Last summer I built an underground sprinkler system for my lawn. As I flipped the switch for the first time and beamed with pride at my wonderful creation, I couldn’t help but notice the loud * thud *whenever one of the valves closed. Lo and behold (insert joke about engineer building projects here), this is a problem that is applicable to any engineered piping system.

* Water hammer* is an increase in the fluid pressure in a pipe caused by a sudden velocity decrease (usually a valve closing). Like a good engineer, I researched it and will report the results to you here.

Analysis of water hammer theory depends on whether the pipe material is rigid or elastic.

## Rigid Pipe

For steel pipe, the time required for the water hammer pressure wave to travel from the suddenly closed valve to a point of interest depends only on the speed of sound in the fluid and the distance between the two points. This is also the time that is required for the fluid to come to rest.

$latex t = \frac{L}{a}&s=2$

*Where:*

* t = time for fluid to come to rest (s)*

* L = length of pipe (ft or m)*

* a = speed of sound within the fluid (ft/s or m/s)*

Thus, when the water hammer pressure wave reaches the original source of the water, the pressure wave will dissipate. A * rarefaction wave* will return to the valve at velocity

*a*. The time required for the compression wave to travel to the source, and the rarefaction wave to travel back to the valve is given by:

$latex t = \frac{2L}{a}&s=2$

This is also be length of time that the pressure is constant at the valve.

The fluid pressure of the wave can be calculated by equating the kinetic energy change of the fluid with the average pressure during the compression process.

$latex \Delta p = \rho a \Delta v&s=2$

*Where:*

*Δp = Change in pressure due to valve closure (lb/ft ^{2} or Pa)*

*ρ = Density of fluid (lb/ft*

^{3}or kg/m^{3})*a = Speed of sound in fluid (ft/s or m/s)*

*Δv = Change in velocity of fluid (ft/s or m/s)*

Isn’t it interesting that the pressure increase at the valve does not depend on the time it takes to close the valve. As long as the valve is closed when the wave returns, there is no difference in pressure. Also, the shorter the pipe, the faster the valve must close, because a long pipe with a slow valve closure has the same effect as a short pipe with a fast valve closure. The important variable is whether the valve is fully closed when the wave returns.

If the valve does not close by the time the rarefaction wave returns to it, the calculation method is very complex. The pressure within the wave will be less than predicted by the above equation, but there is no simple method for calculating it.

## Elastic Pipe

For plastic pipe, the analysis method is the same except that the calculation of the speed of sound in the fluid must account for the elasticity of the pipe material. Thus, the speed of sound, *a*, is modified as follows:

$latex E = \frac{E_{water}\cdot t_{pipe}\cdot E_{pipe}}{t_{pipe}\cdot E_{pipe} + D_{pipe}\cdot E_{water}}\newline\newline

a = \sqrt{\frac{E}{\rho}}&s=2$

*Where:*

* a = speed of sound in fluid (ft/s or m/s)*

* E = Modified modulus of elasticity (aka bulk modulus) of pipe wall (lb/in ^{2} or GPa)*

*ρ = Density of fluid (lb/ft*

^{3}or kg/m^{3})*E*

_{water}= 3.12 lb/in^{2}, or 2.15 x 10^{9}Pa*t*

_{pipe}= thickness of pipe (in or mm)*E*

_{pipe}= Modulus of elasticity of pipe material (lb/in^{2}or GPa)*D*

_{pipe}= Diameter of pipe (ft or m)Interestingly, the effect of the water hammer is reduced by a larger diameter pipe.

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