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Chemical Engineering 378

Science of Materials Engineering

Lecture 35Thermal Properties

Spiritual Thought2

Proverbs 3:5-6

5 Trust in the Lord with all thine heart; and lean not unto thine own understanding.6 In all thy ways acknowledge him, and he shall direct thy paths.

Materials Roadmap3

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OEP-10Nuclear Emergency

Group Work Okay, Due 12/9/2020 at beginning of class(Don't be afraid to "Google" for reasonable assumptions; just provide references!)

The year is 2040, and we have got some serious problems! A foreign country has infected your local light water nuclear reactor with their version of a Stuxnet worm. Unlike Stuxnet however, this virus targets the reactor cooling and reactivity control systems, resulting in a slow heat-up and eventual meltdown of the fuel. Thankfully, you, as a successful chemical engineer with expertise in material science, are available to help! The reactor manager is interested in placing some type of shell around the containment building dome, which could prevent the release of radionuclides to the nearby town in which you and many others live as a result of the meltdown accident. In order for this to work, the shell should not fail at an internal pressure of 500 psia, and a temperature of 400 °C. Further, it should be able to conduct 30 MW of heat from the inside of the shell to the outside, so as to prevent further core damage. Finally, this shell should be made of a material that will be able to be machined into the desired dome shape. Alternatively, the manager is willing to consider a smaller shield constructed directly around the pressure vessel of the reactor. In this case, the material should be able to resist a pressure of 2500 psi and a temperature of 700 °C. The conduction requirement for this shield is the same as for the larger shield dome. However, none of the alloying components in this shield should be smaller in atomic number than 9, so as to avoid neutron activation in the material. Pick a material and configuration that will allow you to protect the public from the effects of this disaster.

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• 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

The ability of a material to absorb heat

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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.-- As temperature increases, the average energy of

atomic vibrations increases.

Dependence of Heat Capacity on Temperature

Adapted from Fig. 19.2, Callister & Rethwisch 10e.

R = gas constant 3R= 8.31 J/mol-K

Cv = constant

Debye temperature (usually less than Troom )

T (K)θD0

0

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Atomic Vibrations

Atomic vibrations are in the form of lattice waves or phonons

Fig. 19.1, Callister & Rethwisch 10e. (Adapted from “The Thermal Properties of Materials” by J. Ziman. Copyright © 1967 by Scientific American, Inc. All rights reserved.)

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incr

easi

ng c

p

• Why is cp significantlylarger for polymers?

Selected values from Table 19.1, Callister & Rethwisch 10e.

• PolymersPolypropylenePolyethylene Polystyrene Teflon

cp (J/kg-K)at room T

• CeramicsMagnesia (MgO)Alumina (Al2O3)Glass

• MetalsAluminum Steel Tungsten Gold

19251850 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)

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Thermal Expansion

Materials change size when temperature is changed

linear coefficient ofthermal expansion (1/K or 1/°C)

Tinitial

Tfinal

initial

final

Tfinal > Tinitial

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Atomic Perspective: Thermal Expansion

Fig. 19.3, Callister & Rethwisch 10e. (Adapted from R. M. Rose, L. A. Shepard, and J. Wulff, The Structure and Properties of Materials, Vol. IV, Electronic Properties, John Wiley & Sons, 1966. Reproduced with permission of Robert M. Rose.)

Asymmetric curve:-- increase temperature, -- increase in interatomic

separation -- thermal expansion

Symmetric curve:-- increase temperature, -- no increase in interatomic

separation -- no thermal expansion

11Coefficient of Thermal Expansion: Comparison

• Q: Why does αgenerally decreasewith increasingbond 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

α (10-6/°C)at room T

Material

Selected values from Table 19.1, Callister & Rethwisch 10e.

Polymers have larger α values because of weak secondary bonds

incr

easi

ng α

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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

rearranging Equation 17.3b

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The ability of a material to transport heat.

temperaturegradient

thermal conductivity (J/mKs)heat flux(J/m2s)

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

T2 T2 > T1T1

x1 x2heat flux

Thermal Conductivity

Fourier’s Law

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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 19.1, Callister & Rethwisch 10e.

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

dimensional changes

Thermal Stresses

Thermal stress

-- 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 ∆

Step 2: Compress specimen back to original length 0

σ σ

∆

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Example Problem (cont.)

0

σ σ

The thermal stress can be directly calculated as

Noting that εcompress = -εthermal and substituting gives

20 x 10-6/°CAnswer: 106°C 100 GPa

20°C

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 σ = σf)

set equal

• Large TSR when is large

Thermal Shock Resistance

Temperature difference thatcan be produced by cooling:

σrapid quench

resists contraction

tries to contract during cooling T2T1

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Chemical Engineering 378��Science of Materials Engineering��Spiritual ThoughtMaterials RoadmapOEP-10Heat CapacityDependence of Heat Capacity on TemperatureAtomic VibrationsSpecific Heat: ComparisonThermal ExpansionAtomic Perspective: Thermal ExpansionCoefficient of Thermal Expansion: ComparisonThermal Expansion: ExampleThermal ConductivityThermal Conductivity: ComparisonThermal StressesExample ProblemExample Problem (cont.)Thermal Shock Resistance