Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

[email protected] • ENGR-45_Lec-13_Thermal_Props.ppt1

Bruce Mayer, PE Engineering-45: Materials of Engineering

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 45

ThermalThermalPropertiesProperties



Learning Goals – Thermal PropsLearning Goals – Thermal Props

Learn How Materials Respond to Elevated Temperatures

How to Define and Measure• Heat Capacity and/or Specific Heat

• Coefficient of Thermal Expansion

• Thermal Conductivity

• Thermal Shock Resistance

How Ceramics, Metals, and Polymers rank in Hi-Temp Applications



Heat Capacity (Specific Heat)Heat Capacity (Specific Heat) Concept Ability of a Substance to

Absorb/Supply Heat Relative to its Change in Temperature

Quantitatively

C

dQdT

heat capacity(J/mol-K)

energy input (J/mol or J/kg)

temperature change (K)

C typically Specified by the Conditions of the Measurement• Cp → Constant PRESSURE on the Specimen

• Cv → Specimen Held at Constant VOLUME



Measure Specific HeatMeasure Specific Heat

Battery

Insulation

t VIw q or • Where

q & w Heat or Work or Energy (Joules or Watt-Sec)

– V Electrical Potential (Volts)

– I Electrical Current (Amps)

t time sec

Recall from ENGR43

The specific Heat for a Solid at Rm-P



Measure Specific Heat contMeasure Specific Heat cont

Battery

Insulation

ifp TT

mtVI

T

mq

dT

dQ c

To Find cp, Measure

• Block Mass, m (kg)

• Voltage, V (Volts)

• Current, I (Amps)

• Initial Temperature, Ti (K or °C)

• Final Temperature, Tf (K or °C)

• Run Time, t (s)



Specific Heats ComparedSpecific Heats Compared cp and C for Some Common Substances At 298K



CCpp vs C vs Cvv Measurements Measurements

GASES are Almost Always Measured at Constant VOLUME• e.g., Fill a Sealed

container with Silane (SiH4) Add Heat, and Measure T

Solids & Liquids Typically Measured at Constant PRESSURE

• Set the Solid or Liquid-Container on the table at ATMOSPHERIC Pressure (101.325 kPa), Add Heat & Measure T

Example = Co

E = 208.6 GPa

= 0.31

= 1.3×10-5 K-1

100 mm



CCpp or C or Cvv for Co for Co

Heat the Block by 20K

Then the Change in Volume, V

E = 208.6 GPa

= 0.31

= 1.3×10-5 K-1

100 mm

ppTVV

mmmmV

KKmmV

56.1710756.1

10758.1026.0

20/103.1100

11

353

35

A VERY Small Difference

The HydroStatic (all-over) Stress Required to Maintain constant V

ATMan of%003.021.3

31.0213

10756.1106.208

213119

Pa

Pa

VVE

• Very Hard to Control to Maintain Const-V



CCvv as Function of Temperature as Function of Temperature

Cv

• Increases with Increasing T

• Tends to a limiting Value of 3R = 24.93 J/mol-k

Quantitatively3R=24.93



CCvv as Function of Temp cont as Function of Temp cont

For Many Crystalline Solids

D

pTHivv

Dv

TTconst

CC C

TTAT C

:

:

,

3

• Where– A Material

Dependent CONSTANT

– TD Debye Temperature, K

Atomic Physics• Energy is Stored in

Lattice Vibration Waves Called Phonons– Analogous to Optical

PHOTONS



4

• Why is cp significantly larger for polymers?

CCpp Comparison Comparison

• PolymersPolypropylene Polyethylene Polystyrene Teflon

cp (J/kg-K) at room T

• CeramicsMagnesia (MgO) Alumina (Al2O3)Glass (SiO2)

• MetalsAluminum Steel

Tungsten Gold

1925 1850 1170 1050

900 486 128 138

incr

easi

ng

cp

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

material

940 775

840



Thermal ExpansionThermal Expansion Concept Materials Change Size When

Heated Tinit

TfinalLfinal

Linit initfinalinit

initfinal TT L

LL

Coefficient of Thermal Expansion

due to Asymmetry of PE InterAtomic Distant Trough• T↑ E↑

• ri is at the Statistical Avg of the Trough Width

Bond energy

Bond length (r)

incr

easi

ng

T

T1

r(T5)

r(T1)

T5Bond-energy vs bond-lengthcurve is “asymmetric”



6

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

Thermal Expansion: ComparisonThermal Expansion: Comparisonin

creasi

ng

• 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

Material



Thermal ConductivityThermal Conductivity Concept Ability of a Substance Tranfer

Heat Relative to Temperature Differences Quantitatively, Consider a Cold←Hot Bar

Characterize the Heat Flux as

T2 > T1 T1

x1 x2heat flux

q k

dTdx

temperatureGradient (K/m)

thermal conductivity (W/m-K)

heat flux(W/m2)

• Q: Why theNEGATIVE Sign before k?



Thermal ConductivityThermal Conductivity: : ComparisonComparisonin

crea

sin

g k

• PolymersPolypropylene 0.12Polyethylene 0.46-0.50 Polystyrene 0.13 Teflon 0.25

By vibration/ rotation of chain molecules

• CeramicsMagnesia (MgO) 38Alumina (Al2O3) 39 Soda-lime glass 1.7 Silica (cryst. SiO2) 1.4

By vibration of atoms

• MetalsAluminum 247Steel 52 Tungsten 178 Gold 315

By vibration of atoms and motion of electrons

k (W/m-K) Energy TransferMaterial



Thermal StressesThermal Stresses As Noted Previously a Material’s Tendency to

Expand/Contract is Characterized by α If a Heated/Cooled Material is Restrained to

its Original Shape, then Thermal Stresses will Develop within the material

For a Solid Material

• Where Stress (Pa or typically MPa)

– E Modulus of Elasticity; a.k.a., Young’s Modulus (GPa) l Change in Length due to the Application of a force (m)

– lo Original, Unloaded Length (m)

ollE /



Thermal Stresses cont.Thermal Stresses cont. From Before

Sub l/l into Young’s Modulus Eqn To Determine the Thermal Stress Relation

Tll o /

TE Eample: a 1” Round 7075-T6 Al

(5.6Zn, 2.5Mg, 1.6Cu, 0.23Cr wt%’s) Bar Must be Compressed by a 8200 lb force when restrained and Heated from Room Temp (295K)• Find The Avg Temperature for the Bar



Thermal Stress ExampleThermal Stress Example Find Stress

MPapsiA

F

in

lb

d

lb

A

F

7244010

41

8200

4

820022

Recall the Thermal Stress Eqn

E = 10.4 Mpsi = 71.7 GPa

α = 13.5 µin/in-°F = 13.5 µm/m-°F

8200 lbs

0 lbs

TE Need E & α

• Consult Matls Ref



Thermal Stress Example contThermal Stress Example cont Solve Thermal

Stress Reln for ΔT

Since The Bar was Originally at Room Temp

8200 lbs

0 lbs

KFTFPa

PaT

ET

3.414.74/105.13107.71

107269

6

FCT

KT

TTT

f

f

if

15065

33841297

• Heating to Hot-Coffee Temps Produces Stresses That are about 2/3 of the Yield Strength (15 Ksi)



Thermal Shock ResistanceThermal Shock Resistance 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

)( 21 TTE

E

TT ffracture )( 21

Temperature difference thatcan be produced by cooling:

k

ratequenchTT )( 21

set equal

Critical temperature differencefor fracture (set = f)

10

• Result: TSRE

kratequench f

fracturefor

)(



TSR – Physical MeaningTSR – Physical Meaning The Reln

For Improved (GREATER) TSR want f↑ Material can withstand higher

thermally-generated stress before fracture

• k↑ Hi-Conductivity results in SMALLER Temperature Gradients; i.e., lower ΔT

• E↓ More FLEXIBLE Material so the thermal stress from a given thermal strain will be reduced (σ = Eε)

• α↓ Better Dimensional Stability; i.e., fewer restraining forces developed

E

kTSR f



WhiteBoard WorkWhiteBoard Work Problem 19.5 – Debye Temperature

Charles Kittel, “Introduction to Solid State Physics”, 6e, John Wiley & Sons, 1986. pg-110

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

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