Lecture 18

Today’s objectives-Thermal Properties
How do atoms in crystals respond to temperature?
What is the atomic mechanism that leads to the coefficient of thermal expansion?
How do we define and measure:
Coefficient of thermal expansion?
Heat capacity?
Thermal conductivity?
Thermoelectric heating and cooling?
Thermal shock resistance?
Be able to calculate the thermal expansion for materials if provided the details.

Scientists do stupid looking things sometimes (though not too unsafe if they made the material carefully enough)

THERMAL EXPANSION • Materials change size when heating. CTE: coefficient of thermal expansion (units: 1/K)

Flashback: PROPERTIES FROM BONDING: Energy versus bond length • Bond length , r • Bond energy , E o

PROPERTIES FROM BONDING: T M • Melting Temperature , T m T m is larger if E o is larger.

PROPERTIES FROM BONDING: Elastic Properties • Elastic modulus , E • E ~ curvature at r o E is larger if curvature is larger. E similar to spring constant

PROPERTIES FROM BONDING: CTE or  • Coefficient of thermal expansion ,  •  ~ symmetry at r o
is larger if E o is smaller
and very asymmetric.

Atomic positions and vibrations
The minimum in an atomic energy vs. interatomic distance curve yields the near neighbor distance (bond length).
The width of the curve is proportional to the amplitude of thermal vibrations for an atom .
If the curve is symmetric, there is no shift in the average position of the atom (the center of the thermal vibrations at any given T).
The coefficient of thermal expansion is negligible for symmetric energy wells.
T 0 T 2 T 3

Thermal Expansion
If the curve is not symmetric, the average position in which the atom sits shifts with temperature.
Bond lengths therefore change (usually get bigger for increased T).
Thermal expansion coefficient is nonzero.

THERMAL EXPANSION: COMPARISON Why does  generally decrease with increasing bond energy? Selected values from Table 19.1, Callister 6e .
Thermal expansion mismatch is a major problem for design of everything from semiconductors to bridges.
Particularly an issue in applications where temperature changes greatly (esp. engines).

Thermal expansion example
Example: 19.9:
An Al wire is 10 m long and is cooled from 38 to -1 degree Celsius. How much change in length will it experience?
-9.2 mm

Small/Negative thermal expansion
Invar (Ni-Fe alloy) is the most common low thermal exp material: α = 1.6*10 -6 / degree
Some materials have α <0 in one dimension and >0 in others.
It is possible, though not intuitive, for materials to have a negative thermal expansion in all dimensions.
An increase in temperature causes the crystal to shrink.
ZrW 2 O 8 : contracts continuously and linearly from 2 to 1050K
Composites could allow zero thermal expansion components (superb for optics, engine parts, etc).
http://www.dur.ac.uk/john.evans/webpages/research_nteintro.html

Heat and Atoms
Heat causes atoms to vibrate.
Vibrating in synch is often a low energy configuration (preferred).
Generates waves of atomic motion.
Often called phonons , similar to photons but atomic motion instead of optical quanta.

HEAT CAPACITY Capacity at constant volume = C V Capacity at constant pressure = C P C P is typically > C V , but the difference is small for solids. When heated, materials experience an increase in T. This means that heat is absorbed. Heat capacity represents the amount of energy required to produce a unit temperature rise. H 2 O has a higher heat capacity

HEAT CAPACITY 3 N 0 k b  D =Temperature at which  D (Cu)  D (Al)  D (Pb)

THERMAL CONDUCTIVITY • General: The ability of a material to transfer heat. • Quantitative: Atomic view: Electronic and/or Atomic vibrations in hotter region carry energy (vibrations) to cooler regions. In a metal, electrons are free and thus dominate thermal conductivity. In a ceramic, phonons are more important. Fick’s First Law temperature gradient k=thermal conductivity (J/m-K-s): Defines material’s ability to transfer heat. heat flux (J/m 2 -s)

THERMAL CONDUCTIVITY Fick’s Second Law • Non-Steady State: dT/dt is not constant.

Selected values from Table 19.1, Callister 6e . K=k l +k e : Again think about band gaps: metals have lots of free electrons (k e is large), while ceramics have few (only k l is active). THERMAL CONDUCTIVITY

Good heat conductors are usually good electrical conductors .
(Wiedemann & Franz, 1853)
Thermal conductivity changes by 4 orders of magnitude (~ 25 for electrical conductivity ).
Metals & Alloys: free e- pick up energy due to thermal vibrations of atoms as T increases and lose it when it decreases .
Insulators (Dielectrics): no free e- . Phonons (lattice vibration quanta) are created as T increases , eliminated as it decreases .
THERMAL CONDUCTIVITY

Thermal conductivity is temperature dependent.
Analagous to electron scattering.
Usually first decreases with increasing temperature
Higher Temp=more scattering of electrons AND phonons, thus less transfer of heat.
Then increases at still higher temperatures due to other processes we haven‘t considered in this class (radiative heat transfer—eg. IR lamps).
THERMAL CONDUCTIVITY

Thermal conductivity optimization
To maximize thermal conductivity, there are several options:
Provide as many free electrons (in the conduction band) as possible
free electrons conduct heat more efficiently than phonons.
Make crystalline instead of amorphous
irregular atomic positions in amorphous materials scatter phonons and diminish thermal conductivity
Remove grain boundaries
gb’s scatter electrons and phonons that carry heat
Remove pores (air is a terrible conductor of heat)

THERMAL STRESSES • Occurs due to: --uneven heating/cooling --mismatch in thermal expansion. • Example Problem 19.1, p. 666, Callister 6e . --A brass rod is stress-free at room temperature (20C). --It is heated up, but prevented from lengthening. --At what T does the stress reach -172MPa? Answer: 106C -172MPa 100GPa 20 x 10 -6 /C 20C Strain ( ε ) due to ∆ T causes a stress ( σ ) that depends on the modulus of elasticity (E):

THERMAL SHOCK RESISTANCE • Thermal shock is fracture of brittle ceramics due to asymmetric thermal expansion. • Occurs due to: uneven heating/cooling. • The change in T with position leads to built in strain and thus stress. • Ex: Assume top thin layer is rapidly cooled from T 1 to T 2 : Tension develops at surface Critical temperature difference for fracture (set  =  f ) Temperature difference that can be produced by cooling: • Result: • Large thermal shock resistance when is large. set equal

THERMAL PROTECTION SYSTEM • Application: Space Shuttle Orbiter Fig. 23.0, Callister 5e . (Fig. 23.0 courtesy the National Aeronautics and Space Administration. 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.) • Silica tiles (400-1260C) : --large scale application Fig. 19.3W, Callister 5e . (Fig. 19.3W courtesy the National Aeronautics and Space Administration. --microstructure: ~90% porosity! Si fibers bonded to one another during heat treatment. Fig. 19.4W, Callister 5e . (Fig. 219.4W courtesy Lockheed Aerospace Ceramics Systems, Sunnyvale, CA.)

THERMOELECTRIC COOLING & HEATING Two different materials are connected at the their ends and form a loop. One junction is heated up. There exists a potential difference that is proportional to the temperature difference between the ends.

THERMOELECTRIC COOLING & HEATING Reversion of the Seebeck effect is the Peltier Effect. A direct current flowing through heterojunctions causes one junction to be cooled and one junction to be heated up. Lead telluride and or bismuth telluride are typical materials in thermoelectric devices that are used for heating and refrigeration .

THERMOELECTRIC COOLING & HEATING Why does this happen? When two different electrical conductors are brought together, e- are transferred from the material with higher E F to the one with the lower E F until E F (material 1)= E F (material 2). Material with smaller E F will be (-) charged. This results in a contact potential which depends on T. e- at higher E F are caused by the current to transfer their energy to the material with lower E F , which in turn heats up . Material with higher E F loses energy and cools down .

THERMOELECTRIC COOLING & HEATING Peltier–Seebeck effect, or the thermoelectric effect , is the direct conversion of thermal differentials to electric voltage and vice versa. The effect for metals and alloys is small , microvolts/K . For Bi 2 Te 3 or PbTe ( semiconductors ), it can reach up to millivolts/K . Applications: Temperature measurement via thermocouples (copper/constantan, Cu-45%Ni, chromel, 90%Ni-10%Cr,…); thermoelectric power generators (used in Siberia and Alaska); thermoelectric refrigerators ; thermal diode in microprocessors to monitor T in the microprocessors die or in other thermal sensor or actuators.

THERMOELECTRIC COOLING & HEATING http://www.sii.co.jp/info/eg/thermic_main.html

Review of thermal conductivity:
19.18 (a) Is thermal conductivity better for a single crystal or a polycrystal?
The thermal conductivity of a single crystal is greater than a polycrystalline specimen of the same material because both phonons and free electrons are scattered at grain boundaries, thus decreasing the efficiency of thermal transport.
19.19 Is thermal conductivity better for crystalline or amorphous?
Thermal conductivities are higher for crystalline than for noncrystalline ceramics because, for noncrystalline, phonon scattering, and thus the resistance to heat transport, is much more effective due to the highly disordered and irregular atomic structure.
19.20 Is thermal conductivity better for ceramics or metals?
Metals are typically better thermal conductors than 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.
19.21 Is thermal conductivity better for porous or dense materials?
(a) Porosity decreases the thermal conductivity of ceramic and polymeric materials because the thermal conductivity of a gas phase that occupies pore space is extremely small relative to that of the solid material. Furthermore, contributions from gaseous convection are generally insignificant.

SUMMARY
How do atoms in crystals respond to temperature?
What is the atomic mechanism that leads to the coefficient of thermal expansion?
How do we define and measure:
coefficient of thermal expansion?
Heat capacity?
thermal conductivity?
Thermoelectric heating and cooling
thermal shock resistance?
Be able to calculate the thermal expansion for materials if provided the details.
Reading for next class: Chapter 20

Lecture 18

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