🔀 Pipe Network Calculator

Series and parallel pipe network analysis with Hardy-Cross iterative method, Colebrook-White friction factors, and flow distribution visualization.

Q_total (IN) Q_total (OUT) Branch 1: Q_1, D_1, L_1 Branch 2: Q_2, D_2, L_2 Branch 3: Q_3, D_3, L_3 Q_total = Q_1 + Q_2 + Q_3 | ΔP_1 = ΔP_2 = ΔP_3

Multi-Pipe Network Layouts

Thermofluid networks manage flow distributions by routing fluid through multiple channels. Nodes must satisfy mass conservation, while closed loops balance energy head losses.

• Series Networks (➡️): Fluid flows sequentially. The total resistance is the sum of pipe resistances; flow rate is constant.
• Parallel Networks (⑃): Fluid splits into multiple branches. Flow distributes to match branch head losses iteratively.

📝 Configuration

Network Type

Number of Pipes

Pipe Data

# $L$ [m] $D$ [mm] $\varepsilon$ [mm] Material

Fluid Properties

Flow Rate

Elevation

Positive = uphill flow
Series:
$Q_1 = Q_2 = \dots = Q_n$
$\Delta P_{total} = \sum \Delta P_i$

Parallel:
$\Delta P_1 = \Delta P_2 = \dots = \Delta P_n$
$Q_{total} = \sum Q_i$

Hardy-Cross:
$\Delta Q = -\frac{\sum R Q |Q|}{\sum 2 R |Q|}$

Darcy-Weisbach:
$\Delta P = f \frac{L}{D} \frac{\rho V^2}{2}$

📊 Results & Visualization

Configure inputs and click Analyze Network to view results.

📘 Calculation Methodology

Mathematical Model & Theory

Pipe networks are solved using node flow balance (Kirchhoff's Flow Law) and loop energy balance (head losses around closed paths sum to zero). The parallel coupling distribution is calculated using the Hardy-Cross iterative algorithm:

$$\sum Q_{in} = \sum Q_{out} \quad \text{(nodes)}$$ $$\sum h_f = \sum r Q |Q|^{n-1} = 0 \quad \text{(loops)}$$ $$\Delta Q = -\frac{\sum r Q_0 |Q_0|^{n-1}}{\sum n r |Q_0|^{n-1}}$$

The Darcy-Weisbach friction loss factor ($f$) for each pipe element is solved implicitly using the Newton-Raphson method applied to the Colebrook-White equation. Series loops include potential head terms ($\Delta Z$):

$$\Delta H_{series} = \sum h_{f,i} + \Delta Z$$

Assumptions

  • Steady, incompressible, one-dimensional fluid flow.
  • Standard Moody boundary conditions apply to each pipe section.
  • Elevation changes only apply to series pipelines.

Academic References

  1. White, F. M.: Fluid Mechanics, McGraw-Hill, 8th Edition.
  2. Hardy Cross: "Analysis of Flow in Networks of Conduits or Conductors", Univ. of Illinois Engineering Experiment Station Bulletin (1936).

Worked Engineering Example

Problem Statement:
A parallel flow network consists of two branches. Branch 1 ($L = 100$ m, $D = 50$ mm, $f = 0.02$) and Branch 2 ($L = 150$ m, $D = 75$ mm, $f = 0.02$). Total flow rate entering is 600 L/min. Find the flow distribution.

Step-by-step Solution:
1. Calculate resistance factors $r$ ($h_f = r Q^2$):
$$r_1 = \frac{8 \times 0.02 \times 100}{\pi^2 \times 9.81 \times 0.05^5} = 52,943 \text{ s}^2/\text{m}^5$$ $$r_2 = \frac{8 \times 0.02 \times 150}{\pi^2 \times 9.81 \times 0.075^5} = 10,457 \text{ s}^2/\text{m}^5$$ 2. Set up parallel flow condition ($r_1 Q_1^2 = r_2 Q_2^2$ and $Q_1 + Q_2 = Q_{tot} = 0.01$ m³/s):
$$\frac{Q_1}{Q_2} = \sqrt{\frac{r_2}{r_1}} = \sqrt{\frac{10457}{52943}} = 0.444$$ $$0.444 Q_2 + Q_2 = 0.01 \Rightarrow Q_2 = 0.00692 \text{ m}^3/\text{s} = 415.5 \text{ L/min}$$ $$Q_1 = 0.01 - 0.00692 = 0.00308 \text{ m}^3/\text{s} = 184.5 \text{ L/min}$$
Final Result:
Branch 1 has 184.5 L/min and Branch 2 has 415.5 L/min.