# 978-0123869449 Chapter 5 Part 4

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Book Title

Radiative Heat Transfer 3rd Edition

Authors

Michael F. Modest

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154 RADIATIVE HEAT TRANSFER

5.33 Consider Configuration 33 in Appendix D with h=w. The bottom wall is at constant temperature T1and has

emittance ǫ1; the side wall is at T2=const and ǫ2. Find the exact expression for q1(x) if ǫ2=1.

(2) q1→ǫ1σT4

CHAPTER 5 155

5.34 An infinitely long half-cylinder is irradiated by the sun as shown in the figure,

with qsun =1000 W/m2. The inside of the cylinder is gray and diﬀuse, the

outside is insulated. There is no radiation from the background. What is the

equilibrium temperature distribution along the cylinder periphery,

(a) using four isothermal zones of 45◦each,

(b) using the exact relations. (Hint: Use diﬀerentiation as in the kernel

approximation method).

qsun

ϕ

r

, T( )

ϕ

∋

a

b

c

d

e

A1

A2A3

A4

qs

qs

5.670 ×10−8W/m2K4#=342.8 K

σT4

qs

qs

5.670 ×10−8W/m2K4#=380.1 K

(b) From equation (5.21) it follows with

Ho(ϕ)=qsdA sin ϕ,

H(ϕ)dA =qssin ϕdA +ZA

J(ϕ′)dFdA′−dAdA′

or

H(ϕ)=qssin ϕ+ZA

J(ϕ′)dFdA−dA′.

CHAPTER 5 157

dθ=dA′

p

S=rdϕ′cos θ

2rcos θ=1

2dϕ′

dFdA−dA′=1

4sin

ϕ′−ϕ

2

dϕ′.

equation (as done in the kernel approximation). Diﬀerentiating twice

J′(ϕ)=qscos ϕ+1

8Zϕ

0

J(ϕ′) cos ϕ−ϕ′

2dϕ′−1

8Zπ

ϕ

J(ϕ′) cos ϕ′−ϕ

2dϕ′,

J′′(ϕ)=−qssin ϕ+1

J(ϕ′) sin ϕ−ϕ′

J(ϕ′) sin ϕ′−ϕ

J(0) =C2=1

4Zπ

0

J(ϕ′) sin ϕ′

2dϕ′

=1

4Zπ

03

4qssin ϕ′+C2sin ϕ′

2dϕ′.

Setting ϕ′=2η, with sin ϕ′=2 sin ηcos η,

C2=1

2Zπ/2

03

2qssin ηcos η+C2sin ηdη

=1

4qssin3η

π/2

0+1

2C2−cos η

π/2

0=1

4qs+1

2C2.

Thus C2=1

2qs, and

J(ϕ)=1

4qs(2 +3 sin ϕ).

[However, to avoid errors it is generally preferable to substitute equation (5.34-B) into equation (5.34-A) for

arbitrary ϕ: if the constants C1(=0) and C2are determined without contradictions, one is almost certain that no

158 RADIATIVE HEAT TRANSFER

errors were made!] For an insulated wall, radiosity is equal to blackbody emissive power [cf. equation (5.31)],

and

σT4(ϕ)=J(ϕ)=1

4qs(2 +3 sin ϕ)

In this particular example, the exact solution is actually simpler than the net radiation approach. This is, of

course, not generally the case. Evaluating the exact solution for ϕ=π/8 and 3π/8 [center of the nodes in part

(a)], we obtain for comparison

σT4

qsπ

8=σT4

qs7π

8=0.7870,

σT4

qs3π

8=σT4

qs5π

8=1.1929,

which is in excellent agreement with the approximate solution.

CHAPTER 5 159

5.35

w1 =1 m

h = 1 m

w2 = 2 m

0 K

1 = 0.1

∋

0 K

To calculate the net heat loss from a part of a spacecraft, this part may

be approximated by an infinitely long black plate at temperature

T2=600 K, as shown. Parallel to this plate is another (infinitely

long) thin plate that is gray and emits/reflects diﬀusely with the

same emittance ǫ1on both sides. You may assume the surroundings

to be black at 0 K. Calculate the net heat loss from the black plate.

=0.874,

F2−1=w1

F1−2=1

160 RADIATIVE HEAT TRANSFER

5.36

Tc, c

qsun

ϕ

Tw , w

∋

∋

A large isothermal surface (exposed to vacuum, temperature Tw,

diﬀuse-gray emittance ǫw) is irradiated by the sun. To reduce the

heat gain/loss from the surface, a thin copper shield (emittance ǫc

and initially at temperature Tc0) is placed between surface and sun as

shown in the figure.

(a) Determine the relationship between Tcand time t(it is suﬃcient

to leave the answer in implicit form with an unsolved integral).

(b) Give the steady state temperature for Tc(i.e., for t→ ∞).

(c) Briefly discuss qualitatively the following eﬀects:

(i) The shield is replaced by a moderately thick slab of styrofoam coated on both sides with a very thin

layer of copper.

(ii) The surfaces are finite in size.

CHAPTER 5 161

5.37

q1

Insulation

y

x

L

h

A2, 2

A1, 1

0 K

∋

∋

Consider two infinitely long, parallel, black plates of width Las shown.

The bottom plate is uniformly heated electrically with a heat flux of

q1=const, while the top plate is insulated. The entire configuration is

placed into a large cold environment.

(a) Determine the governing equations for the temperature variation

across the plates.

(b) Find the solution by the kernel substitution method. To avoid

tedious algebra, you may leave the final result in terms of two

constants to be determined, as long as you outline carefully how

these constants may be found.

(c) If the plates are gray and diﬀuse with emittances ǫ1and ǫ2, how can the temperature distribution be

determined, using the solution from part (b)?

2.

Because of symmetry Φi(ξ)= Φi(−ξ), which requires that C1=C4=0. The other two constants must be found

by substitution into the undiﬀerentiated integral equation at any value for ξ. Choosing ξ=0 and using the

symmetry condition [which shows that R0

−W/2( )dξ=RW/2

0( )dξ], this leads to

C2=1+ZW/2

0 C2−ξ2

2!e−ξdξ=1+C21−e−W/2+ W2

8+W

2+1!e−W/2−1,

C2=W2

8+W

2+1.

2. cosh W

2.

Φ1−Φ2=C3cosh √2ξ+1

2,

Φ1=C3

2cosh √2ξ+C2

2+1

4(1 −ξ2)

164 RADIATIVE HEAT TRANSFER

5.38

2

A1

r2 = r3r4

r1

A2

A3

A4

θ

To reduce heat transfer between two infinite concentric cylin-

ders a third cylinder is placed between them as shown in the

figure. The center cylinder has an opening of half-angle θ. The

inner cylinder is black and at temperature T1=1000 K, while

the outer cylinder is at T4=300 K. Outer cylinder and both

sides of the shield are coated with a reflective material, such

that ǫc=ǫ2=ǫ3=ǫ4Determine the heat loss from the inner

cylinder as function of coating emittance ǫc, using

(a) the net radiation method,

(b) the network analogy.

Solution

CHAPTER 5 165

5.39

A1

A2

A4

R3R

R

2RA3

Consider the two long concentric cylinders as shown in the figure.

Between the two cylinders is a long, thin flat plate as also indicated.

The inner cylinder is black and generating heat on its inside in the

amount of Q′

1=1 kW/m length of the cylinder, which must be removed

by radiation. The plate is gray and diﬀuse with emittance ǫ2=ǫ3=0.5,

while the outer cylinder is black and cold (T4=0 K). Determine the

temperature of the inner cylinder, using

(a) the net radiation method,

(b) the network analogy.

166 RADIATIVE HEAT TRANSFER

5.40

R2 =?

h = ?

R1 = 0.1m

T1, 1 = 1

0 K

2 = 0.1

∋

∋

0 K

An isothermal black disk at T1=500Kis flush with the outer surface

of a spacecraft and is thus exposed to outer space. To minimize heat

loss from the disk a disk-shaped radiation shield is placed coaxially

and parallel to the disk as shown; the shield radius is R2(which may

be smaller or larger than R1), and its distance from the black disk is a

variable h. Determine an expression for the heat loss from the black

disk as a function of shield radius and distance, using

(a) the net radiation method,

(b) the network analogy.

Solution

But q2b+q2t=0 and, therefore, after adding the equations for i=2band 2twe obtain

Eb2=1

2F2−1Eb1;q2b=−ǫ2

2F2−1Eb1,

(b) Thermal resistance analogy:

1/A2F2t–0K

J2t

1

R1−∞,tot

=1

R1−∞

+1

R1−2−∞

;R1−2−∞ =R1−2b+R2b−∞,tot;

CHAPTER 5 167

and

1

=A2

1−F2b−1+1

2

ǫ2−1

=2A2

2−ǫ2−A1F1−2;

and for the grand total resistance

The heat flux per unit area is then determined from

Eb1−,=0

Eb∞

1

=1−F1−2+F1−2

1+(2−ǫ2)F1−2

2A2/A1−(2−ǫ2)F1−2

=1−F1−2+F1−2h2A2

2A2

A1−(2 −ǫ2)F1−2+(2 −ǫ2)F1−2

=1−F1−2+F1−21−1

2

A1

A2

(2 −ǫ2)F1−2=1−1−ǫ2

2F1−2F2−1

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