PartI BasicEnergyPhysicsandUses
Problems
1.1 Confirmeq.(1.14)andtheestimatefortherhinoceros’s kineticenergybeloweq.(1.11)byexplicitcalculation.
Themassofaprotonis Mp = 1 673 × 10 27 kg.Whenaproton andanti-proton,whichhasthesamemassastheproton,annihilate, theenergyreleasedisequaltotheenergyequivalenttothetotal massofthesystem.So,thisenergyis
E = (Mp + Mp )c 2 = 2Mpc 2 = 2(1 673 × 10 27 kg)(2 998 × 108 m/s)2 = 3 007 × 10 10 J
InunitsofMeV,theenergyis
E = 3 007 × 10 10 J × 1MeV 1 602 × 10 13 J = 1877MeV
A3000kgrhinoceroschargingat50km/hourhasatotalkinetic energyof
E = 1 2 mv 2 = 0 5(3000kg) (50km/h) 1000m km 1h 3600s 2 × 1kJ 1000kgm2 /s2 = 289kJ
1.2 Howmuchenergywoulda100Wlightbulbconsume ifleftonfor10hours?
Thelightbulbconsumes
(100W)(10h) = (100J/s)(10h) 3600s 1h = 3 6 × 106 J = 3 6MJ
1.3 Inatypicalmid-latitudelocation,incidentsolarenergy averagesaround200W/m2 overa24-hourcycle.Computethelandareaneededtosupplytheaverageperson’senergyuseof200MJ/dayifsolarpanelsconvert anet5%oftheenergyincidentoveralargeareaof landtousefulelectricalenergy.Multiplybyworldpopulationtogetanestimateoftotallandareaneededto supplytheworldenergyneedsfromsolarpowerunder theseassumptions.Comparethislandareatosomerelevantreferenceareas–theSaharaDesert,yournative country,etc.
Thetotallandareaneededtosupplyanaverageperson’senergy useis
A1person = 200MJ/day (0 05)(200W/m2) 106 W MJ/s 1day 86400s = 231m2
Earthhasatotalpopulationofabout7 × 109 ,sothetotalland areaneededtosupplyworldenergyneedsis
Atotal = (231m2 /person)(7 × 109 persons) ≈ 1 6 × 106 km2 ThisisapproximatelythesizeofAlaskaorIran,roughly17% of theareaoftheSaharaDesert.
1.4 TheUStotalenergyconsumptionin2014was98.0 quads.Whatisthisquantityinjoules?About83%of USenergycomesfromfossilfuels.Ifthewhole83% camefromoil,howmanytonsofoilequivalent(toe) wouldthishavebeen?Howmanybarrelsofoil(bbl)is thisequivalentto?
TheUStotalenergyconsumptioninjoulesis
98 0quads 1 055 × 1018 J 1quad = 1 03 × 1020 J
If83%ofthiswerefromoil,thisisequivalentto
0 83(1 03 × 1020 J) 1toe 41 868 × 109 J = 2 04 × 109 toe
Thisisequivalentto
0 83(1 03 × 1020 J) 1bbl 6 118 × 109 J = 1 40 × 1010 bbl
1.5 Thegravitationalpotentialenergyofanobjectofmass m atadistance h abovegroundisgivenby E = mgh Usedimensionalanalysistocomputetheunitsof g Whatdoesthisresultsuggestaboutthebehaviorof objectsnearEarth’ssurface?
Usingdimensionalanalysis,
[g] = [energy] [mass][distance] = [mass][distance]2 /[time]2 [mass][distance] = [distance] [time]2
Thisisanacceleration,suggestingthatnearthesurfaceallobjects undergothesameaccelerationduetogravity.
1.6 Theenergyemittedorabsorbedinchemicalprocessesisoftenquotedinkilojoulesper mole (abbreviatedmol)ofreactants,whereamolecontains NA 6 022 × 1023 (Avogadro’snumber)molecules(§5). DerivetheconversionfactorfromeV/moleculeto kJ/mol.
1 eV molecule = 1 eV molecule 1 602 × 10 19 J eV 1kJ 1000J × 6 022 × 1023 molecules 1mol = 96 47 kJ mol
1.7 Theenergyavailablefromonekilogramof 235Uis 82TJ.Energyisreleasedwheneachuraniumnucleus splits,or fissions.(Thefissionprocessisdescribedin §16,butyoudonotneedtoknowanythingaboutthe processforthisproblem.)235gramsof 235Ucontain approximatelyAvagadro’snumber(seeProblem1.6) atomsof 235U.Howmanymillionsofelectronvolts (MeV)ofenergyarereleased,onaverage,whena 235Unucleusfissions?Howmanykilogramsofgasolinehavethesameenergycontentasonekilogramof 235U?
TheenergyreleasedpernucleusinMeVis
E 82 TJ kg 1012 J TJ 0 235kg 1mol 1mol 6 022 × 1023 atoms × 1eV 1 602 × 10 19 J 1MeV 106 eV 200 MeV atom
Weneedtheenergycontentofgasoline,120MJ/gallon(US), andthedensityofgasoline,0.755kg/L.Onekgof 235Uis equivalentto
M(gasoline) = 82 × 1012 J 1gallon(US) 120 × 106 J 3 785L gallon(US) 0 755kg L = 1 834 × 106 kg
1.8 TheUStotalelectricalpowerconsumptionin2010was 3.9TkWh.Utilitiestrytomaintainacapacitythatis twicetheaveragepowerconsumptiontoallowforhigh demandonhotsummerdays.Whatinstalledgenerating capacitydoesthisimply?
Theinstalledgeneratingcapacityis
2(3 9 × 1015 Wh/y)/(8766h/y) 890GW
Problems
2.1 Estimatethekineticenergyoftheoceanliner Queen Mary2,withmass(displacement)76000tons,movingatacruisingspeedof26knots.Comparewiththe changeinpotentialenergyofthe QueenMary2 when liftedbyathreefoottide.
Theknotisanauticalunitofspeed,1knot = 0.5144m/s,so 26knots = 13.4m/s.TheQueenMary2’skineticenergyisthen
Ekin = 1 2 mv 2 = 0 5(76 × 106 kg) (13 4m/s)2 = 6 82GJ
Thechangeinpotentialenergywhenliftedbyathreefoottideis
V = mgh = (76 × 106 kg)(9 81m/s 2)(3ft) 0 3048m 1ft = 663MJ , orabout10%ofthekineticenergy.
2.2 [T]Amass m movesundertheinfluenceofaforce derivedfromthepotential V(x) = V0 cosh ax,where thepropertiesofcosh x andother hyperbolicfunctions aregiveninAppendixB.4.2.Whatistheforceon themass?Whatisthefrequencyofsmalloscillations abouttheorigin?Asthemagnitudeoftheoscillations grows,doestheirfrequencyincrease,decrease,orstay thesame?
Theforceonthemassis F(x) = dV/dx = V0a sinh ax, withsinh x = 1 2 (ex e x ).Forsmall x,theexponentialfunction isapproximatedusingaTaylorexpansionby
ey = 1 + y + y2 2 + y3 6 + ...,
socosh y = 1+y2 /2+ andsinh y = y+y3 /6+ .Forsmalloscillations, i.e. when ax ≪ 1,soweonlyneedkeepthefirstterminthe expansionofsinh ax,theforceis F(x) = V0 a2 x.Theequationof motionforsmalloscillationsis
mx = V0a 2 x ,
whichistheequationfora harmonicoscillator (seeeq.(2.10))with frequency ω = a √V0/m
Thisfrequencyisindependentofmagnitudeofoscillations.If, however,theamplitudeoftheoscillationbecomeslargecompared to1/a,thenthenon-lineartermsintheexpansionofsinh x cannot beignoredandtheequationofmotionbecomes
mx = V0a ax + 1 6 a 3 x 3 + ,.
Allofthetermsintheexpansionhavethesamesign,soasthe amplitudeofoscillationsincreases,themagnitudeoftheforce increasescomparedtojustasimpleharmonicoscillatorand the frequencyoftheoscillationsincreases.
2.3 [T]Anobjectofmass m isheldneartheoriginbyaspring, F = kx.Whatisitspotentialenergy?Showthat x(t) = (x0, y0, z0)cos ωt is apossibleclassicalmotionforthemass.Whatis theenergyofthissolution?Isthisthemostgeneralmotion?Ifnot,giveanexampleofanother solution.
Accordingtoeq.(2.18),theforceisthenegativegradientofthe potentialenergy.Setting F = kx = ∇V,thesolutionis V(x) = 1 2 kx2 + C ,
whichcanbecheckedbydifferentiatingwithrespecttoeachcoordinate.Wesettheintegrationconstant C = 0sothatthepotential energyiszeroat x = 0.
Theequationofmotionis mx = F = kx,andsuggested solutioncanbewritten x(t) = x0 cos ωt,where x0 = (x0 , y0 , z0 ). Differentiatingtwicewithrespectto t wesee¨x = ω2 x,which satisfiestheequationofmotionif ω = √k/m
Thetotalenergyisthesumofkinetic, Ekin = 1 2 mx2 ,and potential, V(x).
wherewehaveused mω2 = k andsin2 + cos2 = 1. Thissolutionisnotthemostgeneralsinceitrequirestheoscillatortobemaximallystretchedat t = 0.Itiseasytoshowthat x(t) = x0 cos ω(t t0 )isasolutionforany t0 .Usingtheidentity cos(θ φ) = cos θ cos φ + sin θ sin φ,itisclearthatthissolutionis asumofsineandcosinefunctions.
2.4 Estimatethepotentialenergyoftheinternationalspace station(mass370000kg)relativetoEarth’ssurface wheninorbitataheightof350km.Computethevelocityofthespacestationinacircularorbitandcompare thekineticandpotentialenergy.
Fromeq.(2.20),thepotentialenergyrelativetothesurfaceis
V(h) = GmM⊕ R⊕ + h + GmM⊕ R⊕ = (6 67 × 10 11 Nm2 /kg2 )(3 7 × 105 kg)(6
Inacircularorbit,thegravitationalforceisequaltothemass timesthecentripetalacceleration, mv2 r = GM⊕ m r2
Solvingforvelocity,
v = GM⊕ r = (6 67 × 10 11 )(6 × 1024 ) (6 37 × 106 + 3 5 × 105 ) m/s = 7 7km/s
Thekineticenergyisequalto
Ekin = 1 2 mv 2 = 1 2 (3 7 × 105 kg)(7 7 × 103 m/s)2 = 11TJ
Thus,thekineticenergyis9 1timesthepotentialenergyrelativeto Earth’ssurface.
2.5 [T]Relateeq.(2.20)toeq.(2.9)andcompute g interms of G, M⊕, R⊕ asdescribedinthetext.
Thepotentialyieldingthe r 2 forcelawis V(r) = GMm/r Intheapproximationthatthegravitationalforceisconstant,the potentialis V(z) = mgz,where z = r R⊕ istheheightfromEarth’s surface, R⊕ isEarth’sradiusand z ≪ R⊕.Bydefinition,
V(z) V(0) = GM⊕ m R⊕ + z + GM⊕ m R⊕
Useaseriesexpansionfor1/(R⊕ + z),(seeeq.(B.63)), 1 R⊕ + z = 1 R⊕ z R2 ⊕ + ..., giving
V(z) V(0) = GM⊕ m R2 ⊕ (z + O(z 2 /R3 ⊕))
Comparingthisto V(z) = mgz,andsetting V(0) = 0,wegetto leadingorderin z/R⊕, g = GM⊕ R2 ⊕ = (6 67 × 10 11 Nm2 /kg2)(5 97 × 1024 kg) (6 37 × 106 m)2 = 9 81m/s 2
2.6 Makearoughestimateofthemaximumhydropower availablefromrainfallintheUSstateofColorado.LookuptheaverageyearlyrainfallandaverageelevationofColoradoandestimatethepotentialenergy(withrespecttosealevel)ofallthe waterthatfallsonColoradooverayear.How doesthiscomparewiththeUSyearlytotalenergy consumption?
Coloradohasanaverageyearlyprecipitationof 0.4m.SupposethisistypicalofthewholestateofColorado.Colorado hasa meanelevationofapproximately2070mandatotalsurfacearea of270000km2 .Alltheprecipitationfallingonthestateinoneyear thenhasatotalpotentialenergyof mgh ≈ ρVgh (1000kg/m3)(0 4m)(2 7 × 1011 m2 )
Thisisabout2%ofthetotalenergyconsumedintheUSeachyear.
2.7 Chooseyourfavoritelocalmountain.Estimatehow muchenergyittakestohiketothetopofthemountain (ignoringallthelocalupsanddownsofatypicaltrail). Howdoesyourresultcomparetoatypicalday’sfood energyintakeof10MJ,takingintoaccountthefactthat humanmusclesarenot100%efficientatconverting energyintowork?
MountWashingtonisthetallestmountainintheStateofNew Hampshire?ThepeakofMountWashingtonisat1917meters (6288feet)abovesealevel.Atypicalroute,however,beginsat about600metersabovesealevel,sotheelevationgainisroughly 1300m.Forahikerof80kg(includingfood,water,extraclothing, etc.),thepotentialenergygainonthehikeis
V = mgh (80kg)(9 8m/s 2 )(1300m) 1MJ (2.45)
At20-30%muscleefficiency,thisrepresentsaboutonethirdtoone halfoftheusefulenergyoutputfromthetypicalhiker’s10MJ/day foodenergyintake.Thisiscomputedjustfromtheelevation gain, withoutincludinglocalupsanddownsofthetrail,horizontaldistancecovered,orthehikedown.Sodon’tfeelbadaboutconsuming alotofhigh-calorietrailfoodnexttimeyouareonastrenuous hike!
2.8 Useanymeansatyourdisposal(includingtheinternet)tojustifyestimatesofthefollowing(anorderof-magnitudeestimateissufficient):(a)Thekinetic energyoftheMooninitsmotionaroundEarth.(b) Thekineticenergyofaraindropbeforeithitsthe ground.(c)ThepotentialenergyofthewaterinLake Powell(relativetotheleveloftheColoradoriver directlybeneaththedam).(d)Energyneededforyou topowerabicycle15kmat15km/honflatterrain. (e)TheenergylosttoairresistancebyanOlympic athleterunninga100mrace.(f)Thegravitational potentialenergyofaclimberatthetopofMount Everest.
(a)Themoonorbitstheearthaboutonceamonthandhasanorbital radiusofabout380000km,soitsspeedis v = 2πR/T 0 91km/s.Themoonhasamassof7 3 × 1022 kg,sothis correspondstoakineticenergyof
(b)Largeraindropsareabout5mmindiameterandfallat ∼9m/s, sotheyhaveakineticenergyof
2.9
(c)LakePowellhasavolumeofapproximately33km3 andthe GlenCanyonDamhasaheightof Z = 220m.Treatingthelake asaboxofarea A andvolume V = ZA,itspotentialenergyis V = ρgA Z 0 dzz = ρgAZ2 /2 = (1/2)ρgVZ (0 5)(1000kg/m3 )(9 8m/s 2)(33 × 109 m3 )(220m)
(d)Thecrosssectionalareaofapersononabikeisaround0.5 m2 withadragcoefficientaround1.Theenergyoutputismostlyto counteractairresistance.Theenergylostis
1 2 ρaircd Av2 d = 0 5(1 2kg/m3 )(0 5m2 ) × (4 2m/s)2 (1 5 × 104 m) 79kJ
Assuminganefficiencyof ∼25%forthehumanbody,around 340kJwillwillhavetobeexpended
(e)Assumetherunnerhasacross-sectionalareaofapproximately 0.5m3 andadragcoefficientaround1.Goodrunnersfinishin around10s,sotheyhaveaspeedof10m/s.Thetotalenergy losttoairresistancewillbe(1/2)cd ρair Av2 d ∼ 3kJ. (f)Estimatethataclimberandgearhaveamassofapproximately 90kg.ThetopofMt.Everestis8848mabovesealevel,sothe climberwillhave mgh (90kg)(9 8m/s 2)(8 9 × 103 m) 7 8MJ ofgravitationalpotentialenergywithrespecttosealevel
Verifytheclaimthatconversionofthepotentialenergy atthetopofa15mhilltokineticenergywould increasethespeedoftheToyotaCamryfrom62to 74mph.
ACamryhasamassof1800kg,soitgains265kJofkinetic energywhenitselevationfallsby15m.Conservationofenergy requiresthat
1 2 mv 2 + mgh = constant
Usingthis,thefinalvelocityifitstartsat62mi/hris
v f = v2 i + 2gh = 32 6m/s = 73mi/hr
2.10 Consideracollisionbetweentwoobjectsofdifferentmasses M, m movinginonedimensionwithinitialspeeds V, v.Inthe center-of-massframe,the totalmomentumiszerobeforeandafterthecollision.Inan elastic collisionbothenergyandmomentumareconserved.Computethevelocitiesofthetwo objectsafteranelasticcollisioninthecenter-of-mass frame.Showthatinthelimitwhere m/M → 0,the
moremassiveobjectremainsatrestinthecenter-ofmassframe(V = 0beforeandafterthecollision), andthelessmassiveobjectsimplybouncesoff the moremassiveobject(likeatennisballoff aconcrete wall).
Inthecenter-of-massframetheobjectshavevelocities V ′ = V + u and v′ = v + u,where u isthevelocityofthecenter-of-mass viewedfromtheoriginalframe.Thetotalmomentumofthetwo objectsinthecenter-of-massvanishes: ptot = M(V +u)+m( v+u) = 0,whichdeterminesthecenter-of-massvelocity
u = mv MV m + M and
V ′ = V + u = m m + M (v + V)
v ′ = v + u = M m + M (v + V)
aretheobjects’velocitiesbeforethecollisioninthecenter-ofmass.Ifenergyandmomentumarebothconservedinthecollision, theobjects’mustreversetheircenter-of-massvelocities afterthe collision,
V ′ f = V ′ = m m + M (v + V)
v ′ f = v ′ = M m + M (v + V)
Inthelimit m/M → 0, V ′ → 0and V ′ f → 0,theheavierobject beginsandremainsatrest,while v ′ f = v ′ = (v + V),thelighter objectbouncesoff
2.11 Asweexplainin§34.2.1,thedensityofairdecreases withaltituderoughlyas ρ ρ0e z/H ,where z isthe heightabovetheEarth’ssurfaceand H 8 5kmnear thesurface.Computetheratioofairresistancelosses foranairplanetravelingat750km/hatanaltitudeof 12000mcomparedtoanaltitudeof2000m.Whathappenstothisratioasthespeedoftheairplanechanges? Howdoautomobileairresistancelossesat2000m comparetolossesatsealevel?
Airresistancelossesarelinearlyproportionaltothedensity ofair,sotheratiooflosesat12000mcomparedto2000mis r = e ∆z/H e 10/8 5 31%.Althoughthelosses(dEair /dt)are proportionaltospeedcubed,thespeedcancelsoutintheratioof lossesattwodifferentairdensities,sotheratioisindependentof theairplane’sspeed.Similarlyforacarat2000mcomparedtosea level r e 2/8 5 79%.
2.12 Consideranairplanewithmass70000kg,crosssectionalarea12m2,anddragcoefficient0.03.Estimatetheenergyneededtogettheplanemovingat 800km/handlifttheplaneto10000m,andestimate air-resistancelossesforaflightof2000kmusingthe formulainthepreviousproblem.Doaroughcomparisonoftheenergyusedperpersontodoasimilar tripinanautomobile,assumingthattheplanecarries50passengersandtheautomobilecarriestwo people.
Forsimplicity,assumethataccelerationoccursquicklyso thatair resistanceduringaccelerationcanbeneglected.Theplane requires atotalkineticenergyof
Ekin = 1 2 mv 2 = 0 5(70000kg)(800 × 103 m/3600s)2 = 1 73GJ ,
andat10000mhasapotentialenergyof V = mgh = (70000kg)(9 8m/s 2)(10000m) = 6 86GJ
Thedensityofairat10000misapproximately e 10/8 5 = 31%of thedensityatsealevel ρair = 0 31(1 17kg/m3 ) 0 36kg/m3
Travelingat v = 800km/hfor d = 2000kmat10000m,theplane loses
Eair = 1 2 ρaircd Av2 d 0 5(0 36kg/m3 )(0 03) × (12m2 )(222m/s)2 (2000 × 103 m) 6 4GJ
duetoairresistance.Thetotalenergyusedbytheplaneover theflightisapproximately15 3GJ.Thisis306MJ/passengeror 153kJ/passenger-km.Atypicalcarusesaround210MJofmechanicalenergyoverthe340kmtripbetweenNewYorkandBoston. With2passengers,thisis310kJ/passenger-km,ornearlytwicethe energyusageoftheairplane.So,beforeengineefficienciesareconsidered,automobilesuseroughlytwiceasmuchenergyovera long trip.
2.13 IntheAmericangameofbaseball,a pitcher throws a baseball,whichisaroundsphereofdiameter b = 0 075m,adistanceof18.4m(60.5feet),toa batter, whotriestohittheballasfarashecan.Abaseball hasamasscloseto0.15kg.Aradargunmeasuresthe speedofabaseballatthetimeitreachesthebatterat 44.7m/s(100mph).Thedragcoefficient cd ofabaseballisabout0.3.Giveasemi-quantitativeestimateof thespeedoftheballwhenitleftthepitcher’shand by(a)assumingthattheball’sspeedisnevertoodifferentfrom100mphtocomputeroughlyhowlongit takestogofromthepitchertothebatter,(b)using(a) toestimatetheenergylosttoairresistance,and(c) using(b)toestimatetheoriginalkineticenergyand velocity.
Theballhasaneffectiveareaof A = πR2 = 0 00442m2 .Forair atsealeveland25◦ C, ρair = 1 17kg/m3 .Theenergylossrateis
dE
dt = 1 2 cd Aρairv 3 0 5(0 3)(0 00442m2 )
× (1 17kg/m3 )(44 7m/s)3 69J/s
Iftheball’sspeeddoesnotdiffersignificantlyfrom100mph,it takes(a) t = (18 4m)/(44 7m/s) 0 41secfortheballtoreach thebatter.Thetotalamountofenergylostis(b) ∆E = dE dt × t 28J.
At100mph,theballhasakineticenergyof
Ekin f = 1 2 mv 2 150J
(c)Theinitialkineticenergyoftheballisjustthedifferenceofthis andtheenergyloss,
Ekin i = 150J + 28J = 178J
Thiscorrespondstoavelocityof
v = Ekin im 2(178J) (0 15kg) m/s 49m/s 110mph
2.14
Estimatethepoweroutputofanelitecyclistpedaling abicycleonaflatroadat14m/s.Assumeallenergy islosttoairresistance,thecross-sectionalareaofthe cyclistandthebicycleis0.4m2,andthedragcoefficientis0.75.Nowestimatethepoweroutputofthe sameelitecyclistpedalingabicycleupahillwithslope 8%at5m/s.Computetheairresistanceassumingthe
dragcoefficienttimescross-sectionalareais cd A = 0.45m2 (riderisinalessaerodynamicposition).Computetheratioofthepoweroutputtopotentialenergy gain.Assumethatthemassoftheriderplusbicycleis 90kg.
Ontheflatroad,thepoweroutputofthecyclistisequaltothe powerlostduetoairresistance,
P = 1 2 ρcd Av3 (0 5)(1 2kg/m3)(0 75)(0 4m2 )(14m/s)3 490W
Thepoweroutputofthecyclistontheslopeisequaltothe powerlostduetoairresistanceandduetotherateofchangeof potentialenergy,
P = Pair + Pg
Thecontributionfromairresistanceis
Pair = 1 2 ρcd Av3 = (0 5)(1 2kg/m3 )(0 45m2 )(5m/s)3 = 34W
Thecontributionfromgravitationalpotentialenergy
Pg = mg dz dt (90kg)(9 8m/s)(5m/s)sin(arctan(8/100) 350W
sothetotalpoweroutputofthecyclistontheslopeis
P = 380W
2.15 Comparetherateofpowerlosttoairresistanceforthe followingtwovehiclesat60km/hand120km/h:(a)
GeneralMotorsEV1with cd A 0 37m2,(b)Hummer H2with cd A 2 45m2
(a).At60km/h(16.67m/s)
dEloss dt = 1 2 ρcd Av3 (0 5)(1 2kg/m3)(0 37m2 )(16 7m/s)3 = 1 03kW
At120km/h,thepoweris8timesashigh,or8.2kW.
(b).ThepowerfortheHummerH2scaleswiththe cd A,sotheH2 losesenergyatarateof(2 45/.37)(1 03kW) = 6 8kWat 60km/hand(2 45/.37)(8 2kW) = 54kWat120km/h.Inboth cases,theH2losesenergyatarate6.6timesthatoftheEV1.
2.16 [T]Consideranidealizedcylinderofcross-sectional area Amovingalongitsaxis throughanidealized diffusegasofairmoleculeswithvanishinginitial velocity.Assumethattheairmoleculesarepointlike anddonotinteractwithoneanother.Computethe velocitythateachairmoleculeacquiresafteracollisionwiththecylinder,assumingthatthecylinder ismuchmoremassivethantheairmolecules.[Hint: assumethatenergyisconservedinthereferenceframe ofthemovingcylinderandusetheresultofProblem2.10.]Usethisresulttoshowthatthedragcoefficientofthecylinderinthisidealizedapproximationis cd = 4.
Supposethecylinderistravelingatavelocity v.Inthecylinder’s restframe,theairmoleculesaretravelingatavelocity v perpendiculartothesurfaceofthecylinder.Takingthelimitthatthe cylinderisinfinitelymoremassivethantheairmolecules,thisis thecenterofmassframe.Inthislimit,acollisionmaychangethe directionoftheairmoleculebutnotthespeed.Sincethevelocity oftheairmoleculesisperpendiculartotheendofthecylinder,the finalvelocitymustbe +v inthecylinderrestframe.Goingbackto
thelabframe,weaddavelocity +v toeverything,sothatthecylinderonceagainistravelingatavelocityof +v andtheairmolecule hasavelocityof +2v Tofind cd ,wemustnowcomputetherateofenergylossofthe cylinder.Intime dt,thefrontendofthecylinderpassesthrougha volume Avdt ofair,correspondingtoamassof dm = Aρvdt.This massisacceleratedfromresttoavelocityof2v.So,thecolumnof airgainsanenergyof
Fromconservationofenergy,thecylinderlosesanequalamountof energy,sothepoweroutputduetoairresistanceis
where cd = 4.
2.17
Onewaytoestimatetheeffectivearea(seeeq.(2.31)) ofanobjectistomeasureitslimitingvelocity v∞ fallinginair.Explainhowthisworksandfindthe expressionfor Aeff asafunctionof m (themassofthe object), v∞, g,andthedensityofair.Thedragcoefficientofasoccerball(radius11cm,mass0.43kg)is cd ≈ 0.25.Whatisitslimitingvelocity?
Becausetheforceduetoairresistanceincreaseswithvelocity, ifanobjectisallowedtofallinair,thereissomevelocityatwhich theupwardforceduetoairresistanceisequalinmagnitudetothe downwardgravitationalforce.Whenthishappens,thereisnonet forceontheobject,sotheobjectcontinuesfallingatthisvelocity, whichis v∞ .Thus, mg = 1 2
Theeffectiveareaofthesoccerballis Aeff = cd A = cd
r2 = 0 0095m2 .Thelimitingvelocityis
v∞ = 2mg ρair Aeff 2(0 43kg)(9 8m/s2 ) (1 2kg/m3 )(0 0095m2 ) 27m/s
2.18 Ifthevehicleusedasanexampleinthischapteracceleratesto50km/hbetweeneachstoplight,findthemaximumdistancebetweenstoplightsforwhichtheenergy usedtoacceleratethevehicleexceedstheenergylostto airresistance.(Youmayassumethatthetimeforaccelerationisnegligibleinthiscalculation.)Howdoesthe resultchangeifthevehicletravelsat100km/hbetween lights?Why?
Thevehicleinthischapterhas A = 2 7m2 , cd = 1/3,and m = 1800kg.Thetotalenergylosttofrictionafteradistance d at avelocity v is
Elost = 1 2 ρcd Av2 d
Settingthisequaltothekineticenergyandsolvingfor d togetthe maximumdistanceforwhichthekineticenergyexceedstheenergy lostduetoairresistance,
1 2 mv 2 = 1 2 ρcd Av2d d = m ρcd A 1800kg (1 2kg/m3 )(1/3)(2 7m2 ) 1 67km
Theresultisthesameifthevehicletravelsat100km/h.Both thekineticenergyandtheenergylossareproportionalto v2 sothe distanceatwhichtheyareequalisindependentofthevelocity.
2.19 Estimatetherotationalkineticenergyinaspinningyoyo(aplastictoythatyoucanassumeisacylinder ofdiameter5.7cmandmass52g,whichrotatesat 100Hz).Comparetothegravitationalpotentialenergy oftheyo-yoatheightof0.75m.
Themomentofinertiaforacylinderofmass m,length z and radius R rotatingarounditsaxisis(seeExample2.4),
Theenergyoftherotatingyo-yois
Hz)(100Hz)
Itsgravitationalpotentialenergyataheightof0.75mis mgh 0 38J,almostafactoroftensmallerthanitsrotationalkinetic energyat10Hz.(Mostoftheenergyisimpartedintheinitial “throw,”andnotfromgravitationalpotentialenergy.)
2.20 Verifytheassertion(seeExample2.3)that Ekin = 1 2 V fortheMooninacircularorbitaround Earth.[Hint:themagnitudeofthecentripetal accelerationforcircularmotion(2.35)canberewritten a = v2/r.]
Assumethroughoutthat M⊕ ≫ mmoon,sowemaytakeEarth tobeatrest.Foracircularorbit,thecentripetalaccelerationis a = v2/r.Theaccelerationduetogravityis GM⊕ /r2 where M⊕ isEarth’smass,whichismuchgreaterthanthatofthemoon.The gravitationalpotentialenergyofthemoonis V = GM⊕ m/r
Thecentripetalaccelerationandgravitationalaccelerationare identical,so
v2 r = GM⊕ r2 or v 2 = GM⊕ r , so
Ekin = 1 2 mv 2 = 1 2 GM⊕ m r = 1 2 V
2.21 EstimateEarth’skineticenergyofrotation(themoment ofinertiaofauniformsphereis 2 5 MR2).
Earthhasamassof5 972 × 1024 kgandameanradiusof 6371km.Itrotatesaboutitsaxisoncepersiderealday(23.9345h), givingitanangularvelocityof ω = 2π/(1siderealday) = 7 292 × 10 5 s 1 .Itskineticenergyofrotationis
E = MR2 5 ω 2 = (0 2)(5 972 × 1024 kg)(6371 × 103 m)2
× (7 292 × 10 5 rad/s)2 = 2 578 × 1029 J
Problems
3.1 Theelectricfieldoutsideachargedconductingsphere isthesameasifthechargewerecenteredatitsorigin.Usethisfacttocalculatethecapacitanceofa sphereofradius R,takingthesecondconductortobe locatedatinfinity.Whatisthemostchargeyoucan storeonanotherwiseisolatedsphericalconductorof radius R withoutabreakdownofthesurroundingair asdiscussedinExample3.2?Whatisthemaximum energyyoucanstoreonaconductingsphereofradius 1mm?
Thecapacitanceisdefinedby
= Q V
Foraconductingsphereofradius R andcharge Q, V isjustthe potentialofapointcharge Q ataradius R,so
= 4πε
Breakdowninairoccursat E
= 3 3 × 106 V/m.Thefield aroundthesphereishighestattheradius R,whereitis
4 C/m2) R2
Forasphereofradius1mm,themaximumchargethatcanbe storedis Qmax 3 7 × 10 10 C anditscapacitanceis0 111pF.Thetotal(maximum)energystored bythecapacitor
EEM = Q2
3.2
[T]ProveGauss’slawfromCoulomb’slawforstatic chargedistributionsbyshowingthattheelectricfield ofasinglechargesatisfiestheintegralformofGauss’s lawandtheninvokinglinearity.
TheintegralformofGauss’slawisgivenineq.(3.10).Takea sphericalsurfacesurroundingapointcharge Q.Theelectricfield is E = Qˆ r/(4πǫ0 r2 ),sointegratingthefieldoverthesurface,
dΩr 2 Q 4πǫ0 r2 = Q ǫ0
Theresultisthesameforasphereofanyradiusincludingasphere ofinfinitesimallysmallradius.Now,consideraspherenotenclosingthecharge.Inthelimitthattheradiusisverysmall,the electric fieldisconstantacrosstheentiresphere.Inthiscase,thereisno netfluxintooroutoftheinfinitesimallysmallsphere,sothe same integralgivesavalueofzero.
Anyclosedsurfacecanbecreatedbycombiningthesurfacesof infinitesimallysmallspheres,assurfacessharedbetweenthedifferentsphereshavenonetcontribution.Sinceonlytheinfinitesimally smallspherecenteredonthechargehasanonzerocontribution,for anyclosedsurface S surroundingapointcharge Q,
S dS E = Q ǫ0 (3.75)
Duetolinearity,thefieldofmanychargesorevenacontinuouschargedistributionisidenticaltothesumofthefieldsofthe
individualpointchargesorinfinitesimalchargeelements. Since Gauss’slawholdsforapointcharge,itmustthenholdforany chargedistribution.
3.3 Howmuchenergycanyoustoreonaparallelplate capacitorwith d = 1 µm, A = 10cm2,and ǫ = 100ǫ0, assumingthatthebreakdownfieldofthedielectricis thesameasforair?
Theenergystoredinthecapacitoris EEM = CV 2 /2.Thecapacitanceis C = ǫA/d andthemaximumvoltageis Vmax = |E|max d, so
3.4 [T]StartingfromGauss’slawandignoringedgeeffects (i.e.assumethattheplatesareverylargeandtheelectricfieldisuniformandperpendiculartotheplates), derivetheformulaforthecapacitanceofaparallelplate capacitor, C = ǫ0 A/d.RefertoFigure3.8.Youcan assumethattheelectricfieldvanishesoutsidetheplates definingthecapacitor.Showthattheenergystoredin thecapacitor, 1 2 CV2,canbewrittenastheintegral of ǫ0|E|2/2overtheregionwithinthecapacitor(as assertedineq.(3.20)).
Ignoringedgeeffects,thecharge Q isuniformlydistributedover theplatescorrespondingtoasurfacechargedensity Q/A E points inthenegativeˆz direction.ApplyGauss’slaw(3.10)toasurface withsidesparallelto E.Thetopofthesurface,ofarea a isinside theupperconductingplate,thebottom,alsoofarea a isinsidethe capacitor.Aslicethroughthesurfaceisdenotedbythedottedline inFigure3.8.Gauss’slawgivesnocontributiononthetopor sides,
aE = a Q
A or E = Q
A z
Thevoltagedifferencebetweentheupperandlowerplateisgiven byeq.(3.7),
V = d 0 dx E = Ed = Qd ǫ0 A ,
(Note,thedifferentiallineelement dx =+d z,so E dx = Edz). Fromthedefinition Q = CV,wefind
C = ǫ0 A/d
Integratingoverthevolumeofthecapacitor
3.5 Supposethatacapacitorwithcapacitance C ischarged tosomevoltage V andthenallowedtodischarge througharesistance R.Writeanequationgoverning therateatwhichenergyinthecapacitordecreaseswith timeduetoresistiveheating.Showthatthesolution ofthisequationis EEM(t) = EEM(0)e 2t/RC .Youcan ignoretheinternalresistanceofthecapacitor.Show thattheheatproducedintheresistorequalstheenergy originallystoredinthecapacitor.
Energyconservationrequiresthattherateofchangeofthe energyinthecapacitor d(Q2 /2C)/dt mustequalthepower dissipatedintheresistor, I2 R.Since I = Q,
QQ C = RQ2 or Q = Q/RC , since Q 0.Thisdifferentialequationis Q(t) = Q0 e
,sothe energystoredinthecapacitoris E
Thepowerdissipatedintheresistoris
, andthetotalpowerdissipatedintheresistoris
whichistheenergyoriginallystoredinthecapacitor.
3.6 Thedielectricsincapacitorsallowsome leakagecurrent topassfromoneplatetotheother.Theleakage canbeparameterizedintermsofa leakageresistance RL.Thislimitstheamountoftimeacapacitorcanbe usedtostoreelectromagneticenergy.Thecircuitdiagramdescribinganisolatedcapacitor,slowlyleaking charge,isthereforesimilartotheoneanalyzedinProblem3.5.TheMaxwellBCAP0310ultracapacitor(see §37.4.2)islistedashavingacapacitanceof310Fwith avoltageupto2.85V,andamaximumleakagecurrentof0.45mAwhenfullycharged.Takethistobe thecurrentat t = 0.Whatistheleakageresistanceof theBCAP0310?Estimatethetimescaleoverwhich thechargeonthecapacitorfallsto1/e ofitsinitial value.
Usingtheresultsofthepreviousproblem,thecurrentasa functionoftimeis
I(t) = Q(0) RC e t/RC = V(0) R e t/RC , sotheinitialcurrentis I(0) = V(0)/R.Solvingforresistance, R = V(0) I(0) = 2 85V 0 45 × 10 3 A = 6 33kΩ
Thetimescaleoverwhichthechargefallsto1/e ofitsinitialvalue is
τ = RC = (6 33kΩ)(310F) = 1 96 × 106 s = 22 7days
3.7 Acloud-to-groundlightningboltcanbemodeledas aparallelplatecapacitordischarge,withEarth’ssurfaceandthebottomofthecloudformingthetwo plates(seeExample3.2).Aparticularboltoflightningpassestothegroundfromacloudbottomat aheightof300m.Thebolttransfersatotalcharge of5Candatotalenergyof500MJtotheground, withanaveragecurrentof50kA.Howlongdid thelightningboltlast?Whatwastheelectricfield strengthinthe cloud–earthcapacitor justbeforeitdischarged?Howdoesthiselectricfieldcomparewith
thebreakdownfieldofair(3MV/m)?(Itisnow knownthatvariouseffectscauselightningtobeginat fieldsthatareconsiderablysmallerthanthebreakdown field.)
TheboltlastedatimeT = 5C/(50 × 103 A) = 10 4 s.Wedo notknowthecapacitance, C.Howeverweknowboththecharge onthecapacitorandtheenergystored.Using EEM = 1 2 Q2 /C and Q = CV,wecaneliminate C, EEM = 1 2 QV , or V = 2EEM /Q = 2 × (5 × 108 J)/(5C) = 200MV
Inanidealizedparallelplatecapacitor,theelectricfield isconstant andequalto |E| = V/d = 200MV/(300m) = 0 67MV/m.So thelightningoccurredwhentheelectricfieldwasaboutafactorof 3 3/(0 67) ≈ 5belowthebreakdownvalueforair.
3.8 [T]Consideranelectricdipolecomposedoftwo charges ±Q atpositions ± ξ/2.Writetheexactelectric fieldfromthetwochargesandshowthattheleading terminanexpansionin1/r matchesthe E-fieldquoted inBox3.2.[Hint:considerthebinomialexpansion, eq.(B.66).]
Writethefieldfromthetwochargeslocatedat r = ±
/2, E =
Wewantthefirsttermsinanexpansionin ξ as ξ → 0and Q →∞ withtheproduce d = Qξ fixed.Weneedtoexpandthe denominatorstofirstorderin ξ,
whereinthelaststepwehaveusedthebinomialexpansion(B.66). Nextsubstituteintotheexpressionfor E,
Setting d = ξQ givesthedesiredresult.
3.9 IfeachofthebatteriesusedintheflashlightinExample3.3hasaninternalresistanceof0.5 Ω (inseries withthecircuit),whatfractionofpowerislosttoJoule heatingwithinthebatteries?
Theflashlightusestwo1.5Vbatteries,eachwithaninternal resistanceof0.5Ω.Thebulbhasaresistanceof5Ω,sothetotal resistanceis6Ω.Thepowerdissipatedis P = I2 R.Thesamecurrentgoesthrougheachresistor.Thetotalinternalresistanceofthe batteriesis1Ω,so1/6ofthepowerislosttoJouleheatingwithin thebatteries.
3.10 [T]Considertworesistorsplacedinseries,oneafterthe other,inanelectriccircuitconnectedtoabatterywith voltage V.Showthattheeffectiveresistanceofthepair is R1 +R2 byusingthefactthatthecurrentthroughboth resistorsisthesamewhilevoltagesadd.Nowconnect theresistorsinparallel,sothatthevoltageacrossboth resistorsis V.Computethetotalcurrentandshowthat theeffectiveresistancesatisfies1/Reff = 1/R1 + 1/R2
Inseries,thecurrentthrougheachresistoristhesame,whilethe sumofthevoltagesacrosseachresistorisequaltothetotal voltage V.So,
V = R1 I + R2 I = I(R1 + R2) = IReff ,
where Reff = R1 + R2 .Thustheeffectofthetworesistorsequals thatofasingleresistorwithresistance R1 + R2
Inparallel,thevoltagedropacrosseachresistormustbethe totalvoltage,so V = R1 I1 = R2 I2
Thetotalcurrent I is I = I1 + I2 = V 1 R1 + 1 R2 = V Reff , where 1 Reff = 1 R1 + 1 R2
3.11 Anappliancethatuses1000Wofpowerisconnected by12gauge(diameter2.053mm)copperwiretoa 120V(RMS)ACoutlet.Estimatethepowerlostper meter, dPlost/dL,(inW/m)asresistiveheatinginthe wire.(Rememberthatthewiretotheappliancehastwo separatecurrent-carryingwiresinasinglesheath.)
Thereare2wireswitha2.053mmdiameterineachcableto carrythecurrentandtheresistivityofcopperis ρ = 1 68 × 10 8 Ωm.Theresistanceperunitlengthofthecableis
dR dℓ = 2 ρ
Theaveragecurrentthroughtheapplianceis
IRMS = P VRMS = 1000W 120V = 8 33A
Theheatinglossesinthecableshouldbeinsignificantcomparedto the1000W.Thepowerlossinthecableisgivenby
dP
dℓ = I2 RMS dR dℓ = (8
Forareasonablecablelength,thisisverysmallcomparedto the powerusageoftheappliance.
3.12 Electricalpowerisoftenusedtoboilwaterforcooking. Herearetheresultsofanexperiment:aliterofwater initiallyat30 ◦Cwasboiledonanelectricstovetop burner.Theburnerisratedat6.67A(maximuminstantaneouscurrent)and240VRMS.Ittook7minutes40 secondstoreachtheboilingpoint.Theexperimentwas repeatedusingan“electrickettle.”Thekettleisratedat 15A(maximuminstantaneouscurrentagain)anduses ordinary(US)linevoltageof120VRMS.Thistimeit took4minutesand40secondstoreachboiling.What arethepoweroutputsandresistancesoftheburnerand thekettle?Comparetheefficiencyforboilingwaterof thestovetopburnerandthekettle.Towhatdoyou attributethedifference?
Thepoweroutputofthestovetopburneris PS = IRMS VRMS = IVRMS/ √2 = 1132Wwhilethepoweroutputoftheelectrickettle is PK = 1273W.Theresistanceoftheburneris RS = VRMS/IRMS = 50 9Ω,whiletheresistanceofthekettleis RK = 11 3Ω.Sincethe sameamountofwaterwasboiled,thesameamountofenergy E
wasaddedtothewaterforeachofthese.Let ηS betheefficiency forthestovetopand ηK betheefficiencyforthekettle.Then
E = ηS PS tS = ηK PK tK ηK ηS = PS tS PK tK = (1132W)(460s) (1273W)(280s) = 1 46
Thekettleisnearly50%moreefficientatdeliveringenergytothe waterasthestovetop.Thisdifferenceislikelybecausetheheating elementsofthestovearelocatedinairratherthancontainedwithin thestove,somuchofthepowerislosttoheatingtheenvironment. Thekettle’sheatingelementsarefullycontainedwithinitsvolume, somuchlessheatislost.
3.13 [T]Usethemagneticforcelaw(3.42)andthedefinition ofwork W = b a dx F toshowthatmagneticforces donowork.[Hint:Considerthevectoridentity(B.6).]
Themagneticforceis F = qv × B,sotheworkdoneis
W = q b a dx v × B
Parameterize x as x(t),where t istime,andlet x(0) = a and x(T ) = b.Then dx = xdt (and v = x)andthelineintegralbecomes
W = q T 0 dt x (x × B) = q T 0 dt B (x × x) = 0,
wherewehaveusedeq.(B.6)andthefactthatthecrossproductof anyvectorwithitselfvanishes.
3.14 [T]Showthatachargedparticlemovinginthe xy-plane inthepresenceofamagneticfield B = Bˆ z willmovein acircle.Computetheradiusofthecircleandfrequency ofrotationintermsofthespeed v oftheparticle.Show thataddingaconstantvelocity u ˆ z inthe z-directionstill givesasolutionwithnofurtheracceleration.
FromtheresultofProblem3.13orfromthefactthattheforce is atrightanglestothevelocity,thespeedoftheparticleisconstant. Anobjectmovingatconstantspeedwithconstantperpendicular(centripetal)accelerationisundergoingcircularmotion.The centripetalforceontheparticle mv2/R mustbesuppliedbythe magneticforce F = qvB,thus
R = mv/qB
Tocomputethefrequencyofrotation,notethattheparticle completesanorbitintime T = 2πR/v = 2πm/qB sothefrequencyis
ν = 1/T = qB/2πm Iftheparticlegainsavelocitycomponentinthe z directionparallelto B,thereisnochangeintheforcebecausethecrossproduct ofparallelvectorsvanishes.Theparticlecontinuestomoveinacircleinthe x y plane,butwithaconstantvelocityinthe z direction. Thistypeofmotionisintheshapeofahelix.
3.15 [T]Reviewhowmagneticfieldsarecalculatedfrom Ampere’slawbycomputing(a)themagneticfielddue toastraightwireand(b)themagneticfieldintheinteriorofaverylongsolenoid.Thecontours C1, C2 shown inredinFigure3.18willhelp.Youmayusethefact thatthemagneticfieldvanishesjustoutsidetheouter boundaryofthesolenoidin(b).
(a).ConsiderthegeometryinFigure3.18(a).Ona C1 ,acircleof constantradius ρ,symmetryaroundthe z-axisandmirrorsymmetryintheplaneperpendiculartothe z-axis,themagnetic
fieldhasaconstantmagnitudeandisorientedinthe ˆ φ direction, B = B ˆ φ foracurrent I intheˆz direction.Ampere’slaw thengives:
µ0 I = dx B = 2πρB
Thus,solvingfor B,
(b).Inthiscase,thecurve C2 inFigure3.18(b)applies.Therecan benomagneticfieldintheradialdirectionorintheazimuthal directionduetomirrorsymmetryandrotationalsymmetry aboutthesolenoid’saxis.Onlyfieldsalongtheaxiscanbe non-zeroandthefieldisafunctionofradius ρ only, B = B(ρ)ˆz UsingAmpere’slawbyintegrating B around C2 ,andusingthe factthat B vanishesoutsidethesolenoid,
2 dx B = B(ρ)l = µ0 N(l)I
where l isthelengthofthepathinsidethesolenoidand N(l) isthenumberofturnsofwireenclosedby C2 .Note B isindependentofradiusinsidethesolenoid.Let n = N/L equalthe numberofturnsperunitlengthinthesolenoid’swindings. Then N(l) = ln and B = µ0 nI ˆ z
3.16 [T]Showthattheforceperunitareaonthewindings ofanair-coresolenoidfromthemagneticfieldofthe solenoiditselfisoforder F/A B2/µ0.Checkthatthe dimensionsofthisexpressionarecorrectandestimate F/A inpascalsif |B| = 1T.
Fromeq.(3.71),theenergyperunitvolumeinaaircore solenoidis dEmagnetic /dV = |B|2 /(2µ0 ).Thepressureistherateof changeofenergywithvolume.Sinceforasolenoid |B| isindependentofthevolume, p ∼|B|2 /2µ0.Tochecktheunits,useeq.(3.45) andeq.(3.52),
[p] = [E]/[V] = ml 1 t 2
[B] = [F]/[Qv] = mt 1 Q 1
[µ0 ] = [F]/[I2 ] = ml/Q 2 [B2 /µ
For |B| = 1T,
3.17 [T]Deriveeq.(3.52)fromeq.(3.43)andeq.(3.50). Makesureyougettheboththedirectionandmagnitude.
Themagneticfieldofawirewithacurrent I1 along +ˆz is
where ρ isthedistance(radius)fromthewire.Theforceonan infinitesimalsegment dx = dl z ofasecondwireparalleltoanother atadistance d andhavingcurrent I2 alongthe +z directionis d
So,theforceperunitlengthisalongtheradialdirectionand,letting F =
3.18 Anelectricmotoroperatesat1000rpmwithanaverage torqueof τ = 0 1Nm.Whatisitspoweroutput?Ifit isrunningon1.2Aofcurrent,estimatetheback-EMF fromtherotor.
Themotor’spoweroutputis
P = τω = (0 1Nm)(1000rpm) 1Hz 60rpm 2π s 1 1Hz = 10 5W
Thetime-averagedback-EMFfromtherotorisjust V = P/I = 8 7V.
3.19 ConsiderthemotordescribedinBox3.4.Iftheresistanceinthewirewrappingtherotoris1 Ω,computethe energylostundertheconditionsdescribed.WhatfractionofenergyislosttoJouleheatinginthissituation? Ifthecurrentisdoubledbuttherotationrateiskept fixed,howdotheoutputpower,Jouleheatinglosses, andfractionoflostenergychange?
Themotordraws300mAofcurrent,sothepowerlosttoJoule heatingis P = I2 R = 90mW.Themotorhasanaveragepower outputof1.8W,sotheenergylostisequalto5%oftheoutput energy.
Ifthecurrentisdoubledbuteverythingelsestaysthesame, the outputpowerdoublesto3.6W.ThepowerlosttoJouleheating increasesto360mW.Inthiscase,thepowerlostisequalto10% oftheoutputpower.
3.20 [T]In§3.4.2wederivedtheEMFonawirelooprotatinginamagneticfieldusingtheLorentzforcelawto computetheforcesonthemobilecharges.Although Faraday’slawofinduction(3.64)doesnotapplyina rotatingreferenceframe,showthatthesamerelation (3.62)followsfromFaraday’slaw.
AccordingtoFaraday’slaw E = dΦ/dt.Thefluxthroughthe currentloopis Φ= AB cos θ so E = ( AB sin θθ) = wlBθ sin θ inagreementwitheq.(3.62).
3.21 [T]Explainwhytheintegral S dS B thatappearsin eq.(3.65)isindependentofthechoiceofsurface S [Hint:makeuseofeq.(3.46).]
S dS B = 0overanyclosedsurface.Formaclosedsurfaceby takingtheunionoftwosurfaces S 1 and S 2 , S = S 1 ∪ S 2 sharing thesameedge,whichisaclosedcurve C.Ifthenormalson S 1 and S 2 wereinitiallydefinedbytheorientationofthecurve C (e.g. by theright-handrule),thenthenormalononeofthesurfacesmustbe reversedtoconsistentlydefineanormaloverthewholeof S.Then usingeq.(3.46),
S S 1 ∪S 2 dS B = 0 = S 1 dS B S 2 dS B Thusforanytwosurfaces,
S 1 dS B = S 2 dS B , sothechoiceofsurfacedoesnotmatter.
3.22 [T]Consideralong,hollowsolenoidofvolume V Showthat,ignoringendeffects,itsinductanceis L = n2Vµ0,andthatthemagneticenergyitstores EEM = LI2/2canbewrittenintheformofeq.(3.71).
Supposethesolenoidhascross-sectionalarea A andcurrent I Ithasalength l = V/A.Themagnitudeofthemagneticfieldis
Theelectromotiveforcefromoneloopisequaltochangein magneticfluxthroughasingleloopofthesolenoidis
Thereare ln totalloops,sothetotalchangeinfluxintegratedover theentiresolenoidis
Fromthis,theinductanceis
Thetotalenergyheldinthesolenoidis
3.23 [T]Provethatthemutualinductanceisasymmetric relation, M12 = M21,bycomputingtheenergystored whencurrent I1 isestablishedinloop1andthencurrent I2 isestablishedinloop2.Thensetupthesame currentsintheoppositeorder.
Beginwith I1 = I2 = 0.Withcircuit2open,raisethecurrentin loop1to I1 .Thepowerrequiredis P1 = V1 I1
Thetotalenergyrequiredisgivenby
(seeeq.(3.70)). Nextraisethecurrentinloop2to I2 ,holdingthecurrentinloop1 constant.Thepowerrequiredis
Integratingtofindthetotalenergy(rememberthat I1 isbeingheld fixed) E
+ M
.Sothetotalenergystoredinthetwo circuitsis
Nextperformthesamestepsstartingintheoppositeorder,first raisingthecurrentinloop2,andthen,holding I2 fixed,raisingthe currentinloop1.Thetotalenergystoredis
Thefinalconfiguration(thestateofthesystem)isthesamein both cases,sotheenergystoredmustbethesame,hence M12 = M21 3.24 Designatransmissionsystemtocarrypowerfrom windfarmsinNorthDakotatothestateofIllinois (about1200km).ThesystemshouldhandleIllinois’s summertimeelectricitydemandof42GW.Landfor transmissiontowersisatapremium,sothesystemusesveryhighvoltage(VRMS = 765kV). Assumethattheelectricityistransmittedasordinary alternatingcurrent,althoughinpractice three-phase power (see§38.3.1)wouldbeused.Assumethat eachtowercancarry36aluminumcables(ninelines, eachconsistingoffourconductorsseparatedbynonconductingspacers).Theconductorsare750mm2 in
crosssection.Howmanyseparatestringsoftransmissiontowersareneededifthetransmissionlosses aretobekeptbelow5%?Assumeapurelyresistive load.
Accordingtoeq.(3.41)thepowerlostisgivenby
Thetotalcableareais NA = 36N(750mm2 ) = 0 027m2 N,where N isthenumberofstringsoftransmissiontowers.Theresistivity ofaluminumis2 82 × 10 8 Ω m.Solvingfor N andcombiningall thefactors
N > PL ̺l AV 2 RMS(Plost /PL) (42 × 109 W)(2 82 × 10 8 Ω m)(1 2 × 106 m) (0 027m2 )(7 65 × 105 V)2 (0 05) 1 8
Sotwostringsoftransmissiontowerswouldsuffice.
3.25 [T]ConsiderthetransformerinFigure3.25.Suppose theloadisaresistor R andthatthetransformerisideal, with M2 = LS LP andallmagneticfluxlinespassingthroughbothinductors.Showthatthevoltagedrop acrosstheresistoris VS = √LS /LPVP = NS VP/NP and thatthetime-averagedpowerconsumedintheresistor is P = (LS /LP)(V2 P /2R) = N2 S V2 P/2N2 P R
Sincebothcircuitsareloopedbythesameflux
EP = VP = NP (dΦ/dt)
ES = VS = NS (dΦ/dt)
Thus VS /VP = NS /NP.Nextconsiderthecasewherethesecondary circuitisopenso IS = 0.Then VP = LP IP and VS = MIP , so VS /VP = M/LP = √LS /LP .Now,closethesecondarycircuit throughtheresistorsothat
ES = NS dΦ dt = IS R
Substitutingfor dΦ/dt = VP/NP,wefind
IS = NS NP VP R
Finally,using P = (1/2)I2 S R foranACcircuit,weobtainthedesired result.
3.26 [T]Takethedivergenceofbothsidesofeq.(3.74),use Coulomb’slawontheleftandcurrentconservationon righttoshowthattheequationisconsistent.
UseGauss’slaw(3.12)andtheidentityofeq.(B.21),
Problems
4.1 Soundwavestravelinairatroughly340m/s.The humanearcanhearfrequenciesrangingfrom20Hz to20000Hz.Determinethewavelengthsofthecorrespondingsinewavemodesandcomparetohumanscalephysicalsystems.
Thewavelengthisrelatedtothefrequencyby λ = v/ν,sofor 20Hz, λ = 17mandfor20000Hz, λ = 17mm.Notethatboth ofthesearescalesthatareeasytovisualize.17mmiscomparabletothewidthofafingeror,morerelevant,thesizeoftheear. 17mismuchlargerbutstillcomparableinsizetothingslike trees andbuildings.Thisisincontrasttolight,wherethewavelengthof visiblelightismuchsmallerthancanbeseenbytheeyes.
4.2 AviolinA-stringoflength L = 0 33mwithtotal mass0 23ghasafundamentalfrequency(forthelowestmode)of440Hz.Computethetensiononthestring. Ifthestringvibratesatthefundamentalfrequencywith maximumamplitude2mm,whatistheenergyofthe vibrationalmotion?
Thefundamentalfrequencyisgivenby(eq.(4.12)andbelow)
Solvingforthetension, τ = 4MLν 2 = 4(0 23 × 10 3 kg)(0 33m)(440Hz)2 = 59N
Tocomputethetotalenergyeitherintegratetheenergydensity (4.10)over x,
oruseeq.(4.16)beingcarefultoidentify
Substitutingfor M, ν,and A, E = 1 76mJ.
4.3 [T]Derivetheequationofmotionforthestring(4.9) fromamicroscopicmodel.Assumeasimplemodelof astringasasetofmasses ∆m spacedevenlyonthe x axisatregularintervalsof ∆x,connectedbyspringsof springconstant k = τ/∆x.Computetheleadingtermin theforceoneachmassandtakethelimit ∆x → 0to getthestringwaveequation,where ρ =∆m/∆x.Inthe samelimitshowthattheenergydensityofthestringis givenbyeq.(4.10).
Labeltheverticaldisplacementofeachmassasy y j .Somerelationsbetweenderivativesandthelimitsoffinitedifferencesare useful.As ∆x → 0, y j+1 y j → y′ j ∆
and
j y
j 1 → y′′ j ∆x ApplyHooke’slawtothe jth mass.Theforceonthe jth massinthe y-directionexertedbythe( j + 1)st massisgivenby (F
wheretan θ j = (y j+1 y j )/∆x,sosin θ j = (y j+1 y j )/∆l
Substitutingintotheequationfor(Fy ) j+1 j , (Fy ) j+1 j = k(y j+1 y j )
Combiningtheforcesonthe jth massfromthemassesonitsleft andright, my j = (Fy ) j = k((y j+1 y j ) (y j y j 1 ))
Set m = ρ∆x, k = τ/∆x andtakethelimit ∆x → 0,toobtain ∆xρy j = (τ/∆x)(y ′′ j (∆x))2
Labelingthemassesby x ≡ j∆x,weobtaineq.(4.9). Theenergystoredinthearrayofmassesandspringsconsists of kineticandpotentialenergy
E = ∆m 2 j y 2 j + k 2 j (∆x 2 + (y j+1 y j )2 )
Substituting ∆m = ρ∆x, k = τ/∆x, y j+1 y j → y′ j ∆x,and j ∆x( ) → dx ( ),weobtain
E = dx ρ 2 y 2 + τ 2 y ′2 + dxτ/2
Sincethelasttermisaconstantindependentofthedisplacement y(x, t),wedropit.
4.4 [T]Showthattheenergydensityonastring u(x, t), definedineq.(4.10),obeystheconservationlaw ∂u/∂t + ∂S /∂x = 0,where S (x, t) = τyy′ isthe energy flux,theenergyperunittimepassingapoint x.Forthe travelingwave y(x, t) = f (x vt),find u(x, t)and S (x, t) andshowthatenergyflowstotheright(for v > 0)as thewavepassesapoint x.Showthatthetotalenergy passingeachpointisequaltothetotalenergyinthe wave.
Startingfromeq.(4.10)
∂u ∂t = ρyy + τy ′ y ′ , substitutefor¨ y fromthewaveequation,eq.(4.9), ρy = τy′′
∂u ∂t = τyy ′′ + τy ′ y ′ = ∂ ∂x (τyy ′) = ∂S ∂x with S = τyy′.For y = f (x vt), y′ = f ′ and˙ y = vf ′ ,where f ′ = df (η)/dη,where η = x vt.Substitutinginto S , S = τ( vf ′ )( f ′ ) = τvf ′2 > 0 , whichispositive,indicatingafluxofenergytotheright.
Theintegralof S overalltimeatfixed x is
I = dt( τyy ′) = dtτvf ′2 = τ dη( f ′(η))2
Theintegralof u overall x atfixedtimeis
I′ = dx ρ 2 y 2 + τ 2 y ′2 = dx 1 2 ρv 2 f ′2 + 1 2 τ f ′2 = I
wherewehaveused v = τ/ρ.Thusthetotalenergyfluxpassing apointequalsthetotalenergyinthewave.
4.5 Computethemaximumenergyfluxpossibleforelectromagneticwavesinairgiventheconstraintthat theelectricfieldcannotexceedthebreakdownfield describedinExample3.2.
Thebreakdownfieldofairis Emax = 3 3 × 106 V/m.Inan electromagneticwave, |B| = |E|/c.Themagnitudeoftheenergy
fluxofanelectromagneticwaveis
|E|2 = ǫ0c|E|2 .Themaximumpossibleenergyfluxinair,then,is
|
|
4.6 ThestrongestradiostationsintheUSbroadcastata powerof50kW.Assumingthatthepowerisbroadcast uniformlyoverthehemisphereaboveEarth’ssurface, computethestrengthoftheelectricfieldintheseradio wavesatadistanceof100km.
Themagnitudeoftheenergyfluxofanelectromagneticwave is |S| =
.Theenergyfluxina hemisphericalwaveatdistance R
),so
4.7 [T]Derivethewaveequationfor B analogousto eq.(4.20).
FromMaxwell’sequations(§3.6),takethetimederivativeof Faraday’sLawwith
0,
(4.32) andthensubstitutefromtheAmpere-MaxwellLaw,
4.9 Ithasbeenproposedthatsolarcollectorscouldbe deployedinspace,andthatthecollectedpowercould bebeamedtoEarthusingmicrowaves.Apotentiallimitingfactorforthistechnologywouldbethepossible hazardofhumanexposuretothemicrowavebeam.One proposalinvolvesacircularreceivingarrayofdiameter 10kmforatransmittedpowerof750MW.Compute theenergyfluxinthisscenarioandcomparetothe energyfluxofsolarradiation.
Thisproposalwouldhavethearrayreceiveapowerdensityof |S| = 750 × 106 W π × 108 m
Thepowerdensityofsolarradiationis1366W/m2 ,about500 timesgreater,whilethatofamicrowaveisapproximately1 5 × 104 W/m2 ,about5000timesgreater.Thispowerdensityisactually quitemodestcomparedtoothercommonsourcesofelectromagneticradiation.
4.10 [T,H]Considertwoelectromagneticplanewaves(see eq.(4.21))onewithamplitude Ea 0 andwavevector ka andtheotherwithamplitude Eb 0 andwavevector kb Thesewavesaresaidtoaddcoherentlyiftheaverage energydensity u intheresultingwaveisproportional to |Ea 0 + Eb 0 |2 orincoherentlyiftheaverageenergy densityisproportionalto |Ea 0 |2 + |Eb 0 |2.Showthattwo electromagneticplanewavesarecoherentonlyifthey arepropagatinginthesamedirectionwiththesame frequencyandthesamepolarization.
Aplanewaveoftheformofeq.(4.21)has E = E0 sin(k x ωt), where ˆ k definesthedirectionofpropagationand E0 definesthe directionofpolarization.Notethat k E0 = 0.
Theenergydensityisgivenbyeq.(4.22).Weconsideronly theterm(u1 )proportionalto |E|2 ;thesecondtermcanbehandled analogously.Forasingleplanewave,
wherewehaveusedthevectoridentity(B.22)and ∇ B = 0.
4.8 Supposeanelectromagneticplanewaveisabsorbed onasurfaceorientedperpendiculartothedirectionof propagationofthewave.Showthatthepressureexerted bytheradiationonthesurfaceis prad = W/c,where W isthepowerabsorbedperunitarea.SolarradiationatthetopofEarth’satmospherehasanenergyflux |S| = 1366W/m2.WhatisthepressureofsolarradiationwhentheSunisoverhead?Whatisthetotalforce onEarthexertedbysolarradiation?
Thepressure p istheforceperunitareaperpendicularto thesurface,whileforcecanbedefinedasthetimederivative of momentum d| p|/dt.Ifasurfaceisabsorbingelectromagneticradiation,thepressurethenistherateperunitareaofmomentum reachingthesurface.So, p = d2| p| dtdA = 1 c d2 E dtdA = W c Whenthesunisoverhead,Earthfeelsapressureof
p = 1366W/m2 3 × 108 m/s = 4 56 µPa
ThetotalforceonEarthexertedbysolarradiationis
F = πR2 ⊕ P = π(6
u1 = ǫ0 4 |E0 |2
wherewehaveused sin2 ( ) = 1/2.
Firstconsiderthecasewherethetwowavespropagateinthe samedirectionwiththesamefrequency,butpossiblydifferent polarizations.Then
E = (Ea 0 + Eb 0 )sin(k x ωt)
Then
u1 = ǫ0 4 |Ea 0 + Eb 0 |2 ,
whichisproportionalto |Ea 0 + Eb 0 |2 ifandonlyif Eb 0 isproportional to Ea 0 , i.e. theymusthavethesamepolarization.Thisprovesthat twowaveswiththesamefrequency,directionofpropagation and polarization are coherent.
Ifthefrequenciesaredifferentand/ordirectionofpropagation aredifferentoneobtainscrosstermsinthecomputationof |E|2 proportionalto
sin(ka x ωat)sin(kb x ω bt)
Thetimeaveragevanishesunless ωa = ωb because
lim T →∞ 1 T T 0 sin ωa t sin ω bt → 0
unless ωa = ωb ,andsimilarlyfor sincos and coscos .Ifthefrequenciesarethesamebut ka and kb arenotproportional,thetime averageyieldsaresultproportionaltocos(ka kb) x,whichvanisheswhenaveragedoverspace.Soallcrosstermsvanishunless
thefrequency,directionandpolarizationofthetwowavesarethe same.Whenthecrosstermsvanish u1 = ua 1 + ub 1 ,whichis proportionalto |Ea 0 |2 + |Eb 0 |2
4.11 [T]Deriveeq.(4.23)bytakingthetimederivativeof eq.(4.22)andusingMaxwell’sequations.[Hint:see eq.(B.23).]
Substitutefor E and B fromMaxwell’sequations,
whereinthelaststepwehaveusedeq.(B.23).
4.12 [T]Astringoflength L beginsintheconfiguration y(x) = A[ 1 3 sin k1 x + 2 3 sin k2 x]withnoinitialvelocity. Writetheexacttime-dependentsolutionofthestring y(x, t).Computethecontributiontotheenergyfrom eachmodeinvolved.
Noinitialvelocitymeansthat˙ y(x, 0) = 0forall x,sothetime dependencerequirescosineterms.Addinginthetimedependence foreachmode,andreplacing kn = nπ/L,
Notethat˙ y(x, 0) = 0.Theenergydensityis
Theenergyineachmodedoesnotchangewithtime,sotakethe energydensityat t = 0,where˙ y = 0.Thenthetotalenergyis
Theinitialshapeisantisymmetricabout L/2,soonlyeven n modes havenon-zerocoefficients.Calculatingthecoefficients,
with U1 = π2 τA2/36L fromthefirstmodeand
fromthesecond.
4.13 [TH]Astringoflength L isinitiallystretchedinto a“zigzag”profile,withlinearsegmentsofstring connectingthe(x, f (x))points(0, 0), (L/4, a), (3L/4, a), (L, 0).ComputetheFourierseriescoefficients cn andthetime-evolutionofthestring y(x, t).Compute thetotalenergyinthetensionoftheinitiallystretched string.Computetheenergyineachmodeandshowthat thetotalenergyagreeswiththeenergyofthestretched string.y
At T = 0,theFouriercoefficientsare:
Forallothers, cn = 0.Writingthisinsummationnotation, y(x, 0) = f (x) = ∞ n=0 ( 1)n 4a √2L [(1 + 2n)π]2 y2+4n(x) , where
yn(x) = 2 L sin nπx L
Toaddthetimedependence,eachtermgetsafactorofcos(ωnt φn).Sincethestringisstationaryat t = 0, φn = 0.Thefrequencyis
ωn = nvπ/L (where v = τ/ρ),sothetimeevolutionofthestring is
y(x, t) = ∞ n=0 ( 1)n 8a [(1 + 2n)π]2 cos 2π(1 + 2n)vt L sin 2
Thetotalenergyintensioninitiallyis
L
Theenergyinmode n is
Thetotalenergyis
+
Thesumisequalto π2 /6,sothetotalenergyis E = 8τa2/L,justas wefoundbyintegratingovertheinitialtensionofthestring.
4.14 [T]Whatisthepressureexertedbyabeamoflightona perfectmirrorfromwhichitreflectsatnormal(perpendicular)incidence?Generalizethistolightincidentat anangle θ tothenormalonanimperfectmirror(which reflectsafraction r(θ)ofthelightincidentatangle θ).
Aperfectmirrorreflectsthelightwithoutalteringitsfrequency (energy),sothepressureatnormalincidenceisjusttwicethatof aperfectabsorber.Thisdoublesthechangeinmomentum,sothe pressureis2W/c
Atanangle θ,thepressureisdecreasedbyafactorofcos θ becauseonlythenormalforcecontributestopressure.Aperfect absorberwouldfeelapressureof(W/c)cos θ whileaperfectmirror wouldfeelapressureof(2W/c)cos θ.Animperfectmirrorreflectingafraction r(θ)ofthelightandabsorbingtherestwouldfeela totalpressureof
(θ) = (2r(θ) + (1 r(θ)) W cos θ c = (r(θ) + 1) cos θ W c
4.15 [T]Consideracylindricalresistorofcross-sectional area A andlength L.Assumethattheelectricfield E andcurrentdensity j areuniformwithintheresistor. Provethattheintegratedpowertransferredfromelectromagneticfieldsintotheresistor d3 xj E isequal to IV.Computetheelectricandmagneticfieldson thesurfaceoftheresistor,andshowthatthepower transferisalsogivenbythesurfaceintegralofthe Poyntingvector,sothatallenergydissipatedinthe resistoristransferredinthroughelectricandmagnetic fields.
Usecylindricalcoordinateswithˆz pointingalongthewire,ˆρ pointingradiallyoutwardfromthecenterofthewire,and ˆ φ,the angularunitvectorwithˆ
Forthefirstpart: E = (V/L)ˆz and j = (I/A)ˆz,so d3 xj E = d3 x(IV/AL) = IV , because AL isthevolumeoftheresistor.
Forthesecondpartweneedthemagneticfieldatthesurfaceof thewire.Fromeq.(3.50),
NotethatthePoyntingvectorpointsinwardeverywhereonthesurfaceofthecylindricalresistor,soenergyintheEMfieldsis flowing intotheresistor.Theintegrated(overthesurfaceoftheresistor) fluxofenergygivesthetotalpowerdeliveredtotheresistor bythe EMfields,
P = | d A S| = IV ,
equaltotheresistivepowerdissipatedintheresistor.
4.16 Asstatedinthetext,the dispersionrelation relatingthe wavenumberandangularfrequencyofoceansurface wavesis ω = gk,where g 9 8m/s2 .Computethe wavelengthandspeedofpropagation(phasevelocity) foroceansurfacewaveswithperiods6sand12s.
Theperiodisrelatedtothefrequencyby T = 2π/ω,whilethe wavenumberisrelatedtothewavelengthby k = 2π/λ.Rewriting thedispersionrelationintermsof T and λ,
Thephasevelocityis v
So,when T = 6s, λ = 56mand v = 9 4m/s.When T = 12s, λ = 225mand v = 18 7m/s.
4.17 [T]Awavesatisfyingeq.(4.2)passesfromone mediuminwhichthephasevelocityforallwavelengths is v1 toanothermediuminwhichthephasevelocity is v2 .Theincidentwavegivesrisetoareflectedwave thatreturnstotheoriginalmediumandarefractedwave thatchangesdirectionasitpassesthroughtheinterface. Supposethattheinterfaceistheplane z = 0andthe incomingwaveispropagatinginadirectionatanangle θ1 tothenormalˆz.Provethelawofspecularreflection, whichstatesthatthereflectedwavepropagatesatan angle π θ1 withrespecttothenormalˆz.Alsoprove Snell’slaw,whichstatesthatthewaveinthesecond mediumpropagatesatanangle θ2 fromthenormalˆz, wheresin θ2/ sin θ1 = v2/v1 .Usethefactthatthewave mustbecontinuousacrosstheinterfaceat z = 0.
Let ki, ωi ,and Ai (i = 1, 2, 3)bethewavenumber,frequency, andamplitudeoftheincident,refracted,andreflectedwaves.The waveequationrelates ki = |ki | to ωi andthephasevelocity vi, ki = ωi /vi , where v1 = v3 v2 .Theformoftheincidentwaveis
ψ1 (x, z, t) = A1 cos(k1 x sin θ1 + k1 z cos θ1 ω1 t) withanalogousexpressionsfortherefractedandreflectedwaves. At z = 0,continuityofthewaveformrequiresthatthesumofthe incidentandreflectedwavesequalstherefractedwave, A1 cos(k1 x sin θ1 ω1 t)+A3 cos(k3 x sin θ3 ω3 t) = A2 cos(k2 x
2
2 t) Forthisfunctionalequalitytoholdatall t,allthe ωsmustbeequal, leadingto k1 = k3 = v2 k2 /v1.Forthefunctionalequalitytoholdat allpoints x alongtheinterface,werequire k1 sin θ1 = k3 sin θ3 = k2 sin θ2.Since k1 = k3 ,wehavesin θ1 = sin θ3.Theonlypossibilitydistinctfromtheincidentwaveis θ1 = π θ3,whichprovesthe lawofspecularreflection.Since k1 = (v2 /v1)k2 ,wehave sin θ2 sin θ1 = k1 k2 = v2 v1 , whichisSnell’slaw.
4.18 Awavetravelstotherightonastringwithconstant tension τ andamassdensitythatslowlyincreasesfrom ρ onthefarleftto ρ′ onthefarright.Themassdensity changesslowlyenoughthatitsonlyeffectistochange thespeedwithwhichthewavepropagates.Thewaveformonthefarleftis A cos(kx ωt)andonthefarright is A′ cos(k′ x ωt).Findtherelationbetween k′ and k andthenuseconservationofenergytofindtheamplitude A′ onthefarright.Youmayfindithelpfultouse theresultofProblem4.4(S (x, t) = τyy′ ).
Thesesolutionsobeythewaveequationwithdifferentdensities ontheleftandright.Ontheleft, ρy = τy′′ with y = A cos(kx ωt),so ρω2 = τk2,and k = ω ρ/τ.Ontheright,ananalogous calculationgives k′ = ω ρ′/τ.Thus
k′ = ρ′ ρ k
Thetimeaveragedrateofflowofenergypastapointinthefarleft mustbethesameastherateonthefarright.
S L = τyy ′ = τωkA2 sin2 (kx ωt)
S R = τyy ′ = τωk′ A′2 sin2 (k′ x ωt)
Averagingthesin2 gives1/2,andequatingtheenergyfluxonleft andright,wefind A2 k = A′2 k′ or A′ = 4 ρ/ρ′A
Problems
5.1 Acylinderinitiallycontains V0 = 1Lofargonat temperature T0 = 0 ◦Candpressure p0 = 1atm.Supposethattheargonissomehowmadetoexpandtoa finalvolume V = 2Linsuchawaythatthepressurerisesproportionallytothevolume,finallyreaching p = 2atm.Howmuchworkhastheargondone?What isitsfinaltemperature T andhowmuchthermalenergy hasbeenaddedtothegas?
Theworkdonebythegascanbecomputedfrom W = dVp(V). Since(p0 , V0)and(p, V)arebothknownandthequantityofgas n doesnotchange,wecancompute n from n = p0 V0 /RT0 andthen find T using T = pV/nR.Knowing T T0 andtheheatcapacityof argon(amonatomicidealgas)wecanfindtheincreaseinthermal energycontentofthegas.Thisplustheworkdonebythegasmust sumtothetotalenergyaddedtothegas.
Firstcomputetheworkdone:wearegiven p = p0 V/V0,so
1 01 × 105 Pa 2 × 10 3 m3 (3 × 10 6 m 6 ) = 150 J
Thetemperature T isobtainedfromtheidealgaslaw: T = (pV/p0 V0 )T0 = 4(273K) = 1092K
Theinternal(thermal)energyofargon(amonatomic,idealgas)at temperature T is U = (3/2)nRT .Weget n from n = p0 V0 /RT0 = 0 0446mol.So
∆U = 3 2 nR∆T = 3 2 (0 0446mol)(8 31J/Kmol)(1092 273)K = 455J
musthavebeenaddedtoincreasethetemperatureofthegas.Since thegasalsodid W = 150Jofwork,thetotalthermalenergyadded tothegasmusthavebeen Qtot = 455 + 150 = 605J.
5.2 Non-rigidairshipsknownas blimps haveoccasionally beenusedfortransportation.Ablimpisessentiallya balloonofvolume V filledwithhelium.Theblimp experiencesabuoyancyforce F = (ρatm ρHe)Vg, where ρatm isthedensityofthesurroundingairand g 9 8m/s2 .Theblimpmaintainsitsshapebecause thepressureofthehelium pHe iskepthigherthanthe pressureofthesurroundingatmosphere patm .Amodernblimpwithavolume V = 5740m3 canliftatotal mass m = 5825kgattemperature0 ◦Cand patm = 1atm.Assumingbothairandheliumbehaveasideal gases,estimatethepressureoftheheliumgasinsidethe blimp.Assumingtheblimp’svolumeiskeptconstant, howmuchmasscanitliftat20 ◦ C?
At0 ◦ Cand1atm(STP), ρair = 1 275kg/m3 .Inordertolift m = 5825kg, mg = (ρair ρHe)Vg
Sothedensityofheliumis
ρHe = ρair m/V = 1 275kg/m3 5825kg/5740m3 = 0 260kg/m3
Atanygiven T and p, ρHe/ρair = (4/29),theratiooftheirmolecular masses,soat p = 1atm, ρHe = (4/29)1 275kg/m3 = 0 176kg/m3
5.3
Usingtheidealgaslaw,wecanthensolveforthepressureatwhich thedensityofheliumis0 260kg/m3 ,
pHe = 0 260kg/m3 0 176kg/m3 1atm = 1 48atm
Ifthetemperatureincreasesto20 ◦ Cwhilevolumeremainsconstant,thenthedensityofheliumdoesnotchange,butthedensity oftheoutsideairdropsto ρair = (273/293)1 275 = 1 188kg/m3 , sothemassthatcanbeliftedis
m = (ρair ρHe)V = (1 188kg/m3 0 260kg/m3 )5740m3 = 5330kg
Whenairisinhaled,itsvolumeremainsconstantand itspressureincreasesasitiswarmedtobodytemperature Tbody = 37 ◦C.Assumingthatairbehavesasan idealgasandthatitisinitiallyatapressureof1atm, whatisthepressureofairinthelungsafterinhalation iftheairisinitially(a)atroomtemperature, T = 20 ◦C; (b)atthetemperatureofacoldwinterdayinBoston, T = 15 ◦C;(c)comingfromoneperson’smouthto another’sduringcardiopulmonaryresuscitation(CPR). ForCPR,assumetheairisinitiallyatbodytemperature andat p = 1atm.
Asvolumeisassumedconstant,theidealgaslawimplies
plung = Tlung Tair pair ,
Inallcases Tlung = 310K.Forcasea) pair = 1atmand Tair = 293K,hence plung = 310 294 atm = 1 05atm.Forcaseb) Tair = 268K, plung = 1 20atm.Forcasec) Tair = Tlung,so plung = 1atm.
5.4 [H] Everydayexperienceindicatesthatitismuch easiertocompressgasesthanliquids.Thisproperty ismeasuredbytheisothermalcompressibility, β = 1 V ∂V ∂ p T ,thefractionalchangeinasubstance’svolumewithpressureatconstanttemperature.Whatis theisothermalcompressibilityofanidealgas?The smaller β is,thelessworkmustbedonetopressurizeasubstance.Comparetheworknecessaryat20 ◦ C toraisethepressureofakilogramofairandakilogramofwater(βwater 4 59 × 10 10 Pa 1)from1atm to104 atm.
Firstcomputetheworkdonetocompressair,anidealgas. βair = 1 V ∂V ∂p T = nRT 1 V ∂ ∂p 1 p = pV 1 Vp2 = 1 p Then
Themolecularweightofairis0.029kg/mol,so1kg = 1/0 029 34 5mol,soat T = 298K, Wair = 9 21(34 5mol/kg)(8 31J/molK)(298K) = 787KJ/kg
Nextcomputetheworkdonetocompress1kgwaterwith βwater = 4 6 × 10 10 Pa 1 .Theworkis Wwater = V f V0 dVp = p f p0 (dp βwater V(p)) p
Thefractionalchangeinthevolumeofthewaterwhencompressed from1atmto104 atmis
∆V/V ≈−βwater ∆p (4 6 × 10 10 )(104 × 1 01 × 105 ) = 0 46 ,
sothevariationof V intheintegrationover p cannotbeignored. Fromthedefinitionof β, dV/V = β dp,soln V = βp + const, and V(p) = V0 exp( β(p p0 )).Substitutingintotheintegraland ignoring p0 since p0 ≪ p f ,
Theworknecessarytopressurizewateris170/790 ≈ 22%ofthe workneededtopressurizeair.Notethatthisratiogetsmuch smaller if p f isreduced.
5.5 [HT]AcylindricaltubeorientedverticallyonEarth’s surface,closedatthebottomandopenatthetop(height 100m,cross-sectionalarea1m2)initiallycontainsair atapressureofoneatmosphereandtemperature300K. Adiscofmass m plugsthecylinder,butisfreetoslide upanddownwithoutfriction.Inequilibrium,theairin thecylinderiscompressedto p = 2atmwhiletheair outsideremainsat1atm.Whatisthemassofthedisc?
Next,thediscisdisplacedslightlyinthevertical(z) directionandreleased.Assumingthatthetemperature remainsfixedandthatdissipationcanbeignored,what isthefrequencywithwhichthediscoscillates?
Inmechanicalequilibrium,thenetforcefromthepressureofthe gasinsideandoutsideofthecylindermustbalancethedownwards forceofgravityfromthemass.Thatis, ∆pA = mg, or
m (1 0 × 105 Pa)(1m2 ) 10m/s2 = 104 kg
Sincethegasiscompressedtoapressuretwoatmospheresand since pV isaconstantatfixed T ,equilibriumisreachedwhenthe heightofthecylinderis50m.
Ifthediscisdisplacedbyasmallamount z,thenagainusing pV = constant,
(p +∆p)(V ∆V) = pV ,
where ∆V = Az.Sincethedisplacementandtheresultingpressure changearesmall,weignore ∆V∆p,so V∆p = p∆V,where,in termsofthe z-coordinateoftheoscillator, ∆V/V = z/50m.Hence,
∆p = z[m] 50 (2 × 105 Pa) = 4 × 103 z[m]Pa
Theforceonthediscthenbecomes mz = F = A∆p = kz, where k = 4 × 103 N/m.Thisisthedifferentialequationofaharmonicoscillator,withangularfrequency ω = √k/m 0 6s 1 ,and frequency ν = ω/2π 0 1Hz.
5.6 Howmuchenergydoesittaketoheat1literofsoup fromroomtemperature(20 ◦C)to65 ◦C?(Youmay assumethattheheatcapacityofthesoupisthesame asforwater.)
Waterhasaheatcapacityof4.18kJ/kgK.Oneliterof soup,assumedtohaveapproximatelythesamepropertiesas water,hasamassof1kg.So,ittakes4.18kJtoheatthe soupby1K.Raisingthetemperatureby45Kwouldrequire Q = (4 18kJ/kgK)(45K)(1kg/L) 188kJ/L.
5.7 A“low-flow”showerheadaverages4.8L/min.Taking otherdatafromExample5.3,estimatetheenergysavings(inJ/y)ifallthepeopleinyourcountryswitched fromUScodetolow-flowshowerheads.(Youwill
needtoestimatetheaveragenumberofshowersper personperyearforthepeopleofyourcountry.)
Fromtheexample,aregularshowerheadaverages9.5L/min. Thewaterisheatedupbyapproximately ∆T = 30K,andthe averageshowertakes8min.Waterhasaheatcapacityof cV = 4 18kJ/kgK,sothetotalenergysavingsfortheUS(population 3 × 108 )peryearis
∆E = cV ∆M∆T ,
where ∆M isthedifferenceinthetotalmassofwaterusedin showersbetweenthenormalandlow-flowshowerheads.
∆M = (4 8L/min 9 5L/min) (8min/shower) × (365shower/y)(3 × 108 people)(1kg/L) = 4 12 × 1012 kg/y
Thetotalenergysavingsisthen
∆E = (4 18kJ/kgK)( 4 12 × 1012 kg/y)(30K) = 5 2 × 1017 J/y = 0 52EJ/y, whichisabout0.5%ofthetotalUSyearlyenergyusage.
5.8 Asolarthermalpowerplantcurrentlyunderconstructionwillfocussolarraystoheatamoltensaltworking fluidcomposedofsodiumnitrateandpotassiumnitrate. Themoltensaltisstoredatatemperatureof300 ◦ C, andheatedinapowertowerto550 ◦C.Thesalthas aspecificheatcapacityofroughly1500J/kgKinthe temperaturerangeofinterest.Thesystemcanstore 6000metrictonnesofmoltensaltforpowergeneration whennosolarenergyisavailable.Howmuchenergyis storedinthissystem?
Thetotalamountofenergystoredis
E = (1500J/kgK)(6 × 106 kg)(550 ◦ C 300 ◦ C) = 2 25TJ
5.9 AnewsolarthermalplantbeingconstructedinAustraliawillcollectsolarenergyandstoreitasthermalenergy,whichwillthenbeconvertedtoelectricalenergy.Theplantwillstoresomeofthethermalenergyingraphiteblocksfornighttimepower distribution.Accordingtothecompanyconstructingtheplant,theplantwillhaveacapacityof 30MkWh/year.Graphitehasaspecificheatcapacityofabout700J/kgKatroomtemperature(see[26] fordataonthetemperaturedependenceofthespecificheatofgraphite).Thecompanyclaimsastorage capacityof1000kWh/tonneat1800 ◦C.Isthisplausible?Explain.Iftheyareplanningenoughstorage sothattheycankeepupthesamepoweroutputat nightasintheday,roughlyhowmuchgraphitedo theyneed?Estimatethenumberofpeoplewhoseelectricalenergyneedswillbesuppliedbythispower plant.
Thecompanyclaimsthatthestoragecapacityofgraphiteis 1000kWh/t = (103 kWh/t)(3 6MJ/kWh)(10 3 t/kg) = 3 6 × 106 J/kg , at1800◦ C.Roomtemperatureisapproximately20◦ C,sothis correspondstoaspecificheatcapacityof
C = 3 6 × 106 J/kg
1780 K = 2000J/kgK ,
whichisapproximatelythreetimesashighastheheatcapacityofgraphiteatroomtemperature.Thisdoesnotincludeany lossofenergytotheenvironmentduringstorageortransfer.The heatcapacityofgraphiteshowsastrongtemperaturedependence, increasingbyafactorofthreebetweenroomtemperatureand around1500◦ C[26],sothisdoesseemfairlyreasonable.
Thepowerplantwillreceivenearlyallofitsenergyduringthe day.Weassumethatnightcoversabouthalfthetimethattheplant isoperating.Thusofthe30MkWh/ygenerated,halfcanbedistributedduringthedayandhalf, ≈ 15 MkWh/y,isdistributed atnightfromstoredenergy.Thismeansthattheplantwillneed tostorearound41000kWh/dayofenergyforuseatnight.The graphiteisheatedduringthedayandcoolsoff atnightastheplant suppliesenergy,sothismeansthattheplantwillneedtohaveat least
M = 41000kWh 1000kWh/t = 41tonnes ofgraphitetoprovidethismuchpower.Inreality,iftheplant istoprovideafairlyconstantamountofenergythroughoutthe year,muchmoregraphitewouldlikelyberequiredsincetherewill belargevariationsintheamountofincidentsolarenergydueto seasonalvariationsandlocalweatherconditions.
ThedailyenergyneedsofanaverageAustralianislikelytobe muchclosertothepercapitaenergyusageofwesternEuropeor theUS( 0.5–1GJ/day)thantotheworldaverage(200MJ/day), ofwhichabout14%isintheformofsuppliedelectricalpower (seeFigure1.2).So,thepercapitaelectricalenergyusage is70–140MJ/dayTheplantsuppliesatotalof8 2 × 104 kWh/day,or 3 × 105 MJ/day.At70to140MJperdayperperson,thisplantwill supplytheenergyneedsofroughly2000–4000people.
5.10 Astart-upcompanyismarketing steel“icecubes” to beusedinplaceofordinaryicecubes.Howmuch wouldaliterofwater,initiallyat20 ◦C,becooled bytheadditionof10cubesofsteel,each2.5cmon aside,initiallyat 10 ◦C?Compareyourresultwith theeffectof10ordinaryicecubesofthesamevolume thesameinitialtemperature.Wouldyouinvestinthis start-up?Takethedensityofsteeltobe8.0gm/cm3 and assumetheheatcapacitiestobeconstants,independent oftemperature.
Theheatcapacityof10steelcubesofvolume V = (2 5cm)3 thenis
CS = 10(2 5cm)3 (0 008kg/cm3)(0 52kJ/kgK) 0 65kJ/K
Oneliterofwaterweighs1kg,andhasaheatcapacityof CW = 4 2kJ/K Asenergyisconservedandthesystemwillcomeinto equilibriumatasometemperature T ,wehave
CS (T + 10) + CW (T 20) = 0, andasthesystemwillcomeintothermalequilibrium.Solvingfor T ,wefind T = 16 0 ◦ C,sothesteelicecubescoolthewaterby 4.0 ◦ C.
Thedensityoficeatfreezingpointisapproximately 0.92gm/cm3 .Theheatcapacityof10icecubesofvolume V then is
CI = 10(2 5cm)3 (0 00092kg/cm3)(2 05kJ/kgK) 0 29kJ/K
Whentheicemeltsat0 ◦C,thereisalatentheatabsorptionof
∆Q = 10(2 5cm)3 (0 00092kg/cm3 )(330kJ/kg) 47kJ
Itiseasiesttodotheproblemintwosteps.Firstconsiderthe coolingofthewaterbyheatingtheiceto0 ◦Candmeltingit:
CW (20 T ) = 47kJ + (10K)(0 29kJ/K) 50kJ
Solvingfor T ,wefind T 8 1 ◦ C.Atthispointwehave1kg ofwaterat8 ◦ Cand0.14kgofwaterat0 ◦C.Thesecomeinto equilibriumwhen(8 T ) = 0 14T ,or T = 8/1 14 7 ◦ C.Sothe watercoolsby13 ◦ C.
5.11
Weseethattheiceismuchmoreeffectiveatreducingthetemperature.Theheatcapacityperunitvolumeofthesteelcubesis greaterthanthatofice,butlessthanthatofwater.Thelatentenergy absorbedbythemeltingicealsosignificantlylowersthetemperatureofthewater.Thesteelcubes,however,don’tundergoany phasetransitionsintherelevanttemperaturerange,andso they arereusable,andwillnotdiluteotherdrinks.Theyaremuch more massivethanice,however,andcouldbreakdelicateglassif treated casually.Thesteelcubeswouldlikelyappealtocertainpeopleasa luxurygood.
Roughly70%ofthe5 × 1014 m2 ofEarth’ssurfaceis coveredbyoceans.Howmuchenergywouldittake tomeltenoughoftheiceinGreenlandandAntarcticatoraisesealevels1meter?Supposethat1% ofenergyusedbyhumansbecamewasteheatthat meltsice.Howlongwouldittaketomeltthisquantityofice?IfanincreaseinatmosphericCO2 led toanetenergyfluxof2000TWofsolarenergy absorbedintotheEarthsystemofwhich1%melts ice,howlongwouldittakeforsealevelstoriseone meter?
Theenthalpyoffusionoficeis334kJ/kg.Liquidwaterhasa densityofapproximately1000kg/m3 ,sotheenthalpyoffusionis 334MJ/m3 .Toraisesealevelsby1m,atotalof
V (0 7)(1m)(5 × 1014 m2 ) = 3 5 × 1014 m3 ofwatericemustbemelted.Thisrequires
E (3 5 × 1014 m3 )(334MJ/m3 ) = 1 2 × 1023 J ofenergy.Humansuseapproximately580EJofenergyperyear.If 1%(5.8EJ)contributestoheatmeltingtheice,thenitwould take (1 2 × 1023J)/(5 8 × 1018 J/y) = 21000yearstomeltthismuchice. However,ifanincreaseinCO2 increasestheamountofsolar energyabsorbedbytheearthby2000TWand1%ofthis contributestowarmingitwouldonlytake
T 1 2 × 1023 (0 01)(2 × 1015 J/s) × 1y (3 15 × 107 s) = 190years tomeltthatmuchice.
5.12 Carbondioxidesublimesatpressuresbelowroughly 5atm(seeFigure5.12).Atapressureof2atmthis phasetransitionoccursatabout 69 ◦Cwithan enthalpyofsublimationofroughly26kJ/mol.Suppose akilogramofsolidCO2 atatemperatureof 69 ◦Cis confinedinacylinderbyapistonthatexertsaconstantpressureof2atm.Howmuchheatmustbeadded tocompletelyconvertittogas?AssumingCO2 tobe anidealgas,howmuchworkwasdonebytheCO2 in thecourseofvaporizing?Iftheambientpressureoutsidethecylinderis1atm,howmuchusefulworkwas done?Finally,byhowmuchdidtheinternalenergyof theCO2 changewhenitvaporized?
ThemolecularweightofCO2 is0.044kg/mol,so1kg = 1/0 044 = 22 7mol.TheCO2 sublimationphasetransitiontakes placeatfixed T and p.Theenthalpyofsublimationfor1kgis ∆H = (22 7mol)(26kJ/mol) = 590kJ/kg.Thisisthethermal energy(heat)thatmustbeaddedtovaporizetheCO2 .Thework donebythegasis p∆V ∆V mustbecomputedfromtheidealgas
law,
∆V = ∆NRT
p = (22 7mol)(8 31J/molK)(273K 69K) 2(101325Pa) = 0 190m3
Then W = p∆V = 2(101325Pa)(0 19m3 ) = 38 5kJ.Iftheoutside pressureis1atm,thenhalfofthisisusefulwork, Wuseful 19kJ, and ∆U =∆H p∆V = (590 38 5)kJ/kg = 551kJ/kg.
Problems
6.1 Supposeasmallstonewarekilnwithsurfacearea5m2 sitsinaroomthatiskeptat25 ◦Cbyventilation. The15cmthickwallsofthekilnaremadeofspecial ceramicinsulation,whichhas k = 0 03W/mK.The kilniskeptat1300 ◦ Cformanyhourstofirestoneware. Theroomventilatingsystemcansupply1.5kWof cooling.Isthisadequate?(Youcanassumethatthe wallsofthekilnarethincomparedtotheirareaand ignorecurvature,corners,etc.)
Wecomputetheheatemitted(thermalpower)bythekilnat temperature T = 1300 ◦ C,assumingtheroommaintainsitstemperatureof Tr = 25 ◦ C,andseehowthiscompareswiththe1.5kW ofcoolingavailable.Thewallsofthekilnarethincompared toits area,sowecantreatisasaplanarslabofsurfacearea A = 5m2 and thickness ℓ = 15cm,with k = 0 03W/mK.AccordingtoFourier’s law,thethermalpoweremittedbythekilnis P = kA(T Tr )/ℓ,
P = (0 03W/mK)(5 0m2 )(1300 ◦ C 298 ◦C)/(0 15m) 1 0kW
So,thecoolingventilationsystemissufficienttopreventtheroom fromheatingup.
6.2 Tworigidboardsofinsulatingmaterial,eachwitharea
A,havethermalconductances U1 and U2.Supposethey arecombinedinseriestomakeasingleinsulatorof area A.Whatistheheatfluxacrossthisinsulatorasa functionof ∆T ?Nowsupposetheyareplacedside-byside.Whatistheheatfluxnow?
Thermalconductancesarethereciprocalsofthermalresistances, U j = 1/R j,for j = 1, 2.Resistancesinseriesadd, so Reff = 1/Ueff = 1/U1 + 1/U2 , or Ueff = U1 U2 U1 + U2 , and Q = A
Thermalresistancesinparalleladdasreciprocalsmultipliedby theirareas,
Theexpressionfortheheatfluxis
, whichstatesthatthetotalheatfluxisthesumofthefluxesthrough thetwoinsulators.Notethatthefactorofonehalfintheeffectivethermalconductancecompensatesforthefactoroftwointhe effectivearea.
6.3 Theheattransfercoefficient h forairflowingat30m/s overa1mlongflatplateismeasuredtobe80W/m2 K. Estimatetherelativeimportanceofheattransferby convectionandconductionforthissituation.
TheratioofconvectionheattransfertoconductionheattransferisgivenbytheNusseltnumber,NuL = hL/k.Taking k = 0 026W/mKfromTable6.1,
/m2 K)(1m)
026W/Km 3100
Convectionisover3000timesasimportantasconductionfor heat transferthroughairinthisexample.
6.4 GiventheSun’spoweroutputof384YWandradius 695500km,computeitssurfacetemperatureassuming ittobeablackbodywithemissivityone.
Thesurfacetemperatureofthesunwiththeseassumptionsis
6.5 Humansradiateenergyatanetrateofroughly100W; thisisessentiallywasteheatfromvariouschemical processesneededforbodilyfunctioning.Considerfour humansinaroughlysquarehutmeasuring5m × 5m, withaflatroofat3mheight.Theexteriorwallsofthe hutaremaintainedat0 ◦Cbytheexternalenvironment. Foreachofthefollowingconstructionmaterialsfor thewallsandceiling,(i)computethe R-value (∆T /q) forthematerial,(ii)computetheequilibriumtemperatureinthehutwhenthefourpeopleareinit:(a) 0.3m(1foot)concretewalls/ceiling;(b)10cm(4") softwoodwalls/ceiling;and(c)2.5cm(1")softwood, with0.09m(3.5")fiberglassinsulationoninteriorof wallsandceiling.Takethermalconductivitiesand/or R-valuesfromTables6.1and6.3.
The R valueisjust R = j R j = j L j /k j forinsulatorsconnectedinseries.Heatisconductedawayfromthehouseatarateof Q = A(Teq T0 )/R.Theoutsidetemperatureis T0 = 0 ◦ C,while thesurfaceareaofthehutis A = 75m2 ,ignoringthefloor.Atequilibrium, Q isjusttheheatradiatedfromthefouroccupantsofthe hut, Q = 400W.Solvingfortheequilibriumtemperature, Teq = RQ A + T0
a.Concretehas k ≈ 0 9W/mK.So R ≈ (0 3/0 9) = 0 33Km2 /W. Thus, Teq ≈ (0 33)(400)/75 ≈ 1 8 ◦ C.
b.Softwoodhas k ≈ 0 11W/mK. R ≈ (0 10/0 11) = 0 91 Km2 /W.Thus, Teq ≈ (0 91)(400)/75 ≈ 4 9 ◦ C
c.Thefiberglasshas R = 2 0W/mK. R = (0 025/0 11 + 2 0) ≈ 2 2Km2 /W.Thus Teq ≈ (2 2)(400)/75 ≈ 12 5 ◦C.
6.6 Considerabuildingwith3200ft2 ofwalls.Assumethe ceilingiswell-insulatedandcomputetheenergyloss throughthewallsbasedonthefollowingmaterials, assuminganindoortemperatureof70 ◦Fandanoutdoortemperatureof30 ◦F:(a)wallsarecomposedof 4" thickhardwood;(b)wallscomposedof3/4" ofplywoodontheinsideandoutside,with8" offiberglass insulationbetween;and(c)sameas(b)butwith18 single-paneR-1windows(USunits),eachwitharea 0.7m2 (remainingareaissurroundedbywallsasin (b)).Takethermalconductivitiesand/orR-valuesfrom Tables6.1and6.3.
Theenergylossis Q = A∆T /R where A 300m2 and ∆T = 40◦ F 22◦ C,so Q 6600/R[kmm2 /W].
(a).Inthiscase, R = L/k ≈ 0 10/0 18 ≈ 0 56Km2 /W,so Q ≈ 6600/0 56 11 8kW.
(b).Inthiscase, R = 2(0 19) + 4 6 5 0Km2 /W,so Q 1 32kW.
(c).Inthiscase,mostofthewallismadeof R = 5 0Km2 /Wmaterial,whilethewindowsaremadeof R = 1 0ft ◦Fhr/BTU = 1/5 68Km2 /W = 0 176Km2 /Wmaterial.Thewindowstake up12.6m2 ,whiletheresttakesup287.4m2 ofarea.Therate ofenergylossis
6.7 EstimatetheR-valueofthewallshowninthefigurein Example6.2.
Sincethethermalresistanceofthefoamandfiberglassare giveninUSR-values,weuseUS-unitsthroughout.Combining thetwolayersoffiberglassandfoamwith2.5cmofsheathing (plywood)with R ≈ 1ft ◦Fhr/BTUand1cmofdrywallwith R ≈ 0 5ft ◦ Fhr/BTU,allinseries,
eff = 0 5 + 13 + 10 + 13 + 1 ≈ 37 5ft ◦ Fhr/BTU
6.8 Assumethata60m2 wallofahouseisinsulatedtoan R-valueof5.4(SIunits),butsupposetheinsulationwas omittedfroma1m2 gapwhereonly6cmofwoodwith anR-valueof0.37remains.ShowthattheeffectiveRvalueofthewalldropsto R = 4 4leadingtoa23% increaseinheatlossthroughthewall.
Theinsulatedportionofthewallhas R = 5 4andthegaphas R = 0 375.Theseareinparallel,soinverse R valuesadd.The insulatedportionofthewallis59m2 andtheuninsulatedportion isjust1m2 .Theresistanceofthewallisgivenby
60 R 59 5 4 + 1 0 37 13 6W/K
So R 60m2 ÷ 13 6W/K 4 4m2 K/W,whichrepresentsa23% increase(5 4/4 4)intheheatlossratethroughthewall.
6.9 InExample6.1,afilmofstillairwasidentifiedas thesourceofalmostallofthethermalresistanceofa single-paneglasswindow.Determinethethicknessof thestillairlayersonbothsidesofthewindowifradiativeheattransferisignored.Itisestimatedthatabout halfthethermalconductanceofthewindowisdueto radiativeheattransfer;estimatethethicknessofthestill airlayersifthisisthecase.
Fromtheexample,theR-valueoftheglassaloneis Rglass 0 0031(SIunits).TheeffectiveR-valueofthewindowis Reff = Rglass + Rair 0 17W/m2 K,sotothisaccuracywecanignore Rglass entirely.Using k = 0 026W/mKforairand R = L/k = 0 17W/m2 Kgives L = (0 17)(0 026) 4 4mm.Accordingto thediscussionintheexample,80%ofthestillairisontheinside surfaceand20%isontheoutside.Thus Lairin 3 5mmand Lairout 0 9mm.IfhalfthethermalconductanceisduetoradiativeheattransferthentheR-valueofthestillairmustbetwicewhat wejustcomputed.Thusamoreaccurateestimateofthethickness ofthestillairlayersis7.0mmontheinsideand1.8mmonthe outside.
6.10 Accordingto[28],adoublepanewindowwithanemissivity ε = 0 05coatinganda1/4" airgaphasa measured(centerofglass)U-factorof2 33W/m2 K. Assumethatthiscoatingissufficienttostopallradiativeheattransfer,sothatthethermalresistanceofthis windowcomesfromconductionintheglass,theair gap,andthestillairlayersonbothsidesofthewindow.Taking Rstillair ≈ 0 17m2 K/WfromExample6.1, computetheU-factorofthiswindowignoringconvectionintheairgap.Howdoesyouranswercompare
Figure6.17 (a)Asectionthroughawallinawood-frame building.(b)Ausefulequivalentcircuitdiagram.
withthevaluequotedin[28]?Usethesamemethodto computetheU-factorofasimilarwindowwitha1/4" argon-filled gap(measuredvalue U = 1 87W/m2 K). Repeatyourcalculationfora1/2" airgap(measuredUfactorl.70W/m2 K).Canyouexplainwhyyouranswer agreeslesswellwiththemeasuredvalueinthiscase?
WecomputetheR-valueassumingconductionaloneandthen use U = 1/R.Ingeneral
Reff = Rglass + Rstillair + Rairgap
Forthe1/4inchairgap
Reff (1/4) = 2(0 0031) + 0 17 + 0253 4( 026) 0 42m2 K/W
and Ueff (1/4) = 1/R 2 4,within4%oftheobservedvalue.
Forthe1/4inargongap
Reff (1/4argon) = 2(0 0031) + 0 17 + 0253 4( 018) 0 53m2 K/W
and Ueff (1/4argon) = 1/R 1 9,equaltotheobservedvalue withintheaccuracyofthecalculation.
Forthe1/2inchairgap
Reff (1/2) = 2(0 0031) + 0 17 + 0253 2( 026) 0 66m2 K/W
and Ueff (1/2) = 1/R 1 5.Inthiscasethecalculatedthermal conductivityis12%lowerthanthemeasuredvalue.Possibly the largermeasuredvalueisduetoconvectionwithinthewiderair gap,asdiscussedinthetext.
6.11 zAbuildingwallisconstructedasfollows:startingfromtheinsidethematerialsusedare(a)1/2" gypsumwallboard;(b)wall-cavity,80%ofwhichis occupiedbya3.5" fiberglassbattand20%ofwhich isoccupiedbywoodstuds,headers,etc.;(c)rigid foaminsulation/sheathing(R = 0 7m2 K/W);and(d) hollow-backedvinylsiding.Thewallisillustratedin Figure6.17alongwithanequivalentcircuitdiagram. CalculatetheequivalentR-valueofthewall.Donot forgetthelayersofstillaironbothinsideandoutside.
Allthethermalresistancesaddinseriesexceptforthewood andfiberglass,whichaddinparalleltogiveaneffectiveresistance
Reff .TakingR-valuesandthermalconductivitiesfromTables6.1 and6.3,
Reff = 0 8 2 0 + 0 2 0 81 1 1 54m2 K/W ,
Figure6.18 Crosssectionthroughaninsulatedcopperpipe. Thepipecarriesfluidat100 ◦ C.Theradiusoftheinsulation mustbe3.6timestheradiusofthepipeinordertokeepthe powerlosstolessthan10W/m.
where(3 5)( 0253)/0 11 = 0 81m2 K/WistheR-valueofthe3.5 inchthicksoftwood.Combiningallthecontributions
Rtotal = Rstillair + Rdrywall + Reff + Rfoam + Rsiding = 0 17 + 0 079 + 1 54 + 0 7 + 0 11 2 6m2 K/W
or,inUSunits, R 15ft2 ◦ Fhr/BTU.Notethatifthewoodwere ignored,thecalculationwouldgive R = 3 1m2 K/W,18%higher thantheactualvalue.
6.12 [H]Aninsulatedpipecarriesahotfluid.Thesetup isshowninFigure6.18.Thecopperpipehasradius R0 = 1cmandcarriesaliquidat T0 = 100 ◦C.The pipeisencasedinacylindricallayerofinsulationof outerradius R1;theinsulationhasbeenchosentobe closed-cellpolyurethanesprayfoam withanR-value of1.00m2 K/Wperinchofthickness.Howlargemust R1 besothattheheatlosstothesurroundingroom(at 20 ◦C)islessthan10W/m?[Hint:Firstexplainwhy theheatfluxmustbe q = c ˆ r/r.Youmayfindituseful tomaketheanalogybetweenheatpropagationthrough theinsulatorandcurrentflowingoutwardthrougha cylindricalconductor.Notealsothat ∇ f (r) = ˆ r f ′(r).]
Theconductivityofcopperisveryhighcomparedtothatofthe insulation,sothecopperitselfcanbeneglected.Duetocylindrical symmetry,theheatfluxmustbeintheradialdirectionfromthe copper.Atthermalequilibrium,thefluxofthermalenergyperunit lengthalongthepipemustbethesameatallradii.Foracylinder, theareaisproportionaltotheradius,soif Q isthetotalenergy fluxperunitlength, Q = 2πq(r)r.Since Q isindependentof r, q = Qr/2πr.Thefluxisinverselyproportionaltotheradiusand alignedalongtheradialdirection.Solvingfor T (r),
80Kwith Q = 10W/m.So, R1 = R0 exp 2πk(T0 T (R1 ))/Q = (1cm)exp 2π(80K)(0 0254W/mK) 10W/m = 3 6cm
6.13 Ignoringconvectiveheattransfer,estimatethechange in U-factorbyreplacingargonbykryptoninthe quadrupleglazedwindowsdescribedinTable6.3. Youcanignoreradiativeheattransfersincetheseare low-emissivitywindows.
ThecontributionstotheR-valuefromthestillair,fourpanesof glass,andthe3/4inchofargonorkryptonaddupto
Argon: Reff = 0 17 + 4(0 0031) + 3 4 0 0254 0 018 1 24m2 K/W ,
Krypton: Reff = 0 17 + 4(0 0031) + 3 4 0 0254 0 0094 2 21m2 K/W
Thedifferencein U-factorsistherefore ∆U = 1/1 24 1/2 21 0 35W/m2 K.Theactualchangeis ∆Umeasured 0 29W/m2 K.
6.14 Inareaswherethesoilroutinelyfreezesitisessentialthatbuildingfoundations,conduits,andthelike beburiedbelowthefrostlevel.CentralMinnesota isacoldpartoftheUSwithyearlyaveragetemperatureof T0 5 ◦Candvariation ∆T 20 ◦C. Typicalpeatysoilsinthisregionhavethermaldiffusivity a 1 0 × 10 7 m2/s.Buildingregulationsrequire foundationfootingstoexceedadepthof1.5m.Doyou thinkthisisadequate?
Wemakeuseofeqs.(6.26)and(6.27).Theminimumtemperatureoccurswhenthesine-functionis 1,so Tmin (z) = T0 ∆Te αz , where T0 = 5 ◦ C, ∆T = 20 ◦ Candwewant Tmin (1 5m) > 0 ◦C. Firstwecompute
1
1 , andthenevaluate, Tmin(1 5m) = 5 20e (1 0)(1 5) 5 4 5 0 5 ◦ C Sothedepthof1.5mseemsmarginallysufficient.
6.15 Consideraregionwithaveragesurfacetemperature T0 = 10 ◦ C,annualfluctuationsof ∆T = 30 ◦C,and surfacesoilwith a 2 4 × 10 7 m2/s, k 0 3W/mK. Ifthelocalupwardheatfluxfromgeothermalsources is80mW/m2,computethedepthatwhichtheheatflux fromsurfacefluctuationsiscomparable.Whatisthe answeriftheareaisgeothermallyactiveandhasa geothermalheatfluxof160mW/m2?
Theconstantofintegrationisseparatedintointo T0 and R0 terms sothatthelogarithmtermis0attheinnerradiusofthepipe(where T0 = 100 ◦ C).The R valueis1.00m2 K/Wperinch,giving k = 0 0254W/mK.Wewanttofindtheradius R1 atwhich T (R1) T0 =
Thedownwardheatfluxfromsurfacefluctuationsisproportionaltothegradientofthetemperature, q = k|dT /dz|,wherewe havesuppressedthesigns.Thetemperaturegradientis dT dz = α∆Te αz (sin(ωt αz) + cos(ωt αz)) , andthishasmaximummagnitudewhenthesineandcosinefunctionsareequalandtheirsumis √2.Substitutingfor dT /dz and solvingfor z,
z = 1 α ln q √2αk∆T
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