Pressure Drop

Friction and fittings cost pressure. Learn the Darcy–Weisbach equation, friction factor (laminar 64/Re, Colebrook, Moody chart), minor losses (K-values), system curves and how to size pipes and pumps together.

Every metre of pipe, every elbow, valve and filter takes energy from the fluid. Predicting that head loss tells you the pump size, the pipe diameter and the running cost of a system for decades.

The Darcy–Weisbach equation with a friction factor from the Moody chart is the universal method.

An analogy

Walking through a crowded corridor: the longer and narrower it is, and the faster you try to go, the more effort you waste pushing through.

What causes loss

Wall shear friction (major losses) and turbulence in fittings, bends, valves, entries and exits (minor losses). Loss rises roughly with V², so doubling the flow in a turbulent system roughly quadruples the head loss.

Friction factor and the Moody chart

For laminar flow f = 64/Re. For turbulent flow f depends on Re and relative roughness ε/D: smooth plastic (ε ≈ 0.0015 mm), commercial steel (0.045 mm), cast iron (0.26 mm), concrete (0.3–3 mm). The Moody chart plots Colebrook; Haaland's explicit formula is convenient for calculators.

Minor losses

Typical K: sharp entrance 0.5, exit 1.0, 90° long elbow 0.3–0.5, gate valve open 0.2, globe valve open ≈ 10. In short systems with many fittings, 'minor' losses can dominate.

System curves and networks

System head = static lift + kQ² (friction). Plotting it against the pump curve gives the operating point. Networks use Hardy Cross or Newton–Raphson; long pipelines are optimised on total life-cycle cost (pipe capital vs pumping energy).

Formulas

• Darcy–Weisbach: h_f = f · (L/D) · V²/2g
• Laminar friction factor: f = 64 / Re
• Colebrook (turbulent): 1/√f = −2 log₁₀(ε/3.7D + 2.51/(Re√f))
• Minor losses: h_m = ΣK · V²/2g
• Pressure drop: Δp = ρ g (h_f + h_m)

Worked example: Friction loss in a main

Given: 200 m of 100 mm pipe; V = 2 m/s, f = 0.02

Find: Head loss.

• h_f = 0.02 × (200/0.1) × 2²/19.62 = 8.15 m
• Δp = 998 × 9.81 × 8.15 = 79.8 kPa

Answer: 8.15 m (≈ 0.8 bar)

Worked example: Friction factor from Haaland

Given: Re = 2 × 10⁵; ε/D = 0.00046 (steel, 100 mm)

Find: f.

• 1/√f = −1.8 log[(0.00046/3.7)^1.11 + 6.9/2 × 10⁵]
• f ≈ 0.0184

Answer: f ≈ 0.018

Common mistakes

• Wrong: Ignoring fittings in short systems. — Right: Sum K for all elbows, valves, entries and exits.
• Wrong: Using the Fanning factor in the Darcy equation. — Right: Darcy f = 4 × Fanning f.

Applications

• Building services: Heating/cooling water circuits sized for ~100–300 Pa/m pressure gradient.
• Oil & gas pipelines: Pump/compressor station spacing from friction losses over hundreds of km.
• Hydraulic systems: Hose and valve losses reduce available actuator pressure and create heat.

Key takeaways

• h_f = f(L/D)V²/2g.
• Laminar f = 64/Re; turbulent from Moody/Colebrook.
• Add minor losses ΣK·V²/2g.
• h_f ∝ Q²/D⁵ — diameter is powerful.

Going further

• Hazen–Williams: An empirical alternative used in water distribution (C-factor), valid only for water near room temperature.
• Non-circular ducts: Use hydraulic diameter D_h = 4A/P for rectangular ducts and annuli.
• Two-phase flow: Gas–liquid and slurry flows need special correlations (Lockhart–Martinelli, Durand).