SME
University Physics I: Mechanics · Problem type
Heat flow through a wall
Use this for “How much heat flows through a concrete wall 10 m² in area and 20 cm thick when inside and outside differ by 20 °C?”
Diagram
R″
=
L₁
k₁
=
0.1429
=
0.1429
m²·K/W.
q″
=
Tᵢₙ − Tₒᵤₜ
R
″
=
20
0.1429
q″
=
140
W/
m
2
Each
drop
is
q″
×
R:
20
K;
the
steepest
line
is
through
the
wall,
the
layer
with
the
most
L
k
.
Heat flow P = 1,400 W
Thermal resistance R = 0.01429 K/W
This is an example. Type your own values to start a new problem.
P (
)
=
k (W/(m·K))
×
A (m²)
×
ΔT (
)
L (
)
Tᵢₙ
Inside face temperature
example · °C
Tₒᵤₜ
Outside face temperature
example · °C
R
Thermal resistance
worked out (not typed) · K/W
Formulas
ΔT = Tᵢₙ − Tₒᵤₜ
+20 = 20 − 0
P
=
k
×
A
×
ΔT
L
+1,400
=
1.4
×
10
×
20
0.2
R
=
L
k
×
A
0.01429
=
0.2
1.4 × 10
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
Steady, one-dimensional conduction through a flat wall of one material; Tᵢₙ and Tₒᵤₜ are its two faces.
•
Heat flows from hot to cold: P > 0 is outward, from the inside face to the outside face.
•
R = L ÷ (kA), so P = ΔT ÷ R; per square metre of wall it is R″ = L ÷ k. Concrete: k ≈ 1.4 W/(m·K).
Step-by-step
WE KNOW
k = 1.4 W/(m·K), A = 10 m², L = 20 cm, Tᵢₙ = 20 °C, Tₒᵤₜ = 0 °C
FIND
temperature difference (ΔT), heat flow (P), thermal resistance (R)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
L = 20 cm = 0.2 m (1 m = 100 cm)
Step 2 · Find temperature difference (ΔT)
Use ΔT = Tᵢₙ − Tₒᵤₜ
The inside face less the outside face; a difference of 1 °C is 1 K.
ΔT = 20 − 0
ΔT = +20 K
Step 3 · Find heat flow (P)
Use P = k × A × ΔT ÷ L
Fourier’s law: the heat flow grows with the conductivity, the area and the temperature difference, and falls with the thickness.
P
=
1.4
×
10
×
20
0.2
P
=
14
×
20
0.2
P
=
280
0.2
P
=
+1,400
W
Step 4 · Find thermal resistance (R)
Use R = L ÷ (k × A)
The
wall’s
thermal
resistance:
P
=
ΔT
R
,
as
a
current
is
a
voltage
over
a
resistance.
R
=
0.2
1.4 × 10
R
=
0.2
14
R
=
0.01429
K/W
Check
Put the numbers back in. Both sides should match.
+20 = 20 − 0 ✓
+1,400
=
1.4
×
10
×
20
0.2
✓
0.01429
=
0.2
1.4 × 10
✓
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The final temperature of a mixture
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First and second laws of thermodynamics
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