Chapter 17: Thermal PropertiesA 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.

ISSUES TO ADDRESS...

• How does a material respond to heat?

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• How do we define and measure... --heat capacity --coefficient of thermal expansion --thermal conductivity --thermal shock resistance

• How do ceramics, metals, and polymers rank?

CHAPTER 17:THERMAL PROPERTIES

17.1 IntroductionThermal property: Response of materials to

the application of heat

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• General: The ability of a material to absorb heat.• Quantitative: The energy required to increase the temperature of the material.

C

dQdT

heat capacity(J/mol-K)

energy input (J/mol)

temperature change (K)

• Two ways to measure heat capacity: -- Cp : Heat capacity at constant pressure.

-- Cv : Heat capacity at constant volume.

17.2 HEAT CAPACITY

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Vibrational Heat CapacityGeneration of lattice waves in a crystal by atomic vibrations.The phonon versus photon

The temperature dependence of the heat capacity at constant volume.

D = Debye temperature D = ħmax/kD < Troom

Cv = constant = ~3R

Heat Capacity vs T --increases with temperature--reaches a limiting value of 3R

• Atomic view: --Energy is stored as atomic vibrations. --As T goes up, energy of atomic vibration goes up too

Debye temperature (usually less than Troom)

T (K)

Heat capacity, Cv3R

D

Cv= constant

gas constant = 8.31 J/mol-K

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• Why is cp significantly larger for polymers?

Selected values from Table 19.1, Callister 6e.

HEAT CAPACITY: COMPARISON

• PolymersPolypropylene Polyethylene Polystyrene Teflon

cp (J/kg-K) at room T

• CeramicsMagnesia (MgO) Alumina (Al2O3) Glass

• MetalsAluminum Steel Tungsten Gold

1925 1850 1170 1050

900 486 128 138

incr

easi

ng c

p

cp: (J/kg-K) Cp: (J/mol-K)

material

940 775

840

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• Materials change size when heating.

• Atomic view: Mean bond length increases with T.

Lfinal LinitialLinitial

(Tfinal Tinitial)

coefficient ofthermal expansion (1/K)

Tinit

TfinalLfinal

Linit

Bond energy

Bond length (r)

incr

easi

ng

T

T1

r(T5)

r(T1)

T5bond energy vs bond length curve is “asymmetric”

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.)

17.3 THERMAL EXPANSION

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Potential 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).

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• PolymersPolypropylene Polyethylene Polystyrene Teflon

145-180 106-198 90-150 126-216

(10-6/K) at room T

• CeramicsMagnesia (MgO) Alumina (Al2O3) Soda-lime glass Silica (cryst. SiO2)

13.5 7.6 9 0.4

• MetalsAluminum Steel Tungsten Gold

23.6 12 4.5 14.2

incr

easi

ng

Material

• Q: Why does generally decrease with increasing bond energy?

Selected values from Table 19.1, Callister 6e.

THERMAL EXPANSION: COMPARISON

For thermal expansion of fractional volumeFor isotropic materials v = ~3l

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• General: The ability of a material to transfer heat.• Quantitative:

q k

dTdx

temperaturegradient

thermal conductivity (J/m-K-s)

heat flux(J/m2-s)

• Atomic view: Atomic vibrations in hotter region carry energy (vibrations) to cooler regions.

T2 > T1 T1

x1 x2heat flux

17.4 THERMAL CONDUCTIVITY

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• PolymersPolypropylene Polyethylene Polystyrene Teflon

0.12 0.46-0.50 0.13 0.25

k (W/m-K)

• CeramicsMagnesia (MgO) Alumina (Al2O3) Soda-lime glass Silica (cryst. SiO2)

38 39 1.7 1.4

• MetalsAluminum Steel Tungsten Gold

247 52 178 315

incr

easi

ng k

By vibration/ rotation of chain molecules

Energy Transfer

By vibration of atoms

By vibration of atoms and motion of electrons

Material

Selected values from Table 19.1, Callister 6e.

THERMAL CONDUCTIVITY

for pure metals

Thermal conductivity versus composition for copper–zinc alloys.

Impurities decrease thermal conductivity (scattering centers in solid solutions)

Dependence of thermal conductivity on temperature for ceramics

Nonmetallic materialsThermal insulatorsPhonons for thermal conductionPhonon scattering by imperfectionsAt higher T, radiant heat transfer

Porosity: increasing pore volume reduces thermal conductivity also gaseous convection ineffective

• Modulus of Elasticity, E: (also known as Young's modulus)

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• Hooke's Law:

= E

F

Fsimple tension test

Linear- elastic

1E

E: [GPa] or [psi]

17.5 Thermal Stresses

REVIEW OF ELASTIC PROPERTIES

= E

= stressE = modulus of elasticity= displacement

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• Occurs due to: --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?

LLroom

thermal(T Troom)

Troom

LroomT

L

compressive keeps L = 0 E( thermal) E(T Troom)

100GPa 20 x 10-6 /C

20CAnswer: 106C-172MPa

17.5 THERMAL STRESSES

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• Occurs due to: uneven heating/cooling.

• Ex: Assume top thin layer is rapidly cooled from T1 to T2:

rapid quench

doesn’t want to contract

tries to contract during coolingT2T1

Tension develops at surface

E(T1 T2)

Critical temperature differencefor fracture (set = f)

(T1 T2)fracture

fE

Temperature difference thatcan be produced by cooling:

(T1 T2)

quenchratekset equal

• Result:

• Large thermal shock resistance when is large.

fkE

(quenchrate)for fracture

fkE

THERMAL SHOCK RESISTANCE

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• Application:Space Shuttle Orbiter

• Silica tiles (400-1260C):--large scale application --microstructure:

100m

~90% porosity!Si fibersbonded to oneanother duringheat treatment.

Fig. 23.0, Callister 5e. (Fig. 23.0 courtesy the National Aeronautics and Space Administration.

reinf C-C (1650°C)

Re-entry T Distribution

silica tiles (400-1260°C)

nylon felt, silicon rubber coating (400°C)

Fig. 19.2W, Callister 6e. (Fig. 19.2W adapted from L.J. Korb, C.A. Morant, R.M. Calland, and C.S. Thatcher, "The Shuttle Orbiter Thermal Protection System", Ceramic Bulletin, No. 11, Nov. 1981, p. 1189.)

Fig. 19.3W, Callister 5e. (Fig. 19.3W courtesy the National Aeronautics and Space Administration.

Fig. 19.4W, Callister 5e. (Fig. 219.4W courtesy Lockheed Aerospace CeramicsSystems, Sunnyvale, CA.)

THERMAL PROTECTION SYSTEM

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Low expansion alloys

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• A material responds to heat by: --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. --polymers have the largest values.• 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 not crack. Maximize fk/E.

SUMMARY

T = Q

m cp

cp = (375 J/kg - K)2.39 10 4 Btu /lbm - F

1 J /kg - K

= 0.090 Btu/lbm - F

T = 65 Btu

(10 lbm)(0.090 Btu /lbm - F)= 72.2 F

Tf = T0 + T = 77 F + 72.2 F = 149.2 F (65.1C)

Heat CapacityTo 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 25C would be raised if 65 Btu of heat is supplied. This is accomplished by utilization of a modified form of Equation 17.1 as

in which Q 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 just

Thus

and

C=dQ/dT c=(1/m) dQ/dT

(a) Briefly explain why Cv rises with increasing temperature at temperatures near

0 K.

(b) 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.

l = l0lT = l0l (Tf T0)

= (15 m) 17.0 10 6 (C)-1 ( 9C 40C)

= 1.25 10-2 m = 12.5 mm ( 0.49 in.)

Thermal Expansion

A 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 l taken from Table 17.1 [17.0 10-6 (C)-1] as

Briefly explain why metals are typically better thermal conductors than ceramic materials.

Solution

Metals 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.

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.

M w

M w

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 (

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.

= 5 × 105 g/mol).

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, f and l are most amenable to alteration, usually be changing the composition and/or the microstructure.