SME
Heat Transfer · Problem type
The radiation heat transfer coefficient
Use this for “A surface with ε = 0.8 is at 400 K in surroundings at 300 K. Find its radiation coefficient hᵣ and the net radiation flux.”
Diagram
hᵣ
=
εσ(Tₛ
+
Tₛᵤᵣᵣ)(
Tₛ
2
+
Tₛᵤᵣᵣ
2
)
=
7.938
W/(
m
2
·K)
in
kelvins;
radiated
q″
=
hᵣ(Tₛ
−
Tₛᵤᵣᵣ)
=
793.8
W/
m
2
hᵣ writes the radiation like a film, so it can sit beside h for the combined loss.
This is an example. Type your own values to start a new problem.
hᵣ (W/(m²·K))
=
ε
×
5.67
×
10
−8
×
(
Tₛ (K)
+
Tₛᵤᵣᵣ (K)
)
×
(
T
s
²
+
T
surr
²)
q″ (W/m²)
=
h
r
×
(
T
s
−
T
surr
)
Formulas
hᵣ
=
ε
×
5.67 ×
10
-8
×
(Tₛ
+
Tₛᵤᵣᵣ)
×
(
Tₛ
2
+
Tₛᵤᵣᵣ
2
)
7.938
=
0.8
×
5.67 ×
10
-8
×
(400
+
300)
×
(
400
2
+
300
2
)
q″ = hᵣ × (Tₛ − Tₛᵤᵣᵣ)
793.8 = 7.938 × (400 − 300)
Assumptions
•
A small gray surface in large surroundings; σ = 5.67 × 10⁻⁸ W/(m²·K⁴).
•
hᵣ writes radiation like convection, q″ = hᵣ(Tₛ − Tₛᵤᵣᵣ), so it adds to h for combined loss.
•
Temperatures in kelvins; q″ is positive from the surface to the surroundings.
Step-by-step
WE KNOW
Tₛ = 400 K, Tₛᵤᵣᵣ = 300 K, ε = 0.8
FIND
radiation coefficient (hᵣ), net radiation flux (q″)
Step 1 · Find radiation coefficient (hᵣ)
Use hᵣ = ε × 5.67 × 10⁻⁸ × (Tₛ + Tₛᵤᵣᵣ) × (Tₛ² + Tₛᵤᵣᵣ²)
Tₛ
4
−
Tₛᵤᵣᵣ
4
factors
as
(Tₛ
−
Tₛᵤᵣᵣ)(Tₛ
+
Tₛᵤᵣᵣ)(
Tₛ
2
+
Tₛᵤᵣᵣ
2
);
all
but
the
difference
is
hᵣ.
hᵣ
=
0.8
×
5.67 ×
10
-8
×
(400
+
300)
×
(
400
2
+
300
2
)
hᵣ
=
0.8
×
(5.67
×
10
-8
)
×
700
×
(160,000
+
90,000)
hᵣ
=
0.8
×
(5.67
×
10
-8
)
×
700
×
250,000
hᵣ
=
(4.536
×
10
-8
)
×
700
×
250,000
hᵣ
=
(3.175
×
10
-5
)
×
250,000
hᵣ
=
7.938
W/(
m
2
·K)
Step 2 · Find net radiation flux (q″)
Use q″ = hᵣ × (Tₛ − Tₛᵤᵣᵣ)
Like convection: the coefficient times the temperature difference.
q″ = 7.938 × (400 − 300)
q″ = 7.938 × 100
q″
=
793.8
W/
m
2
Check
Put the numbers back in. Both sides should match.
7.938
=
0.8
×
5.67 ×
10
-8
×
(400
+
300)
×
(
400
2
+
300
2
)
✓
793.8 = 7.938 × (400 − 300) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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