Presentation on theme: "Chapter 17: Thermal Properties"— Presentation transcript:
1 Chapter 17: Thermal Properties A white-hot cube of a silica fiber insulation material, which, only seconds after having been removed from a hot furnace, can be held by its edges with the bare hands. Initially, the heat transfer from the surface is relatively rapid; however, the thermal conductivity of this material is so small that heat conduction from the interior [maximum temperature approximately 1250C (2300F)] is extremely slow. This material was developed especially for the tiles that cover the Space Shuttle Orbiters and protect and insulate them during their fiery reentry into the atmosphere. Other attractive features of this high-temperature reusable surface insulation (HRSI) include low density and a low coefficient of thermal expansion.Chapter 17: Thermal Properties
2 CHAPTER 17: THERMAL PROPERTIES 17 CHAPTER 17: THERMAL PROPERTIES 17.1 Introduction Thermal property: Response of materials to the application of heatISSUES TO ADDRESS...• How does a material respond to heat?• How do we define and measure...--heat capacity--coefficient of thermal expansion--thermal conductivity--thermal shock resistance• How do ceramics, metals, and polymers rank?1
3 17.2 HEAT CAPACITY• General: The ability of a material to absorb heat.• Quantitative: The energy required to increase thetemperature of the material.energy input (J/mol)heat capacity(J/mol-K)temperature change (K)• Two ways to measure heat capacity:-- Cp : Heat capacity at constant pressure.-- Cv : Heat capacity at constant volume.2
4 Vibrational Heat Capacity Generation of lattice waves in a crystal by atomic vibrations.The phonon versus photonc17f01c17f01
5 Heat Capacity vs T c17f02 Cv = constant = ~3R • Atomic view: --increases with temperature--reaches a limiting value of 3Rc17f02Cv = constant = ~3R• Atomic view:--Energy is stored as atomic vibrations.--As T goes up, energy of atomic vibration goes up tooThe temperature dependence of the heat capacity at constant volume.qD = Debye temperature qD = ħnmax/kqD < Troom
6 HEAT CAPACITY: COMPARISON • Why is cp significantlylarger for polymers?Selected values from Table 19.1, Callister 6e.4
7 17.3 THERMAL EXPANSION • Materials change size when heating. coefficient ofthermal expansion (1/K)• Atomic view: Mean bond length increases with T.Adapted from Fig. 19.3(a), Callister 6e. (Fig. 19.3(a) adapted from R.M. Rose, L.A. Shepard, and J. Wulff, The Structure and Properties of Materials, Vol. 4, Electronic Properties, John Wiley and Sons, Inc., 1966.)5
8 c17f03c17f03Potential energy versus interatomic distance. Interatomic separation increases with rising temperature. With heating, the interatomic separation increases from r0 to r1 to r2, and so on.For a symmetric potential energy-versus-interatomic distance curve, there is no increase in interatomic separation with rising temperature (i.e., r1 r2 r3).
9 THERMAL EXPANSION: COMPARISON • Q: Why does agenerally decreasewith increasingbond energy?Selected values from Table 19.1, Callister 6e.For thermal expansion of fractional volumeFor isotropic materials av = ~3al6
10 17.4 THERMAL CONDUCTIVITY• General: The ability of a material to transfer heat.• Quantitative:temperaturegradientheat flux(J/m2-s)thermal conductivity (J/m-K-s)• Atomic view: Atomic vibrations in hotter region carryenergy (vibrations) to cooler regions.7
11 THERMAL CONDUCTIVITY for pure metals Selected values from Table 19.1, Callister 6e.for pure metals8
12 Thermal conductivity versus composition for copper–zinc alloys. c17f04Impurities decrease thermal conductivity (scattering centers in solid solutions)
13 Dependence of thermal conductivity on temperature for ceramics c17f05Nonmetallic materialsThermal insulatorsPhonons for thermal conductionPhonon scattering by imperfectionsAt higher T, radiant heat transferPorosity: increasing pore volume reduces thermal conductivity also gaseous convection ineffective
14 17.5 Thermal Stresses REVIEW OF ELASTIC PROPERTIES • Modulus of Elasticity, E:(also known as Young's modulus)• Hooke's Law:s = E eE: [GPa] or [psi]s = E es = stressE = modulus of elasticity= displacement10
15 17.5 THERMAL STRESSES • Occurs due to: • Example: --uneven heating/cooling--mismatch in thermal expansion.• Example:--A brass rod is stress-free at room temperature (20°C).--It is heated up, but prevented from lengthening.--At what T does the stress reach -172 MPa?100GPa20 x 10-6 /C20C-172MPaAnswer: 106C9
17 THERMAL SHOCK RESISTANCE • Occurs due to: uneven heating/cooling.• Ex: Assume top thin layer is rapidly cooled from T1 to T2:Tension develops at surfaceTemperature difference thatcan be produced by cooling:Critical temperature differencefor fracture (set s = sf)set equal• Result:• Large thermal shock resistance when is large.10
21 SUMMARY • A material responds to heat by: • Heat capacity: --increased vibrational energy--redistribution of this energy to achieve thermal equil.• Heat capacity:--energy required to increase a unit mass by a unit T.--polymers have the largest values.• Coefficient of thermal expansion:--the stress-free strain induced by heating by a unit T.• Thermal conductivity:--the ability of a material to transfer heat.--metals have the largest values.• Thermal shock resistance:--the ability of a material to be rapidly cooled and notcrack. Maximize sfk/Ea.12
22 C=dQ/dT c=(1/m) dQ/dT Heat Capacity To what temperature would 10 lbm of a brass specimen at 25°C (77°F) be raised if 65 Btu of heat is supplied?SolutionWe are asked to determine the temperature to which 10 lbm of brass initially at 25°C would be raised if 65 Btu of heat is supplied. This is accomplished by utilization of a modified form of Equation 17.1 asC=dQ/dT c=(1/m) dQ/dTin which DQ is the amount of heat supplied, m is the mass of the specimen, and cp is the specific heat. From Table 17.1, cp = 375 J/kg-K for brass, which in Customary U.S. units is justThusand
23 Briefly explain why Cv rises with increasing temperature at temperatures near 0 K. Briefly explain why Cv becomes virtually independent of temperature at temperatures far removed from 0 K.Solution(a) Cv rises with increasing temperature at temperatures near 0 K because, in this temperature range, the allowed vibrational energy levels of the lattice waves are far apart relative to the available thermal energy, and only a portion of the lattice waves may be excited. As temperature increases, more of the lattice waves may be excited by the available thermal energy, and, hence, the ability of the solid to absorb energy (i.e., the magnitude of the heat capacity) increases.(b) At temperatures far removed from 0 K, Cv becomes independent of temperature because all of the lattice waves have been excited and the energy required to produce an incremental temperature change is nearly constant.
24 Thermal ExpansionA copper wire 15 m (49.2 ft) long is cooled from 40 to –9°C (104 to 15°F). How much change in length will it experience?SolutionIn order to determine the change in length of the copper wire, we must employ a rearranged form of Equation 17.3b and using the value of al taken from Table 17.1 [17.0 10-6 (°C)-1] as
25 Briefly explain why metals are typically better thermal conductors than ceramic materials. SolutionMetals are typically better thermal conductors than are ceramic materials because, for metals, most of the heat is transported by free electrons (of which there are relatively large numbers). In ceramic materials, the primary mode of thermal conduction is via phonons, and phonons are more easily scattered than are free electrons.
26 For some ceramic materials, why does the thermal conductivity first decrease and then increase with rising temperature?SolutionFor some ceramic materials, the thermal conductivity first decreases with rising temperature because the scattering of lattice vibrations increases with temperature. At higher temperatures, the thermal conductivity will increase for some ceramics that are porous because radiant heat transfer across pores may become important, which process increases with rising temperature.
27 = 5 × 105 g/mol). = 106 g/mol); isotactic polypropylene ( For each of the following pairs of materials, decide which has the larger thermal conductivity. Justify your choices.(a) Fused silica; polycrystalline silica.(b) Atactic polypropylene (= 106 g/mol); isotactic polypropylene (= 5 × 105 g/mol).Solution(a) Polycrystalline silica will have a larger conductivity than fused silica because fused silica is noncrystalline and lattice vibrations are more effectively scattered in noncrystalline materials.(b) The isotactic polypropylene will have a larger thermal conductivity than the atactic polypropylene because isotactic polymers have a higher degree of crystallinity. Since heat transfer is accomplished by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.
28 What measures may be taken to reduce the likelihood of thermal shock of a ceramic piece? SolutionAccording to Equation 17.9,the thermal shock resistance of a ceramic piece may be enhanced by increasing the fracture strength and thermal conductivity, and by decreasing the elastic modulus and linear coefficient of thermal expansion. Of these parameters, sf and al are most amenable to alteration, usually be changing the composition and/or the microstructure.