transfer through the wall of a house can be modelled as steady and one-dimensional. The temperature of the wall in this case depends on one direction only (say the x-direction) and can be expressed as T(x). } ⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of change ⎞ ⎜ heat transfer ⎟ − ⎜ heat transfer ⎟ = ⎜ of the energy ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ into the wall⎠ ⎝ out of the wall⎠ ⎝ of the wall ⎠ & − Qout = dEwall Qin & dtHeat transfer through a wallis one-dimensional when thetemperature of the wallvaries in one direction only. KEITH VAUGH
    10. 10. STEADY HEAT CONDUCTION IN PLANE WALLS Heat transfer through the wall of a house can be modelled as steady and one-dimensional. The temperature of the wall in this case depends on one direction only (say the x-direction) and can be expressed as T(x). } ⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of change ⎞ ⎜ heat transfer ⎟ − ⎜ heat transfer ⎟ = ⎜ of the energy ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ into the wall⎠ ⎝ out of the wall⎠ ⎝ of the wall ⎠ & − Qout = dEwall Qin & dt In steady operation, the rate of heat transferHeat transfer through a wall through the wall is constant.is one-dimensional when thetemperature of the wallvaries in one direction only. & dT Fourier’s law of Qcond,wall = −kA (W) dx heat conduction KEITH VAUGH
    11. 11. KEITH VAUGH
    12. 12. Under steady conditions, thetemperature distribution in aplane wall is a straight line:dT/dx = const. KEITH VAUGH
    13. 13. & dT Qcond,wall = −kA dxUnder steady conditions, thetemperature distribution in aplane wall is a straight line:dT/dx = const. KEITH VAUGH
    14. 14. & dT Qcond,wall = −kA dx & L T2 ∫ Qcond,wall dx = − ∫ kA dT x=0 T =T1Under steady conditions, thetemperature distribution in aplane wall is a straight line:dT/dx = const. KEITH VAUGH
    15. 15. & dT Qcond,wall = −kA dx & L T2 ∫ Qcond,wall dx = − ∫ kA dT x=0 T =T1 & T1 − T2 Qcond,wall = kA (W) LUnder steady conditions, thetemperature distribution in aplane wall is a straight line:dT/dx = const. KEITH VAUGH
    16. 16. & dT Qcond,wall = −kA dx & L T2 ∫ Qcond,wall dx = − ∫ kA dT x=0 T =T1 & T1 − T2 Qcond,wall = kA (W) L The rate of heat conduction through a plane wall is proportional to the average thermal conductivity, the wall area, and the temperature difference, but is inversely proportional to the wall thickness.Under steady conditions, the Once the rate of heat conduction is available, thetemperature distribution in a temperature T(x) at any location x can be determinedplane wall is a straight line:dT/dx = const. by replacing T2 by T, and L by x. KEITH VAUGH
    17. 17. THERMAL RESISTANCECONCEPTAnalogy between thermal andelectrical resistance concepts. KEITH VAUGH
    18. 18. THERMAL RESISTANCECONCEPT & T −T Q= 1 2 R V1 − V2 I= ReAnalogy between thermal andelectrical resistance concepts. KEITH VAUGH
    19. 19. THERMAL RESISTANCECONCEPT & Q = kA T −T cond,wall 1 2 L & T −T Q= 1 2 R V1 − V2 I= ReAnalogy between thermal andelectrical resistance concepts. KEITH VAUGH
    20. 20. THERMAL RESISTANCECONCEPT & Q = kA T −T cond,wall 1 2 L & T −T & T1 − T2 Q= 1 2 Qcond,wall = (W) R Rwall V1 − V2 I= ReAnalogy between thermal andelectrical resistance concepts. KEITH VAUGH
    21. 21. THERMAL RESISTANCECONCEPT & Q = kA T −T cond,wall 1 2 L & T −T & T1 − T2 Q= 1 2 Qcond,wall = (W) R Rwall Conduction resistance of the wall: Thermal resistance of the wall against heat V1 − V2 conduction. I= Re L Rwall = kA (°C W ) Thermal resistance of a medium depends onAnalogy between thermal and the geometry and the thermal properties ofelectrical resistance concepts. the medium. KEITH VAUGH
    22. 22. THERMAL RESISTANCECONCEPT & Q = kA T −T cond,wall 1 2 L & T −T & T1 − T2 Q= 1 2 Qcond,wall = (W) R Rwall Conduction resistance of the wall: Thermal resistance of the wall against heat V1 − V2 conduction. I= Re L Rwall = kA (°C W ) Thermal resistance of a medium depends onAnalogy between thermal and the geometry and the thermal properties ofelectrical resistance concepts. the medium.rate of heat transfer → electric current V1 − V2 Re = L σ Athermal resistance → electrical resistance I= e Retemperature difference → voltage difference Electrical resistance KEITH VAUGH
    23. 23. KEITH VAUGH
    24. 24. & Newton’s lawQconv = hAs (Ts − T∞ ) of cooling Schematic for convection resistance at a surface. KEITH VAUGH
    25. 25. & Newton’s lawQconv = hAs (Ts − T∞ ) of cooling& = Ts − T∞Qconv (W) RconvConvection resistance of the surface:Thermal resistance of the surface against heatconvection. 1Rwall = hAs (°C W ) Schematic for convection resistance at a surface.When the convection heat transfer coefficient is very large (h → ∞), theconvection resistance becomes zero and Ts ≈ T.That is, the surface offers no resistance to convection, and thus it doesnot slow down the heat transfer process.This situation is approached in practice at surfaces where boiling andcondensation occur. KEITH VAUGH
    26. 26. KEITH VAUGH
    27. 27. Schematic for convectionand radiation resistances ata surface. KEITH VAUGH
    28. 28. & = εσ As Ts4 − Tsurr = hrad As (Ts − Tsurr ) = Ts − TsurrQrad ( 4 ) RradRadiation resistance of the surface:Thermal resistance of the surface against heat radiation. Schematic for convection and radiation resistances at a surface. KEITH VAUGH
    29. 29. & = εσ As Ts4 − Tsurr = hrad As (Ts − Tsurr ) = Ts − TsurrQrad ( 4 ) RradRadiation resistance of the surface:Thermal resistance of the surface against heat radiation. 1Rrad = hrad As (K W ) Schematic for convection and radiation resistances at a surface. KEITH VAUGH
    30. 30. & = εσ As Ts4 − Tsurr = hrad As (Ts − Tsurr ) = Ts − TsurrQrad ( 4 ) RradRadiation resistance of the surface:Thermal resistance of the surface against heat radiation. 1Rrad = hrad As ( K W ) & Qradhrad = As (Ts − Tsurr ) ( ) = εσ Ts2 + Tsurr (Ts + Tsurr ) 2 (W m ⋅ K ) 2 Note: K (Kelvins), for radiation analysis Ts and Tsurr must be in Kelvins Schematic for convection and radiation resistances at a surface. KEITH VAUGH
    31. 31. & = εσ As Ts4 − Tsurr = hrad As (Ts − Tsurr ) = Ts − TsurrQrad ( 4 ) RradRadiation resistance of the surface:Thermal resistance of the surface against heat radiation. 1Rrad = hrad As ( K W ) & Qradhrad = As (Ts − Tsurr ) ( ) = εσ Ts2 + Tsurr (Ts + Tsurr ) 2 (W m ⋅ K ) 2 Note: K (Kelvins), for radiation analysis Ts and Tsurr must be in Kelvins When Tsurr ≈ T∞ hcombined = hconv + hrad Combined heat transfer coefficient Schematic for convection and radiation resistances at a surface. KEITH VAUGH
    32. 32. STEADY ONE DIMENSIONALHEAT TRANSFER KEITH VAUGH
    33. 33. STEADY ONE DIMENSIONALHEAT TRANSFER⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of ⎞⎜ heat convection ⎟ = ⎜ heat conduction ⎟ = ⎜ heat convection ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ into the wall ⎠ ⎝ through the wall⎠ ⎝ from the wall ⎠ KEITH VAUGH
    34. 34. STEADY ONE DIMENSIONALHEAT TRANSFER⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of ⎞⎜ heat convection ⎟ = ⎜ heat conduction ⎟ = ⎜ heat convection ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ into the wall ⎠ ⎝ through the wall⎠ ⎝ from the wall ⎠⎛ Rate of ⎞⎜ heat convection ⎟ = h A (T − T ) = T∞1 − T1 = T∞1 − T1 1 ∞1 1⎜ ⎟ ⎛ 1 ⎞ Rconv,1⎝ into the wall ⎠ ⎜ h A ⎟ ⎝ 1 ⎠ KEITH VAUGH
    35. 35. STEADY ONE DIMENSIONALHEAT TRANSFER⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of ⎞⎜ heat convection ⎟ = ⎜ heat conduction ⎟ = ⎜ heat convection ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ into the wall ⎠ ⎝ through the wall⎠ ⎝ from the wall ⎠⎛ Rate of ⎞⎜ heat convection ⎟ = h A (T − T ) = T∞1 − T1 = T∞1 − T1 1 ∞1 1⎜ ⎟ ⎛ 1 ⎞ Rconv,1⎝ into the wall ⎠ ⎜ h A ⎟ ⎝ 1 ⎠ ⎛ Rate of ⎞ ⎜ heat conduction ⎟ = kA T1 − T2 = T1 − T2 = T1 − T2 ⎜ ⎟ L ⎛ L ⎞ Rwall ⎝ through the wall⎠ ⎜ ⎟ ⎝ kA ⎠ KEITH VAUGH
    36. 36. STEADY ONE DIMENSIONALHEAT TRANSFER⎛ Rate of ⎞ ⎛ Rate of ⎞ ⎛ Rate of ⎞⎜ heat convection ⎟ = ⎜ heat conduction ⎟ = ⎜ heat convection ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ into the wall ⎠ ⎝ through the wall⎠ ⎝ from the wall ⎠⎛ Rate of ⎞⎜ heat convection ⎟ = h A (T − T ) = T∞1 − T1 = T∞1 − T1 1 ∞1 1⎜ ⎟ ⎛ 1 ⎞ Rconv,1⎝ into the wall ⎠ ⎜ h A ⎟ ⎝ 1 ⎠ ⎛ Rate of ⎞ ⎜ heat conduction ⎟ = kA T1 − T2 = T1 − T2 = T1 − T2 ⎜ ⎟ L ⎛ L ⎞ Rwall ⎝ through the wall⎠ ⎜ ⎟ ⎝ kA ⎠ ⎛ Rate of ⎞ ⎜ heat convection ⎟ = h A (T − T ) = T2 − T∞2 = T2 − T∞2 2 2 ∞2 ⎜ ⎟ ⎛ 1 ⎞ Rconv,2 ⎝ from the wall ⎠ ⎜ h A ⎟ ⎝ 2 ⎠ KEITH VAUGH
    37. 37. KEITH VAUGH
    38. 38. KEITH VAUGH
    39. 39. KEITH VAUGH
    40. 40. Thermal resistance network KEITH VAUGH
    41. 41. Thermal resistance network KEITH VAUGH
    42. 42. &= T∞1 − T∞2 Q RtotalThermal resistance network KEITH VAUGH
    43. 43. &= T∞1 − T∞2 Q Rtotal Thermal resistance network 1 L 1Rtotal = Rconv,1 + Rwall + Rconv,2 = + + h1 A kA h2 A (°C W ) KEITH VAUGH
    44. 44. The temperature drop across a layer isproportional to its thermal resistance. KEITH VAUGH
    45. 45. The ratio of temperature drop to thermal resistanceacross the layer is proportional, therefore: & ΔT = QR (°C ) The temperature drop across a layer is proportional to its thermal resistance. KEITH VAUGH
    46. 46. The ratio of temperature drop to thermal resistanceacross the layer is proportional, therefore: & ΔT = QR (°C )Heat transfer can be expressed in a similar mannerto Newtons Law of cooling: & Q = UAΔT ( W)U overall heat transfer coefficient The temperature drop across a layer is proportional to its thermal resistance. KEITH VAUGH
    47. 47. The ratio of temperature drop to thermal resistanceacross the layer is proportional, therefore: & ΔT = QR (°C )Heat transfer can be expressed in a similar mannerto Newtons Law of cooling: & Q = UAΔT ( W)U overall heat transfer coefficientComparing Q &= T∞1 − T∞2 and Q = UAΔT , reveals & Rtotal 1 UA = Rtotal (°C K) The temperature drop across a layer is proportional to its thermal resistance. KEITH VAUGH
    48. 48. The ratio of temperature drop to thermal resistanceacross the layer is proportional, therefore: & ΔT = QR (°C )Heat transfer can be expressed in a similar mannerto Newtons Law of cooling: & Q = UAΔT ( W)U overall heat transfer coefficientComparing Q &= T∞1 − T∞2 and Q = UAΔT , reveals & Rtotal 1 UA = Rtotal (°C K) The temperature drop across a layer isOnce Q is evaluated, the surface temperature T1 proportional to its thermal resistance.can be determined from KEITH VAUGH
    49. 49. The ratio of temperature drop to thermal resistanceacross the layer is proportional, therefore: & ΔT = QR (°C )Heat transfer can be expressed in a similar mannerto Newtons Law of cooling: & Q = UAΔT ( W)U overall heat transfer coefficientComparing Q &= T∞1 − T∞2 and Q = UAΔT , reveals & Rtotal 1 UA = Rtotal (°C K ) The temperature drop across a layer isOnce Q is evaluated, the surface temperature T1 proportional to its thermal resistance.can be determined from &= T∞1 − T1 = T∞1 − T1 Q Rconv,1 ⎛ 1 ⎞ ⎜ h A ⎟ ⎝ 1 ⎠ KEITH VAUGH
    50. 50. MULTILAYER PLANE WALLS The thermal resistance network for heat transfer through a two-layer plane wall subjected to convection on both sides. KEITH VAUGH
    51. 51. MULTILAYER PLANE WALLS The thermal resistance network for heat transfer through a two-layer plane wall subjected to convection on both sides. &= T∞1 − T∞2 Q Rtotal KEITH VAUGH
    52. 52. MULTILAYER PLANE WALLS The thermal resistance network for heat transfer through a two-layer plane wall subjected to convection on both sides. &= T∞1 − T∞2 Q Rtotal 1 L1 L2 1Rtotal = Rconv,1 + Rwall,1 + Rwall,2 + Rconv,2 = + + + h1 A k1 A k2 A h2 A (°C W ) KEITH VAUGH
    53. 53. KEITH VAUGH
    54. 54. An unknown surface temperature Tj on any surface orinterface j can be calculate when T∞1 and T∞2 are givenand the rate heat transfer is known T − Tj where Ti is a known temperature at&= iQ location i and Rtotal, i-j is the total thermal Rtotal, i− j resistances between location i and j. KEITH VAUGH
    55. 55. An unknown surface temperature Tj on any surface orinterface j can be calculate when T∞1 and T∞2 are givenand the rate heat transfer is known T − Tj where Ti is a known temperature at&= iQ location i and Rtotal, i-j is the total thermal Rtotal, i− j resistances between location i and j.& T∞1 − T∞2 T∞1 − T∞2Q= = Rconv, 1 + Rwall, 1 ⎛ 1 ⎞ ⎛ L1 ⎞ ⎜ h A ⎟ + ⎜ k A ⎟ ⎝ 1 ⎠ ⎝ 1 ⎠ KEITH VAUGH
    56. 56. An unknown surface temperature Tj on any surface orinterface j can be calculate when T∞1 and T∞2 are givenand the rate heat transfer is known T − Tj where Ti is a known temperature at&= iQ location i and Rtotal, i-j is the total thermal Rtotal, i− j resistances between location i and j.& T∞1 − T∞2 T∞1 − T∞2Q= = Rconv, 1 + Rwall, 1 ⎛ 1 ⎞ ⎛ L1 ⎞ ⎜ h A ⎟ + ⎜ k A ⎟ ⎝ 1 ⎠ ⎝ 1 ⎠To find T1 : &= T∞1 − T1 Q Rconv, 1 & T∞1 − T2To find T2 : Q= Rconv, 1 + Rwall, 1To find T3 : &= T3 − T∞2 Q Rconv, 2 Refer to pages 659-663 in text book KEITH VAUGH
    57. 57. QUESTIONA double pane glass window 0.8 m high and 1.5 mwide consisting of two 4 mm thick layers of glasseach having a thermal conductivity k = 0.78W·m-1·K-1 and separated by a 10 mm wide stagnant }air space is used in a building. Determine the steadyrate of heat transfer through this window and thetemperature of its inner surface for a day duringwhich the room is maintained at 20°C and the outsidetemperature is -10°C. Assume the heat transfercoefficients on the inner and outer surfaces of thewindow to be h1 = 10 W/m2 °C and h2 = 40 W·m-2·°C-1, which includes the effects of radiation. KEITH VAUGH
    58. 58. Question marked approximately as:Identification of problem: 0-39% (39% band)Formulation of problem: 40-69% (29% band)Solution of problem: 70-100% (30% band) KEITH VAUGH
    59. 59. KEITH VAUGH
    60. 60. ASSUMPTIONSHeat transfer through the mediums is steadyHeat transfer is one dimensionalThermal conductivity is constant KEITH VAUGH
    61. 61. ASSUMPTIONSHeat transfer through the mediums is steadyHeat transfer is one dimensionalThermal conductivity is constant KEITH VAUGH
    62. 62. ASSUMPTIONS ANALYSISHeat transfer through the mediums is steady A double pane window is considered. The rate ofHeat transfer is one dimensional heat transfer through the window and the innerThermal conductivity is constant surface temperature are to be determined The diagram illustrates this problem. We do not know the values of T2, T3 therefore must employ the concept of thermal resistance, KEITH VAUGH
    63. 63. ASSUMPTIONS ANALYSISHeat transfer through the mediums is steady A double pane window is considered. The rate ofHeat transfer is one dimensional heat transfer through the window and the innerThermal conductivity is constant surface temperature are to be determined The diagram illustrates this problem. We do not know the values of T2, T3 therefore must employ the concept of thermal resistance, &= ΔTmediums L Q where, Rmediums = ΔRmediums kA The thermal resistance network comprises of 5 elements, Ri, Ro, representing the inner and outer elements, R1, R3, representing the inner and outer glass panes, and R2 represents the Air cavity. This problem comprises both conduction and convection. KEITH VAUGH
    64. 64. KEITH VAUGH
    65. 65. A = 0.8m × 1.5m = 1.2m 2 KEITH VAUGH
    66. 66. A = 0.8m × 1.5m = 1.2m 2 1 1Ri = Rconv,1 = = = 0.0833 °C W ( 2 )( h1 A 10W / m × °C 1.2m ) 2 KEITH VAUGH
    67. 67. A = 0.8m × 1.5m = 1.2m 2 1 1Ri = Rconv,1 = = = 0.0833 °C W ( 2 )( h1 A 10W / m × °C 1.2m 2 ) L1 0.004mR1 = R3 = Rglass = = = 0.00437 °C W k1 A ( )( ) 0.78W / m 2 × °C 1.2m 2 KEITH VAUGH
    68. 68. A = 0.8m × 1.5m = 1.2m 2 1 1Ri = Rconv,1 = = = 0.0833 °C W ( 2 )( h1 A 10W / m × °C 1.2m 2 ) L1 0.004mR1 = R3 = Rglass = = = 0.00437 °C W k1 A ( )( 0.78W / m 2 × °C 1.2m 2 ) L2 0.01mR2 = Rair = = = 0.3205 °C W k2 A ( 2 0.026W / m × °C 1.2m 2 )( ) KEITH VAUGH
    69. 69. A = 0.8m × 1.5m = 1.2m 2 1 1Ri = Rconv,1 = = = 0.0833 °C W ( 2 )( h1 A 10W / m × °C 1.2m 2 ) L1 0.004mR1 = R3 = Rglass = = = 0.00437 °C W k1 A ( )( 0.78W / m 2 × °C 1.2m 2 ) L2 0.01mR2 = Rair = = = 0.3205 °C W k2 A ( 2 0.026W / m × °C 1.2m 2 )( ) 1 1Ro = Rconv,2 = = = 0.02083 °C W h2 A ( )( 40W / m 2 × °C 1.2m 2 ) KEITH VAUGH
    70. 70. A = 0.8m × 1.5m = 1.2m 2 1 1Ri = Rconv,1 = = = 0.0833 °C W (2 h1 A 10W / m × °C 1.2m 2 )( ) L1 0.004mR1 = R3 = Rglass = = = 0.00437 °C W k1 A ( 0.78W / m 2 × °C 1.2m 2 )( ) L2 0.01mR2 = Rair = = = 0.3205 °C W k2 A ( 2 0.026W / m × °C 1.2m 2 )( ) 1 1Ro = Rconv,2 = = = 0.02083 °C W h2 A ( 40W / m 2 × °C 1.2m 2)( )RTotal = Rconv,1 + Rglass,1 + Rair + Rglass,2 + Rconv,2 KEITH VAUGH
    71. 71. A = 0.8m × 1.5m = 1.2m 2 1 1Ri = Rconv,1 = = = 0.0833 °C W (2 h1 A 10W / m × °C 1.2m 2 )( ) L1 0.004mR1 = R3 = Rglass = = = 0.00437 °C W k1 A ( 0.78W / m 2 × °C 1.2m 2 )( ) L2 0.01mR2 = Rair = = = 0.3205 °C W k2 A ( 2 0.026W / m × °C 1.2m 2 )( ) 1 1Ro = Rconv,2 = = = 0.02083 °C W h2 A ( 40W / m 2 × °C 1.2m 2)( )RTotal = Rconv,1 + Rglass,1 + Rair + Rglass,2 + Rconv,2RTotal = 0.0833 + 0.00437 + 0.3205 + 0.00437 + 0.02083 KEITH VAUGH
    72. 72. A = 0.8m × 1.5m = 1.2m 2 1 1Ri = Rconv,1 = = = 0.0833 °C W (2 h1 A 10W / m × °C 1.2m 2 )( ) L1 0.004mR1 = R3 = Rglass = = = 0.00437 °C W k1 A ( 0.78W / m 2 × °C 1.2m 2 )( ) L2 0.01mR2 = Rair = = = 0.3205 °C W k2 A ( 2 0.026W / m × °C 1.2m 2 )( ) 1 1Ro = Rconv,2 = = = 0.02083 °C W h2 A ( 40W / m 2 × °C 1.2m 2)( )RTotal = Rconv,1 + Rglass,1 + Rair + Rglass,2 + Rconv,2RTotal = 0.0833 + 0.00437 + 0.3205 + 0.00437 + 0.02083RTotal = 0.4332 °C W KEITH VAUGH
    73. 73. A = 0.8m × 1.5m = 1.2m 2 1 1Ri = Rconv,1 = = = 0.0833 °C W (2 h1 A 10W / m × °C 1.2m 2 )( ) L1 0.004mR1 = R3 = Rglass = = = 0.00437 °C W k1 A ( 0.78W / m 2 × °C 1.2m 2 )( ) L2 0.01mR2 = Rair = = = 0.3205 °C W k2 A ( 2 0.026W / m × °C 1.2m 2 )( ) 1 1Ro = Rconv,2 = = = 0.02083 °C W h2 A ( 40W / m 2 × °C 1.2m 2)( )RTotal = Rconv,1 + Rglass,1 + Rair + Rglass,2 + Rconv,2RTotal = 0.0833 + 0.00437 + 0.3205 + 0.00437 + 0.02083RTotal = 0.4332 °C WQ ⎣ 20 − ( −10 ) ⎤ °C = 69.2W&= T∞1 − T∞2 = ⎡ ⎦ RTotal 0.4332 °C W KEITH VAUGH
    74. 74. A = 0.8m × 1.5m = 1.2m 2 1 1Ri = Rconv,1 = = = 0.0833 °C W (2 h1 A 10W / m × °C 1.2m 2 )( ) L1 0.004mR1 = R3 = Rglass = = = 0.00437 °C W k1 A ( 0.78W / m 2 × °C 1.2m 2 )( ) L2 0.01mR2 = Rair = = = 0.3205 °C W k2 A ( 2 0.026W / m × °C 1.2m 2 )( ) 1 1Ro = Rconv,2 = = = 0.02083 °C W h2 A ( 40W / m 2 × °C 1.2m 2 )( )RTotal = Rconv,1 + Rglass,1 + Rair + Rglass,2 + Rconv,2RTotal = 0.0833 + 0.00437 + 0.3205 + 0.00437 + 0.02083RTotal = 0.4332 °C WQ ⎣ 20 − ( −10 ) ⎤ °C = 69.2W&= T∞1 − T∞2 = ⎡ ⎦ RTotal 0.4332 °C W & (T1 = T∞1 − QRconv,1 = 20 − ( 69.2W ) 0.0833 °C W = 14.2°C ) KEITH VAUGH
    75. 75. THERMAL CONTACT RESISTANCE } KEITH VAUGH
    76. 76. THERMAL CONTACT RESISTANCEWhen two such surfaces are pressed against each other,the peaks form good material contact but the valleys formvoids filled with air. } KEITH VAUGH
    77. 77. THERMAL CONTACT RESISTANCEWhen two such surfaces are pressed against each other,the peaks form good material contact but the valleys formvoids filled with air. }These numerous air gaps of varying sizes act asinsulation because of the low thermal conductivity of air. KEITH VAUGH
    78. 78. THERMAL CONTACT RESISTANCEWhen two such surfaces are pressed against each other,the peaks form good material contact but the valleys formvoids filled with air. }These numerous air gaps of varying sizes act asinsulation because of the low thermal conductivity of air.Thus, an interface offers some resistance to heat transfer,and this resistance per unit interface area is called thethermal contact resistance, Rc. KEITH VAUGH
    79. 79. Temperature distribution and heat flow lines alongtwo solid plates pressed against each other for thecase of perfect and imperfect contact. KEITH VAUGH
    80. 80. Temperature distribution and heat flow lines alongtwo solid plates pressed against each other for thecase of perfect and imperfect contact. KEITH VAUGH
    81. 81. The value of thermal contact resistance depends on: ✓ surface roughness, ✓ material properties, ✓ temperature and pressure at the interface ✓ type of fluid trapped at the interface. KEITH VAUGH
    82. 82. The value of thermal contact resistance depends on: ✓ surface roughness, ✓ material properties, ✓ temperature and pressure at the interface ✓ type of fluid trapped at the interface. & & & Q = Qcontact + Qgap KEITH VAUGH
    83. 83. The value of thermal contact resistance depends on: ✓ surface roughness, ✓ material properties, ✓ temperature and pressure at the interface ✓ type of fluid trapped at the interface. & & & Q = Qcontact + Qgap & Q = hc AΔTinterface hc thermal contact conductance KEITH VAUGH
    84. 84. The value of thermal contact resistance depends on: ✓ surface roughness, ✓ material properties, ✓ temperature and pressure at the interface ✓ type of fluid trapped at the interface. & & & Q = Qcontact + Qgap & Q = hc AΔTinterface hc thermal contact conductance & ⎛ Q ⎞ ⎜ A ⎟ ⎝ ⎠ hc = ΔTinterface (W m ⋅°C ) 2 KEITH VAUGH
    85. 85. The value of thermal contact resistance depends on: ✓ surface roughness, ✓ material properties, ✓ temperature and pressure at the interface ✓ type of fluid trapped at the interface. & & & Q = Qcontact + Qgap & Q = hc AΔTinterface hc thermal contact conductance & ⎛ Q ⎞ ⎜ A ⎟ ⎝ ⎠ hc = ΔTinterface (W m ⋅°C ) 2 1 ΔTinterface Rc = = hc & ⎛ Q ⎞ ( m ⋅ °C W ) 2 ⎜ A ⎟ ⎝ ⎠ KEITH VAUGH
    86. 86. The value of thermal contact resistance depends on: ✓ surface roughness, ✓ material properties, ✓ temperature and pressure at the interface ✓ type of fluid trapped at the interface. & & & Q = Qcontact + Qgap & Q = hc AΔTinterface hc thermal contact conductance & ⎛ Q ⎞ ⎜ A ⎟ ⎝ ⎠ hc = ΔTinterface (W m ⋅°C ) 2 1 ΔTinterface Rc = = hc & ⎛ Q ⎞ ( m ⋅ °C W ) 2 ⎜ A ⎟ ⎝ ⎠Thermal contact resistance is significant and can even dominatethe heat transfer for good heat conductors such as metals, but canbe disregarded for poor heat conductors such as insulations. KEITH VAUGH
    87. 87. GENERALISED THERMALRESISTANCE NETWORKS KEITH VAUGH
    88. 88. GENERALISED THERMALRESISTANCE NETWORKS &= Q1 + Q2 = T1 − T2 + T1 − T2 = (T1 − T2 ) ⎛ 1 + 1 ⎞ Q & & R1 R2 ⎜ R R ⎟ ⎝ 1 2 ⎠ &= T1 − T2 Q Rtotal 1 1 1 R1 R2 = + → Rtotal = Rtotal R1 R2 R1 + R2 KEITH VAUGH
    89. 89. KEITH VAUGH
    90. 90. Thermal resistance network forcombined series-parallel arrangement. KEITH VAUGH
    91. 91. &= T1 − T∞Q Rtotal R1 R2Rtotal = R12 + R3 + Rconv = + R3 + Rconv R1 + R2 L1 L2 L1 1R1 = R2 = R3 = R conv = k1 A1 k2 A2 k1 A1 hA3 Thermal resistance network for combined series-parallel arrangement. KEITH VAUGH
    92. 92. &= T1 − T∞Q Rtotal R1 R2Rtotal = R12 + R3 + Rconv = + R3 + Rconv R1 + R2 L1 L2 L1 1R1 = R2 = R3 = R conv = k1 A1 k2 A2 k1 A1 hA3Two assumptions in solving complex multidimensionalheat transfer problems by treating them as one-dimensional using the thermal resistance network are:1. Any plane wall normal to the x-axis is isothermal (i.e. to assume the temperature to vary in the x-direction only)2. any plane parallel to the x-axis is adiabatic (i.e. to Thermal resistance network for assume heat transfer to occur in the x-direction only) combined series-parallel arrangement. KEITH VAUGH
    93. 93. HEAT CONDUCTION INCYLINDERS & SPHERES } KEITH VAUGH
    94. 94. HEAT CONDUCTION INCYLINDERS & SPHERES Heat transfer through the pipe can be modelled as steady and one-dimensional. } KEITH VAUGH
    95. 95. HEAT CONDUCTION INCYLINDERS & SPHERES Heat transfer through the pipe can be modelled as steady and one-dimensional. The temperature of the pipe depends on one } direction only (the radial r-direction) and can be expressed as T = T(r). KEITH VAUGH
    96. 96. HEAT CONDUCTION INCYLINDERS & SPHERES Heat transfer through the pipe can be modelled as steady and one-dimensional. The temperature of the pipe depends on one } direction only (the radial r-direction) and can be expressed as T = T(r). The temperature is independent of the azimuthal angle or the axial distance. KEITH VAUGH
    97. 97. HEAT CONDUCTION INCYLINDERS & SPHERES Heat transfer through the pipe can be modelled as steady and one-dimensional. The temperature of the pipe depends on one } direction only (the radial r-direction) and can be expressed as T = T(r). The temperature is independent of the azimuthal angle or the axial distance. This situation is approximated in practice in long cylindrical pipes and spherical containers. KEITH VAUGH
    98. 98. A long cylindrical pipe (or sphericalshell) with specified inner and outersurface temperatures T1 and T2. KEITH VAUGH
    99. 99. & cyl = −kA dTQcond, ( W) dr A long cylindrical pipe (or spherical shell) with specified inner and outer surface temperatures T1 and T2. KEITH VAUGH
    100. 100. & cyl = −kA dTQcond, ( W) dr r2 & Qcond, cyl T2∫r=r1 A dr = − ∫T =T1 k dT A long cylindrical pipe (or spherical shell) with specified inner and outer surface temperatures T1 and T2. KEITH VAUGH
    101. 101. & cyl = −kA dTQcond, ( W) dr r2 & Qcond, cyl T2∫r=r1 A dr = − ∫T =T1 k dTA = 2π rL A long cylindrical pipe (or spherical shell) with specified inner and outer surface temperatures T1 and T2. KEITH VAUGH
    102. 102. & cyl = −kA dT Qcond, ( W) dr r2 & Qcond, cyl T2 ∫r=r1 A dr = − ∫T =T1 k dT A = 2π rL& cyl = 2π Lk T1 − T2Qcond, ( W) ⎛ r2 ⎞ ln ⎜ ⎟ ⎝ r1 ⎠ A long cylindrical pipe (or spherical shell) with specified inner and outer surface temperatures T1 and T2. KEITH VAUGH
    103. 103. & cyl = −kA dT Qcond, ( W) dr r2 & Qcond, cyl T2 ∫r=r1 A dr = − ∫T =T1 k dT A = 2π rL& cyl = 2π Lk T1 − T2Qcond, ( W) ⎛ r2 ⎞ ln ⎜ ⎟ ⎝ r1 ⎠ & cyl = T1 − T2Qcond, ( W) Rcyl A long cylindrical pipe (or spherical shell) with specified inner and outer surface temperatures T1 and T2. KEITH VAUGH
    104. 104. & cyl = −kA dT Qcond, ( W) dr r2 & Qcond, cyl T2 ∫r=r1 A dr = − ∫T =T1 k dT A = 2π rL & cyl = 2π Lk T1 − T2 Qcond, ( W) ⎛ r2 ⎞ ln ⎜ ⎟ ⎝ r1 ⎠ & cyl = T1 − T2 Qcond, ( W) Rcyl A long cylindrical pipe (or spherical shell) with specified inner and outer surface temperatures T1 and T2. ⎛ r2 ⎞ ⎛ outer radius ⎞ ln ⎜ ⎟ ln ⎜ ⎝ r1 ⎠ ⎝ inner radius ⎟ ⎠Rcyl = = 2π Lk 2π × Length × Thermal conductivity KEITH VAUGH
    105. 105. & cyl = −kA dT Qcond, ( W) dr r2 & Qcond, cyl T2 ∫r=r1 A dr = − ∫T =T1 k dT A = 2π rL & cyl = 2π Lk T1 − T2 Qcond, ( W) ⎛ r2 ⎞ ln ⎜ ⎟ ⎝ r1 ⎠ & cyl = T1 − T2 Qcond, ( W) Rcyl A long cylindrical pipe (or spherical shell) with specified inner and outer surface temperatures T1 and T2. ⎛ r2 ⎞ ⎛ outer radius ⎞ ln ⎜ ⎟ ln ⎜ ⎝ r1 ⎠ ⎝ inner radius ⎟ ⎠ Conduction resistance ofRcyl = = the cylinder layer 2π Lk 2π × Length × Thermal conductivity KEITH VAUGH
    106. 106. A spherical shell with specified inner andouter surface temperatures T1 and T2. KEITH VAUGH
    107. 107. A spherical shell with specified inner and outer surface temperatures T1 and T2.& sph = T1 − T2Qcond, Rsph KEITH VAUGH
    108. 108. A spherical shell with specified inner and outer surface temperatures T1 and T2. & sph = T1 − T2 Qcond, Rsph r2 − r1 outer radius - inner radiusRsph = = 2π r1r2 k 2π ( outer radius ) ( inner radius ) ( Thermal conductivity ) KEITH VAUGH
    109. 109. KEITH VAUGH
    110. 110. &= T∞1 − T∞2 Q RtotalThe thermal resistance network for a cylindrical(or spherical) shell subjected to convection fromboth the inner and the outer sides. KEITH VAUGH
    111. 111. &= T∞1 − T∞2 Q Rtotal for a cylindrical layer Rtotal = Rconv, 1 + Rcyl + Rconv, 2 ⎛ r2 ⎞ ln ⎜ ⎟ 1 ⎝ r1 ⎠ 1 = + + ( 2π r1L ) h1 2π Lk ( 2π r2 L ) h2The thermal resistance network for a cylindrical(or spherical) shell subjected to convection fromboth the inner and the outer sides. KEITH VAUGH
    112. 112. &= T∞1 − T∞2 Q Rtotal for a cylindrical layer Rtotal = Rconv, 1 + Rcyl + Rconv, 2 ⎛ r2 ⎞ ln ⎜ ⎟ 1 ⎝ r1 ⎠ 1 = + + ( 2π r1L ) h1 2π Lk ( 2π r2 L ) h2 for a spherical layerThe thermal resistance network for a cylindrical Rtotal = Rconv, 1 + Rsph + Rconv,2(or spherical) shell subjected to convection from 1 r −r 1both the inner and the outer sides. = + 2 1 + ( ) ( ) 4π r12 h1 4π r1r2 k 4π r22 h2 KEITH VAUGH
    113. 113. MULTILAYER CYLINDER &SPHERESThe thermal resistance network for heattransfer through a three-layered compositecylinder subjected to convection on bothsides. KEITH VAUGH
    114. 114. MULTILAYER CYLINDER &SPHERESThe thermal resistance network for heattransfer through a three-layered compositecylinder subjected to convection on bothsides. &= T∞1 − T∞2 Q Rtotal KEITH VAUGH
    115. 115. MULTILAYER CYLINDER &SPHERESThe thermal resistance network for heattransfer through a three-layered compositecylinder subjected to convection on bothsides. &= T∞1 − T∞2 Q Rtotal ⎛ r3 ⎞ ⎛ r4 ⎞ ⎛ r2 ⎞ ln ⎜ ⎟ ln ⎜ ⎟ ln ⎜ ⎟ 1 ⎝ r2 ⎠ ⎝ r3 ⎠ ⎝ r1 ⎠ 1 Rtotal = Rconv, 1 + Rcyl, 1 + Rcyl, 2 + Rcyl, 3 + Rconv, 2 = + + + + ( 2π r1L ) h1 2π Lk1 2π Lk2 2π Lk3 ( 2π r2 L ) h2 KEITH VAUGH
    116. 116. The ratio ΔT/R across any layer is equal to the heat transfer rate, which remains constant in one-dimensional steady conduction KEITH VAUGH
    117. 117. Once heat transfer rate has been calculated,any intermediate temperature Tj can bedetermined by: The ratio ΔT/R across any layer is equal to the heat transfer rate, which remains constant in one-dimensional steady conduction KEITH VAUGH
    118. 118. Once heat transfer rate has been calculated,any intermediate temperature Tj can bedetermined by: Ti − T j & Q= The ratio ΔT/R across any layer is equal to the Rtotal, i− j heat transfer rate, which remains constant inwhere Ti is a known temperature at location i one-dimensional steady conductionand Rtotal, i-j is the total thermal resistancesbetween location i and j. KEITH VAUGH
    119. 119. Once heat transfer rate has been calculated,any intermediate temperature Tj can bedetermined by: Ti − T j & Q= The ratio ΔT/R across any layer is equal to the Rtotal, i− j heat transfer rate, which remains constant inwhere Ti is a known temperature at location i one-dimensional steady conductionand Rtotal, i-j is the total thermal resistancesbetween location i and j. &= T∞1 − T1 = T∞1 − T2 = T1 − T3 = T2 − T3 = T2 − T∞2 = T3 − T∞2 Q Rconv, 1 Rconv, 1 + R1 R1 + R2 R2 R1 + Rconv, 2 Rconv, 2 KEITH VAUGH
    120. 120. Once heat transfer rate has been calculated,any intermediate temperature Tj can bedetermined by: Ti − T j & Q= The ratio ΔT/R across any layer is equal to the Rtotal, i− j heat transfer rate, which remains constant inwhere Ti is a known temperature at location i one-dimensional steady conductionand Rtotal, i-j is the total thermal resistancesbetween location i and j. &= T∞1 − T1 = T∞1 − T2 = T1 − T3 = T2 − T3 = T2 − T∞2 = T3 − T∞2 Q Rconv, 1 Rconv, 1 + R1 R1 + R2 R2 R1 + Rconv, 2 Rconv, 2For example, the interface temperature T2 betweenthe first and second layer can be determined from: & T∞1 − T2 T∞1 − T2 Q= = Rconv,1 + Rcyl,1 ⎛ r2 ⎞ ln ⎜ ⎟ 1 ⎝ r1 ⎠ + h1 ( 2π r1 L ) 2π Lk1 Refer to pages 674-678 in text book KEITH VAUGH
    121. 121. CRITICAL RADIUS OF INSULATION } KEITH VAUGH
    122. 122. CRITICAL RADIUS OF INSULATION Adding more insulation to a wall or to the attic always decreases heat transfer since the heat transfer area is constant, and adding insulation always increases the thermal resistance of the wall without increasing the convection resistance. } KEITH VAUGH
    123. 123. CRITICAL RADIUS OF INSULATION Adding more insulation to a wall or to the attic always decreases heat transfer since the heat transfer area is constant, and adding insulation always increases the thermal resistance of the wall without increasing the convection resistance. } In a a cylindrical pipe or a spherical shell, the additional insulation increases the conduction resistance of the insulation layer but decreases the convection resistance of the surface because of the increase in the outer surface area for convection. KEITH VAUGH
    124. 124. CRITICAL RADIUS OF INSULATION Adding more insulation to a wall or to the attic always decreases heat transfer since the heat transfer area is constant, and adding insulation always increases the thermal resistance of the wall without increasing the convection resistance. } In a a cylindrical pipe or a spherical shell, the additional insulation increases the conduction resistance of the insulation layer but decreases the convection resistance of the surface because of the increase in the outer surface area for convection. The heat transfer from the pipe may increase or decrease, depending on which effect dominates. KEITH VAUGH
    125. 125. An insulated cylindrical pipe exposed toconvection from the outer surface and thethermal resistance network associated with it. KEITH VAUGH
    126. 126. An insulated cylindrical pipe exposed toconvection from the outer surface and thethermal resistance network associated with it. &= T1 − T∞ = Q T1 − T∞ Rins + Rconv ⎛ r2 ⎞ ln ⎜ ⎟ ⎝ r1 ⎠ 1 + 2π Lk h ( 2π r2 L ) KEITH VAUGH
    127. 127. KEITH VAUGH
    128. 128. The critical radius of insulationfor a cylindrical body: k rcr,cylinder = (m) h The variation of heat transfer rate with the outer radius of the insulation r2 when r1 < rcr. KEITH VAUGH
    129. 129. The critical radius of insulationfor a cylindrical body: k rcr,cylinder = (m) hThe critical radius of insulation fora spherical shell: 2k rcr,sphere = (m) h The variation of heat transfer rate with the outer radius of the insulation r2 when r1 < rcr. KEITH VAUGH
    130. 130. The critical radius of insulationfor a cylindrical body: k rcr,cylinder = (m) hThe critical radius of insulation fora spherical shell: 2k rcr,sphere = (m) hThe largest value of the critical radius we arelikely to encounter is kmax, insulation 0.05 W m ⋅°C rcr,max = ≈ hmin 5W 2 m ⋅°C = 0.01 m = 1 cm The variation of heat transfer rate with the outer radius of the insulation r2 when r1 < rcr. KEITH VAUGH
    131. 131. The critical radius of insulationfor a cylindrical body: k rcr,cylinder = (m) hThe critical radius of insulation fora spherical shell: 2k rcr,sphere = (m) hThe largest value of the critical radius we arelikely to encounter is kmax, insulation 0.05 W m ⋅°C rcr,max = ≈ hmin 5W 2 m ⋅°C = 0.01 m = 1 cm The variation of heat transfer rate with theWe can insulate hot-water or steam pipes freely outer radius of the insulation r2 when r1 < rcr.without worrying about the possibility ofincreasing the heat transfer by insulating the pipes. KEITH VAUGH
    132. 132. QUESTIONA 3 mm diameter and 5 m long electric wire is tightlywrapped with a 2 mm thick plastic cover whosethermal conductivity is k = 0.15 W/m2 °C. Electricalmeasurements indicate that a current of 10 A passes }through the wire and there is a voltage drop of 8 Valong the length. If the insulated wire is exposed to amedium at T∞ = 30 °C with a heat transfer coefficientof h = 12 W/m2 °C, determine the temperature at theinterface of the wire and the plastic cover in steadyoperation. Also determine whether doubling thethickness of the plastic cover will increase ordecrease this interface temperature. KEITH VAUGH
    133. 133. KEITH VAUGH
    134. 134. ASSUMPTIONS✓ Heat transfer is steady since there is no indication of any change with time✓ Heat transfer is one dimensional since there is thermal symmetry about the centerline and no variation in the axial direction✓ Thermal conductivities are constant✓ The thermal contact resistance at the interface is negligible✓ Heat transfer coefficient incorporates the radiation effects KEITH VAUGH
    135. 135. KEITH VAUGH
    136. 136. ANALYSIS KEITH VAUGH
    137. 137. ANALYSISHeat is generated in the wire and its temperature rises as a resultof resistance heating. We assume heat is generated uniformlythroughout the wire and is transferred to the surroundingmedium in the radial direction. In steady operation, the rate ofheat transfers becomes equal to the heat generated within thewire, which is determined to be: KEITH VAUGH
    138. 138. ANALYSISHeat is generated in the wire and its temperature rises as a resultof resistance heating. We assume heat is generated uniformlythroughout the wire and is transferred to the surroundingmedium in the radial direction. In steady operation, the rate ofheat transfers becomes equal to the heat generated within thewire, which is determined to be: & & Q = We = VI = ( 8V ) (10A ) = 80W KEITH VAUGH
    139. 139. ANALYSISHeat is generated in the wire and its temperature rises as a resultof resistance heating. We assume heat is generated uniformlythroughout the wire and is transferred to the surroundingmedium in the radial direction. In steady operation, the rate ofheat transfers becomes equal to the heat generated within thewire, which is determined to be: & & Q = We = VI = ( 8V ) (10A ) = 80WThe thermal resistance network for this problem involves aconduction resistance for the plastic cover and a convectionresistance for the outer surface in series. The values of these tworesistances is determined: KEITH VAUGH
    140. 140. KEITH VAUGH
    141. 141. A2 = ( 2π r2 ) L = 2π ( 0.0035m ) ( 5m ) = 0.110m 2 KEITH VAUGH
    142. 142. A2 = ( 2π r2 ) L = 2π ( 0.0035m ) ( 5m ) = 0.110m 2 1 1Rconv = = = 0.76 °C W hA2 (12 W m 2 ⋅°C ) ( 0.110m ) 2 KEITH VAUGH
    143. 143. A2 = ( 2π r2 ) L = 2π ( 0.0035m ) ( 5m ) = 0.110m 2 1 1Rconv = = = 0.76 °C W hA2 ( 12 W m 2 ⋅°C ) ( 0.110m ) 2 ⎛ r2 ⎞ ⎛ 3.5 ⎞ ln ⎜ ⎟ ln ⎜ ⎝ r1 ⎠ ⎝ 1.5 ⎟ ⎠Rplastic = = = 0.18 °C W ( 2π kL 2π 0.15 W 2 m ⋅°C ( 5m )) KEITH VAUGH
    144. 144. A2 = ( 2π r2 ) L = 2π ( 0.0035m ) ( 5m ) = 0.110m 2 1 1Rconv = = = 0.76 °C W hA2 ( 12 W m 2 ⋅°C ) ( 0.110m ) 2 ⎛ r2 ⎞ ⎛ 3.5 ⎞ ln ⎜ ⎟ ln ⎜ ⎝ r1 ⎠ ⎝ 1.5 ⎟ ⎠Rplastic = = = 0.18 °C W ( 2π kL 2π 0.15 W 2 m ⋅°C ( 5m ))Therefore KEITH VAUGH
    145. 145. A2 = ( 2π r2 ) L = 2π ( 0.0035m ) ( 5m ) = 0.110m 2 1 1Rconv = = = 0.76 °C W hA2 ( 12 W m 2 ⋅°C ) ( 0.110m ) 2 ⎛ r2 ⎞ ⎛ 3.5 ⎞ ln ⎜ ⎟ ln ⎜ ⎝ r1 ⎠ ⎝ 1.5 ⎟ ⎠Rplastic = = = 0.18 °C W ( 2π kL 2π 0.15 W 2 m ⋅°C ( 5m ))ThereforeRTotal = Rplastic + RConv = 0.76 + 0.18 = 0.94 °C W KEITH VAUGH
    146. 146. A2 = ( 2π r2 ) L = 2π ( 0.0035m ) ( 5m ) = 0.110m 2 1 1Rconv = = = 0.76 °C W hA2 ( 12 W m 2 ⋅°C ) ( 0.110m ) 2 ⎛ r2 ⎞ ⎛ 3.5 ⎞ ln ⎜ ⎟ ln ⎜ ⎝ r1 ⎠ ⎝ 1.5 ⎟ ⎠Rplastic = = = 0.18 °C W ( 2π kL 2π 0.15 W 2 m ⋅°C ( 5m ))ThereforeRTotal = Rplastic + RConv = 0.76 + 0.18 = 0.94 °C WThen the interface temperature can be determined from KEITH VAUGH
    147. 147. A2 = ( 2π r2 ) L = 2π ( 0.0035m ) ( 5m ) = 0.110m 2 1 1Rconv = = = 0.76 °C W hA2 ( 12 W m 2 ⋅°C ) ( 0.110m ) 2 ⎛ r2 ⎞ ⎛ 3.5 ⎞ ln ⎜ ⎟ ln ⎜ ⎝ r1 ⎠ ⎝ 1.5 ⎟ ⎠Rplastic = = = 0.18 °C W ( 2π kL 2π 0.15 W 2 m ⋅°C ( 5m ))ThereforeRTotal = Rplastic + RConv = 0.76 + 0.18 = 0.94 °C WThen the interface temperature can be determined from&= T1 − T∞ → T1 = T∞ + QRTotal = 30°C + ( 80W ) 0.94 °CQ & ( ) RTotal W = 105°C KEITH VAUGH
    148. 148. KEITH VAUGH
    149. 149. To answer the second part of the question, we need to know thecritical radius of insulation of the plastic cover. It is determined as; KEITH VAUGH
    150. 150. To answer the second part of the question, we need to know thecritical radius of insulation of the plastic cover. It is determined as; k 0.15 W m ⋅°C rcr = = = 0.0125m = 12.5mm h 12 W 2 m ⋅°C KEITH VAUGH
    151. 151. To answer the second part of the question, we need to know thecritical radius of insulation of the plastic cover. It is determined as; k 0.15 W m ⋅°C rcr = = = 0.0125m = 12.5mm h 12 W 2 m ⋅°CWhich is larger than the radius of the plastic cover. Therefore,increasing the thickness of the plastic cover will enhance the heattransfer until the outer radius of the cover reaches 12.5 mm. As aresult the rate of heat transfer will increase when the interfacetemperature T1 is held constant, or T1 will decrease when heattransfer is held constant. KEITH VAUGH
    152. 152. Steady Heat Conduction in Plane Walls  Thermal Resistance Concept  Thermal Resistance Network  Multilayer Plane WallsThermal Contact ResistanceGeneralised Thermal Resistance NetworksHeat Conduction in Cylinders and Spheres  Multilayered Cylinders and SpheresCritical Radius of Insulation KEITH VAUGH

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