# 978-0123869449 Chapter 5 Part 2

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Book Title

Radiative Heat Transfer 3rd Edition

Authors

Michael F. Modest

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120 RADIATIVE HEAT TRANSFER

5.15

TC

ss

∋

∋

ss, ho

∋

∋

vacuum

helium

ss, hi

TC, hTC

Tamb

A thermocouple used to measure the temperature of cold, low pressure

helium flowing through a long duct shows a temperature reading of

10 K. To minimize heat losses from the duct to the surroundings the

duct is made of two concentric thin layers of stainless steel with an

evacuated space in between (inner diameter di=2 cm, outer diameter

do=2.5 cm; stainless layers very thin and of high conductivity). The

emittance of the thermocouple is ǫTC =0.6, the convection heat transfer

coeﬃcient between helium and tube wall is hi=5 W/m2K, thermocouple

and helium is hTC =2 W/m2K, and the emittance of the stainless steel

is ǫss =0.2 (gray and diﬀuse, all four surfaces). The free convection

heat transfer coeﬃcient between the outer tube and the surroundings at

Tamb =300 K is ho=5 W/m2K. To determine the actual temperature of the helium,

(a) Prepare an energy balance for the thermocouple.

(b) Prepare an energy balance for the heat loss through the duct wall (the only unknowns here should be

THe and Tw).

(c) Outline how to solve for the temperature of the helium (no need to carry out solution).

(d) Do you expect the thermocouple to be accurate? (hint: check magnitudes of terms in (a)).

hTC

2=0.4593

or

Q=θi−θTC +C3(θ4

THe =9.997 K, TTC =10 K, Ti=20.24 K, and To=293.38 K

.

(d) Linearizing the θHe equation, we obtain

122 RADIATIVE HEAT TRANSFER

5.16

10 cm

0

K

Al shield

Pt sphere

laser

1

cm

During a materials processing experiment on the space shuttle (un-

der microgravity conditions) a platinum sphere of 3 mm diameter

is levitated in a large, cold black vacuum chamber. A spherical

aluminum shield (with a circular cut-out) is placed around the

sphere as shown, to reduce heat loss from the sphere. Initially, the

sphere is at 200 K and is suddenly irradiated with a laser providing

an irradiation of 100 W (normal to beam) to raise its temperature

rapidly to its melting point (2741 K).

Determine the time required to reach the melting point.

You may assume the platinum and aluminum to be gray and dif-

fuse (ǫPt =0.25, ǫAl =0.1), the sphere to be essentially isothermal

at all times, and the shield to have zero heat capacity.

1

ǫ1−(2 −ǫ2)C21

ǫ1−1=1−1.9×8.85 ×10−3

3.950 =0.2489

=3.024 ×10−4s−1−q1

σT4

1,i

0.07087 −1

te=6.48 s

Pt =3.5×106W/m2), and T2≃860 K (mostly due to laser radiation

reflected from the sphere). If the shield is neglected (C2=0), the required heating time increases to 6.73 s.

124 RADIATIVE HEAT TRANSFER

5.17

φ

qsun

d

d

A1

A2

R

Two identical circular disks are connected at one point of their periphery

by a hinge. The configuration is then opened by an angle φas shown in

the figure. Assuming the opening angle to be φ=60◦,d=1 m, calculate

the average equilibrium temperature for each of the two disks, with solar

radiation entering the configuration parallel to Disk 2 with a strength of

qsun =1000 W/m2. Disk 1 is gray and diﬀuse with α=ǫ=0.5, Disk 2 is

black. Both disks are insulated.

1−F1−2F2−1

1.

The view factor F1−2=F2−1is readily determined from the inside sphere method, by placing the disks inside

a sphere of radius R=d/2 cos φ

2(see sketch). Then

cos φ

5.670 ×10−8W/m2K41−1

9=1.7183 ×1010 K4

CHAPTER 5 125

5.18

60°

A2

A1

A3

60°

L = 1 m

ground

qsun

A long greenhouse has the cross-section of an equilateral triangle

as shown. The side exposed to the sun consists of a thin sheet of

glass (A1) with reflectivity ρ1=0.1. The glass may be assumed

perfectly transparent to solar radiation, and totally opaque to ra-

diation emitted inside the greenhouse. The other side wall (A2) is

opaque with emittance ǫ2=0.2, while the floor (A3) has ǫ3=0.8.

All surfaces reflect diﬀusely. For simplicity, you may assume sur-

faces A1and A2to be perfectly insulated, while the floor loses heat

to the ground according to

q3,conduction =U(T3−T∞)

where T∞=280 K is the temperature of the ground, and U=

19.5W/m2K is an overall heat transfer coeﬃcient.

Determine the temperatures of all three surfaces for the case that the sun shines onto the greenhouse with

strength qsun =1000 W/m2in a direction parallel to surface A2.

=789.4 W/m2

or

T1=T2=789.4/5.67 ×10−81/4=343.5 K.

126 RADIATIVE HEAT TRANSFER

5.19

θ

qsol = 1000 W/m2

30

L

0 K

A2

A1

A1

A long, black V-groove is irradiated by the sun as shown. Assuming the

groove to be perfectly insulated, and radiation to be the only mode of

heat transfer, determine the average groove temperature as function of

solar incidence angle θ(give values for θ=0◦,15◦,30◦,60◦,90◦). For

simplicity the V-groove wall may be taken as a single zone.

5.670 ×10−81/4

T1(θ=0◦)=364 K,

T1(θ=15◦)=361 K,

CHAPTER 5 127

5.20

Ho

T,

∋

γ

2Rb

2R

Consider the conical cavity shown (radius of opening R, opening angle

γ=30◦), which has a gray, diﬀusely reflective coating (ǫ=0.6) and is

perfectly insulated. The cavity is irradiated by a collimated beam of

strength H0and radius Rb=0.5R).

(a) Using a single node analysis, develop an expression relating H0

to the average cavity temperature T.

(b) For a more accurate analysis a two-node analysis is to be performed. What nodes would you choose?

Develop expressions for the necessary view factors in terms of known ones (including those given in

Appendix D) and surface areas, then relate the two temperatures to H0.

(c) Qualitatively, what happens to the cavity’s overall average temperature, if the beam is turned away by

an angle α?

nonabsorbed energy is reflected diﬀusely, indistinguishable from emission for gray radiation).

CHAPTER 5 129

5.21

Ho

T,

∋

γ

A1

A1

Ae

A (simplified) radiation heat flux meter consists of a conical cavity

coated with a gray, diﬀuse material, as shown in the figure. To measure

the radiative heat flux, the cavity is perfectly insulated.

(a) Develop an expression that relates the flux, Ho, to the cavity

temperature, T.

(b) If the cavity is turned away from the incoming flux by an angle

α, what happens to the cavity temperature?

130 RADIATIVE HEAT TRANSFER

5.22

l2 = 60 cm

Reflector

Collector plate

l1 = 80cm

T1, 1

q2 = 0, 2

qsun = 1000 W/m2

30°

∋

∋

A very long solar collector plate is to collect energy at a temperature

of T1=350 K. To improve its performance for oﬀ-normal solar in-

cidence, a highly reflective surface is placed next to the collector as

shown in the adjacent figure. How much energy (per unit length)

does the collector plate collect for a solar incidence angle of 30◦? For

simplicity you may make the following assumptions: the collector

is isothermal and gray-diﬀuse with emittance ǫ1=0.8; the reflec-

tor is gray-diﬀuse with ǫ2=0.1, and heat losses from the reflector

by convection as well as all losses from the collector ends may be

neglected.

0.8−1

0.8−1×1

4×1

3

5.23

A2

A3

A3

TC(A1)

L=10 cm

D=10 cm

A thermocouple (approximated by a 1 mm diameter sphere with gray-

diﬀuse emittance ǫ1=0.5) is suspended inside a tube through which

a hot, nonparticipating gas at T1=2000 K is flowing. In the vicinity

of the thermocouple the tube temperature is known to be T2=1000 K

(wall emittance ǫ2=0.5). For the purpose of this problem you may

assume both ends of the tube to be closed with a black surface at the

temperature of the gas, T3=2000 K. Again, for the purpose of this

problem, you may assume that the thermocouple gains a heat flux of

104W/m2of thermocouple surface area, which it must reject again in

the form of radiation. Estimate the temperature of the thermocouple.

Hints:

(a) treat the tube ends together as a single surface A3;

(b) note that the thermocouple is small, i.e., Fx−1≪1.

132 RADIATIVE HEAT TRANSFER

5.24

R = 40 cm R

A2: 2 = 0.1, q2 = 0

∋

r=25 cm

h=30 cm

L=30 cm

A1: 1 = 0.8, Q1 = -0.4 kW

∋

Q

s

=10 kW

A small spherical heat source outputting Qs=10 kW power,

spreading equally into all directions, is encased in a reflector as

shown, consisting of a hemisphere of radius R=40 cm, plus a ring

of radius Rand height h=30 cm. The arrangement is used to heat

a disk of radius =25 cm a distance of L=30 cm below the reflector.

All surfaces are gray and diﬀuse, with emittances of ǫ1=0.8 and

ǫ2=0.1. Reflector A2is insulated.

(a) Determine (per unit area of receiving surface) the irradiation

from heat source to reflector and to disk;

(b) all relevant view factors;

(c) the temperature of the disk, if 0.4 kW of power is extracted

from the disk.

134 RADIATIVE HEAT TRANSFER

5.25

A1: wire

A3: 3 = 0.5, T3 = 300K

∋

R = 2 cm R

h = 3 cm

A2: 2 = 0.2, q2 = 0

∋

A long thin black heating wire radiates 300 W per cm length of

wire and is used to heat a flat surface by thermal radiation. To

increase its eﬃciency the wire is surrounded by an insulated half-

cylinder as shown in the figure. Both surfaces are gray and diﬀuse

with emittances ǫ2and ǫ3, respectively. What is the net heat flux at

Surface 3? How does this compare with the case without cylinder?

Hint: You may either treat the heating wire as a thin cylinder whose

radius you eventually shrink to zero, or treat radiation from the

wire as external radiation (the second approach being somewhat

simpler).

0.5×0.6366 −1

0.5−1×1

2×0.3183

with half−cylinder : q3=−18.72 W/cm2

CHAPTER 5 135

5.26

h

Q2h

A2

A1

2R

Consider the configuration shown, consisting of a cylindrical cavity A2,

a circular disk A1at the bottom, and a small spherical radiation source

(blackbody at 4000K) of strength Q=10,000 W as shown (R=10 cm,

h=10 cm). The cylinder wall A2is covered with a gray, diﬀuse material

with ǫ2=0.1, and is perfectly insulated. Surface A1is kept at a constant

temperature of 400 K. No other external surfaces or sources aﬀect the

heat transfer. Assuming surface A1to be gray and diﬀuse with ǫ1=0.3

determine the amount of heat that needs to be removed from A1(Q1).

20.82845.670 ×10−84004−4.66 ×104−,=2

2πR25.63 ×104

0.3−1

21

0.3−10.8284

q1=−47,200W/m2,

136 RADIATIVE HEAT TRANSFER

5.27

rr

R

qs = 1000W/m2

A2

A1

Determine F1−2for the rotationally symmetric configuration shown in

the figure (i.e., a big sphere, R=13 cm, with a circular hole, r=5 cm,

and a hemispherical cavity, r=5 cm). Assuming Surface 2 to be gray

and diﬀuse (ǫ=0.5) and insulated and Surface 1 to be black and

also insulated, what is the average temperature of the black cavity if

collimated irradiation of 1000 W/m2is penetrating through the hole

as shown?

Ho2=0.

The two equations are readily solved for Eb1, leading to

1−F2−2=F2−3+F2−4=2F2−4=2F2−1,

Eb2=F2−1

2F2−1

Eb1,

CHAPTER 5 137

5.28

Collimated

beam

Entrance

port

Detector

Sample

MgO

An integrating sphere (a device to measure surface properties) is 10 cm

in radius. It contains on its inside wall a 1 cm2black detector, a 1 ×2 cm

entrance port and a 1 ×1 cm sample as shown. The remaining por-

tion of the sphere is smoked with magnesium oxide having a short-

wavelength reflectance of 0.98, which is almost perfectly diﬀuse. A

collimated beam of radiant energy (i.e., all energy is contained within a

very small cone of solid angles) enters the sphere through the entrance

port, falls onto the sample, and then is reflected and interreflected, giv-

ing rise to a sphere wall radiosity and irradiation. Radiation emitted

from the walls is not detected because the source radiation is chopped,

and the detector-amplifier system responds only to the chopped radi-

ation. Find the fractions of the chopped incoming radiation that are

(a) lost out the entrance port,

(b) absorbed by the MgO-smoked wall, and

(c) absorbed by the detector.

[Item (c) is called the “sphere eﬃciency.”]

CHAPTER 5 139

5.29

R2

Liquid

helium

Foil

ho

R1

R3

Tamb

R4

The side wall of a flask holding liquid helium may be approximated as

a long double-walled cylinder as shown in the adjacent sketch. The con-

tainer walls are made of 1 mm thick stainless steel (k=15 W/m K, ǫ =0.2),

and have outer radii of R2=10 cm and R4=11 cm. The space between

walls is evacuated, and the outside is exposed to free convection with the

ambient at Tamb =20◦C and a heat transfer coeﬃcient of ho=10 W/m2K

(for the combined eﬀects of free convection and radiation). It is reason-

able to assume that the temperature of the inner wall is at liquid helium

temperature, or T(R2)=4 K.

(a) Determine the heat gain by the helium, per unit length of flask.

(b) To reduce the heat gain a thin silver foil (ǫ=0.02) is placed midway between the two walls. How does

this aﬀect the heat flux?

For the sake of the problem, you may assume both steel and silver to be diﬀuse reflectors.

=1.2974 m−1.

(a) Without the silver foil the radiative resistance follows from Fig. 5-10cas

R23 =1

+1

R23

86.36 =1.5023 ×10−2.

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