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Thermal Physics METHODS OF HEAT TRANSFER CONDUCTION CONVECTION RADIATION

Thermal Physics METHODS OF HEAT TRANSFER CONDUCTION CONVECTION RADIATION. PHYSICS: FUN EXCITING SIMPLE. ptD_transfer.ppt. Overview of Thermal Physics Module: Thermodynamic Systems: Work, Heat, Internal Energy 0 th , 1 st and 2 nd Law of Thermodynamics Thermal Expansion

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Thermal Physics METHODS OF HEAT TRANSFER CONDUCTION CONVECTION RADIATION

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  1. Thermal Physics METHODS OF HEAT TRANSFER CONDUCTION CONVECTION RADIATION PHYSICS: FUN EXCITING SIMPLE ptD_transfer.ppt

  2. Overview of Thermal Physics Module: • Thermodynamic Systems:Work, Heat, Internal Energy0th, 1st and 2nd Law of Thermodynamics • Thermal Expansion • Heat Capacity, Latent Heat • Methods of Heat Transfer:Conduction, Convection, Radiation • Ideal Gases, Kinetic Theory Model • Second Law of ThermodynamicsEntropy and Disorder • Heat Engines, Refrigerators

  3. METHODS OF HEAT TRANSFER energy transfer (heat, Q) due to a temperature difference, T CONDUCTION CONVECTION RADIATION Qnet §17.7 p591 Environment, TE References: University Physics 12th ed Young & Freedman system, T European heat wave, 2003  ~35 000 deaths in France

  4. Live sheep trade Sunday, October 26, 2003 Sheep to shore ... finally The Labor Opposition will pursue the Government over the cost of the "ship of death" saga, which ended on Friday when 50,000 Australian sheep at sea for three months began being unloading in Eritrea.

  5. Heat Conduction Conduction is heat transfer by means of molecular agitation within a material without any motion of the material as a whole. If one end of a metal rod is at a higher temperature, then energy will be transferred down the rod toward the colder end because the higher speed particles will collide with the slower ones with a net transfer of energy to the slower ones. TC TH conduction though glass Q Q

  6. For conduction between two plane surfaces (eg heat loss through the wall of a house) the rate of heat transfer is steady-state TC TH energy transferred through slab Q Q Q A Thermal conductivity k (W.m-1.K-1) L heat current H = dQ/dt

  7. Thermal Conduction through a uniform slab TH steady-state TC 0 L x

  8. Thermal conductivity, k property of the material kdiamondvery high: perfect heat sink, e.g. for high power laser diodes khuman low: core temp relatively constant (37 oC) kairvery low: good insulator * home insulation * woolen clothing * windows double glazing Metals – good conductors: electrons transfer energy from hot to cold

  9. Heat Convection Convection is heat transfer by mass motion of a fluid such as air or water when the heated fluid is caused to move away from the source of heat, carrying energy with it. Convection above a hot surface occurs because hot air expands, becomes less dense and rises (natural or free). Convection assisted by breeze, pump or fan – forced convection. Hot water is likewise less dense than cold water and rises, causing convection currents which transport energy. Convection coefficient, h DT between surface and air way from surface

  10. http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatra.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatra.html

  11. Sea & Land Breezes, Monsoons 35 oC 20 oC 11 oC 17 oC What is the role of heat capacity, c of water and soil? http://sol.sci.uop.edu/~jfalward/heattransfer/heattransfer.html

  12. RADIATION Energy transferred by electromagnetic waves All materials radiate thermal energy in amounts determined by their temperature, where the energy is carried by photons of light in the infrared and visible portions of the electromagnetic spectrum. Thermal radiation wavelength ranges: IR ~ 100 - 0. 8 m Visible ~ 0.8 - 0.4 m 800 – 400 nm UV ~0.4 - 0.1 m For exam: more detail than in the textbook

  13. Ludwig Boltzmann        (1844-1906) All objects above absolute zero emit radiant energy and the rate of emission increases and the peak wavelength decreases as the temperature of object increases

  14. Thermography

  15. Absorption & Stefan-Boltzmann Law Incident radiation (INTENSITYI - energy passing through a square metre every second Iinc = P / A Iabs = a Iinc Power absorbed by surface of an object Qabs • Surface Area, A • Absorption coefficient, a = 0 to 1 • Stefan-Boltzmann constant σ = 5.67 x 10-8 W.m-2.K-4 A, a Ts

  16. Emission & Stefan-Boltzmann Law Power radiated from the surface of an object A, e, T Qrad • Surface Area, A • Emissivity, e = 0 to 1 • Stefan-Boltzmann constant σ = 5.67 x 10-8 W.m-2.K-4 Pnet > 0 net heat transfer out of system

  17. A blackbody absorbs all the radiation incident upon it and emits the max possible radiation at all wavelengths (e = a = 1) A graybody is a surface that absorbs a certain proportion of the energy of a blackbody, the constant being constant over the entire band of wavelengths (0 e = a < 1) emissivity e absorption coefficient (absorptivity) a Stefan-Boltzmann constant  = 5.6710-8 W.m-2.K-4

  18. Wien’s Displacement Law Wien constant b = 2.89810-3 m.K Blackbody: absorbs ALL the EMR radiation falling on it & emits the max possible energy over all wavelengths

  19. Wien’s Displacement Law Wien constant: b = 2.89810-3 m.K Blackbody radiation curves show different peak wavelengths at various temperatures

  20. Stefan-Boltzmann constant, s = 5.67 x 10-8 W.m-2.K-4* emissivity, e = 0 to 1 Blackbody, e= 1 * Absorption coefficient, a = 0 to 1 * At a temperature Ta = e all wavelengths * T > 700 oC visible radiation (dull red ~ 800 oC white ~ 2000 oC) * Black surface (e ~ 1) – good emitter / absorber * Polished surface (e ~ 0.01) – poor emitter / absorber, good reflector * Hot stars – blue * Cool stars - red Water (e ~ 0.96) Earth (e ~ 0.3)

  21. Sun (6000 K - hot!) Earth (300 K - cold!) Visible radiation Infrared radiation

  22. Sun & Photosynthesis TSun ~ 6000 K peak ~ 480 nm (blue/green) UV (ionization of molecules) ~ 9% Visible (excite molecules) ~ 49 % IR (warm) ~ 42% 0.1% of radiant energy captured by chlorophyll of plants What about life on Earth if the Sun was hotter or colder ?

  23. Emissivity, e – the nature of the surface Summer clothing:  white reflects radiant energy better than black. Wrap an ice-cube in black cloth and another in aluminium foil and  place both in the sunshine.  What will happen? Why is the pupil of the eye black?  e ~ 0.8 e ~ 0.4 http://sol.sci.uop.edu/~jfalward/heattransfer/heattransfer.html

  24. Problem D1:Estimate the Sun’s temperature Assume e = 1 Distance from Sun to Earth:  RSE = 1.5 x 1011 m     Radius of the Sun: RS = 6.9 x 108 mSolar radiation at Earth’s surface: I = 103 W.m-2 s = 5.67 x 10-8 W.m-2.K-4 Solution Power radiated by Sun Prad = I A = I 4pRSE2 = (103)(4p)(1.5x1011)2 W = 2.83 x 1026 W Surface area of the Sun, ASun = 4pRS2 = 5.98 x 1018 m2 T4 = Prad / (ASun es) T = 5.4 x 103 K RS RSE

  25. Problem D.2Estimate the Earth’s surface Temperature TE (assume NO atmosphere) Solar constant Io= 1360 W.m-2 Earth albedo (reflectivity)aE= 0.3 RSE Power absorbed by Earth: Pabs = (1-aE) AdiskIo Earth Adisk = pRE2 Power radiated by Earth: Prad = AEsTE4 Pabs = Prad TE = 255 K = -18 oC e = 1 AE = 4pRE2 radiation emitted from the surface of a sphere What is natural greenhouse effect?

  26. Greenhouse Effect Earth’s albedo (reflectivity) aE= 0.3

  27. American Journal of Physics: Feb 1983

  28. Selective surfaces Emissivity, e – the nature of the surface Value of e is temperature and wavelength dependent e Selective surface used in solar collectors 1 short long 0   Good emitter / absorber at short wavelengths Poor emitter / absorber at long wavelengths

  29. 1 Black paint 1 Black skin e e White paint White skin 0 0 short long short long   Black skin absorbs heat more readily than white skin but also radiates heat more readily - the heat balance favours black skin in the tropics

  30. Problem D.3 Steel plates are placed on a non-conducting opaque surface, normal to incident solar radiation (direct + diffuse) of 750 W.m-2. Neglecting convection, calculate the equilibrium temperature T for a polished steel plate (e = 0.07) and a dull steel plate (e = 0.8). Assume graybody behaviour. Calculate the effect of coating a steel plate with a selective surface e = 0.92 l < 2 mm e = 0.10 l > 2 mm short long

  31. Solution Identify / Setup Pabs= aI A I = 750 W.m-2 e = a Equilibrium: Pabs = Prad Execute The equilibrium temperature is the same for both surfaces – the equilibrium temperature is unaffected by the area and nature of the surface if the emissivity (and absorptivity) remain constant over the range of temperature & wavelength.

  32. Assume Sun a blackbody at 5800 K Absorbed: virtually all the incident radiation lies at < 2 mm thus Pabs = e I A = (0.92)(750)A = 690A Radiated: most of the radiation will be at wavelengths > 2 µm Prad = e s A T4 = 0.1 sA T4 Equil  Pabs = Prad T = 591 K = 317 oC 250 oC hotter e short long 0.92 0.10 l 2 mm

  33. Problem D.4 Semester 1, 2007 Examination An igloo is a hemispherical enclosure built of ice. Elmo’s igloo has an inner radius of 2.55 mm and the thickness of the ice is 0.30 m. This thickness can be considered small compared to the radius. Heat leaks out of the igloo at a rate determined by the thermal conductivity of ice, kice = 1.67 W.m-1.K-1. At what rate must thermal energy be generated inside the igloo to maintain a steady air temperature inside the igloo at 6.5 oC when the outside temperture is -40 oC? Ignore all thermal energy losses by conduction through the ground, or any heat transfer by radiation or convection or leaks.

  34. Solution Identify / Setup -40 oC thickness t = 0.30 m 6.5 oC kice = 1.67 W.m-1.K-1 radius r = 2.55 m The rate of energy production must be equal to the rate of loss of thermal energy by conduction through the hemispherical ice wall. Rate of energy transfer by conduction

  35. dQ/dt = ? W k = 1.67 W.m-1.K-1 dT = {6.5 – (– 40)} oC = 46.5 oC thickness of ice, dx = 0.30 m area, A = surface area of hemisphere = (4 R2) / 2 = 2 R2 Because the thickness of the ice is much smaller than either the inside or outside radius, it does not matter which radius is used – taking the average radius R = (2.55 + 0.15) m = 2.70 m

  36. Execute dQ/dt = – (1.67)(2)(2.70)2(46.5)/0.30 W dQ/dt = – 1.2  104 W Evaluate sensible value units significant figures did I answer the question ?

  37. Problem D.5 Suppose a human could live for 120 min unclothed in air at 8 oC.   How long could they live in water at 8 oC?

  38. Problem D.5 Suppose a human could live for 120 min unclothed in air at 8 oC.  How long could they live in water at 8 oC? Solution Identify / Setup Thermal conductivities kair = 0.024 W.m-1.K-1kwater = 0.6 W.m-1.K-1 Execute Evaluate

  39. Problem D.6 Why do droplets of water dance over the very hot pan ? http://sol.sci.uop.edu/~jfalward/heattransfer/heattransfer.html

  40. Why do droplets of water dance over the very hot pan ? Water at the bottom of the drops is evaporated and provides insulation against further evaporation. http://sol.sci.uop.edu/~jfalward/heattransfer/heattransfer.html

  41. Problem D.7 An aluminium pot contains water that is kept steadily boiling (100 ºC). The bottom surface of the pot, which is 12 mm thick and 1.5104 mm2 in area, is maintained at a temperature of 102 °Cby an electric heating unit. Find the rate at which heat is transferred through the bottom surface. Compare this with a copper based pot. The thermal conductivities for aluminium and copper are kAl = 235 W.m-1.K-1 and kCu = 401 W.m-1.K-1

  42. TC = 100 oC Solution Identify / Setup Base area A = 1.5x104 mm2 = 1.5x10-2 m2 Base thickness L = 12 mm = 12x10-3 m kAl = 235 W.m-1.K-1 kCu = 401 W.m-1.K-1 dT/dx = (TH – TC) / L dQ/dt = ? W TH = 102 oC Execute Al, dQ/dt = 5.9x102 W Cu, dQ/dt = 1.0x103 W Cu pots ~ 2 times more efficient

  43. “Body Heat” Heat lost by convection is very important for humans For a naked person, h ~ 7.1 W.m-2.K-1 Assume the person’s surface area is 1.5 m2, skin temperature of 33 oC and surrounding temperature 29 oC. A = 1.5 m2T = 33 oC TE = 29 oC dQ/dtconvection = h ADT = (7.1)(1.5)(4) W = 43 W If there is a breeze, convection losses are greater – wind chill factor Viscosity of fluids slows natural convection near a stationary surface by producing a boundary layer which has about the same insulating values of 10 mm plywood.

  44. For the naked person, estimate the net rate of energy radiated. A = 1.5 m2T = 33 oC = (33 + 273) K = 306 K TE= (29 + 273) K = 302 K assume e = 1 s = 5.67x10-8 W.m-2.K-4 Pradiation = esA (T4 – TE4) = 39 W dQ/dtloss = dQ/dtradiation + dQ/dtconvection = 39 W + 43 W = 82 W Ans. similar to the rate at which heat is generated by the body when resting

  45. CONVECTION Why not heat the water at the top? http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatra.html

  46. CONVECTION Warm air rises above the ground Forced convection http://sol.sci.uop.edu/~jfalward/heattransfer/heattransfer.html

  47. CONVECTION Why do we get this pollution haze? Temperature inversion prevents air rising and the dispersing the pollution http://sol.sci.uop.edu/~jfalward/heattransfer/heattransfer.html

  48. CONVECTION In bathrooms, the heater is often near the ceiling. Problem ? Why are the cooling coils at the top ? http://sol.sci.uop.edu/~jfalward/heattransfer/heattransfer.html

  49. RADIATION Wien constant bf= 2.83 kB/h s-1.K

  50. RADIATION Why are pipes in solar panels painted black ? http://sol.sci.uop.edu/~jfalward/heattransfer/heattransfer.html

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