Flow in pipes
Water supply mains, pumping lines, sewers flowing full and penstocks are all pipe flow problems. The questions are: for a given pipe, how much head is lost at a given flow, or what diameter is needed to carry a flow with an available head?
Flow is laminar at low Reynolds number (below about 2 000) and turbulent above about 4 000; almost all civil engineering pipe flow is turbulent. Head loss from friction is given by the Darcy-Weisbach equation, h_f = f L V² / (2 g D). The friction factor f depends on Reynolds number and relative roughness and is read from the Moody diagram or calculated from the Colebrook-White or Swamee-Jain equations. Waterworks practice also uses the empirical Hazen-Williams formula with a coefficient C that depends on pipe material and age; older pipes become rougher and lose capacity.
Minor losses at bends, valves, entrances, exits and fittings are expressed as K V² / 2g. In long mains they are small, but in pump stations and treatment plants with many fittings they can be significant. Pipes in series share the same flow and add head losses; pipes in parallel share the same head loss and add flows. Networks are solved by Hardy Cross or by computer models. Economic diameter balances pipe cost against pumping energy over the life of the system.
Reynolds: Re = V D / νDarcy-Weisbach: h_f = f L V² / (2 g D)Swamee-Jain: f = 0.25 / [log₁₀(ε/3.7D + 5.74/Re^0.9)]²Hazen-Williams (SI): V = 0.849 C R^0.63 S^0.54; minor loss h_m = K V²/2g
Mixing diameter units: Darcy-Weisbach needs D in metres when V is in m/s and g in m/s².
- Darcy-Weisbach applies to full, steady, incompressible flow; Hazen-Williams is valid only for water at ordinary temperatures in the turbulent range.
- Friction factor depends on roughness, which grows with pipe age.
Problem. A 300 mm main, 2.4 km long, carries 80 L/s. Take f = 0.018. Find velocity, friction head loss and Reynolds number (ν = 1.0 × 10⁻⁶ m²/s).
- A = π × 0.3² / 4 = 0.07069 m², V = 0.080 / 0.07069 = 1.132 m/s.
- h_f = 0.018 × 2 400 × 1.132² / (2 × 9.81 × 0.3) = 0.018 × 2 400 × 1.281 / 5.886 = 9.40 m.
- Re = 1.132 × 0.3 / 1.0 × 10⁻⁶ = 3.4 × 10⁵: turbulent (the assumed f = 0.018 should be confirmed from the Moody diagram for the pipe's roughness).
Answer. V ≈ 1.13 m/s, head loss ≈ 9.4 m, Re ≈ 3.4 × 10⁵
Quick quiz
1At what Reynolds number does flow become turbulent?
Above about 4 000 (laminar below about 2 000).
2How do flows behave in parallel pipes?
Same head loss, flows add.
3Why does pipe capacity fall with age?
Roughness increases from corrosion and deposits.
- CPHEEO Manual on Water Supply and Treatment (India): Central government manual for planning, design and operation of public water supply.
Clauses and documents to read
- CPHEEO Manual on Water Supply and Treatment, Part A (pipe flow formulas and C values, cited by chapter)
- IS 1172:1993 (basic requirements for water supply)
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- P. N. Modi, S. M. Seth, Hydraulics and Fluid Mechanics including Hydraulic Machines, Standard Book House, latest edition.
- V. T. Chow, Open-Channel Hydraulics, McGraw Hill, 1959.
- K. Subramanya, Flow in Open Channels, McGraw Hill Education (India), latest edition.
Free and official sources
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