978-0123869449 Chapter 5 Part 3

140 RADIATIVE HEAT TRANSFER

(b) With a silver foil the radiative resistance becomes

A2ǫ2

As2

ǫs−1+1

A31

ǫ3−1,

2πLR23 =1

0.100×0.2+1

0.1045 2

0.02 −1+1

0.109 1

0.2−1=1034.1 m−1

and

R3∞σT3

∞

=1.2974

T3=292.6 K.

1034.1/2π3.47 ×10−8−0.9988

CHAPTER 5 141

5.30 Repeat Problem 5.6, breaking up the sidewall into four equal ring elements. Use the view factors calculated

in Problem 4.25 together with program graydiffxch of Appendix F.

!

! From Problem 4.20

Fpd(0)=1.

DO i=1,NR

arg(1)=i*dl

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

ENDDO

! ring element to itself

F40(1)=1.-R/dl*(1.-Fpd(1))

! view factor to ring displaced by (i-1)*dl

DO i=2,NR

F40(i)=.5*R/dl*(Fpd(i)-2.*Fpd(i-1)+Fpd(i-2))

ENDDO

! Surfaces: 1=bottom ring…NR=top ring; NR+1=bottom disk, NR+2=top disk

DO i=1,NR

! ring to ring

DO j=i,NR

F(i,j)=F40(abs(i-j)+1)

ENDDO

! ring to ends

F(i,NR+1)=.5*R/dl*(Fpd(i-1)-Fpd(i))

F(i,NR+2)=.5*R/dl*(Fpd(NR-i)-Fpd(NR+1-i))

ENDDO

! end to end

F(NR+1,NR+2)=Fpd(NR)

! Set up surface data

DO i=1,NR

142 RADIATIVE HEAT TRANSFER

A(i)=2.*pi*R*dl

eps(i)=1.; id(i)=0; q(i)=0.; PIN(i)=q(i)

ENDDO

i=NR+1 ! bottom disk

A(i)=pi*R*R; eps(i)=1.; id(i)=1; T(i)=1500.; PIN(i)=sigma*T(i)**4

i=NR+2 ! top disk

A(i)=pi*R*R; eps(i)=.5; id(i)=1; T(i)= 500.; PIN(i)=sigma*T(i)**4

! Solve system of equations

CALL GRAYDIFF(NN,iclsd,A,EPS,HO,F,ID,PIN,POUT)

! Output

OPEN(UNIT=8,FILE=’problem5_22.dat’,STATUS=’UNKNOWN’)

! Convert emissive powers to temperatures

sumq=0 ! Check total flux=0?

CHAPTER 5 143

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

broken up into 28 1 m ×1 m subareas. The gray-diﬀuse side walls (of dimension 1 m ×2 m and 1 m ×4 m)

have emittances of ǫs=0.7 and are perfectly insulated, the bottom surface has an emittance of ǫb=0.9 and a

temperature Tb=1600 K, while the top’s emittance is ǫt=0.2 and its temperature is Tt=500 K. Using the view

factors calculated in Problem 4.26 and program graydiffxch of Appendix F, calculate the heating/cooling

requirements for bottom and top surfaces, as well as the temperature distribution along the side walls.

f1(1)

n1(13) n2(14)

n3(15) n4(16)

n5(17) n6(18)

n7(19) n8(20)

f2(2)

e1(5)

e2(6)

e3(7)

e4(6)

w1(9)

w2(10)

w3(11)

w4(12) b1(3) b2(4)

s3(23)

s1(21) s2(22)

s4(24)

s5(25)

s7(27) s8(28)

s6(26)

Solution

All view factors for this problem have already been determined in Problem 4.26. All that remains to be done

is to fill the A,HO,EPS,id and PIN vectors and graydiff can be called as shown in the Fortran90 code below.

PROGRAM PARALLELEPIPED

IMPLICIT NONE

INTEGER,PARAMETER :: NN=28

REAL :: A(NN),FS(6,8,6,8),F(NN,NN),EPS(NN),HO(NN),PIN(NN),POUT(NN)

REAL :: PARLPLTF,PERPPLTF,GAUSS,sumq,sigma=5.670E-8,T(NN),q(NN),QA(NN)

REAL :: Y1,Y2,Y3,Z2,Z3

integer :: id(NN),iclsd,i,j,k,l,m,n

! View Factors; since configuration is closed (iclsd=1), diagonal terms are not needed

iclsd=1

! FS-subscripts FS(1,2,3,4): 1=originating surface, 2=originating node, 3=target surface, 4=target node

! surfaces: 1=front, 2=back, 3=east, 4=west, 5=north, 6=south

! Initialize

.

! Initialize areas and external irradiation

A=1.

HO=0.

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

DO m=1,28

IF(m<3) THEN

! Front

i=1; j=m

eps(m)=0.7; id(m)=0; PIN(m)=0.

ELSEIF(m<5) THEN

! Back

i=2; j=m-2

eps(m)=0.7; id(m)=0; PIN(m)=0.

ELSEIF(m<9) THEN

! East

i=3; j=m-4

eps(m)=0.7; id(m)=0; PIN(m)=0.

144 RADIATIVE HEAT TRANSFER

eps(m)=0.2; id(m)=1; T(m)= 500.; PIN(m)=sigma*T(m)**4

ELSE

! South

i=6; j=m-20

eps(m)=0.9; id(m)=1; T(m)=1600.; PIN(m)=sigma*T(m)**4

ENDIF

DO n=m+1,28

IF(n<3) THEN

k=1; l=n

ELSEIF(n<5) THEN

k=2; l=n-2

ELSEIF(n<9) THEN

k=3; l=n-4

ELSEIF(n<13) THEN

k=4; l=n-8

ELSEIF(n<21) THEN

k=5; l=n-12

ELSE

k=6; l=n-20

ENDIF

F(m,n)=FS(i,j,k,l)

ENDDO

! Solve system of equations

CALL GRAYDIFF(NN,iclsd,A,EPS,HO,F,ID,PIN,POUT)

! Output

OPEN(UNIT=8,FILE=’problem5_23.dat’,STATUS=’UNKNOWN’)

! Convert emissive powers to temperatures

sumq=0 ! Check total flux=0?

146 RADIATIVE HEAT TRANSFER

5.32

lid

steak

coal rack BBQ base

grill

30

cm

60

cm

For your Memorial Day Barbecue you would like to broil a steak on your

backyard BBQ, which consists of a base unit in the shape of a hemisphere

(D=60 cm), fitted with a disk-shaped coal rack, and a disk-shaped grill,

as shown in the sketch. Hot coal may be assumed to cover the entire ﬂoor

of the unit, with uniform temperature Tc=1200 K, and an emittance of

ǫc=1. The side wall is soot-covered and black on the inside, but has an

outside emittance of ǫo=0.5.

The steak (modeled as a ds=15 cm disk, 1 cm thick, emittance ǫs=

0.8, initially at Ts=280 K ) is now placed on the grill (assumed to be

so lightweight as to be totally transparent and not participating in the

heat transfer). The environment is at 300 K, and free convection may be

neglected.

(a) Assuming that the lid is not placed on top of the unit, estimate the initial heating rates on the two

surfaces of the steak.

(b) How would the heating rates change, if the lid (also a hemisphere) is put on (ǫi=ǫo=0.5)? Could one

achieve a more even heating rate (top and bottom) if the emittance of the inside surface is increased or

decreased?

1.3285

q2=−27,045 W/m2

With a little thought, it is also possible to use graydiff.f90 for this problem, using a 4 surface closed

enclosure (the ambient being a fourth, black surface at Tamb), or a 3 surface open enclosure (treating Tamb–

radiation as external). To allow for the radiative loss external to A2we note that equation (5.32-A) is identical

R2=0.30

Tamb=300.

A(1)=pi*R1*R1

id(1)=1 ! T specified

A(2)=2.*pi*R2*hh

id(2)=0 ! q=0 specified

A(3)=pi*R3*R3

id(3)=1 ! T specified

T(3)= 280. ! T in K

! Fill PIN array with q and T

DO i=1,N

IF(id(i)==0) THEN

arg(1)=hh; arg(2)=R1; arg(3)=R3

arg(1)=hh; arg(2)=R1; arg(3)=R2

F1amb=VIEW(40,3,arg)-F(1,3)

! F12

HO(3)=0. ! no external irradiation

F(2,2)=F(2,2)-epso ! modified for external radiation loss

! Solve system of equations

CALL GRAYDIFF(N,iclsd,A,EPS,HO,F,ID,PIN,POUT)

! Output ! Convert emissive powers to temperatures

write(*,20) (i,T(i),q(i),QA(i),i=1,N)

10 format(’surface T q [W/m2] Q [W]’)

20 format(i4,3x,f9.1,2e13.4)

stop

end

surface T q [W/m2] Q [W]

ǫ3−1F2−3q3−1

ǫ5−1F2−5q5=σT4

i=3 : q3

ǫ3

=σT4

3−F3−1σT4

1−F3−2σT4

2,

i=4 : q4

4−F4−5σT4

5,

ǫ3−1F1−3q3−1

ǫ5−1F1−5ǫoσT4

| {z }

=Ho1

=σT4

1−F1−2σT4

2−F1−3σT4

3−F1−51+1

ǫ5−1ǫo

σT4

5,

| {z }

F′

2−5

i=3 : q3

=σT4

3−F3−1σT4

1−F3−2σT4

2,

ǫ4−1F5−4q4+1

ǫ5−1

ǫ5−1F5−5ǫoσT4

| {z }

Ho5

=σT4

51−F5−5+1

ǫ5−1

ǫ5−1F5−5ǫo−F5−1σT4

1−F5−2σT4

2−F5−4σT4

4.

150 RADIATIVE HEAT TRANSFER

In order to use graydiff.f90 for this problem, these equations must be brought into the form of equa-

tion (5.32), which can be achieved in three steps:

1. A2and A5are treated as insulated (q2=q5=0), giving rise to artificial “external irradiation” terms Hoi

as indicated (some negative);

2. As in (a) F2−2is modified to F′

2−2=F2−2−ǫo; and the Fi−5are modified to

F′

i−5=Fi−51+1

ǫ5−1ǫo;

similarly,

F′

5−5=F5−51+1

ǫ5−1ǫo−ǫo

ǫ5

.

3. To avoid having graydiff recalculate F2−2and F5−5we set iclsd=2 (open enclosure). Also, since the

F5−iare calculated by reciprocity, we set A′

5=A5h1+1

ǫ5−1ǫoi, so that

F5−i=Ai

Fi−5=Ai

F′

R2=0.30

Tamb=300.

A(1)=pi*R1*R1

id(1)=1 ! T specified

A(2)=2.*pi*R2*hh

id(2)=0 ! q=0 specified

A(3)=pi*R3*R3

id(3)=1 ! T specified

A(4)=pi*R3*R3

id(4)=1 ! T specified

A(5)=2.*pi*R2*R2

id(5)=0 ! q=0 specified

q(5)= 0. ! q in W/m2

! Fill PIN array with q and T

DO i=1,N

CHAPTER 5 151

arg(1)=hh; arg(2)=R1; arg(3)=R3

arg(1)=hh; arg(2)=R1; arg(3)=R2

F(1,5)=VIEW(40,3,arg)-F(1,3)

! F12

F(1,2)=1-F(1,3)-F(1,5)

! F14

HO(2)=(epso-cc1*F(2,5))*sigma*Tamb**4 ! artificial external irradiation

HO(4)=-cc1*F(4,5)*sigma*Tamb**4 ! artificial external irradiation

HO(5)=(epso/eps(5)-cc1*F(5,5))*sigma*Tamb**4 ! artificial external irradiation

F(2,2)=F(2,2)-epso ! modified for external radiation loss

DO i=1,5

F(i,5)=F(i,5)*(1.+cc1)

q(5)=-epso*sigma*(T(5)**4-Tamb**4)

A(5)=A(5)/(1.+cc1)

sumq=0 ! Check total flux=0?

A(2)=2.*pi*R2*hh

id(2)=1 ! T specified

A(5)=2.*pi*R2*R2

id(5)=1 ! T specified

T(5)=Tamb ! first guess

.

! Iterate to find correct T2 and T5

iter=-1

write(9,31)

T(5)=T5new

GOTO 5

T(2)=T2new

T(5)=T5new

1 598.2 548.3 660.8 608.6 122.93

3 688.1 634.7 701.5 647.5 26.14

5 708.4 654.0 712.0 657.4 7.02

7 713.9 659.2 714.9 660.2 1.98

9 715.4 660.7 715.7 661.0 0.56

surface T q [W/m2] Q [W]