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BEATING THE THERMAL CONDUCTIVITY OF AIR USING PACKED NANOPARTICLE BED

Ravi Prasher

Adjunct Professor

Department of Mechanical and Aerospace Engineering

Arizona State University

[email protected]

Proceedings of IMECE2006 2006 ASME International Mechanical Engineering Congress and Exposition

November 5-10, 2006, Chicago, Illinois, USA

IMECE2006-14931

ABSTRACT Thermal conductivity of packed bed of nanoparticles is

calculated in this paper. Results show that effective thermal

conductivity of nanoparticle bed can be very low. Thermal

conductivity of the nanoparticle bed can be smaller than the thermal

conductivity of air. Thermal conductivity depends on pressure,

surface energy of the nanoparticle, and phonon mean free path.

1. INTRODUCTION The determination of the conductivity of packing of spheres

has long been a subject of great importance. Much of the work done

in this domain relates to applications such as cryogenic insulation,

boilers and heat exchangers, breeder blanket about a fusion reactor,

chemical catalysts and powder metallurgy. Industrial processes such

as the hot forming of metal and ceramic powders involve particulate

materials at high temperatures. Knowing the thermal properties of

these materials and its evolution throughout the densification is

essential to allow the correct simulation of any hot forming process.

Thermal conductivity of packed bed is an important parameter in all

the applications mentioned above. Both experimental and theoretical

literature on thermal conductivity of packed bed of spheres is vast1,

however no work has been reported on the thermal conductivity of

packed beds made from nanoparticles. Due to the advent of

nanoscience and nanotechnology now it is possible to make systems

from the nano scale to macro scale. Limited experimental work

already indicate that it is possible to make packed bed of

nanoparticles3. Xia and Brueck2 made nanochennels using bed of

silica nanoparticles. Nanoparticles in fluidized bed are also very

important3.

In this paper effective thermal conductivity of packed bed

of nanoparticles as shown in Fig. 1 is analytically modeled using

phonon transport theory. The contact deformation analysis is

performed using the Hertz theory as well as adhesion theory because

adhesion forces will be very important for nanoparticles3,4. Well

know model by Johnson, Kendall, and Roberts (widely known in

literature as JKR) is used to model the impact of adhesion forces.

2. MODELING OF THERMAL CONDUCTIVITY OF PARTICULATE BED

Figure 1 shows the schematic of a particulate bed. The

particles are arranged in simple cubic fashion. Due to the periodicity

of the particles, thermal problem reduces to the problem of one unit

cell as shown in Fig. 1b and 1c. Figure 1c shows the top view of the

unit cell. The effective conductivity of the nanoparticle bed can be

written as

1

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aircylcylcylbed kkk )1( φφ −+= (1)

ForceHeat Flow

Isothermal Surface

(T1)

Isothermal Surface

(T2)

Cylinder of

effective

conductivity

kcyl

Adiabatic

surface

Adiabatic surface

q Air

Figure 1a

Figure 1b Figure 1c

Nano

particle

Air

Figure1: a) Schematic of the nanoparticle bed b) Side view of the

unit cell 3) Top view of the unit cell

where kcyl is the conductivity of the cylindrical region formed by the

spheres and the air as shown in Fig. 1b and φcyl is the volume fraction

of the cylinder as shown in Fig. 1c and φcyl = π/4.

The main problem at hand is to obtain the value of kcyl.

Figure 2 shows the geometry of the problem in details to obtain kcyl.

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D

R is the radius of the particle and a is the radius of the contact region.

An approximate method is used in the paper to model kcyl, validity of

which is first demonstrated using macroscopic heat conduction theory

based on Fourier law of heat conduction.

Fig. 2: Details of the unit cell to model kcyl.

2.1 MACROSCOPIC MODEL As shown in Fig. 1b, there are two parallel path for heat transfer: 1)

From the solid to the air 2) Through the constriction formed by the

two spheres. The constriction resistance is modeled using the

Maxwell’s formula5. Constriction resistance (Λc) is given as

kac

2

1=Λ (2)

For modeling the thermal resistance of the solid/air combination an

infinitesimally small cylindrical strip of area dA as shown in Fig. 2 is

taken. The differential thermal conductance (dG’) of the is small area

is given as

dAkdAk

y

dG airs

δ22

'

1+= (3)

where ks is the thermal conductivity of the solid and y and δ denote

the thickness of the solid and the air. Therefore total conductance is

given as

∫ +=

R

airs kky

dAG

0/2/2

'δ

(4)

dA can be written in radial coordinates as rdrdA π2=

Therefore G’ is given as

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'

1

/2/2

2'

0Λ

=+

= ∫R

airs kky

rdrG

δ

π (5)

where Λ' is the thermal resistance. Assuming that application of

pressure has not changed the spherical profile too much y and δ can

be written as

22 rRy −= (6)

Figure 3: Solid model of the FEM to model kcyl

Figure 4: Temperature contour for R/a = 0.01

22rRRYR −−=−=δ (7)

Therefore Eq. (6) reduces to ( assuming a << R)

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Do

'

1

/)(/'

02222 Λ

=−−+−

= ∫R

airs krRRkrR

rdrG π

(8)

Therefore total thermal resistance (Λcyl) is

RkRk

R

cylcylc

c

cylππ

22

'

'2

==Λ+Λ

ΛΛ=Λ (9)

Therefore kcyl is given by

a/R

0.01 0.1

Kcyl (

W m

-1 K

-1)

1

10

100

Numerical

Analytical (Eq. 10)

Fig. 5: Comparison between analytical model and the numerical

model

cyl

cylR

kΛ

=π

2 (10)

The validity of the approximate analytical model was checked by

performing a finite element numerical model for different values of

R/a. In the finite element model (FEM) solid was assumed to be

silicon. Solid model for the FEM is shown in Fig. 3. Temperature

profile for R/a = 0.01 is shown in Fig. 4. Figure 5 shows the

comparison between the analytical model developed above and the

numerical model for R/a ranging from 0.01 to 0.1. Figure 5 shows

that analytical model is in excellent agreement with the numerical

model.

2.2 MICROSCOPIC MODEL Prasher6 showed that Eq. (2) is valid only if the mean free path

(m.f.p) of the phonons << smaller than the constriction size, a.

Prasher6 showed that accounting for the size effects the constriction

resistance is given as

3

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+=Λ Kn

kac

π3

81

2

1 (11)

where Kn is the Knudsen number given by, alKn /= where l is

the mean free path of phonons. M.f.p of phonons in silicon at room

temperature is ~250 nm7. The constriction size for nanoparticle will

be a few nanometers. Therefore Kn will be very large causing much

larger resistance at the constriction. Note that for 0→Kn , Eq.

(11) reduces to Eq. 2.

The thermal boundary resistance at the air/solid interface

will be also dominant for the nanoparticles. Thermal boundary

resistance expressed in the units of impedance for the air/solid

interface can be written as8

airvairb lCR )]/(1)][(1/(2][/)2[(2, µγββ +−= (12)

where β is the accommodation coefficient, lair the m.f.p of air

molecules, γ the ratio of specific heats, µ the viscosity of the gas, Cv

is the specific heat of air at constant volume. β for air8 is typically

0.9. Note that in engineering literature the same phenomenon at

gas/solid interface is treated as temperature jump distance.

Dharmadurai9 showed treating the problem from thermal boundary

resistance point of view or temperature jump distance point of view

are equivalent. Putting the properties of air at room temperature in

Eq. (12) 6

, 10885.4 −×=airbR K m2 W-1. (13)

Expressed in terms of conductance, 25.0, ≈airbG MW/m2-K. The

thermal conductance of liquid/solid interface is ~10 MW/m2-K14.

Mismatch between acoustic properties of air and solid is more drastic

than liquid and solid. Therefore thermal conductance of air/solid

interface is one order of magnitude smaller than liquid/solid interface.

Including Rb,air, Eq. (8) can be written as

'

1

/)(/'

0 ,

2222 Λ=

+−−+−= ∫

R

airbairs RkrRRkrR

rdrG π

The contribution of the radiation heat transfer between the two

spheres is also included. For the macroscopic case the radiative

conductance is given as

radrad RTG Λ=×= /124 23 πσ (14)

where T is the temperature and σ is the Stefan-Boltzmann constant.

Due to the closeness of the nanoparticles, near field effects might

become important11. An upper bound estimate11 for the near field can

be made for dielectrics such as intrinsic silicon for which absorption

is negligible. Accounting for near field effects Grad for the

microscopic case is given as

radrad RTnG Λ=×= /124 232 πσ (15)

where n is the refractive index of silicon.

3. DEFORMATION ANALYSIS For the macroscopic treatment the radius of the constriction

is given by Hertzian contact analysis. The Hertzian contact analysis

gives

=

*4

*3

E

FRa (16)



D

where F is the force, )/(* 2121 RRRRR += where R1 and R2 are

the radius of the two spheres and

2

2

2

1

2

1

*

111

EEE

νν −+

−= (17)

where E is the Young’s modulus and υ is the Poisson’s ratio. If the

pressure applied at the top substrate is P then the force on each sphere

is, PRF2π= . Therefore Eq. 10 can be written as

Pressure (PSI)

0.1 1 10 100 1000

aJK

R/a

Hert

z

0

5

10

15

20

25

R = 10 nm

R = 50 nm

R = 250 nm

Figure 6: Radius of the constriction predicted by Hertzian theory

and JKR theory to account for adhesion

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Nanoparticle Diameter (nm)

10 100 1000

Th

erm

al

Co

nd

uc

tiv

ity

(W

m-1

K-1

)

10x10-3

100x10-3

1x100

Macro

Micro, Hertz Contact

Micro, JKR Contact

P = 0

γγγγ1 = 100 mJ/m2

Fig 7: Effective thermal conductivity of the nanoparticle bed for

P = 0.

RE

Pa

3/1

*8

3

=

π (18)

for R1 = R2.


10 100 1000

Th

erm

al

Co

nd

uc

tiv

ity

(W

m-1

K-1

)

0.01

0.10

1.00

10.00

Macro


Micro, JKR Contact

P = 20 PSI



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Figure 8: Effective thermal conductivity of the nanoparticle bed

for P = 20 PSI For small particles microscopic surface forces such as van

der Walls force will play a very dominant role. Radius of the

constriction including the surface forces is modeled using the well

JKR theory3. According to JKR theory the effective force is given by

2*)3(*6*3 RFRRFFeff γπγπγπ +++= (19)

where γ is the work of adhesion per unit area. γ is given by12

1221 γγγγ −+= (20)

where γ1 and γ2 are the surface energy per unit area of material 1 and

2 and γ12 is the interface energy per unit area. If two material are

same then112 γ12 = 0 and

12γγ = (21)

Surface energy of silicon at room temperature ranges from 10 mJ/m2

– 100 mJ/m2 depending on the surface treatment13. Figure 6 shows

the comparison between a obtained from Hertzian contact analysis

and that obtained by JKR model for γ1 = 100 mJ/m2. Figure (6)

shows that surface forces a play very important role for nanoparticles.

Impact of surface forces increases with decreasing particle size.

4. RESULTS FOR THERMAL CONDUCTIVITY

Figures 7 – 9 show the thermal conductivity calculated

using macroscopic approach for thermal transport and Hertzian

contact, microscopic approach for thermal transport and Hertzian

Contact and microscopic approach for thermal transport and JKR

contact. γ1 = 100 mJ/m2 is assumed in Figs. 7-9. Note that for P = 0,

a = 0 in Hertzian contact, whereas it is finite for JKR contact.


10 100 1000

Th

erm

al

Co

nd

uc

tiv

ity

(W

m-1

K-1

)

0.01

0.10

1.00

10.00

Macro


Micro, JKR Contact

P = 100 PSI



for P = 100 PSI

5

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20 40 60 80 100

Th

erm

al

Co

nd

ucti

vit

y (

W m

-1 K

-1)

15x10-3

20x10-3

25x10-3

30x10-3

35x10-3

40x10-3

P = 0

P = 20 PSI

P = 100 PSI

kair = 26x10-3

γγγγ1 = 10 mJ/m

2


for γγγγ1 = 100 mJ/m2.

Hertizian contact gives much smaller value of a as

compared to JKR contact as shown in Fig. 6. Thermal conductivity

given by macroscopic is very high because ballistic component of

thermal resistance is not included. Microscopic approach combined

with Hertzian contact gives a very small thermal conductivity

because constriction radius predicted by Hertzian theory is very small

thereby making thermal constriction resistance very high. Figures 7-

9 show that thermal conductivity of nanoparticle bed can be very

small depending on the size of the nanoparticles.

Figure 10 shows the thermal conductivity assuming JKR

contact for γ1 = 10 mJ/m2. Figure 10 shows that depending on the size

of the nanoparticle, effective thermal conductivity of the nanoparticle

bed can be smaller than air.

5. CONCLUSION Analytical model for effective thermal conductivity of bed

of nanoparticles was developed. Surface forces are very dominant for

nanoparticles. Results also show that depending on the surface

energy of the nanoparticle, the effective thermal conductivity of the

nanoparticle bed can be smaller than the thermal conductivity of air.

ACKNOWLEDGEMENT: The author would like to acknowledge the help of Dr. David Song in

performing the finite element numerical analysis and Mr. Tao Tong

for helpful discussions.

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