SME
Heat Transfer · Problem type
Heat loss through pipe insulation
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.”
Diagram
Rcond
=
ln(r₂ ÷ r₁)
2πkL
=
ln(1.6)
2π × 0.05 W/(m·K) × 10 m
=
0.1496
K/W:
the
insulation’s
drop
follows
ln
r,
steep
inside,
flatter
out.
q = 668.4 W: the temperature difference over the total resistance.
This is an example. Type your own values to start a new problem.
R (K/W)
=
ln(
r₂ (mm)
÷
r₁ (mm)
)
2π
×
k (W/(m·K))
×
L (m)
q (W)
=
ΔT (K)
÷
R
Formulas
R
=
ln(
r₂
÷
r₁
)
2π ×
k
×
L
0.1496
=
ln(80 ÷ 50)
2π × 0.05 × 10
q
=
ΔT
R
668.4188
=
100
0.1496
Assumptions
•
Steady radial conduction; ΔT is the insulation’s inner surface minus its outer surface.
•
The pipe’s own thin metal wall conducts well: its resistance is left out.
•
Only the ratio r₂ ÷ r₁ matters, so the radii may be in any one unit.
Step-by-step
WE KNOW
r₁ = 50 mm, r₂ = 80 mm, k = 0.05 W/(m·K), L = 10 m, ΔT = 100 K
FIND
conduction resistance (R), heat lost (q)
Step 1 · Find conduction resistance (R)
Use R = ln(r₂ ÷ r₁) ÷ (2π × k × L)
Through
a
cylinder’s
wall
the
area
grows
with
r,
so
the
resistance
goes
as
ln(
r₂
r₁
),
not
as
a
thickness.
R
=
ln(80 ÷ 50)
2π × 0.05 × 10
R
=
0.1496068
K/W
Step 2 · Find heat lost (q)
Use q = ΔT ÷ R
Heat flows like current: the temperature difference over the insulation’s resistance.
q
=
100
0.1496068
q
=
668.4188
W
Check
Put the numbers back in. Both sides should match.
0.1496068
=
ln(80 ÷ 50)
2π × 0.05 × 10
✓
668.4188
=
100
0.1496068
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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Conduction
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