SME
Transport Phenomena · Problem type
A liquid film falling down a wall
Use this for “An oil (ρ = 900 kg/m³, μ = 0.05 Pa·s) runs down a vertical wall in a film 0.8 mm thick. How fast does it flow on average?”
Diagram
The film is still at the wall and fastest at its free surface, where nothing drags it: vₘₐₓ = 1.5 vavg = 0.06131 m/s.
This is an example. Type your own values to start a new problem.
vavg (m/s)
=
ρ (kg/m³)
×
9.81
3
×
μ (Pa·s)
×
(
δ (mm)
)
2
×
cos(
β
°)
Formulas
vavg
=
ρ
×
9.81
×
δ
2
×
cos(
β
)
3 ×
μ
0.04088
=
1,000
×
9.81
×
0.0005
2
×
cos(60°)
3 × 0.01
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
Steady laminar flow; the film’s thickness doesn’t change and the air drags nothing at the free surface.
•
A shell balance across the film gives half a parabola, fastest at the free surface.
•
β is the wall’s angle from vertical (0° for a vertical wall); g = 9.81 m/s².
•
The film stays smooth and laminar only for Re = 4ρvavgδ ÷ μ below about 20; ripples appear above it.
Step-by-step
WE KNOW
ρ = 1,000 kg/m³, μ = 0.01 Pa·s, δ = 0.5 mm, β = 60°
FIND
mean speed (vavg)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
δ = 0.5 mm = 0.0005 m (1 m = 1000 mm)
Step 2 · Find mean speed (vavg)
Use vavg = ρ × 9.81 × δ² × cos(β) ÷ (3 × μ)
Gravity along the wall against viscosity: the mean of the half parabola.
vavg
=
1,000
×
9.81
×
0.0005
2
×
cos(60°)
3 × 0.01
vavg
=
1,000
×
9.81
×
(2.5
×
10
-7
)
×
0.5
0.03
vavg
=
0.0012263
0.03
vavg = 0.04088 m/s
Check
Put the numbers back in. Both sides should match.
0.04088
=
1,000
×
9.81
×
0.0005
2
×
cos(60°)
3 × 0.01
✓
Topic check
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