Special case 2—Radiation between two parallel surfaces with middle shields as
shown in Figure 13.4:
A 1 σ(T 1
4 − T 2
4 )
q 1 = q 12 =
(1 − ε 1 )/ε 1 + (1/F 13 ) + (1 − ε 31 )/ε 31
+(1 − ε 32 )/ε 32 + (1/F 32 ) + (1 − ε 2 )/ε 2
(13.10)
A 1 σ(T 1
4 − T 2
4 )
=
= −q 2
(1/ε 1 ) + (1 − ε 31 )/ε 31 + (1 − ε 32 )/ε 32 + (1/ε 2 )
where F 13 = F 32 = 1.
To cut down radiation heat loss, ε 31 and ε 32 should be small, that is, ρ 31 is
large.
Special case 3—Reradiating surfaces (insulated surface, q R = 0): The following
electric furnaces, as shown in Figure 13.5, can be modeled as radiation heat
transfer between two opposite surfaces (hot and cold) with a third reradiation
side surface (perfect reflection and perfect insulation).
q 12
Radiation
shield
q 1
ε 31
ε 32
1
3
2
A 1
A 3
A 2
ε 1
ε 2
σT
4
1
σT
4
1
E b1
J 1
J 31
E b3
J 31
J 2
E b2
1– ε 1
1
1– ε 31 1– ε 32
1
1– ε 2
A 1 ε 1
A 1 F 13
A 3 ε 31
A 3 ε 32
A 3 F 32
A 2 ε 2
1
q
2
q
−
261
Radiation Exchange in a Nonparticipating Medium
FIGURE 13.4
Radiation between two parallel surfaces with middle shield.