SME
Transport Phenomena · Problem type
Heat lost from an insulated pipe; the critical radius
Use this for “A steam line of outer radius 0.03 m at 120 °C gets 0.02 m of insulation (k = 0.05 W/(m·K)) in 25 °C air with h = 8 W/(m²·K). Find the heat lost per metre.”
Diagram
Rcond
=
ln(r₂ ÷ r₁)
2πk
=
ln(1.6)
2π × 0.04 W/(m·K)
=
1.87
K/W
per
m:
the
insulation’s
drop
follows
ln
r,
steep
inside,
flatter
out.
Rconv
=
1
2πr₂h
=
0.1989
K/W
per
m,
in
series
after
it.
q′ = 62.83 W/m: the temperature difference over the total resistance.
rc
=
k
h
=
0.004
m;
r₂
is
past
it,
so
more
insulation
always
cuts
the
loss.
This is an example. Type your own values to start a new problem.
Rcond (m·K/W)
=
ln
r₂ (m)
r₁ (m)
2π
×
k (W/(m·K))
Rconv (m·K/W)
=
1
2π
×
r₂ (m)
×
h (W/(m²·K))
q′ (W/m)
=
Tᵢ (°C)
−
T∞ (°C)
Rcond (m·K/W)
+
Rconv (m·K/W)
rc
Critical radius
calculated · m
Formulas
Rcond
=
ln(
r₂
÷
r₁
)
2π ×
k
1.8701
=
ln(0.08 ÷ 0.05)
2π × 0.04
Rconv
=
1
2π ×
r₂
×
h
0.1989
=
1
2π × 0.08 × 10
q′
=
Tᵢ − T∞
Rcond + Rconv
62.8314
=
150 − 20
1.8701 + 0.1989
rc
=
k
h
0.004
=
0.04
10
Assumptions
•
Steady radial conduction per metre of pipe; the insulation’s inside surface is at Tᵢ.
•
The pipe wall is thin and conducts well, so its own resistance is left out.
•
More insulation cuts q′ only once r₂ is past the critical radius rc = k ÷ h.
Step-by-step
WE KNOW
r₁ = 0.05 m, r₂ = 0.08 m, k = 0.04 W/(m·K), h = 10 W/(m²·K), Tᵢ = 150 °C, T∞ = 20 °C
FIND
conduction resistance (Rcond), film resistance (Rconv), heat lost per metre (q′), critical radius (rc)
Step 1 · Find conduction resistance (Rcond)
Use Rcond = ln(r₂ ÷ r₁) ÷ (2π × k)
The
area
grows
with
r
through
the
wall,
so
the
resistance
goes
as
ln(
r₂
r₁
),
not
as
a
thickness.
Rcond
=
ln(0.08 ÷ 0.05)
2π × 0.04
Rcond
=
0.47
0.25133
Rcond = 1.87009 m·K/W
Step 2 · Find film resistance (Rconv)
Use Rconv = 1 ÷ (2π × r₂ × h)
The
outside
film
is
1
h
spread
over
the
outer
surface,
2πr₂
for
each
metre.
Rconv
=
1
2π × 0.08 × 10
Rconv
=
1
5.0265
Rconv = 0.19894 m·K/W
Step 3 · Find heat lost per metre (q′)
Use q′ = (Tᵢ − T∞) ÷ (Rcond + Rconv)
The two resistances are in series: add them, then divide the temperature difference by the sum.
q′
=
150 − 20
1.87009 + 0.19894
q′
=
130
2.069
q′ = 62.8314 W/m
Step 4 · Find critical radius (rc)
Use rc = k ÷ h
Below the critical radius, more insulation adds surface faster than it adds resistance.
rc
=
0.04
10
rc = 0.004 m
Check
Put the numbers back in. Both sides should match.
1.87009
=
ln(0.08 ÷ 0.05)
2π × 0.04
✓
0.19894
=
1
2π × 0.08 × 10
✓
62.8314
=
150 − 20
1.87009 + 0.19894
✓
0.004
=
0.04
10
✓
Topic check
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