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Water hammer in elastic cast-iron pipe

Short answer

Water hammer is sudden pressure rise when flow stops abruptly; causes pipe burst, cavitation, vibration. For elastic pipe, maximum velocity V = p/√(ρ/(1/K + D/Et)) = 1.57 m/s given p=1.962 MN/m².

What is water hammer? Write down the problems with water hammer. Water is flowing through a cast-iron pipe of diameter 150 mm and thickness 10 mm which is provided with a valve at its end. Water is suddenly stopped with closing the valve. Find the maximum velocity of water, when rise of pressure due to sudden closure of valve is 1.962 MN/m². Take K = 1.962 GN/m² and E = 117.7 GN/m².

Water hammer

Water hammer is the sudden rise (and following fall) of pressure in a pipe when the flowing water is suddenly stopped or its velocity is suddenly changed, e.g. by quick closing of a valve or sudden load rejection at a turbine. The kinetic energy of the moving water is turned into pressure energy, and a pressure wave travels back and forth along the pipe at the speed of sound in water.

Problems caused by water hammer

  • Very high pressure can burst the pipe or penstock.
  • Negative pressure in the return wave can cause collapse of thin pipes and cavitation.
  • Damage to valves, joints, anchors and supports; noise and vibration.
  • Need for thicker (costlier) pipes and protective devices like surge tanks and relief valves.
  • Unstable turbine governing.

Numerical: elastic pipe, sudden closure

Data: D=0.15D = 0.15 m, t=0.01t = 0.01 m, p=1.962p = 1.962 MN/m², K=1.962K = 1.962 GN/m², E=117.7E = 117.7 GN/m², ρ=1000\rho = 1000 kg/m³.

For sudden closure in an elastic pipe:

p=Vρ1K+DE tp = V\sqrt{\frac{\rho}{\dfrac{1}{K} + \dfrac{D}{E\,t}}} 1K+DEt=11.962×109+0.15117.7×109×0.01=5.097×10−10+1.274×10−10=6.371×10−10 m2/N10006.371×10−10=1.2528×106V=1.962×1061.2528×106=1.566 m/s\begin{aligned} \frac{1}{K} + \frac{D}{Et} &= \frac{1}{1.962 \times 10^9} + \frac{0.15}{117.7 \times 10^9 \times 0.01} \\ &= 5.097 \times 10^{-10} + 1.274 \times 10^{-10} = 6.371 \times 10^{-10}\ \text{m}^2/\text{N} \\ \sqrt{\frac{1000}{6.371 \times 10^{-10}}} &= 1.2528 \times 10^6 \\ V &= \frac{1.962 \times 10^6}{1.2528 \times 10^6} = 1.566\ \text{m/s} \end{aligned}

Answer: Maximum velocity of water ≈ 1.57 m/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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