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

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

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

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

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)

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

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 (101.325 kPa), Add Heat & Measure T Example = Co 100 mm E = 208.6 GPa  = 0.31  = 1.3×10-5 K-1

Cp or Cv for Co Heat the Block by 20K Then the Change in Volume, V 100 mm E = 208.6 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

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

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 http://hyperphysics.phy-astr.gsu.edu/hbase/solids/phonon.html#c2

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

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

Thermal Expansion: Comparison increasing a Thermal Expansion: Comparison • Polymers Polypropylene Polyethylene Polystyrene Teflon 145-180 106-198 90-150 126-216 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

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)

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 0.46-0.50 Polystyrene 0.13 Teflon 0.25 By vibration/ rotation of chain molecules

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

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

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 http://hyperphysics.phy-astr.gsu.edu/hbase/solids/phonon.html#c2 Need E & α Consult Matls Ref

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 http://hyperphysics.phy-astr.gsu.edu/hbase/solids/phonon.html#c2 Heating to Hot-Coffee Temps Produces Stresses That are about 2/3 of the Yield Strength (15 Ksi)

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

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

WhiteBoard Work Problem 19.5 – Debye Temperature Charles Kittel, “Introduction to Solid State Physics”, 6e, John Wiley & Sons, 1986. pg-110