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Pipe network systems

Dr. Huidae Cho

Department of Civil and Environmental Engineering...New Mexico State University

1 Learning objectives

Understand what defines a pipe network
Apply continuity at nodes and energy in loops
Recognize why simple methods no longer work
Learn the idea of the Hardy Cross method

2 What is a pipe network?

Multiple pipes connected in loops and nodes
Flow can take multiple paths between points
Unknowns increase significantly
Examples:
- Municipal water distribution systems
- Irrigation networks
- Industrial piping systems

3 Why previous methods are not enough

Series pipes:
- Single flow path → straightforward
Parallel pipes:
- Same head loss → manageable
Networks:
- Multiple loops
- Interdependent flows
- No direct solution

4 Governing principles

Continuity at each node: \[ \sum Q = 0 \]
Energy conservation in each loop: \[ \sum h_L = 0 \]
Head loss relation: \[ h_L = K Q^2 \]

5 Key idea

Unknowns:
- Flow in each pipe
Constraints:
- Node continuity
- Loop energy balance
System becomes nonlinear

6 Loops and nodes

Node:
- Junction where pipes meet
- Apply continuity
Loop:
- Closed path in the network
- Apply energy conservation

7 Sign convention

Assign arbitrary flow directions
Head loss is always in direction of assumed flow
If computed flow is negative:
- Actual flow is opposite

8 Why iteration is needed

Equations are nonlinear: \[ h_L \propto Q^2 \]
Cannot solve directly
Must guess and correct

9 Hardy Cross method (concept)

Iterative method for loop correction
Basic idea:
- Assume initial flows
- Check loop energy balance
- Apply corrections

10 Hardy Cross procedure (one loop)

Steps:
1. Assume initial flows in all pipes
2. Compute head losses $h_L = K Q^2$
3. Compute sum of head losses around loop
4. Compute correction $\Delta Q$
5. Adjust flows
6. Repeat until $\sum h_L \approx 0$

11 Flow correction idea

If $\sum h_L \neq 0$:
- Loop is unbalanced
Apply correction:
- Increase or decrease flows to reduce imbalance

12 Physical interpretation

System redistributes flow
Goal:
- Energy loss around loop equals zero
Similar to:
- Balancing forces in statics

13 Example (conceptual)

Single loop with 3 pipes
Questions:
1. What happens if one pipe has very large resistance?
2. How does flow redistribute?
3. Which pipe carries most flow?

14 Example: single-loop network

A reservoir supplies $40 \ \text{L/s}$ to a single-loop network.
The flow leaves the network at the opposite junction, also at $40 \ \text{L/s}$.
Initial guess:
- Upper path: $20 \ \text{L/s}$
- Lower path: $20 \ \text{L/s}$
Pipe data:
- Pipe AB: $K_{AB} = 4$
- Pipe BC: $K_{BC} = 6$
- Pipe CD: $K_{CD} = 5$
- Pipe DA: $K_{DA} = 3$
Use \[ h_L = K Q^2 \]

14.1 Loop orientation and assumed flows

Take the loop direction as A $\rightarrow$ B $\rightarrow$ C $\rightarrow$ D $\rightarrow$ A.
Initial assumed flows relative to the loop direction:
- $Q_{AB} = +0.020 \ \text{m}^3/\text{s}$
- $Q_{BC} = +0.020 \ \text{m}^3/\text{s}$
- $Q_{CD} = -0.020 \ \text{m}^3/\text{s}$
- $Q_{DA} = -0.020 \ \text{m}^3/\text{s}$
Negative means the actual flow is opposite to the loop direction.

14.2 Step 1: compute $\sum KQ|Q|$

For pipe AB: \[ KQ|Q| = 4(0.020)(0.020) = 0.0016 \]
For pipe BC: \[ KQ|Q| = 6(0.020)(0.020) = 0.0024 \]
For pipe CD: \[ KQ|Q| = 5(-0.020)(0.020) = -0.0020 \]
For pipe DA: \[ KQ|Q| = 3(-0.020)(0.020) = -0.0012 \]
Sum: \[ \sum KQ|Q| = 0.0016 + 0.0024 - 0.0020 - 0.0012 = 0.0008 \]

14.3 Step 2: compute $\sum 2K|Q|$

For pipe AB: \[ 2K|Q| = 2(4)(0.020) = 0.16 \]
For pipe BC: \[ 2K|Q| = 2(6)(0.020) = 0.24 \]
For pipe CD: \[ 2K|Q| = 2(5)(0.020) = 0.20 \]
For pipe DA: \[ 2K|Q| = 2(3)(0.020) = 0.12 \]
Sum: \[ \sum 2K|Q| = 0.16 + 0.24 + 0.20 + 0.12 = 0.72 \]

14.4 Step 3: compute the Hardy Cross correction

Correction formula: \[ \Delta Q = -\frac{\sum KQ|Q|}{\sum 2K|Q|} \]
Substitute: \[ \Delta Q = -\frac{0.0008}{0.72} = -0.00111 \ \text{m}^3/\text{s} \]
Therefore, \[ \Delta Q = -1.11 \ \text{L/s} \]

14.5 Step 4: update the flows

For flows in the same direction as the loop, add $\Delta Q$:
- $Q_{AB,new} = 0.020 - 0.00111 = 0.01889 \ \text{m}^3/\text{s}$
- $Q_{BC,new} = 0.020 - 0.00111 = 0.01889 \ \text{m}^3/\text{s}$
For flows opposite to the loop direction, subtract $\Delta Q$:
- $Q_{CD,new} = -0.020 - (-0.00111) = -0.01889 \ \text{m}^3/\text{s}$ ? No
- It is easier to update the actual lower-path flow directly:
  1. Lower-path actual flow $= 0.020 + 0.00111 = 0.02111 \ \text{m}^3/\text{s}$
Final corrected path flows:
- Upper path: $18.89 \ \text{L/s}$
- Lower path: $21.11 \ \text{L/s}$

14.6 Check the result

Upper-path head loss: \[ h_{L,upper} = (4 + 6)(0.01889)^2 = 0.00357 \]
Lower-path head loss: \[ h_{L,lower} = (3 + 5)(0.02111)^2 = 0.00356 \]
The two path losses are nearly equal, so the corrected flow split is essentially balanced.

14.7 Interpretation

The upper path had larger resistance:
- $K_{AB} + K_{BC} = 10$
The lower path had smaller resistance:
- $K_{DA} + K_{CD} = 8$
Therefore, more flow shifts to the lower path after correction.

14.8 Key point from this example

Initial guess from continuity:
- $20 \ \text{L/s}$ and $20 \ \text{L/s}$
Hardy Cross correction:
- Moves some flow from the higher-resistance path to the lower-resistance path
Corrected split:
- $18.89 \ \text{L/s}$ in the upper path
- $21.11 \ \text{L/s}$ in the lower path

15 Limitations

Can be tedious for large systems
Convergence may be slow
Better numerical methods exist

16 Key points

Pipe networks require:
- Continuity at nodes
- Energy balance in loops
System is nonlinear
Hardy Cross is an iterative solution method
Flow directions can be assumed initially