Pipes in parallel
Department of Civil and Environmental Engineering...New Mexico State University
Contents
1 Learning objectives
- Understand flow distribution in parallel pipe systems
- Apply continuity and energy principles to parallel pipes
- Compute flow split using proportional and iterative methods
- Interpret how diameter, length, and roughness affect flow
- Solve practical engineering problems involving parallel pipelines
Key points
- Equal head loss governs the system
- Flow adjusts based on hydraulic resistance
2 Concept
- Multiple pipes connect the same upstream and downstream nodes
- Flow splits into branches and recombines
- Common in water distribution and bypass systems
Key points
- Same endpoints → same energy drop
- Flow distribution is not equal
3 Governing principles
3.1 Continuity
$$ Q = Q_1 + Q_2 + \cdots + Q_n $$3.2 Energy (head loss equality)
$$ h_{L1} = h_{L2} = \cdots = h_{Ln} $$- All parallel pipes experience the same head loss
Key points
- Equal head loss, not equal discharge
4 Head loss models
4.1 Darcy–Weisbach
$$ h_L = f \frac{L}{D} \frac{V^2}{2g} $$ $$ Q = VA $$ $$ h_L \propto \frac{L}{D^5} Q^2 $$4.2 Hazen–Williams
$$ h_L = K Q^{1.852} $$- Common in water systems
- Nonlinear relation between $h_L$ and $Q$
5 Flow distribution
$$ h_{L1}(Q_1) = h_{L2}(Q_2) $$ $$ Q = Q_1 + Q_2 $$- Larger, shorter, smoother pipes carry more flow
- Smaller, longer, rougher pipes carry less flow
Key points
- Distribution depends on resistance
6 Solution strategy
6.1 Procedure
- Assume $Q_1$
- Compute $Q_2 = Q - Q_1$
- Compute $h_{L1}$ and $h_{L2}$
- Adjust until the two head losses are equal
6.2 Numerical methods
- Trial-and-error
- Spreadsheet solver
- Newton method (advanced)
Key points
- Iteration is standard
7 Worked example 1: same material pipes
- Same $L$, same roughness
- $D_1 = 0.30$ m, $D_2 = 0.15$ m
- $Q = 0.12$ m3/s
7.1 Step 1
$$ \frac{Q_1^2}{D_1^5} = \frac{Q_2^2}{D_2^5} $$7.2 Step 2
$$ \frac{Q_1}{Q_2} = \left( \frac{D_1}{D_2} \right)^{2.5} $$ $$ \frac{Q_1}{Q_2} = 2^{2.5} \approx 5.66 $$7.3 Step 3
$$ Q = Q_1 + Q_2 = 0.12 $$- Let $Q_2 = x$, then $Q_1 = 5.66x$
- \[ 5.66x + x = 0.12 \]
- \[ x = 0.018 \]
- \[ Q_1 = 0.102,\quad Q_2 = 0.018 \]
7.4 Interpretation
- Large pipe dominates flow
8 Worked example 2: iterative solution
- $L_1 = L_2 = 100$ m
- $D_1 = 0.20$ m, $D_2 = 0.15$ m
- $Q = 0.10$ m3/s
8.1 Step 1: initial guess
- Try $Q_1 = 0.05$ and $Q_2 = 0.05$
- The smaller pipe has the larger head loss for the same flow
- So Pipe 2 is carrying too much flow
8.2 Step 2
- Adjust the guess to $Q_1 = 0.07$ and $Q_2 = 0.03$
- Compare the two head losses using $$ \frac{h_{L1}}{h_{L2}} = \frac{Q_1^2 / D_1^5}{Q_2^2 / D_2^5} $$
- The ratio is about $0.88$
- This means $h_{L1} < h_{L2}$, so increase $Q_1$ slightly
8.3 Step 3
- Try $Q_1 \approx 0.072$ and $Q_2 \approx 0.028$
- Now the two head losses are approximately equal
8.4 Final answer
$$ Q_1 \approx 0.072,\quad Q_2 \approx 0.028 $$9 Physical interpretation
- Flow redistributes to equalize energy loss
- The lower-resistance branch carries more flow
- The system naturally balances resistance and discharge
10 Common mistakes
- Assuming equal flow
- Ignoring nonlinear dependence
- Mixing formulas incorrectly
- Ignoring minor losses
11 Quick check
- If Pipe 2 becomes rougher:
- $Q_2$ decreases
- Flow shifts to Pipe 1
12 Key points
- Equal head loss controls flow split
- Diameter strongly affects flow
- Iteration is usually required
- Always satisfy continuity and energy