12-1
12-2
Electromagnetic and Thermal Radiation
12-2C Electromagnetic waves are characterized by their frequency
v
and wavelength
. These two properties in a medium
are related by
vc /=
where c is the speed of light in that medium.
12-3C Thermal radiation is the radiation emitted as a result of vibrational and rotational motions of molecules, atoms and
electrons of a substance, and it extends from about 0.1 to 100
m
in wavelength. Unlike the other forms of electromagnetic
12-6C Light (or visible) radiation consists of narrow bands of colors from violet to red. The color of a surface depends on its
12-7C Because the snow reflects almost all of the visible and ultraviolet radiation, and the skin is exposed to radiation both
from the sun and from the snow.
0.40 m. The human body does not emit any radiation in the ultraviolet region since bodies at room temperature emit
radiation in the infrared region only.
12-3
12–10 A cordless telephone operates at a frequency of 8.5108 Hz. The wavelength of these telephone waves is to be
determined.
Hz(1/s) 105.8
8
v
12-11 Electricity is generated and transmitted in power lines at a frequency of 50 Hz. The wavelength of the electromagnetic
waves is to be determined.
12-4
12-14 A microwave oven operates at a frequency of 2.2109 Hz. The wavelength of these microwaves and the energy of each
microwave are to be determined.
Analysis The wavelength of these microwaves is
m/s 10998.2
8
c
12-5
12-6
Blackbody Radiation
12-19C Spectral blackbody emissive power is the amount of radiation energy emitted by a blackbody at an absolute
temperature T per unit time, per unit surface area and per unit wavelength about wavelength
. The integration of the
spectral blackbody emissive power over the entire wavelength spectrum gives the total blackbody emissive power,
4
12-7
12–23 A thin–walled tube’s inner surface is coated with polypropylene lining. The maximum amount of total radiation
emission rate per unit length of the tube that can be achieved without exceeding the maximum use temperature for
polypropylene lining is to be determined.
Assumptions 1 The tube behaves as blackbody. 2 Uniform surface temperature. 3 Thin-walled tube has negligible thermal
12–24 An isothermal cubical body is suspended in the air. The rate at which the cube emits radiation energy and the spectral
blackbody emissive power are to be determined.
Assumptions The body behaves as a black body.
Analysis (a) The total blackbody emissive power is determined from Stefan-Boltzman Law to be
222 m 24.0)2.0(66 === aAs
W8928=== −)m 24.0(K) 900)(K. W/m1067.5()( 244284
sb ATTE
(b) The spectral blackbody emissive power at a wavelength of 4
m
is determined
from Plank‘s distribution law,
μmkW/m 6.84 2==
−
=
−
=
μm W/m6841
1
K) m)(900 4(
Km 1043878.1
expm) 4(
/mm W1074177.3
1exp
2
4
5
248
2
5
1
T
C
C
Eb
T = 900 K
20 cm
20 cm
20 cm
12-8
12-9
12–26 An ASTM A479 904L stainless steel bar has a maximum use temperature of 260°C set by the ASME Code for
Process Piping. (a) The spectral blackbody emissive powers at the limits of the thermal radiation spectrum for the maximum
use temperature are to be determined and (b) The maximum spectral blackbody emissive power that the bar can reach without
(0.1 μm)5[exp(1.43878×104μm∙K
(0.1 μm)(260+273)K)−1]=𝟐.𝟏𝟖𝟔×𝟏𝟎−𝟏𝟎𝟒W/m2⋅μm ≈𝟎
At λ = 100 μm and the maximum use temperature 260°C,
𝐸𝑏𝜆=3.74177×108W∙μm4/m2
2897.8 μm⋅K)−1]=𝟓𝟓𝟑.𝟓 W/m2⋅μm
12–10
12-27 The blackbody temperature and the total emissive power at a given wavelength and its
corresponding emissive power are to be determined.
Assumptions 1 Blackbody radiation.
Analysis (a) Using the Planck’s law find the blackbody radiation
]1)/[exp(
),(
5
1
1−
=
TC
C
TEb
12–11
12–12
12-29 The spectral blackbody emissive power of the sun versus wavelength in the range of 0.01 m to 1000 m is to
be plotted.
Analysis The problem is solved using EES, and the solution is given below.
“GIVEN”
T=5780 [K]
lambda=0.01[micrometer]
“ANALYSIS”
E_b_lambda=C_1/(lambda^5*(exp(C_2/(lambda*T))-1))
C_1=3.742E8 [W-micrometer^4/m^2]
C_2=1.439E4 [micrometer-K]
[m]
Eb,
[W/m2–m]
0.01
0
10.11
12684
20.21
846.3
30.31
170.8
40.41
54.63
50.51
22.52
60.62
10.91
70.72
5.905
80.82
3.469
90.92
2.17
…
…
…
…
909.1
0.0002198
919.2
0.0002103
929.3
0.0002013
939.4
0.0001928
949.5
0.0001847
959.6
0.000177
969.7
0.0001698
979.8
0.0001629
989.9
0.0001563
1000
0.0001501
0.01 0.1 1 10 100 1000 10000
0.0001
0.001
0.01
0.1
1
10
100
1000
10000
100000
[ m]
Eb [W/m2– m]
12–13
12–14
12–15
12–33 A thin vertical plate, modeled as blackbody, is subjected to uniform heat flux on one side and exposed to radiation and
natural convection on the other side. The plate surface temperature is to be determined.
Assumptions 1 The plate emits radiation as a blackbody. 2 Thermal properties are constant. 3 Plate surface temperature is
uniform. 4 Heat loss from plate’s side surface is negligible. 5 The surroundings are treated as an isothermal surface, Tsurr =
T∞.
12-34 A circular plate that is modeled as a blackbody is heated by an electrical heater with an efficiency of 80%. The electric
power required to keep the plate surface temperature at 200°C is to be determined.
12–16
12-35 An incandescent light bulb emits 15% of its energy at wavelengths shorter than 0.8
m. The temperature of the filament is to be determined.
Assumptions The filament behaves as a black body.
12–36 The temperature of the filament of an incandescent light bulb is given. The fraction of visible radiation emitted by the
Analysis The visible range of the electromagnetic spectrum extends from
m 76.0 tom 40.0 21
==
. Noting that T =
2500 K, the blackbody radiation functions corresponding to
TT 21 and
are determined from Table 12-2 to be
000321.0mK 1000=K) m)(2500 40.0(
1
1
=⎯→⎯=
λ
fT
m 1.16
=
=⎯→⎯= K 2500
Km 8.2897
Km 8.2897)( powermax powermax
T
Discussion Note that the radiation emitted from the filament peaks in the infrared region.
12–37 The temperature of the filament of an incandescent light bulb is given. The fraction of visible radiation emitted by the
Analysis The visible range of the electromagnetic spectrum extends from
m 76.0 tom 40.0 21
==
. Noting that T =
2800 K, the blackbody radiation functions corresponding to
TT 21 and
are determined from Table 12-2 to be
0014088.0mK 1120=K) m)(2800 40.0(
1
1
=⎯→⎯=
λ
fT
m 1.035
=
=⎯→⎯= K 2800
Km 8.2897
Km 8.2897)( powermax powermax
T
T = ?
12–17
12–18
12–19
12–41E The sun is at an effective surface temperature of 10,400 R. The rate of infrared radiation energy emitted by the sun is
to be determined.
Assumptions The sun behaves as a black body.
Analysis Noting that T = 10,400 R = 5778 K, the blackbody radiation functions corresponding to
TT 21 and
are determined from Table 12-2 to be
0.1mK 577,800=K) m)(5778 100(
547370.0mK 4391.3=K) m)(5778 76.0(
2
1
2
1
=⎯→⎯=
=⎯→⎯=
λ
λ
fT
fT
Then the fraction of radiation emitted between these two wavelengths becomes
453.0547.00.1
12 =−=−
ff
(or 45.3%)
The total blackbody emissive power of the sun is determined from Stefan-Boltzmann Law to be
2744284 Btu/h.ft 10005.2R) 400,10)(R.Btu/h.ft 101714.0( === −
TEb
Then,
26 Btu/h.ft 109.08 === )Btu/h.ft 10005.2)(453.0()453.0( 27
infrared b
EE
12–42 A glass window transmits 90% of the radiation in a specified wavelength range and is opaque for radiation at other
wavelengths. The rate of radiation transmitted through this window is to be determined for two cases.
W10775.5)m 9(K) 5800()K. W/m1067.5()( 8244284 === −
sb ATTE
The fraction of radiation in the range of 0.3 to 3.0 m is
97875.0mK 17,400=K) m)(5800 0.3(
03345.0mK 1740=K) m)(5800 30.0(
2
1
2
1
=⎯→⎯=
=⎯→⎯=
λ
λ
fT
fT
9453.003345.097875.0
12 =−=−=
fff
Noting that 90% of the total radiation is transmitted through the window,
=
)(90.0
transmit TfEEb
273232.0mK 3000=K) m)(1000 0.3(
2
2
=⎯→⎯=
λ
fT
0273232.0
12 −=−=
fff
and
Sun
T = 10,400 R
Glass
= 0.9
L = 3 m
Sun
12–20
12–43 The radiation energy emitted within the visible light region by daylight and candlelight is to be determined.
Km 2320)K 5800)(m 40.0(
1==
T
→
124509.0
daylight ,
1=
f
4255.0124509.0550015.0
daylight ,
21 =−=
−
f
Hence, the radiation energy emitted (for daylight) is determined using
=== −
−− )4255.0()K 5800)(K W/m1067.5(),()( 4428
daylight ,
4
, 21
2
21
fTTETE bb
1==
candle ,
1
Km 1368)K 1800)(m 76.0(
2==
T
→
006885.0
candle ,
2=
f
006875.00000096.0006885.0
candle ,
21 =−=
−
f
Hence, the radiation energy emitted (for candlelight) is determined using
=== −
−− )006875.0()K 1800)(K W/m1067.5(),()( 4428
daylight ,
4
, 21
2
21
fTTETE bb