1. Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Engineering 45
Thermal Properties
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer

2. Learning 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

3. Heat Capacity (Specific Heat)
Concept  Ability of a Substance to Absorb/Supply Heat Relative to its Change in Temperature
Quantitatively
energy input (J/mol or J/kg)
heat capacity
(J/mol-K)
temperature change (K)
C typically Specified by the Conditions of the Measurement
Cp → Constant PRESSURE on the Specimen
Cv → Specimen Held at Constant VOLUME
Cp best for SOLIDS & LIQUIDS at room PRESSURE
Cv best for GASES & VAPORS in a FIXED CONTAINER

4. Measure Specific Heat Recall from ENGR43
Battery
Where
q & w  Heat or Work or Energy (Joules or Watt-Sec)
V  Electrical Potential (Volts)
I  Electrical Current (Amps)
t  time sec
Insulation
 The specific Heat for a Solid at Rm-P

5. Measure Specific Heat cont
Battery
Insulation
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)

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

7. Cp vs Cv 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 ( kPa), Add Heat & Measure T
Example = Co
100 mm
E = GPa
 = 0.31
 = 1.3×10-5 K-1

8. Cp or Cv for Co Heat the Block by 20K Then the Change in Volume, V
100 mm
E = GPa
 = 0.31
 = 1.3×10-5 K-1
Heat the Block by 20K
Then the Change in Volume, V
The HydroStatic (all-over) Stress Required to Maintain constant V
VERY S
A VERY Small Difference
Very Hard to Control to Maintain Const-V

9. Cv as Function of Temperature
Increases with Increasing T
Tends to a limiting Value of 3R = J/mol-k
Quantitatively
3R=24.93

10. Cv as Function of Temp cont
For Many Crystalline Solids
Atomic Physics
Energy is Stored in Lattice Vibration Waves Called Phonons
Analogous to Optical PHOTONS
Where
A  Material Dependent CONSTANT
TD  Debye Temperature, K
TD would be defined in a 3rd or 4th yr solid-state physics class; ref. Kittel, Charles, Introduction to Solid State Physics, 7th Ed., Wiley, (1996); pg 106
Ref also

11. Cp Comparison increasing cp • Why is cp significantly
• Polymers
Polypropylene
Polyethylene
Polystyrene
Teflon c p
(J/kg-K)
at room T
• Ceramics
Magnesia (MgO)
Alumina (Al2O3)
Glass (SiO2)
• Metals
Aluminum
Steel
Tungsten
Gold
1925
1850
1170
1050
900
486
128
138
increasing cp
cp:
Cp:
(J/mol-K)
material
940
775
840
• Why is cp significantly
larger for polymers?
Low Densities; it takes a lot of Atoms to make a Kg, so a large lattice which allows for much lattice vibration (phonons) 4

12. Coefficient of Thermal Expansion
Concept  Materials Change Size When Heated T
init
final L
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)
increasing T T 1
r(T 5 )
Bond-energy vs bond-length
curve is “asymmetric”
PE = Potential Energy

13. Thermal Expansion: Comparison
increasing a
Thermal Expansion: Comparison
• Polymers
Polypropylene
Polyethylene
Polystyrene
Teflon
90-150 a
(10 -6
/K)
at room T
• Ceramics
Magnesia (MgO)
Alumina (Al2O3)
Soda-lime glass
Silica (cryst. SiO2)
13.5
7.6 9
0.4
• Metals
Aluminum
Steel
Tungsten
Gold
23.6 12
4.5
14.2
Material
• Q: Why does a
generally decrease
with increasing
bond energy?
The Higher the bond Energy the Deeper and more symmetrical the E vs ri and hence the avg position for ri moves little with increasing temperature 6

14. Thermal Conductivity
Concept  Ability of a Substance Tranfer Heat Relative to Temperature Differences
Quantitatively, Consider a Cold←Hot Bar T 2
> T 1 x
heat flux
Characterize the Heat Flux as
dT/dx = (T2-T1)/(x2-x1) is NEGATIVE as Heat Flows from Hi-T to Lo-T
• Q: Why the NEGATIVE Sign before k?
temperature
Gradient (K/m)
heat flux
(W/m2)
thermal conductivity (W/m-K)

15. Thermal Conductivity: Comparison
k (W/m-K)
Energy Transfer
Material
increasing k
• Metals
Aluminum
247
Steel 52
Tungsten
178
Gold
315
By vibration of
atoms and
motion of
electrons
• Ceramics
Magnesia (MgO) 38
Alumina (Al2O3) 39
Soda-lime glass
1.7
Silica (cryst. SiO2)
1.4
By vibration of
atoms
• Polymers
Polypropylene
0.12
Polyethylene
Polystyrene
0.13
Teflon
0.25
By vibration/
rotation of chain
molecules

16. Thermal 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)
dT/dx = (T2-T1)/(x2-x1) is NEGATIVE as Heat Flows from Hi-T to Lo-T

17. Thermal Stresses cont. From Before
Sub l/l into Young’s Modulus Eqn To Determine the Thermal Stress Relation
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
dT/dx = (T2-T1)/(x2-x1) is NEGATIVE as Heat Flows from Hi-T to Lo-T

18. Thermal Stress Example
Find Stress
8200 lbs
0 lbs
E = 10.4 Mpsi = 71.7 GPa
α = 13.5 µin/in-°F = 13.5 µm/m-°F
Recall the Thermal Stress Eqn
TD would be defined in a 3rd or 4th yr solid-state physics class; ref. Kittel, Charles, Introduction to Solid State Physics, 7th Ed., Wiley, (1996); pg 106
Ref also
Need E & α
Consult Matls Ref

19. Thermal Stress Example cont
Solve Thermal Stress Reln for ΔT
8200 lbs
0 lbs
Since The Bar was Originally at Room Temp
TD would be defined in a 3rd or 4th yr solid-state physics class; ref. Kittel, Charles, Introduction to Solid State Physics, 7th Ed., Wiley, (1996); pg 106
Ref also
Heating to Hot-Coffee Temps Produces Stresses That are about 2/3 of the Yield Strength (15 Ksi)

20. Thermal Shock Resistance
Occurs due to: uneven heating/cooling.
Ex: Assume top thin layer is rapidly cooled from T1 to T2:
Tension develops at surface
Temperature difference that
can be produced by cooling:
Critical temperature difference
for fracture (set s = sf)
k above is thermal conductivity
set equal
• Result: 10

21. TSR – 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
dT/dx = (T2-T1)/(x2-x1) is NEGATIVE as Heat Flows from Hi-T to Lo-T

22. WhiteBoard Work Problem 19.5 – Debye Temperature
Charles Kittel, “Introduction to Solid State Physics”, 6e, John Wiley & Sons, pg-110