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

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

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

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  1. Engineering 45 ThermalProperties Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

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

  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. Battery Insulation Measure Specific Heat cont • 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. 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 Cp vs Cv Measurements 100 mm E = 208.6 GPa  = 0.31  = 1.3×10-5 K-1

  8. Heat the Block by 20K Then the Change in Volume, V Cp or Cv for Co 100 mm E = 208.6 GPa  = 0.31  = 1.3×10-5 K-1 • The HydroStatic (all-over) Stress Required to Maintain constant V • A VERY Small Difference • Very Hard to Control to Maintain Const-V

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

  10. For Many Crystalline Solids Cv as Function of Temp cont • Atomic Physics • Energy is Stored in Lattice Vibration Waves Called Phonons • Analogous to OpticalPHOTONS • Where • A  Material Dependent CONSTANT • TD  Debye Temperature, K

  11. Cp Comparison c (J/kg-K) p material at room T • Polymers 1925 Polypropylene cp: (J/kg-K) 1850 Polyethylene Cp: (J/mol-K) 1170 Polystyrene 1050 Teflon • Ceramics Magnesia (MgO) 940 Alumina (Al2O3) 775 Glass (SiO2) 840 increasing cp • Metals Aluminum 900 Steel 486 Tungsten 128 Gold 138 • Why is cp significantly larger for polymers? 4

  12. T init L init T final L final Bond energy ) ) 1 5 r(T r(T Bond length (r) T 5 Bond-energy vs bond-length curve is “asymmetric” increasing T T 1 Thermal Expansion • Concept  Materials Change Size When Heated Coefficient of Thermal Expansion •  due to Asymmetry of PE InterAtomic Distant Trough • T↑  E↑ • ri is at the Statistical Avg of the Trough Width

  13. Thermal Expansion: Comparison a -6 (10 /K) Material at room T • Polymers 145-180 Polypropylene 106-198 Polyethylene 90-150 Polystyrene 126-216 Teflon • Metals a Aluminum 23.6 Steel 12 Tungsten 4.5 Gold 14.2 increasing • Ceramics Magnesia (MgO) 13.5 Alumina (Al2O3) 7.6 Soda-lime glass 9 Silica (cryst. SiO2) 0.4 • Q: Why does a generally decrease with increasing bond energy? 6

  14. temperature Gradient (K/m) heat flux (W/m2) thermal conductivity (W/m-K) T > T T 2 1 1 x x 1 2 heat flux Thermal Conductivity • Concept  Ability of a Substance Tranfer Heat Relative to Temperature Differences • Quantitatively, Consider a Cold←Hot Bar • Characterize the Heat Flux as • Q: Why theNEGATIVE Sign before k?

  15. Material k (W/m-K) Energy Transfer • Metals Aluminum 247 By vibration of Steel 52 atoms and Tungsten 178 motion of Gold 315 electrons • Ceramics Magnesia (MgO) 38 By vibration of Alumina (Al2O3) 39 increasing k atoms Soda-lime glass 1.7 Silica (cryst. SiO2) 1.4 • Polymers Polypropylene 0.12 By vibration/ Polyethylene 0.46-0.50 rotation of chain Polystyrene 0.13 molecules Teflon 0.25 Thermal Conductivity: Comparison

  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)

  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

  18. Find Stress 0 lbs 8200 lbs Thermal Stress Example • E = 10.4 Mpsi = 71.7 GPa • α= 13.5 µin/in-°F = 13.5 µm/m-°F • Recall the Thermal Stress Eqn • Need E & α • Consult Matls Ref

  19. Solve Thermal Stress Reln for ΔT 0 lbs 8200 lbs Thermal Stress Example cont • Since The Bar was Originally at Room Temp • Heating to Hot-Coffee Temps Produces Stresses That are about 2/3 of the Yield Strength (15 Ksi)

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

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

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