2.58 Spring 2006
Midterm 1 Solutions
Question 1.
The peak wavelength of the solar radiation is around 500 nm. For a particle with a
diameter d less than 50 nm, Rayleigh scattering is a good approximation. The absorption
efficiency of the small particle is given by
Qa = 4x
2
 m
Im  2
m
 ( 2 + 0.1i ) −1  0.133π d
− 1  4π d
Im 
=
=
 ( 2 + 0.1i ) 2 + 2 
+2
λ
λ


2
(1)
The solar irradiance that reaches the particle is
 R 
Gλ =  s 
 d se 
2
eb λ (Ts )
 R 
= s 
 d se 
2
C1
λ
5
(2)


 C2 
exp 
 − 1
 λTs 


where Rs is the radius of the sun, dse the distance between the sun and the earth, and C1
and C2 are constants.
(a) The total solar energy absorbed by the particle is
qa = ∫
∞
0
πd2
4
Gλ Qa d λ
(3)
Plug Eqs. (1) and (2) into Eq. (3) to yield
0.133π 2 d 3  Rs
qa =

4
 d se
Let y =



2
Ts 5 ∫
∞
0
C1
( λTs )
6
d ( λTs )
(4)


 C2 
exp 
 −1
 λTs 


1
. Equation (4) becomes
λTs
2
∞
0.133π 2 d 3  Rs 
0.133π 2 d 3  Rs
y 5−1
5
qa =
dy =

 C1Ts ∫0 C2 y

4
4
e −1
 d se 
 d se
( Γ ( 5 ) = 4! = 24 , and for ν ≥ 5 , ζ
( ) ≈ 1 .)



2
C1
Γ ( 5 ) ζ ( 5 ) Ts 5
5
C2
(5)
(b) The scattering efficiency of the small particle is
2
m 2 −1
8
8πd 
Qs = x 4 2
= 

3 λ 
3 m +2
4
( 2 + 0.1i ) −1
2
( 2 + 0.1i ) + 2
2
2
πd 
= 0.673 

 λ 
4
(6)
Similar to part (a), the scattered energy is given by
2
∞
0.673π 5 d 6  Rs 
8
qa =

 C1Ts ∫0
4
 d se 
1
( λTs )
9


 C2 
exp 
 −1
 λTs 


d ( λTs )
2
0.673π 5 d 6  Rs
 C1
=
Γ ( 8 ) ζ ( 8 ) Ts
8


8
4
 d se  C2
Question 2.
(a)
1
glass
2
d
3
The reflectivity at the interfaces is given by
2
ρ = ρ12 = ρ23 = r12
2
N −1
0.5 + iκ 2
= 2
=
N 2 +1
2.5 + iκ 2
2
(i) For λ ≤ 8µ m , κ 2 = 0
ρ12 = 0.04
The peak wavelength of the radiation from the oven is around 1.5 µ
m, which is much less
than the thickness of the glass. We can use ray tracing to calculate the reflectance of the
glass slab.
Rslab =

ρ 1 +

(1− ρ )
1− ρ
2
1− ρ
= 0.923
1+ ρ
Aslab = 0
ελ = 0
2



=
2ρ
= 0.077
1+ ρ
Tslab =
(ii) For λ > 8µ m , κ 2 << 1
ρ ≈ 0.04
τ =e
− κ 2 d⋅4π λ0
=e
−100 λ0
4π
λ0
⋅d
= 0.285
Rslab =

ρ 1 +

2
(1 − ρ ) τ 2  = 0.043
1 − ρ 2τ 2



2
1− ρ ) τ
(
=
= 0.262
1 − ρ 2τ 2
= 1 − Rslab − Tslab = 0.695
Tslab
Aslab
ε λ = 0.695
(b)
Qe
Qconv
Qe
Qa
The glass view port absorbs irradiance from the oven and loses heat by radiation from
both surfaces as well as convection to the air. Since the area of the oven surface is much
larger than that of the glass, the oven can be treated as a blackbody source. The radiation
absorbed by the bottom surface of the glass is then given by:
4
Qa = Aασ Toven
where A is the area of the glass surface (one side), and α is the total hemispherical
absorptance of the glass, which is given by
∞
α=
∫0
ε λ ebλ dλ
4
σ Toven
∞
=
∫8 µ m ε λ ebλ d λ
4
σ Toven
= 0.695 (1− f ( 8µ m ⋅
oven
) ) = 0.018
where f ( λT ) can be found in the table of blackbody emissive power.
Similarly, the total hemispherical emissivity of the glass is given by
(
ε = 0.695 1− f (8µ m ⋅
glass
))
Apply energy balance to the glass view port:
4
4
ασ Toven
= 2ε Tglass
+ h(
glass
(
− T∞ ) = 1.39 1− f ( 8µ m ⋅
glass
))T
4
glass
+ h (Tglass − T∞ )
where we neglect the absorbed radiation from the low temperature environment. Plug in
numbers and the above equation becomes
(
)
4
1.633 ×104 = 1.39 1 − f ( 8µ m ⋅ Tglass ) Tglass
+ 10 ( Tglass − 293)
(c) Iteration with the table value for f, we get, Tglass ≈ 930K