T
0
∫E λ,b dλ
E (λ,b)
(λT)
λ
0
λ
λ 1 λ 2 λ 3
�
�
�
�
�
�
= ε 1 F 0−λ 1 + ε 2 [F 0−λ 2 − F 0−λ 1 ] + ε 3 [1 − F 0−λ 2 ]
(11.15)
= 0.1 × 0 + 0.5 × 0.634 + 0.8 × (1 − 0.634)
where
λ
0
∞
E λ,b dλ
λT
λ
0 E λ,b dλ
E λ,b d(λT)
F 0−λ =
= f (λT)
(11.16)
=
=
σT 4
σT 5
0
λ 2
E λ,b dλ
0
λ 1
σT 4
From Table 11.1 or from Figure 11.9, F 0−λ is a function of λT(μm K), with
T = T s = 500 K, emission from the brick wall.
E λ,b dλ −
E λ,b dλ
0
0
F λ 1 −λ 2
= F 0−λ 2 − F 0−λ 1
(11.17)
=
1.0
0.8
0.6
F (0→λ)
0.4
0.2
0
λT × 10
−3 (μm K)
0
4
8
1 2
16
2 0
228
Analytical Heat Transfer
FIGURE 11.8
Concept of fraction method from a blackbody.
FIGURE 11.9
Fraction of the total blackbody emission in the spectral band from 0 to λ as a function of λT.
0
∫E λ,b dλ
E (λ,b)
(λT)
λ
0
λ
λ 1 λ 2 λ 3
�
�
�
�
�
�
= ε 1 F 0−λ 1 + ε 2 [F 0−λ 2 − F 0−λ 1 ] + ε 3 [1 − F 0−λ 2 ]
(11.15)
= 0.1 × 0 + 0.5 × 0.634 + 0.8 × (1 − 0.634)
where
λ
0
∞
E λ,b dλ
λT
λ
0 E λ,b dλ
E λ,b d(λT)
F 0−λ =
= f (λT)
(11.16)
=
=
σT 4
σT 5
0
λ 2
E λ,b dλ
0
λ 1
σT 4
From Table 11.1 or from Figure 11.9, F 0−λ is a function of λT(μm K), with
T = T s = 500 K, emission from the brick wall.
E λ,b dλ −
E λ,b dλ
0
0
F λ 1 −λ 2
= F 0−λ 2 − F 0−λ 1
(11.17)
=
1.0
0.8
0.6
F (0→λ)
0.4
0.2
0
λT × 10
−3 (μm K)
0
4
8
1 2
16
2 0
228
Analytical Heat Transfer
FIGURE 11.8
Concept of fraction method from a blackbody.
FIGURE 11.9
Fraction of the total blackbody emission in the spectral band from 0 to λ as a function of λT.