Mechanical Engineering DepartmentAlabama A&M University

Fall 2014, Lecture 18

Mechanical Behavior: Part III

Dr. Aaron L. Adams,

Assistant Professor

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

dQ QC

dT T

• 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:

Cp : Heat capacity at constant pressure.

Cv : Heat capacity at constant volume.

Cp usually > Cv

• Heat capacity has units of: J Btu J

or or mol K lb mol F g-atom K

Is the ability of a material to absorb heat

3

qc

m T

Specific heat(J/kg-K)

energy input (J/mol)

temperature change (K)

Specific Heat

• Specific heat has units of: J

kg K

Common to re-define heat capacity using a basis of unit mass (because we tend to measure mass).

mass

4

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.

which can be re-written as q

c q c m Tm T

3

J1880 2 kg 100 C 25 C

kg×K

282 10 J

282 kJ

Substitution of known values yields:

5

incr

eas

ing

cp • Why is cp significantly

larger for polymers?

Selected values from Table 17.1, Callister & Rethwisch 4e.

• PolymersPolypropylene Polyethylene Polystyrene Teflon

cp (J/kg-K)at room T

• CeramicsMagnesia (MgO)Alumina (Al2O3)Glass

• MetalsAluminum Steel Tungsten Gold

1925 1850 1170 1050

900 486 138 128

cp (specific heat): (J/kg-K)

Material

940 775 840

Specific Heat: Comparison

Cp (heat capacity): (J/mol-K)

Specific heat is per unit mass.

• How would the value for H2O () compare? cp (H2O) = 4186 J/kg-K !

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• 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 3R = 8.31 J/mol-K

Cv = constant

Debye temperature (usually less than Troom )

T (K)qD0

0

Cv

Cv = AT3

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Thermal Expansion

Materials change size when temperature is changed

)(α initialfinalinitial

initialfinal TT

linear coefficient of thermal expansion (units = K-1 or C-1)

Tinitial

Tfinal

initial

final

Tfinal > Tinitial

Coefficient of linear expansionCoefficient of expansion

dT

dL

lo

1

8

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 bondingTypical of metals

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Coefficient of Thermal Expansion: Comparison

• Q: Why does a

generally decrease with increasing bond energy?

Polypropylene 145-180 Polyethylene 106-198 Polystyrene 90-150 Teflon 126-216

• Polymers

• CeramicsMagnesia (MgO) 13.5Alumina (Al2O3) 7.6Soda-lime glass 9Silica (cryst. SiO2) 0.4

• MetalsAluminum 23.6Steel 12 Tungsten 4.5 Gold 14.2

a (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

incr

eas

ing

is typically smaller for ceramics & glasses than for metals

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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

mm 12m 012.0

)]C9(C40[)m 15)](C/1(10 x5.16[ 60

T

rearranging Equation 17.3b

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In class example:

= which can be re-written as: o oo

dL LL L L T

LdT L T

1o o o oL L L T L L T

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.

6 6 14.5 4.810 mm/mm C =4.65 10 C

2avg

6 10.01 m 1 4.65 10 C 500 27 C

0.01002 m 10.02 mm

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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 (oC-

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

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Pendulum and balance wheels in clocks

Directional bending; more deflection for a given temperature change

Invar Applications

Thermostat in your homeWatches

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The ability of a material to transport heat.

temperaturegradient

thermal conductivity (J/m-K-s)

heat flux(J/m2s)

• Atomic perspective: Atomic vibrations and free electrons in hotter regions transport energy to cooler regions.

T2 T2 > T1

T1

x1 x2heat flux

Thermal Conductivity

Fourier’s Law

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Heat transfer by conduction is defined by Fourier’s law .

1dQ dTk

dt A dx

heat flux per unit area

1dQ dTq k

dt A dxdT

q kdx

At steady state: dQ/dt = constantdQ/dt = the heat flux

Like diffusion!

fluxconstant

Temperature gradient

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( / )

( / )

dQ dtk

A dT dx

mm

kg

s

m

sm

kg

x

C-DJ

gradiention concentratx

C

position x s

m ][ diffusion oft coefficien D

s m

kg ][ flux mass J

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2

2

2x

Mass Transfer – Mass Flux - Diffusion all are equivalent and described by Fick’s Law of diffusion

m

mkg 3/ ][

[SI units]

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Molecular Heat Transfer – Conduction is described by Fourier’s Law of Conduction

m

K

Ksm

J

sm

Jx

T-kq

gradient etemperaturx

T

position x Km

W

Km

J ][ty conductivi thermalk

s m

J ][ flux heat q

2

x

2x

s

[SI units]

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What is k?

k = kl + ke

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

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Thermal Conductivity: Comparisonin

crea

sing

k

• PolymersPolypropylene 0.12Polyethylene 0.46-0.50 Polystyrene 0.13 Teflon 0.25

vibration/rotation of chain molecules

• CeramicsMagnesia (MgO) 38Alumina (Al2O3) 39 Soda-lime glass 1.7 Silica (cryst. SiO2) 1.4

atomic vibrations

• MetalsAluminum 247Steel 52 Tungsten 178 Gold 315

atomic vibrations and motion of free electrons

k (W/m-K)Energy Transfer

MechanismMaterial

Selected values from Table 17.1, Callister & Rethwisch 4e. Vice-versa of heat capacity! 20

• Application:

Space Shuttle Orbiter

• Silica tiles (400-1260C):-- 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 CeramicsSystems, Sunnyvale, CA.)

Thermal Protection System

reinf C-C (1650°C)

Re-entry T Distribution

silica tiles(400-1260°C)

nylon felt, silicon rubbercoating (400°C)

~90% porosity!Silica fibersbonded to oneanother duringheat treatment.

100 mm

Chapter-opening photograph, Chapter 23, Callister 5e (courtesy of the National Aeronautics and Space Administration.)

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A comment on k

metal ceramic

radiant heat transfer (IR heat)

why the drop? why the drop?

CuZn

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Outer surface heats up first and lattice values want to expand – but can’t because the interior is at a lower temperature….

heat

Thermal StressesWhat 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

• Occur due to: -- restrained thermal expansion/contraction -- temperature gradients that lead to differential dimensional changes

Thermal Stresses

0( )fE T T E T Thermal stress

http://www.121designs.com/images/wheel.jpg

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)

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-- 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

T0

0

Solution:

Original conditions

Tf

Step 1: Assume unconstrained thermal expansion 0 D

Step 2: Compress specimen back to original length

0

D

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Example Problem (cont.)

0

The thermal stress can be directly calculated as

Noting that compress = -thermal and substituting gives

20 x 10-6/°CAnswer: 106°C 100 GPa

20ºC

Rearranging and solving for Tf gives

-172 MPa (since in compression)

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• Occurs due to: nonuniform heating/cooling• Ex: Assume top thin layer is rapidly cooled from T1 to T2

Tension develops at surface

Critical temperature differencefor fracture (set s = sf)

set equal

• Large TSR when is large

Thermal Shock Resistance

Temperature difference thatcan be produced by cooling:

srapid quench

resists contraction

tries to contract during cooling T2

T1

•

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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

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