Water Hammer (Joukowsky)

Equivalent forms

\Delta H = \frac{c\,\Delta v}{g}

c = \sqrt{\dfrac{K/\rho}{1 + (K D)/(E\,e)}}

\Delta p = \rho c v_0\ (\text{full stop})

Momentum destroyed in an instant reappears as a pressure wave — one clean product of three numbers.

Symbol	Name	SI unit	Dimension	Range
$\rho$	Fluid Density Density of the flowing liquid	kg/m³	M·L⁻³	500 – 1500
$c$	Wave Speed Pressure-wave (acoustic) speed in the fluid-pipe system	m/s	L·T⁻¹	200 – 1500
$\Delta v$	Velocity Change Change in flow speed at valve closure	m/s	L·T⁻¹	0 – 10
$\Delta p$	Surge Pressureoutput Peak transient pressure rise	Pa	M·L⁻¹·T⁻²	0 – 20000000

Unit systems

Symbol	SI	CGS	Imperial
$\rho$	kg/m³	g/cm³	slug/ft³
$c$	m/s	cm/s	ft/s
$\Delta v$	m/s	cm/s	ft/s
$\Delta p$	Pa	ba	psi

Where it holds

Rigid-column/elastic-wave (Joukowsky) limit for rapid closure (closure time < 2L/c) with no column separation. Slow closures and air pockets require full method-of-characteristics transient analysis.

Dimensional analysis

[\rho c \ $\Delta v] = (M\cdot L^{-3})(L\cdot T^{-1})(L\cdot T^{-1}) = M\cdot L^{-1}\cdot T^{-2} = Pa \checkmark$ (pressure)

Discovery

Nikolai Joukowsky · 1898

After burst mains plagued Moscow's water system, Joukowsky ran controlled valve-closure experiments on kilometre-long pipes and derived Δp = ρcΔv. His formula still governs surge-protection design for pipelines, penstocks, and rocket fuel lines.

Try this

Slam a valve and the bang you hear can burst the pipe — how big is that pressure spike?

Water (ρ = 1000 kg/m³) flows at 2 m/s when a valve shuts almost instantly. With a pressure-wave speed of about 1200 m/s in the pipe, estimate the Joukowsky surge pressure.

Research status: stable

Real-world applications

Surge-tank and air-chamber sizing for pipelines
Hydropower penstock and turbine trip protection
Domestic plumbing 'banging pipes'
Rocket and process-plant fluid-line design

Common misconceptions

Water hammer needs an air bubble to make the bang — the noise is the pipe and fluid reacting to the pressure wave itself
Slowly closing a valve gives the same peak — stretching closure beyond 2L/c sharply reduces the surge
Only valves cause it — pump start/stop and check-valve slam are equally common triggers

Experimental verification

Joukowsky's Moscow pipeline tests matched the formula directly; modern pressure transducers on penstocks and pump trip-out events reproduce the predicted surge magnitude and the 2L/c round-trip period.

Derivation

Across the wave front the impulse-momentum theorem on a fluid slug gives

\Delta p\cdot A\cdot dt = \rho \cdot A\cdot c\cdot dt\cdot \Delta v

, where

c\cdot dt

is the length brought to rest in time dt.

Cancelling

A\cdot dt

leaves

\Delta p = \rho c\Delta v

.

The wave speed c itself depends on fluid bulk modulus K and pipe elasticity E.

Limiting cases

\Delta v \to 0⟶ \Delta p \to 0Gradual valve closure produces no sudden momentum change and hence no hammer.

Closure time < 2L/c⟶ Full Joukowsky surgeIf the valve shuts before the wave can return from the far end, the maximum

\Delta p = \rho c\Delta v

develops.

Soft/flexible pipe⟶ Lower c, smaller surgeElastic walls or trapped air drop the wave speed c, cushioning the spike.

What if…

What if you halve the flow speed before closing?

$\Delta v$ halves, so the surge pressure halves — throttling down before shut-off is a real mitigation tactic.

What if the pipe walls are very flexible?

The wave speed c drops, cutting the surge — plastic pipes and air chambers exploit exactly this.

1

Surge from a slammed valve

Given ·

\rho:: 1000
c:: 1200
\Delta v:: 2

Find · \Delta p

Steps

Apply Joukowsky: $\Delta p = \rho c \Delta v$
$\Delta p = 1000 \times 1200 \times 2 = 2$ ,400,000 Pa
$\approx 24\,\mathrm{bar}$ — enough to rupture an unprotected main, hence surge tanks

Result ·

\Delta p = \rho c\Delta v = 1000\times 1200\times 2 = 2.4\times 10^{6} Pa \approx 24\,\mathrm{bar} (\sim 350 \psi )

on top of line pressure.

Bernoulli's Equation→Mach Number→Newton's Law of Viscosity→