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Solved HT problems.ppt

Apr 13, 2016



Ravichandran G
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  • *PHYS1001Physics 1 REGULARModule 2 Thermal Physics




    RADIATION ptD_transfer.ppt

  • *Overview of Thermal Physics Module:Thermodynamic Systems: Work, Heat, Internal Energy 0th, 1st and 2nd Law of ThermodynamicsThermal ExpansionHeat Capacity, Latent HeatMethods of Heat Transfer: Conduction, Convection, RadiationIdeal Gases, Kinetic Theory ModelSecond Law of Thermodynamics Entropy and DisorderHeat Engines, Refrigerators

  • * METHODS OF HEAT TRANSFER energy transfer (heat, Q) due to a temperature difference, T


    17.7 p591system, TQnetEnvironment, TEEuropean heat wave, 2003 ~35 000 deaths in FranceReferences: University Physics 12th ed Young & Freedman

  • *Live sheep tradeSunday, October 26, 2003Sheep to shore ... finallyThe 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.

  • *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. THTCconduction though glassQQ

  • *For conduction between two plane surfaces (eg heat loss through the wall of a house) the rate of heat transfer isenergy transferred through slabQTHTCLThermal conductivity k (W.m-1.K-1)steady-stateheat current H = dQ/dtQQA

  • *steady-stateThermal Conduction through a uniform slab0xLTCTH

  • *Thermal conductivity, k property of the material

    kdiamond very high: perfect heat sink, e.g. for high power laser diodes

    khuman low: core temp relatively constant (37 oC)

    kair very low: good insulator * home insulation * woolen clothing * windows double glazing Metals good conductors: electrons transfer energy from hot to cold

    MaterialThermal conductivity k (W.m-1.K-1)diamond 2450Cu 385Al 205Brick 0.2Glass 0.8Body fat 0.2Water 0.6Wood 0.2Styrofoam 0.01Air 0.024

  • *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, hDT between surface and air way from surface

  • *

  • *Sea & Land Breezes, Monsoons oC20 oC17 oC11 oCWhat is the role of heat capacity, c of water and soil?

  • *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 mVisible ~ 0.8 - 0.4 m 800 400 nmUV ~0.4 - 0.1 mFor exam: more detail than in the textbook

  • *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

  • *Thermography

  • *Power absorbed by surface of an objectA, aQabsTsAbsorption & Stefan-Boltzmann LawIncident radiation (INTENSITY I - energy passing through a square metre every second Iinc = P / A Iabs = a Iinc Surface Area, AAbsorption coefficient, a = 0 to 1Stefan-Boltzmann constant = 5.67 x 10-8 W.m-2.K-4

  • *Power radiated from the surface of an objectA, e, T QradEmission & Stefan-Boltzmann LawPnet > 0 net heat transfer out of system

    Surface Area, AEmissivity, e = 0 to 1Stefan-Boltzmann constant = 5.67 x 10-8 W.m-2.K-4

  • *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 = 5.6710-8 W.m-2.K-4

  • *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 T a = 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)

  • *Problem D.4Semester 1, 2007 Examination

    An igloo is a hemispherical enclosure built of ice. Elmos 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.

  • *SolutionIdentify / Setupthickness t = 0.30 mradius r = 2.55 mkice = 1.67 W.m-1.K-1The 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 conduction6.5 oC-40 oC

  • *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

  • *ExecutedQ/dt = (1.67)(2)(2.70)2(46.5)/0.30 WdQ/dt = 1.2 104 WEvaluatesensible valueunitssignificant figuresdid I answer the question ?

  • *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?

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


    Identify / SetupThermal conductivities kair = 0.024 W.m-1.K-1 kwater = 0.6 W.m-1.K-1



  • *Why do droplets of water dance over the very hot pan ? D.6

  • *Why do droplets of water dance over the very hot pan ? at the bottom of the drops is evaporated and provides insulation against further evaporation.

  • *Problem D.7An aluminium pot contains water that is kept steadily boiling (100 C). The bottom surface of the pot, which is 12 mm thick and 1.5104 mm2in 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

  • *SolutionIdentify / Setup

    TH = 102 oCTC = 100 oCBase 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


    Al, dQ/dt = 5.9x102 W

    Cu, dQ/dt = 1.0x103 W

    Cu pots ~ 2 times more efficient

  • *Body HeatHeat lost by convection is very important for humansFor a naked person, h ~ 7.1 W.m-2.K-1Assume the persons surface area is 1.5 m2, skin temperature of 33 oC and surrounding temperature 29 oC.

    A = 1.5 m2 T = 33 oC TE = 29 oC

    dQ/dtconvection = h A DT = (7.1)(1.5)(4) W = 43 W

    If there is a breeze, convection losses are greater wind chill factorViscosity 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.

  • *For the naked person, estimate the net rate of energy radiated.A = 1.5 m2 T = 33 oC = (33 + 273) K = 306 KTE = (29 + 273) K = 302 K assume e = 1 s = 5.67x10-8 W.m-2.K-4

    Pradiation = e s A (T 4 TE4) = 39 WdQ/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

  • * not heat the water at the top?CONVECTION

  • * air rises above the groundForced convectionCONVECTION

  • *Why do we get this pollution haze?

    Temperature inversion prevents air rising and the dispersing the pollution

  • *Why are the cooling coils at the top ?In bathrooms, the heater is often near the ceiling. Problem ?