188 RADIATIVE HEAT TRANSFER
6.15 Repeat problem 5.7 for the case that the flat part of the rod (A1)is a purely specular reflector.
CHAPTER 6 189
6.16
A1
A3
A2
A2
A2
R=1m
R
h=0.1m
2w=0.1m
A long furnace may, in a simplified scenario, be considered to consist of
a strip plate (the material to be heated, A1:ǫ1=0.2,T1=500K, specular
reflector), unheated refractory brick (flat sides and bottom, A2:ǫ2=0.1,
diffuse reflector), and a cylindrical dome of heated refractory brick (A3:
ǫ3=1,T3=1000K). Heat release inside the heated brick is qh(W/m2). The
total heat release is radiated into the furnace cavity and is removed by
convection, such that the convective heat loss is uniform everywhere (at qc
W/m2on all three surfaces).
(a) Express the net radiative fluxes on all three surfaces in terms of qh.
(b) Determine the qhnecessary to maintain the indicated temperatures.
3(1)−3is obstructed). We
have
3(1)−3is found from the crossed strings
F∗
3(1)−3=diagonalad −sideabc
πR
A1
A3
A2
A2
2w
d
b
A2(1)
A3(1)
Fs
+h2
CHAPTER 6 191
6.17
α
∋
, T
L
∋
, T
A typical space radiator may have a shape as shown in the adjacent sketch,
i.e., a small tube to which are attached a number of flat plate fins, spaced
at equal angle intervals. Assume that the central tube is negligibly small,
and that a fixed amount of specularly-reflecting fin material is available
(ǫ=ρs=0.5), to give (per unit length of tube) a total, one-sided fin area
of A′=N×L. Also assume the whole structure to be isothermal. Develop
an expression for the total heat loss from the radiator as a function of the
number of fins (each fin having length L=A′/N). Does an optimum exist?
Qualitatively discuss the more realistic case of supplying a fixed amount of
heat to the bases of the fins (rather than assuming isothermal fins).
ieven
Fs
1−2=F1−2+ρ2F1(21)−2+···
=1−sin α
iodd
A2
α
A1
α
α
α
α
A1(212)
A2(1212)
192 RADIATIVE HEAT TRANSFER
6.18 Repeat Problem 5.15 for the case that the stainless steel, while being a gray and diffuse emitter, is a purely
specular reflector (all four surfaces).
6.4 and equation (6.26),
Q=σAi(T4
o−T4
i)
,
To=293.92 K.
CHAPTER 6 193
6.19 Repeat Problem 5.16 for the case that, both, the platinum sphere as well as the aluminum shield, while being
a gray and diffuse emitters, are purely specular reflectors.
Fs
1−1=1−ǫ2
=0.9
Fs
2−1=A1
Fs
1−2=3
×3.077 =2.77 ×10−3
=3.537 ×106W
=3.537 ×106W
m2×0.75 ×2.77 ×10−3=7,346 W
m2
=s
0.04001 −1
194 RADIATIVE HEAT TRANSFER
te=1.43 s
For specular surfaces the heat-up is considerably faster than for diffuse ones: any radiation reflected from A1
(diffusely or specularly) travels to A2and, if reflected specularly, travels back to the platinum sphere (rather
CHAPTER 6 195
6.21
w1 =1 m
hb
w2 = 2 m
T2, 2 = 1
0 K
= 0.1
∋
∋
0 K
ht= 0.1
∋
Reconsider the spacecraft of Problem 6.10. To decrease the heat loss
from Surface 2 a specularly reflecting shield, of the same dimensions
as the black surface and with emittance ǫ=0.1, is placed between
the two plates. Determine the net heat loss from the black plate as a
function of shield location. Where would you place the shield?
2−2=ρs
bF2(b)−2=(1 −ǫ)hp4h2+(wb+w1)2−2hi,2w2=(1 −ǫ)s1+ hb
w2!2
w2
Similarly,
Fs
X
(1−ǫ)2nFn,1−t
A1
At
At(1)
side
with
Fn,1−t=q(wt+w1)2+(2n+1)2h2
t−qw2
1+h2
t−2nht2w1.
Also
Fs
1−1=ρsF1(t1)−1+(ρs)3F1(t1t1)−1+. . . =∞
X
n=1
(1 −ǫ)2n−1Fs
n,1−1
Fs
n,1−1=qw2
1+(nht)2−nhtw1.
And, finally,
Fs
t−1=A1
At
Fs
1−t=1
2Fs
1−t.
Because of the infinite series for these view factors, and because we would like to see the variation of q2with
variable shield position (subject to hb+ht=h=const), this is best solved with little computer program given
below. Of course, the minimum heat loss is obtained when hb→0, since only then all radiation is blocked
from directly going to outer space.
PROGRAM POBLEM6.19
IMPLICIT NONE
INTEGER :: i,II,n
REAL :: sigma=5.670e-8,w1,w2,hh,T2,eps,rho,hb,ht,F2b,F22,Ftt,F1t,Ft1,F11,add,denom,q2
! Dimensions
w1=1.
w2=2.
T2=600.
eps=.1
rho=1.-eps
II=20
! Loop over different hb
Ftt=0.
DO n=1,100
add=rho**(2*n-1)*(sqrt(w2**2+(n*ht)**2)-sqrt(w1**+ht**2)-(n-1)*ht)/w2
Ftt=Ftt+add
IF(add<1.e-6) EXIT
0.05 695.18
0.15 1019.81
0.25 1326.99
0.35 1616.97
0.45 1890.11
0.55 2146.83
0.65 2387.36
0.75 2611.72
0.85 2818.67
0.95 3002.27
198 RADIATIVE HEAT TRANSFER
6.22 Evaluate the normal emittance for the V-corrugated surface shown in Fig. 6-10a. Hint: This is most easily
calculated by determining the normal absorptance, or the net heat flux on a cold groove irradiated by parallel
light from the normal direction.
(1 additional reflections if γ < 60◦,
2 additional reflections if γ < 30◦,etc.).
If an n−th additional reflection occurs, it
may be partial or full (i.e. all incoming
beams hitting “0” undergo the n−th ad-
A2(12)
A1(212) 2
γ
4
y
x
γ
2n
αη=ǫ
1+
n=1
(1 −ǫ)nmin 1,sin(2n+1)γ
sin γ!
,N≤1
2 π
γ−1!
6.23 Redo 6.22 for an arbitrary off-normal direction 0 < θ < π/2 in a two-dimensional sense (i.e., what is the
off-normal absorptance for parallel incoming light whose propagation vector is in the same plane as all the
surface normal, namely the plane of the paper in Fig. 6-10).
200 RADIATIVE HEAT TRANSFER
6.24
A2
T2, 2
q1, 1
b/2
A1
h2 = 1cm
∋
∋
S’
S'(2)
A long, thin heating wire, radiating energy in the amount of S′=
300 W/cm (per cm length of wire), is located between two long, parallel
plates as shown in the adjacent sketch. The bottom plate is insulated
and specularly reflecting with ǫ2=1−ρs
2=0.2, while the top plate is
isothermal at T1=300 K and diffusely reflecting with ǫ1=1−ρd
1=0.5.
Determine the net radiative heat flux on the top plate.
1)Fs
1−1iEb1−(1−ρs
2)Fs
1−2Eb2=1−ρd
ǫ1
ǫ2
o1,
−(1−ρs
1)Fs
2−1Eb1+h1−(1−ρs
2)Fs
2−2iEb2=−ρd
1
Fs
2−1q1+1−ρd
2Fs
2−2
q2+Hs
o2.
fraction Fw−1=θ1/π falls into A1, and Fw−2=θ2/π falls into A2. In addition, surface A1sees a mirror image
of the wire behind A2. Thus, A1gets additional, external irradiation of the amount ρs
2Fw(2)−1=ρs
2θ1(2)/π. The
external contributions are then
bHs
2θ1(2)
Fs
1−1=ρs
2F1(2)−1=ρs
2
2b
=0.8×√100+64−8
10 =0.3845,
Fs
2−1=F2−1=2pb2+(h1+h2)2−2(h1+h2)
2b=√100+16−4
10 =0.6770,
Fs