SME
Fluid Mechanics · Problem type
Laminar pipe flow: the pressure drop
Use this for “Oil with μ = 0.1 Pa·s and ρ = 900 kg/m³ flows at 10⁻⁴ m³/s through 10 m of 20 mm tube. Find the pressure drop.”
Diagram
The
energy
grade
line
falls
by
hL
=
2.884
m
along
the
pipe;
the
hydraulic
grade
line
runs
V
2
2g
=
0.005164
m
under
it.
Heads are to the scale at the left; the pipe’s length is not. ΔP = ρghL = 25,460 Pa.
Dynamic viscosity μ = 0.1 Pa·s
Friction factor f = 1.117
This is an example. Type your own values to start a new problem.
ΔP (kPa)
=
128
×
μ (Pa·s)
×
L (m)
π
×
(
D (mm)
)
4
×
Q (m³/s)
ρ
Density
example · kg/m³
V
Mean speed
calculated · m/s
Re
Reynolds number
calculated
f
Friction factor
calculated
hL
Head loss
calculated · m
Formulas
Q
=
π
4
×
D
2
×
V
0.0001
=
π
4
×
0.02
2
×
0.3183
Re
=
ρ
×
V
×
D
μ
57.3
=
900
×
0.3183
×
0.02
0.1
f
=
64
Re
1.117
=
64
57.3
ΔP
=
128
×
μ
×
L
×
Q
π ×
D
4
25,460
=
128
×
0.1
×
10
×
0.0001
π ×
0.02
4
ΔP
=
ρ
×
9.81
×
hL
25,460 = 900 × 9.81 × 2.884
Formulas are worked in m, Pa. The step-by-step shows the conversions.
Assumptions
•
Steady, fully developed laminar flow (Re under 2300) in a round tube.
•
The fluid is Newtonian: its viscosity μ is the same at every shear rate.
•
hL is the drop as a head of the flowing fluid, ΔP ÷ ρg; g = 9.81 m/s².
Step-by-step
WE KNOW
μ = 0.1 Pa·s, ρ = 900 kg/m³, L = 10 m, D = 20 mm, Q = 0.0001 m³/s
FIND
mean speed (V), Reynolds number (Re), friction factor (f), pressure drop (ΔP), head loss (hL)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m, Pa).
D = 20 mm = 0.02 m (1 m = 1000 mm)
Step 2 · Find mean speed (V)
Use Q = π ÷ 4 × D² × V
The
mean
speed
is
the
flow
rate
over
the
tube’s
area
πD²
4
.
V
=
4
×
Q
π ×
D
2
V
=
4
×
0.0001
π ×
0.02
2
V
=
4
×
0.0001
0.0004 × π
V
=
4
×
0.0001
0.0012566
V
=
0.0004
0.0012566
V
=
0.3183
m/s
Step 3 · Find Reynolds number (Re)
Use Re = ρ × V × D ÷ μ
Re weighs inertia against viscosity; under 2300 the flow stays laminar.
Re
=
900
×
0.3183
×
0.02
0.1
Re
=
286.47
×
0.02
0.1
Re
=
5.7294
0.1
Re = 57.3
Step 4 · Find friction factor (f)
Use f = 64 ÷ Re
Laminar
flow’s
Darcy
friction
factor:
Hagen–Poiseuille
written
as
hL
=
f
(
L
D
)
V
2
2g
.
f
=
64
57.3
f
=
1.117
Step 5 · Find pressure drop (ΔP)
Use ΔP = 128 × μ × L × Q ÷ (π × D⁴)
Hagen–Poiseuille:
the
parabolic
profile
gives
a
drop
in
step
with
μ
,
L
and
Q
,
and
with
1
D
4
.
ΔP
=
128
×
0.1
×
10
×
0.0001
π ×
0.02
4
ΔP
=
128
×
0.1
×
10
×
0.0001
(1.6 ×
10
-7
) × π
ΔP
=
128
×
0.1
×
10
×
0.0001
5.027 ×
10
-7
ΔP
=
12.8
×
10
×
0.0001
5.027 ×
10
-7
ΔP
=
128
×
0.0001
5.027 ×
10
-7
ΔP
=
0.0128
5.027 ×
10
-7
ΔP = 25,464.7909 Pa
Step 6 · Find head loss (hL)
Use ΔP = ρ × 9.81 × hL
Divide the pressure by ρg.
hL
=
ΔP
ρ
× 9.81
hL
=
25,460
900 × 9.81
hL
=
25,460
8,829
hL = 2.884 m
Step 7 · Convert the answers
Change each answer back to the unit you chose.
ΔP = 25,460 Pa = 25.46 kPa (1 kPa = 1000 Pa)
Check
Put the numbers back in. Both sides should match.
0.0001
=
π
4
×
0.02
2
×
0.3183
✓
57.3
=
900
×
0.3183
×
0.02
0.1
✓
1.117
=
64
57.3
✓
25,460
=
128
×
0.1
×
10
×
0.0001
π ×
0.02
4
✓
25,460 = 900 × 9.81 × 2.884 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.