CHAPTER 15
15.1 Derive the jump boundary condition for the diffusion approximation, equation (15.26), for the case of concen-
tric cylinders. Assume the heat transfer to be one-dimensional (only radial, no azimuthal or axial dependence).
Hint: Introduce a local Cartesian coordinate system at a point at the boundary, express any other r-location
within the medium in terms of x,y,z, and transform the derivatives in equation (15.26) to r-derivatives; finally
let x,y,zgo to zero (since the derivatives at the boundary are needed).
378 RADIATIVE HEAT TRANSFER
15.2 The gap between two parallel black plates at T1and T2, respectively, is filled with a particle-laden gas.
Radiative equilibrium prevails, the particle loading is a fixed volume fraction, with particles manufactured
from two different materials (one a specular reflector, the other a diffuse reflector, both having the same ǫ).
Sketch nondimensional heat flux Ψ = q/σ(T4
1−T4
2) vs. particle size (but keeping volume-fraction constant).
CHAPTER 15 379
15.3 Consider radiative equilibrium of a gray, absorbing, emitting, and isotropically scattering medium contained
between two isothermal, gray-diffuse, parallel plates spaced a distance Lapart. Determine the nondimen-
sional temperature variation within the medium, Φ = (σT4−J2)/(J1−J2) for the optically thin case (τL≪1).
380 RADIATIVE HEAT TRANSFER
15.4 Consider a gray, absorbing-emitting, linear-anisotropically scattering medium at radiative equilibrium. The
medium is confined between two parallel, isothermal, black plates (at temperatures T1and T2). Determine
an expression for the radiative heat flux between the two plates using the diffusion approximation with jump
boundary conditions.
σ(T4
1−T4
2)=1
1+3
41−A1ω
3τL
382 RADIATIVE HEAT TRANSFER
15.5 Do Problem 15.4 using the Schuster-Schwarzschild (2-flux) approximation.
σ(T4
1−T4
2)=1
1+1−A1ω
4τL
CHAPTER 15 383
15.6 Do Problem 15.4 using the Milne-Eddington (differential) approximation.
σ(T4
1−T4
2)=1
1+3
41−A1ω
3τL
384 RADIATIVE HEAT TRANSFER
15.7 Do Problem 15.4 using the exponential kernel approximation method.
Solution
For linear-anisotropic scattering, instead of starting from equation (15.2), we need to start with equa-
0
−2ZτL
τ
Eb(τ′)E2(τ′−τ)dτ′+A1ω
2q2
3−E4(τ)−E4(τL−τ).
Using the kernel approximation this leads to equation (15.48) with an additional term,
−τ)/2+3
d2q
dτ2=9
4Eb1e−3τ/2−9
4Eb2e−3(τL
−τ)/2+3dEb
dτ
−9
4×A1ω
6qe−3τ/2−e−3(τL
=9
4q−A1ω
3q+3dEb
dτ.
Since, for radiative equilibrium, q=const and d2q/dτ2=0 this implies
q=−4
dEb
0
Eb(τ′)e−3(τ−τ′)/2dτ′=2
3Eb(τ′)e−3(τ−τ′)/2
τ
0−2
3Zτ
0
dEb
dτ′e−3(τ−τ′)/2dτ′
=2
3Eb(τ)−2
3Eb(0) e−3τ/2+1
21−A1ω
3qZτ
0
e−3(τ−τ′)/2dτ′
3Eb(τ)−2
3Eb(τL)e−3(τL
31−A1ω
3q1−e−3(τL
CHAPTER 15 385
Substituting these relations into equation (15.7-A)
Note that the terms without exponentials cancel each other; similarly, the terms in brackets must be zero, i.e.,
Subtracting the two from one another leads to
σ(T4
1−T4
2)=1
1+3
41−A1ω
3τL
386 RADIATIVE HEAT TRANSFER
15.8 Do Problem 14.7 using the Milne-Eddington (differential) approximation.
0 K (since all radiation escapes toward z→ ∞, no radiation enters from z→ ∞ toward smaller z). Thus, from
CHAPTER 15 387
15.9 Do Problem 14.11 using the Milne-Eddington (differential) approximation.
2+√3 cosh(1
2√3 20)
| {z }
=1
=2.1436 ×[1 −0.73778] ×5.670 ×10−8W
388 RADIATIVE HEAT TRANSFER
15.10 Consider a space enclosed by infinite, diffuse-gray, parallel plates 1 m apart filled with a gray, nonscattering
medium (κ=5 m−1). The surfaces are isothermal (both at Tw=500 K with emittance ǫw=0.6), and there
is uniform and constant heat generation within the medium per unit volume, ˙
Q′′′ =106W/m3. Conduction
and convection are negligible such that ∇·q=˙
Q′′′. Determine the radiative heat flux to the walls as well
as the maximum temperature within the medium, using the diffusion approximation with jump boundary
conditions.
Eb(0) −Jw=2
dEb
d2Eb
3Φ′
b(0) −1
2Φ′′
b(0),
C=2
3×3
8τL−1
2×−3
4=3
8+τL
4.
Thus,
The maximum temperature occurs at the center, τ=τL/2:
Φmax =1
ǫw−1
2τL
2+3
8 1+τ2
L
4!.
CHAPTER 15 389
Substituting the given values: τL=5 m−1×1 m =5,
390 RADIATIVE HEAT TRANSFER
15.11 Do Problem 15.10 using the Schuster-Schwarzschild approximation.