by Lev Nelik, Ph.D., P.E. Pumping Machinery, LLC

If, however, there is no dampener, then the spring will chatter when the pressure is at 122 psi. It will close the release valve, and at 277 psi it will obviously open. In such a case, the user would observe this happening and set the spring to open at a lower value, such as 122 psi. This way, the release valve never closes. In fact, the user will likely not notice the piston pump pulsation, as you described, but will only be watching the practical result to get the spring set. When his gauges show the (average) pressure, he will stop turning down the spring cap and leave it at that point so that the release valve will be open when he (practically) had it set.

PD pump manufacturers follow the same process—setting the spring to open and fully bypass not at calculated instantaneous values but at the actual gauge readings (time-averaged damped). They, of course, deal with the internal release valves, but external release valve manufacturers follow similar steps. Similar logic can be used for the pure static pressure (injection) or a combined system. I did not bother getting the numbers calculated too exactly, nor drafting a quick sketch of this, but I think you get the main point.

Tom Morrison responds:
This was an awesome response, and I appreciate your not getting down to the “nth” decimal place with your calculations. You addressed your response at the exact level of detail I was hoping for and provided enough detail that I could follow along precisely.

If you would allow me to summarize back to you what I think I read, I would say the following. The concern I asked you about is real, and the manufacturers incorporate this knowledge into the somewhat-iterative setting of the spring in the internal release valve to dampen out the pulsation effect.

For good design practice, one or more pulsation dampeners are necessary to properly dampen this swing for this type pump. In my revalidation of the external relief valve on the pump discharge, I should check the sizing against the normal design flow of the pump, not the instantaneous flow indicated by the Q = Qmax x sin (2πx) term you showed in your response.