Water hammer
When the flow in a full pipe is suddenly stopped, by closing a valve fast or by a pump tripping, the moving water cannot stop all at once. Its momentum is converted into a pressure wave that travels along the pipe at close to the speed of sound in the pipe-water system, reflects at reservoirs and closed ends, and oscillates until friction damps it. This surge, water hammer, can burst pipes, collapse them under negative pressure, damage valves and lift joints.
The size of the pressure rise for a rapid closure is given by the Joukowsky equation: water density times wave speed times the change in velocity. Wave speed depends on the bulk modulus of water and on how much the pipe wall stretches: around 1 000 to 1 300 m/s in steel and ductile iron, a few hundred metres per second in plastic pipes. A velocity change of only 1 m/s in a steel main can raise pressure by over 100 m of head.
A closure is rapid if it takes less than the time for the wave to travel to the reservoir and back, two lengths over the wave speed. Slower closure reduces the peak roughly in proportion. Pump trips are often worse than valve closures because the flow stops quickly and the pressure first falls, sometimes to vapour pressure, causing column separation and a violent rejoining.
Protection includes slow-closing valves, surge tanks and air vessels, pressure relief and air valves, flywheels on pumps and careful start-up and shut-down procedures. Transient analysis software models these effects for long mains.
Check the pump-trip case as well as valve closure; low pressure can be as damaging as high.
Joukowsky: Δp = ρ c Δv; head rise Δh = c Δv / gWave speed: c = √[(K / ρ) / (1 + K D / (E e))]Rapid closure when closure time < 2 L / c
Ignoring negative pressures that collapse pipes or cause column separation.
Problem. A steel pipe has D = 0.5 m, wall e = 10 mm, E = 200 GPa, length 1 200 m, carrying water (K = 2.2 GPa, ρ = 1 000 kg/m³) at 2 m/s. Find the wave speed, the pressure rise for instant closure and the critical closure time.
- K D / (E e) = 2.2 × 10⁹ × 0.5 / (2 × 10¹¹ × 0.01) = 0.55.
- c = √[(2.2 × 10⁹ / 1 000) / 1.55] = √(1.419 × 10⁶) = 1 191 m/s.
- Δp = 1 000 × 1 191 × 2 = 2.38 × 10⁶ Pa = 2.38 MPa, a head rise of 1 191 × 2 / 9.81 = 243 m.
- 2 L / c = 2 × 1 200 / 1 191 = 2.0 s; any closure faster than this gives the full rise.
Answer. c ≈ 1 190 m/s; Δp ≈ 2.4 MPa (about 243 m head); closures faster than about 2 s are rapid
Quick quiz
1What does the Joukowsky equation give?
The pressure rise for a rapid change of velocity: Δp = ρ c Δv.
2Why do plastic pipes have lower wave speeds?
Their walls are much more flexible.
3Name two surge protection devices.
Surge tank and air vessel (also relief valves, air valves, flywheels).
- CPHEEO Manual on Water Supply and Treatment (India): Central government manual for planning, design and operation of public 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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