1. Mechanical Engineering Department Alabama A&M University Fall 2014, Lecture 18 Mechanical Behavior: Part III Dr. Aaron L. Adams, Assistant Professor

2. Learning Objectives... How do materials respond to the application of heat? How do we define and measure... -- heat capacity? -- thermal expansion? -- thermal conductivity? -- thermal shock resistance? How do the thermal properties of ceramics, metals, and polymers differ? Chapter 17, pp. 733-750 Relevant Reading for this Lecture... 2

3. Quantitatively: The energy required to produce a unit rise in temperature for one mole of a material. heat capacity (J/mol-K) energy input (J/mol) temperature change (K) Heat Capacity Two ways to measure heat capacity: C p : Heat capacity at constant pressure. C v : Heat capacity at constant volume. C p usually > C v Heat capacity has units of: Is the ability of a material to absorb heat 3

4. Specific heat (J/kg-K) energy input (J/mol) temperature change (K) Specific Heat Specific heat has units of: Common to re-define heat capacity using a basis of unit mass (because we tend to measure mass). mass 4

5. In class example: How to apply specific heat. Estimate the amount of heat (in J) required to raise 2 kg of polypropylene from room temperature (i.e., 25°C) to 100°C. The specific heat for polypropylene is 1880 J/kg·K. Express your answer in kJ. Substitution of known values yields: 5

6. increasing c p Why is c p significantly larger for polymers? Selected values from Table 17.1, Callister & Rethwisch 4e. Polymers Polypropylene Polyethylene Polystyrene Teflon c p (J/kg-K) at room T Ceramics Magnesia (MgO) Alumina (Al 2 O 3 ) Glass Metals Aluminum Steel Tungsten Gold 1925 1850 1170 1050 900 486 138 128 c p (specific heat): (J/kg-K) Material 940 775 840 Specific Heat: Comparison C p (heat capacity): (J/mol-K) Specific heat is per unit mass. How would the value for H 2 O ( ) compare? c p (H 2 O) = 4186 J/kg-K ! 6

7. Heat capacity... -- increases with temperature -- for solids it reaches a limiting value of 3R From atomic perspective: -- Energy is stored as atomic vibrations… more to come -- As temperature increases, the average energy of atomic vibrations increases. Dependence of Heat Capacity on Temperature Adapted from Fig. 17.2, Callister & Rethwisch 4e. R = gas constant 3R3R = 8.31 J/mol-K C v = constant Debye temperature (usually less than T room ) T (K)  D 0 0 CvCv C v = AT 3 7

8. Thermal Expansion Materials change size when temperature is changed linear coefficient of thermal expansion (units = K -1 or  C -1 ) T initial T final initial final T final > T initial Coefficient of linear expansion Coefficient of expansion 8

9. Atomic Perspective: Thermal Expansion Adapted from Fig. 17.3, Callister & Rethwisch 4e. Asymmetric curve: -- increase temperature, -- increase in interatomic separation -- thermal expansion Symmetric curve: -- increase temperature, -- no increase in interatomic separation -- no thermal expansion mean atomic distance This is typical of ceramics, which have larger bonding energy due to ionic & covalent-type bonding Typical of metals 9

10. Coefficient of Thermal Expansion : Comparison Q: Why does  generally decrease with increasing bond energy? Polypropylene145-180 Polyethylene106-198 Polystyrene90-150 Teflon126-216 Polymers Ceramics Magnesia (MgO)13.5 Alumina (Al 2 O 3 )7.6 Soda-lime glass9 Silica (cryst. SiO 2 )0.4 Metals Aluminum23.6 Steel12 Tungsten4.5 Gold14.2  (10 -6 /  C) at room T Material Selected values from Table 17.1, Callister & Rethwisch 4e. Polymers have larger  values because of weak secondary bonds increasing   is typically smaller for ceramics & glasses than for metals 10

11. Thermal Expansion: Example Ex: A copper wire 15 m long is cooled from 40 to -9°C. How much change in length will it experience? Answer: For Cu rearranging Equation 17.3b 11

12. In class example: We need to use the average CTE for this problem A 0.01 m long bar of tungsten (W) is placed in a laboratory furnace and heated from room temperature (25°C) to 500°C. What will be the length of the bar at 500°C? The coefficient of thermal expansion (α) for W at 27°C is 4.5 mm/(mm·°C)  10 -6 ) while the value at 527°C is 4.8 mm/(mm·°C)  10 -6. 12

13. Invar – a material with little to no expansion! Invar, also known generically as FeNi36 (64FeNi in the US), is a nickel iron alloy notable for its uniquely low coefficient of thermal expansion. (CTE or α), 1.6x10 -6 ( o C - 1 ). The name, Invar, comes from the word invariable, referring to its lack of expansion or contraction with temperature changes Mechanism - would it be the bonding potential wells? But it’s a metal? It is a metal and one would expect it to expand, but both metals are magnetic. The ferromagnetic behavior causes a magneto-restriction effect that significantly limits the expansion. Once the metal is heated above the Curie temperature (future lecture), it is no longer magnetic and the alloy expands Discovered in 1896 by Charles-Edouard Guillaume of France 1920 Noble Prize in Physics 13

14. Pendulum and balance wheels in clocks Directional bending; more deflection for a given temperature change Invar Applications Thermostat in your home Watches 14

15. The ability of a material to transport heat. temperature gradient thermal conductivity (J/m-K-s) heat flux (J/m 2  s) Atomic perspective: Atomic vibrations and free electrons in hotter regions transport energy to cooler regions. T2T2 T2 > T1 T2 > T1 T1T1 x1x1 x2x2 heat flux Thermal Conductivity Fourier’s Law 15

16. Heat transfer by conduction is defined by Fourier’s law. At steady state: dQ/dt = constant dQ/dt = the heat flux Like diffusion! flux constant Temperature gradient 16

17. Mass Transfer – Mass Flux - Diffusion all are equivalent and described by Fick’s Law of diffusion [SI units] 17

18. Molecular Heat Transfer – Conduction is described by Fourier’s Law of Conduction [SI units] 18

19. What is k? k = k l + k e Lattice vibrations (phonons)Free electrons Electrons gain kinetic energy from the heat; high velocities Which mechanism of conduction dominates in ceramics? in metals? Why? Atomic vibrations are in the form of lattice waves or phonons 19

20. Thermal Conductivity : Comparison increasing k Polymers Polypropylene0.12 Polyethylene0.46-0.50 Polystyrene0.13 Teflon0.25 vibration/rotation of chain molecules Ceramics Magnesia (MgO)38 Alumina (Al 2 O 3 )39 Soda-lime glass1.7 Silica (cryst. SiO 2 )1.4 atomic vibrations Metals Aluminum247 Steel52 Tungsten178 Gold315 atomic vibrations and motion of free electrons k (W/m-K) Energy Transfer Mechanism Material Selected values from Table 17.1, Callister & Rethwisch 4e. Vice-versa of heat capacity! 20

21. Application: Space Shuttle Orbiter Silica tiles (400-1260  C) : -- large scale application-- microstructure: Fig. 19.2W, Callister 6e. (Fig. 19.2W adapted from L.J. Korb, C.A. Morant, R.M. Calland, and C.S. Thatcher, "The Shuttle Orbiter Thermal Protection System", Ceramic Bulletin, No. 11, Nov. 1981, p. 1189.) Fig. 19.3W, Callister 5e. (Fig. 19.3W courtesy the National Aeronautics and Space Administration.) Fig. 19.4W, Callister 5e. (Fig. 219.4W courtesy Lockheed Aerospace Ceramics Systems, Sunnyvale, CA.) Thermal Protection System reinf C-C (1650°C) Re-entry T Distribution silica tiles (400-1260°C) nylon felt, silicon rubber coating (400°C) ~90% porosity! Silica fibers bonded to one another during heat treatment. 100  m Chapter-opening photograph, Chapter 23, Callister 5e (courtesy of the National Aeronautics and Space Administration.) 21

22. A comment on k metal ceramic radiant heat transfer (IR heat) why the drop? Cu Zn 22

23. Outer surface heats up first and lattice values want to expand – but can’t because the interior is at a lower temperature…. heat Thermal Stresses What happens when a block of material is placed in a furnace? tension compression Consequently, lattice parameter on outer surface is squeezed (compressed) to match the inner lattice parameter as the inner material is stretch (tension) to match the expanding outer material…generates a stress gradient in the material. Only when part at the same temperature does this stress go to zero. 23

24. Occur due to: -- restrained thermal expansion/contraction -- temperature gradients that lead to differential dimensional changes Thermal Stresses Thermal stress  http://www.121designs.com/images/wheel.jp g When you heat up a part, the outer surface heats up faster than inner and vice versa on cooling. These differences can result in thermal stresses (a load on part) 24

25. -- A brass rod is stress-free at room temperature (20°C). -- It is heated up, but prevented from lengthening. -- At what temperature does the stress reach -172 MPa? Example Problem T0T0 0 Solution: Original conditions TfTf Step 1: Assume unconstrained thermal expansion 0  Step 2: Compress specimen back to original length 0    25

26. Example Problem (cont.) 0   The thermal stress can be directly calculated as Noting that  compress = -  thermal and substituting gives 20 x 10 -6 /°C Answer: 106°C 100 GPa 20ºC Rearranging and solving for T f gives -172 MPa (since in compression) 26

27. Occurs due to: nonuniform heating/cooling Ex: Assume top thin layer is rapidly cooled from T 1 to T 2 Tension develops at surface Critical temperature difference for fracture (set  =  f ) set equal Large TSR when is large Thermal Shock Resistance Temperature difference that can be produced by cooling:  rapid quench resists contraction tries to contract during cooling T2T2 T1T1 27

28. The thermal properties of materials include: Heat capacity: -- energy required to increase a mole of material by a unit T -- energy is stored as atomic vibrations Coefficient of thermal expansion: -- the size of a material changes with a change in temperature -- polymers have the largest values Thermal conductivity: -- the ability of a material to transport heat -- metals have the largest values Thermal shock resistance: -- the ability of a material to be rapidly cooled and not fracture -- is proportional to Summary 28