Advanced Heat and Mass Transfer by Amir Faghri, Yuwen Zhang, and John R. Howell 9.2 The Blackbody as the Ideal Radiator A material that absorbs 100 percent of the energy incident on it from all directions and at all wavelengths (i.e., has no reflection at all) is defined as a blackbody. Consider such a blackbody element suspended within an evacuated enclosure at uniform temperature T (Fig. 9.3). The blackbody will absorb all energy incident on the blackbody that originates by emission from the surface of the enclosure. This absorbed thermal radiant energy will increase the internal energy (and thus temperature) of the blackbody until the blackbody reaches the temperature of the enclosure surface. The perfectly absorbing blackbody must also be the best possible emitter of thermal radiation.
G T Figure 9.3 Radiating evacuated enclosure containing a blackbody.
9.2.1 The Planck Distribution and its Consequences Planck's fundamental relation for the rate of energy emission (into all directions) from an ideal blackbody is 2π C1 E (, ) (9.1) λ T = 5 C2 / λ T λ e 1 ( ) A more general relation for E can be formed by dividing Eq. (9.1) by T 5, resulting in Eλ b 2π C1 = (9.2) 5 5 C2 / λ T T λ T e 1 ( ) ( ) The maximum in the blackbody function occurs at a particular value of the λt product, given by ( λ T ) = C max 3 = 2897.8( µ m K) (9.3)
10 8 10 7 E (W/m 2 -sr-µ m) 10 6 10 5 10 4 1000 100 400 K 2000 K 5780 K Figure 9.4: Planck blackbody spectral emissive power from Eq. (9.1) 10 1 0 5 10 15 20 Wavelength, λ, (µ m)
1.4 10-11 E /T 5 (W/m 2 -µ m-sr-k 4 ) 1.2 10-11 1 10-11 8 10-12 6 10-12 4 10-12 Figure 9.5 Plot of generalized spectral blackbody emissive power. 2 10-12 0 500 1000 2898 10 4 5 10 4 λ T, (µ m-k)
For engineering calculation, the total energy emitted by a blackbody surface at all wavelengths is generally of interest, and can be found by integrating Eq. (9.1) over all wavelengths. This can be done through a change of variables using ξ = C / λ T 2 and results in 4 3 2π C1 2π C1T ξ 4 Eb = d λ = ( ) dξ = σ T λ = 0 5 C 4 2 / λ T 0 1 C ξ = ξ λ e 2 e 1 (9.4) where s is the Stefan-Boltzmann constant, again made up of a combination of the more fundamental constants, 5 4 σ 2 C1π /(15 C2 ). It has a numerical value in SI units of σ -8 2 4 = 5.6704 10 W/m K.
ds dθ da s R R dω Figure 9.6 Relation of angle and radius for (a): Planar angle and (b) Solid angle ds = R dθ da s = R 2 dω (a) (b)
I λb R θ da s θ da φ da da p = da cos θ Figure 9.7 Projected area and intensity
Spectral blackbody intensity is defined as the rate of energy de λ emitted per unit solid angle subtended by da s, per unit projected area da cosθ, per unit wavelength interval, or deλ I = dacosθ dω d λ (9.5) The energy leaving da in the direction of da s is, from Eq. (9.5) das deλ, da da = I ( ) cos ( ) cos s θ da θ d Ω d λ = I θ da θ d 2 R λ (9.6) Similarly, the energy from da s that is incident on da is da cosθ deλ, da ( 0) ( 0) s da = I θ s = das d Ω d λ = I θ s = das d λ 2 R (9.7) Equating (9.6) and (9.7) results in I ( θ ) = I ( θ = 0) = I s (9.8)
Examination of Fig. 9.8 allows an alternate definition of the solid angle dω subtended by the area da s on the surface of the hemisphere in terms of the angles θ and φ as sinθ φ θ Using Eq. (9.6) gives E = de = I cosθ d Ω ( ) ( ) das ds1 ds R d Rd 2 d Ω = = = = 2 2 2 R R R A λ, da das ω = 2π s 2 π π / 2 π / 2 sinθ dθ dφ = I cosθ sinθ dθ dφ = 2π I cosθ sinθ dθ = π I φ = 0 θ = 0 θ = 0 (9.9) (9.10)
I λb ds 1 R R sin θ da s θ ds 2 da φ ds 1 = R sin θ dφ ds 2 = R dθ Figure 9.8 Solid angle in terms of angles dθ and dφ
I λb 0 θ π / 2 0 φ 2 π and. θ dθ φ dφ 0 φ 2π Figure 9.9: Hemisphere for integration of intensity over 0 θ π / 2 and
9.2.2 The Blackbody Fraction 0 λ T * ( ) Integration of equation (9.2) over the range results in the fraction of blackbody emission in that range λ T as E Eλ b F0 λ T = d( λ T ) * / d( λ T ) * ( λ T )* = 0 5 ( T ) * 0 5 T λ = T 1 2π C λ T 1 = d ( λ T ) * ( T )* 0 5 C2 /( λ T ) * σ λ = ( λ T ) * ( e 1) (9.11) This can be integrated by parts using the same change of variables as for Eq. (9.4) F nξ 2 15 e 3 3ξ 6ξ 6 0 λ T = 4 ξ + + + 2 3 π n= 1 n n n n (9.12) λ T
Given Eq. (9.11), the blackbody emissive power in any interval (λ T) 1 to (λ T) 2 can be found from 1 ( λ T ) 2 2π C1 F( ) 2 ( ) = d T λ T λ T 1 ( λ T ) * ( λ T ) 5 C 1 2 λ T σ = * 1 0 ( λ T ) 0 ( λ T ) 2 1 ( λ T ) / ( )* ( e ) ( λ ) 1 ( λ T ) 2 2π C ( λ T ) 1 1 2π C1 = d( λ T )* d( λ T )* ( T )* 0 5 C2 /( λ T )* ( )* 0 5 C2 /( T )* ( T )* [ e 1] T λ σ λ = λ λ = ( λ T )* [ e 1] = F F * (9.13)
1 0.8 F 0-λ T 0.6 0.4 0.2 Figure 9.10 Plot of the blackbody fraction vs. λt product. 0 1000 10 4 10 5 λ T (µ m-k)
1.4 10-11 1.2 10-11 E /T 5 W/(m 2 -µ m-sr-k 5 ) 1 10-11 8 10-12 6 10-12 4 10-12 2 10-12 Figure 9.11 Blackbody energy in the ranges 0 < λt < (λt) 1 and 0 < λt < (λt) 2 0 1000 (λ T) (λ T) 10 4 1 2 λ T (µ m-k)
Example 9.1 Find the fraction of emission from blackbodies at T = 1500 K and at T = 5780 K (the apparent temperature of the Sun) that lie in the range 1 λ 5 μm.
Solution: Using Eq. (9.12), for T = 1500K, at (λt) 1 = 1500 μm-k, F 0-1500 = 0.01285, and at (λt) 2 = 7500 μm-k, F 0-7500 = 0.83436, so F 1500-7500 = F 0-7500 - F 0-1500 = 0.83436-0.01285 = 0.82151. Thus, 82.2 percent of the blackbody emission lies in the prescribed range for a 1500K blackbody. For T = 5780 K, F 0-5780 = 0.71828 and F 0-28900 = 0.99390, so F 5780-28900 = 0.99390-0.71828 = 0.27562, or 27.6 percent of the solar energy lies in the range 1 λ 5 μm. This smaller fraction than for the 1500K blackbody reflects the fact that the peak of the blackbody curve at the solar effective temperature is in the shorter wavelength range and thus so is a large fraction of the solar energy.