SME
Hydraulics & Hydrology · Problem type
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.”
Diagram
Clockwise
flows
are
+.
ΔQ
=
−
Σhf
Σ(2 hf ÷ Q)
=
−
0.27
82
=
−0.00329
m³/s,
added
to
every
pipe’s
flow
(the
arrows
show
the
new
direction
where
one
changes
sign).
This is an example. Type your own values to start a new problem.
K₁
Pipe 1 constant
example · s²/m⁵
K₂
Pipe 2 constant
example · s²/m⁵
K₃
Pipe 3 constant
example · s²/m⁵
K₄
Pipe 4 constant
example · s²/m⁵
Q₁
Pipe 1 flow
example · m³/s
Q₂
Pipe 2 flow
example · m³/s
Q₃
Pipe 3 flow
example · m³/s
Q₄
Pipe 4 flow
example · m³/s
ΔQ
Correction
worked out (not typed) · m³/s
Formulas
ΔQ
=
−(
K₁
×
Q₁
×
|
Q₁
|
+
K₂
×
Q₂
×
|
Q₂
|
+
K₃
×
Q₃
×
|
Q₃
|
+
K₄
×
Q₄
×
|
Q₄
|)
÷
(2
×
(
K₁
×
|
Q₁
|
+
K₂
×
|
Q₂
|
+
K₃
×
|
Q₃
|
+
K₄
×
|
Q₄
|))
−0.003293 = −(200 × 0.06 × |0.06| + 300 × 0.03 × |0.03| + 200 × (−0.02) × |−0.02| + 400 × (−0.04) × |−0.04|) ÷ (2 × (200 × |0.06| + 300 × |0.03| + 200 × |−0.02| + 400 × |−0.04|))
Assumptions
•
Each pipe loses hf = KQ|Q| (Darcy with a fixed f); clockwise flows are +.
•
The assumed flows already balance at every node.
•
One correction is the page: add ΔQ to every pipe and repeat until ΔQ is small.
Step-by-step
WE KNOW
K₁ = 200 s²/m⁵, K₂ = 300 s²/m⁵, K₃ = 200 s²/m⁵, K₄ = 400 s²/m⁵, Q₁ = 0.06 m³/s, Q₂ = 0.03 m³/s, Q₃ = −0.02 m³/s, Q₄ = −0.04 m³/s
FIND
correction (ΔQ)
Step 1 · Find correction (ΔQ)
Use ΔQ = −(K₁ × Q₁ × |Q₁| + K₂ × Q₂ × |Q₂| + K₃ × Q₃ × |Q₃| + K₄ × Q₄ × |Q₄|) ÷ (2 × (K₁ × |Q₁| + K₂ × |Q₂| + K₃ × |Q₃| + K₄ × |Q₄|))
Round a loop the head losses must add to 0. Each hf = KQ|Q| changes by 2K|Q| per unit of flow, so the correction is −Σhf ÷ Σ2K|Q|.
ΔQ = −(200 × 0.06 × |0.06| + 300 × 0.03 × |0.03| + 200 × (−0.02) × |−0.02| + 400 × (−0.04) × |−0.04|) ÷ (2 × (200 × |0.06| + 300 × |0.03| + 200 × |−0.02| + 400 × |−0.04|))
Σhf = 0.72 + 0.27 − 0.08 − 0.64 = 0.27 m
Σ2K|Q|
=
2
×
(12
+
9
+
4
+
16)
=
82
s/
m
2
ΔQ
=
−0.003293
m³/s
New
flows:
0.05671,
0.02671,
(−0.02329),
(−0.04329)
m
3
/s
Check
Put the numbers back in. Both sides should match.
−0.003293 = −(200 × 0.06 × |0.06| + 300 × 0.03 × |0.03| + 200 × (−0.02) × |−0.02| + 400 × (−0.04) × |−0.04|) ÷ (2 × (200 × |0.06| + 300 × |0.03| + 200 × |−0.02| + 400 × |−0.04|)) ✓
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?”
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.”