SME
Heat Transfer · Problem type
Laminar flow over a flat plate
Use this for “Air at 5 m/s flows along a 1 m plate; ν = 1.6 × 10⁻⁵ m²/s, Pr = 0.71, k = 0.026 W/(m·K). Find the average h.”
Diagram
ReL
=
VL
ν
=
5
×
1
1.6
×
10
-5
=
312,500,
laminar
Nu = 0.664√ReL ∛Pr = 331.1
h
=
Nu
k
L
=
8.61
W/(
m
2
·K)
The layer grows as √x from the leading edge, so h is highest there and falls along the plate.
This is an example. Type your own values to start a new problem.
ReL
=
V (m/s)
×
L (m)
ν (m²/s)
=
0.664
×
Re
×
3
Pr
h (W/(m²·K))
=
Nu
×
k (W/(m·K))
÷
L
Formulas
ReL
=
V
×
L
ν
312,500
=
5
×
1
1.6 ×
10
-5
Nu = 0.664 × √ReL × ∛Pr
331.1407
=
0.664
×
√
312
,500
×
∛0.71
h
=
Nu
×
k
L
8.6097
=
331.1407
×
0.026
1
Assumptions
•
Laminar boundary layer the whole length: ReL from about 10³ to 500,000.
•
Plate at one temperature; h is the average over the length L.
•
ν, Pr and k are read at the film temperature (Tₛ + T∞) ÷ 2.
Step-by-step
WE KNOW
V = 5 m/s, L = 1 m, ν = 1.6 × 10⁻⁵ m²/s, Pr = 0.71, k = 0.026 W/(m·K)
FIND
Reynolds number (ReL), average Nusselt number (Nu), average film coefficient (h)
Step 1 · Find Reynolds number (ReL)
Use ReL = V × L ÷ ν
Re
weighs
inertia
against
viscosity:
speed
times
L
,
over
the
kinematic
viscosity.
ReL
=
5
×
1
1.6 ×
10
-5
ReL = 312,500
Step 2 · Find average Nusselt number (Nu)
Use Nu = 0.664 × √ReL × ∛Pr
The laminar plate’s average Nu grows as the square root of Re and the cube root of Pr.
Nu
=
0.664
×
√
312
,500
×
∛0.71
Nu = 0.664 × 559.017 × 0.8921
Nu = 331.1407
Step 3 · Find average film coefficient (h)
Use h = Nu × k ÷ L
Nu
is
hL
k
:
multiply
it
by
k
,
then
divide
by
L
.
h
=
331.1407
×
0.026
1
h
=
8.6097
W/(
m
2
·K)
Check
Put the numbers back in. Both sides should match.
312,500
=
5
×
1
1.6 ×
10
-5
✓
331.1407
=
0.664
×
√
312
,500
×
∛0.71
✓
8.6097
=
331.1407
×
0.026
1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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