issues to address... how does a material respond to heat ? 1 how do we define and measure... --heat...
TRANSCRIPT
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
• Motivation– Many applications necessitate a thorough understanding of how
materials respond to temperature changes– How does a material respond to heat?
• Much of this will be a refresher for you (especially if you have taken/are taking heat transfer)!
Chapter 17 – Thermal Properties
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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.
HEAT CAPACITY
• Vibrational Heat capacity– In most solids the principal mode of heat absorption
is by increasing the vibrational energy of the atoms– Atoms are constantly vibrating at high frequencies– Atomic vibrations are coupled because of bonding– Waves have very high frequencies and short
wavelengths, and travel at the velocity of sound– From quantum mechanics – the vibrational energies
must have well-defined (quantized) values• A single quantum of vibration energy is called a phonon
– Tie to Ch 12 – the thermal scattering of free electrons during conduction is by phonons
Chapter 17 – Thermal Properties
How does Cv vary with temperature?The heat capacity is zero at zero 0 KStrongly nonlinear below the Debye temperaturePlateau above the Debye temperature
3ATCv
Vibrational contributionto heat capacity
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• Heat capacity... --increases with temperature --reaches a limiting value of 3R
• Atomic view: --Energy is stored as atomic vibrations. --As T goes up, so does the avg. energy of atomic vibr.
Debye temperature (usually less than Troom)
T (K)
Heat capacity, Cv3R
D
Cv= constant
gas constant = 8.31 J/mol-K
Adapted from Fig. 19.2, Callister 6e.
HEAT CAPACITY VS T
• Other heat capacity contributions– Other mechanisms for heat absorption are possible, though are
usually minor compared to the vibrational contribution– Electronic contribution – electrons increase their KE with heat
absorption (do you think this is significant?)• Must be free electrons to contribute to this
– Randomization of spins in a ferromagnetic material produces heat at T > Curie temperature
Chapter 17 – Thermal Properties
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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.)
THERMAL EXPANSION
• Thermal expansion– How does a sample’s length change upon heating/cooling?
Chapter 17 – Thermal Properties
ofloo
of TTl
l
l
ll
• lf and lo are the final and initial lengths, l – linear coefficient of thermal expansion l [=] 1/K
• Could also express this in terms of volume
• Molecular picture – expansion is due to increase in spacing of atoms
Asymmetric curvature of the potential wellThe shape/depth of the potential well is related to bonding strength – stronger bonds, smaller l
Increase of interatomic separation with increasing T
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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
• Thermal properties (relate to bonding strength)– Metals: l typically between 5 – 25 x 10-6 1/C
• Can make alloys with lower values
– Ceramics – low expansion coefficients (0.5–15 x 10-6 1/C); coefficient can be anisotropic
– Polymers – can have very high expansion coefficients
Chapter 17 – Thermal Properties
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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
THERMAL CONDUCTIVITY
Similar to Fick’sfirst law
• Thermal conductivity– Mechanism for transport of heat from high to
low temperature domains
Chapter 17 – Thermal Properties
dx
dTkq q – heat flux (W/m2)
k – thermal conductivity (W/m-K)
• Mechanisms of heat conduction– Two contributions: lattice vibration waves (phonons) and free
electrons
el kkk
Thermal conductivity in metals
• Metals – Free electron
mechanism dominates (k ~ 20 – 400 W/m-K)
– Free electrons participate both in thermal and electrical conduction – Wiedemann-Franz law
– L is a constant
T
kL
Effect of impurities on thermal conductivity
• Ceramics– Very few free electrons … phonon processes dominate the heat
conduction process (k ~ 2 – 50 W/m-K)– Amorphous materials have lower thermal conductivities than
crystalline solids (why?)– Most ceramics conductivity goes down when temperature
increases (why?) But at very high temperature, radiation takes place also
– Porosity leads to reduction (dramatic) in thermal conductivity (why?)
• Polymers– k is small ~ 0.3 W/m-K– Energy transfer due to vibration and rotation of the chains– k depends on polymer crystallinity – Typically viewed as thermal insulators
Chapter 17 – Thermal Properties
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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: COMPARISON
• Thermal stresses– Induced as a result of temperature changes– Consider an isotropic solid rod heated or cooled uniformly; no T
gradient imposed. However if the axial motion of the rod is restrained, introduces stress (think back to thermal expansion!)
– Magnitude of the stress is given by
Chapter 17 – Thermal Properties
TETTE lfol
Stresses caused by temperature gradients:Heating – compressive stressCooling – tensile stress
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• Occurs due to: --uneven heating/cooling --mismatch in thermal expansion.
• Example Problem 17.1, p. 724, Callister 2e. --A brass rod is stress-free at room temperature (20C). --It is heated up, but prevented from lengthening (ends are kept rigid). --At what T does the stress reach -172MPa?
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
EX: THERMAL STRESS
• Thermal shock (brittle materials)– So, change in temperature can
cause stress– If these are large enough can
have plastic deformation (for metals)
– But brittle materials can’t deform plastically!
– Brittle materials can fracture due to large temperature changes (thermal shock)
– Cooling more problematic – why?– This is a problem because
ceramics are very good insulators– How to know when thermal shock
can happen – thermal shock resistance parameter
Chapter 17 – Thermal Properties
l
f
E
kTSR
High fracture strengthsHigh thermal conductivitiesLow elasticity modulusLow coeff. thermal expansion
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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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• 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
Reading: Chapter 17 (pdf sent to you by e-mail)
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