SME
Hydraulics & Hydrology
Pipe networks
Use this for “0.1 m³/s of water flows through 500 m of 0.3 m steel pipe (ε = 0.045 mm). What is the friction head loss?”
Diagram
The
energy
grade
line
falls
by
hL
=
2.596
m
along
the
pipe;
the
hydraulic
grade
line
runs
V
2
2g
=
0.102
m
under
it.
Heads are to the scale at the left; the pipe’s length is not.
Roughness ε = 0.045 mm
This is an example. Type your own values to start a new problem.
D
Diameter
example
L
Length
example
Q
Flow rate
example
ε
Roughness
example
V
Mean velocity
calculated · m/s
ν
Kinematic viscosity
example · m²/s
Re
Reynolds number
calculated
f
Friction factor
calculated
hf
Head loss
calculated · m
Formulas
V
=
Q
π ×
D
2
÷ 4
1.4147
=
0.1
π ×
0.3
2
÷ 4
Re
=
V
×
D
ν
424,413.1816
=
1.4147
×
0.3
1 ×
10
-6
f
=
0.25
÷
(log(
ε
3.7 ×
D
+
5.74
R
e
0.9
))²
0.01527
=
0.25
÷
(log(
4.5 ×
10
-5
3.7 × 0.3
+
5.74
424,413
.
1816
0.9
))²
hf
=
f
×
L
D
×
V
2
2 × 9.81
2.5965
=
0.01527
×
500
0.3
×
1.4147
2
2 × 9.81
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
Turbulent flow (Re > 4000) of water at 20 °C, ν = 10⁻⁶ m²/s.
•
Swamee–Jain gives f directly, within 1% of Colebrook.
•
Minor losses at bends and fittings are ignored; g = 9.81 m/s².
Step-by-step
WE KNOW
D = 0.3 m, L = 500 m, Q = 0.1 m³/s, ε = 0.045 mm, ν = 1 × 10⁻⁶ m²/s
FIND
mean velocity (V), Reynolds number (Re), friction factor (f), head loss (hf)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
ε = 0.045 mm = 4.5 × 10⁻⁵ m (1 m = 1000 mm)
Step 2 · Find mean velocity (V)
Use V = Q ÷ (π × D² ÷ 4)
The mean velocity is the flow over the pipe’s area.
V
=
0.1
π ×
0.3
2
÷ 4
V
=
0.1
π × 0.09 ÷ 4
V
=
0.1
0.28274 ÷ 4
V
=
0.1
0.070685
V
=
1.4147106
m/s
Step 3 · Find Reynolds number (Re)
Use Re = V × D ÷ ν
Inertia
over
viscosity:
velocity
times
diameter
over
ν
.
Re
=
1.4147106
×
0.3
10
-6
Re
=
0.4244
10
-6
Re = 424,413.2
Step 4 · Find friction factor (f)
Use f = 0.25 ÷ (log(ε ÷ (3.7 × D) + 5.74 ÷ Re0.9))²
Rougher
pipes
(
ε
D
)
and
slower,
more
viscous
flow
(smaller
Re)
both
raise
f
.
f
=
0.25
÷
(log(
4.5 ×
10
-5
3.7 × 0.3
+
5.74
424,413
.
2
0.9
))²
f
=
0.25
÷
(log(
4.5 ×
10
-5
1.11
+
5.74
424,413
.
2
0.9
))²
f
=
0.25
÷
(log(
4.5 ×
10
-5
1.11
+
5.74
116,100
))²
f
=
0.25
÷
(log((4.054
×
10
-5
)
+
(4.944
×
10
-5
)))²
f
=
0.25
(log(8.998 ×
10
-5
))²
f
=
0.25
÷
(−4.046)
2
f
=
0.25
16.37
f
=
0.015272
Step 5 · Find head loss (hf)
Use hf = f × L ÷ D × V² ÷ (2 × 9.81)
Darcy–Weisbach:
f
times
the
pipe’s
length
in
diameters
times
the
velocity
head
V
2
2g
.
hf
=
0.015272
×
500
0.3
×
1.4147106
2
2 × 9.81
hf = 0.01527 × 1,666.67 × 0.102
hf = 2.5965 m
Check
Put the numbers back in. Both sides should match.
1.4147106
=
0.1
π ×
0.3
2
÷ 4
✓
424,413.2
=
1.4147106
×
0.3
10
-6
✓
0.015272
=
0.25
÷
(log(
4.5 ×
10
-5
3.7 × 0.3
+
5.74
424,413
.
2
0.9
))²
✓
2.5965
=
0.015272
×
500
0.3
×
1.4147106
2
2 × 9.81
✓
← Previous lesson
Hydraulic jump: sequent depth and head loss
Next lesson →
Head loss by Hazen–Williams
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Head loss by Hazen–Williams
Use this for “0.1 m³/s of water runs through 500 m of 0.3 m pipe with C = 130. What is the head loss?”
Flow split between two parallel pipes
Use this for “0.15 m³/s splits between a 0.3 m pipe 500 m long and a 0.2 m pipe 400 m long (C = 130). Find each flow and the head loss.”
One Hardy Cross correction of a loop
Use this for “Round a loop, K = 200, 300, 200, 400 and the assumed flows are +0.06, +0.03, −0.02, −0.04 m³/s. Find the correction ΔQ.”