Introduction to Engineering Materials ENGR2000 Chapter 19: Thermal Properties. Dr. Coates
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1 Introduction to Engineering Materials ENGR2000 Chapter 19: Thermal Properties Dr. Coates
2 Chapter 19: Thermal Properties ISSUES TO ADDRESS... 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? 2
3 Heat Capacity The ability of a material to absorb heat Quantitatively: The energy required to produce a unit rise in temperature for one mole of a material. energy input (J/mol) heat capacity dq (J/mol-K) C dt Two ways to measure heat capacity: C p : Heat capacity at constant pressure. C v : Heat capacity at constant volume. temperature change (K) C p usually > C v Heat capacity has units of J molk lb Btu molf 3
4 Vibrational heat capacity Increase in thermal energy is assimilated in solids as the increase in vibrational energy of the atoms. Vibration of adjacent atoms are coupled via atomic bonding. Traveling lattice waves
5 Phonons Vibrational thermal energy for a material consists of a series of lattice waves. These vibrational wave packets are known as phonons.
6 Dependence of Heat Capacity on Temperature Heat capacity increases with temperature -- for solids it reaches a limiting value of 3R R = gas constant 3R = 8.31 J/mol-K C v = constant 0 0 T (K) qd Debye temperature (usually less than Troom ) From atomic perspective: -- Energy is stored as atomic vibrations. -- As temperature increases, the average energy of atomic vibrations increases. Adapted from Fig. 19.2, Callister & Rethwisch 8e. 6
7 Atomic vibrations are in the form of lattice waves or phonons Atomic Vibrations Adapted from Fig. 19.1, Callister & Rethwisch 8e. 7
8 increasing c p Specific Heat: Comparison Material Polymers Polypropylene Polyethylene Polystyrene Teflon Ceramics Magnesia (MgO) Alumina (Al 2 O 3 ) Glass Metals Aluminum Steel Tungsten Gold c p (J/kg-K) at room T c p (specific heat): (J/kg-K) C p (heat capacity): (J/mol-K) Why is c p significantly larger for polymers? Selected values from Table 19.1, Callister & Rethwisch 8e. 8
9 Thermal Expansion Materials change size when temperature is changed initial final T initial T final T final > T initial l final l initial l initial l (T final T initial ) linear coefficient of thermal expansion (1/K or 1/ºC) 9
10 Atomic Perspective: Thermal Expansion Asymmetric curve: -- increase temperature, -- increase in interatomic separation -- thermal expansion Symmetric curve: -- increase temperature, -- no increase in interatomic separation -- no thermal expansion Adapted from Fig. 19.3, Callister & Rethwisch 8e. 10
11 increasing Coefficient of Thermal Expansion: Comparison Material Polymers (10-6 /C) at room T Polypropylene Polyethylene Polystyrene Teflon Metals Aluminum 23.6 Steel 12 Tungsten 4.5 Gold 14.2 Ceramics Magnesia (MgO) 13.5 Alumina (Al 2 O 3 ) 7.6 Soda-lime glass 9 Silica (cryst. SiO 2 ) 0.4 Polymers have larger values because of weak secondary bonds Q: Why does generally decrease with increasing bond energy? Selected values from Table 19.1, Callister & Rethwisch 8e. 11
12 Thermal Expansion: Example Ex: A copper wire 15 m long is cooled from 40 to -9ºC. How much change in length will it experience? Answer: For Cu 16.5 x 10 6 ( C) 1 rearranging Equation 19.3b T 0 [16.5 x 10 6 (1/ C)](15 m)[40c ( 9C)] 0.012m 12mm 12
13 At steady state, the heat flux : is 19.4 Thermal conductivity - ability to transport heat dt q k dx where k is the thermal conductivity and dt dx the temperature gradient.
14 Thermal Conductivity The ability of a material to transport heat. Fourier s Law heat flux (J/m 2 -s) q k dt dx temperature gradient thermal conductivity (J/m-K-s) T 1 x1 T 2 T 2 > T 1 heat flux x2 Atomic perspective: Atomic vibrations and free electrons in hotter regions transport energy to cooler regions. 14
15 Mechanisms of heat conduction Heat is transported by Phonons (are easily scattered) -quantum of vibrational energy Free Electrons Total conductivity: k k l k where k l and electron thermal e k e represent the lattice vibration and conductivities.
16 increasing k Thermal Conductivity: Comparison Material k (W/m-K) Metals Aluminum 247 Steel 52 Tungsten 178 Gold 315 Ceramics Magnesia (MgO) 38 Alumina (Al 2 O 3 ) 39 Soda-lime glass 1.7 Silica (cryst. SiO 2 ) 1.4 Polymers Polypropylene 0.12 Polyethylene Polystyrene 0.13 Teflon 0.25 Selected values from Table 19.1, Callister & Rethwisch 8e. Energy Transfer Mechanism atomic vibrations and motion of free electrons atomic vibrations vibration/rotation of chain molecules 16
17 19.5 Thermal Stresses For a material with a nonzero th, a temperature change results in a dimensional change or thermal strain th. Under certain conditions, the thermal strain can cause a thermal stress which may result in material failure.
18 Thermal Stresses Occur due to: -- restrained thermal expansion/contraction -- temperature gradients that lead to differential dimensional changes Thermal stress E (T 0 T f ) E T 18
19 Example Problem -- A brass rod is stress-free at room temperature (20ºC). -- It is heated up, but prevented from lengthening. -- At what temperature does the stress reach -172 MPa? Solution: T 0 Original conditions 0 0 Step 1: Assume unconstrained thermal expansion T f thermal (T f T 0 ) room 0 Step 2: Compress specimen back to original length compress room thermal 19
20 Example Problem (cont.) 0 The thermal stress can be directly calculated as E( compress ) Noting that compress = - thermal and substituting gives E( thermal ) E (T f T 0 ) E (T 0 T f ) Rearranging and solving for T f gives 20ºC T f T 0 E -172 MPa (since in compression) Answer: 106ºC 100 GPa 20 x 10-6 /ºC 20
21 Bimetallic strip A bimetallic strip used as a switch in a thermostat the differential expansion induces stresses in the two metals
22 Bimetallic strip A bimetallic strip used as a switch in a thermostat the induced stresses cause the strip to bend
23 Thermal expansion of a rod showing (a) unconstrained expansion Totalstrain : l l 0 Mechanical and t th m t T 0 th m th 0 thermal strain :
24 Thermal expansion of a rod showing (b) completely constrained expansion Totalstrain : t th t l l T m 0 th 0 m T th th 0 Mechanical and thermal strains :
25 Thermal expansion of a rod showing (b) completely constrained expansion Assuming elastic deformation, the mechanical stress: m th th m m E ET th th The thermal stress: ET
26 Thermal expansion of two rods showing (a) different amounts of thermal expansion when rods are isolated
27 Thermal expansion of two rods showing (b) an intermediate amount of thermal expansion when rods are rigidly connected Totalstrain in each rod : A t B t A t l l 0 l l 0 B t
28 Thermal expansion of two rods showing (b) an intermediate amount of thermal expansion when rods are rigidly connected B m B th B t A m A th A t B t A t T T Totalstrain in each rod :
29 The difference in the thermal expansion coefficients of the materials leads to thermal stresses ' Resultant'strain and stress: T th th T E th
30 Thermal Shock Resistance Occurs due to: nonuniform heating/cooling Ex: Assume top thin layer is rapidly cooled from T 1 to T 2 Temperature difference that can be produced by cooling: quench rate ( T T ) 1 2 k set equal Tension develops at surface E (T 1 T 2 ) Critical temperature difference for fracture (set = f ) (T 1 T 2 ) fracture f E (quench rate) for fracture Thermal Shock Resistance (TSR) f k E Large TSR when rapid quench tries to contract during cooling T 2 resists contraction T 1 f k E is large 30
31 Thermal shock of brittle materials Ductile metals and polymers Thermal stresses are alleviated by plastic deformation Brittle ceramics Fracture due to thermal stress Thermal shock
32 Thermal shock resistance - capacity of a material to withstand failure due to thermal stress Thermal shock resistance parameter (high TSR prefered): f k TSR E where is the fracture strength of the material, k is the thermal conductivity, E is the elastic modulus and is thecoefficien t of thermal expansion. l f l
33 Application: Space Shuttle Orbiter Thermal Protection System Re-entry T Distribution Chapter-opening photograph, Chapter 23, Callister 5e (courtesy of the National Aeronautics and Space Administration.) Silica tiles ( ºC): 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 ) -- large scale application -- microstructure: Fig. 19.3W, Callister 5e. (Fig. 19.3W courtesy the National Aeronautics and Space Administration.) reinf C-C (1650ºC) silica tiles ( ºC) 100mm nylon felt, silicon rubber coating (400ºC) ~90% porosity! Si fibers bonded to one another during heat treatment. Fig. 19.4W, Callister 5e. (Fig W courtesy Lockheed Aerospace Ceramics Systems, Sunnyvale, CA.) 33
34 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 f k E Summary 34
35 Example What would be your choice of material for applications that require a high degree of dimensional stability with temperature fluctuations?
36 Why does fused silica have a small expansion coefficient? Low atomic packing density Inter-atomic expansion produces relatively small macroscopic dimensional changes.
37 Thermal expansion of polymers Linear and branched polymers Weak secondary bonds Higher coefficient of thermal expansion Crosslinked polymers Lower coefficient of thermal expansion
38 Thermal conductivity of metals Large number of free electrons Phonon contribution is lower as they are easily scattered.
39 Thermal conductivity of ceramics Thermal insulators No free electrons Thermal conduction via phonons
40 Why does glass (amorphous) have lower conductivity than crystalline ceramics? Phonon scattering is more effective when the structure is disordered & irregular
41 Thermal conductivity of polymers Energy is transferred via the vibration & rotation of the chain molecules.
42 Example An Al rod and a nylon rod have equal unstressed lengths at 20 C. If each rod is subjected to a tensile stress of 5x10 6 Pa, find the temperature at which the two stressed rods will have equal lengths. E E Al nylon th, Al 7010 th, nylon Pa Pa 0 6 C 0 1 C 1
43 ? Given : , 1 0 6, 9 9 T Pa same L same L C T C C Pa E Pa E f nylon th Al th nylon Al C T T T C T T E T E nylon th nylon Al th Al nylon th nylon m Al th Al m nylon t Al t : lengths & final initial have equal The rods
44 Soda-lime glass Vs Pyrex glass Pyrex is suitable for kitchen oven heating and cooling cycles
45 ANNOUNCEMENTS Reading: Core Problems: 19.1, 19.4, 19.8, 19.13, Self-help Problems: Concept Check 19.1, 19.2,19.5: 45
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