SME
Transport Phenomena
Momentum transport
Use this for “Water flows through a tube of radius 1 mm and length 1 m under a 1000 Pa pressure drop. Find Q, vₘₐₓ and τw.”
Diagram
vₘₐₓ
=
2 vavg
=
2
×
0.125
=
0.25
m/s
at
the
centre,
0
at
the
wall.
τw
=
ΔPR
2L
=
0.5
Pa.
Re = 250: laminar.
Not to scale: the tube is far longer than it is wide.
This is an example. Type your own values to start a new problem.
Q (m³/s)
=
π
×
ΔP (Pa)
×
(
R (mm)
)
4
8
×
μ (Pa·s)
×
L (m)
τw (Pa)
=
ΔP (Pa)
×
R (mm)
2
×
L (m)
μ is the liquid’s viscosity, about 0.001 Pa·s for water near 20 °C.
vavg
Mean speed
calculated
vₘₐₓ
Centre speed
calculated
Re
Reynolds number
calculated
Formulas
Q
=
π
×
ΔP
×
R
4
8 ×
μ
×
L
3.927 ×
10
-7
=
π
×
1,000
×
0.001
4
8 × 0.001 × 1
vavg
=
Q
π ×
R
2
0.125
=
3.927 ×
10
-7
π ×
0.001
2
vₘₐₓ = 2 × vavg
0.25 = 2 × 0.125
τw
=
ΔP
×
R
2 ×
L
0.5
=
1,000
×
0.001
2 × 1
Re
=
1,000
×
vavg
×
2
×
R
μ
250
=
1,000
×
0.125
×
2
×
0.001
0.001
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
Steady, laminar (Re < 2100), Newtonian and fully developed flow in a straight, round tube.
•
A shell balance on a thin cylinder of fluid gives the parabola v = vₘₐₓ(1 − r² ÷ R²).
•
Re uses ρ = 1000 kg/m³ (water); for another liquid, scale Re by its density ÷ 1000.
Step-by-step
WE KNOW
ΔP = 1,000 Pa, L = 1 m, R = 1 mm, μ = 0.001 Pa·s
FIND
flow rate (Q), mean speed (vavg), centre speed (vₘₐₓ), wall shear stress (τw), Reynolds number (Re)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
R = 1 mm = 0.001 m (1 m = 1000 mm)
Step 2 · Find flow rate (Q)
Use Q = π × ΔP × R⁴ ÷ (8 × μ × L)
Hagen–Poiseuille: the shell balance’s parabola, its speeds added over the cross-section.
Q
=
π
×
1,000
×
0.001
4
8 × 0.001 × 1
Q
=
π
×
1,000 ×
10
-12
0.008
Q
=
3.927 ×
10
-7
m
3
/s
Step 3 · Find mean speed (vavg)
Use vavg = Q ÷ (π × R²)
The flow spread over the cross-section.
vavg
=
3.927 ×
10
-7
π ×
0.001
2
vavg
=
3.927 ×
10
-7
3.1416 ×
10
-6
vavg = 0.125 m/s
Step 4 · Find centre speed (vₘₐₓ)
Use vₘₐₓ = 2 × vavg
For a parabola the centre is twice the mean.
vₘₐₓ = 2 × 0.125
vₘₐₓ = 0.25 m/s
Step 5 · Find wall shear stress (τw)
Use τw = ΔP × R ÷ (2 × L)
A force balance: ΔP × πR² pushes, τw × 2πRL holds back.
τw
=
1,000
×
0.001
2 × 1
τw
=
1
2
τw = 0.5 Pa
Step 6 · Find Reynolds number (Re)
Use Re = 1,000 × vavg × 2 × R ÷ μ
Inertia
over
viscosity
on
the
diameter
2
R
,
with
ρ
=
1000
kg/
m
3
;
under
2100
it
stays
laminar.
Re
=
1,000
×
0.125
×
2
×
0.001
0.001
Re
=
125
×
2
×
0.001
0.001
Re
=
250
×
0.001
0.001
Re
=
0.25
0.001
Re = 250
Check
Put the numbers back in. Both sides should match.
3.927 ×
10
-7
=
π
×
1,000
×
0.001
4
8 × 0.001 × 1
✓
0.125
=
3.927 ×
10
-7
π ×
0.001
2
✓
0.25 = 2 × 0.125 ✓
0.5
=
1,000
×
0.001
2 × 1
✓
250
=
1,000
×
0.125
×
2
×
0.001
0.001
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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