# Solution manual fundamentals of heat and mass transfer 6th edition ch12

PROBLEM 12.1
KNOWN: Rate at which radiation is intercepted by each of three surfaces (see (Example 12.1).
2

FIND: Irradiation, G[W/m ], at each of the three surfaces.
SCHEMATIC:

ANALYSIS: The irradiation at a surface is the rate at which radiation is incident on a surface per unit
area of the surface. The irradiation at surface j due to emission from surface 1 is

Gj =

q1− j
Aj

.
-3

2

With A1 = A2 = A3 = A4 = 10 m and the incident radiation rates q1-j from the results of Example

12.1, find
12.1×10−3 W
G2 =
= 12.1 W / m 2
−3 2

<

10

G3 =

G4 =

m

28.0 ×10−3 W
10−3 m 2
19.8 × 10−3 W
10−3 m 2

= 28.0 W / m 2

<

= 19.8 W / m 2 .

<

COMMENTS: The irradiation could also be computed from Eq. 12.13, which, for the present
situation, takes the form

G j = I1 cos θ j ω1− j
2

where I1 = I = 7000 W/m ⋅sr and ω1-j is the solid angle subtended by surface 1 with respect to j. For
example,

G 2 = I1 cosθ 2 ω1− 2
G 2 = 7000 W / m 2 ⋅ sr ×

cos 30°

10−3 m 2 × cos 60°

( 0.5m )2

G 2 = 12.1 W / m 2 .
Note that, since A1 is a diffuse radiator, the intensity I is independent of direction.

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PROBLEM 12.2
KNOWN: A diffuse surface of area A1 = 10-4m2 emits diffusely with total emissive power E = 5 × 104
W/m2 .
FIND: (a) Rate this emission is intercepted by small surface of area A2 = 5 × 10-4 m2 at a prescribed
location and orientation, (b) Irradiation G2 on A2, and (c) Compute and plot G2 as a function of the
separation distance r2 for the range 0.25 ≤ r2 ≤ 1.0 m for zenith angles θ2 = 0, 30 and 60°.
SCHEMATIC:

ASSUMPTIONS: (1) Surface A1 emits diffusely, (2) A1 may be approximated as a differential surface
area and that A 2 r22 << 1.
ANALYSIS: (a) The rate at which emission from A1 is intercepted by A2 follows from Eq. 12.6 written
on a total rather than spectral basis.

q1→2 = Ie,1 (θ , φ ) A1 cosθ1dω 2 −1 .

(1)

Since the surface A1 is diffuse, it follows from Eq. 12.11 that

Ie,1 (θ , φ ) = Ie,1 = E1 π .

(2)

The solid angle subtended by A2 with respect to A1 is

dω 2 −1 ≈ A 2⋅ cos θ 2 r22 .

(3)

Substituting Eqs. (2) and (3) into Eq. (1) with numerical values gives
q1→ 2 =

E1

π

⋅ A1 cos θ1 ⋅

A 2 cos θ 2
2

r2

=

(

5 × 104 W m 2

π sr

)

(

× 10

−4

⎡ 5 × 10−4 m 2 × cos 30D ⎤
⎥ sr (4)
m × cos 60 × ⎢
⎢⎣
⎥⎦
( 0.5m )2
2

D

)

q1→ 2 = 15, 915 W m 2sr × 5 × 10 −5 m 2 × 1.732 × 10−3 sr = 1.378 × 10−3 W .

<

(b) From section 12.2.3, the irradiation is the rate at which radiation is incident upon the surface per unit
surface area,
q
1.378 × 10−3 W
= 2.76 W m 2
G 2 = 1→ 2 =
(5)
−4 2
A2
5 × 10 m
(c) Using the IHT workspace with the foregoing equations, the G2 was computed as a function of the
separation distance for selected zenith angles. The results are plotted below.

<

Continued...

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PROBLEM 12.2 (Cont.)

10

5

0
0.2

0.4

0.6

0.8

1

Separation distance, r2 (m)
theta2 = 0 deg
theta2 = 30 deg
theta2 = 60 deg

For all zenith angles, G2 decreases with increasing separation distance r2 . From Eq. (3), note that dω2-1
and, hence G2, vary inversely as the square of the separation distance. For any fixed separation distance,
G2 is a maximum when θ2 = 0° and decreases with increasing θ2, proportional to cos θ2.
COMMENTS: (1) For a diffuse surface, the intensity, Ie, is independent of direction and related to the
emissive power as Ie = E/ π. Note that π has the units of [sr ] in this relation.

(2) Note that Eq. 12.7 is an important relation for determining the radiant power leaving a surface in a
prescribed manner. It has been used here on a total rather than spectral basis.
(3) Returning to part (b) and referring to Figure 12.9, the irradiation on A2 may be expressed as

A cos θ1
G 2 = Ii,2 cos θ 2 1
r22
2

Show that the result is G2 = 2.76 W/m . Explain how this expression follows from Eq. (12.13).

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PROBLEM 12.3
KNOWN: Intensity and area of a diffuse emitter. Area and rotational frequency of a second surface,
as well as its distance from and orientation relative to the diffuse emitter.
FIND: Energy intercepted by the second surface during a complete rotation.
SCHEMATIC:

ASSUMPTIONS: (1) A1 and A2 may be approximated as differentially small surfaces, (2) A1 is a
diffuse emitter.
ANALYSIS: From Eq. 12.7, the rate at which radiation emitted by A1 is intercepted by A2 is

(

q1− 2 = Ie A1 cos θ1 ω 2 −1 = Ie A1 A 2 cosθ 2 / r 2

)

where θ1 = 0 and θ2 changes continuously with time. The amount of energy intercepted by both sides
of A2 during one rotation, ΔE, may be grouped into four equivalent parcels, each corresponding to
rotation over an angular domain of 0 ≤ θ2 < π/2. Hence, with dt = dθ2/ θ2 , the radiant energy
intercepted over the period T of one revolution is
T
0

ΔE = ∫ qdt =

ΔE =

π /2
4Ie A1 ⎛ A 2 ⎞ π / 2
4Ie A1 ⎛ A 2 ⎞
=
cos
d
sin
θ
θ
θ

2
2
2
0
θ2 ⎝ r 2 ⎠ ∫0
θ2 ⎝ r 2 ⎠

4 ×100 W / m 2 ⋅ sr × 10−4 m 2 ⎡⎢ 10−4 m 2 ⎤⎥
sr = 8 × 10−6 J
2
⎢ ( 0.50m ) ⎥

COMMENTS: The maximum rate at which A2 intercepts radiation corresponds to θ2 = 0 and is qmax
2
-6
= Ie A1 A2/r = 4 × 10 W. The period of rotation is T = 2π/ θ2 = 3.14 s.

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PROBLEM 12.4
KNOWN: Furnace with prescribed aperture and emissive power.
FIND: (a) Position of gauge such that irradiation is G = 1000 W/m2, (b) Irradiation when gauge is tilted
θd = 20o, and (c) Compute and plot the gage irradiation, G, as a function of the separation distance, L, for
the range 100 ≤ L ≤ 300 mm and tilt angles of θd = 0, 20, and 60o.
SCHEMATIC:

ASSUMPTIONS: (1) Furnace aperture emits diffusely, (2) Ad << L2.
ANALYSIS: (a) The irradiation on the detector area is defined as the power incident on the surface per
unit area of the surface. That is
G = q f →d Ad
q f → d = Ie A f cos θ f ω d − f
(1,2)

where q f → d is the radiant power which leaves Af and is intercepted by Ad. From Eqs. 12.2 and 12.7,
ω d − f is the solid angle subtended by surface Ad with respect to Af,

ω d − f = A d cos θ d L2 .

(3)

Noting that since the aperture emits diffusely, Ie = E/π (see Eq. 12.12), and hence

(

G = ( E π ) A f cos θ f A d cos θ d L2

(4)

Solving for L2 and substituting for the condition θf = 0o and θd = 0o,
L2 = E cos θ f cos θ d A f π G .

L = ⎢3.72 × 105 W m 2 ×

(5)
1/ 2

π

(20 × 10−3 ) 2 m 2 π × 1000 W m 2 ⎥
4

<

= 193 mm .

(b) When θd = 20o, qf→d will be reduced by a factor of cos θd since ωd-f is reduced by a factor cos θd.
Hence,

<

G = 1000 W/m2 × cos θd = 1000W/m2 × cos 20o = 940 W/m2 .
(c) Using the IHT workspace with Eq. (4), G is computed and plotted as a function of L for selected θd.
Note that G decreases inversely as L2. As expected, G decreases with increasing θd and in the limit,
approaches zero as θd approaches 90o.

3000

2000

1000

0
100

200

300

Separation distance, L (mm)

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PROBLEM 12.5
5

2

KNOWN: Radiation from a diffuse radiant source A1 with intensity I1 = 1.2 × 10 W/m ⋅sr is
incident on a mirror Am, which reflects radiation onto the radiation detector A2.
FIND: (a) Radiant power incident on Am due to emission from the source, A1, q1→m (mW), (b)
2
Intensity of radiant power leaving the perfectly reflecting, diffuse mirror Am, Im (W/m ⋅sr), and (c)
Radiant power incident on the detector A2 due to the reflected radiation leaving Am, qm→2 (μW), (d)
Plot the radiant power qm→2 as a function of the lateral separation distance yo for the range 0 ≤ yo ≤
0.2 m; explain features of the resulting curve.
SCHEMATIC:

ASSUMPTIONS: (1) Surface A1 emits diffusely, (2) Surface Am does not emit, but reflects perfectly
and diffusely, and (3) Surface areas are much smaller than the square of their separation distances.
ANALYSIS: (a) The radiant power leaving A1 that is incident on Am is

q1→ m = I1 ⋅ A1 ⋅ cosθ 1 ⋅ Δω m-1
where ωm-1 is the solid angle Am subtends with respect to A1, Eq. 12.2,

Δω m-1 ≡

dA n
r2

=

A m cos θ m
x 2o + y 2o

=

2 × 10−4 m2 ⋅ cos 45°
0.12 + 01
. 2 m2

= 7.07 × 10−3 sr

with θ m = 90°−θ 1 and θ 1 = 45° ,

q1→ m = 1.2 × 105 W / m2 ⋅ sr × 1 × 10-4 m2 × cos 45°×7.07 × 10-3 sr = 60 mW

<

(b) The intensity of radiation leaving Am, after perfect and diffuse reflection, is

b

g

I m = q1→ m / A m / π =

60 × 10−3 W

π × 2 × 10-4 m2

= 955
. W / m2 ⋅ sr

(c) The radiant power leaving Am due to reflected radiation leaving Am is

q m→2 = q 2 = I m ⋅ A m ⋅ cos θ m ⋅ Δω 2 − m
where Δω2-m is the solid angle that A2 subtends with respect to Am, Eq. 12.2,
Continued …..

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PROBLEM 12.5 (Cont.)
Δω 2 − m ≡

dA n
r2

=

b

A 2 cos θ 2
1 × 10−4 m2 × cos 45°
=
= 354
. × 10−3 sr
2
Lo − x o + y o2
0.12 + 01
. 2 m2

g

with θ2 = 90° - θm
q m → 2 = q 2 = 95.5 W/m 2 ⋅ sr × 2 × 10-4 m 2 × cos 45° × 3.54 × 10-3 sr = 47.8 μW

<

(d) Using the foregoing equations in the IHT workspace, q 2 is calculated and plotted as a function of
yo for the range 0 ≤ yo ≤ 0.2 m.

100

Em itted power from A1 reflected from Am onto A2

q2 (uW)

80

60

40

20

0
0

0.05

0.1

0.15

0.2

yo (m )

From the relations, note that q2 is dependent upon the geometric arrangement of the surfaces in the
following manner. For small values of yo, that is, when θ1 ≈ 0°, the cos θ1 term is at a maximum, near
unity. But, the solid angles Δωm-1 and Δω2-m are very small. As yo increases, the cos θ1 term doesn’t
diminish as much as the solid angles increase, causing q2 to increase. A maximum in the power is
reached as the cos θ1 term decreases and the solid angles increase. The maximum radiant power
occurs when yo = 0.058 m which corresponds to θ1 = 30°.

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PROBLEM 12.6
KNOWN: Flux and intensity of direct and diffuse components, respectively, of solar irradiation.
SCHEMATIC:

ANALYSIS: Since the irradiation is based on the actual surface area, the contribution due to the

G dir = q′′dir ⋅ cosθ .
From Eq. 12.17 the contribution due to the diffuse radiation is

Gdif = π Idif .
Hence

G = Gdir + Gdif = q′′dir ⋅ cosθ + π Idif
or

G = 1000 W / m 2 × 0.866 + π sr × 70 W / m 2 ⋅ sr
G = ( 866 + 220 ) W / m 2
or

G = 1086 W / m 2 .

<

COMMENTS: Although a diffuse approximation is often made for the non-direct component of
solar radiation, the actual directional distribution deviates from this condition, providing larger
intensities at angles close to the direct beam.

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PROBLEM 12.7
7

2

KNOWN: Daytime solar radiation conditions with direct solar intensity Idir = 2.10 × 10 W/m ⋅sr
-5
within the solid angle subtended with respect to the earth, ΔωS = 6.74 × 10 sr, and diffuse intensity
2
Idif = 70 W/m ⋅sr.
FIND: (a) Total solar irradiation at the earth’s surface when the direct radiation is incident at 30°, and
9
(b) Verify the prescribed value of ΔωS recognizing that the diameter of the earth is DS = 1.39 × 10 m,
11
and the distance between the sun and the earth is re-S = 1.496 × 10 m (1 astronomical unit).
SCHEMATIC:

Δ

ANALYSIS: (a) From Eq. 12.17 the diffuse irradiation is

G dif = π I dif = π sr × 70 W / m2 ⋅ sr = 220 W / m2
The direct irradiation follows from Eq. 12.13, expressed in terms of the solid angle

G dir = I dir cosθ Δω S

G dir = 2.10 × 107 W / m2 ⋅ sr × cos 30°×6.74 × 10-5 sr = 1226 W / m2
The total solar irradiation is the sum of the diffuse and direct components,

b

g

G S = G dif + G dir = 220 + 1226 W / m2 = 1446 W / m2

<

(b) The solid angle the sun subtends with respect to the earth is calculated from Eq. 12.2,

9 m 2 /4
139
10
π
.
×
2
dA n π DS / 4
=
=
= 6.74 × 10−5 sr
Δω S =
2
2
2
r
re-S
1496
.
× 1011 m

e
e

j

<

j

where dAn is the projected area of the sun and re-S, the distance between the earth and sun. We are
2 >> D 2 .
assuming that re-S
S
COMMENTS: Can you verify that the direct solar intensity, Idir, is a reasonable value, assuming that
4
the solar disk emits as a black body at 5800 K? I b,S = σTS4 / π = σ 5800 K / π

FH

b

g

j

= 2.04 × 107 W / m2 ⋅ sr . Because of local cloud formations, it is possible to have an appreciable
diffuse component. But it is not likely to have such a high direct component as given in the problem
statement.

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PROBLEM 12.8
KNOWN: Directional distribution of solar radiation intensity incident at earth’s surface on an
overcast day.
FIND: Solar irradiation at earth’s surface.
SCHEMATIC:

ASSUMPTIONS: (1) Intensity is independent of azimuthal angle θ.
ANALYSIS: Applying Eq. 12.15 to the total intensity

2π π / 2
Ii (θ ) cosθ sin θ d θ d φ
0
0

G=∫

G = 2π In ∫

π /2

0

cos 2 θ sin θ d θ

⎛ 1
⎞ π /2
G = ( 2 π sr ) × 80 W / m 2 ⋅ sr ⎜ − cos3 θ ⎟
⎝ 3
⎠ 0

π

G = −167.6 W / m 2 ⋅ sr ⎜ cos3 − cos3 0 ⎟
2

G = 167.6 W / m 2 .

<

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PROBLEM 12.9
KNOWN: Emissive power of a diffuse surface.
FIND: Fraction of emissive power that leaves surface in the directions π/4 ≤ θ ≤ π/2 and 0 ≤ φ ≤ π.
SCHEMATIC:

ASSUMPTIONS: (1) Diffuse emitting surface.
ANALYSIS: According to Eq. 12.10, the total, hemispherical emissive power is

∞ 2π π / 2
Iλ ,e ( λ ,θ , φ ) cos θ sin θ d θ d φ dλ .
0 0
0

E=∫

∫ ∫

For a diffuse surface Iλ,e (λ, θ, φ) is independent of direction, and as given by Eq. 12.12,

E = π Ie .
The emissive power, which has directions prescribed by the limits on θ and φ, is

⎡ π ⎤⎡ π /2

Iλ ,e ( λ ) dλ ⎢ dφ ⎥ ⎢
cos θ sin θ d θ ⎥
0
⎣ 0
⎦⎣ π /4

ΔE = ∫

π /2
⎡ sin 2 θ ⎤
⎡1

ΔE = Ie [φ ]0 × ⎢
= Ie [π ] ⎢ 1 − 0.707 2 ⎥

⎣2

⎢⎣ 2 ⎥⎦
π /4
π

(

)

ΔE = 0.25 π Ie .
It follows that

ΔE 0.25 π Ie
=
= 0.25.
E
π Ie

<

COMMENTS: The diffuse surface is an important concept in radiation heat transfer, and the
directional independence of the intensity should be noted.

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PROBLEM 12.10
KNOWN: Spectral distribution of Eλ for a diffuse surface.
FIND: (a) Total emissive power E, (b) Total intensity associated with directions θ = 0o and θ = 30o,
and (c) Fraction of emissive power leaving the surface in directions π/4 ≤ θ ≤ π/2.
SCHEMATIC:

ASSUMPTIONS: (1) Diffuse emission.
ANALYSIS: (a) From Eq. 12.9 it follows that

5

10

15

20

E = ∫ E λ (λ ) dλ = ∫ (0) dλ + ∫ (100) dλ + ∫ (200) dλ + ∫ (100) dλ + ∫ (0) dλ
0
0
5
10
15
20

E = 100 W/m2 ⋅μm (10 − 5) μm + 200W/m2 ⋅μm (15 − 10) μm + 100 W/m2 ⋅μm (20−15) μm

<

E = 2000 W/m2
(b) For a diffuse emitter, Ie is independent of θ and Eq. 12.12 gives

Ie =

E

π

=

2000 W m 2

π sr

<

Ie = 637 W m 2⋅ sr

(c) Since the surface is diffuse, use Eqs. 12.8 and 12.12,
E(π 4 → π 2)
E
E(π 4 → π 2)
E
E(π 4 → π 2)
E

π /2

∫ ∫π / 4
= 0

Ie cos θ sin θ dθ dφ

π Ie

π /2

∫ cos θ sin θ dθ ∫0
= π /4
π

=

=

π /2
1 ⎡ sin 2 θ ⎤

π ⎢⎣

2

φ 0

⎥⎦π / 4

1 ⎡1 2

(1 − 0.707 2 )(2π − 0) ⎥ = 0.50

π ⎣2

<

COMMENTS: (1) Note how a spectral integration may be performed in parts.

(2) In performing the integration of part (c), recognize the significance of the diffuse emission
assumption for which the intensity is uniform in all directions.

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PROBLEM 12.11
KNOWN: Diffuse surface ΔAo, 5-mm square, with total emissive power Eo = 4000 W/m2.
FIND: (a) Rate at which radiant energy is emitted by ΔAo, qemit; (b) Intensity Io,e of the radiation field
emitted from the surface ΔAo; (c) Expression for qemit presuming knowledge of the intensity Io,e beginning
with Eq. 12.10; (d) Rate at which radiant energy is incident on the hemispherical surface, r = R1 = 0.5 m,
due to emission from ΔAo; (e) Rate at which radiant energy leaving ΔAo is intercepted by the small area
ΔA2 located in the direction (40o, φ) on the hemispherical surface using Eq. 12.5; also determine the
irradiation on ΔA2; (f) Repeat part (e), for the location (0o, φ); are the irradiations at the two locations
equal? and (g) Irradiation G1 on the hemispherical surface at r = R1 using Eq. 12.5.
SCHEMATIC:

ASSUMPTIONS: (1) Diffuse surface, ΔAo, (2) Medium above ΔAo is also non-participating, (3)
R12 >> ΔA o , ΔA2.

ANALYSIS: (a) The radiant power leaving ΔAo by emission is

<

qemit = Eo⋅ΔAo = 4000 W/m2 (0.005 m × 0.005 m) = 0.10 W
(b) The emitted intensity is Io,e and is independent of direction since ΔAo is a diffuser emitter,

<

Io,e = E o π = 1273 W m 2⋅ sr

The intensities at points P1 and P2 are also Io,e and the intensity in the directions shown in the schematic
above will remain constant no matter how far the point is from the surface ΔAo since the space is nonparticipating.
(c) From knowledge of Io,e, the radiant power leaving ΔAo from Eq. 12.8 is,
2π π / 2
q emit = ∫ Io,e ΔA o cos θ sin θ dθ dφ = Io,e ΔAo ∫
cos θ sin θ dθ dφ = π Io,e ΔA o = 0.10 W
φ = 0 ∫θ = 0
h

<

(d) Defining control surfaces above ΔAo and on A1, the radiant power leaving ΔAo must pass through A1.
That is,

<

q1,inc = Eo ΔAo = 0.10 W
Recognize that the average irradiation on the hemisphere, A1, where A1 = 2π R12 , based upon the
definition, Section 12.2.3,
G1 = q1,inc A1 = E o Δ A o 2π R12 = 63.7 mW m 2

where q1,inc is the radiant power incident on surface A1.
Continued...

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PROBLEM 12.11 (Cont.)
(e) The radiant power leaving ΔAo intercepted by ΔA2, where ΔA2 = 4×10-6 m2, located at (θ = 45o, φ) as
per the schematic, follows from Eq. 12.7,

q ΔA →ΔA = Io,e ΔA o cos θ o Δω 2 − o
o
2
where θo = 45o and the solid angle ΔA2 subtends with respect to ΔAo is
Δω 2 − o = ΔA 2 cos θ 2 R12 = 4 × 10 −6 m 2 ⋅ 1 (0.5m) 2 = 1.60 × 10 −5 sr

where θ2 = 0o, the direction normal to ΔA2,

q ΔA →ΔA = 1273 W m2 ⋅ sr × 25 × 10−6 m2 cos 45o × 1.60 × 10−5 sr = 3.60 × 10−7 W
o
2

<

From the definition of irradiation, Section 12.2.3,

G 2 = q ΔA →ΔA ΔA 2 = 90 mW m 2
o
2
(f) With ΔA2, located at (θ = 0o, φ), where cosθo = 1, cosθ2 = 1, find
Δω 2 − o = 1.60 × 10 −5 sr

q ΔA →ΔA = 5.09 × 10−7 W
o
2

G 2 = 127 mW m 2

<

Note that the irradiation on ΔA2 when it is located at (0o, φ) is larger than when ΔA2 is located at (45o, φ);
that is, 127 mW/m2 > 90 W/m2. Is this intuitively satisfying?
(g) Using Eq. 12.13, based upon Figure 12.19, find
G1 = ∫ I1,i dA1 ⋅ dω 0 −1 A1 = π Io,e ΔA o ΔA1 = 63.7 mW m 2
h

<

where the elemental area on the hemispherical surface A1 and the solid angle ΔAo subtends with respect to
ΔA1 are, respectively,
dA1 = R12 sin θ dθ dφ

dω o −1 = ΔA o cos θ R12

From this calculation you found that the average irradiation on the hemisphere surface, r = R1, is
G1 = 63.7 mW m 2 . From parts (e) and (f), you found irradiations, G2 on ΔA2 at (0o, φ) and (45o, φ) as

127 mW/m2 and 90 mW/m2, respectively. Did you expect G1 to be less than either value for G2? How
do you explain this?
COMMENTS: (1) Note that from Parts (e) and (f) that the irradiation on A1 is not uniform. Parts (d)
and (g) give an average value.
(2) What conclusions would you reach regarding G1 if ΔAo were a sphere?

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PROBLEM 12.12
KNOWN: Hemispherical and spherical arrangements for radiant heat treatment of a thin-film material.
Heater emits diffusely with intensity Ie,h = 169,000 W/ m2⋅sr and has an area 0.0052 m2.
FIND: (a) Expressions for the irradiation on the film as a function of the zenith angle, θ, and (b) Identify
arrangement which provides the more uniform irradiation, and hence better quality control for the
treatment process.
SCHEMATIC:

ASSUMPTIONS: (1) Heater emits diffusely, (2) All radiation leaving the heater is absorbed by the thin
film.
ANALYSIS: (a) The irradiation on any differential area, dAs, due to emission from the heater, Ah ,
follows from its definition, Section 12.2.3,

q
G = h →s
dAs

(1)

Where qh→s is the radiant heat rate leaving Ah and intercepted by dAs. From Eq. 12.7,

q h →s = Ie,h ⋅ dA h cosθ h ⋅ ωs − h

(2)

where ωs-h is the solid angle dAs subtends with respect to any point on Ah. From the definition, Eq. 12.2,

ω=

dA n

(3)

r2

where dAn is normal to the viewing direction and r is the separation distance.
For the hemisphere: Referring to the schematic above, the solid angle is

ωs − h =

dAs
R2

and the irradiation distribution on the hemispheric surface as a function of θh is
G = Ie,h A h cos θ h R 2

(1)

<

For the sphere: From the schematic, the solid angle is

ωs,h =

dAs cos θ s
R o2

=

dAs
4R 2 cos θ h

where Ro, from the geometry of sphere cord and radii with θs = θh, is
Continued...

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PROBLEM 12.12 (Cont.)

R o = 2R cosθ h
and the irradiation distribution on the spherical surface as a function of θh is
G = Ie,h d A h 4R 2

(2)

<

(b) The spherical shape provides more uniform irradiation as can be seen by comparing Eqs. (1) and (2).
In fact, for the spherical shape, the irradiation on the thin film is uniform and therefore provides for better
quality control for the treatment process. Substituting numerical values, the irradiations are:

G hem = 169, 000 W m2⋅ sr × 0.0052m2 cosθ h ( 2m ) = 219.7 cosθ h W m2
2

(3)

2

Gsph = 169, 000 W m2⋅ sr × 0.0052 m 2 4 ( 2m ) = 54.9 W m 2

(4)

COMMENTS: (1) The radiant heat rate leaving the diffuse heater surface by emission is

q tot = π Ie,h A h = 2761W
The average irradiation on the spherical surface, Asph = 4πR2,

G sph = q tot Asph = 2761W 4π ( 2m ) = 54.9 W m 2
2

while the average irradiation on the hemispherical surface, Ahem = 2πR2 is

G hem = 2761W 2π ( 2m ) = 109.9 W m2
2

(2) Note from the foregoing analyses for the sphere that the result for Gsph is identical to that found as
Eq. (4). That follows since the irradiation is uniform.
(3) Note that G hem > Gsph since the surface area of the hemisphere is half that of the sphere.
Recognize that for the hemisphere thin film arrangement, the distribution of the irradiation is quite
variable with a maximum at θ = 0° (top) and half the maximum value at θ = 30°.

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PROBLEM 12.13
KNOWN: Hot part, ΔAp, located a distance x1 from an origin directly beneath a motion sensor at a
distance Ld = 1 m.
FIND: (a) Location x1 at which sensor signal S1 will be 75% that corresponding to x = 0, directly beneath
the sensor, So, and (b) Compute and plot the signal ratio, S/So, as a function of the part position x1 for the
range 0.2 ≤ S/So ≤ 1 for Ld = 0.8, 1.0 and 1.2 m; compare the x-location for each value of Ld at which
S/So = 0.75.
SCHEMATIC:

ASSUMPTIONS: (1) Hot part is diffuse emitter, (2) L2d >> ΔAp, ΔAo.
ANALYSIS: (a) The sensor signal, S, is proportional to the radiant power leaving ΔAp and intercepted

S ~ q p →d = I p,e ΔA p cos θ p Δω d − p

(1)

L
cos θ p = cos θ d = d = Ld (L2d + x12 )1/ 2
R

(2)

when

Δω d − p =

R2

= ΔAd ⋅ Ld (L2d + x12 )3 / 2

(3)

Hence,
q p → d = I p,e ΔA p ΔA d

L2d

(4)

(L2d + x12 ) 2

It follows that, with So occurring when x= 0 and Ld = 1 m,
L2 (L2d + x12 ) 2 ⎡ L2d ⎤
= d
=⎢

So L2 (L2 + 02 ) 2 ⎢ L2 + x 2 ⎥
⎣ d 1⎦
d
d

2

S

(5)

so that when S/So = 0.75, find,

<

x1 = 0.393 m
(b) Using Eq. (5) in the IHT workspace, the signal ratio, S/So, has been computed and plotted as a
function of the part position x for selected Ld values.

Continued...

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PROBLEM 12.13 (Cont.)
1

Signal ratio, S/So

0.8

0.6

0.4

0.2

0
0

1

2

Part position, x (m)
Sensor position, Ld = 0.8 m
Ld = 1 m
Ld = 1.2 m

When the part is directly under the sensor, x = 0, S/So = 1 for all values of Ld. With increasing x, S/So
decreases most rapidly with the smallest Ld. From the IHT model we found the part position x
corresponding to S/So = 0.75 as follows.
S/So
0.75
0.75
0.75

Ld (m)
0.8
1.0
1.2

x1 (m)
0.315
0.393
0.472

If the sensor system is set so that when S/So reaches 0.75 a process is initiated, the technician can use the
above plot and table to determine at what position the part will begin to experience the treatment process.

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PROBLEM 12.14

KNOWN: Surface area, and emission from area A1. Size and orientation of area A2.
FIND: (a) Irradiation of A2 by A1 for L1 = 1 m, L2 = 0.5 m, (b) Irradiation of A2 over the range 0
≤ L2 ≤ 10 m.

I1 = 1000 W/m2·sr

SCHEMATIC:

A1
x
θ1
L2 = 0.5 m
θ2

L1 = 1 m

A2

ASSUMPTIONS: Diffuse emission.
ANALYSIS: (a) The irradiation of Surface 1 is G1-2 = q1-2/A2 and from Example 12.1,
q1-2 = I1A1cosθ1ω2-1 = I1A1cosθ1A2cosθ2/r2
Since θ1 = θ2 = θ = tan-1(L1/L2) = tan-1(1/0.5) = 63.43° and r2 = L12 + L22 = (1m)2 + (0.5m)2 = 1.25 m2,
G1-2 = I1A1cos2θ/r2 = 1000W/m2⋅sr × 2 × 10-4 m2 × cos2(63.43°)/1.25m2 = 0.032 W/m2

<

(b) The preceding equations may be solved for various values of L2. The irradiation over
the range 0 ≤ L2 ≤ 10 m is shown below.
Irradiation of Surface 2 vs. Distance L2
0.06

G2(W/m^2)

0.04

0.02

0
0

2

4

6

8

10

L2(m)

COMMENTS: The irradiation is zero for L2 = 0 and L2 → ∞.

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PROBLEM 12.15

KNOWN: Intensities of radiating various surfaces of known areas.
FIND: Surface temperature and emitted energy assuming blackbody behavior.
E, qe

SCHEMATIC:

I(W/m2·sr)

A

1/ 4

⎛ ⎞
ANALYSIS: For blackbody emission, T = ⎜ ⎟
⎝σ⎠
E

and E = πI. Therefore,

1/ 4

⎛ πI ⎞
T=⎜ e⎟
⎝ σ ⎠

; qe = AE = AπIe

(1,2)

<

Equations (1) and (2) may be used to find T and qe as follows.

Problem

Ie (W/m2⋅sr)

A(m2)

T(K)

qe(W)

Example 12.1
Problem 12.3
Problem 12.5
Problem 12.12
Problem 12.14

7000
100
1.2 × 105
169,000
1000

10-3
10-4
10-4
0.0052
2 × 10-4

789
273
1606
1750
485

22
0.031
37.7
2761
0.628

COMMENTS: If the surface is not black, the intensity leaving the surface will include a
reflected component.

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courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
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PROBLEM 12.16
KNOWN: Diameter and temperature of burner. Temperature of ambient air. Burner efficiency.
FIND: (a) Radiation and convection heat rates, and wavelength corresponding to maximum spectral
emission. Rate of electric energy consumption. (b) Effect of burner temperature on convection and
SCHEMATIC:

ASSUMPTIONS: (1) Burner emits as a blackbody, (2) Negligible irradiation of burner from
surrounding, (3) Ambient air is quiescent, (4) Constant properties.
-6

2

PROPERTIES: Table A-4, air (Tf = 408 K): k = 0.0344 W/m⋅K, ν = 27.4 × 10 m /s, α = 39.7 ×
-6 2
-1
10 m /s, Pr = 0.70, β = 0.00245 K .
ANALYSIS: (a) For emission from a blackbody

(

)

q rad = A s E b = π D 2 / 4 σ T 4 = ⎡π ( 0.2m ) / 4 ⎤ 5.67 × 10−8 W / m 2 ⋅ K 4 ( 523 K ) = 133 W

⎣⎢

2

4

⎦⎥

3

2

<

-1

With L = As/P = D/4 = 0.05m and RaL = gβ(Ts - T∞) L /αν = 9.8 m/s × 0.00245 K (230 K)
3
-12 4 2
5
(0.05m) /(27.4 × 39.7 × 10 m /s ) = 6.35 × 10 , Eq. (9.30) yields

h=

(

)

k
⎛k⎞
4 = ⎛ 0.0344 W / m ⋅ K ⎞ 0.54 6.35 × 105 1/ 4 = 10.5 W / m 2 ⋅ K
Nu L = ⎜ ⎟ 0.54 Ra1/

L
L
0.05m
⎝L⎠

2
q conv = h As ( Ts − T∞ ) = 10.5 W / m 2 ⋅ K ⎡π ( 0.2m ) / 4 ⎤ 230 K = 75.7 W
⎥⎦
⎣⎢

<

The electric power requirement is then

q + q conv (133 + 75.7 ) W
=
= 232 W
0.9
η

<

The wavelength corresponding to peak emission is obtained from Wien’s law, Eq. (12.25)

λmax = 2898μ m ⋅ K / 523K = 5.54μ m

<

(b) As shown below, and as expected, the radiation rate increases more rapidly with temperature than

(

)

the convection rate due to its stronger temperature dependence Ts4 vs. Ts5 / 4 .
Continued …..

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PROBLEM 12.16(Cont.)
500

Heat rate (W)

400
300
200
100
0
100

150

200

250

300

350

Surface temperature (C)
qconv
Pelec

COMMENTS: If the surroundings are treated as a large enclosure with isothermal walls at Tsur = T∞
2

4
= 293 K, irradiation of the burner would be G = σ Tsur
= 418 W/m and the corresponding heat rate

would be As G = 13 W. This input is much smaller than the energy outflows due to convection and

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PROBLEM 12.17
KNOWN: Evacuated, aluminum sphere (D = 2m) serving as a radiation test chamber.
FIND: Irradiation on a small test object when the inner surface is lined with carbon black and at
600K. What effect will surface coating have?
SCHEMATIC:

ASSUMPTIONS: (1) Sphere walls are isothermal, (2) Test surface area is small compared to the
enclosure surface.
ANALYSIS: It follows from the discussion of Section 12.3 that this isothermal sphere is an enclosure
behaving as a blackbody. For such a condition, see Fig. 12.11(c), the irradiation on a small surface
within the enclosure is equal to the blackbody emissive power at the temperature of the enclosure.
That is

G1 = E b ( Ts ) = σ Ts4

G1 = 5.67 ×10−8 W / m 2 ⋅ K 4 ( 600K ) = 7348 W / m 2 .
4

<

The irradiation is independent of the nature of the enclosure surface coating properties.
COMMENTS: (1) The irradiation depends only upon the enclosure surface temperature and is
independent of the enclosure surface properties.
(2) Note that the test surface area must be small compared to the enclosure surface area. This allows
for inter-reflections to occur such that the radiation field, within the enclosure will be uniform
(diffuse) or isotropic.
(3) The irradiation level would be the same if the enclosure were not evacuated since, in general, air
would be a non-participating medium.

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PROBLEM 12.18
KNOWN: Isothermal enclosure of surface area, As, and small opening, Ao, through which 70W
emerges.
FIND: (a) Temperature of the interior enclosure wall if the surface is black, (b) Temperature of the
wall surface having ε = 0.15.
SCHEMATIC:

ASSUMPTIONS: (1) Enclosure is isothermal, (2) Ao << As.
ANALYSIS: A characteristic of an isothermal enclosure, according to Section 12.3, is that the radiant
power emerging through a small aperture will correspond to blackbody conditions. Hence

q rad = A o E b ( Ts ) = A o σ Ts4
where qrad is the radiant power leaving the enclosure opening. That is,
1/ 4

⎛q

⎝ Ao σ ⎠

1/ 4

70W
=⎜

⎜ 0.02m 2 × 5.670 × 10−8 W / m 2 ⋅ K 4 ⎟

= 498K.

<

Recognize that the radiated power will be independent of the emissivity of the wall surface. As long
as Ao << As and the enclosure is isothermal, then the radiant power will depend only upon the
temperature.
COMMENTS: It is important to recognize the unique characteristics of isothermal enclosures. See
Fig. 12.11 to identify them.

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PROBLEM 12.19
KNOWN: Sun has equivalent blackbody temperature of 5800 K. Diameters of sun and earth as well
as separation distance are prescribed.
FIND: Temperature of the earth assuming the earth is black.
SCHEMATIC:

ASSUMPTIONS: (1) Sun and earth emit as blackbodies, (2) No attenuation of solar irradiation
enroute to earth, and (3) Earth atmosphere has no effect on earth energy balance.
ANALYSIS: Performing an energy balance on the earth,

E in − E out = 0

Ae,p ⋅ GS = A e,s ⋅ E b ( Te )

(π De2 / 4) GS = π De2σ Te4
Te = ( GS / 4σ )

1/ 4

where Ae,p and Ae,s are the projected area and total surface area of the earth, respectively. To
determine the irradiation GS at the earth’s
surface, equate the rate of emission from the
sun to the rate at which this radiation passes
through a spherical surface of radius RS,e – De/2.

E in − E out = 0
2

π DS2 ⋅ σ TS4 = 4π ⎡⎣ R S,e − De / 2 ⎤⎦ GS

(

π 1.39 × 109 m

)

2

× 5.67 ×10−8 W / m 2 ⋅ K 4 ( 5800 K )

4
2

= 4π ⎡⎢1.5 × 1011 − 1.29 ×107 / 2 ⎤⎥ m 2 × GS

G S = 1377.5 W / m 2 .
Substituting numerical values, find

(

Te = 1377.5 W / m 2 / 4 × 5.67 × 10−8 W / m 2 ⋅ K 4

)

1/ 4

= 279 K.

<

COMMENTS: (1) The average earth’s temperature is greater than 279 K since the effect of the
atmosphere is to reduce the heat loss by radiation.

(2) Note carefully the different areas used in the earth energy balance. Emission occurs from the total
spherical area, while solar irradiation is absorbed by the projected spherical area.

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