# 978-0123869449 Chapter 6 Part 2

Document Type

Homework Help

Book Title

Radiative Heat Transfer 3rd Edition

Authors

Michael F. Modest

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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

d

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).

Fs

1−2=F1−2+ρ2F1(21)−2+···

=1−sin α

A2

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?

Similarly,

Fs

X

(1−ǫ)2nFn,1−t

A1

At(1)

side

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-

A1(2)

A2(12)

A1(212) 2

γ

4

y

x

γ

γ

2n

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)

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

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