1、材料物理材料物理課課件件ch17 ch17 thermal thermal propertiesspropertiess2ISSUES TO ADDRESS. How do materials respond to the application of heat? How do we define and measure. - heat capacity? - thermal expansion? - thermal conductivity? - thermal shock resistance? How do the thermal properties of ceramics, meta

2、ls, and polymers differ?Chapter 17:Thermal Properties3 Quantitatively: The energy required to produce a unit rise in temperature for one mole of a material.heat capacity(J/mol-K)energy input (J/mol)temperature change (K)17.2 Heat Capacity Two ways to measure heat capacity:Cp : Heat capacity at const

3、ant pressure.Cv : Heat capacity at constant volume.Cp usually Cv Heat capacity has units of KmolJdTdQC The ability of a material to absorb heat6increasing cp Why is cp significantly larger for polymers? PolymersPolypropylene Polyethylene Polystyrene Tefloncp (J/kg-K)at room T CeramicsMagnesia (MgO)A

4、lumina (Al2O3)Glass MetalsAluminum Steel Tungsten Gold1925 1850 1170 1050900 486 138 128cp (specific heat): (J/kg-K)Material940 775 840Specific Heat: ComparisonCp (heat capacity): (J/mol-K)717.3 Thermal ExpansionMaterials change size when temperature is changed lfinallinitiallinitiall(TfinalTinitial

5、)linear coefficient ofthermal expansion (1/K or 1/C)TinitialTfinal initial finalTfinal Tinitial8Atomic Perspective: Thermal ExpansionAdapted from Fig. 19.3, Callister & Rethwisch 8e.Asymmetric curve: - increase temperature, - increase in interatomic separation - thermal expansionSymmetric curve: - i

6、ncrease temperature, - no increase in interatomic separation - no thermal expansion9Coefficient of Thermal Expansion: Comparison Q: Why does generally decrease with increasing bond energy?Polypropylene145-180 Polyethylene 106-198 Polystyrene 90-150 Teflon126-216 Polymers CeramicsMagnesia (MgO)13.5Al

7、umina (Al2O3)7.6Soda-lime glass 9Silica (cryst. SiO2)0.4 MetalsAluminum 23.6Steel 12 Tungsten 4.5 Gold14.2 (10-6/C)at room TMaterialPolymers have larger values because of weak secondary bondsincreasing 10Thermal Expansion: ExampleEx: A copper wire 15 m long is cooled from 40 to -9C. How much change

8、in length will it experience? 16.5 x106(oC)1 Answer: For Cu mm 12m 012.0 )C9(C40 )m 15)(C/1(10 x5.16 60Trearranging Equation 17.3b11The ability of a material to transport heat.temperaturegradientthermal conductivity (J/m-K-s)heat flux(J/m2-s) Atomic perspective: Atomic vibrations and free electrons

9、in hotter regions transport energy to cooler regions.T2 T2 T1 T1x1x2heat flux17.4 Thermal ConductivitydxdTkqFouriers Law12Thermal Conductivity: Comparisonincreasing k PolymersPolypropylene0.12Polyethylene 0.46-0.50 Polystyrene 0.13 Teflon0.25vibration/rotation of chain molecules CeramicsMagnesia (Mg

10、O)38Alumina (Al2O3)39 Soda-lime glass 1.7 Silica (cryst. SiO2)1.4atomic vibrations MetalsAluminum 247Steel 52 Tungsten 178 Gold315atomic vibrations and motion of free electronsk (W/m-K)Energy TransferMechanismMaterial1314 Occur due to: - restrained thermal expansion/contraction - temperature gradien

11、ts that lead to differential dimensional changes17.5 Thermal Stresses E(T0Tf) ETThermal stress - A brass rod is stress-free at room temperature (20C). - It is heated up, but prevented from lengthening. - At what temperature does the stress reach -172 MPa?Example ProblemT0 0Solution:Original conditio

12、ns room thermal (TfT0)TfStep 1: Assume unconstrained thermal expansion 0 Step 2: Compress specimen back to original length 0 compressroom thermal 1516Example Problem (cont.) 0The thermal stress can be directly calculated as E(compress) E(thermal) E(TfT0) E(T0Tf)Noting that compress = -thermal and su

13、bstituting gives20 x 10-6/CAnswer: 106C100 GPa TfT0E20CRearranging and solving for Tf gives-172 MPa (since in compression)17 Occurs due to: nonuniform heating/cooling Ex: Assume top thin layer is rapidly cooled from T1 to T2Tension develops at surface E(T1T2)Critical temperature differencefor fractu

14、re (set = f) (T1T2)fracturefEset equal Large TSR when is large fkEThermal Shock ResistanceTemperature difference thatcan be produced by cooling:kTTrate quench)(21rapid quenchresists contractiontries to contract during coolingT2T1 (quench rate)for fracture Thermal Shock Resistance (TSR)fkE18 Applicat

15、ion:Space Shuttle Orbiter Silica tiles (400-1260C):- large scale application- microstructure:Thermal Protection Systemreinf C-C (1650C)Re-entry T Distributionsilica tiles(400-1260C)nylon felt, silicon rubbercoating (400C)90% porosity!Si fibersbonded to oneanother duringheat treatment.100 mm1920The thermal properties of materials include: Heat capacity: - energy required to increase a mole of material by a unit T - energy is stored as atomic vibrations Coefficient of thermal expansion: - the size of a material changes with a change in temperature - polymers have the largest values Therma