Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Ch2 Heat transfer - conduction by eky047 11116 views
- SSL3 Heat Transfer by Keith Vaugh 37728 views
- SSL2 Basic Concepts by Keith Vaugh 37749 views
- SSL6 Properties of Pure Substances by Keith Vaugh 34594 views
- Transient heat conduction by Ahmadreza Aminian 1612 views
- SSL5 Energy Transfer and Analysis by Keith Vaugh 33699 views

38,830 views

Published on

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

No Downloads

Total views

38,830

On SlideShare

0

From Embeds

0

Number of Embeds

33,148

Shares

0

Downloads

0

Comments

0

Likes

21

No embeds

No notes for slide

- 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. OBJECTIVES } KEITH VAUGH
- 3. OBJECTIVESUnderstand the concept of thermal resistance and itslimitations, and develop thermal resistance networksfor practical heat conduction problems } KEITH VAUGH
- 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. 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. 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. 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. 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. 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. 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. KEITH VAUGH
- 12. Under steady conditions, thetemperature distribution in aplane wall is a straight line:dT/dx = const. KEITH VAUGH
- 13. & dT Qcond,wall = −kA dxUnder steady conditions, thetemperature distribution in aplane wall is a straight line:dT/dx = const. KEITH VAUGH
- 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. & 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. & 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. THERMAL RESISTANCECONCEPTAnalogy between thermal andelectrical resistance concepts. KEITH VAUGH
- 18. THERMAL RESISTANCECONCEPT & T −T Q= 1 2 R V1 − V2 I= ReAnalogy between thermal andelectrical resistance concepts. KEITH VAUGH
- 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. 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. 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. 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. KEITH VAUGH
- 24. & Newton’s lawQconv = hAs (Ts − T∞ ) of cooling Schematic for convection resistance at a surface. KEITH VAUGH
- 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. KEITH VAUGH
- 27. Schematic for convectionand radiation resistances ata surface. KEITH VAUGH
- 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. & = 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. & = 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. & = 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. STEADY ONE DIMENSIONALHEAT TRANSFER KEITH VAUGH
- 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. 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. 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. 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. KEITH VAUGH
- 38. KEITH VAUGH
- 39. KEITH VAUGH
- 40. Thermal resistance network KEITH VAUGH
- 41. Thermal resistance network KEITH VAUGH
- 42. &= T∞1 − T∞2 Q RtotalThermal resistance network KEITH VAUGH
- 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. The temperature drop across a layer isproportional to its thermal resistance. KEITH VAUGH
- 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. 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. 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. 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. 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. 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. 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. 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. KEITH VAUGH
- 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. 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. 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. 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. 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. KEITH VAUGH
- 60. ASSUMPTIONSHeat transfer through the mediums is steadyHeat transfer is one dimensionalThermal conductivity is constant KEITH VAUGH
- 61. ASSUMPTIONSHeat transfer through the mediums is steadyHeat transfer is one dimensionalThermal conductivity is constant KEITH VAUGH
- 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. 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. KEITH VAUGH
- 65. A = 0.8m × 1.5m = 1.2m 2 KEITH VAUGH
- 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. 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. 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. 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. 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. 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. 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. 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. 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. THERMAL CONTACT RESISTANCE } KEITH VAUGH
- 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. 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. 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. Temperature distribution and heat ﬂow lines alongtwo solid plates pressed against each other for thecase of perfect and imperfect contact. KEITH VAUGH
- 80. Temperature distribution and heat ﬂow lines alongtwo solid plates pressed against each other for thecase of perfect and imperfect contact. KEITH VAUGH
- 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. 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. 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. 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. 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. 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. GENERALISED THERMALRESISTANCE NETWORKS KEITH VAUGH
- 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. KEITH VAUGH
- 90. Thermal resistance network forcombined series-parallel arrangement. KEITH VAUGH
- 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. &= 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. HEAT CONDUCTION INCYLINDERS & SPHERES } KEITH VAUGH
- 94. HEAT CONDUCTION INCYLINDERS & SPHERES Heat transfer through the pipe can be modelled as steady and one-dimensional. } KEITH VAUGH
- 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. 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. 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. A long cylindrical pipe (or sphericalshell) with speciﬁed inner and outersurface temperatures T1 and T2. KEITH VAUGH
- 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. & 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. & 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. & 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. & 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. & 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. & 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. A spherical shell with speciﬁed inner andouter surface temperatures T1 and T2. KEITH VAUGH
- 107. A spherical shell with speciﬁed inner and outer surface temperatures T1 and T2.& sph = T1 − T2Qcond, Rsph KEITH VAUGH
- 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. KEITH VAUGH
- 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. &= 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. &= 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. MULTILAYER CYLINDER &SPHERESThe thermal resistance network for heattransfer through a three-layered compositecylinder subjected to convection on bothsides. KEITH VAUGH
- 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. 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. 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. 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. 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. 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. 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. CRITICAL RADIUS OF INSULATION } KEITH VAUGH
- 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. 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. 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. An insulated cylindrical pipe exposed toconvection from the outer surface and thethermal resistance network associated with it. KEITH VAUGH
- 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. KEITH VAUGH
- 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. 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. 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. 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. 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. KEITH VAUGH
- 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. KEITH VAUGH
- 136. ANALYSIS KEITH VAUGH
- 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. 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. 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. KEITH VAUGH
- 141. A2 = ( 2π r2 ) L = 2π ( 0.0035m ) ( 5m ) = 0.110m 2 KEITH VAUGH
- 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. 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. 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. 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. 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. 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. KEITH VAUGH
- 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. 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. 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. 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

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment