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Multi-scale Heat Conduction Solution of the EPRT

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- Dec. 6th , 2011

Multi-scale Heat ConductionSolution of the EPRT

Hong goo, Kim

1st year of M.S. course

- Introduction
- Two-Flux Method
- Modeling of Thin Layer
- Solution
- Example 7-4

- Scheme
- Thermal Resistance Network
- Three-Layer Structure

Two-Flux Method

Modeling of Thin Layer

- Assumption

Scheme

- 1-D, Steady state
- Medium is gray

Emission

- Absorption coefficient is independent of phonon frequency

T1

T2

- Walls are diffuse and gray

ε1

ε2

- Absorption coefficient is independent of direction and phonon frequency
- Emission at wall is independent of direction

- Governing Equation

Steady-state

EPRT

Gray medium

; Positive direction

; Negative direction

x = L

x = 0

Two-Flux Method

- Modeling of Thin Layer

- Boundary Conditions

- Temperature at the walls

(7.46)

- Intensity at the walls

(7.47)

Emission

Emission

(7.48)

Reflection

Reflection

T1

T2

Irradiation

Irradiation

x = L

x = 0

Two-Flux Method

- Solution

- Derivation of the Solutions

- From the governing equation (EPRT)

(7.45a)

- Integrating from 0 to x, after multiplying on both sides

LHS

RHS

LHS

RHS

Two-Flux Method

- Solution

- Derivation of the Solutions (continued)

- For positive directions (from left to right)

(7.49)

(7.50)

Attenuation of the intensity originated from the left surface (x = 0)

Generation term

- For negative directions (from right to left)

Two-Flux Method

- Solution

- Spectral Net Heat Flux (in x-direction)

(7.37)

Appendix (1)

(7.51a)

- For a diffuse surface (x = 0, x = L)

Diffuse Surface

(7.51b)

Two-Flux Method

- Solution

- Energy Balance

- For 1-D steady-state,

- Differentiation of heat flux (diffuse surface) from (7.51b)

Appendix (2)

(7.52)

Two-Flux Method

- Solution

- Energy Balance

- Total blackbody emissive power = total radiosities at 1 and 2

- Blackbody emissive power

- Total radiosity

- Energy balance, from (7.52)

→ Radiative equilibrium condition

Two-Flux Method

- Example 7-4

- Objectives

- Heat flux
- Thermal conductivity
- Temperature distribution

- Assumptions

- Medium is gray
- Surfaces are diffuse and gray
- Radiative thicklimit : Kn =Λ/L<< 1

Two-Flux Method

- Example 7-4

- Spectral Heat Flux

- In the radiative thick limit : Λ << x, Λ << L − x

- Local equilibrium holds if location of x is not too close to either surfaces
- Flux originating from the left/right surfaces are attenuated to ‘0’

(7.51a)

Two-Flux Method

- Example 7-4

- Spectral Heat Flux (continued)

- Exponential terms are significant only in the neighbor of x

- Taylor series 1st order approximation is valid

(7.54)

Two-Flux Method

- Example 7-4

- T << θD

- Net heat flux obtained from integrating (7.54) over frequency

- Thermal conductivity

Thermal conductivity

(7.55a)

- From kinetic theory

- Integration of (7.55a) overx from 0 to L :

(7.55b)

LHS

RHS

(7.56a)

Net heat flux

Two-Flux Method

- Example 7-4

- T << θD(continued)

- Temperature distribution

- By comparing (7.55a) and (7.56a)

(7.56b)

- Thermal resistance

- By definition of the thermal resistance and (7.56a)

(7.57)

Two-Flux Method

- Example 7-4

- T > θD

- Spectral heat flux

(7.54)

(7.34)

- Total intensity

- Net heat flux

(7.58)

Two-Flux Method

- Example 7-4

- T > θD

- Thermal conductivity

- Net heat flux

- From kinetic theory,

(7.58)

(7.59)

- Assuming small temperature difference

- Thermal conductivity can be approximated as a constant

- Thermal resistance

- Temperature distribution

Two-Flux Method

- Example 7-4

- Temperature Profiles

- T

- (T > θD)

- (T << θD)

- x

Two-Flux Method

- Example 7-4

- Radiative Equilibrium Conditions

- Local equilibrium condition, gray medium

(7.40)

- is the average of and

- Assuming T1 > T2: net heat flux in positive x

- should be greater than

- Local equilibrium is not a stable state

- Heat flux in the radiative thin limit

(7.60)

Thermal Resistance Network

- Scheme

- Internal Thermal Resistance

- Diffusion process: classical Fourier law

- Boundary Thermal Resistance

- When medium is not in radiative thick limit

- Due to radiation slip

- Does not exist in classical Fourier law

- Temperature jump approaches to zero in the radiative thick limit ( Kn << 1 )

- Restrictions

- Applicable to one-dimensional problem

- Results in temperature jump at the boundaries

Thermal Resistance Network

- Energy Transport

- T << θD

- Heat flux in thermal network resistance

(7.61)

Bulk

Radiation slip

- For blackbody walls ( ε1 = ε2= 1 )

- Temperature difference between T1 and T2 is small (T1, T2≈ T )

Effective

thermal conductivity

(7.63)

Bulk

thermal conductivity

(7.55b)

Thermal Resistance Network

- Energy Transport

- T > θD

- Heat flux in thermal network resistance

(7.64)

- Effective vs. bulk heat conductivity ratio is the same as in low temperature for blackbody walls

- Discussion

- Fourier law can be applied inside the medium
- Heat flux for thermal resistance network can be applied between the diffusion and ballistic extremes

Thermal Resistance Network

- Three Layer Structure

- Thermal Resistance Network

TH

TL

TH

T1a

T1b

T2a

T2b

T3a

T3b

TL

Thermal Resistance Network

- Three Layer Structure

- Internal Resistance

- Due to diffusion (Fourier’s law)

- Interface Resistance

- Transmission of phonon through the interface

Γij : transmissivity from ito j

- Boundary Resistance

- Transmission of phonon transport considered

Thermal Resistance Network

- Three Layer Structure

- Total Resistance

- Heat Flux

- Effective Thermal Conductivity

- Appendix (1)

- Appendix (2)