SME
Fluid Mechanics
Pipe flow
Use this for “Water at 2 m/s flows through 100 m of 0.1 m steel pipe (ε = 0.045 mm). Find the head loss and the pressure drop.”
Diagram
Re
=
200,000:
turbulent.
ε
D
=
0.045
mm
100
mm
=
0.00045.
On that curve, Colebrook gives f = 0.0186 at this Re: the ringed point.
Every curve is computed from Colebrook, not read off a printed chart.
Mean speed V = 2 m/s
Pipe length L = 100 m
Kinematic viscosity ν = 1 × 10⁻⁶ m²/s
Head loss hL = 3.784 m
Density ρ = 1,000 kg/m³
Pressure drop ΔP = 37.12 kPa
This is an example. Type your own values to start a new problem.
hL (m)
=
f
×
L (m)
D (mm)
×
(
V (m/s)
)
2
2
×
9.81
ΔP (kPa)
=
ρ (kg/m³)
×
9.81
×
hL (m)
ν
Kinematic viscosity
example · m²/s
ε
Pipe roughness
example
Re
Reynolds number
calculated
Formulas
Re
=
V
×
D
ν
200,000
=
2
×
0.1
1 ×
10
-6
1
÷
√
f
=
−2
×
log(
ε
3.7 ×
D
+
2.51
Re × √
f
)
1
÷
√
0.01856
=
−2
×
log(
4.5 ×
10
-5
3.7 × 0.1
+
2.51
200,000 ×
√
0.01856
)
hL
=
f
×
(
L
D
)
×
V
2
2 × 9.81
3.784
=
0.01856
×
(
100
0.1
)
×
2
2
2 × 9.81
ΔP
=
ρ
×
9.81
×
hL
37,120 = 1,000 × 9.81 × 3.784
Formulas are worked in m, Pa. The step-by-step shows the conversions.
Assumptions
•
Steady, fully developed turbulent flow (Re above 4000) in a round pipe; for laminar flow use the laminar page.
•
Colebrook has f on both sides: it is solved by repeating it until f settles.
•
Minor losses at fittings are left out; g = 9.81 m/s².
Step-by-step
WE KNOW
V = 2 m/s, D = 100 mm, L = 100 m, ν = 1 × 10⁻⁶ m²/s, ε = 0.045 mm, ρ = 1,000 kg/m³
FIND
Reynolds number (Re), friction factor (f), head loss (hL), pressure drop (ΔP)
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 = 100 mm = 0.1 m (1 m = 1000 mm)
ε = 0.045 mm = 4.5 × 10⁻⁵ m (1 m = 1000 mm)
Step 2 · Find Reynolds number (Re)
Use Re = V × D ÷ ν
Re weighs inertia against viscosity: speed times diameter, over the kinematic viscosity.
Re
=
2
×
0.1
10
-6
Re
=
0.2
10
-6
Re = 200,000
Step 3 · Find friction factor (f)
Use 1 ÷ √f = −2 × log(ε ÷ (3.7 × D) + 2.51 ÷ (Re × √f))
Guess
f
=
0.02,
put
it
in
on
the
right,
and
repeat
until
f
stops
changing;
the
last
round
is
shown.
f
=
(1
÷
(−2
×
log(
ε
3.7 ×
D
+
2.51
Re ×
√
0.0185602
)))²
f
=
(1
÷
(−2
×
log(
4.5 ×
10
-5
3.7 × 0.1
+
2.51
200,000 ×
√
0.0185602
)))²
f
=
(1
÷
((−2)
×
log(
4.5 ×
10
-5
0.37
+
2.51
200,000 × 0.13624
)))²
f
=
(1
÷
((−2)
×
log(
4.5 ×
10
-5
0.37
+
2.51
27,248
)))²
f
=
(1
÷
((−2)
×
log(0.00012162
+
(9.2117
×
10
-5
))))²
f
=
(
1
(−2) × log(0.00021374)
)²
f
=
(
1
(−2) × (−3.6701)
)²
f
=
(
1
7.3402
)
2
f
=
0.13624
2
f
=
0.01856
Step 4 · Find head loss (hL)
Use hL = f × (L ÷ D) × V² ÷ (2 × 9.81)
Darcy–Weisbach:
f
times
the
length
in
diameters
times
the
velocity
head
V
2
2g
.
hL
=
0.01856
×
(
100
0.1
)
×
2
2
2 × 9.81
hL
=
0.01856
×
1,000
×
2
2
19.62
hL
=
0.01856
×
1,000
×
4
19.62
hL
=
18.56
×
4
19.62
hL
=
74.24
19.62
hL = 3.784 m
Step 5 · Find pressure drop (ΔP)
Use ΔP = ρ × 9.81 × hL
A head of hL metres of the fluid is a pressure of ρghL.
ΔP = 1,000 × 9.81 × 3.7839
ΔP = 9,810 × 3.7839
ΔP = 37,120.3045 Pa
Step 6 · Convert the answers
Change each answer back to the unit you chose.
ΔP = 37,120 Pa = 37.12 kPa (1 kPa = 1000 Pa)
Check
Put the numbers back in. Both sides should match.
200,000
=
2
×
0.1
10
-6
✓
1
÷
√
0.01856
=
−2
×
log(
4.5 ×
10
-5
3.7 × 0.1
+
2.51
200,000 ×
√
0.01856
)
✓
3.784
=
0.01856
×
(
100
0.1
)
×
2
2
2 × 9.81
✓
37,120 = 1,000 × 9.81 × 3.784 ✓
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Laminar pipe flow: the pressure drop
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