Does anyone know of a formula for calculating the minimum closing time of a butterfly valve to prevent water hammer?
I have a vague recollection of there being such a beast but cannot find any references.

Marks handbook has a short bit on waterhammer

Some rules of thumb on water hammer
For every 1 ft/sec of velocity there is a 50 psi surge on a sudden stoppage,  so that give you the magentude of the surge.  As you close the valve a pressure wave develops as a higher pressure upstream and a lower pressure down stream, the time it takes this wave to reach a point where the water does not have velocity is based upon the speed of sound in water (about 2000 ft/sec) For example, two tanks with water flowing from the upper tank to lower tank through a 3000 ft pipe. (diameter, pressure and flow do not matter for amount of surge pressure) A valve is 1000 ft from the upper tank, the water is flowing at 10 ft/sec. 
Lets look at upstream for now, assume the system can withstand an additional 50 psi upstream of the valve (static pressure 100 psi and system rated for 150).  The fastest you can slow the water down is 1 ft/sec in a sudden change in flow, it will take 1/2 second for the pressure wave to reach the upper tank and another 1/2 second later the wave will return while that wave is propigating the pressure will read 150 just upstream of the valve, since the valve is still mostly open the wave will then travel through the valve and a small amount will be reflected back, at this point you can close the valve to decrease the flow another ft/sec.  but as the valve continues to close more of each pressure wave will be reflected instead of passing through, this is when the closing rate needs to be slowed down so the waves to do not stack upon each other. When the valve is fully closed the wave will bounce back and forth between the valve and the upper tank until it is disapated in 10 to 100 cycles. 
Downstream has the same thing but the pressure wave is negative, for the example the cycle is 1 second down and 1 second back and the pressure will read 50 psi for those 2 seconds.  The down stream case does not seam as damaging but if the static pressure was 35 psi, the pressure will fall below the vapor pressure of the fluid at minus 10 psi for cold water, then the water will undergo volumetric boiling also known as cavitation, and cavitation can destroy anything with enough time, other considerations: thin wall steel pipes that are not designed to take a vaccumm can colapse under even mild negative pressures, and leaks in the system can suck back in contaimination.
So for this example the operator should not close the valve in less than one minute, and to be safe take five minutes...

As explained by Hydrae "2 * L/a" is the time for the pressure wave to travel from the valve to the tank and back again to the tank.

If the valve is closed faster than this time it is termed a rapid closure and the pressure wave generated is equal to that generated by an instantaneous closure.

For water the pressure rise due to instantaneous closure is V.a/g (in SI units for water in a rigid pipe - v = initial velocity m/sec, a is the wave speed 1100 m/sec and g is 9.81 m/sec/sec).

If the valve is closed slower than "2 * l/a" then it is termed a slow closure and the pressure rise will be less than v.a/g. For a slow closure the reflected pressure wave will return to the valve before it is closed. The sign of the reflected pressure wave is opposite to the propagated wave. On valve closure a positive pressure wave is propagated upstream. When it reaches the tank it is reflected back at an equal magnitude but opposite sign (negative). It is this fact that results in the pressure rise being less for slow valve closure.

If the pump or valve closure time is rapid (less than 2 * L/a) then you really don't need to do a complex waterhammer analyses because the pressure rise can be calculated from v * a/g. If the closure time is slow then you need a complex analyses which examines the interaction of the propagating and reflected pressure waves is needed.

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