💰 Economic Pipe Diameter Sizer

Determine the optimum, economic pipe diameter that minimizes total annual cost (fixed capital investment + pumping energy costs) for a piping system.

⚡ Fortran 90 Engine Double Precision (IEEE 754) ✓ ISO / ASME Validated

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📅 Released Jun 2026

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📝 Configuration

📊 Results

Configure and click Optimize.

📘 Methodology

Economic Optimization

Larger pipes reduce friction loss and pumping cost, but increase capital cost. The economic optimum minimizes Total = Energy_annual + Capital_annual over the pipe lifetime.

Capital Recovery Factor

CRF = i(1+i)^n / [(1+i)^n − 1] annualizes the pipe investment over its lifetime n at interest rate i. This converts a one-time capital cost into an equivalent annual payment.

Friction & Pumping

dP = f·L/D·ρV²/2 (Darcy-Weisbach). Power = Q·dP/η. The friction factor f is computed automatically using the Swamee-Jain approximation for smooth pipes.

📘 Calculation Methodology: Optimum Economic Pipe Diameter Sizing

Mathematical Model & Theory

Economic pipe sizing balances amortized capital expenditure (pipe material and installation cost $\propto D^n$) against lifetime operating pumping energy costs ($\propto 1/D^5$):

$$D_{opt} = C \cdot Q^{0.45} \rho^{0.13} \mu^{0.03} \approx K \cdot Q^{0.48} \quad (\text{Generaux / Peters-Timmerhaus})$$

$$\text{Recommended Velocity: } V_{econ} \approx 1.2 - 2.5\text{ m/s (Liquids)}, \quad 15 - 30\text{ m/s (Gases)}$$

Assumptions

Continuous operation ($7000-8000 ext{ hrs/year}$) with standard electrical power costs.
Turbulent flow in commercial carbon/stainless steel pipelines.

Academic References

Peters, M. S., & Timmerhaus, K. D.: Plant Design and Economics for Chemical Engineers, McGraw-Hill.
Generaux, M. V. (1937): Fluid Flow Design Methods, Ind. Eng. Chem.

Worked Engineering Example

Problem Statement:
Water ($\rho = 1000\text{ kg/m}^3$) is pumped at $Q = 120\text{ m}^3/\text{h}$ ($0.0333\text{ m}^3/\text{s}$). Find the optimum economic pipe diameter using target economic velocity $V = 1.8\text{ m/s}$.

Step-by-step Solution:
1. Flow cross section: $A = Q / V = 0.0333 / 1.8 \approx 0.01852\text{ m}^2$.
2. Diameter: $D = \sqrt{4 A / \pi} = \sqrt{4 \times 0.01852 / \pi} = \sqrt{0.02358} \approx 0.1536\text{ m} = 153.6\text{ mm}$.
3. Standard nominal size: Select **DN150 / 6\" NPS** (ID $\approx 154.1\text{ mm}$).
Final Result:
Optimal pipe diameter is $\mathbf{DN150\ (6\text{'' NPS})}$ with actual velocity $V = 1.79\text{ m/s}$.