Chapter 11
Fundamentals of Thermal Radiation
1185
A small surface emits radiation. The rate of radiation energy emitted through a band is to be
determined.
A
= 2 cm
2
T
= 600 K
40
°
50
°
Assumptions
Surface
A
emits diffusely as a blackbody.
Analysis
The rate of radiation emission from a surface per unit surface
area in the direction (
θ
,
φ
) is given as
φθθ
θφ
d
d
I
dA
Q
d
dE
e
e
sin
cos
)
,
(
=
=
&
The total rate of radiation emission through the band
between 40
°
and 50
°
can be expressed as
4
4
2
0
50
40
1736
.
0
)
1736
.
0
(
)
1736
.
0
(
sin
cos
)
,
(
T
T
I
d
d
I
E
b
e
σ
π
φθ
θθ
=
=
=
=
∫∫
==
since the blackbody radiation intensity is constant (
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 Fall '10
 Dr.DanielArenas
 Thermodynamics, Energy, Mass, Radiation, Black body, Infrared, Thermal radiation

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