SME
Heat Transfer
Radiation
Use this for “A gray surface with ε = 0.8 and area 2 m² is at 400 K in a large room whose walls are at 300 K. Find the net heat it radiates.”
Diagram
Out:
εσTₛ⁴
=
1,161
W/
m
2
;
in
(absorbed):
εσTₛᵤᵣᵣ⁴
=
367.4
W/
m
2
;
the
arrows
are
to
one
scale,
3.16
:
1
=
(
Tₛ
Tₛᵤᵣᵣ
)
4
.
q = εσA(Tₛ⁴ − Tₛᵤᵣᵣ⁴) = 2 m² × 793.8 W/m² = 1,588 W, in kelvins.
This is an example. Type your own values to start a new problem.
q (W)
=
ε
×
5.67
×
10
−8
×
A (m²)
×
(
(
Tₛ (K)
)
4
−
(
Tₛᵤᵣᵣ (K)
)
4
)
Formulas
q
=
ε
×
5.67 ×
10
-8
×
A
×
(
Tₛ
4
−
Tₛᵤᵣᵣ
4
)
1,587.6
=
0.8
×
5.67 ×
10
-8
×
2
×
(
400
4
−
300
4
)
Assumptions
•
A gray surface (it absorbs as well as it emits, ε) inside surroundings much larger than it.
•
σ = 5.67 × 10⁻⁸ W/(m²·K⁴); both temperatures in kelvins (add 273.15 to °C).
•
The net heat q is positive from the surface to the surroundings.
Step-by-step
WE KNOW
Tₛ = 400 K, Tₛᵤᵣᵣ = 300 K, ε = 0.8, A = 2 m²
FIND
net heat radiated (q)
Step 1 · Find net heat radiated (q)
Use q = ε × 5.67 × 10⁻⁸ × A × (Tₛ⁴ − Tₛᵤᵣᵣ⁴)
The
surface
sends
out
εσTₛ⁴
and
takes
in
εσTₛᵤᵣᵣ⁴
per
square
metre;
the
net,
times
A
.
q
=
0.8
×
5.67 ×
10
-8
×
2
×
(
400
4
−
300
4
)
q
=
0.8
×
(5.67
×
10
-8
)
×
2
×
((2.56
×
10
10
)
−
(8.1
×
10
9
))
q
=
0.8
×
(5.67
×
10
-8
)
×
2
×
(1.75
×
10
10
)
q
=
(4.536
×
10
-8
)
×
2
×
(1.75
×
10
10
)
q
=
(9.072
×
10
-8
)
×
(1.75
×
10
10
)
q
=
1,587.6
W
Check
Put the numbers back in. Both sides should match.
1,587.6
=
0.8
×
5.67 ×
10
-8
×
2
×
(
400
4
−
300
4
)
✓
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A blackbody’s peak wavelength and emissive power
Topic check
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