14-1 The tungsten filament of a light bulb is at 2600 K.

Assuming the filament is black, calculate the wavelength

at which the maximum power is emitted. In what range of the electromagnetic spectrum does this
wavelength fall?

Approach:
Apply Wien’s displacement law.

Assumptions:
1. The filament is black.

Solution:
From Wien's displacement law
λmax T = 2898 µm ⋅ K
2898µm ⋅ K 2898µm ⋅ K
λmax = = = 1.12 µm Answer
T 2600 K
From Fig. 14-1, this is in the infrared range.

Comments:
Incandescent bulbs are inefficient, producing more heat (infrared) than light (visible).

14-2 Measurements of the spectrum of the star Vega show that radiation peaks at 0.29 µm . Neglecting the
Doppler shift of the star, estimate Vega's effective surface temperature.

Approach:
Apply Wien’s displacement law.

Assumptions:
1. The Doppler shift from Vega is small.
2. Vega radiates like a black body.

Solution:
From Wien's displacement law
λmaxT = 2898 µm ⋅ K
2898µm ⋅ K 2898µm ⋅ K
T= = = 9993 K Answer
λmax 0.29µm

Comments:
By comparison, our sun is cold at only 5800 K.

14 - 1
14-3 The planet Mars has an average radius of 3380 km and an average distance from the sun of 2.28 × 108 km.
The sun radiates like a black sphere with a radius of 6.95 × 108 m and a temperature of 5800 K. Assume
Mars is black and isothermal, with no atmosphere and no internal heat generation and calculate its average
surface temperature.

Approach:
Draw an imaginary sphere around the sun and
distribute the energy radiated from the sun
uniformly in all directions. Apply the first law to
Mars, balancing absorbed and emitted energy.

Assumptions:
1. The temperature of Mars does not change with
time.
2. The kinetic energy of Mars is constant.
3. Mars’ orbit is circular.
4. The sun radiates like a black body.
5. The sun is isothermal and spherical.
6. Mars does not produce internal heat.
7. The sun emits uniformly in all directions.
8. The sun’s rays are parallel in Mars orbit.
9. Mars is spherical.
10. Mars radiates like a black body.

Solution:
The total energy emitted by the sun in all directions is:
Q s = Asσ Ts 4 = 4π Rs 2σ Ts 4
where Rs is the radius of the sun.
2⎛ W ⎞
Q s = 4π ( 6.95 × 108 m ) ⎜ 5.67 × 10−8 2 4 ⎟ ( 5800 K ) = 3.89 × 1026 W
4

⎝ m ⋅ K ⎠
Draw an imaginary sphere of radius S around the sun, where S is the distance between the sun and Mars. The sun's
rays are essentially parallel by the time they reach Mars orbit. The interaction of Mars and the imaginary sphere
around the sun is a circle with the radius of Mars. To find the heat flux on this circle (also called the projected area
of Mars), compute the heat flux on the inner surface of the imaginary sphere as
Q s A s 3.89 × 1026 W W
qs " = = = = 5.95 × 108
Asphere 4π S 4π ( 2.28 × 10 km )
2 2 2
8 km
The energy absorbed by Mars is:
Q abs = π Rm 2 qs "
Mars emits over its actual surface area, not its projected area. The energy emitted by Mars is
Q emit = Amσ Tm 4 = 4π Rm 2σ Tm 4
Since there is no heat generated by Mars, energy absorbed equals energy emitted and
Qabs = 4π Rm 2σ Tm 4
1
⎡ 12 W ⎛ 1km ⎞
⎤
2 4
1 1 1 ⎢ 5.95 × 10 ⎜ ⎟ ⎥

⎡ Qabs ⎤ 4 ⎡ π Rm qs " ⎤
2 4
⎡ qs " ⎤ 4 ⎢
2
km ⎝ 1000 m ⎠ ⎥
Tm = ⎢ 2 ⎥
=⎢ ⎥ =⎢ ⎥ =⎢
⎣ 4π Rm σ ⎦ ⎣ 4π Rm σ ⎦
2
⎣ 4σ ⎦ ⎛ W ⎞ ⎥
⎢ 4 ⎜ 5.67 × 10−8 2 4 ⎟ ⎥
⎢⎣ ⎝ m ⋅K ⎠ ⎥
⎦
Tm = 226K = −46.7 o C Answer

Comments:
The actual average surface temperature of Mars is about –55oC. There is a feeble greenhouse effect on Mars, which
has a thin atmosphere and there is evidence of significant volcanic activity.

14 - 2
14-4 Radiation from the sun incident on the earth's outer atmosphere has been measured as 1353 W/m2. The
average distance between the sun and the earth is 1.5x1011 m. If the planet Pluto is at a distance from the
sun of 5.87x109 km, find the solar flux incident on Pluto.

Approach:
Draw two imaginary spheres around the sun, one
intersecting the earth and the other intersecting Pluto.
Equate the total energy impinging on the earth sphere
with that impinging on the Pluto sphere.

Assumptions:
1. Pluto’s orbit is circular.
2. Earth’s orbit is circular.
3. Space is a perfect vacuum.
4. The sun emits uniformly in all directions.

Solution:
Draw two imaginary spheres around the sun —one at sun-earth radius and the other at sun-Pluto radius.
Assuming space is a perfect vacuum, the total solar energy impinging on the smaller imaginary sphere equals that
impinging on the larger imaginary sphere. Therefore
qe " Ae = q p " Ap
where qe" is the solar flux at the earth orbit and Ae is the surface area of the smaller imaginary sphere around the
sun, while qp" and Ap are the corresponding values in Pluto orbit.
qe "4π Re 2 = q p "4π R p 2
2
⎛R ⎞
q p " = qe " ⎜ e ⎟⎟
⎜R
⎝ p ⎠
2
⎡ ⎛ 1km ⎞ ⎤
⎢1.5 × 10 m ⎜
11
⎟⎥
⎛ W⎞ ⎝ 1000 m ⎠ ⎥
q p = ⎜ 1353 2 ⎟ ⎢
⎝ m ⎠⎢ 5.87 × 109 km ⎥
⎢ ⎥
⎣⎢ ⎦⎥
W
= 0.884 Answer
m2

Comments:
The orbit of Pluto is quite elliptic, so this value varies considerably over the Plutonian year.

14 - 3
14-5 A furnace has an inside area of 320 ft2 and black walls. A radiant power of 95 W issues from a rectangular
opening in the furnace wall. If the opening measures 5 in. by 6 in., determine the interior wall temperature.
If the emissivity of the walls is 0.88, what is the wall temperature?

Approach:
Model the furnace as a cavity and treat the
opening as a black surface at the
temperature of the interior walls.

Assumptions:
1. The radiation issuing from the furnace is
black.

Solution:
The opening is small compared to the size of the furnace interior; therefore, the furnace acts like a black cavity.
The heat flux escaping from the furnace is
Q = Aσ T 4
1
⎡ ⎛ 39.37 in. ⎞
2
⎤4
1 ⎢ 95 W ⎜ ⎟ ⎥
⎡ Q ⎤ 4 ⎢ ⎝ 1m ⎠ ⎥
T =⎢ ⎥ =⎢ ⎥ = 542 K = 269 C
o
Answer
⎣ σ ⎦ ⎛ W ⎞
⎢ ( 5in.)( 6in.) ⎜ 5.67 × 10
A −
⎟⎥
8

⎢⎣ ⎝ m2 ⋅ K ⎠ ⎥
⎦

Comments:
If the emissivity of the furnace walls is 0.88, the temperature is unchanged. The furnace still acts like a black
emitter, since the interior is so large compared to the opening.

14-6 A square window of side length 12 cm is located in the center of an oven wall measuring 35 cm by 45 cm.
The depth of the oven is 50 cm. Considering all the oven walls as one surface and the glass window as a
second surface, find the view factor from the oven walls to the window.

Approach:
The view from the window to the walls is
unity by inspection. Reciprocity can be used
to find the remaining view factor.

Assumptions:
1. The radiosity is uniform over each surface.
2. All surfaces are diffuse.

Solution:
Choose the interior of the oven as the enclosure. The view factor from surface 1, the window, to surface 2, all the
other walls of the enclosure, is unity by inspection. Therefore
F1→2 = 1
From reciprocity
A1 F1→2 = A2 F2→1

F2→1 =
A1 F1→2
=
(12 )(12 )(1) = 9.14 × 10−6 Answer
A2 ( 35 )( 45 )( 2 ) + ( 35 )( 50 )( 2 ) + ( 45)( 50 )( 2 )
Comments:
The view factor is very small. If the window were replaced by an opening, the radiation issuing from the opening
would be black radiation.

14 - 4
14-7 Two square plates of size 8 cm by 8 cm are directly opposite each other in parallel planes. What should the
spacing between the plates be so that the view factor from one to the other is 0.5?

Approach:
Use the equation in Fig. 14-16. Iterate to find the
correct value of plate spacing.

Assumptions:
1. The radiosity is uniform over each surface.
2. Both surfaces are diffuse.

Solution:
Use the equation in Fig. 14-16 with
X 8cm Y 8cm
X= = Y = =
L L L L
⎧ 1
⎫
⎪ (⎡ + )( + ) ⎤
2 2
⎪
2
1 X 1 Y 1 1
⎥ + X (1 + Y ) tan + Y (1 + X ) tan
X Y
2 ⎪ln ⎢
2 2 −1 2 2 −1
1 ⎪
F1→2 = 0.5 = ⎨ ⎣⎢ 1 + X + Y ⎥⎦
1

(1 + X 2 ) 2 ⎪⎬
2 2

π XY ⎪ (1 + Y 2 ) 2
⎪ − X tan −1 X − Y tan −1 Y ⎪
⎩ ⎭
The value of L needed so that F1→2 = 0.5 must be found iteratively. It is recommended that equation-solving
software be used for this purpose. The result is
L = 3.04 cm Answer

Comments:
It is good practice to check the result using the graph in Fig. 14-16. Note that the output of the arctangent must be
in radians.

14 - 5
14-8 A cryogenic dewer consists of two concentric cylinders. The inner cylinder has a length of 2.2 ft and a
diameter of 8 in. The outer cylinder has a length of 2.6 ft and a diameter of 10 in. The space between the
cylinders is evacuated. Find the view factor from the outer to the inner cylinder. Include the end area in the
calculation.

Approach:
Note that F1→2 = 1 by inspection. Find the
remaining view factor by reciprocity.

Assumptions:
1. The radiosity is uniform over each
surface.
2. All surfaces are diffuse.

Solution:
We must find F2→1 , where surface 2 is the
inner surface of the large cylinder and surface
1 is the outer surface of the small cylinder.
By inspection
F1→2 = 1
since all the energy leaving 1 arrives at 2. From reciprocity
A1 F1→2 = A2 F2→1
⎡ ⎛8 ⎞ ⎛ 8 ⎞ ⎤
2

⎢π ⎜ ft ⎟ ( 2.2 ft ) + 2π ⎜
⎜ 2 (12 ) ft ⎟⎟ ⎥ (1)
⎥
⎢ ⎝ 12 ⎠ ⎝ ⎠
=⎣ ⎦
A1 F1→2
F2→1 =
A2 ⎡ ⎛ 10 ⎞ ⎛ 10 ⎞ ⎤
2

⎢π ⎜ ft ⎟ ( 2.6 ft ) + 2π ⎜ ft ⎟⎟ ⎥
⎜
⎢ ⎝ 12 ⎠
⎣ ⎝ 2 (12 ) ⎠ ⎦⎥
F2→1 = 0.672 Answer

Comments:
The view factor from the outer cylinder to itself is F2→2 = 1 − F1→2 = 0.328

14 - 6
14-9 The inside of a sphere is divided into two hemispherical surfaces. Find the view factor from one
hemisphere to the other.

Approach:
Construct an imaginary surface bisecting the
sphere. Determine view factor using
inspection and reciprocity.

Assumptions:
1. The radiosity is uniform over each
surface.
2. All surfaces are diffuse.

Solution:
Construct imaginary surface 1which divides the sphere in half, as shown. Surface 1 is an infinitesimally thin disk.
Surfaces 2 and 3 are the insides of the two hemispheres. We need F2→3 . All the radiation that leaves 2 and arrives
at 3 must cross 1. Therefore
F2→3 = F2→1
Consider an enclosure including the top of surface 1 and surface 2. By inspection
F1→2 = 1
since all the energy leaving the top of 1 arrives at 2. From reciprocity
A1 F1→2 = A2 F2→1
A1 F1→2 π R (1)
2

F2→1 = = = 0.5
A2 2π R 2
Therefore F2→1 = F2→3 = 0.5 Answer

Comment:
One could also deduce the same result by symmetry.

14 - 7
14-10 A room in an art gallery has a floor area of 16 ft by 16 ft and 10 ft high ceilings. One wall is part of the
exterior of the building and the other three are interior walls. In winter, the exterior wall is cooler than the
other three and looses heat to the environment by radiation and convection. Calculate the view factor from
the floor to the exterior wall and from each interior wall to the exterior wall.

Approach:
Use Fig. 14-17 and 14-16 to find some of the view
factors. Then use summation to find the rest.

Assumptions:
1. The radiosity is uniform over each surface.
2. All surfaces are diffuse.

Solution:
Let surface 1 be the exterior wall, surface 2 be the
floor, surface3 be the ceiling, surface 4 be the wall
opposite the exterior wall, and surfaces 5 and 6 be
the two walls adjacent to the exterior wall, as
shown. First find F1→2 . Use Fig 14-17, defining
X = 16 Y = 16 Z = 10
Z 10 Y 16
Then H = = = 0.625 W = = =1
X 16 X 16
From the chart in Fig. 14-17a, F2→1 ≈ 0.16. By symmetry F3→1 = F2→1 = 0.16.
Next, find the view factor from the opposite wall to the exterior wall, F4→1. Use Figure 14-16 with
X 10 Y 16
X= = = 0.625 Y = = =1
L 16 L 16
From the graph, F4→1 ≈ 0.14. To find the last view factor, note that
F1→2 + F1→3 + F1→4 + F1→5 + F1→6 = 1
By symmetry F1→5 = F1→6 , so
F1→2 + F1→3 + F1→4 + 2 F1→5 = 1
Apply reciprocity to each term to get
A2 F2→1 A3 F3→1 A4 F4→1 2 A5 F5→1
+ + + =1
A1 A1 A1 A1
Solving for F5→1

F5→1 =
(16 )(10 ) − (16 )(16 )( 0.16 ) − (16 )(16 )( 0.16 ) − (10 )(16 )( 0.14 ) = 0.174
2 (16 )(10 )
In summary:
F2→1 = 0.16
F3→1 = 0.16
F4→1 = 0.14 Answer
F5→1 = 0.174
F6→1 = 0.174

Comments:
For greater accuracy, use the equations in Figs. 14-16 and 14-17. Note that the output of the arc tangent must be
in radians.

14 - 8
14-11 During a chemical-vapor deposition process, a disk-shaped silicon substrate 2.3 cm in diameter is placed in
a reaction chamber. The chamber is cylindrical with a height of 13 cm and a diameter of 5 cm. The silicon
substrate rests on the bottom surface, as shown. Find the view factor from the curved side wall of the
chamber (surface 1) to the substrate (surface 2).
Approach:
Use Figure 14-18 to find F2→3 . Then use
summation to calculate F2→1. Finally use
reciprocity to determine F1→2 .

Assumptions:
1. The radiosity is uniform over each
surface.
2. All surfaces are diffuse.

Solution:
By the summation relation:
F2→1 + F2→3 = 1
Note that surface 2 cannot “see” any portion of the bottom surface, so views to that surface need not be included
in the summation relation. To find F2→3 , use the equation in Fig. 14-18 with
r2 1.15 cm
Ri = R2 = = = 0.0885
L 13 cm
r3 2.5 cm
R j = R3 = = = 0.192
L 13 cm
1 + ( 0.192 )
2
1 + R32
S =1+ = 1+ = 133
( 0.0885 )
2 2
R2
where L is the height of the cylinder. The view factor equation is
1 ⎧⎪ ⎫⎪
1

F2→3 = ⎨ S − ⎡ S 2 − 4 ( r3 / r2 ) ⎤ 2 ⎬
2

2 ⎩⎪ ⎣ ⎦ ⎪
⎭
Substituting values:
1 ⎪⎧ ⎪⎫
1

F2→3 = ⎨133 − ⎡1332 − 4 ( 2.5 /1.15 ) ⎤ 2 ⎬ = 0.0354

2

2 ⎪⎩ ⎣ ⎦ ⎪⎭
Thus
F2→1 = 1 − F2→3 = 1 − 0.0354 = 0.965
From reciprocity
A2 F2→1 π r22 F2→1 r22 F2→1 (1.15cm ) ( 0.965 )
2

F1→2 = = = =
A1 2π r3 L 2r3 L 2 ( 2.5cm )(13 cm )

F1→2 = 0.0196 Answer

Comments:
Because the view factor F2→3 was small, it would have been difficult to read it accurately from the chart. Using the
equation was a better approach.

14 - 9
14-12 Find the view factors between all the surfaces within a cubical enclosure.

Approach:
Use Figure 14-16 to find the view factor
between two opposite walls. Then use
summation and symmetry to determine all
other view factors.

Assumptions:
1. The radiosity is uniform over each
surface.
2. All surfaces are diffuse.

Solution:
Define surfaces 1 and 2 to be two directly opposite sides of the cube, as shown. Define surface 3 to be a side
adjacent to surface 1, shown as the bottom surface in the figure. Surfaces 4, 5,and 6 (not shown) are the remaining
three sides. To find F1→2 , use the graph in Fig. 14-16, with
X H Y H
= = 1 and = =1
L H L H
From the graph
F1→2 ≈ 0.2
From the summation relation
F1→2 + F1→3 + F1→4 + F1→5 + F1→6 = 1
Furthermore, by symmetry
F1→3 = F1→4 = F1→5 = F1→6
Thus
F1→2 + 4 F1→3 = 1
1 − F1→2 1 − 0.2
F1→3 = = = 0.2
4 4
In summary
F1→2 = F1→3 = F1→4 = F1→5 = F1→6 = 0.2 Answer

14 -10
14-13 A flat panel heater is suspended over a warming tray, as shown. Calculate the view factor from the heater to
the tray.

Approach:
Create two artificial surfaces, 1b and 1c, as
shown. Break up the warming tray into three
surfaces. Use shape decomposition to find the
desired view factor in terms of the view factor
for two aligned rectangles in parallel planes.

Assumptions:
1. The radiosity is uniform over each surface.
2. All surfaces are diffuse.

Solution:
By shape decomposition:
F1a→2 = F *1a→2 a + F1a→2b + F1a→2 c
By symmetry F1a→2b = F1a→2c , therefore
F1a→2 = F *1a→2 a + 2 F1a→2 c Eq.(1)
An additional shape decomposition will be helpful:
F *( 2 a+2 c )→(1a+1c ) = F( 2 a+2 c )→1a + F( 2 a+2 c )→1c
By symmetry F( 2 a+2 c )→1a = F( 2 a+2 c )→1c , so
F *( 2 a+2 c )→(1a+1c ) = 2 F( 2 a+2 c )→1a Eq.(2)
By reciprocity
A1a F1a→( 2 a+2 c )
F( 2 a+ 2c )→1a =
A( 2 a+ 2c )
Since A1a = A( 2 a+ 2c ) / 2
F1a→( 2 a+ 2c )
F( 2 a+ 2c )→1a =
2
Substituting into Eq.(2) yields
F *( 2 a+2 c )→(1a+1c ) = F1a→( 2 a+2c )
Again applying shape decomposition
F *( 2 a+2 c )→(1a+1c ) = F *1a→2 a + F1a→2 c
F1a→2 c = F *( 2 a+2 c )→(1a+1c ) − F *1a→2 a
Substituting this into Eq.(1) gives
F1a→2 = F *1a→2 a + 2 ⎡⎣ F *( 2 a+ 2c )→(1a+1c ) − F *1a→2 a ⎤⎦ = 2 F *( 2 a+2 c )→(1a+1c ) − F *1a→2 a
To find F *( 2 a+2 c )→(1a+1c ) , use Fig. 14-16 with
16in. 10in.
X= = 2.67 Y= = 1.67
6in. 6in.
F *( 2 a+2 c )→(1a+1c ) ≈ 0.4
To find F *1a→2 a , again use Fig 14-16, this time with
8in. 10in.
X= = 1.33 Y= = 1.67
6in. 6in.
F *1a→2 a ≈ 0.3
Substituting:
F1a→2 = 2 ( 0.4 ) − 0.3 = 0.5 Answer

14 -11
14-14 Using data on the figure below, find the view factor, F3→1 . Surfaces 2 and 3 are perpendicular at their
common edge.

Approach:
Construct surface 1+2, which is the combination
of 1 and 2. The view factor from 1+2 to 3 can
then be calculated using Fig. 14-17. Fig. 14-17
can also be used to find F2→3 . Apply shape
decomposition and reciprocity to find the desired
view factor, F3→1.

Assumptions:
1. The radiosity is uniform over each surface.
2. All surfaces are diffuse.

Solution:
Designating 1+2 as the combination of surfaces l and 2, the view factor from 3 to 1+2 may be decomposed into
F *3→(1+2) = F3→1 + F *3→2
F3→1 = F *3→(1+2) − F *3→2
To find F *3→2 , use Fig. 14-17 with
1.2 3
H= = 0.3 W= = 0.75
4 4
From the plot
F *3→2 ≈ 0.14
To find F *3→(1+2) , again use Fig. 14-17 with
1.2 + 1.8 3
H= = 0.75 W= = 0.75
4 4
From the plot
F *3→(1+2) ≈ 0.23
Finally
F3→1 = 0.23 − 0.14 = 0.09 Answer

Comment:
Greater accuracy could have been obtained by using the equation in Fig. 14-20 rather than the plot.

14 -12
14-15 Two flat rings of equal size in parallel planes exchange heat by radiation. Using data on the figure,
calculate the view factor from one ring to the other.

Approach:
Construct surfaces 3 and 4 as shown. Use
shape decomposition and the formula for
two disks in parallel planes from Fig. 14-18.

Assumptions:
1. The radiosity is uniform over each
surface.
2. All surfaces are diffuse.

Solution:
Designating 1+3 as the combination of surfaces 1 and 3, and indicating 2+4 similarly, we may use shape
decomposition to obtain
F *(1+3)→(2+4) = F(1+3)→2 + F *(1+3)→4
Where the * superscript indicates a quantity that can be calculated from Fig. 14-18. Rearranging:
F(1+3)→2 = F *(1+3)→(2+4) − F *(1+3)→4
From reciprocity
A(1+3) F(1+3)→2 A(1+3) ( F *(1+3)→(2+4) − F *(1+3)→4 )
F2→(1+3) = = eq.(1)
A2 A2
Again applying shape decomposition:
F2→(1+3) = F2→1 + F2→3 eq.(2)
We can find the desired view factor, F2→1 , if we can calculate F2→3 . To accomplish that, begin with
F *3→(2+4) = F3→2 + F *3→4
F3→2 = F *3→(2+4) − F *3→4
By reciprocity
A3 F3→2 A3 ( F 3→(2+4) − F 3→4 )
* *

F2→3 = = eq.(3)
A2 A2
F *3→4 may be calculated using the equation in Fig. 14-18 with
1 + (1.667 )
2
5 5
R3 = = 1.667 R4 = = 1.667 S =1+ = 2.36
(1.667 )
2
3 3
⎧ 2 2⎫
1

1 ⎪ ⎡ ⎛ 5 ⎞ ⎤ ⎪
F *3→4 = ⎨2.36 − ⎢ 2.362 − 4 ⎜ ⎟ ⎥ ⎬ = 0.554
2⎪ ⎢⎣ ⎝ 5 ⎠ ⎥⎦ ⎪
⎩ ⎭
*
Similarly for F 3→2+4
1 + ( 2.667 )
2
5 8
R3 = = 1.667 R(2+4) = = 2.667 S =1+ = 3.92
(1.667 )
2
3 3
⎧ 2 2⎫
1

1⎪ ⎡ ⎛ 8 ⎞ ⎤ ⎪
F *3→(2+ 4) = ⎨3.92 − ⎢3.922 − 4 ⎜ ⎟ ⎥ ⎬ = 0.828
2⎪ ⎢⎣ ⎝ 5 ⎠ ⎥⎦ ⎪
⎩ ⎭
From eq.(3),
π ( 5 ) ( 0.828 − 0.554 )
2

F2→3 = = 0.176
π ⎡( 8 ) − ( 5 ) ⎤
2 2
⎣ ⎦
*
We must also evaluate F (1+3)→(2+4) . In this case

14 -13
 1 + ( 2.667 )
2
8 8
R(1+3) = = 2.667 R(2+4) = = 2.667 S =1+ = 2.14
( 2.667 )
2
3 3
⎧ 1
⎫
1⎪ ⎡ ⎛ 8 ⎞ ⎤2 ⎪
2

F (1+3)→(2+ 4) = ⎨2.14 − ⎢ 2.14 − 4 ⎜ ⎟ ⎥ ⎬ = 0.689

* 2

2⎪ ⎣⎢ ⎝ 8 ⎠ ⎦⎥ ⎪
⎩ ⎭
*
The last view factor needed is F (1+3)→4 . It is calculated using
1 + (1.667 )
2
8 5
R(1+3) = = 2.667 R4 = = 1.667 S =1+ = 1.53
( 2.667 )
2
3 3
⎧ 2 2⎫
1

1 ⎪ ⎡ ⎛ 5 ⎞ ⎤ ⎪
F *(1+3)→4 = ⎨1.53 − ⎢1.532 − 4 ⎜ ⎟ ⎥ ⎬ = 0.323
2⎪ ⎢⎣ ⎝ 8 ⎠ ⎥⎦ ⎪
⎩ ⎭
Substituting values into eq. (1):
π ( 8 ) ( 0.689 − 0.323)
2

F2→(1+3) = = 0.600
π ( 82 − 5 2 )
Solving eq. (2) for F2→1 ,
F2→1 = F2→(1+3) − F2→3 = 0.600 − 0.176 = 0.424 Answer
The view factor F1→2 is identical to F2→1 by symmetry.

14 -14
14-16 A tubular enclosure is divided into four surfaces for the purpose of a radiation analysis. Surfaces 1 and 2
are on the inner wall of the tube and surfaces 3 and 4 are on the ends. Find the view factor F1→ 2 .

Approach:
The secret of this problem is to construct
surface 5 as shown. Then consider first the
enclosure formed from 1, 3 and 5 and
calculate F1→3 and F1→1 . Next, remove
surface 5 and find the remaining view
factors in the enclosure formed from 1, 2, 3,
and 4.

Assumptions:
1. The radiosity is uniform over each
surface.
2. All surfaces are diffuse.

Solution:
Construct an artificial surface, 5, as shown, and consider the enclosure formed by surfaces 1, 3, and 5. To
find F3→5 , use the equation in Fig. 14-18 with
L
R3 = 2 = 0.5 = R5
L
1 + ( 0.5 )
2
1 + R52
S =1+ = 1+ =6
( 0.5)
2 2
R3
⎧ 2 2⎫
1

1⎪ ⎡ ⎛ r ⎞ ⎤ ⎪
F3→5 = ⎨ S − ⎢ S 2 − 4 ⎜ 5 ⎟ ⎥ ⎬
2⎪ ⎢⎣ ⎝ r3 ⎠ ⎥⎦ ⎪
⎩ ⎭
1⎪ ⎧ 1
⎫
⎪
F3→5 = ⎨6 − ⎡62 − 4 (1) ⎤ 2 ⎬ = 0.1716
2

2 ⎩⎪ ⎣ ⎦ ⎪
⎭
The view factor F3→1 may now be found by summation as
F3→1 = 1 − F3→5 = 1 − 0.1716 = 0.8284
Applying reciprocity
2
⎛L⎞
π ⎜ ⎟ F3→1
F1→3 =
A3 F3→1
= ⎝ ⎠
2
=
F3→1
= 0.2071
A1 π L ( L) 4
Now apply summation to surface 1 to get (note that surface 1 can see itself)
F1→1 = 1 − F1→3 − F1→5
Because of symmetry
F1→3 = F1→5
F1→1 = 1 − 2 F1→3 = 1 − 2 ( 0.2071) = 0.5858
Next, remove surface 5 and consider the enclosure formed by surfaces 1, 2, 3, and 4. The view factor F3→4 can be
found using Fig. 14-18 with
L
R3 = 2 = 0.25 = R4
2L
1 + ( 0.25 )
2

S =1+ = 18
( 0.25 )
2

14 -15
 ⎧ 2 2⎫
1

1⎪ ⎡ ⎛ r ⎞ ⎤ ⎪
F3→4 = ⎨ S − ⎢ S 2 − 4 ⎜ 4 ⎟ ⎥ ⎬
2⎪ ⎢⎣ ⎝ r3 ⎠ ⎥⎦ ⎪
⎩ ⎭
1 ⎪⎧ ⎪⎫
1

F3→4 = ⎨18 − ⎡182 − 4 (1) ⎤ 2 ⎬ = 0.05573

2

2 ⎪⎩ ⎣ ⎦ ⎪⎭
From summation
F3→2 = 1 − F3→4 − F3→1
= 1 − 0.05573 − 0.8284
= 0.1158
By reciprocity
2
⎛L⎞
π ⎜ ⎟ F3→2
F2→3 =
A3 F3→2
= ⎝ ⎠
2
=
F3→2
= 0.02896
A2 π L ( L) 4
By symmetry
F2→2 = F1→1 = 0.5858
F1→4 = F2→3 = 0.02896
By summation
F1→1 + F1→2 + F1→3 + F1→4 = 1
Finally
F1→2 = 1 − F1→1 − F1→3 − F1→4
F1→2 = 1 − 0.5858 − 0.2071 − 0.02896 = 0.1781 Answer

14 -16
14-17 Two rectangular surfaces face each other in parallel planes as shown in the figure. The top surface is
aligned directly above the bottom surface and both surfaces have the same size. Each surface is divided into
two equal parts, so that A1 = A2 = A3 = A4 . Find the view factor F1→4 .

Approach:
The view factor from the combination of 1
and 2 to 3 and 4 can be found from Fig. 14-
16. Repeatedly apply shape decomposition,
symmetry, and reciprocity to determine the
unknown view factor.

Assumptions:
1. The radiosity is uniform over each
surface.
2. All surfaces are diffuse.

Solution:
By shape decomposition
F *(1+2)→(3+ 4) = F(1+2)→3 + F(1+ 2)→4
By symmetry, this reduces to
F *(1+2)→(3+4) = 2 F(1+2)→3
Applying reciprocity and noting that A3 = A(1+ 2) / 2,
2A3 F3→(1+2)
F *(1+2)→(3+4) = = F3→(1+2)
A(1+2)
Again using shape decomposition
F *(1+2)→(3+4) = F *3→1 + F3→2
By symmetry
F3→2 = F1→4 = F *(1+ 2)→(3+4) − F *3→1
To find F *(1+2)→(3+ 4) , use Fig. 14-16 with
X 3 Y 2
X= = = 0.75 Y = = = 0.5
L 4 L 4
From the chart, F *(1+2)→(3+4) ≈ 0.095. Similarly for F *3→1 , use
X 3 Y 1
X= = = 0.75 Y = = = 0.25
L 4 L 4
From the chart, F *3→1 ≈ 0.05. Therefore
F1→4 = 0.095 − 0.05 = 0.045 Answer

14 -17
14-18 Three radiation zones are used in an oven analysis. Surfaces 1 and 2 are placed on the side wall and surface
3 is placed on the bottom of the oven. Using dimensions on the figure, calculate F1→3 .

Approach:
Use shape decomposition, symmetry, and
reciprocity.

Assumptions:
1. The radiosity is uniform over each
surface.
2. All surfaces are diffuse.

Solution:
By shape decomposition
F *3→(1+2) = F3→1 + F3→2
By symmetry,
F *3→(1+ 2) = 2 F3→1
Using reciprocity
2A1 F1→3
F *3→(1+ 2) =
A3
Solving for F1→3
A3 F *3→(1+2)
F1→3 =
2 A1
Use Fig. 14-17 to evaluate F *3→(1+2) with
Z 2.5
H= = = 0.625
X 4
Y 3
W= = = 0.75
X 4
From the graph, F *3→(1+2) ≈ 0.18. Therefore

F1→3 =
( 3)( 4 )( 0.18 ) = 0.216 Answer
2 ( 2 )( 2.5 )

14 -18
14-19 A planar black surface at 850oC is parallel to a second planar black surface at 300oC. The two surfaces are
separated by a small gap. If a third black surface is inserted in the gap, calculate
a. the net radiative heat flux between the outer surfaces with and without the insert
b. the temperature of the inserted surface

Approach:
Without the insert, use the formula for radiation
between two infinite black parallel plates. With the
insert in place, the net radiation from the left
surface to the insert equals the net radiation from
the insert to the right surface. Equate the heat rates
and solve for the unknown insert temperature.
Finally, calculate the net radiation from one of the
plates to the insert.

Assumptions:
1. The surfaces are very large compared to the size
of the gap.
2. The surfaces are black.
3. Radiation is the only mode of heat transfer.

Solution:
Assume the surfaces behave like infinite parallel plates because of the small gap. Without the insertion of surface
3, the net heat flux is
Q1→2
= F1→2σ (T14 − T2 4 )
A1
Since the plates are assumed infinite, F1→2 = 1 .
T1 = 850°C = 1123 K T2 = 300°C = 573 K
Q1→2 ⎛ W ⎞
= ⎜ 5.67 × 10−8 2 4 ⎟ (11234 − 5734 ) = 84, 066 2
W
A1 ⎝ m ⋅K ⎠ m
If surface 3 is inserted in the gap, then the net radiation from 1 to 3 must equal the net radiation from 3 to 2, or
Q1→3 = Q 3→2
F1→3σ (T14 − T34 ) = F3→2σ (T34 − T2 4 )
Because plates 1, 2, and 3 are all assumed infinite, F1→3 = F3→2 = 1 . Thus
T14 − T34 = T34 − T2 4
1 1
⎛ T 4 + T2 4 ⎞ 4 ⎛ 11234 + 5734 ⎞ 4
T3 = ⎜ 1 ⎟ =⎜ ⎟ = 960 K = 687°C
⎝ 2 ⎠ ⎝ 2 ⎠
The net radiative flux with the insert is
Q1→3
= F1→3σ (T14 − T34 )
A1
Q1→3 ⎛ W ⎞
= ⎜ 5.67 × 10−8 2 4 ⎟ (11234 − 9604 ) = 42, 033 2
W
Answer
A1 ⎝ m ⋅K ⎠ m

Comment:
Note that inserting the plate had the effect of impeding heat transfer. The insulating effect would have been
greater if plate 3 had a lower emissivity.

14 -19
14-20 A wire coated with insulation runs through the center of an evacuated cylindrical channel in a space station.
Both the wire and the channel wall may be approximated as black surfaces. The channel wall is maintained
at 50 K and the wire dissipates 75 mW per cm of length. The outer diameter of the wire is 0.3 cm and the
inner diameter of the channel is 1.5 cm. Calculate the wire temperature.

Approach:
Use the formula for radiation between two
infinitely-long parallel cylinders.

Assumptions:
1. The wire is very long compared to the gap
between wire and channel.
2. The surfaces are black.
3. Radiation is the only mode of heat transfer.

Solution:
If the wire and channel are assumed to be very long compared to the spacing between wire and channel, end
effects may be ignored. For two infinitely-long concentric cylinders, from Fig. 14-35,
σ A1 (T14 − T2 4 )
Q1→2 =
1 1 − ε 2 ⎛ r1 ⎞
+ ⎜ ⎟
ε1 ε 2 ⎝ r2 ⎠
Since ε1 = ε 2 = 1, this reduces to
Q = σ A (T 4 − T 4 )
1→2 1 1 2

We know the heat generated in a 1-cm length of wire. The wire surface area is
A1 = π DL = π ( 0.3cm )(1cm ) = 0.9425cm 2
Solving the rate equation for T1 yields
1
⎡ ⎛ 1W ⎞ ⎤4
⎢ 75 mW ⎜ ⎟ ⎥
⎛ Q1→2 4⎞ ⎢ ⎝ 1000 mW ⎠ ⎥
T1 = ⎜ + T2 ⎟ = ⎢ 2⎥
= 344 K Answer
⎝ σ A ⎠ ⎢⎛ W ⎞ 2 ⎛ 1m ⎞ ⎥
⎟ ( 0.9425cm ) ⎜
1 -8
⎢⎣ ⎜⎝
5.67×10 ⎟
m2 ⋅ K 4 ⎠ ⎝ 100 cm ⎠ ⎥⎦

14 -20
14-21 A generator is constructed of an inner rotating cylinder (the rotor) and an outer, stationary cylindrical shell
(the stator). During testing, the rotor develops a thermally-sensitive vibration and the engineer suggests that
one side of the rotor is hotter than the other. To correct the problem, a stripe of black paint is applied to the
hot side. The unpainted rotor surface has an emissivity of 0.2 and the painted surface has an emissivity of
0.98. The rotor outer surface is as 160oC and the stator inner surface is at 95oC. The diameter of the rotor is
1.2 m and the gap between rotor and stator is 5 cm. Calculate the net radiative flux from the rotor on the
painted portion and on the unpainted portion.

Approach:
Given that the rotor-stator gap is small
compared to both the rotor size and the width
of the paint stripe, we may use the formula
for radiation between two infinitely-long
parallel plates.

Assumptions:
1. The rotor diameter and length are large
compared to the gap between rotor and
stator.
2. The paint stripe is wide compared to the
gap between rotor and stator.
3. All surfaces are gray and diffuse.

Solution:
The gap between rotor and stator is very
small compared to the diameter of the rotor;
therefore, we may assume that the rotor and
stator act like two infinite flat plates. The net
radiative heat flux is given by
Q1→2 σ (T1 − T2 )
4 4

=
A 1 1
+ −1
ε1 ε2
T1 = 160 C = 433K
o
T2 = 95 o C = 368 K
On the unpainted portion:
⎛ W ⎞
5.67 × 10−8 2 4 ⎟ ( 4334 − 3684 ) K 4
Q1→2 ⎜⎝ m ⋅K ⎠ W
= = 168 2 Answer
A 1 1 m
+ −1
0.2 0.6
On the painted portion
⎛ W ⎞
5.67 × 10−8 2 4 ⎟ ( 4334 − 3684 ) K 4
Q1→2 ⎜⎝ m ⋅K ⎠ W
= = 565 2 Answer
A 1 1 m
+ −1
0.98 0.6

Comments:
The simple act of painting a metal surface can alter the radiative heat transfer dramatically. In this case, if the
radiation is a significant fraction of total heat transfer (radiation plus convection) across the gap, then altering the
surface emissivity could have a big effect on rotor surface temperature. Painting the hot side increases the heat
transfer from that side and reduces temperature there, resulting in more thermal uniformity over the entire rotor
surface.

14 -21
14-22 Two large flat plates in parallel planes are at 60oF and 420oF. The cold plate has an emissivity of 0.6 and
the hot plate has an emissivity of 0.2. Find the net rate of heat transfer.

Approach:
Use the formula for radiation between two
infinite parallel plate.

Assumptions:
1. Radiation is the only mode of heat transfer.
2. All surfaces are gray and diffuse.

Solution:
For two parallel flat plates, the net radiative
heat flux is given by
Q1→2 σ (T2 − T1 )
4 4

=
A 1 1
+ −1
ε1 ε2
T1 = 60 F = 520 R
o
T2 = 420 o F = 880 R
Substituting values,
⎛ Btu ⎞
0.171 × 10−8 ⎟ ( 880 − 420 ) R
4 4 4
Q1→2 ⎜⎝ h ⋅ ft 2 ⋅ R 4 ⎠ Btu
= = 159 Answer
A 1
+
1
−1 h ⋅ ft 2
0.6 0.2

Comments:
The answer is the same if the emissivities are switched, as is evident from the rate equation.

14 -22
14-23 A thermocouple is used to measure the temperature of a hot gas flowing in a pipe whose wall is at 350oC.
A cylindrical radiation shield, large compared to the size of the thermocouple, encloses the thermocouple,
as shown. The shield has an emissivity of 0.13. The emissivity of the thermocouple bead is 0.68 and the
emissivity of the pipe wall is 0.94. The convective heat transfer coefficient on the thermocouple is 70
W/m2·oC and the thermocouple reads 500oC. The convective heat transfer coefficient on the shield is 35
W/m2·oC. The shield has a diameter of 3.5 cm and the pipe has a diameter of 6 cm. Calculate the actual gas
temperature.

Approach:
Perform two energy
balances – one on the
thermocouple and the other
on the shield, including
both radiation and
convection. Solve the
resulting non-linear
equations simultaneously.

Assumptions:
1. The thermocouple is
small relative to the
shield.
2. The shield is long
relative to its radius.
2. All surfaces are gray and
diffuse.

Solution:
Begin with an energy balance on the thermocouple. Designate T1 as the gas temperature, T2 as the thermocouple
temperature, and T3 as the shield temperature. The shield is a large enclosure compared to the size of the
thermocouple; therefore, it appears black. The net radiative heat transfer leaving the thermocouple equals the
convective heat transfer to the thermocouple; i.e.,
A2ε 2σ (T2 4 − T34 ) = h2 A2 (T1 − T2 )
Simplifying,
ε 2σ (T2 4 − T34 ) = h2 (T1 − T2 ) eq.(1)
Next, perform an energy balance on the radiation shield. The shield and pipe may be approximated as two
infinitely-long cylinders. Note that convection occurs on both sides of the shield. Using Fig. 14-36,
A3σ (T34 − T4 4 )
= 2h3 (T1 − T3 ) eq.(2)
1 1 − ε 4 ⎛ r3 ⎞
+ ⎜ ⎟
ε3 ε 4 ⎝ r4 ⎠
Eq. (1) and (2) are two equations in the two unknowns T1 and T3. The equations are non-linear and must be solved
by trial and error. Known parameters are
W
ε 2 = 0.68 T2 = 500 o C = 773K h2 = 70 2 o r3 = 3.5cm
m ⋅ C
W
ε 3 = 0.13 T4 = 350 o C = 623K h3 = 35 2 o r4 = 6.0 cm
m ⋅ C
W
ε 4 = 0.94 σ = 5.67 × 10−8 2 4
m ⋅K
Using equation solving software,
T1 = 783K T3 = 763K Answer

Comments:
Even with the shield, the measured thermocouple temperature is 10 degrees below the actual gas temperature.
Occasionally, multiple shields are used to further improve accuracy.

14 -23
14-24 A hot potato with a surface temperature of 375oF and emissivity of 0.93 radiates to a large room with walls
at 70oF. To keep the potato warm, someone covers it loosely with a single layer of aluminum foil, which
has an emissivity of 0.2. Calculate the net rate of radiation loss from the potato with and without the foil.
Assume the potato is spherical with a diameter of 4.5 in. and the foil forms a spherical shell around the
potato.

Approach:
Equate radiation from the potato to the foil to radiation from
the foil to the room.

Assumptions:
1. Radiation is the only mode of heat transfer.
2. The foil is very close to the potato.
3. All surfaces are gray and diffuse.
4. The potato is spherical.

Solution:
To begin, recognize that the radiative flux from the potato to the foil must equal the flux from the foil to the walls
of the room. (We are assuming that radiation is the only mode of heat transfer.) The potato is assumed to be
spherical and the foil is assumed to form a spherical shell around the potato. The net radiative flux between two
concentric spheres is
A1σ (T14 − T2 4 )
Q1→2 = 2
1 1 − ε 2 ⎛ r1 ⎞
+ ⎜ ⎟
ε1 ε 2 ⎝ r2 ⎠
The foil is very close to the potato, so we assume r1 ≈ r2 and Q1→2 becomes
A1σ (T14 − T2 4 )
Q1→2 =
1 1− ε2
+
ε1 ε2
The net radiative flux from the foil to the walls of the room is
Q 2→3 = A2ε 2σ (T2 4 − T34 )
Equating the net fluxes:
A1σ (T14 − T2 4 )
Q1→2 = Q 2→3 → = A2ε 2σ (T2 4 − T34 )
1 1− ε2
+
ε1 ε2
Since r1 ≈ r2 , A1 ≈ A2 . Simplifying and solving for T2
⎡ 4 4 ⎛ ε2 ⎞⎤ ⎡ 4 ⎛ 0.2 ⎞⎤
⎢ T1 + T3 ⎜ + 1 − ε 2 ⎟ ⎥ ⎢ ( 835 R ) + ( 530 R ) ⎜ + 1 − 0.2 ⎟ ⎥
4

⎝ ε ⎠ ⎝ ⎠ ⎥ = 728 R
T2 = ⎢ ⎥ =⎢
1 0.93
⎢ ε2 ⎥ ⎢ 0.2 ⎥
⎢ + 2 − ε2 ⎥ ⎢ + 2 − 0.2
ε ⎥
⎣⎢ 1
⎦⎥ ⎣ 0.93 ⎦
The net radiative flux with the foil is
2
⎛ 4.5 ⎞ ⎛ Btu ⎞ ⎡
Q 2→3 = A2ε 2σ (T2 4 − T34 ) = 4π ⎜
Btu
ft ⎟ ( 0.2 ) ⎜ 0.171 × 10−8 4 ⎟ ⎣(
728 ) − ( 530 ) ⎤ R 4 = 30.5
4 4
⎜ ( 2 )(12 ) ⎟ ⋅ 2 ⎦
⎝ ⎠ ⎝ h ft R ⎠ h
The net radiative flux without the foil is
2
⎛ 4.5 ⎞ ⎛ Btu ⎞ ⎡
Q1→3 = A1ε1σ (T14 − T34 ) = 4π ⎜
Btu
ft ⎟ ( 0.93) ⎜ 0.171 × 10−8 4 ⎟ ⎣(
835 ) − ( 530 ) ⎤ R 4 = 286
4 4
⎜ ( 2 )(12 ) ⎟ ⋅ 2 ⎦
⎝ ⎠ ⎝ h ft R ⎠ h

Comments:
Adding the foil reduced the radiative flux by nearly an order of magnitude. In the actual situation, natural
convection should also be considered.

14 -24
14-25 A motorist leaves a car parked on a driveway with the engine on. The driveway is covered with a layer of
ice 0.5 in. thick at 32oF. The undercarriage of the car in the vicinity of the catalytic converter is at a
temperature of 190oF. Assume the undercarriage radiates like a black surface and the ice has an emissivity
of 0.75. Ignoring heat transfer by convection, calculate the time required to melt the ice. The latent heat of
fusion and density of ice are 143.5 Btu/lbm and 62.4 lbm/ft3, respectively.

Approach:
Consider the undercarriage and the ice as two infinite
parallel plates and calculate the net rate of radiative transfer
from the undercarriage to the ice. Calculate the mass of ice
to be melted per unit area. Use the heat transfer rate, the
mass of ice, and the heat of fusion to determine the melting
time.

Assumptions:
1. Radiation is the only mode of heat transfer.
2. The undercarriage is a black surface.
3. The ice is gray and diffuse.
4. As the ice melts, water drains off and exposes more ice.
5. The undercarriage and ice layer are very large compared to
the distance between them and compared to the ice
thickness; therefore they can be treated as infinite parallel
plates.

Solution:
Considering the undercarriage and the ice to be two infinite parallel plates, the net rate of radiative transfer is,
from Fig. 14-35
Q1→2 σ (T1 − T2 )
4 4

=
A 1 1
+ −1
ε1 ε2
Using T1 = 190 F = 650 R and T2 = 32 o F = 492 R
o

⎛ Btu ⎞
0.171 × 10−8 ⎟ ( 650 − 492 )
4 4
 ⎜
Q1→2 ⎝ h ⋅ ft 2 ⋅ R 4 ⎠ Btu
= = 154
A 1
+
1
−1 h ⋅ ft 2
1 0.75
The mass of 1 ft2 of ice is
⎛ lbm ⎞ ⎛ 1ft ⎞
m = ρV = ⎜ 62.4 3 ⎟ ( 0.5in.2 ) ⎜ 2 ⎟(
1ft 2 ) = 2.6 lbm
⎝ ft ⎠ ⎝ 12in. ⎠
Designating the latent heat of fusion of ice as L and the rate of radiative transfer for 1 ft2 of ice as q, the
time to melt the ice is
⎛ Btu ⎞
Lm ⎜⎝
143.5 ⎟ ( 2.6 lbm )
lbm ⎠
t= = = 2.43h Answer
q Btu
154
h

Comments:
The car would have to be idling a long time for all the ice to melt.

14 -25
14-26 An astronaut with a surface area of 1.8 m2 generates 150 W of body heat during a space walk. The exterior
of the spacesuit is exposed to outer space, which is black at 4 K. Both exterior and interior surfaces of the
suit are silvered with an emissivity of 0.4. It is necessary to keep the surface of the astronaut’s body, which
has an emissivity of 0.85, at no less than 16oC in steady state. Should the suit be made of one layer of
silvered material or two layers?

Approach:
The suit follows the contours of the astronaut’s body closely; therefore
assume radiation between the body and the suit is like radiation
between two infinite parallel plates. Equate the generated heat to the
net radiation from the body to the inside of the suit. Similarly, equate
the net radiation on the inside to that on the outside of the suit. Solve
for body temperature.

Assumptions:
1. Conduction in the suit is neglected.
2. The suit and body may be represented as two infinite parallel planes.
3. The suit and body are gray, diffuse surfaces.
4. Space is black at 4 K.
5. Reflections between locations on the suit exterior are negligible.
6. No radiation is incident on the astronaut from the sun or the space
vehicle.

Solution:
Let the body be surface 1 and the inside of the suit be surface 2. The radiation from the body to the inside of the
suit is given by (see Fig. 14-35)
σ A (T14 − T2 4 )
Q =
1 1
+ −1
ε1 ε2
In steady state, the body heat is equal to the heat leaving the outside of the suit. If the suit has a single layer of
silvered material, then the net radiation from the exterior of the suit to the environment, which is black at
temperature T3, is
Q = Aε 2σ (T2 4 − T34 )
Solving for T2 4
Q
T2 4 = + T34
Aε 2σ
Substituting this into the first equation gives
⎛ Q ⎞
σ A ⎜ T14 − − T34 ⎟
⎝ A ε σ ⎠
Q = 2
1 1
+ −1
ε1 ε2
Solving for T1 produces
1 1
⎡ ⎛1 1 ⎞ Q ⎤4 ⎡ ⎛ 1 ⎞ (150 W ) ⎛ W ⎞ 4⎤
⎢ Q ⎜ + − 1⎟ + + Aσ T3 ⎥ + (1.8 m 2 ) ⎜ 5.67 × 10−8 2 4 ⎟ ( 4 K ) ⎥
4
1
(150 W ) ⎜ + − 1⎟ +
4

ε ε2 ⎠ ε2 ⎢
m ⋅K ⎠
T1 = ⎢ ⎝ 1 ⎥ =⎢ ⎝ 0.85 0.4 ⎠ ⎝
0.4 ⎥
⎢ Aσ ⎥ ⎢ 2 ⎛ W ⎞ ⎥
⎢ ⎥ ⎢ (1.8 m ) ⎜⎝ 5.67 ×10 m2 ⋅ K 4 ⎟⎠
−8
⎥
⎣⎢ ⎦⎥ ⎣ ⎦

T1 = 295 K = 22 o C

This is above the limit of 16ºC. There is no need for a second layer of silvered material, which would produce a
body temperature higher than 22ºC. Answer

14 -26
14-27 A cryogenic dewer is constructed of two concentric spheres 45 cm and 53 cm in diameter. Liquid nitrogen
at 100 K is stored in the inner sphere and the space between the spheres is evacuated. The outer sphere has
a temperature of 220 K and the emissivity of both spheres is 0.023. If the latent heat of vaporization of the
nitrogen is 210 kJ/kg, determine the number of kilograms of nitrogen evaporated per hour.

Approach:
Use the formula for net radiation between two concentric spheres from
Fig. 14-35. Find the rate of evaporation using the latent heat and the net
radiative heat.

Assumptions:
1. Radiation is the only mode of heat transfer.
2. Both surfaces are gray and diffuse.

Solution:
From Fig. 14-35, the net rate of radiative transfer between two spheres is
A1σ (T14 − T2 4 )
Q1→2 = 2
1 1 − ε 2 ⎛ r1 ⎞
+ ⎜ ⎟
ε1 ε 2 ⎝ r2 ⎠
Substituting values (note that the surface area of a sphere is 4π r 2 )
W ⎞ ⎛ 1m 2 ⎞
2
⎛ 45 ⎞ ⎛
4π ⎜ cm ⎟ ⎜ 5.67 × 10−8 2 4 ⎟ ⎜ 4 2 ⎟ (1004 − 2204 ) K 4
⎝ 2 ⎠ ⎝ m ⋅ K ⎠ ⎝ 10 cm ⎠
Q1→2 = 2
= −1.09 W
1 1 − 0.023 ⎛ 45cm ⎞
+ ⎜ ⎟
0.023 0.023 ⎝ 53cm ⎠
The heat is negative since heat leaks into the dewer from the environment. The rate of evaporation of nitrogen is
⎛ J ⎞
⎜ 1 ⎟ ⎛ 3600s ⎞
( −1.09 W ) ⎜ s ⎟ ⎜ ⎟
⎜ 1W ⎟ ⎝ 1h ⎠
m = ⎝ ⎠ = −0.0187
kg
Answer
⎛ kJ ⎞ ⎛ 1000 J ⎞ h
⎜ 210 ⎟ ⎜ ⎟
⎝ kg ⎠ ⎝ 1kJ ⎠

Comments:
In actuality, some structural supports would be needed to hold the two spheres apart. There would be conduction
losses through the structural supports.

14 -27
14-28 During a deposition process, a long strip heater is centered above a long, well-insulated plate, as shown in
the figure. The heater is maintained at 400ºC and the surroundings are black at 25ºC. The geometry is
effectively two-dimensional and radiation is the only mode of heat transfer. Using data on the figure, find
the plate temperature.

Approach:
Consider the heater, the plate , and the
surroundings to be a three-surface
radiation enclosure and apply Eq. 14-39
and 14-40. View factors may be found
using Table 14-2.

Assumptions:
1. Radiation is the only mode of heat
transfer.
2. Both surfaces are gray and diffuse.
3. The geometry is two-dimensional.
4. The plate is perfectly insulated.

Solution:
Let surface 1 be the heater, surface 2 be the
plate, and surface 3 be the surroundings.
Then apply Eq. 14-39 to surfaces 1 and 2
to get
J − J 2 J1 − J 3
Q1 = 1 + Eq. (1)
R1→2 R1→3
J − J 3 J 2 − J1
Q 2 = 2 + Eq. (2)
R2→3 R2→1
Also apply Eq. 14-40 to surface 1, yielding
J1 = σ T14 − R1Q1 Eq. (3)
Surface 2 is perfectly insulated; therefore, it is a reradiating surface and
J 2 = σ T24
Surface 3 is black, therefore
J 3 = σ T34
Substituting these expressions for J2 and J3 into Eqs. (1)-(3) and noting that Q 2 = 0, since 2 is insulated, gives
J − σ T24 J1 − σ T34
Q1 = 1 + Eq. (4)
R1→2 R1→3
σ T24 − σ T34 σ T24 − J1
0= + Eq. (5)
R2→3 R2→1
J1 = σ T14 − R1Q1 Eq. (6)
Once all the radiative resistances are determined, Eqs. (4)-(6) will be three equations in the three unknowns J1,
Q1 , and T2. The radiative resistances are:
1
R1→2 =
A1 F1→2
1
R1→3 =
A1 F1→3
1
R2→3 =
A2 F2→3
By reciprocity,

14 -28
 1 1
R2→1 = = = R1→2
A2 F2→1 A1 F1→2
1 − ε1
R1 =
A1ε1
The view factor F1→2 may be calculated using the first geometry in Table 14-2 with
w1 30 w2 50
W1 = = = 1.5 W2 = = = 2.5
L 20 L 20
1 1 1 1
⎡(W1 + W2 )2 + 4 ⎤ 2 − ⎡(W2 − W1 )2 + 4 ⎤ 2 ⎡(1.5 + 2.5 )2 + 4 ⎤ 2 − ⎡( 2.5 − 1.5 )2 + 4⎤ 2
F1→2 =⎣ ⎦ ⎣ ⎦ =⎣ ⎦ ⎣ ⎦ = 0.745
2W1 2 (1.5 )
All the radiation leaving surface 1 arrives either at 2 or at 3 (the surroundings). Therefore
F1→3 = 1 − F1→2 = 1 − 0.745 = 0.255
The view factor F2→1 may be found using reciprocity
A1 F1→2 ( 30 )( 0.745 )
F2→1 = = = 0.447
A2 50
Finally, F2→3 is by summation
F2→3 = 1 − F2→1 = 1 − 0.447 = 0.553
All these view factors are now substituted into the radiative resistances to get
1
R1→2 = = 4.47 m -2
( 0.3m )(1m )( 0.745 )
1
R1→3 = = 13.09 m -2
( 0.3m )(1m )( 0.255 )
1
R2→3 = = 3.62 m -2
( 0.5 m )(1m )( 0.553)
By reciprocity,
R2→1 = R1→2 = 4.47 m -2
1 − 0.87
R1 = = 0.498 m -2
( 0.3m )( )( )
1m 0.87
These are the resistances for an area of depth 1 m. In addition to these resistance values, use the given data:
T1 = 400 o C = 673K
T3 = 25 o C = 298 K
W
σ = 5.67 × 10−8
m ⋅ K4
2

and solve Eqs. (4)-(6) simultaneously for J1 , Q1 , and T2 . It is recommended that equation-solving software be used
for this purpose. The result for T2 is
T2 = 545 K = 272 o C Answer

14 -29
14-29 A chemical reaction chamber is in the shape of a cylinder with a height of 2.6 ft and a diameter of 0.6 ft. A
disk-shaped heater with an emissivity of 0.92 entirely covers the bottom of the chamber and generates 1.3
kW of heat. The side wall is at 86oF and the top end is at 65oF. Both the side wall and top have an
emissivity of 0.73. The chamber is partially evacuated and the only significant mode of heat transfer
present is radiation. The back of the heater is well-insulated, so that all the heater power is removed by
radiation into the chamber. Find the heater temperature.

Approach:
Model the reaction chamber as a three-
surface enclosure. Apply Eqs. 14-39 and
14-40. View factors may be found using
Fig. 14-18 and the summation relation.

Assumptions:
1. Radiation is the only mode of heat
transfer.
2. All surfaces are gray and diffuse.

Solution:
Apply Eq. 14-39 to the three surfaces in
the enclosure
J − J 2 J1 − J 3
Q1 = 1 + Eq. (1)
R1→2 R1→3
J − J 3 J 2 − J1
Q 2 = 2 + Eq. (2)
R2→3 R2→1
J − J J − J2
Q 3 = 3 1 + 3 Eq. (3)
R3→1 R3→2
Also apply Eq. 14-40
J1 = σ T14 − R1Q1 Eq. (4)
J = σ T 4 − R Q
2 2 2 2 Eq. (5)
J 3 = σ T − R3Q 3
3
4
Eq. (6)
The radiative resistances are:
1
R1→2 =
A1 F1→2
1
R1→3 =
A1 F1→3
1
R2→3 =
A2 F2→3
By reciprocity,
1 1
R2→1 = = = R1→2
A2 F2→1 A1 F1→2
1 1
R3→1 = = = R1→3
A3 F3→1 A1 F1→3
1 1
R3→2 = = = R2→3
A3 F3→2 A2 F2→3
1 − ε1 1− ε2 1 − ε3
R1 = R2 = R3 =
A1ε1 A2ε 2 A3ε 3
The view factor F1→3 may be calculated using the equation in Fig. 14-18 with
r1 0.3 r2
R1 = = = 0.115 R2 = = 0.115
L 2.6 L

14 -30
 1 + ( 0.115 )
2
1 + R22
S =1+ = 1+ = 77.1
( 0.115)
2 2
R1

1 ⎪⎧ ⎫ 1 ⎪⎧ ⎫
1 1
2 2⎪ 2 2⎪
F1→3 = ⎨ S − ⎡⎣ S − 4 ( r3 / r1 ) ⎤⎦ ⎬ = ⎨77.1 − ⎡⎣( 77.1) − 4 ( 0.3 / 0.3) ⎤⎦ ⎬ = 0.013
2 2

2 ⎪⎩ ⎪⎭ 2 ⎪⎩ ⎪⎭
From summation
F1→2 = 1 − F1→3 = 1 − 0.013 = 0.987
To find F3→2 , start with
F3→2 = F1→2 = 0.987
From reciprocity
2
⎛ 0.6 ⎞
π⎜ ⎟ ( 0.987 )
F2→3 =
A3 F3→2
= ⎝ 2 ⎠ = 0.0569
A2 π ( 0.6 )( 2.6 )
The radiative resistances may now be calculated as
1
R1→2 = 2
= 3.58ft -2 = R2→1
⎛ 0.6 ⎞ 2
π⎜ ⎟ ft ( 0.987 )
⎝ 2 ⎠
1
R1→3 = 2
= 272 ft -2 = R3→1
⎛ 0.6 ⎞ 2
π⎜ ⎟ ft ( 0.013)
⎝ 2 ⎠
1
R2→3 = = 3.59 ft -2 = R3→2
π ( 0.6 )( 2.6 ) ft 2 ( 0.0569 )

1 − 0.92
R1 = 2
= 0.308ft -2
⎛ 0.6 ⎞ 2
π⎜ ⎟ ft ( 0.92 )
⎝ 2 ⎠
1 − 0.73
R2 = = 0.0755ft -2
π ( 0.6 )( 2.6 ) ft 2 ( 0.73)
1 − 0.73
R3 = 2
= 1.31ft -2
⎛ 0.6 ⎞ 2
π⎜ ⎟ ft ( 0.73)
⎝ 2 ⎠
The final step is to solve Eqs. (1)-(6) simultaneously for the unknowns
J J J T Q Q
1 2 3 1 2 3

Note that Q1 = 1.3kW = 4436 Btu/h, T2 = 546 R, and T3 = 525 R. Using equation-solving software, the result is
Q = −4322 Btu/h
2

Q 3 = −114 Btu/h
T1 = 1789 R = 1329 o F Answer

Comments:
In practice, the curved side wall would probably not be isothermal, but would be hotter near the heater and cooler
farther away. A more sophisticated analysis would divide the side wall into a number of separate radiation zones.

14 -31
14-30 A very long enclosure is formed from two perpendicular, equal-width plates and a slanted cover plate, as
shown. The cross-section of the enclosure is in the shape of an isosceles, right triangle. Assuming gray and
diffuse surfaces, calculate the net radiative heat transfer from the hottest plate.

Approach:
Use Eqs. 14-39 and 14-40 for a three-surface
enclosure. Find view factors from Table 14-2.

Assumptions:
1. Radiation is the only mode of heat transfer.
2. Surfaces 1 and 2 are gray and diffuse.
3. Surface 3 is black.

Solution:
Apply Eq. 14-39 to surfaces 1 and 2
J − J 2 J1 − J 3
Q1 = 1 + Eq. (1)
R1→2 R1→3
J − J 3 J 2 − J1
Q 2 = 2 + Eq. (2)
R2→3 R2→1
From Eq. 14-40,
J1 = σ T14 − R1Q1 Eq. (3)
J = σ T 4 − R Q
2 2 2 2 Eq. (4)
Surface 3 is black, therefore
J 3 = σ T34 Eq. (5)
Eqs. (1) – (5) are 5 equations in the 5 unknowns J1 , J 2 , J 3 , Q1 , and Q 2 . The radiative resistances are:

1 1 1
R1→2 = R1→3 = R2→3 =
A1 F1→2 A1 F1→3 A2 F2→3
1 1 1 − ε1 1− ε2
R2→1 = = = R1→2 R1 = R2 =
A2 F2→1 A1 F1→2 A1ε1 A2ε 2
To find F1→2 , use Table 14-2 with
w1 + w2 − w3 10 + 10 − 102 + 102
F1→2 = = = 0.293
2w1 2 (10 )
By summation
F1→3 = 1 − F1→2 = 1 − 0.293 = 0.707
By symmetry
F2→3 = F1→3 = 0.707
The radiative resistances may now be calculated as
1 1
R1→2 = = 34.1m -1 = R2→1 R1→3 = = 14.1m -1
( 0.1m )( 0.293) ( 0.1m )( 0.707 )
1 1 − 0.5
R2→3 = = 14.1m -1 R1 = = 10 m -1
( 0.1m )( 0.707 ) ( 0.1m )( 0.5 )
1 − 0.75
R2 = = 3.33m -1
( 0.1m )( 0.75 )
The final step is to solve Eqs. (1)-(5) using T1 = 500 o C = 773K, T2 = 400 o C = 673K, and T3 = 100 o C = 373K.
Using equation-solving software, the result is
Q1 = 830 W/m Answer

This is the net heat leaving by radiation from the hottest plate per meter of enclosure depth.

14 -32
14-31 A short, thin-walled tube of length 2.5 in. and diameter 0.75 in. is open at both ends. The tube is made of
copper ( ε =0.25) and is maintained at 155oF. The surroundings surfaces are at 65oF. Including radiation
from both the outside and the inside of the tube, find the net rate of radiative transfer.

Approach:
Add two imaginary black surfaces, 2 and 3, to
cover the ends of the tube. Find the view factor
from 2 to 3 using Fig. 14-18. Find the view factor
from 1 to 2 and from 1 to 3 using summation and
reciprocity. Find the net heat leaving surface 1 in
the enclosure formed by 1, 2, and 3. Add this to the
net heat leaving the outside of the tube.

Assumptions:
1. All surfaces are gray and diffuse.
2. Surrounding surfaces are far from the tube.

Solution:
Construct two imaginary black surfaces which cover the ends of the tube, as shown. To calculate F2→3 , use Fig.
14-18 with
0.375
Ri = = 0.15 R j = Ri = 0.15
2.5
1+ Rj2 1 + ( 0.15 )
2

S =1+ = 1+ = 46.4
( 0.15 )
2
Ri 2
To calculate the view factor, use the formula:
⎧ 2 2⎫
1
⎧ 1
⎫
1⎪ ⎡ ⎤
⎛ rj ⎞ ⎪ 1⎪ ⎡ ⎛ 0.375 ⎞ ⎤ 2 ⎪
2

F2→3 = ⎨ S − S − 4 ⎜ ⎟ ⎬ = ⎨46.6 − ⎢( 46.4 ) − 4 ⎜

⎢ ⎥ ⎟ ⎥ ⎬ = 0.0215
2 2

2⎪ ⎢⎣ ⎝ ri ⎠ ⎥⎦ ⎪ 2⎪ ⎣⎢ ⎝ 0.375 ⎠ ⎦⎥ ⎪
⎩ ⎭ ⎩ ⎭
By summation
F2→1 = 1 − F2→3 = 1 − 0.0215
= 0.978
By reciprocity
2
⎛D⎞
π⎜ ⎟ F2→1 DF
F1→2 =
A2 F2→1
= ⎝2⎠ = 2→1
=
( 0.75in )( 0.978 ) = 0.0734
A1 π DL 4L 4 ( 2.5in )
The thermal circuit for the enclosure is shown to the right.
The surface resistance, R1 , is
1 − ε1
R1 =
A1ε1
⎛ 1ft 2 ⎞
A1 = π DL = π ( 0.75 in )( 2.5in ) ⎜ 2 ⎟
= 0.0409 ft 2
⎝ 144in ⎠
Therefore
1 − 0.25
R1 = = 73.3 ft -2
( 0.0409 ft 2 ) ( 0.25)
The space resistance between surfaces 1 and 2 is
1 1
R1→2 = = = 333 ft -2
A1 F1→2 ( 0.0409 ft 2 ) ( 0.0734 )

14 -33
 By symmetry
R1→3 = R1→2 = 333 ft -2
Since T2 = T3 , the thermal circuit reduces to the circuit
shown to the right. The total resistance is
R R
Rtot = R1 + 1→2 1→3 = 73.3 +
( 333)( 333) = 240 ft -2
R1→2 + R1→3 ( 333 + 333)
The net radiative energy leaving surface 1 out the ends of the
tube is calculated as
E − Eb 2 σ (T1 − T2 )
4 4

Q1 = b1 =
Rtot Rtot

4 (
6154 − 5254 ) R 4
Btu
0.171 × 10−8
h ⋅ ft R
2
Btu
= -2
= 0.478
240 ft h
From the outside of the tube to the environment
Q out = ε1σ A1 (T14 − T2 4 )
⎛ Btu ⎞
Q out = ( 0.25 ) ⎜ 0.171 × 10−8 ⎟ ( 0.0409 ft )( 615 − 525 ) R
2 4 4 4

⎝ h ⋅ ft 2 ⋅ R 4 ⎠
Btu
= 1.17
h
Finally,
Btu
Q tot = Q1 + Q out = 0.478 + 1.17 = 1.65 Answer
h

14 -34
14-32 A ceramic cooktop has a heating element 9 in. in diameter. A person’s hands are placed over the heating
element at a distance of 4.5 in. Model the hands as a disk with an emissivity of 0.90 and radius of 9 in. and
the heating element as a parallel disk with an emissivity of 0.97. The heating element is at 350oF, and the
walls of the kitchen are at 70oF. If the hands are reradiating surfaces, what is their temperature?

Approach:
Consider the heating element, the hands,
and the walls of the room to be a three-
surface enclosure and apply Eqs. 14-39
and 14-40. View factors may be found
using Fig. 14-18.

Assumptions:
1. Radiation is the only mode of heat
transfer.
2. The hands and heating element are gray
and diffuse.
3. No heat is conducted into the hands
(they are reradiating).
4. The hands are shaped approximately like
a disk of 9 in. diameter.
5. The walls of the room are far enough
away so that they may be considered
black.

Solution:
Apply Eq. 14-39 to the heater and the hands; i.e.,
J − J 2 J1 − J 3
Q1 = 1 + Eq. (1)
R1→2 R1→3
J − J 3 J 2 − J1
Q 2 = 2 + Eq. (2)
R2→3 R2→1
Also apply Eq. 14-40 to the heater to get
J1 = σ T14 − R1Q1 Eq. (3)
The hands are a reradiating surface, therefore
J 2 = σ T24
The surroundings may be considered black, therefore
J 3 = σ T34
Substituting these expressions for J2 and J3 into Eqs.(1)-(3) and noting that Q 2 = 0, since 2 is insulated gives
J − σ T24 J1 − σ T34
Q1 = 1 + Eq. (4)
R1→2 R1→3
σ T24 − σ T34 σ T24 − J1
0= + Eq. (5)
R2→3 R2→1
Once all the radiative resistances are determined, Eqs.(3)-(5) will be three equations in the three unknowns
J1 , Q1 , and T2 . The radiative resistances are:
1 1 1
R1→2 = R1→3 = R2→3 =
A1 F1→2 A1 F1→3 A2 F2→3
By reciprocity,
1 1 1 − ε1
R2→1 = = = R1→2 R1 =
A2 F2→1 A1 F1→2 A1ε1
The view factor F1→2 may be calculated using the equation in Fig. 14-18 with

14 -35
 r1 4.5in. r2 4.5in.
R1 = = =1 R2 = = =1
L 4.5in. L 4.5in.
1 + (1)
2
1 + R22
S =1+ = 1+ =3
(1)
2
R12

1 ⎧⎪ 2 2⎫ ⎪ 1 ⎧⎪ ⎡ 2 2 2⎫
1 1
⎪
F1→2 = ⎨ S − ⎡
⎣
S 2
− 4 ( r2 / r1 ) ⎤ =
⎦ ⎬⎪ 2 ⎨⎪ ⎣
3 − ( 3 ) − 4 ( 4.5 / 4.5 ) ⎤
⎦ ⎬⎪
= 0.382
2 ⎩⎪ ⎭ ⎩ ⎭
All the radiation leaving surface 1 arrives either at 2 or at 3 (the surroundings). Therefore
F1→3 = 1 − F1→2 = 1 − 0.382 = 0.618
By symmetry
F2→3 = F1→3 = 0.618
The areas are
2
⎛ 4.5 ⎞ 2
A1 = A2 = π ⎜ ⎟ ft = 0.442 ft
2

⎝ 12 ⎠
Substituting values into the radiative resistances
1
R1→2 = = 5.93ft -2
( 0.442 ft 2 ) ( 0.382 )
1
R1→3 = = 3.66 ft -2
( 0.442 ft 2 ) ( 0.618)
1
R2→3 = = 3.66 ft -2
( 0.442 ft ) ( 0.618)
2

R2→1 = R1→2 = 5.93ft -2

1 − 0.97
R1 = = 0.07 ft -2
( 0.442 ft ) ( 0.97 )
2

The given data is

W
T1 = 350 o F = 810 R T3 = 70 o F = 530 R σ = 5.67 × 10−8
m ⋅ K4 2

Using the radiative resistances and these data in Eqs. (3)-(5) and solving simultaneously gives
T2 = 677 R = 217 o F
It is recommended that the equations be solved using equation solving software.

Comments:
Ouch! In reality, some heat would be conducted into the hands and convected from the hands, so this high
temperature would not be reached. Also, hands are usually warmed in a transient process rather than a steady-state
one.

14 -36
14-33 An attic space is in the shape of an isosceles triangle, with dimensions as shown. The floor of the attic is at
50oF, the two ceiling surfaces under the roof are at 30oF, and the air in the attic is at 40oF. The convective
heat transfer coefficient is 1.4 Btu/ h·ft2·oF on the floor and 0.66 Btu/ h·ft2·oF on the ceiling surfaces. The
emissivity of the ceiling surfaces is 0.93. Calculate the heat transferred from the floor by radiation and by
convection per foot of depth if the floor is covered with
a. unsilvered insulation ( ε 1= 0.87 )
b. silver-backed insulation ( ε 1= 0.19 )

Approach:
The attic space is a three-surface
enclosure. Apply Eqs. 14-39 and
14-40. View factors may be
found using Table 14-2.

Assumptions:
1. The attic space may be modeled
as two-dimensional.
2. All surfaces are gray and
diffuse.

Solution:
Apply Eq. 14-39 to the three surfaces in the enclosure
J − J 2 J1 − J 3
Q1 = 1 + Eq. (1)
R1→2 R1→3
J − J 3 J 2 − J1
Q 2 = 2 + Eq. (2)
R2→3 R2→1
J − J J − J2
Q 3 = 3 1 + 3 Eq. (3)
R3→1 R3→2
Also apply Eq. 14-40
J = σ T 4 − R Q
1 1 1 1 Eq. (4)
J 2 = σ T24 − R2Q 2 Eq. (5)
J = σ T 4 − R Q
3 3 3 3 Eq. (6)
The radiative resistances are:
1 1 1
R1→2 = R1→3 = R2→3 =
A1 F1→2 A1 F1→3 A2 F2→3
1 1 1 1 1 1
R2→1 = = = R1→2 R3→1 = = = R1→3 R3→2 = = = R2→3
A2 F2→1 A1 F1→2 A3 F3→1 A1 F1→3 A3 F3→2 A2 F2→3
1 − ε1 1− ε2 1 − ε3
R1 = R2 = R3 =
A1ε1 A2ε 2 A3ε 3
To find F1→2 , use Table 14-2:
w1 + w2 − w3 30 + 20 − 20
F1→2 = = = 0.5
2w1 2 ( 30 )
This result could also have been deduced by inspection. By symmetry,
F1→3 = F1→2 = 0.5
From reciprocity
AF
F2→1 = 1 1→2 =
( 30 )( 0.5) = 0.75
A2 20
By summation
F2→3 = 1 − F2→1 = 1 − 0.75 = 0.25

14 -37
 The radiative resistances may now be calculated as
1
R1→2 = = 0.0667 ft -1 = R2→1
( 30 ft )( 0.5)
1
R1→3 = = 0.0667 ft -1 = R3→1
( 30 ft )( 0.5)
1
R2→3 = = 0.2 ft -1 = R3→2
( 20 ft )( 0.25 )
1 − 0.87
R1 = = 0.00498ft -1
( )( )
30 ft 0.87
1 − 0.93
R2 = R3 = = 0.00376 ft -1
( 20 ft )( 0.93)
The final step is to solve Eqs. (1)-(6) using
Btu
T1 = 50 + 460 = 510 R T2 = T3 = 30 + 460 = 490 R σ = 0.171 × 10−8
h ⋅ ft 2 ⋅ R 4
It is recommended that the calculation be done using equation-solving software. The result for the bottom surface
(with unsilvered insulation, ε1 = 0.87 )
Q = 426 Btu/h
1 Answer
If the calculation is repeated with silver-backed insulation, ( ε1 = 0.19 ) ,
Q1 = 96.5 Btu/h Answer
By comparison, natural convection from the bottom surface per foot of depth is given by
⎛ Btu ⎞ Btu
Q conv = hA (T1 − Tair ) = ⎜ 1.4 2 o ⎟(
30 ft )(1ft )( 50 − 40 ) o F = 420 Answer
⎝ h ⋅ ft ⋅ F ⎠ h

Comments:
Silver-backed insulation significantly lowers heat loss by radiation from the building. Heat transfer by natural
convection is comparable in size to heat transfer by radiation when unsilvered insulation is used. It is often the
case that radiation is important when a surface is cooled or heated by natural convection in air.

14 -38
14-34 Two diffuse, gray flat plates of size 60 cm by 45 cm face each other in parallel planes. One plate has an
emissivity of 0.66 and a temperature of 460oC while the other has an emissivity of 0.31 and a temperature
of 120oC. The spacing between the plates is 14 cm. The surroundings are black at 20oC. Calculate the net
rate of radiation heat transfer from the hot plate.

Approach:
Consider the two plates and the
surroundings to be a three-surface
radiation enclosure and apply Eqs.
14-39 and 14-40. View factors may
be found using Fig. 14-16.

Assumptions:
1. Radiation is the only mode of
heat transfer.
2. Each plate is gray and diffuse.
3. Only consider radiation from one
side of each plate–the side that
faces the opposing plate.

Solution:
Apply Eq. 14-39 to plates 1 and 2 to get
J − J 2 J1 − J 3
Q1 = 1 + Eq. (1)
R1→2 R1→3
J − J 3 J 2 − J1
Q 2 = 2 + Eq. (2)
R2→3 R2→1
Also apply Eq. 14-40 to the two plates, yielding
J = σ T 4 − R Q
1 1 1 1 Eq. (3)
J 2 = σ T24 − R2Q 2 Eq. (4)
Since the surroundings (surface 3) are black,
J 3 = σ T34
The radiative resistances are:
1
R1→2 =
A1 F1→2
1
R1→3 =
A1 F1→3
1
R2→3 =
A2 F2→3
By reciprocity,
1 1
R2→1 = = = R1→2
A2 F2→1 A1 F1→2
1 − ε1 1− ε2
R1 = R2 =
A1ε1 A2ε 2
The view factor F1→2 may be calculated using Fig. 14-16 with
X 60 Y 45
= = 4.29 = = 3.21
L 14 L 14
From the plot
F1→2 ≈ 0.6
By summation
F1→3 = 1 − F1→2 = 1 − 0.6 = 0.4

14 -39
By symmetry
F2→3 = F1→3 = 0.4
The radiative resistances may now be calculated using
⎛ 45 ⎞ ⎛ 60 ⎞
A1 = A2 = ⎜ ⎟m⎜ ⎟ m = 0.27 m
2

⎝ 100 ⎠ ⎝ 100 ⎠
1
R1→2 = = 6.17 m -2
( 0.27 m 2 ) ( 0.6 )
1
R1→3 = = 9.26 m -2
( 0.27 m ) ( 0.4 )
2

1
R2→3 = = 9.26 m -2
( 0.27 m2 ) ( 0.4 )
R2→1 = R1→2 = 6.17 m -2
1 − 0.66 1 − 0.31
R1 = = 1.91m -2 R2 = = 8.24 m -2
( 0.27 m ) ( 0.66 )
2
( 0.27 m ) ( 0.31)
2

We are now prepared to solve Eqs.(1)-(5) simultaneously with the following additional inputs
T1 = 460 o C = 733K
T2 = 120 o C = 393K
W
σ = 5.67 × 10−8
m ⋅ K4 2

Using equation-solving software, the result is

Q1 = 2300 W Answer

14 -40
14-35 In the middle of a blizzard, a homeowner’s furnace fails. In desperation, the homeowner turns on the oven
of the electric range and opens the oven door. The thermostat for the oven is set at 450oF and the inside
walls of the oven are black. Assume the oven door acts like a reradiating surface (i.e., neglect convection)
with an emissivity of 0.89. The walls of the kitchen are at 55oF. Calculate the temperature of the oven door
and the electric power consumed by the oven.

Approach:
The oven cavity may be covered by
an imaginary black surface at the
temperature of the interior of the
oven. The door is represented by a
gray, reradiating surface
perpendicular to the imaginary
oven surface. A three-surface
enclosure analysis may then be
used, taking the environment as the
third surface.

Assumptions:
1. Radiation is the only mode of
heat transfer.
2. The oven door is gray and diffuse.
3. The oven is powerful enough so
that the interior walls can be kept
at 450 oF while the door is open.
4. The oven walls are black.

Solution:
Apply Eq. 14-39 to plates 1 and 2 to
get
J − J 2 J1 − J 3
Q1 = 1 + Eq. (1)
R1→2 R1→3
J − J 3 J 2 − J1
Q 2 = 2 + Eq. (2)
R2→3 R2→1
Surfaces 1 and 3 are black, therefore
J1 = σ T14 J 3 = σ T34
Surface 2 is a reradiating surface, therefore
J 2 = σ T24 Q 2 = 0
Eqs. (1) and (2) become
σ (T14 − T24 ) σ (T14 − T34 )
Q1 = + Eq. (3)
R1→2 R1→3
σ (T24 − T34 ) σ (T24 − T14 )
0= + Eq. (4)
R2→3 R2→1
The radiative resistances are:
1
R1→2 =
A1 F1→2
1
R1→3 =
A1 F1→3
1
R2→3 =
A2 F2→3
By reciprocity,

14 -41
 1 1
R2→1 = = = R1→2
A2 F2→1 A1 F1→2
The view factor F1→2 may be calculated using Fig. 14-17 with
Z 2 ft Y 2 ft
= = 0.8 = = 0.8
X 2.5ft X 2.5ft
From the plot in Fig. 14-17,
F1→2 ≈ 0.22
By summation
F1→3 = 1 − F1→2 = 1 − 0.22 = 0.78
By symmetry
F2→3 = F1→3 = 0.78
The radiative resistances are
1
R1→2 = = 0.909 ft -2
( 2 )( 2.5) ft 2 ( 0.22 )
1
R1→3 = = 0.256 ft -2
( )( ) ( )
2 2.5 ft 2
0.78
1
R2→3 = = 0.256 ft -2
( 2 )( 2.5) ft 2 ( 0.78 )
R2→1 = R1→2 = 0.909 ft -2
The given temperatures are
T1 = 450 o F = 910 R
T3 = 55 o F = 515 R
Solving Eq. (4) for T2
1

⎧ 4 R T 4
⎫
1
4
⎧
⎪ 5154 R 4 +
( 0.256 ft -2 ) 9104 R 4 ⎫4
⎪
⎪ T3 +
2→3 1

⎪
T2 = ⎨
R2→1
⎪
⎪ ⎪
=⎨
( 0.909 ft -2 ) ⎪
⎬ ⎬ = 673R
R
⎪ 1 + 2→3 ⎪ ⎪
1 +
( 0.256 ft -2 ) ⎪
⎪⎩ ⎪⎭ ⎪ ⎪
R2→1
⎩ ( 0.909 ft -2 ) ⎭
Substituting values into Eq. (3)
⎛ Btu ⎞ ⎛ Btu ⎞
4 ⎟(
⎜ 0.171 × 10
−8
9104 − 6734 ) R 4 ⎜ 0.171× 10−8 4 ⎟(
9104 − 5154 ) R 4
⎝ h ⋅ ft 2
⋅ R ⎠ ⎝ h ⋅ ft 2
⋅ R ⎠
Q1 = +
0.909 ft -2 0.256 ft -2
Btu
Q1 = 5007 Answer
h

14 -42
14-36 An evacuated enclosure is in the shape of a cube with a side length of 12 cm. The top has an emissivity of
0.85 and a temperature of 800 oC while the bottom has an emissivity of 0.47 and a temperature of 210 oC.
The remaining four sides are perfectly insulated. Find the side wall temperature.

Approach:
Model the inside of the cube as a
three-surface radiation enclosure.
Apply Eqs. 14-39 and 14-40. View
factors may be found using Fig. 14-
16.

Assumptions:
1. Radiation is the only mode of heat
transfer.
2. All surfaces are gray and diffuse.

Solution:
Apply Eq. 14-39 to the three surfaces in the enclosure
J − J 2 J1 − J 3
Q1 = 1 + Eq. (1)
R1→2 R1→3
J − J 3 J 2 − J1
Q 2 = 2 + Eq. (2)
R2→3 R2→1
J − J J − J2
Q 3 = 3 1 + 3 Eq. (3)
R3→1 R3→2
Also apply Eq. 14-40
J = σ T 4 − R Q
1 1 1 1 Eq. (4)
J 2 = σ T24 − R2Q 2 Eq. (5)
J = σ T 4 − R Q
3 3 3 3 Eq. (6)
Because Q 3 = 0,
J 3 = σ T34 Eq. (7)
The radiative resistances are:
1 1 1
R1→2 = R1→3 = R2→3 =
A1 F1→2 A1 F1→3 A2 F2→3
1 1 1 1 1 1
R2→1 = = = R1→2 R3→1 = = = R1→3 R3→2 = = = R2→3
A2 F2→1 A1 F1→2 A3 F3→1 A1 F1→3 A3 F3→2 A2 F2→3
1 − ε1 1− ε2
R1 = R2 =
A1ε1 A2ε 2
The view factor F1→2 may be calculated using Fig. 14-16 with
X 12 cm Y 12 cm
= =1 = =1
L 12 cm L 12 cm
From the plot
F1→2 ≈ 0.2
By summation
F1→3 = 1 − F1→2 = 1 − 0.2 = 0.8
By symmetry
F2→3 = F1→3 = 0.8
The areas and radiative resistances are now calculated using
⎛ 12 ⎞ ⎛ 12 ⎞
A1 = A2 = ⎜ ⎟m⎜ ⎟ m = 0.0144 m
2

⎝ 100 ⎠ ⎝ 100 ⎠

14 -43
 A3 = 4 A1 = 0.0576 m 2
1
R1→2 = = 347 m -2 = R2→1
( 0.0144 m2 ) ( 0.2 )
1
R1→3 = = 86.8 m -2 = R3→1
( 0.0144 m ) ( 0.8)
2

1
R1→3 = = 86.8 m -2 = R3→1
( 0.0144 m ) ( 0.8)
2

1 − 0.85 1 − 0.47
R1 = = 12.25 m -2 R2 = = 78.3m -2
( 0.0144 m ) ( 0.85)
2
( 0.0144 m ) ( 0.47 )
2

We are now prepared to solve Eqs.(1)-(5) simultaneously with the following additional inputs
T1 = 800 o C = 1073K
T2 = 210 o C = 483K
Q 3 = 0
W
σ = 5.67 × 10−8
m2 ⋅ K 4
Using equation-solving software, the result is
Q1 = −Q 2 = 349 W Answer
T3 = 972 K = 699 C o
Answer

14 -44
14-37 A selective surface has a spectral, hemispherical absorptivity which varies with wavelength as shown in the
figure. Find the total hemispherical absorptivity at 185oC.

Approach:
Use Eq. (14-54), obtaining blackbody fractions
from Table 14-3.

Assumptions:
1. The surface is diffuse.

Solution:
Define
λ1 = 2µm α1 = 0.15
λ2 = 3.5µm α 2 = 0.4
α 3 = 0.87
From Eq.(14-54) applied to three bands:
α = α1 F0−λ1T + α 2 Fλ1T −λ2T + α 3 Fλ2T −∞
T = 185 o C = 458 K
λ1T = ( 2µm )( 458 K ) = 916 µm ⋅ K
λ2T = ( 3.5µm )( 458 K ) = 1603 µm ⋅ K
By interpolation in Table 14-3,
F0−λ1T = 0.000193
F0−λ2T = 0.0200
Fλ1T −λ2T = F0−λ2T − F0−λ1T = 0.02 − 0.000193 = 0.0198
Fλ2T −∞ = 1 − F0−λ2T = 1 − 0.02 = 0.98
Therefore
α = ( 0.15 )( 0.000193) + ( 0.4 )( 0.0198 ) + ( 0.87 )( 0.98 ) = 0.861 Answer

Comment:
At this low temperature, very little of the radiation is at wavelengths less than 3.5µm.

14 -45
14-38 The spectral emissivity of a polished aluminum plate is 0.7 for λ < 6.6µm and 0.4 for λ > 6.6µm. Calculate
the power emitted by this plate at 300 K and at 900 K.

Approach:
Multiply spectral emissivity in each band by
the blackbody fraction for that band. Total
emissivity is the sum of all the spectral
contributions.

Assumptions:
1. The plate is diffuse.

Solution:
Define λc = 6.6µm , ε1 = 0.7 , ε 2 = 0.4 , TA = 300 K , and TB = 900 K . The total, hemispherical emissivity of the
plate at TA is
ε A = ε1 F0−λcTA + ε 2 FλcTA −∞
λcTA = ( 6.6µm )( 300 K ) = 1980 µm ⋅ K
By interpolation in Table 14-3,
F0−λcTA = 0.064
FλcTA −∞ = 1 − F0−λcTA = 1 − 0.064 = 0.936
Therefore
ε A = ( 0.7 )( 0.064 ) + ( 0.4 )( 0.936 ) = 0.419
At TB , the total, hemispherical emissivity is
ε B = ε1 F0−λcTB + ε 2 FλcTB −∞
λcTB = ( 6.6µm )( 900 K ) = 5940 µm ⋅ K
By interpolation in Table 14-3,
F0−λcTB = 0.7325
FλcTB −∞ = 1 − F0−λcTB = 1 − 0.7325 = 0.2675
Therefore
ε B = ( 0.7 )( 0.7325 ) + ( 0.4 )( 0.2675 ) = 0.6198
The emissive power at 300 K is
⎛ W ⎞ W
E A = ε Aσ TA 4 = ( 0.419 ) ⎜ 5.67 × 10−8 2 4 ⎟ ( 300 K ) = 192 2
4
Answer
⎝ mK ⎠ m

14 -46
14-39 The tungsten filament of an incandescent light bulb is heated to 2600 K. What fraction of the energy
emitted is in the visible range ( 0.4 to 0.7 µm )? What is the wavelength of maximum emission?

Approach:
Use blackbody fractions to find the total
energy emitted in the visible. Use Wien’s
law to find the wavelength of maximum
emission.

Assumptions:
none

Solution:
The fraction of radiation between λ1 and λ2 is
Fλ1T −λ2T = F0−λ2T − F0−λ1T
λ1T = ( 0.4µm )( 2600 K ) = 1040 µm ⋅ K
λ2T = ( 0.7 µm )( 2600 K ) = 1820 µm ⋅ K
By interpolation in Table 14-3,
F0−λ1T = 0.000684
F0−λ2T = 0.0421
Fλ1T −λ2T = 0.0421 − 0.000684 = 0.0414 Answer
To find the wavelength of maximum emission, apply Wien’s law
λmaxT = 2898µm ⋅ K
2898µm ⋅ K
λmax = = 1.12µm Answer
2600 K

Comments:
Most of the energy emitted by an incandescent light bulb is in the infrared range. Eyes are very sensitive
instruments and can detect light even at low intensities.

14 -47
14-40 A glowing coal can be seen in a darkened room at the Draper point, which is 798 K. What fraction of the
energy emitted by the coal at this temperature is in the visible range (0.4 to 0.7 µm )?

Approach:
Use blackbody fractions to find the total
energy emitted in the visible.

Assumptions:
None

Solution:
The fraction of radiation between λ1 and λ2 is
Fλ1T −λ2T = F0−λ2T − F0−λ1T
λ1T = ( 0.4µm )( 798 K ) = 319 µm ⋅ K
λ2T = ( 0.7 µm )( 798 K ) = 559 µm ⋅ K
Checking in Table 14-3, we find that blackbody fractions for these values of λT are virtually zero. Effectively, we
are out of range of the table. To make further progress, use the equation
15 ∞ e − nz ⎡ 3 3 z 2 6 z 6 ⎤
F0 − λT = 4 ∑ z + + 2 + 3⎥
π n =1 n ⎢⎣ n n n ⎦
where
14387.69µm ⋅ K
z=
λT
The infinite series is best evaluated using appropriate computer software. The series converges very rapidly, so
only one term is needed at these values of λT . The results are
F0−λ1T = 4.009 × 10−16
F0−λ2T = 1.93 × 10−8
The fraction of energy emitted in the visible by the glowing coal is
Fλ1T −λ2T = 1.93 × 10−8 − 4.009 × 10−16 = 1.93 × 10−8 Answer

Comments:
The eyes are incredibly sensitive organs, able to detect radiation even when only 2 parts in 100 million are in the
visible range.

14 -48
14-41 A roughly spherical satellite with a diameter of 1.6 m is in orbit around the sun at an average distance of
6.7 × 108 km. The sun radiates like a black sphere with a radius of 6.95 × 108 m and a temperature of 5800
K. The surface of the satellite is coated with a material having an absorptivity of 0.88 for λ < 2.5 µm and
an absorptivity of 0.12 for λ > 2.5 µm. The satellite rotates and may be assumed to be isothermal at 310 K.
Determine the amount of internal heat generation in the satellite.

Approach:
Calculate the heat flux from the sun striking
the satellite (see example 14-1). Also
determine the total absorptivity and
emissivity of the satellite. Use these to
perform an energy balance on the satellite.

Assumptions:
1. The satellite is isothermal.
2. The satellite is not in the shade of a planet
or moon.
3. The system is in steady state.
4. Outer space is at approximately 0 K.
5. The satellite is diffuse.
6. Solar radiation is emitted uniformly in all
directions.

Solution:
The total energy emitted by the sun is
Q s = Asσ Ts4 = 4π Rs2σ Ts4
where As is the surface area of the sun and Rs is the radius. Substituting values
2⎛ W ⎞
Q s = 4π ( 6.95 ×108 m ) ⎜ 5.67 × 10−8 2 4 ⎟ ( 5800K ) = 3.89 × 1026 W
4

⎝ mK ⎠
To determine the amount of solar radiation impinging on the satellite, draw an imaginary sphere with the sun at
the center. The radius of the sphere is the distance between the sun and the satellite. The heat flux due to solar
radiation on the inside surface of the sphere is
Q Q s 3.89 × 1026 W W
qs" = s = = = 69 2
Asph 4π S 2
⎡ ⎛ 10 m ⎞ ⎤
2
m
4π ⎢( 6.7 × 108 km ) ⎜
3

⎟⎥
⎣ ⎝ 1km ⎠ ⎦
To find the total absorptivity of the satellite, use
α = α1 F0−λT + α 2 (1 − F0−λT
s s
)
We use the solar temperature in this calculation because the energy originates at the sun and has the spectrum of
the sun.
λTs = ( 2.5µm )( 5800 K ) = 14,500 µm ⋅ K
By interpolation in Table 14-3,
F0−λTs = 0.966
The total absorptivity becomes
α = ( 0.88 )( 0.966 ) + ( 0.12 )(1 − 0.966 ) = 0.854
From Kirchoff’s law,
ελ = αλ
Therefore the total emissivity may be calculated using
ε = ε1 F0−λT + ε 2 (1 − F0−λT
sat sat
) =α F
1 0−λTsat (
+ α 2 1 − F0−λTsat )
where Tsat is the temperature of the satellite. The emission occurs at the satellite temperature, so use
λTsat = ( 2.5µm )( 310 K ) = 775 µm ⋅ K
From Table 14-3

14 -49
 F0−λTsat ≈ 0
ε = ( 0.88 )( 0 ) + ( 0.12 )(1) = 0.12
To find the heat absorbed, use the projected area of the satellite in the direction of the sun’s rays
2⎛ W⎞
Q abs = απ rsat qs = ( 0.854 ) π ( 0.8 m ) ⎜ 69 2 ⎟ = 119 W
2 "

⎝ m ⎠
To find the heat emitted, use the actual surface area of the satellite
2⎛ W ⎞
Q emit = ε 4π rsat2 σ Tsat4 = ( 0.12 ) 4π ( 0.8 m ) ⎜ 5.67 × 10−8 2 4 ⎟ ( 310 K ) = 505 W
4

⎝ m ⋅K ⎠
The heat generated is
Q gen = Q emit − Q abs = 505 − 119 = 387 W Answer

14 -50
14-42 The windshield of an automobile is made of glass with a transmissivity of 0.91 between 0.3 and 3 µm.
Outside this range, the transmissivity is virtually zero. Calculate the total transmissivity for the windshield
for solar radiation ( Tsun ≈ 5800 K ) and for radiation from the car seats which are at 20oC.

Approach:
Use an equation similar to Eq. (14-54), with
α replaced by τ .

Assumptions:
none

Solution:
Define
λ1 = 0.3µm λ2 = 3µm
Also define
τ = τ1 λ < λ1
τ =τ2 λ1 < λ < λ2
τ = τ3 λ2 < λ
The total transmissivity for solar radiation is
τ = τ 1 F0−λ1Tsun + τ 2 Fλ1Tsun −λ2Tsun + τ 3 Fλ2Tsun −∞
From the problem statement, τ 1 = τ 3 = 0, therefore
τ = τ 2 Fλ Tsun −λ Tsun = τ 2 ( F0−λ Tsun − F0−λ Tsun )
1 2 2 1

λ1Tsun = ( 0.3µm )( 5800 K ) = 1740 µm ⋅ K

λ2Tsun = ( 3µm )( 5800 K ) = 17, 400 µm ⋅ K
By interpolation in Table 14-3,
F0−λ1Tsun = 0.0335
F0−λ2Tsun = 0.979
τ = 0.91( 0.979 − 0.0335 ) = 0.86 Answer
For radiation from the car seats
λ1Tcar = ( 0.3µm )( 293K ) = 87.9 µm ⋅ K
λ2Tcar = ( 3µm )( 293K ) = 879 µm ⋅ K
By interpolation in Table 14-3,
F0−λ1Tcar ≈ 0
F0−λ2Tcar = 0.000136
τ = 0.91( 0 − 0.000136 ) = 0.000124 Answer

Comment:
This selective transmissivity is called the “greenhouse effect.” Radiation from the sun can get in, but radiation
from interior surfaces cannot get out.

14 -51
14-43 An engineer sends an oxidized metal test sample to the laboratory to determine the emissivity. The lab
reports a value of 0.72 for the total, normal emissivity. What value of total, hemispherical emissivity can
the engineer expect?

Approach:
Use Fig. 14-40b.

Assumptions:
1. The surface is diffuse.

Solution:
Fig. 14-40b gives the rate of hemispheric to
normal emissivity for metals. In this case
ε n ' = 0.72
From Fig. 14-40b, for metals
ε
= 0.93
εn '
Therefore
ε = ( 0.93)( 0.72 ) = 0.67 Answer

Comment:
The oxide layer is a dielectric, while the underlying metal is a conductor. The surface actually behaves somewhere
between these two extremes. The insulator plot in Fig. 14-40 would also give a reasonable estimate.

14 -52
14-44 The normal, spectral emissivity of silicon oxide on aluminum is 0.96 for λ < 1.4 µm and 0.04 for
λ > 1.4 µm. Calculate the total, hemispherical emissivity at 300oC for this surface.

Approach:
Use Fig. 14-40 to determine hemispherical emissivity from normal
emissivity. Use an equation similar to Eq.(14-52) to find total
emissivity.

Assumptions:
1. If λ < 1.4µm, the surface behaves like an insulator, since
emissivity is high.
2. If λ > 1.4µm, the surface behaves like a metal, since emissivity is
low.

Solution:
For λ < 1.4µm, treat the surface as an insulator. From Fig. 14-40a, if ε n ' = 0.96
ε
≈ 0.945
εn '
ε = ( 0.945 )( 0.96 ) = 0.907 = ε1
where ε1 is the hemispherical emissivity for λ < 1.4µm. For λ > 1.4µm, treat the surface as a metal. From Fig. 14-
40b, if ε n ' = 0.04
ε
= 1.28
εn '
ε = (1.28 )( 0.04 ) = 0.0512 = ε 2
where ε 2 is the hemispherical emissivity for λ > 1.4µm. The total, hemispherical emissivity is
ε = ε1 F0−λT + ε 2 FλT −∞
ε = ε1 F0−λT + ε 2 (1 − F0−λT )
λT = (1.4µm )( 573K ) = 802 µm ⋅ K
By interpolation in Table 14-3,
F0−λT ≈ 0
Therefore
ε = ε 2 = 0.0512 Answer

14 -53