Добавил:

okley Опубликованный материал нарушает ваши авторские права? Сообщите нам.

Вуз:

Национальный исследовательский ядерный университет (МИФИ)

Предмет:

Тепломассообмен

Файл:

Chapter-8---Thermal-radiation_2021_Heat-Transfer-Engineering

.pdf

Скачиваний:

Добавлен:

08.05.2026

Размер:

4.3 Mб

Скачать

☆

<<< < Предыдущая 1 2 34 / 74 5 6 7 > Следующая >>>

8.8 Radiation view factor and its determination 263

FIGURE 8.17

Figure showing two surfaces Ai and Aj with respective orientations in space, with the goal being the determination of the view factor from ith surface to the jth surface, denoted by Fij.

where

Ii = Ii,emitted +reflected

(8.75)

dω j −i =

dAj cosθ j

(8.76)

For a diffuse surface

Ii =

(8.77)

Ji cosθi cosθ j dAi dAj

(8.78)

−dA

π R

The radiation leaving dAi is

Ji .dAi .

The fraction of the radiation leaving dAi that is intercepted by dAj, known as the

elemental view factor dFdA −dA

, is given by

dQdAi

−dAj

Ji cosθi cosθ j dAi dAj

(8.79)

dA −dA

Ji dAi

Ji dAiπ R

dFdA −dA

cosθi cosθ j dAj

(8.80)

π R

264CHAPTER 8 Thermal radiation

The view factor from Ai to Aj or from i to j is then given by

∫∫ dQdAi −dAj

= F

Aj Ai

− A

i− j

∫ Ji dAi

For uniform radiosity

∫ ∫ cosθi cosθ j dAi dAj

Aj Ai

π R2

− Aj

Similarly,

∫ ∫ cosθi cosθ j dAi dAj

Ai Aj

π R2

− Ai

From Eqs. 8.82 and 8.83, it is evident that

Ai Fij = Aj Fji

Eq. (8.84) is frequently referred to as the reciprocity or reciprocal rule.

(8.81)

(8.82)

(8.83)

(8.84)

8.8.1 View factor algebra

Consider an evacuated enclosure with N surfaces, as shown in Fig. 8.18. There will be N2 view factors for this problem arranged in the matrix form below. Unless we calculate N2 view factors, we will not be able to solve the radiation problem of finding the net radiation from any given surface. The view factor matrix for this enclosure is given as,

FIGURE 8.18

A depiction of an evacuated enclosure with N surfaces.

8.8 Radiation view factor and its determination 265

F11	F11	… … … F1N
F	F	… … …	F
21	22		2 N
		… … …
FN1	FN2	… … …	FNN

For any surface, it is intuitively apparent that the sum of the view factors should be 1.

N
∑Fij = 1, 1 ≤ i ≤ N	(8.85)

j =1

In a N surface enclosure there are N such sum rules.

The number of reciprocal rules Eq. (8.84) available for N surface enclosure is NC2.The total number of view factors to be independently determined are


= N2 −N C2 − N				(8.86)
= N2 −		N(N − 1)	− N	(8.87)
	2

= 2N2 − N2 + N − 2N				(8.88)
	2
=	N(N − 1)			(8.89)
	2

The total number of view factors to be independently determined are NC2. In the view of the preceding discussion, for a N surface enclosure, the number of independent view factors that need to be obtained by using the tedious view factor expression (Eq. 8.82) is NC2 while this number for a N surface enclosure with all plane or convex surface is NC2-N. The self-view factor for any plane/convex surface is Fii = 0. Suppose all the surfaces in the N-surface enclosure are plane or convex, then N view factor values are zero.

For example, consider a triangular enclosure with all the surfaces being plane, as shown in Fig. 8.19. According to Eq. (8.89), the number of independent view factors to be determined from the fundamental view factor is three.

FIGURE 8.19

Triangular enclosure.

266 CHAPTER 8 Thermal radiation

Total number of view factors = N2 = 32 = 9.

Total number of view factors that can be determined from reciprocal rules = NC2 = 3C2 = 3.

Total number of view factors that can be determined from summation rule = N = 3. Total number of view factors need to be independently determined = 9 – 3 – 3 = 3. Total number of self-view factors = 3.

Therefore, the total number of view factors to be independently determined is 3 – 3 = 0.

So, we should be able to get all the view factors by just using view factors algebra. Assuming unit depth in the direction perpendicular to the plane of the paper, we have the following:

Area of the surface AB = c. Area of the surface BC = a. Area of the surface AC = b. From the summation rule

Faa + Fab + Fac = 1	(8.90)
Fba + Fbb + Fbc = 1	(8.91)
Fca + Fcb + Fcc = 1	(8.92)

Since all the surfaces are plane, the self-view factors for all the surfaces are zero. That is,

Faa = Fbb = Fbb = 0

Eqs. (8.90–8.92) can now be written as

Fab + Fac = 1	(8.93)
Fba + Fbc = 1	(8.94)
Fcb + Fca = 1	(8.95)
From the reciprocal rule,
aFab = bFba	(8.96)
aFac = cFca	(8.97)
bFbc = cFcb	(8.98)
Multiplying Eq. (8.93) by a, Eq. (8.94) by b, and Eq. (8.96) by c, we have
aFab + aFac = a	(8.99)
bFba + bFbc = b	(8.100)

8.8 Radiation view factor and its determination 267

cFca + cFcb = c

(8.101)

Adding Eqs. (8.99) and (8.100) and subtracting Eq. (8.101) from the sum, we have

aFab + aFac + bFba + bFbc − cFca − cFcb = a + b − c

(8.102)

Now, using Eqs. (8.97 and 8.98) rearranging Eq. (8.102), we have

aFab + (aFac − cFca ) + bFba + (bFbc − cFcb ) = a + b − c

(8.103)

However, aFac = cFca and bFbc = cFcb by the reciprocal rule.

Eq. (8.103) simplifies to

aFab + bFba = a + b − c

(8.104)

Again applying Eq. (8.96), i.e., the reciprocal rule on (8.104), we have

2aFab = a + b + c

(8.105)

Fab =

a + b − c

(8.106)

By induction,

Fbc =

b + c − a

(8.107)

Fca =

c + a − b

(8.108)

Fba =

a + b − c

(8.109)

Fcb =

b + c − a

(8.110)

Fac =

c + a − b

(8.111)

Example 8.9: Determine all the view factors for the following two dimensional (i.e., very long) surfaces (refer to Fig. 8.20). The enclosed spaces in all the problems are evacuated.

Solution:

(1)

	F11 = 0 and F12 = 1
					80		7
A1	= 2rL and A2	= 2π	1	−		rL =		2π rL
A1	= 2rL and A2	= 2π	1	−	360	rL =	9	2π rL
					360		9

268 CHAPTER 8 Thermal radiation

FIGURE 8.20

(A) Long duct of radius r, (B) long cylindrical annulus, and (C) sphere within a sphere under consideration in example 8.9.

A F = A F ; F =

2rL

7π

9 2πrL

F21 = 0.409 and F22 = 0.591

Viewfactormatrix =

0.409 0.591

(2)

F11 = 0

and

F12

= 1

A1 = 2π r1L

and

A2 = 2π r2 L

A F = A F ; F =

2πr1L

2πr2 L r2

F22

= 1 −

Viewfactormatrix =

1 −

(3)

F11 = 0

and

F12

= 1

A1 = 4πr12

and

A2 = 4πr22

A F = A F ; F =

4πr12

r12

4πr22

1 12

2 21

r22

= 1 −

8.8 Radiation view factor and its determination 269

FIGURE 8.21

Infinitely deep wedge under consideration in example 8.10.

	0	1
Viewfactormatrix =	r2	1 −	r2
Viewfactormatrix =	1		1
	r22		r22
	r22		r22

Example 8.10: Consider an infinitely deep wedge with surfaces 1 and 2, as shown in Fig. 8.21.

1.Determine F12.

2.What will be the value of F12 when α = 90°?

3.What is the value of F12 when α = 180°? What do you infer from this result?

Solution:

(1) We know that, for a triangular enclosure with sides a, b, and c


			Fab =		a + b − c
			Fab =
						2a
		F12 =		L1 + L2 − L3
		F12 =
						2L1
	L1 = L2 = L					and L3 = x
F12	=		L + L − x			= 2L − x	= 1 −	x
F12	=		2L			= 2L − x	= 1 −	2L
			2L			2L		2L

From Fig 8.22

FIGURE 8.22

View factors for an infinitely deep wedge under consideration in example 8.10.

270 CHAPTER 8 Thermal radiation

sin	α		=	x /2	=	x
sin	2		=	L	=	2L
	2			L		2L
F12						α
F12		= 1 − sin
						2

(2)

Whenα = 90

F12 = 1− sin45 = 1− 12

F12 = 0.293

(3)

When α = 180°

F12 = 1− sin90°

= 1−1 = 0

F12 = 0

As α = 180°, both the surfaces are on the same plane, and they do not see each other. Hence, it stands to reason that the resulting view factor equals zero.

Example 8.11: Consider an evacuated two-dimensional quadrilateral enclosure, as shown in Fig. 8.23.

1.Determine Fac.

2.If all the four sides of a quadrilateral are equal, that is, it becomes a square, what will be the value of Fac?

FIGURE 8.23

Two-dimensional quadrilateral enclosure under consideration in example’ 8.11.

8.8 Radiation view factor and its determination 271

FIGURE 8.24

Two-dimensional quadrilateral enclosure under consideration in example 8.11.

Solution:

(1) Let us join the opposite vertices so that we now have two diagonals with lengths L1 and L2 as shown in Fig. 8.24.

From the sum rule for the quadrilateral, we have,

Faa + Fab + Fac + Fad = 1

As ‘a’ is a plane surface Faa = 0

Fac = 1 − (Fab + Fad )

From the view factor relations for a triangle, we have

a + b − L Fab = 2a 2

a + d − L Fad = 2a 1

Substuting for Fab and Fad in the expression for Fac we get the following,


Fac = 1− a + d − L1		− a + b − L2
	2a	2a
Fac =	(L1 + L2 ) − (b + d)
	2a

This is known as the Hottel’s crossed string method. The view factor between the opposite sides of a two-dimensional quadrilateral enclosure is given by the difference between the sum of the crossed strings and the sum of the uncrossed strings divided by twice the length of the side of the enclosure from which the view factor to the opposite side is sought.

Fac =	(Sum of the crossed strings) − (Sum of the uncrossed strings)
	2a

272 CHAPTER 8 Thermal radiation

(2) For a square a = b = c	= d and L1 = L2 =	2a
Fac =	( 2a + 2a) − (a + a) =		2 2 − 2	= 2 − 1
	2a		2
	Fac	= 0.414

Evaluation of the view factors for three-dimensional surfaces using the Eq. (8.82) can be very tedious. So, researchers over the years developed many techniques to reduce the complexity of the view factor evaluation (see Balaji, 2014) for a full discussion on this. For regular three-dimensional objects like cylinders and cuboidal surfaces, charts are available to determine the view factors. Many techniques like contour integration have also been developed to aid in the evaluation of view factors from three dimensional surfaces.

8.9 The radiosity-irradiation method

Now, we will use all of what we have studied so far in our pursuit of radiation and employ the radiosity irradiation method to determine the net radiative heat transfer from a surface.

Consider an evacuated enclosure with N surfaces, as shown in Fig. 8.25. The radiation leaving from the ith surface of the enclosure is given by

Ai Ji = Emitted radiation + Reflected radiation	(8.112)
Ai Ji = Ai εi σ Ti4 + Ai ρi Gi	(8.113)

In the above equation Ji is the leaving radiation or radiosity and Gi is the incoming radiation or irradiation, as already introduced.

Let all surfaces be opaque.

FIGURE 8.25

Depiction of an evacuated enclosure with N surfaces.

<<< < Предыдущая 1 2 34 / 74 5 6 7 > Следующая >>>

Соседние файлы в папке Книги

#
08.05.20264.08 Mб0Chapter-3---Conduction--One-dimensional-transient-and-_2021_Heat-Transfer-En.pdf
#
08.05.2026656.23 Кб0Chapter-4---Fundamentals-of-convection_2021_Heat-Transfer-Engineering.pdf
#
08.05.20262.69 Mб0Chapter-5---Forced-convection_2021_Heat-Transfer-Engineering.pdf
#
08.05.2026821.19 Кб0Chapter-6---Natural-convection_2021_Heat-Transfer-Engineering.pdf
#
08.05.20263.02 Mб0Chapter-7---Heat-exchangers_2021_Heat-Transfer-Engineering.pdf
#
08.05.20264.3 Mб0Chapter-8---Thermal-radiation_2021_Heat-Transfer-Engineering.pdf
#
08.05.2026967.4 Кб0Chapter-9---Numerical-heat-transfer_2021_Heat-Transfer-Engineering.pdf
#
08.05.2026174.57 Кб0Copyright_2021_Heat-Transfer-Engineering.pdf
#
08.05.202618.95 Кб0Dedication_2021_Heat-Transfer-Engineering.pdf
#
08.05.2026185.55 Кб0Front-matter_2021_Heat-Transfer-Engineering.pdf
#
08.05.2026114.33 Кб0Index_2021_Heat-Transfer-Engineering.pdf