NUMERICAL SOLUTION FOR THE RADIATIVE HEAT

DISTRIBUTION IN A CYLINDRICAL ENCLOSURE

Cosmin Dan, Gilbert De Mey, Erik Dick

University of Ghent, Belgium

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Overview

• Introduction• The net radiation method• The configuration factors

computation• Results• Conclusions

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Introduction

• Motivation

Electronic Package

Heat source

Reflected heat

Incident heat Transmitted heat

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Introduction

ZPk

Transmission line – Wave guide

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Introduction

• Assumptions:– Cylindrical enclosure made of aluminium– Very well polished surfaces– Known temperature distribution – Known emissivity of the inner walls– The surfaces are diffuse gray– The cylinder is divided in small finite

areas (ring shape)

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• The cavity is divided in small areas in which:– The surfaces are isothermal– The surfaces are diffuse emitters and reflectors– The surfaces are gray

The net radiation method

qk

qik

q0k

Sk

ikkk qqq 0

ikkkkk qTq 10

N

jjkjik qFq

10

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• Two energy balance equations

The net radiation method

)(1 0

4kk

k

kk qTq

N

jjkjkk qFqq

100

)()(1

1

44

1

4

1

N

jjkkj

N

jkjkjjj

N

j j

jkj

j

kj TTFFTqF

kjkj CqA

-Gaussian elimination-configuration factors

computation

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R

L

Twall, εwall

Surface i

Tsource, εsource

Tambient

The net radiation method

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The configuration factors

2

1

2221 4

21

RR

XXF

A1

A2

r2

r1

h

R

x

Ai1 Ai2 Aj1 Aj2

Ai

Aj

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The configuration factors

)()(11222211211 ijijjijijj

iji FFAFFA

AF

)()(1

1

44

1

4

1

N

jjkkj

N

jkjkjjj

N

j j

jkj

j

kj TTFFTqF

j

jkj

j

kjkj Fa

1

N

jjkkjk TTFc

1

44 )(

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Results

0

2

4

6

8

10

12

14

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Cylinder length [m]

q/qi

[-]

q/qi-analytical

q/qi-numerical

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Results

00.5

11.5

22.5

33.5

44.5

5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Cylinder length [m]

Rel

ativ

e er

ror

[%]

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Results

00.5

11.5

22.5

33.5

44.5

5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Cylinder length [m]

Rel

ativ

e er

ror

[%]

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Results

0.00E+00

5.00E+01

1.00E+02

1.50E+02

2.00E+02

2.50E+02

3.00E+02

3.50E+02

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Cylinder length [m]

Hea

t Flo

w [W

/m^2

]

N=20N=100N=200

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Results

0

10

20

30

40

50

60

70

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Cylinder length [m]

Hea

t Flo

w [W

/m^2

]

emissivity 0.1emissivity 0.5emissivity 0.9

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Results

-350

-300

-250

-200

-150

-100

-50

00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Cylinder length [m]

Rad

iativ

e flu

x [W

/m^2

]

emissivity 0.1emissivity 0.5emissivity 0.9

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Results

-250

-200

-150

-100

-50

0

50

100

150

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Cylinder length [m]

Rad

iativ

e flu

x [W

/m^2

]

0.10.50.9

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Conclusions

• A software was developed in order to calculate the heat flow distribution along cylinder wall was realised

• The results obtained using the numerical solution were validated against an analytical case from the literature (R. Siegel, J. R. Howell, “Thermal Radiation Heat Transfer”, 1992, pp. 428-430, pp. 464-467, pp. 477)

• In the future work the obtained results are to be used together with experimental data

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Thank You