🔧 Pipe Flow & Pressure Drop Calculator
Darcy-Weisbach pressure drop with Colebrook-White friction factor, minor losses, and Moody chart visualization.
📝 Configuration
Darcy-Weisbach:
ΔP = f × (L/D) × (ρU²/2)
Colebrook-White:
1/√f = -2 log₁₀(ε/D/3.7 + 2.51/(Re√f))
Minor losses:
ΔP_minor = ΣK × (ρU²/2)
ΔP = f × (L/D) × (ρU²/2)
Colebrook-White:
1/√f = -2 log₁₀(ε/D/3.7 + 2.51/(Re√f))
Minor losses:
ΔP_minor = ΣK × (ρU²/2)
📊 Results & Visualization
Configure the inputs and click Calculate to see results.
ℹ️ About Pipe Flow
The Darcy-Weisbach equation is the most general equation for calculating pressure drop in fully developed pipe flow. Combined with the Colebrook-White equation, it accurately predicts friction factors for all flow regimes.
Key factors:
• Re < 2300: Laminar flow (f = 64/Re)
• Re > 4000: Turbulent flow (f from Colebrook-White)
• Surface roughness significantly affects turbulent friction
• Minor losses from fittings can dominate in short pipes
The Darcy-Weisbach equation is the most general equation for calculating pressure drop in fully developed pipe flow. Combined with the Colebrook-White equation, it accurately predicts friction factors for all flow regimes.
Key factors:
• Re < 2300: Laminar flow (f = 64/Re)
• Re > 4000: Turbulent flow (f from Colebrook-White)
• Surface roughness significantly affects turbulent friction
• Minor losses from fittings can dominate in short pipes
📘 Calculation Methodology
Mathematical Model & Theory
Flow in pipes experiences pressure drops due to viscous friction and minor losses (valves, bends). Friction factor is determined using the Colebrook-White equation for turbulent flow ($Re > 4000$):
$$\Delta P = \left( f \frac{L}{D} + \sum K \right) \frac{\rho V^2}{2}, \quad Re = \frac{\rho V D}{\mu}$$
$$\text{Laminar: } f = \frac{64}{Re}, \quad \text{Turbulent: } \frac{1}{\sqrt{f}} = -2\log_{10}\left(\frac{\varepsilon/D}{3.7} + \frac{2.51}{Re\sqrt{f}}\right)$$
Worked Engineering Example
Problem Statement:
Water ($\rho = 1000$ kg/m³, $\mu = 1.0 \times 10^{-3}$ Pa·s) flows through a 100 m long pipe of diameter 50 mm (roughness $\varepsilon = 0.046$ mm) at a flow rate of 300 L/min. Calculate the pressure drop.
Step-by-step Solution:
1. Calculate velocity $V$ and Reynolds number:
$$Q = 300 \text{ L/min} = 0.005 \text{ m}^3\text{s}$$ $$V = \frac{4Q}{\pi D^2} = \frac{4 \times 0.005}{\pi \times 0.05^2} = 2.546 \text{ m/s}$$ $$Re = \frac{1000 \times 2.546 \times 0.05}{1.0 \times 10^{-3}} = 127,300 \quad \text{(Turbulent flow)}$$ 2. Solve Colebrook-White equation ($\varepsilon/D = 0.00092$):
$$f \approx 0.0211$$ 3. Calculate pressure drop (neglecting minor losses):
$$\Delta P = f \frac{L}{D} \frac{\rho V^2}{2} = 0.0211 \times \frac{100}{0.05} \times \frac{1000 \times 2.546^2}{2} = 136,770 \text{ Pa} = 136.8 \text{ kPa}$$
Final Result:
The pressure drop is 136.8 kPa.
Water ($\rho = 1000$ kg/m³, $\mu = 1.0 \times 10^{-3}$ Pa·s) flows through a 100 m long pipe of diameter 50 mm (roughness $\varepsilon = 0.046$ mm) at a flow rate of 300 L/min. Calculate the pressure drop.
Step-by-step Solution:
1. Calculate velocity $V$ and Reynolds number:
$$Q = 300 \text{ L/min} = 0.005 \text{ m}^3\text{s}$$ $$V = \frac{4Q}{\pi D^2} = \frac{4 \times 0.005}{\pi \times 0.05^2} = 2.546 \text{ m/s}$$ $$Re = \frac{1000 \times 2.546 \times 0.05}{1.0 \times 10^{-3}} = 127,300 \quad \text{(Turbulent flow)}$$ 2. Solve Colebrook-White equation ($\varepsilon/D = 0.00092$):
$$f \approx 0.0211$$ 3. Calculate pressure drop (neglecting minor losses):
$$\Delta P = f \frac{L}{D} \frac{\rho V^2}{2} = 0.0211 \times \frac{100}{0.05} \times \frac{1000 \times 2.546^2}{2} = 136,770 \text{ Pa} = 136.8 \text{ kPa}$$
Final Result:
The pressure drop is 136.8 kPa.