Heat Transfer

Convection

Use this for “Water at 1 m/s is heated in a 25 mm tube; ν = 0.8 × 10⁻⁶ m²/s, Pr = 5.4, k = 0.615 W/(m·K). Find h.”

Diagram

DV = 1 m/s, D = 0.025 mh = 4,380 W/(m²·K) at the wallWarm film k ÷ h ≈ 0.14 mm (drawn thicker)
Re = 31,250, Pr = 5.4, Nu = 178.1
h
=
Nu
k
D
=
178.1
×
0.615
0.025
=
4,380
W/(
m
2
·K):
the
heat
crosses
a
thin
film
at
the
wall,
where
the
water
barely
moves.
This is an example. Type your own values to start a new problem.
Re
=
V (m/s)
×
D (m)
ν (m²/s)
=
0.023
×
Re
0.8
×
Pr
h (W/(m²·K))
=
Nu
×
k (W/(m·K))
÷
D

Formulas

Re
=
V
×
D
ν
31,250
=
1
×
0.025
8 ×
10
-7
Nu
=
0.023
×
Re0.8
×
Pr^
n
178.0596
=
0.023
×
31,
250
0.8
×
5.4
0.4
h
=
Nu
×
k
D
4,380.2658
=
178.0596
×
0.615
0.025

Assumptions

•
Fully developed turbulent flow in a smooth tube, Re above 10,000.
•
Prandtl exponent n = 0.4 when the fluid is heated, 0.3 when it is cooled.
•
ν, Pr and k are read at the fluid’s mean temperature.

Step-by-step

WE KNOW
V = 1 m/s, D = 0.025 m, ν = 8 × 10⁻⁷ m²/s, Pr = 5.4, n = 0.4, k = 0.615 W/(m·K)
FIND
Reynolds number (Re), Nusselt number (Nu), film coefficient (h)
Step 1 · Find Reynolds number (Re)
Use Re = V × D ÷ ν
Re
weighs
inertia
against
viscosity:
speed
times
D
,
over
the
kinematic
viscosity.
Re
=
1
×
0.025
8 ×
10
-7
Re = 31,250
Step 2 · Find Nusselt number (Nu)
Use Nu = 0.023 × Re0.8 × Prⁿ
Dittus–Boelter:
the
turbulent
tube’s
Nu
grows
as
Re
to
the
0.8
and
Pr
to
the
n
.
Nu
=
0.023
×
31,
250
0.8
×
5.4
0.4
Nu = 0.023 × 3,943.5 × 1.9632
Nu = 90.7005 × 1.9632
Nu = 178.0596
Step 3 · Find film coefficient (h)
Use h = Nu × k ÷ D
Nu
is
hD
k
:
multiply
it
by
k
,
then
divide
by
D
.
h
=
178.0596
×
0.615
0.025
h
=
109.5
0.025
h
=
4,380.266
W/(
m
2
·K)
Check
Put the numbers back in. Both sides should match.
31,250
=
1
×
0.025
8 ×
10
-7
✓
178.0596
=
0.023
×
31,
250
0.8
×
5.4
0.4
✓
4,380.266
=
178.0596
×
0.615
0.025
✓
Topic check
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