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

• Materials change size when heating.

Lfinal LinitialLinitial

(Tfinal Tinitial)

CTE: coefficient ofthermal expansion (units: 1/K)

Tinit

TfinalLfinal

Linit

THERMAL EXPANSION

• Bond length, r

• Bond energy, Eo

F F

r

Eo=

“bond energy”

Energy (r)

ro r

unstretched length

Flashback: PROPERTIES FROM BONDING:Energy versus bond length

• Melting Temperature, Tm

r

larger Tm

smaller Tm

Energy (r)

ro

Tm is larger if Eo is larger.

PROPERTIES FROM BONDING: TM

• Elastic modulus, E

• E ~ curvature at ro

cross sectional area Ao

L

length, Lo

F

undeformed

deformed

L F Ao

= E Lo

Elastic modulus

r

larger Elastic Modulus

smaller Elastic Modulus

Energy

ro unstretched length

E is larger if curvature is larger.

PROPERTIES FROM BONDING: Elastic Properties

E similar to spring constant

• Coefficient of thermal expansion,

• ~ symmetry at ro

is larger if Eo is smaller and very asymmetric.

L

length, Lo

unheated, T1

heated, T2

= (T2-T1) L Lo

coeff. thermal expansion

r

smaller

larger

Energy

ro

PROPERTIES FROM BONDING: CTE or

T0

T2

T3

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.

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.

Bond energy

Bond length (r)

incr

easi

ng

T

T1

r(T5)

r(T1)

T5bond energy vs bond length curve is “asymmetric”

• 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

Why does generally decrease with increasing

bond energy?

Selected values from Table 19.1, Callister 6e.

THERMAL EXPANSION: COMPARISON•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• An Al wire is 10 m long and is cooled from 38 to -1

degree Celsius. How much change in length will it experience? l = lo lT

= (10 m) 23.6 x 10 6(C)-1 ( 1C 38C)

-9.2 mm

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.

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

q k

dTdx

temperaturegradient

k= thermal conductivity (J/m-K-s): Defines material’s ability to transfer heat.

heat flux(J/m2-s)

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.

T2 > T1 T1

x1 x2heat flux

THERMAL CONDUCTIVITY

Fick’s First Law

T2 > T1 T1

x1 x2heat flux

THERMAL CONDUCTIVITY

2

nd2 ' 2 L

T T T Tk if K f T k Fick s aw

t x x t x

Fick’s Second

Law

• Non-Steady State: dT/dt is not constant.

• 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.

K=kl+ke: Again think about band gaps: metals have lots of free electrons (ke is large), while ceramics have few (only kl 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

)sK/J(10443.23

282

22

e

kL

T

k B

• 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

• Occurs due to: --uneven heating/cooling --mismatch in thermal expansion.

• Example Problem --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?

Troom

LroomT

L

compressive keeps L = 0Answer: 106C

THERMAL STRESSES

-172MPa

100GPa 20 x 10-6 /C

20C

)()( othermal TTEE

)( othermalo

TTL

L

Strain (ε) due to ∆T causes a stress (σ) that depends on the modulus of elasticity (E):

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.

)V/K(tCoefficienSeebeck dT

dVS

THERMOELECTRIC COOLING & HEATING

Reverse 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.

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

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 Bi2Te3 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