978-0123869449 Chapter 4 Part 2

CHAPTER 4 83

4.16

2d

A2

A1

2d

d/2 d/2

Consider the configuration shown; determine the view factor F1−2assuming the

configuration

a) to be axisymmetric (1 conical, 2 a disk with hole), or

b) to be two-dimensional Cartesian (1 is a V-groove, 2 two infinitely long strips).

2d

84 RADIATIVE HEAT TRANSFER

4.17

r

A2

Find F1−2for the configuration shown in the figure (infinitely long perpen-

dicular to paper).

CHAPTER 4 85

4.18 Calculate the view factor between two infinitely long cylinders as shown in the figure. If a radiation shield is

placed between them to obstruct partially the view (dashed line), how does the view factor change?

Solution

The view factor is most easily determined by the crossed strings method,

F1−2=diagonals −sides

2×base .

R

L

l

P1

P2

a

b

c

d

e

f

g

To evaluate the lengths, two “pins” are placed right next to each other on the backside of each cylinder, say

points P1and P2. For both cases, (a) and (b), the strings of diagonals and sides run along the cylinder walls

from P1to aand d, and from P2to fand 1, thus cancelling each other. Therefore:

(a)

ac −4ab

⌢

ac −ab

F1−2=1

πsin−11

X+√X2−1−X

(b) The view from A1to A2is now partially obstructed, and the bottom string now runs along the first cylinder

CHAPTER 4 87

4.19

H

R1A1R2

a1a2

O1

O2

P1

P2

Find the view factor between spherical caps as shown in

the figure, for the case of

H≥R2

1

qR2

1−a2

1

+R2

2

qR2

2−a2

2

,

where H=distance between sphere centers, R=sphere

radius, a=radius of cap base. Why is this restriction

necessary?

88 RADIATIVE HEAT TRANSFER

4.20 Determine the view factor for Configuration 18 in Appendix D, using the unit sphere method.

h2

4.21 Consider the axisymmetric configuration

shown in the figure. Calculate the view factor

F1−3.

4cm

R1=1

5,R2=2

5,X=1+1+4/25

1/25 =30,

F5−4=1

230 −√302−4×4=0.134.

Thus

F1−3=π12

2π12×(1−0.134)=0.433

4.22 Find Fd1−2from the infinitesimal area

to the disk as shown in the figure, with

0≤β≤π.

dA1.

(i) All of A2visible: β < β1, where

cos β1=sin π

2−β1=r

R=r

√r2+h2,

Fd1−2=A′′

2

πR2=πr2cos β

π(r2+h2)

Fd1−2=r2cos β

r2+h2,for 0 < β < β1=cos−1r

√r2+h2

=tan−1h

r

r< β < π

(iii) A2is partially visible: β1< β < β2.

Recalling the principle of the unit sphere method: A′is the projection of (the visible part of) Aonto the sphere,

and A′′ is the projection of A′onto the bottom disk. Therefore, A′′

2consists of two parts: the projection of the

upper part of the disk (indicated by “y” in the sketch), and a part of the bottom disk (indicated by “z”).

When two disks, such as disk A2and the base disk

containing dA1, intersect each other, the cut is a

straight line as indicated in the sketch on the right

(looking from the bottom of A2towards dA1). The

sliver of A2not visible by dA1,Ac, is

Ac

γ

ρ

CHAPTER 4 91

rcot β, B=h

92 RADIATIVE HEAT TRANSFER

4.23

2r

2R



A1b

A1t

A2

A3



Consider the configuration shown (this could be a long cylindrical BBQ with a

center shelf/hole; or an integrating sphere). Determine the view factors F2−2and

F2−3assuming the configuration

(a) to be axisymmetric (sphere),

(b) to be two-dimensional Cartesian (cylinder), using view factor algebra,

(c) to be two-dimensional Cartesian (cylinder), using the string rule (F2−3only).

CHAPTER 4 93

4.24

R

4R

6R

R

concentrator

oil

tube

In the solar energy laboratory at UC Merced parabolic concentrators are

employed to enhance the absorption of tubular solar collectors as shown

in the sketch. Calculate the view factor from the parabolic concentrator

A1to collecting cylinder A2, using

(a) view factor algebra,

(b) Hottel’s string rule.

πtan−1π

2=0.3195;

from view factor algebra

F1−2=A2

(1 −F2−3)=2πR

2

6−1

6tan−1π

61−1

πtan−1π

94 RADIATIVE HEAT TRANSFER

4.25 The interior of a right-circular cylinder of length L=4R, where

Ris its radius, is to be broken up into 4 ring elements of equal

width. Determine the view factors between all the ring ele-

ments, using

(a) view factor algebra and the view factors of Configuration

40,

(b) Configuration 9 with the assumption that this formula can

be used for rings of finite widths.

Assess the accuracy of the approximate view factors. What

would be the maximum allowable value for ∆Xto ensure that

all view factors (within a distance of 4R) are accurate to at least

5%? Use program viewfactors or function view in your calcu-

lations.

L

/4

R

A2

A3

A4

A12

A34

A23

A45

Fpd(0)=1.

arg(1)=i*dl

Fpd(i)=view(40,3,arg)

ENDDO

arg(1)=R

arg(2)=(i-1)*dl

F9(i)=view(9,2,arg)*dl/(2.*R)

Er(i)=100.*(1.-F9(i)/F40(i))

ENDDO

!

! Output

9 0.50000 0.18695 0.05806 0.01997

40 0.24570 0.17398 0.09359 0.04801 0.02498 0.01359 0.00780

9 0.28571 0.17094 0.09053 0.04618 0.02407 0.01316 0.00759

96 RADIATIVE HEAT TRANSFER

4.26 The inside surfaces of a furnace in the shape of a parallelepiped with dimensions 1 m ×2 m ×4 m is to be

FS(1,1,6,i)=FS(1,1,5,i)

FS(1,2,6,i)=FS(1,2,5,i)

ENDDO

! FSb-e, Fb-w

DO i=1,4

FS(2,2,5,2*i) =FS(1,2,5,10-2*i)

FS(2,1,6,2*i-1)=FS(2,1,5,2*i-1)

FS(2,1,6,2*i) =FS(2,1,5,2*i)

FS(2,2,6,2*i-1)=FS(2,2,5,2*i-1)

FS(2,2,6,2*i) =FS(2,2,5,2*i)

ENDDO

! FSe-n (perpendicular plates with x1=0)

DO i=1,4

Z2=i-1.

! FSe-s

DO j=1,4

DO i=1,8

FS(3,j,6,i)=FS(3,j,5,i)

ENDDO

DO i=1,8

FS(4,j,6,i)=FS(4,j,5,i)

ENDDO

! FSn-s (parallel plates with c=1)

FS(5,2*j-1,6,2*i) =FS(5,1,6,2*abs(i-j)+2)

ENDDO

DO j=1,4

DO i=1,4

A=1.

! Convert FS to 2D array; set surface emittances and temperatures/heat fluxes

DO m=1,28

IF(m<3) THEN

! Front

ELSEIF(m<21) THEN

! North

i=5; j=m-12

ELSE

! South

k=4; l=n-8

ELSEIF(n<21) THEN

k=5; l=n-12

ELSE

k=6; l=n-20

ENDDO

! Make sure summation rule is fulfilled

do i=1,NN

sum(i)=0.

do j=1,NN

10 format(’Fij ’,28i7,’ Sum’)

20 format(i3,3x,28f7.4,f8.4)

close(9)

! Printable output

open(unit=10,FILE=’4_22p.dat’,STATUS=’UNKNOWN’)

15 format(’east to west’/(i3,3x,4f7.4))

16 format(’east to north’/(i3,3x,8f7.4))

close(10)

stop

end

Fij 12345678

1 0.0191 0.0171

1 0.0328 0.0329 0.0159 0.0078

1 0.2000 0.0406 0.0328 0.0189 0.0089 0.0069 0.0035 0.0030

1 0.0686 0.0481 0.0206 0.0079

3 0.0206 0.0481 0.0686 0.0481

1 0.0328 0.2000 0.0189 0.0406 0.0059 0.0043 0.0019 0.0009

3 0.0059 0.0043 0.0189 0.0406 0.0328 0.2000 0.0189 0.0406