bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege
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BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt1
Bruce Mayer, PE Engineering-45: Materials of Engineering
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering 45
ThermalThermalPropertiesProperties
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt2
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt3
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt4
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt5
Bruce Mayer, PE Engineering-45: Materials of Engineering
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)
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt6
Bruce Mayer, PE Engineering-45: Materials of Engineering
Specific Heats ComparedSpecific Heats Compared cp and C for Some Common Substances At 298K
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt7
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt8
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt9
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt10
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt11
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt12
Bruce Mayer, PE Engineering-45: Materials of Engineering
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”
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt13
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt14
Bruce Mayer, PE Engineering-45: Materials of Engineering
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?
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt15
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt16
Bruce Mayer, PE Engineering-45: Materials of Engineering
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 /
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt17
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt18
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt19
Bruce Mayer, PE Engineering-45: Materials of Engineering
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)
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt20
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
)(
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt21
Bruce Mayer, PE Engineering-45: Materials of Engineering
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
BMayer@ChabotCollege.edu • ENGR-45_Lec-13_Thermal_Props.ppt22
Bruce Mayer, PE Engineering-45: Materials of Engineering
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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