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- 1. Heat transfer - Source of heat - Heat transfer - Steam and electricity as heating media - Determination of requirement of amount of steam/electrical energy - Steam pressure - Mathematical problems on heat transfer 1
- 2. What is Heat? 2
- 3. What is Heat? Heat is energy in transit. 3
- 4. Units of Heat• The SI unit is the joule (J), which is equal to Newton-metre (Nm).• Historically, heat was measured in terms of the ability to raise the temperature of water.• The calorie (cal): amount of heat needed to raise the temperature of 1 gramme of water by 1 C0 (from 14.50C to 15.50C)• In industry, the British thermal unit (Btu) is still used: amount of heat needed to raise the temperature of 1 lb of water by 1 F0 (from 630F to 640F) 4
- 5. Conversion between different units of heat:1 J = 0.2388 cal = 0.239x10-3 kcal = 60.189 Btu 1 cal = 4.186 J = 3.969 x 10-3 Btu 5
- 6. Sensible Heat• What is sensible heat‘? Sensible heat is associated with a temperature change 6
- 7. Specific Heat Capacity• To raise the temperature by 1 K, different substances need different amount of energy because substances have different molecular configurations and bonding (eg: copper, water, wood)• The amount of energy needed to raise the temperature of 1 kg of a substance by 1 K is known as the specific heat capacity• Specific heat capacity is denoted by c 7
- 8. Calculation of Sensible HeatQ is the heat lost or gained by a substancem is the mass of substancec is the specific heat of substance which changes with temperatureT is the temperatureWhen temperature changes causes negligible changes in c,where ΔT is the temperature change in the substance 8
- 9. Calculation of Sensible HeatWhen temperature changes causes significant changes in c, Q = m c ∆T cannot be used.Instead, we use the following equation: Q = ∆H = m ∆h where ΔH is the enthalpy change in the substance and ∆h is the specific enthalpy change in the substance. To apply the above equation, the system should remain at constant pressure and the associated volume change must be negligibly small. 9
- 10. Calculation of Sensible HeatCalculate the amount of heat required to raise the temperature of 300 g Al from 25oC to 70oC. Data: c = 0.896 J/g oC for Al Q = m c ΔT (since c is taken as a constant) = (300 g) (0.896 J/g oC)(70 - 25)oC = 12,096 J = 13.1 kJ 10
- 11. Exchange of HeatCalculate the final temperature (tf), when 100 g iron at 80oC is tossed into 53.5g of water at 25oC. Data: c = 0.452 J/g oC for iron and 4.186 J/g oC for water Heat lost by iron = Heat gained by water (m c ΔT)iron = (m c ΔT)water (100 g) (0.452 J/g oC)(80 - tf)oC = (53.5 g) (4.186 J/g oC)(tf - 25)oC 80 - tf = 4.955 (tf -25) tf = 34.2oC 11
- 12. Latent Heat• What is ‘latent heat‘? Latent heat is associated with phase change of matter 12
- 13. Phases of Matter 13
- 14. Phase Change• Heat required for phase changes: » Melting: solid liquid » Vaporization: liquid vapour » Sublimation: solid vapour• Heat released by phase changes: » Condensation: vapour liquid » Fusion: liquid solid » Deposition: vapour solidProf. R. Shanthini 14 5&
- 15. Phase Diagram: Water 15
- 16. Phase Diagram: Water Compressed liquid Saturated liquid Superheated steam Saturated steam 16
- 17. Phase Diagram: Water Explain why water is at liquid state at atm pressure 17
- 18. Phase Diagram: Carbon Dioxide Explain why CO2 is at gas state at atm pressure Explain why CO2 cannot be made a liquid at atm pressure 18
- 19. Latent HeatLatent heat is the amount of heat added per unit mass ofsubstance during a phase changeLatent heat of fusion is the amount of heat added to melta unit mass of ice OR it is the amount of heat removedto freeze a unit mass of water.Latent heat of vapourization is the amount of heat addedto vaporize a unit mass of water OR it is the amount ofheat removed to condense a unit mass of steam. 19
- 20. Water:Specific Heat Capacities and Latent Heats Specific heat of ice ≈ 2.06 J/g K (assumed constant) Heat of fusion for ice/water ≈ 334 J/g (assumed constant) Specific heat of water ≈ 4.18 J/g K (assumed constant) Latent heat of vaporization cannot be assumed aconstant since it changes significantly with the pressure, and could be found from the Steam Table How to evaluate the sensible heat gained (or lost) by superheated steam? 20
- 21. Water: Specific Heat Capacities and Latent Heats How to evaluate the sensible heat gained (or lost) by superheated steam? Q = m c ∆T cannot be used since changes in c with changing temperature is NOT negligible.Instead, we use the following equation: Q = ∆H = m ∆h provided the system is at constant pressure and the associated volume change is negligible. Enthalpies could be referred from the Steam 21 Table
- 22. Properties of SteamLearnt to refer to Steam Table to find properties ofsteam such as saturated (or boiling point) temperatureand latent heat of vapourization at give pressures, andenthalpies of superheated steam at various pressures andtemperatures.Reference:Chapter 6 of “Thermodynamics for Beginners with workedexamples” by R. Shanthini (published by Science Education Unit, Faculty of Science, University of Peradeniya) (also uploaded at http://www.rshanthini.com/PM3125.htm) 22
- 23. Warming curve for waterWhat is the amount of heat required to change 2 kg of ice at -20oC to steam at 150oC at 2 bar pressure? -20oC ice 23
- 24. Warming curve for waterWhat is the amount of heat required to change 2 kg of ice at -20oC to steam at 150oC at 2 bar pressure? 0oC melting point of ice -20oC ice 24
- 25. Warming curve for waterWhat is the amount of heat required to change 2 kg of ice at -20oC to steam at 150oC at 2 bar pressure? 120.2oC boiling point of water at 2 bar Boiling point of water at 1 atm pressure is 100oC. Boiling point of water at 2 bar is 120.2oC. [Refer the Steam Table.] 0oC melting point of ice -20oC ice 25
- 26. Warming curve for waterWhat is the amount of heat required to change 2 kg of ice at -20oC to steam at 150oC at 2 bar pressure? 150oC superheated steam Specific heat 120.2oC boiling point of water at 2 bar Latent heat Specific heat 0oC melting point of ice Latent heat Specific heat -20oC ice 26
- 27. Warming curve for waterWhat is the amount of heat required to change 2 kg of ice at -20oC to steam at 150oC at 2 bar pressure?Specific heat required to raise the temperature of ice from -20 oCto 0oC = (2 kg) (2.06 kJ/kg oC) [0 - (-20)]oC = 82.4 kJLatent heat required to turn ice into water at 0oC = (2 kg) (334 kJ/kg) = 668 kJSpecific heat required to raise the temperature of water from 0oC to120.2oC = (2 kg) (4.18 kJ/kg oC) [120.2 - 0)]oC = 1004.9 kJ 27
- 28. Warming curve for waterWhat is the amount of heat required to change 2 kg of ice at -20oC to steam at 150oC at 2 bar pressure?Latent heat required to turn water into steam at 120.2oC and at 2 bar = (2 kg) (2202 kJ/kg) = 4404 kJ [Latent heat of vapourization at 2 bar is 2202 kJ/kg as could be referred to from the Steam Table]Specific heat required to raise the temperature of steam from 120.2oCto 150oC = (2 kg) (2770 – 2707) kJ/kg = 126 kJ [Enthalpy at 120.2oC and 2 bar is the saturated steam enthalpy of 2707 kJ/kg and the enthalpy at 150oC and 2 bar is 2770 kJ/kg as could be referred to from the Steam Table] 28
- 29. Warming curve for waterWhat is the amount of heat required to change 2 kg of ice at -20oC to steam at 150oC at 2 bar pressure?Total amount of heat required = 82.4 kJ + 668 kJ + 1004.9 kJ + 4404 kJ + 126 kJ = 6285.3 kJ 29
- 30. Application: Heat ExchangerIt is an industrial equipment in which heat is transferred from a hotfluid (a liquid or a gas) to a cold fluid (another liquid or gas) withoutthe two fluids having to mix together or come into direct contact. Cold fluid Cold fluid at TC,out at TC,in Hot fluid at TH,in Heat lost by the hot fluid Hot fluid = Heat gained by the cold fluid at TH,out 30
- 31. Application: Heat Exchanger 31
- 32. Heat Exchanger Heat lost by the hot fluid = Heat gained by the cold fluid .m hot chot (TH,in – TH,out . )=m cold ccold (TC,out – TC,in)mass flow rate mass flow rateof hot fluid of cold fluid Specific heat Specific heat of hot fluid of cold fluid Temperature Temperature decrease in the increase in the hot fluid cold fluid 32
- 33. Heat Exchanger Heat lost by the hot fluid = Heat gained by the cold fluid .m hot chot (TH,in – TH,out . )=m cold ccold (TC,out – TC,in)The above is true only under the following conditions:(1) Heat exchanger is well insulated so that no heat is lost to the environment(2) There are no phase changes occurring within the heat exchanger. 33
- 34. Heat Exchanger If the heat exchanger is NOT well insulated, thenHeat lost by the hot fluid = Heat gained by the cold fluid + Heat lost to the environment 34
- 35. Worked Example 1 in Heat Exchanger High pressure liquid water at 10 MPa (100 bar) and 30oC enters a series of heating tubes. Superheated steam at 1.5 MPa (15 bar) and 200oC is sprayed over the tubes and allowed to condense. The condensed steam turns into saturated water which leaves the heat exchanger. The high pressure water is to be heated up to 170oC. What is the mass of steam required per unit mass of incoming liquid water? The heat exchanger is assumed to be well insulated (adiabatic). 35
- 36. Solution to Worked Example 1 in Heat Exchanger 36
- 37. Solution to Worked Example 1 in Heat Exchanger contd. High pressure (100 bar) water enters at 30oC and leaves at 198.3oC. Boiling point of water at 100 bar is 311.0oC. Therefore, no phase changes in the high pressure water that is getting heated up in the heater. Heat gained by high pressure water = ccold (TC,out – TC,in) = (4.18 kJ/kg oC) x (170-30)oC = 585.2 kJ/kg [You could calculate the above by taking the difference in enthalpies at the 2 given states from tables available.] 37
- 38. Solution to Worked Example 1 in Heat Exchanger contd. Superheated steam at 1.5 MPa (15 bar) and 200oC is sprayed over the tubes and allowed to condense. The condensed steam turns into saturated water which leaves the heat exchanger. Heat lost by steam = heat lost by superheated steam to become saturated steam + latent heat of steam lost for saturated steam to turn into saturated water = Enthalpy of superheated steam at 15 bar and 200oC – Enthalpy of saturated steam at 15 bar + Latent heat of vapourization at 15 bar = (2796 kJ/kg – 2792 kJ/kg) + 1947 kJ/kg = 1951 kJ/kg 38
- 39. Solution to Worked Example 1 in Heat Exchanger contd. Since there is no heat loss from the heater, Heat lost by steam = Heat gained by high pressure water Mass flow rate of steam x 1951 kJ/kg = Mass flow rate of water x 585.2 kJ/kg Mass flow rate of steam / Mass flow rate of water = 585.2 / 1951 = 0.30 kg stream / kg of water 39
- 40. Assignment Give the design of a heat exchanger which has the most effective heat transfer properties.Learning objectives:1) To be able to appreciate heat transfer applications in pharmaceutical industry2) To become familiar with the working principles of various heat exchangers3) To get a mental picture of different heat exchangers so that solving heat transfer problems in class becomes more interesting 40
- 41. Worked Example 2 in Heat Exchanger Steam enters a heat exchanger at 10 bar and 200oC and leaves it as saturated water at the same pressure. Feed- water enters the heat exchanger at 25 bar and 80oC and leaves at the same pressure and at a temperature 20oC less than the exit temperature of the steam. Determine the ratio of the mass flow rate of the steam to that of the feed-water, neglecting heat losses from the heat exchanger. If the feed-water leaving the heat exchanger is fed directly to a boiler to be converted to steam at 25 bar and 300oC, find the heat required by the boiler per kg of feed- water. 41
- 42. Solution to Worked Example 2 in Heat Exchanger - Steam enters at 10 bar and 200oC and leaves it as saturated water at the same pressure. - Saturation temperature of water at 10 bar is 179.9oC. - Feed-water enters the heat exchanger at 25 bar and 80oC and leaves at the same pressure and at a temperature 20oC less than the exit temperature of the steam, which is 179.9oC. - Boiling point of water at 25 bar is (221.8+226.0)/2 = 223.9oC. - Therefore, no phase changes in the feed-water that is being heated. Heat lost by steam = Heat gained by feed-water (with no heat losses) Mass flow rate of steam x [2829 – 2778 + 2015] kJ/kg = Mass flow rate of feed-water x [4.18 x (179.9-20-80) ] kJ/kg Mass flow of steam / Mass flow of feed-water = 333.98 / 2066 = 0.1617 kg stream / kg of water 42
- 43. Solution to Worked Example 1 in Heat Exchanger contd. If the feed-water leaving the heat exchanger is fed directly to a boiler to be converted to steam at 25 bar and 300oC, find the heat required by the boiler per kg of feed-water. - Temperature of feed-water leaving the heat exchanger is 159.9oC - Boiling point of water at 25 bar is (221.8+226.0)/2 = 223.9oC - The feed-water is converted to superheated steam at 300oC Heat required by the boiler per kg of feed-water = {4.18 x (223.9-159.9) + (1850+1831)/2 + [(3138+3117)/2 – (2802+2803)/2]} kJ/kg = {267.52 + 1840.5 + [3127.5 – 2802.5]} kJ/kg = 2433 kJ/kg of feed-water 43
- 44. Heat Transferis the means by which energy moves from a hotter object to a colder object 44
- 45. Mechanisms of Heat Transfer Conduction is the flow of heat by direct contact between a warmer and a cooler body. Convectionis the flow of heat carried by moving gas or liquid. (warm air rises, gives up heat, cools, then falls) Radiationis the flow of heat without need of an intervening medium. (by infrared radiation, or light) 45
- 46. Mechanisms of Heat Transfer Latent heat ConductionConvection Radiation Prof. R. Shanthini 46 5&
- 47. Conduction HOT COLD(lots of vibration) (not much vibration) Heat travels along the rod 47
- 48. Conduction Conduction is the process whereby heat is transferred directly through a material, any bulk motion of the material playing no role in the transfer. Those materials that conduct heat well are calledthermal conductors, while those that conduct heat poorly are known as thermal insulators. Most metals are excellent thermal conductors, while wood, glass, and most plastics are common thermal insulators. The free electrons in metals are responsible for the excellent thermal conductivity of metals. 48
- 49. Conduction: Fourier’s Law Cross-sectional area A L Q = heat transferredQ =k A ( )t ΔT L k = thermal conductivity A = cross sectional area ∆T = temperature difference between two endsWhat is the unit of k? L = length t = duration of heat transfer 49
- 50. Thermal ConductivitiesSubstance Thermal Substance Thermal Conductivity Conductivity k [W/m.K] k [W/m.K]Syrofoam 0.010 Glass 0.80 Air 0.026 Concrete 1.1 Wool 0.040 Iron 79 Wood 0.15 Aluminum 240 Body fat 0.20 Silver 420 Water 0.60 Diamond 2450 50
- 51. Conduction through Single Wall Use Fourier’s Law:T1 Q =k A ( )t ΔT L . .Q Q T2 < T1 . k A (T1 – T2) Q = x Δx Δx 51
- 52. Conduction through Single WallT1 . k A (T1 – T2) Q = Δx . .Q Q T1 – T 2 = T2 < T1 Δx/(kA) x Δx Thermal resistance (in K/W) (opposing heat flow) 52 52
- 53. Conduction through Composite Wall T1 A B C. T2 .Q T3 Q kA kB kC T4 x ΔxA ΔxB ΔxC . T1 – T 2 T2 – T 3 T3 – T 4 Q = = = (Δx/kA)A (Δx/kA)B (Δx/kA)C 53 53
- 54. Conduction through Composite Wall . T1 – T 2 T2 – T 3 T3 – T 4 Q = = = (Δx/kA)A (Δx/kA)B (Δx/kA)C . [ Q (Δx/kA)A + (Δx/kA)B + (Δx/kA)C ] = T 1 – T 2 + T2 – T 3 + T3 – T 4 . T1 – T 4 Q= (Δx/kA)A + (Δx/kA)B + (Δx/kA)C 54 54
- 55. Example 1An industrial furnace wall is constructed of 21 cm thickfireclay brick having k = 1.04 W/m.K. This is covered onthe outer surface with 3 cm layer of insulating materialhaving k = 0.07 W/m.K. The innermost surface is at 1000oCand the outermost surface is at 40oC. Calculate the steadystate heat transfer per area. Solution: We start with the equation . Tin – Tout Q= (Δx/kA)fireclay + (Δx/kA)insulation 55
- 56. Example 1 continued . (1273.15 – 313.15) AQ = (0.21/1.04) + (0.03/0.07).Q = 1522.6 W/m2A 56
- 57. Example 2We want to reduce the heat loss in Example 1 to 960 W/m2.What should be the insulation thickness?Solution: We start with the equation . Tin – Tout Q= (Δx/kA)fireclay + (Δx/kA)insulation . Q (1000 – 40) = = 960 W/m2 A (0.21/1.04) + (Δx)insulation /0.07) (Δx)insulation = 5.6 cm 57
- 58. Conduction through hollow-cylinder ro Ti ri To L . Ti – T o Q = [ln(ro/ri)] / 2πkL 58
- 59. Conduction through the compositer3 wall in a hollow-cylinder r2 To Material ATi r1 Material B. Ti – T oQ = [ln(r2/r1)] / 2πkAL + [ln(r3/r2)] / 2πkBL 59
- 60. Example 3A thick walled tube of stainless steel ( k = 19 W/m.K) with2-cm inner diameter and 4-cm outer diameter is coveredwith a 3-cm layer of asbestos insulation (k = 0.2 W/m.K).If the inside-wall temperature of the pipe is maintained at600oC and the outside of the insulation at 100oC, calculatethe heat loss per meter of length.Solution: We start with the equation . Ti – T o Q = [ln(r2/r1)] / 2πkAL + [ln(r3/r2)] / 2πkBL 60
- 61. Example 3 continued. 2 π L ( 600 – 100)Q = [ln(2/1)] / 19 + [ln(5/2)] / 0.2 . Q = 680 W/m L 61
- 62. Mechanisms of Heat Transfer Conduction √ is the flow of heat by direct contact between a warmer and a cooler body. Convectionis the flow of heat carried by moving gas or liquid. (warm air rises, gives up heat, cools, then falls) Radiationis the flow of heat without need of an intervening medium. (by infrared radiation, or light) 62
- 63. ConvectionConvection is the process in which heat is carried fromplace to place by the bulk movement of a fluid (gas orliquid). Convection currents are set up when a pan of water is heated. 63
- 64. ConvectionIt explains why breezes come from the ocean in the dayand from the land at night 64
- 65. Convection: Newton’s Law of CoolingFlowing fluid at Tfluid Heated surface at Tsurface . Qconv. = h A (Tsurface – Tfluid) Area exposed Heat transfer coefficient (in W/m2.K) 65
- 66. Convection: Newton’s Law of CoolingFlowing fluid at Tfluid Heated surface at Tsurface . Tsurface – Tfluid Qconv. = 1/(hA) Convective heat resistance (in K/W) 66
- 67. Example 4The convection heat transfer coefficient between a surface at 50oC andambient air at 30oC is 20 W/m2.K. Calculate the heat flux leaving thesurface by convection.Solution: Use Newton’s Law of cooling : . Flowing fluid at Tfluid = 30oC Q = h A (Tsurface – Tfluid) conv. = (20 W/m2.K) x A x (50-30)oC Heated surface at Tsurface = 50oC Heat flux leaving the surface: . Q conv. = 20 x 20 = 400 W/m2 h = 20 W/m2.K A 67
- 68. Example 5Air at 300°C flows over a flat plate of dimensions 0.50 m by 0.25 m.If the convection heat transfer coefficient is 250 W/m2.K, determinethe heat transfer rate from the air to one side of the plate when theplate is maintained at 40°C.Solution: Use Newton’s Law of cooling :Flowing fluid at Tfluid = 300oC . Q = h A (Tsurface – Tfluid) conv.Heated surface at Tsurface = 40oC = 250 W/m2.K x 0.125 m2 x (40 - 300)oC = - 8125 W/m2 h = 250 W/m2.K Heat is transferred from A = 0.50x0.25 m2 the air to the plate. 68
- 69. Forced ConvectionIn forced convection, a fluid is forced by external forcessuch as fans. In forced convection over external surface: Tfluid = the free stream temperature (T∞), or a temperature far removed from the surface In forced convection through a tube or channel: Tfluid = the bulk temperature 69
- 70. Free ConvectionIn free convection, a fluid is circulated due to buoyancyeffects, in which less dense fluid near the heated surface risesand thereby setting up convection. In free (or partially forced) convection over external surface: Tfluid = (Tsurface + Tfree stream) / 2 In free or forced convection through a tube or channel: Tfluid = (Tinlet + Toutlet) / 2 70
- 71. Change of Phase ConvectionChange-of-phase convection is observed withboiling or condensation.It is a very complicated mechanism andtherefore will not be covered in this course. 71
- 72. Overall Heat Transfer through a Plane Wall Fluid A at TA > T1 T1 . . Q Q T2 Fluid B at TB < T2 x Δx . T A – T1 T 1 – T2 T2– TB Q = = = 1/(hAA) Δx/(kA) 1/(hBA) 72
- 73. Overall Heat Transfer through a Plane Wall . T A – T1 T 1 – T2 T2– TB Q = = = 1/(hAA) Δx/(kA) 1/(hBA) . TA – TB Q = 1/(hAA) + Δx/(kA) + 1/(hBA) . Q = U A (TA – TB) where U is the overall heat transfer coefficient given by 1/U = 1/hA + Δx/k + 1/hB 73
- 74. Overall heat transfer through hollow-cylinder Fluid A is inside the pipe ro Fluid B is outside the pipe TA > TB Ti r i To L . Q = U A (TA – TB) where 1/UA = 1/(hAAi) + ln(ro/ri) / 2πkL + 1/(hBAo) 74
- 75. Example 6Steam at 120oC flows in an insulated pipe. The pipe ismild steel (k = 45 W/m K) and has an inside radius of 5cm and an outside radius of 5.5 cm. The pipe is coveredwith a 2.5 cm layer of 85% magnesia (k = 0.07 W/m K).The inside heat transfer coefficient (hi) is 85 W/m2 K, andthe outside coefficient (ho) is 12.5 W/m2 K. Determine theheat transfer rate from the steam per m of pipe length, ifthe surrounding air is at 35oC.Solution: Start with . Q = U A (TA – TB) = U A (120 – 35) What is UA? 75
- 76. Example 6 continued1/UA = 1/(hAAi) + ln(ro/ri) / 2πkL + … + 1/(hBAo)1/UA = 1/(85Ain) + ln(5.5/5) / 2π(45)L + ln(8/5.5) / 2π(0.07)L + 1/(12.5Aout) Ain = 2π(0.05)L and Aout = 2π(0.08)L1/UA = (0.235 + 0.0021 +5.35 + 1) / 2πL 76
- 77. Example 6 continuedUA = 2πL / (0.235 + 0.0021 +5.35 + 1).Q = U A (120 – 35) steel air = 2πL (120 – 35) / (0.235 + 0.0021 +5.35 + 1) steam insulation = 81 L . Q / L = 81 W/m 77
- 78. Mechanisms of Heat Transfer Conduction √ is the flow of heat by direct contact between a warmer and a cooler body. Convection √is the flow of heat carried by moving gas or liquid. (warm air rises, gives up heat, cools, then falls) Radiationis the flow of heat without need of an intervening medium. (by infrared radiation, or light) 78
- 79. RadiationRadiation is the process in which energy is transferred bymeans of electromagnetic waves of wavelength band between0.1 and 100 micrometers solely as a result of the temperatureof a surface.Heat transfer by radiationcan take place throughvacuum. This is becauseelectromagnetic wavescan propagate throughempty space. 79
- 80. The Stefan–Boltzmann Law of Radiation Q = ε σ A T4 t ε = emissivity, which takes a value between 0 (for an ideal reflector) and 1 (for a black body). σ = 5.668 x 10-8 W/m2.K4 is the Stefan-Boltzmann constant A = surface area of the radiator T = temperature of the radiator in Kelvin. 80
- 81. Why is themother shielding her cub? Ratio of the surface area of a cub to itsvolume is much larger than for its mother. 81
- 82. What is the Sun’s surface temperature?The sun provides about 1000 W/m2 at the Earths surface. Assume the Suns emissivity ε = 1 Distance from Sun to Earth = R = 1.5 x 1011 m Radius of the Sun = r = 6.9 x 108 m 82
- 83. What is the Sun’s surface temperature? Q = ε σ A T4 t (4 π 6.92 x 1016 m2)(4 π 1.52 x 1022 m2)(1000 W/m2) = 5.98 x 1018 m2 = 2.83 x 1026 W 2.83 x 1026 W T4 = (1) (5.67 x 10-8 W/m2.K4) (5.98 x 1018 m2) ε σ T = 5375 K 83
- 84. If object at temperature T is surrounded byan environment at temperature T 0, the netradioactive heat flow is: Q = ε σ A (T4 - To4 ) t Temperature of the radiating surface Temperature of the environment 84
- 85. Example 7What is the rate at which radiation is emitted by a surfaceof area 0.5 m2, emissivity 0.8, and temperature 150°C?Solution: Q [(273+150) K]4 = ε σ A T4 t 0.5 m2 0.8 5.67 x 10-8 W/m2.K4 Q = (0.8) (5.67 x 10-8 W/m2.K4) (0.5 m2) (423 K)4 t = 726 W 85
- 86. Example 8If the surface of Example 7 is placed in a large, evacuated chamberwhose walls are maintained at 25°C, what is the net rate at whichradiation is exchanged between the surface and the chamber walls?Solution: Q = ε σ A (T4 - To4 ) t [(273+25) K]4 [(273+150) K]4Q = (0.8) x (5.67 x 10-8 W/m2.K4) x (0.5 m2)t x [(423 K)4 -(298 K)4 ] = 547 W 86
- 87. Example 8 continuedNote that 547 W of heat loss from the surface occursat the instant the surface is placed in the chamber. Thatis, when the surface is at 150oC and the chamber wallis at 25oC.With increasing time, the surface would cool due tothe heat loss. Therefore its temperature, as well as theheat loss, would decrease with increasing time.Steady-state conditions would eventually be achievedwhen the temperature of the surface reached that of thesurroundings. 87
- 88. Example 9Under steady state operation, a 50 W incandescent light bulb has asurface temperature of 135°C when the room air is at a temperatureof 25°C. If the bulb may be approximated as a 60 mm diametersphere with a diffuse, gray surface of emissivity 0.8, what is theradiant heat transfer from the bulb surface to its surroundings?Solution: Q = ε σ A (T4 - To4 ) t [(273+25) K]4 [(273+135) K]4 Q = (0.8) x (5.67 x 10-8 J/s.m2.K4) x [π x (0.06) m2] t x [(408 K)4 -(298 K)4 ] = 10.2 W (about 20% of the power is dissipated by radiation) 88

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