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1、?46?10?Vol.46No.102010?10?12441249?ACTA METALLURGICA SINICAOct.2010 pp.12441249?(?,?150001)?,?3?Tersoff?,?C,BN?SiC?.?3?,?3?.?:?k?L?,?k L?,?;?;?,?,?3?;?3?:C,BN?SiC.?,?,?O482.22?A?04121961(2010)10124406MOLECULAR DYNAMICS SIMULATION ON THERMALCONDUCTIVITY OF ONE DIMENISONNANOMATERIALSGAO Yufei,MENG Qin
2、gyuanSchool of Astronautics,Harbin Institute of Technology,Harbin 150001Correspondent:MENG Qingyuan,professor,Tel:(0451)86414143,E-mail:Supported by National Natural Science Foundation of China(No.10772062)Manuscript received 20100407,in revised form 20100630ABSTRACTThe Nonequilibrium molecular dyna
3、mics(NEMD)simulation method which is basedon the linear response theory is applied to simulate the thermal conduction process of C,BN and SiCnanotubes.The threebody Tersoffpotential is used to simulate the interactions among atoms.Theeffects of axial length,temperature and tensile strain on the axia
4、l thermal conductivity of the threekinds of nanotubes are investigated,and their thermal conductivities are compared and analyzed.Thesimulation results show that the axial thermal conductivity increases as the axial length increases,andexhibits a relationship k Lthat is in agreement with the solutio
5、n of Boltzmann-Peierls phonontransport equation(BP equation).It is found that the thermal conductivity of nanotube decreaseswith the increase of temperature.As the tensile strain increases,the thermal conductivity of nanotubesshow an slight increase first,and then decreases.But,the corresponding ten
6、sile strains at whichthe tendency of thermal conductivity of the three nanotubes changes are different.Under the sameconditions,the sequence of thermal conductivity from the biggest to the smallest is in the order ofcarbon nanotubes,boron nitride nanotubes and carbon silicon nanotubes.KEY WORDS none
7、quilibrium molecular dynamics simulation,nanotube,thermal conductivity?,?,?,?.?,?,?*?10772062?:20100407,?:20100630?:?,1984?,?DOI:10.3724/SP.J.1037.2010.00164?.?1,2?,?,?.MingoBroido3?Boltzmann?,?4?,?.?10?:?1245?Pop?5Yu?6?C?,?C?Wm1K1.?,?C?(102106Wm1K1)59,?.?,ZhangLi10?C?,?;Nanda?11?(PAO)?C?,?C?;Maruya
8、ma?12?C?,?,?;Bi?13?C?,?C?.?,?C?,?BN?SiC?.?,?,?14?,?.?C,BNSiC?,?;?BNSiC?,?BoltzmannPeierls?.1?15?10 10?C,BNSiC?.?Tersoff?16?.Terrsoff?E=12XiXj6=iVij(1)Vij=fC(rij)fR(rij+bijfA(rij)(2)fC(r)=1:r R D1212sin(2rRD):R D r R+D(3)fR(r)=Aexp(1r)(4)fA(r)=Bexp(2r)(5)bij=(1+nnij)12n(6)ij=Xk6=i,jfC(rik)g(ijk)expm3
9、(rij rik)m(7)g()=rijk?1+c2d2c2d2+(cos cos 0)2?(8)?,i,jk?;E?;Vij?;rij?;fC?;fR?;fA?;bij?;ij?;ijk?;RD?;AB?;12?;0?;?;3?;n?;mr?;g()?;c?;d?.?,Tersoff?.?C?SiC?Tersoff?Lammps?.?Lammps?NB?3?Tersoff?,?NB?2?Tersoff?17?3?Tersoff?,?1?.?BN?,?1NB 3?Tersoff?Table 1The parameters of three body Tersoffpotential funct
10、ion of NBcdhn2BRD1Anm1eVnmnmnm1eVBBB0.5261.5871030.53.991.61062.1810543.130.20.012.34810540.05BBN0.5261.5871030.53.991.61062.1810543.130.20.012.34810540.05BNB0.5261.5871030.53.991.61061.5110489.070.20.011.63410482.71BNN0.5261.5871030.53.991.61061.5110489.070.20.011.63410482.71NBB17.85.948400.620.019
11、31.5110489.070.20.011.63410482.71NBN17.85.948400.620.01931.5110489.070.20.011.63410482.71NNB17.85.948400.620.01930.2624142.90.20.010.28252133.0NNN17.85.948400.620.01930.2624142.90.20.010.28252133.0Note:1and 2lattice constant,A and Bcohesive energy,R and Dcutoffparameters,hcosine of the energetically
12、optimal angle,bulk modulus,ncorrspond to simple cubic phase,ddetermine the dependence on angle,cdetermine the strength of angular effect?1246?46?BN?0.146 nm,?BN?,?3?Tersoff?BN?.?C,BNSiC?,?.?,?0 K?4104?,?,?100 K,?,?1600 K,?.?C?0.142 nm,BN?0.146 nm,SiC?0.18 nm,?TersoffBrenner?18?,?Tersoff?3?.C,BNSiC?1
13、?.?,?,?,?19,?0.5 fs.?,?N?,?N=0?,N=N/2?,?(?2).?1C,BN?SiC?Fig.1Scketchs of C(a),BN(b)and SiC(c)nanotubes?2?Fig.2Schematic of nonequilibrium molecular dynamics(NEMD)simulation?,?(?),?,?,?,?,?Fourier?J=T,?15,19.2?3?:?,?NVT?(?N?V?T?)?1105step,?;?,?NVE?(?N?V?E?)?1106step,?;?NVE?2106step,?,?,?.?,?.?,?,?,?F
14、ourier?20.?,?1200 K?,10 step?,?5 step?.?10 step?.2.1?3,?20,30,40,50,60,80100 nm?10 10?C,BNSiC?300 K?,?3?.?3?,3?,?.?20?.?:1leff1l+4LZ(9)?,leff?,l?,LZ?.?=13cvl(10)c=32kBn(11)?10?:?1247?3?Fig.3The relationship between nanotubes thermal con-ductivity and axial length?,l?,v?,c?,n?,kB?Boltzmann?.?,?(911)?
15、12kBnv?1l+4LZ?(12)?(12)?LZ?,?,?,?.?BoltzmannPeierls?(BP?)3?.BP?vpdnpdx=cnp(13)?,vp?,np?,n?,npn?np=n+n(n+1)g(x)(14)n (e h/kBT 1)1(15)?,BP?cnp/(n(n+1)=Xdir(p)(n n)pp,p(g g g)+12Xrev(p)(1+n+n)p,pp(g+g g)(16)?dir(p)rev(p)?,?:p+p pp p+p?.p,pp?;pp,pp,pp?;g,gg?p,pp?.?=1SXZvp hpgpnp(np+1)dq(17)?,h?Planck?,p
16、?,q?,S?.?gp?gp hkBT2vppTL(18)?,T?,L?.?1p=Xdir(p)(nn)pp,p+Xrev(p)(1+n+n)p,pp+2vpL(19)?,n,n?pp?.?,?,?1pdirXppp(nn)+12revXppp(1+n+n)+aL(20)?a?;,?p,pp?.?0?n n 1/1/()(21)1+n+n 1+1/+1/1+/()/()(22)?Z(q2+a/L)1dq L1/2(23)?,?Z(q3/2+a/L)1dq L1/3(24)?,?BP?,?L?,?0.5?.?L?,?0.250.45?,?.?413?,20,80100 nm?(10 10)C?3
17、00 K?247,405456 W/mK(?0.34 nm?),?,?3?.2.2?,?3?10 10?(1001200 K)?,?3?4.?4?,?,?,3?.?.?1248?46?4?Fig.4The relationship between nanotubes thermal con-ductivity and temperature?,?Debye?D,?Einstein?21?eDT(25)?,?Debye?h kBT,?n?n=1e hkBT 1kBT h(26)?(26)?,?,?n?,?,?.?(9)?,?,?1T(27)?(2527)?4?.?13?,?300,600800
18、K?,C?247,148120 W/(mK)(?),?.2.3?,?,?,?.?,?20 nm?10 10?,?300 K?,?,?,?5000 steps?0.0025 nm?.?3?5.?5?,3?,?3?,?,C?,BN?,SiC?.?5?Fig.5The relationship between nanotubes thermal con-ductivity and axial strain?,?,?,?,?,?,?,?,?,?(10)?;?,?,?,?,?,?,?,?(10)?.3?:CC?BN?BN?SiC?.?.?,?,?,?22.?3?:CC?BN?SiC?,?,?,?,?.3
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