SME
Hydraulics & Hydrology · Problem type
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.”
Diagram
The flow splits so both pipes lose the same head from A to B: Q₁ + Q₂ = 0.108 + 0.04196 = 0.15 m³/s.
Pipe widths are to scale with each other; the lengths are labelled.
This is an example. Type your own values to start a new problem.
Q₁ (m³/s)
+
Q₂ (m³/s)
=
Q (m³/s)
hf (m)
=
10.67
×
L₁ (m)
×
Q₁
1.852
C
1.852
×
(
D₁ (m)
)
4.87
h
f
=
10.67
×
L₂ (m)
×
Q₂
1.852
C
1.852
×
(
D₂ (m)
)
4.87
Formulas
Q₁
=
Q
÷
(1
+
(
L₁
L₂
×
(
D₂
D₁
)
4.87
)
1 ÷ 1.852
)
0.108
=
0.15
÷
(1
+
(
500
400
×
(
0.2
0.3
)
4.87
)
1 ÷ 1.852
)
Q₁
+
Q₂
=
Q
0.108 + 0.04196 = 0.15
hf
=
10.67
×
L₁
×
Q₁
1.852
C
1.852
×
D₁
4.87
3.7044
=
10.67
×
500
×
0.108
1.852
130
1.852
×
0.3
4.87
hf
=
10.67
×
L₂
×
Q₂
1.852
C
1.852
×
D₂
4.87
3.7044
=
10.67
×
400
×
0.04196
1.852
130
1.852
×
0.2
4.87
Assumptions
•
Both pipes join the same two nodes, so each loses the same head hf.
•
Hazen–Williams in SI with the same C in both pipes; minor losses are ignored.
•
Whatever enters the first node leaves it: Q₁ + Q₂ = Q.
Step-by-step
WE KNOW
Q = 0.15 m³/s, D₁ = 0.3 m, L₁ = 500 m, D₂ = 0.2 m, L₂ = 400 m, C = 130
FIND
pipe 1 flow (Q₁), pipe 2 flow (Q₂), head loss (hf)
Step 1 · Find pipe 1 flow (Q₁)
Use Q₁ = Q ÷ (1 + (L₁ ÷ L₂ × (D₂ ÷ D₁)4.87)1 ÷ 1.852)
Set
the
two
losses
equal:
Q₂
Q₁
=
((
L₁
L₂
)
(
D₂
D₁
)
4.87
)
1 ÷ 1.852
,
so
Q₁
is
Q
over
1
plus
that.
Q₁
=
0.15
÷
(1
+
(
500
400
×
(
0.2
0.3
)
4.87
)
1 ÷ 1.852
)
Q₁
=
0.15
÷
(1
+
(
500
400
×
0.66667
4.87
)
1 ÷ 1.852
)
Q₁
=
0.15
÷
(1
+
(
500
400
× 0.13882)
0.53996
)
Q₁
=
0.15
÷
(1
+
(1.25 × 0.13882)
0.53996
)
Q₁
=
0.15
1 +
0.17352
0.53996
Q₁
=
0.15
1 + 0.3884
Q₁
=
0.15
1.3884
Q₁
=
0.108038
m³/s
Step 2 · Find pipe 2 flow (Q₂)
Use Q₁ + Q₂ = Q
What does not take pipe 1 takes pipe 2.
Q₂
=
Q
−
Q₁
Q₂
=
0.15
−
0.108038
Q₂
=
0.04196
m³/s
Step 3 · Find head loss (hf)
Use hf = 10.67 × L₁ × Q₁1.852 ÷ (C1.852 × D₁4.87)
Longer,
fuller
and
narrower
pipes
lose
more
head;
a
smoother
pipe
(larger
C
)
loses
less.
hf
=
10.67
×
500
×
0.108038
1.852
130
1.852
×
0.3
4.87
hf
=
10.67
×
500
×
0.01623
23.367
hf = 3.7044 m
Check
Put the numbers back in. Both sides should match.
0.108038
=
0.15
÷
(1
+
(
500
400
×
(
0.2
0.3
)
4.87
)
1 ÷ 1.852
)
✓
0.108038 + 0.04196 = 0.15 ✓
3.7044
=
10.67
×
500
×
0.108038
1.852
130
1.852
×
0.3
4.87
✓
3.7044
=
10.67
×
400
×
0.041962
1.852
130
1.852
×
0.2
4.87
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Pipe networks
Main lesson
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?”
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.”