SME
Transport Phenomena · Problem type
Couette flow between two plates
Use this for “Oil of viscosity 0.3 Pa·s fills a 1 mm gap under a 0.5 m² plate moving at 2 m/s. Find the shear stress and the force.”
Diagram
A
straight
line
from
0
at
the
still
plate
to
V
at
the
moving
one;
τ
=
μV
h
=
600
Pa
everywhere
in
the
gap.
This is an example. Type your own values to start a new problem.
τ (Pa)
=
μ (Pa·s)
×
V (m/s)
h (mm)
F (N)
=
τ (Pa)
×
A (m²)
Formulas
τ
=
μ
×
V
h
600
=
0.3
×
2
0.001
F
=
τ
×
A
300 = 600 × 0.5
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
Steady laminar flow of a Newtonian fluid; the bottom plate is still and no pressure drop acts along the gap.
•
The fluid sticks to both plates (no slip), so the speed rises in a straight line from 0 to V.
Step-by-step
WE KNOW
μ = 0.3 Pa·s, V = 2 m/s, h = 1 mm, A = 0.5 m²
FIND
shear stress (τ), force on the plate (F)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
h = 1 mm = 0.001 m (1 m = 1000 mm)
Step 2 · Find shear stress (τ)
Use τ = μ × V ÷ h
Newton’s
law
of
viscosity
with
a
straight
profile:
the
slope
dv/dy
is
V
h
.
τ
=
0.3
×
2
0.001
τ
=
0.6
0.001
τ
=
600
Pa
Step 3 · Find force on the plate (F)
Use F = τ × A
The stress acts over the whole plate.
F
=
600
×
0.5
F
=
300
N
Check
Put the numbers back in. Both sides should match.
600
=
0.3
×
2
0.001
✓
300 = 600 × 0.5 ✓
Topic check
See what you know: 5 questions on this topic
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