SlideServe
Télécharger Obtenir Pro
Menu
Accueil Pro
Préférences
ApparenceLight Aide et commentaires
© SlideServe · v2026

Understanding Thermal Properties in Engineering

zerlina-mcgowan avatar
zerlina-mcgowan
22 diapositives · 344 Views · Nov 29, 2014

Explore how materials react to heat, measure heat capacity, thermal conductivity, shocks, and ranks in high-temp applications. Learn specific heat concepts and insulation methods from an experienced engineer.

#understanding thermal properties in engineering #understanding thermal properties in engineering ppt #powerpoint #presentation
Understanding Thermal Properties in Engineering - slide 1
01 / 22Understanding Thermal Properties in Engineering
1 / 17

Télécharger la présentation

Understanding Thermal Properties in Engineering
Une image / un lien ci-dessous est fourni (tel quel) pour télécharger la présentation

Politique de téléchargement : le contenu du site vous est fourni TEL QUEL pour votre information et votre usage personnel ; il ne peut être vendu, concédé sous licence ni partagé sur d’autres sites sans l’accord de son auteur.Le contenu vous est fourni TEL QUEL, pour votre information et votre usage personnel uniquement. Téléchargez la présentation en cliquant sur ce lien. Lors du téléchargement, si vous ne parvenez pas à télécharger une présentation, l’éditeur a peut-être supprimé le fichier de son serveur.Pendant le téléchargement, si vous ne parvenez pas à obtenir une présentation, le fichier a peut-être été supprimé par l’éditeur.

F I N

Plus de contenu associé

Transcription de la présentation

  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