SME
Fluid Mechanics · Problem type
A turbulent boundary layer on a flat plate
Use this for “Water at 3 m/s flows along a 2 m plate 1 m wide. How thick is the turbulent layer at the end, what is Cf, and what is the drag on one side?”
Diagram
ReL
=
VL
ν
=
6,000,000.
δ
=
0.37x
Reₓ⁰
·²
grows
as
x
0
·⁸:
0.0326
m
at
the
end,
0.0187
m
halfway.
Heights are stretched 29 times to be seen.
Skin-friction coefficient Cf = 0.003263
Density ρ = 1,000 kg/m³
Plate width b = 1 m
Drag force FD = 29.37 N
This is an example. Type your own values to start a new problem.
δ (mm)
=
0.37
×
L (m)
ReL
0.2
Cf
=
0.074
ReL
0.2
FD (N)
=
½
×
ρ (kg/m³)
×
(
V (m/s)
)
2
×
Cf
×
b (m)
×
L (m)
ν
Kinematic viscosity
example · m²/s
Formulas
ReL
=
V
×
L
ν
6,000,000
=
3
×
2
1 ×
10
-6
δ
=
0.37
×
L
Re
_L0.2
0.03263
=
0.37
×
2
6,000
,
000
0.2
Cf
=
0.074
Re
_L0.2
0.003263
=
0.074
6,000
,
000
0.2
FD
=
1
2
×
ρ
×
V
2
×
Cf
×
b
×
L
29.37
=
1
2
×
1,000
×
3
2
×
0.003263
×
1
×
2
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
A smooth flat plate with no pressure change along it, turbulent from the leading edge.
•
The one-seventh-power velocity profile: the fits hold for ReL from 5 × 10⁵ to about 10⁷.
•
Cf is averaged over the whole plate; the drag is Cf times ½ρV² times the wetted area.
Step-by-step
WE KNOW
V = 3 m/s, L = 2 m, ν = 1 × 10⁻⁶ m²/s, ρ = 1,000 kg/m³, b = 1 m
FIND
Reynolds number at L (ReL), layer thickness at L (δ), skin-friction coefficient (Cf), drag force (FD)
Step 1 · Find Reynolds number at L (ReL)
Use ReL = V × L ÷ ν
Re
at
the
trailing
edge:
the
stream
speed
times
the
plate’s
length,
over
ν
.
ReL
=
3
×
2
10
-6
ReL
=
6
10
-6
ReL = 6,000,000
Step 2 · Find layer thickness at L (δ)
Use δ = 0.37 × L ÷ ReL0.2
The
turbulent
fit
δ
=
0.37x
Re
_x0.2
grows
as
x
0.8
,
faster
than
a
laminar
layer.
δ
=
0.37
×
2
6,000
,
000
0.2
δ
=
0.37
×
2
22.68
δ
=
0.74
22.68
δ
=
0.03263
m
Step 3 · Find skin-friction coefficient (Cf)
Use Cf = 0.074 ÷ ReL0.2
The turbulent skin friction averaged over the whole plate.
Cf
=
0.074
6,000
,
000
0.2
Cf
=
0.074
22.68
Cf = 0.003263
Step 4 · Find drag force (FD)
Use FD = ½ × ρ × V² × Cf × b × L
Drag
is
Cf
times
the
dynamic
pressure
1
2
ρV²
times
the
wetted
area
bL.
FD
=
1
2
×
1,000
×
3
2
×
0.003263
×
1
×
2
FD
=
1
2
×
1,000
×
9
×
0.003263
×
2
FD = 500 × 9 × 0.003263 × 2
FD = 4,500 × 0.003263 × 2
FD = 14.6835 × 2
FD = 29.37 N
Step 5 · Convert the answers
Change each answer back to the unit you chose.
δ = 0.03263 m = 32.63 mm (1 m = 1000 mm)
Check
Put the numbers back in. Both sides should match.
6,000,000
=
3
×
2
10
-6
✓
0.03263
=
0.37
×
2
6,000
,
000
0.2
✓
0.003263
=
0.074
6,000
,
000
0.2
✓
29.37
=
1
2
×
1,000
×
3
2
×
0.003263
×
1
×
2
✓
Topic check
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