1. Thermal conductivity (k) of CNTs Have a look at C 60 Heat transfer media : phonons and electrons C 60 has a perfect symmetry and  -electrons move freely On C60 surface, so it has a high k = 3000-4000 W/mK  -e - heat

2. Heat resistance at interball Strong inter-ball interaction gives better thermal conduction Thermal conductivity of C 60 crystal (=0.4 W/mK at room temp) C 60 crystal

3. T 260 K f.c.c phases.c phase C 60 rotation within crystal begins at 260 K 所以 F.C.C 相之 orientational disordering 比 S.C. 相高

4. What is orientation ordering? N S N S N S N S Stronger interaction between C 60 so better thermal conduction. Weaker interaction between C 60, so poor thermal conduction S.C phase f.c.c phase ordering disordering

5. K (W/mk) T (K) 0.4 26085 Temp dep Temp indep 0.4 + 0.4  (25 %) Thermal conductivity of C 60 crystal 1.Why K increases by 25% at 260 K,? 2.Why K becomes temp indep above 260 K? 3.Why K becomes temp dep at 85 K?

6. Thermal conductivity, K K = 1/3  C v. .l C v : specific heat  : sound speed l: phonon mean free path 260 KWhy K increases by 25 % 85 K f.c.c s.c K = 1/3  C v. .l invariables 增加 phonon scattering propagation > 260 K, l = 50 Å, C v = 50 j/kmol

7. 260 K Why K is temp indep above 260 K phonon Strong scattering above 260 K, so phonons only get luck with propagation, and temp makes no difference here!

8. Debye temp  D of C 60 crystal ~ 75 K (very low) because weak inter-ball interaction (or low density of inter-ball phonon mode).  D = hν m /K K: boltzmann constant h: planck constant ν m : Debye frequency Debye temp: the temp of a crystal’s highest normal mode of vibration Normal mode: an oscillation in which all particles move with the same frequency and phase. ν m = (3N/4  V) 1/2 ·V s N/V: number density of atom V s : effective speed of sound

9. Why K becomes temp dep at 85 K There still has some mis-orientated C 60 Duration cooling, needs a time for C 60 alignement. 260 K 85 K

10. Thermal conductivity of CNTs Diamond (C-sp 3 bond): stiff (faster conduction: K = 1500 W/mK) soft bonding (slow conduction) p.s. diamond has perfect lattice, less defects, so phonon density is high !

11. CNTs are sp 2 bonds (C-C=C-C),stronger than diamond Double bonds So higher K is expected for CNTs? Yes, but only when CNTs zero defects and perfect crystallized K of a single graphite layer is very high (>2000 W/mK), phonon only moves on in-plane phonon Layer spacing (0.335 nm)  Weak inter-layer interaction weakens the thermal conductivity K. Phonon also has to move along the c-axis

12. CNTs have higher inter-layer spacing (0.34 nm), so K is similar to a single graphene sheet c-axis In-plane direction (a-axis) Another calculations of K 1. (1/A)(dQ/dt) = - K(dT/dz) A: cross section area dQ: thermal energy dt: time interval dT/dz: temp gradient along z direction 2. K = (1/3vk B T 2 )  dt 0  v: volume k B : boltzmann constant T: sample temp : mean value of heat flux vector

13. 40000 W/mK K Temp (k) (10, 10) tube (ideal model) 100 K 400 K50 K 6000 W/mK 1. Between 50-400 k, K is temp dep 2. Why a change-over emerges at 100 k? K = C v · ·l <100 k, l  constant, so K is dominated by C v. >100 k, C v  constant, so K is dominated by l, and l decreases as temp increase, due to umklapp process,

14. What is umklapp process ? umklapp process 熱阻 ( 類似電阻 ) k1k1 k2k2 k3k3 1 st BZ k 1 + k 2 = k 3 + G k 1 : wave vector of phonon 1 k 2 : wave vector of phonon 2 k 3 : k 1 and k 2 碰撞後合向量 G = 0, no heat resistance, and phonon wave vector moving forward (normal process) G  0, heat resistance, and phonon moving backward (umklapp process) G: reciprocal lattice vector k1k1 k2k2 k3k3 k 3 +G 1 st BZ

15. Thermal expansion and contraction of CNTs Zero thermal expansion expansion contraction compensation e.g. Invar alloy (Fe 65 Ni 35, Zero thermal expansion) Polymers: rubber, polyethylene Layered mateirals: graphite and BN 3D oxides: NaTi 2 P 3 O 12, ZrW 2 O 8 Thermal contraction materials

16. Low dimension system (nanowires, nanotubes…) Thermal expansion or contraction is determined by competition between internal energy and entropy. 高溫區中低溫區 Entropy dominates, harmonic regime contraction Internal energy plays crucial role, anharmonic regime,expansion VoVo V1V1  V = V 1 -V o  V/V o < 0, contraction  V/V o > 0, expansion

17. PRL, 92,015901, 2004 260 K   200 cm -1 ellipsoidal deformation Softest vibration mode of C 60

18. Thermal contraction of CNTs Tube length Tube volume  : thermal linear expansion coefficient  : thermal volumetric expansion