Materials Engineering (Week 5)

Question 1

Why Study Thermal Properties of
Materials?

Answer 1

Materials susceptible to thermal shock
brittle fracture from internal stress
due to cooling

Or, fracture from thermal gradients across coatings
E.g. Case Study 1 (Jet engine turbine blades)

• Function of thermal and mechanical properties
  Coefficient of thermal expansion, thermal conductivity,
  elasticity, fracture strength

Question 2

Heat Capacity:

Answer 2

General: The ability of a material to absorb heat
Quantitative: The energy required to increase the
temperature of the material.

Question 3

Heat Capacity Equation:

Answer 3

C = dQ / dT

C, heat capacity (J/mol - K)
dQ, energy input (J/mol)
dT, Temperature change (K)

Question 4

Two ways to measure heat capacity: (2)

Answer 4

Cp : Heat capacity at constant pressure.
Cv : Heat capacity at constant volume.

Cp > Cv

Specific heat has typical units of
J / kg x K

Question 5

Energy Storage
How is the (thermal) energy stored?

Answer 5

Phonons – thermal waves - vibrational modes

Question 6

Energy Storage

Debye temperature:

Answer 6

linked to highest
frequency of phonons

(usually less than Troom )

Question 7

Heat Capacity vs T

Heat capacity increases with what ?

Answer 7

Temp
SHC reaches a limiting value of 3R (R, gas constant)

Cv= AT^3 at low temp

Question 8

Heat Capacity vs T
Atomic view:

Answer 8

– Energy is stored as atomic vibrations.
– As T goes up, so does the avg. energy of atomic vibr.

Question 9

Heat Capacity: Comparison between metals and ceramics

Answer 9

Ceramics tend to have much higher SHC when compared to metals

Question 10

Energy Storage
* Other small contributions to energy storage

Answer 10

– Electron energy levels
* Contribution is relatively large in metals at low
temperature.

Question 11

Thermal Expansion

Answer 11

Materials change size when heating

Question 12

Thermal Expansion
Equation

Answer 12

L(final) - L(initial) / L(initial) = alpha x (Tfinal -Tinitial)

L, length
alpha, coefficient of thermal expansion (1/K or 1/°C)

Question 13

Thermal Expansion
Atomic View

Answer 13

Atomic view: Mean bond length increases with T.
bond energy vs bond length
curve is “asymmetric”

Question 14

Thermal Expansion: Comparison

Answer 14

Is different for each element in both metals and ceramics

Question 15

What is Thermal Conductivity

Answer 15

General: The ability of a material to transfer heat.

Question 16

Thermal Conductivity: Equation
Fourier’s Law

Answer 16

q = -k (dT / dx)

q, heat flux, (J/m2-s)
k, thermal conductivity (J/m-K-s)
dT / dx, temperature gradient

Question 17

Thermal Conductivity
Atomic View:

Answer 17

Atomic view: Atomic vibrations in hotter region carry
energy (vibrations) to cooler regions.

k = ke + kl
e – electronic, l - lattice k - thermal conductivity

Question 18

Thermal Conductivity: Comparison of energy transfer for metals and ceramics

Which type of material have a higher thermal Conductivity

Answer 18

Metals: By vibration of atoms and motion of electrons

Ceramics: By vibration of atoms

Metals have much higher Thermal Conductivity

Question 19

Explain the Thermal Conductivity trend for Ceramics

Answer 19

Intially decreases because of phonon scattering. Then
increases because of radiative heat transfer.

Question 20

Thermal Stress, occurs due to what? (2)

Answer 20

– uneven heating/cooling
– mismatch in thermal expansion.

Question 21

A material responds to heat by: (2)

Answer 21

– increased vibrational energy
– redistribution of this energy to achieve thermal equil.

Question 22

Heat capacity: (2)

Answer 22

– energy required to increase a unit mass by a unit T.
– metals somewhat higher than ceramics.

Question 23

Coefficient of thermal expansion: (2)

Answer 23

– the stress-free strain induced by heating by a unit T.
– metals higher than ceramics.

Question 24

Thermal conductivity: (2)

Answer 24

– the ability of a material to transfer heat.
– metals have the largest values

Question 25

Thermal shock resistance: (1)

Answer 25

– the ability of a material to be rapidly cooled and not
crack. Maximize sigma(f) k/E(alpha).