HEAT TRANSFER 843

dharm

\M-therm\Th15-4.pm5

= 5.67 ×

5914

100

F^4

HG

I

KJ = 6.936 ×^10

(^7) W/m 2
Example 15.25. Calculate the following for an industrial furnance in the form of a black body
and emitting radiation of 2500°C :
(i)Monochromatic emissive power at 1.2 μm length,
(ii)Wavelength at which the emission is maximum,
(iii)Maximum emissive power,
(iv)Total emissive power, and
(v)Total emissive power of the furnance if it is assumed as a real surface with emissivity equal
to 0.9.
Solution. Given : T = 2500 + 273 = 2773 K ; λ = 12 μm, ε = 0.9
(i)Monochromatic emissive power at 1.2 μm length, (Eλ)b :
According to Planck’s law, ()
exp
E C
C
T
λb
λ
λ

F
HG
I
KJ−
1 −^5
(^21)
...[Eqn. (15.77)]
where, C 1 = 3.742 × 10^8 W. μm^4 /m^2 = 0.3742 × 10–15 W.m^4 /m^2 and
C 2 = 1.4388 × 10–2 mK
Substituting the values, we get
()
exp
Eλb= ×××
×
××
F
HG
I
KJ
−
= ×
−−−
−
0.3742 10 (1.2 10 )
1.4388 10
1.2
(^1565) 1.5 10
–4
14
10 2773
1
74.48
6
= 2.014 × 1012 W/m^2. (Ans.)
(ii)Wavelength at which the emission is maximum, λmax :
According to Wien’s displacement law,
λmax==^28982898
T 2773
= 1.045 μm. (Ans.)
(iii)Maximum emissive power, (E )λbmax :
()Eλbmax = 1.285 × 10–5 T^5 W/m^2 per metre length ...[Eqn. (15.81)]
= 1.285 × 10–5 × (2773)^5 = 2.1 × 10^12 W/m^2 per metre length. (Ans.)
[Note. At high temperature the difference between ()Eλb and ()Eλbmax is very small].
(iv)Total emissive power, Eb :
Eb = σT^4 = 5.67 × 10–8 (2273)^4 = 5.67^2773
100
F^4
HG
I
KJ = 3.352 ×^10
(^6) W/m 2
(v)Total emissive power, E with emisivity (ε) = 0.9
E = ε σT^4 = 0.9 × 5.67
2773
100
F^4
HG
I
KJ = 3.017 ×^10
(^6) W/m (^2). (Ans.)
15.5.10. Radiation Exchange Between Black Bodies Separated by a Non-absorbing
Medium
Refer Fig. 15.52. Let us consider heat exchange between elementary areas dA 1 and dA 2 of two
black radiating bodies, separated by a non-absorbing medium, and having areas A 1 and A 2 and
temperatures T 1 and T 2 respectively. The elemenary areas are at a distance r apart and the normals
to these areas make angles θ 1 and θ 2 with the line joining them. Each elemental area subtends a