SME
Heat Transfer · Problem type
Radiation between two parallel gray plates
Use this for “Two large parallel plates face each other: one at 500 K with ε = 0.8, the other at 300 K with ε = 0.6. Find the net radiation flux between them.”
Diagram
1
ε₁
+
1
ε₂
−
1
=
1.25
+
1.667
−
1
1
ε₁
+
1
ε₂
−
1
=
1.917
q″
=
σ(
T₁
4
−
T₂
4
)
1.917
=
3,084
1.917
q″
=
1,609
W/
m
2
Black plates would trade σ(T₁⁴ − T₂⁴); each gray surface adds a resistance in series.
This is an example. Type your own values to start a new problem.
q″ (W/m²)
=
5.67
×
10
−8
×
(
T₁ (K)
)
4
−
(
T₂ (K)
)
4
1
ε₁
+
1
ε₂
−
1
Formulas
q″
=
5.67 ×
10
-8
×
T₁
4
−
T₂
4
1 ÷
ε₁
+ 1 ÷
ε₂
− 1
1,609.2939
=
5.67 ×
10
-8
×
500
4
−
300
4
1 ÷ 0.8 + 1 ÷ 0.6 − 1
Assumptions
•
Both plates are gray and diffuse, and so large that each sees only the other (F₁₂ = 1).
•
σ = 5.67 × 10⁻⁸ W/(m²·K⁴); temperatures in kelvins.
•
The net flux q″ is positive from plate 1 to plate 2.
Step-by-step
WE KNOW
T₁ = 500 K, T₂ = 300 K, ε₁ = 0.8, ε₂ = 0.6
FIND
net radiation flux (q″)
Step 1 · Find net radiation flux (q″)
Use q″ = 5.67 × 10⁻⁸ × (T₁⁴ − T₂⁴) ÷ (1 ÷ ε₁ + 1 ÷ ε₂ − 1)
Black
plates
would
trade
σ(
T₁
4
−
T₂
4
);
each
gray
surface
adds
a
resistance,
so
divide
by
1
ε₁
+
1
ε₂
−
1.
q″
=
5.67 ×
10
-8
×
500
4
−
300
4
1 ÷ 0.8 + 1 ÷ 0.6 − 1
q″
=
5.67 ×
10
-8
×
5.44 ×
10
10
1.9167
q″
=
3,084.5
1.9167
q″
=
1,609.294
W/
m
2
Check
Put the numbers back in. Both sides should match.
1,609.294
=
5.67 ×
10
-8
×
500
4
−
300
4
1 ÷ 0.8 + 1 ÷ 0.6 − 1
✓
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