Chapter 4 Radiation Heat Transfer_Part 2

8/13/2019 Chapter 4 Radiation Heat Transfer_Part 2

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BKF2422

Chapter 4

Principles of Steady-state

Heat Transfer in Radiation(Part 2)


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Topic Outcomes

It is expected that students will be able to: Explain the principles and mechanism of heat transfer

by radiation Calculate the energy transfer from combined heat of

radiation and convection Derive the view factors for various geometries


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Content Advanced Radiation Heat Transfer Principles

Introduction View factor for infinite parallel black planes View factor for infinite parallel gray planes

View factor for finite planes View factors when black bodies surfaces areconnected by reradiating walls

View factors when gray bodies surfaces are

connected by reradiating walls Radiation in Absorbing Gas Characteristic Mean Beam Length of Absorbing

Gas

G, G and radiation of a gas


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Advanced Radiation Heat

Transfer Principles Exchange of radiation between two faces/surfaces

depends on size, shape, orientation, and Medium in between surfaces absorb the radiation

energy at certain amount. Normally can beneglected, e.g. air. CO 2, H2O vapor absorb more.

Radiant energy electromagnetic wavese.g. gamma, = 10 -13 10 -10 m, cosmic < 10 -13 m, thermal rays = 10 -7 10 -4 m

Thermal rays solely from temperaturevisible radiation 3.8 x 10 -7 7.6 x 10 -7 m, i.e. (liesin thermal ray range)


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Introduction

When two different surfaces are heated at the sametemperature, they do not emit and absorb energy atthe same amount.

Black body absorbs and emits the maximumenergy.

Emissive power of a black body at T and :

The wavelength at which emissive power ismaximum:

Wiens Displacement Law

1107418.3

/104388.15

16

2 T x B

e

x E

K mT .10898.2 3

max,


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Introduction

Total emissive power: Amount of radiation energyper unit area leaving a surface at temp. T over allwavelengths.

Total emissive power for black body, E B

where = 5.676 x 10 -8 W/m 2.K 4

Emissivity of a surface, is defined as the totalemitted energy of a surface divided by total emitenergy of black body at the same temperature.

For most material,

4T E B

4T

E

E

E

B

0.1


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Introduction

Relationship between emissivity and absorptivity: Heating a body in an enclosure at same temperature allows

thermal equilibrium. If G = irradiation the body.

For black body,

since

11 E G

B E G

2

B E

E 1

2

1

11

1 B E

E 11

12


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Introduction

Gray body a surface for which the monochromaticproperties are constant over all wavelength.

Even if the body is not in thermal equilibrium with itssurroundings.

In actual system, T of various surfaces may differentand varies with . In this case is evaluated basedon temperature of the source.

For engineering calculation, this assumption givereasonable accuracy.

const const

, ,


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View factor for various geometries

If two surfaces are arranged so that the radiantenergy can be exchanged, net flow of energy willoccur from hotter to colder surface.

The size, shape, orientation of two surfacesdetermining the net heat flow between them.

It is assumed that the surfaces are separated by non-absorbing medium such as air.


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View factor for infinite parallel blackplanes

If two parallel planes are radiating each other, forplane 1, net radiation from plane 1 2:

F12 is the fraction of radiation leaving plane 1 that iscaptured by plane 2. The factor F 12 is called viewfactor.

For parallel planes, F 12 =F 21 =1.0.

4

24

1112 T T Aq

424111212 T T A F q 4

24

122121 T T A F q


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View factor for infinite parallel grayplanes

View factor for infinite parallel gray planes:since parallel planes, F 12 =F 21= 1.

Surface A 1 emits 1 A 1T14

radiation to surface A 2. Ofthis, a fraction of 2 is absorbed. Energy absorbed by surface A 2= 2 ( 1 A 1T14 ) Energy reflected by A 2 to A 1 = (1- 2) ( 1 A 1T14 ). Of

this, surface A1 reflected back (1- 1) (1- 2) ( 1 A 1T14

). Energy further absorbed by surface A 2= 2 (1- 1) (1- 2) ( 1 A 1T14 ) and so on..


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The total amount of heat absorbed by surface A 2:

In geometric series, total amount of heat absorbedby surface A 2 from radiation of surface A 1:

Simplified form:

.....1111 222

121212121

4

1121 T Aq

21214

1121

111

T Aq

1111

21

41121

T Aq


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In geometric series, total amount of heat absorbedby surface A 1 from radiation of surface A 2:

Net radiation from plane A 1 to A 2 for gray planes:

For black bodies, the equation will be reduced to:

1111

21

42112

T Aq

1

111

21

42

41112

T T Aq

42

41112 T T Aq


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Try This

Two parallel gray planes which are very large haveemissivities of 1 = 0.8 and 2 = 0.7; surface 1 is at900K and surface 2 at 600K.

a) What is the net radiation of both planes?b) If the surface are both black, what is the netradiation?

Ans: a) 17.8 kW/m 2 ; b) 29.9 kW/m 2


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View factor for finite planes

For planes that are finite is size, we must considertwo parameters:- Solid angle, (unit: sr)- Intensity of radiation, I (unit: W/m 2.sr)

Solid angle: dimensionless, measure a quantity of anangle in solid geometry.

Intensity of radiation: rate of radiation emitted per

unit area projected in a normal direction to thesurface


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View factor for finite planes-cont.

Total net heat flow between finite areas of blackbodies;

where,F12 = view factor fraction of radiation from A 1 thatstrikes A 2.F

21 = view factor fraction of radiation from A

2 that

strikes A 1.

Valid for black and non-black surface.

4241212424112112 T T F AT T F Aq

212121 F A F A


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View factor for finite planes-cont.

Considering the geometrical arrangement, view factorcan be calculate using

The integration of above equation has beenperformed for various configuration and tabulated.

Figure 4.11-7 (pg 316) gives view factors for parallelplanes directly opposed and Figure 4.11-8 (pg 317)gives view factors for adjacent perpendicularrectangles.

2

2121

112

coscos112 r

dAdA

A F

A A


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Try This

Two adjacent rectangles are perpendicular to eachother. The first rectangle is 1.55 x 2.44 m and thesecond 1.83 x 2.44 m, with 2.44 m side common toboth. The temperature of the first surface is 699 Kand that of the second is 478. Both surfaces areblack. Calculate the radiant heat transfer betweenboth surfaces.

(answer: q= 9.424 kW)


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View factors when black bodies surfacesare connected by reradiating walls

Example: furnace

A larger fraction of the radiation from surface 1 isintercepted by 2. View factor =

For the case of two black bodies surfaces A 1 & A 2

connected with reradiating walls:

1 2

Refractory reradiating wall

12 F

122121

2

1221

12121

2

122112

/21

/1

2 F A A A A F A A

F A A A F A A

F

212121 F A F A 4

24

111212 T T A F q


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View factors when gray bodies surfaces areconnected by reradiating walls

For the case of two gray bodies surfaces A 1 & A 2 which cannot see themselves and connected withreradiating walls:

F 2

q12 =F

12

A1

(T14

T24

)A1F 12 = A2 F 21 For black or gray bodies, with no reradiating walls,

again .

1

11

111

122

1

12 A

A F

1212 F F


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Radiation in Absorbing Gas Most mono atomic & diatomic gases are transparent to

thermal radiation. (He, H 2, Ar, O 2, N2).

Dipole gases and higher polyatomic gases emit andabsorb radiant energy significantly. (CO 2, CO, H 2O, SO 2,NH3).

Absorption of thermal radiant by gas depend on pressureand temperature.i.e. partial pressure of gases, amount of absorption.If the gas is heated, it radiates energy to surrounding.


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Characteristic Mean Beam Length of Absorbing Gas

The absorption of energy by gas T1, P and characteristiclength, L (mean beam length). Mean beam length, L specific geometry.

The mean beam length has been evaluated for variousgeometries in Table 4.11-1 (pg. 321).

For other shapes,

where,L = m, V = m 3 of gases A = surface area of enclosure m 2.

A

V L 6.3


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G , G and radiation of a gas

q = G TG4 - (rate of radiation emitted) from thegas.

q = G T14

- (rate of absorption of energy as it isradiated back to the gas from midpoint of surfaceelement.)

The net rate of radiant transfer between a gas at T G and a black surface of finite area A 1 at T 1.

41

4 T T Aq GGG


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Figure 4.11-10 gives the gas emissivity, G of CO2 ata total pressure of the system = 1.0 atm abs using

at T G.pG = partial pressure of CO 2 in atm.L = mean beam length in m (Table 4.11-1 ).

G of CO 2 is determined also from Figure 4.11-10but at T 1.(T1 =temperature of midpoint of surface element.)

But instead of using p G L, need to use:

The value obtained from the chart is then multipliedby correction factor .

G

G

T

T L p 1

65.0

1

GT T

L p G


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For the case when wall of enclosure are not black. Anapproximation when the emissivity is greater than0.7, effective emissivity, can be used

Hence the radiant transfer between a gas at T G andnot black surface at T 1

414' T T Aq GGG

2

0.1'


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Try This

A furnace is in the form of a cube 0.3 m on a sideinside and these interior walls can be approximatedas black surfaces. The gas inside at 1.0 atm total

pressure at 1100K contains 10 mol% CO 2 and therest is O 2 and N 2. The small amount of water vaporpresent will be neglected. The walls of the furnaceare maintained at 600K by external cooling. Calculatethe total heat transfer to the walls neglecting heat

transfer by convection. Ans: 2.6 kW

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Chapter 4 Radiation Heat Transfer_Part 2