ch-19 compatibility mode

7/28/2019 Ch-19 Compatibility Mode

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CHAPTER - 19

THERMAL PROPERTIES

1

Problems to be practiced :

Problems related to thermal stress & strain, heat flux andheat capacity



ISSUES TO ADDRESS...

• How does a material respond to heat?

• How do we define and measure...

2

-- eat capac ty--coefficient of thermal expansion

--thermal conductivity

--thermal shock resistance

• How do ceramics, metals, and polymers rank?



General: The ability of a material to absorbheat.

HEAT CAPACITY

3

increase the temperature of the material.

energy input (J/mol)

C =

dQ

dT

heat capacity

(J/mol-K)temperature change (K)



Two ways to measure heat capacity:

-- Cp : Heat capacity at constant pressure.

4

-- Cv : Heat capacity at constant volume.



Heat capacity...

--

Heat Capacity Vs Temp

5

--reaches a limiting value of

3R



T K

Heat capacity, C v3R Cv= constant

gas constant = 8.31 J/mol-K

6

Debye temperature(usually less than T room )

• Atomic view:--Energy is stored as atomic vibrations.

--As Temperature goes up, so does the avg. energy of

atomic vibration.



HEAT CAPACITY: COMPARISON

• PolymersPolypropylenePolyethylenePolystyreneTeflon

cp (J/kg-K)

at room T

•

1925185011701050

cp: (J/kg-K)

Cp: (J/mol-K)

material

7

• cp significantly larger for polymers.

Magnesia (MgO)Alumina (Al2O3)

Glass

• MetalsAluminum

SteelTungstenGold

900

486128138

i n

c r e a s i n g c

p 940775

840



•

Materials change size when heating.T init

L init

THERMAL EXPANSION

8

L final − L initial

L initial= α ( T final − T initial )

coefficient of

thermal expansion (1/K)

finalL final



• Atomic view:

Mean bond length increases with Temp.

Bond energy

T 5 )

T 1 )

9

Bond length (r)

i n c r e

a s i n g T

T1

r

(

r

(

T5bond energy vs bond lengthcurve is “asymmetric”




145-180106-198

90-150126-216

α (10-6/K)

at room T

• Metals

Material

THERMAL EXPANSION:COMPARISON

10

• CeramicsMagnesia (MgO)Alumina (Al 2O3)

Soda-lime glassSilica (cryst. SiO 2)

13.5

7.6 9 0.4

SteelTungstenGold

.

124.5

14.2

i n c r e a s i n g

• Q: Why does α generally decrease with increasing bond

energy?



• General: The ability of a material to transfer heat.

• Quantitative:

q = −k dT

dx

temperature

gradient

thermal conductivity (J/m-K-s)

heat flux

(J/m2-s)

THERMAL CONDUCTIVITY

11

• Atomic view: Atomic vibrations in hotter region carry

energy (vibrations) to cooler regions.

K = Kl+K

e

T2 > T1 T1

x1 x2heat flux



k (W/m-K)

• Metals

AluminumSteelTungstenGold

24752

178315

Energy Transfer

By vibration of atoms andmotion of electrons

Material

THERMAL CONDUCTIVITY:

COMPARISON

12


0.120.46-0.500.130.25

•Magnesia (MgO)Alumina (Al2O3)

Soda-lime glassSilica (cryst. SiO 2)

3839

1.71.4

i n

c r e a s i n g

By vibration/rotation of chainmolecules

By vibration of atoms



• Occurs due to:

THERMAL STRESS

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----mismatch in thermal expansion.



• Example Problem 19.1, p. 666, Callister 6e.--A brass rod is stress-free at room temperature (20C).

--It is heated up, but prevented from lengthening.

--At what T does the stress reach -172MPa?

Troom

14

∆L

Lroom= εthermal = α(T − Troom )

LroomT

∆L

compressive σ keeps ∆L = 0 σ = E(−ε thermal ) = −Eα(T − Troom )

100GPa 20 x 10-6 /C

20CAnswer: 106C-172MPa



• Occurs due to: uneven heating/cooling.

• Ex: Assume top thin layer is rapidly cooled from T1 to T2:

rapid quench

THERMAL SHOCK RESISTANCE

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doesn’t want to contract

tries to contract during cooling T2

T1

Tension develops at surface

σ = − E α ( T 1 − T 2 )



• Reducing the thermal gradient by

– changing its temperature more slowly

– increasing the material's thermalconductivity

'

Thermal shock can be prevented by:

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thermal expansion

• Increasing its strength

• Increasing its toughness



Temperature difference that

can be produced by cooling:

(T1 − T2 )fracture =

σ f

Eα (T1 − T2 ) =

quench rate

k

Critical temperature

difference

for fracture (set σ = σ

f

)

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• Result:

• Large thermal shock resistance when is large.

σfk

Eα

(quench rate ) for fracture ∝σ f k

Eα



• Application:

Space Shuttle Orbiter

THERMAL PROTECTION SYSTEM

18



Re-entry TDistribution

19

reinf C-C(1650°C)

silica tiles(400-1260°C)

nylon felt, silicon rubber

coating (400°C)



• Silica tiles (400-1260C):--large scale application

20



--

microstructure:

21

100 µ m

~90% porosity Si fibers bonded to one another during

heat treatment.



• A material responds to heat by:

--increased vibrational energy

--redistribution of this energy to

achieve thermal e uilibrium.

SUMMARY

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Heat capacity:

--energy required to increase a unit

mass by a unit Temp.

--polymers have the largest values.



• Coefficient of thermal expansion:--the stress-free strain induced by

heating by a unit Temp.--polymers have the largest

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.• Thermal conductivity:

--the ability of a material totransfer heat.

--metals have the largest values.



Thermal shock resistance:

--the ability of a material to be

rapidly cooled and not crack.Maximize

σfk

Eα

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