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Heat transfer for CNT Jiayun Shao

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Page 1: research outcome

Heat transfer for CNT

Jiayun Shao

Page 2: research outcome

1.Graphene and CNT structure

2.CNT code by Matlab

3.run with lammps

4.Calculate the thermal conductivity 

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Sample matlab result

For a n=6 m=6 Height =100*10-10 m ;

Result:T(Translation vector) = -1*a1+1*a2 =2.46 * 10-10 mCh(Chiral vector) = 6*a1+6*a2=25.56 * 10-10 mDt(The nanotube diameter) = = 8.14 *10-10 m

Figure 1. General Graphene structure

Constructed parametrical model for Graphene and Carbon nanotube base on spatial operation based on Graphene surface with MATLAB

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Figure 2. 6*6 Graphene unit in 10-10 m

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Figure 3. 6*6 CNT Side view

Figure 4. 6*6 CNT top view

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20 30 40 50 60 70 80290

295

300

305

310

315

f(x) = − 0.0343278571428571 x + 305.789005765306

T VS Position

T VS PositionLinear (T VS Position)

Z direction (*10^-10m)

Tem

pera

ture

(K)

Figure 5.

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A (Cross section area)= Dt*3.4*Pi = 86.96 * 10-20 m2

Q = 0.1eV = 1.6*10-20 J

Based on the Fourier’s Law:

k=

Figure 6. cross section area

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Stone wales defect

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  position  Inter face start from 496.68  end at 511.165  middle point  503.9225       First part from 278.214  end at 476.161       second part from 567.894  end at 767.048       

T1 at middle point 300.9966398 K

T2 at middle point 298.515463 Kdel T 2.48117675 K     q 0.00000008  R=DELT/q 3.10E+07 K/ WA 8.696E-19m^2Boundary resistance 2.70E-11

K*m^2/W

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Vacancy defect

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  position  Interface start from 606.5  end at 615.5  middle point  611       First part from 328.5  end at 545.5       second part from 652.5  end at 904.5       T1 at middle point 297.4304 KT2 at middle point 301.9068 Kdel T 4.4764 K     q 0.00000016  R=DELT/q 2.80E+07 K/ WA 8.696E-19M^2Boundary resistance 2.43E-11 K*M^2/W

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Unzipped CNT

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  position  Interface start from 505.13  end at 522.028  middle point  513.579       First part from 270.972  end at 406.156       second part from 585.999  end at 721.183       T1 at middle point 301.3752297 KT2 at middle point 298.734735 Kdel T 2.6404947 K     q 0.00000008  R=DELT/q 3.30E+07 K/ WA 8.696E-19M^2Boundary resistance 2.87E-11 K*M^2/W

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  position  Interface start from 505.13  end at 522.028  middle point  513.579       First part from 270.972  end at 406.156       second part from 585.999  end at 721.183       T1 at middle point 300.4431247 KT2 at middle point 297.5127663 Kdel T 2.9303584 K     q 0.00000008  R=DELT/q 3.66E+07 K/ WA 8.696E-19M^2Boundary resistance 3.19E-11

K*M^2/W