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140 RADIATIVE HEAT TRANSFER
(b) With a silver foil the radiative resistance becomes
A2ǫ2
As2
ǫs−1+1
A31
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.
1034.1/2π3.47 ×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
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]
ǫ3−1F2−3q3−1
ǫ5−1F2−5q5=σT4
i=3 : q3
ǫ3
=σT4
3−F3−1σT4
1−F3−2σT4
2,
i=4 : q4
4−F4−5σT4
5,
ǫ3−1F1−3q3−1
ǫ5−1F1−5ǫoσT4
| {z }
=Ho1
=σT4
1−F1−2σT4
2−F1−3σT4
3−F1−51+1
ǫ5−1ǫo
σT4
5,
| {z }
F′
2−5
i=3 : q3
=σT4
3−F3−1σT4
1−F3−2σT4
2,
ǫ4−1F5−4q4+1
ǫ5−1
ǫ5−1F5−5ǫoσT4
| {z }
Ho5
=σT4
51−F5−5+1
ǫ5−1
ǫ5−1F5−5ǫo−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−51+1
ǫ5−1ǫo;
similarly,
F′
5−5=F5−51+1
ǫ5−1ǫo−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
ǫ5−1ǫoi, 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]
