SME
Heat Transfer
Conduction
Use this for “A wall is 0.2 m of brick (k = 0.72 W/(m·K)) and 0.05 m of foam (k = 0.04 W/(m·K)). Inside air is 20 °C with h = 10 W/(m²·K); outside is −10 °C with h = 25. Find the heat flux.”
Diagram
R″
=
1
hᵢ
+
L₁
k₁
+
L₂
k₂
+
1
hₒ
R″ = 0.1 + 0.2778 + 1.25 + 0.04
R″ = 1.668 m²·K/W
q″
=
Tᵢₙ − Tₒᵤₜ
R
″
=
30
1.668
q″
=
17.99
W/
m
2
Each
drop
is
q″
×
R:
1.8
K,
5
K,
22.5
K,
0.72
K;
the
steepest
line
is
through
the
foam,
the
layer
with
the
most
L
k
.
This is an example. Type your own values to start a new problem.
R″ (m²·K/W)
=
1
hᵢ (W/(m²·K))
+
L₁ (m)
k₁ (W/(m·K))
+
L₂ (m)
k₂ (W/(m·K))
+
1
hₒ (W/(m²·K))
q″ (W/m²)
=
Tᵢₙ (°C)
−
(
Tₒᵤₜ (°C)
)
R″
Ts,i
Inside surface temperature
worked out (not typed) · °C
Formulas
R″
=
1
hᵢ
+
L₁
k₁
+
L₂
k₂
+
1
hₒ
1.6678
=
1
10
+
0.2
0.72
+
0.05
0.04
+
1
25
q″
=
Tᵢₙ − Tₒᵤₜ
R
″
17.988
=
20 − (−10)
1.6678
Ts,i = Tᵢₙ − q″ ÷ hᵢ
18.2012
=
20
−
17.988
10
Assumptions
•
Steady, one-dimensional conduction; the layers touch perfectly, with no contact resistance.
•
Per square metre of wall: the films and layers are resistances in series.
•
The heat flux q″ is positive from the inside to the outside.
Step-by-step
WE KNOW
Tᵢₙ = 20 °C, Tₒᵤₜ = −10 °C, hᵢ = 10 W/(m²·K), L₁ = 0.2 m, k₁ = 0.72 W/(m·K), L₂ = 0.05 m, k₂ = 0.04 W/(m·K), hₒ = 25 W/(m²·K)
FIND
total resistance per m² (R″), heat flux (q″), inside surface temperature (Ts,i)
Step 1 · Find total resistance per m² (R″)
Use R″ = 1 ÷ hᵢ + L₁ ÷ k₁ + L₂ ÷ k₂ + 1 ÷ hₒ
Resistances
in
series
add,
like
resistors:
each
film
is
1
h
,
each
layer
L
k
.
R″
=
1
10
+
0.2
0.72
+
0.05
0.04
+
1
25
R″ = 0.1 + 0.2778 + 1.25 + 0.04
R″ = 1.6678 m²·K/W
Step 2 · Find heat flux (q″)
Use q″ = (Tᵢₙ − Tₒᵤₜ) ÷ R″
Heat flows like current: the temperature difference over the total resistance gives the flux through every layer.
q″
=
20 − (−10)
1.6678
q″
=
30
1.6678
q″
=
17.988
W/
m
2
Step 3 · Find inside surface temperature (Ts,i)
Use Ts,i = Tᵢₙ − q″ ÷ hᵢ
The flux crosses the inside film first: it drops q″ ÷ hᵢ below the room air.
Ts,i
=
20
−
17.988
10
Ts,i = 20 − 1.7988
Ts,i = 18.2012 °C
Check
Put the numbers back in. Both sides should match.
1.6678
=
1
10
+
0.2
0.72
+
0.05
0.04
+
1
25
✓
17.988
=
20 − (−10)
1.6678
✓
18.2012
=
20
−
17.988
10
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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Use this for “A 10 m pipe has insulation from radius 50 mm to 80 mm, k = 0.05 W/(m·K). Its inner surface is 100 K hotter than its outer. Find the heat lost.”
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Use this for “A 10 mm steel ball (ρ = 7800 kg/m³, c = 450 J/(kg·K), k = 40 W/(m·K)) at 300 °C cools in 25 °C air with h = 50 W/(m²·K). What is its temperature after 120 s?”
Heat from a pin fin
Use this for “An aluminum pin fin (k = 200 W/(m·K)) is 0.005 m across and 0.05 m long, in air with h = 20 W/(m²·K). Its base is 60 K above the air. How much heat does it shed?”