SME
Heat Transfer · Problem type
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?”
Diagram
m
=
√(
hP
kAc
)
=
√(
4h
kD
)
=
8.944
m⁻¹,
so
mL
=
0.447.
q
=
√
hPkAc
θb
tanh(mL)
=
0.8843
W:
the
fin
cools
along
its
length,
θ
=
θb
cosh
m(L − x)
cosh(mL)
(adiabatic
tip).
This is an example. Type your own values to start a new problem.
m (m⁻¹)
=
4
×
h (W/(m²·K))
k (W/(m·K))
×
D (m)
q (W)
=
π
2
×
h
×
k
×
D³
×
θb (K)
×
tanh(m
×
L (m)
)
For a pin the picture’s √(hPkAc) becomes (π ÷ 2) × √(hkD³), the factor in the second line.
ηf
Fin efficiency
worked out (not typed)
Formulas
m
=
√(4
×
h
k
×
D
)
8.9443
=
√(4
×
20
200 × 0.005
)
q
=
(
π
2
)
×
√
h
×
k
×
D
3
×
θb
×
tanh(
m
×
L
)
0.8843
=
(
π
2
)
×
√
20 × 200 ×
0.005
3
×
60
×
tanh(8.9443
×
0.05)
ηf
=
tanh(
m
×
L
)
m
×
L
0.9383
=
tanh(8.9443 × 0.05)
8.9443 × 0.05
Assumptions
•
Steady conduction along the fin only; h is the same all along it.
•
The tip loses no heat (adiabatic tip); θb is the base’s temperature above the air.
•
For a pin, P = πD and Ac = πD² ÷ 4, so P ÷ Ac = 4 ÷ D.
Step-by-step
WE KNOW
h = 20 W/(m²·K), k = 200 W/(m·K), D = 0.005 m, L = 0.05 m, θb = 60 K
FIND
fin parameter (m), heat from the fin (q), fin efficiency (ηf)
Step 1 · Find fin parameter (m)
Use m = √(4 × h ÷ (k × D))
m
=
√(
hP
kAc
),
and
for
a
pin
P
Ac
=
4
D
,
so
m
=
√(
4
h
kD
).
m
=
√(4
×
20
200 × 0.005
)
m
=
√
4 × 20
m
=
√
80
m
=
8.9443
m⁻¹
Step 2 · Find heat from the fin (q)
Use q = (π ÷ 2) × √(h × k × D³) × θb × tanh(m × L)
For
a
pin
√
hPkAc
=
(
π
2
)
√
hkD
³
;
times
θb,
times
tanh(mL)
for
the
tip
that
loses
nothing.
q
=
(
π
2
)
×
√
20 × 200 ×
0.005
3
×
60
×
tanh(8.9443
×
0.05)
q
=
1.5708
×
0.02236
×
60
×
0.4196
q
=
0.8843
W
Step 3 · Find fin efficiency (ηf)
Use ηf = tanh(m × L) ÷ (m × L)
The fin’s heat over what it would shed if it were all at the base temperature.
ηf
=
tanh(8.9443 × 0.05)
8.9443 × 0.05
ηf
=
tanh(0.4472)
0.4472
ηf
=
0.4196
0.4472
ηf = 0.9383
Check
Put the numbers back in. Both sides should match.
8.9443
=
√(4
×
20
200 × 0.005
)
✓
0.8843
=
(
π
2
)
×
√
20 × 200 ×
0.005
3
×
60
×
tanh(8.9443
×
0.05)
✓
0.9383
=
tanh(8.9443 × 0.05)
8.9443 × 0.05
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Conduction
Main lesson
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.”
Lumped cooling and the Biot check
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?”