材料科学基础课后习题答案13.docx

材料科学基础课后习题答案13.docx

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材料科学基础课后习题答案13

SolutionsforChapter13

1.FIND:

Cite5applicationswherethermalpropertiesarekey.

SOLUTION:

Herearebutafewofthemanyapplicationswherethermalpropertiesarecritical:

gasketsinspaceshipsparticularlynearliquidoxygen;oven-to-freezerdishes;insulationforhomesandcoats;solarstoragemedium;bi-metallicstrip;andacatalyticconvertersupport.

2.FIND:

Calculatetheenergyofaphoton.

GIVEN:

Itswavelengthis1micrometer.

DATA:

Plank'sconstant,h=6.63x10-33J-s.Thespeedoflight,c=3x108m/s.

SOLUTION:

Theequationthatisusedtocalculatetheenergyofawaveis:

E=hc/=（6.63x10-33J-s）（3x108m/s）/10-6m=2.0x10-18J.

3.Find:

Verifythatxv3xth.

Data:

Equation13.2-3v/vo=vT

Equation13.2-1th=th/T

Solution:

vT=v/vo=（（Lo+L）3-Lo3）/Lo3=（3Lo2L+3LoL2+L3）/Lo3

WeneglectthetermsoforderL2andL3becausetheyareverysmallcomparedtotheterm3Lo2L,so

vT（3Lo2L）/Lo3=3（L/Lo）.

Recognizingthatth=L/Lo=thT,wecanwrite

vT3（thT）orv3th.

4.Find:

Relationshipsbetweenbond-energycurvesandthermalproperties.

Solution:

A.SincematerialIIhasthemoreasymmetricbond-energywellitwillhavethelargercoefficientofthermalexpansionand,therefore,thelargerthermalstrainforagivenT.

B.SincematerialIhasthedeeperbond-energycurveitwillhaveahighermeltingtemperature.

5.FIND:

Doceramicshavelowthermalexpansioncoefficientsbecausetheyhaveopenstructures?

DATA:

AlookatthelistofthermalexpansioncoefficientsofmetalsinTable13.2-1showsarangeof5to83x10-6/C.Ceramicsrangefrom0.5to14x10-6/C.Ingeneral,thepolymersarehigherthanthemetalsorceramics.

SOLUTION:

Thehugerangeofthermalexpansioncoefficientinmetals,manybeinglessthanthatofMgO,andthefactthatpolymersareperhapsthemostopenstructuressuggesttheargumentistentative.

6.FIND:

Whydopolymerfibersgenerallyhavenegativeaxialthermalexpansioncoefficients?

SOLUTION:

Polymermoleculeswanttocoilonthemselves.Intextilefibers,andevenmoresoinhighperformancefibers,themoleculesarestretchedoutalongthefiberaxis.Whenthepolymerisheatedthemoleculesgainsomemobility,theyseektolowertheirenergystatebyshrinkingtomovetowardscoiling.

7.

Find:

MaximumTforv~3thtogive<1%errorforcopper.

Data:

th（Cu）=17x10-6oC-1fromTable13.2-1.

Solution:

Inthesolutiontoproblem#3theterm（3Lo2L/Lo3）wasretainedwhiletheterms（3LoL2/Lo3）and（L3/Lo3）wereassumedtobenegligible.Theerrorintroducedviathisassumptionis:

[（3LoL2/Lo3）+（L3/Lo3）]/[（3Lo2L/Lo3）]=L/Lo+3L2/Lo2=th+3th2.Nowthiserrorisequalto1%whenth+3th2=0.01.Solvingforthestraingivesth=[-1±（1-4（3）（-0.01））]/6=[-1±（1.12）]/6=9.72x10-3.SolvingforTandusingth（Cu）=17x10-6oC-1givesT=th/th=（9.72x10-3）/（17x10-6oC-1）=572oC.

8.Find:

thfromtheinformationgivenbelow.

Given:

MetalrodwithL=2mat25oCandL=2.002mat1000oC.E=20.7GPa,D=4cm.

Data:

Equation13.2-1th=th/T.

Solution:

th=th/T=（L/Lo）/T=[（Lf-Lo）/Lo]/T

th=[（2.002m-2m）/2m]/（1000-25）oC

th=1.025x10-6oC-1.

9.FIND:

Howcanapolyimidehavelowplanarthermalexpansioncoefficients?

GIVEN:

SomeJapanesescientistshaveallegedlydevelopednewpolyimideswithalowthermalexpansioncoefficient.

SOLUTION:

Takeanindividualpolymermolecule.Itwillhaveathermalexpansioncoefficientparalleltoamerandanotherthermalexpansioncoefficientnormaltothemer.Byputtingthelowthermalexpansioncoefficientdirectioninaplane,thein-planethermalexpansioncoefficientisminimalandtheoutofplanethermalexpansioncoefficientishigh.Alternatively,ifthestretchedoutpolymerisintheplane,thenitcontributesanegativethermalexpansioncoefficienttothecomposite.

COMMENTS:

Suchpolyimidesareusefulinintegratedcircuits,whereagoodthermalexpansioncoefficientmatchbetweenceramic,metal,andpolymerisdesirable.

10.FIND:

Calculatetheamountofexpansion-contractionthatmustbedesignedforinanexpansiongateonabridge.

GIVEN:

Eachspanisabout100feet.

ASSUMPTIONS:

Theminimumtemperatureisabout-40Candthemaximumtemperatureisabout40C.Thebridgestructureisplaincarbonsteel.

DATA:

Thethermalexpansioncoefficientof1020steel,whichmaybethestructuralmaterialusedinbridgeconstruction,is12x10-6/C.

SOLUTION:

Thetemperatureextremesareperhaps-40to40CinWisconsin.Thus,T=80C.Thethermalstrainisgivenby:

=thT=（12x10-6/C）（80C）=9.6x10-4.

But,

=9.6x10-4=l/l=l/100ft.

Hence,l=0.096ft.x12in./ft.=1.15inches.

11.Find:

Temperaturechangerequiredtocausea1mmlengthchangeinAl,PC,andSiCrods.

Given:

LAl=LPC=LSiC=1.0mat25oC

Data:

FromTable13.2-1:

Al=25x10-6oC-1PC=66x10-6oC-1SiC=4.8x10-6oC-1

Equation13.2-1th=th/TT=th/th

Solution:

Thestrainiscalculatedasth=L/Lo=0.001m/1.0m=1x10-3.TherequiredtemperaturechangeisT=1x10-3/th.

ForAl:

T=1x10-3/25x10-6oC-1=40oC

ForPC:

T=1x10-3/66x10-6oC-1=15.2oC

ForSiC:

T=1x10-3/4.8x10-6oC-1=208oC.

12.Find:

Relativethermalexpansioncoefficientsformetals,ceramics,thermoplasticsandthermosets.

Solution:

Ingeneral,ceramicshavethelowestvalue,metalshavetheintermediatevalues,andpolymershavethehighestvalues（with（TP）>（TS））.Thereasonforthistrendis,ofcourse,theshapesoftherespectivebond-energycurves.Deepcurves,suchasthoseforceramics,tendtobesymmetricwhileshallowcurves,suchasthoseforthermoplasticpolymers,areveryasymmetric.

13.Find:

Whichhasahighermeltingtemperature:

FeorW?

Data:

FromTable13.2-1:

（Fe）=12x10-6oC-1and（W）=5x10-6oC-1

Solution:

ThelowervalueforWsuggeststhatthismetalhasacomparativelysymmetricbond-energywell.Sincesuchwellsareusuallydeep（indicatinghighTm）anddisplayhighcurvature（indicatinghighE）wesuspectthatWhasahigherTmandahigherEthanFe.

Comment:

Tm（Fe）=1535oCwhileTm（W）=3410oC

E（Fe）~170GpawhileE（W）=408Gpa

14.FIND:

Calculatethesaginthewire.

GIVEN:

Thesagat-28C,theminimumtemperature,is20feet.Themaximumallowablesagissignificantlylessthan75feet.Themaximumtemperatureis35C.thepolespanis1000ft.

ASSUMPTIONS:

Creepisnotaccountedfor.

APPROXIMATIONS:

Theshapeiscatenary.Idon'tknowtheequationofacatenary.Iwillthereforeapproximate（crudely）theshapeasastraightlinefromthepoletothemidpoint.Atthelowesttemperature,itis500feetacrossand20feetdowntothelowestpoint.Perhapsyoudidtheproblemmoreaccurately.

DATA:

ThethermalexpansioncoefficientofCuis17x10-6/C,accordingtoTable13.2-1.ThethermalexpansioncoefficientofKevlaris-4x10-6/C（S.B.Warner,FiberScience,Prentice-Hall,1995,p.241）.

SKETCH:

SOLUTION:

Let'sjustprobetheproblemwithouttryingtogetaccuratevalues.WebeginbyusingthePythagoreantheoremtodeterminethelengthofthewirethatenablesasagof20ftasperthesketch:

5002+202=length2length,l=500.4ft.

Onlya0.4ft.=5inchgrowthcausesthedeflectionshowninthefigure!

（Thecatenaryismoreforgivingthanisthetriangleshape.）

Nextweusetheequation:

=l/l=thT.

Here,

T=35C-（-28C）=63Candl=500.4ft,thehalf-length.

Hence,

l=500.4ftxthx63C.

SubstitutionforCu,Kevlar,andglassgives:

l（Cu）=500.4ft17x10-6/Cx63C=0.54ft.

l（Kevlar）=500.4ft-4x10-6/Cx63C=-0.12ft.

l（glass）=500.4ft0.55x10-6/Cx63C=0.005ft.

UsingthePythagoreantheoremagain:

500.942-5002=sag2sag=31ftwithcopper

ThesagwithKevlaralonewouldbereduced,buttheglasswantstoincreasethesagslightly.Thenetchangeinlengthwiththeincreasedtemperature,whichisdifficulttocompute,wouldbesmall.

15.Find:

Anexampleofamaterialthatcontractswhenheated.

Solution:

AsdiscussedinSection6.6.5,elastomerscontractwhenheated.

16.Find:

Influenceofamorphous/crystallinestructureonthecoefficientofthermalexpansion.

Solution:

Sinceamorphousstructurearemore"open"thantheircrystallinecounterpartstheyareabletoaccommodatesomeofthetemperatureinducedthermalvibrationswithoutexpansion.Thus,forTTg,however,（amorphous）isapproximatelyequaltothatofthecorrespondingliquidandisthereforeaboutthreetimeslargerthan（crystalline）.

17.

Find:

WhatTisrequiredtoincreasetheroomtemperaturevolumeofSiby1%?

Whataboutadecreaseof1%?

Data:

FromTable13.2-1:

（Si）=2.6x10-6oC-1

FromEqn.13.2-2:

V/Vo=vT

FromEqn.13.2-3:

v3

Solution:

v（Si）3（（Si））=3（2.6x10-6oC-1）=7.8x10-6oC-1

Fora1%expansion:

V/Vo=0.01T=0.01/v=

0.01/7.8x10-6oC-1,thereforeT=1282oCT=1282oC+25oC=1307oC.

Fora1%contraction:

T=-0.01/v=0.01/7.8x10-6oC-1=-1282oC.

Comment:

The1%volumecontractionisnotpossiblesincetherequiredtemperatureisbelowabsolutezero.

18.

Find:

Shouldyouusehotorcoldwatertoremoveametallidfromaglassjar?

Solution:

Sincemetalsgenerallyhavelargerexpansioncoefficientsthan（oxide）glasses,thedimensionsofthelidwillchangemorethanthedimensionsofthejarforanytemperaturechange.Ifcoldwaterisusedthelidwillcontractmorethanthejarandthefitwillbetighter.If,however,hotwaterisusedthenthelidexpandsmorethanthejaranditwillbeeasiertoremove.

19.

Find:

Influenceofmaterialtypeonaccuracyofameasuringtape.

Solution: