I don’t know how I managed to give that impression, but if I did I apologise. I’m always ready to learn, that’s why I started this thread, and I’m grateful for the responses, which helped me fix the problem. On to the hydraulics! Take a situation where you’re pumping from a tank into a pipe with an inverted U, and the height from the top water level in the tank to the top of the U is say 3m, and it’s 1m from the top of the U to the end of the pipe. If the flow is reasonable, or there is a seal pot on the end of the pipe, (and any air has been expelled) there is 1m siphonic assistance, so the pump sees 2m head (ignoring any losses). In that case the pressure inside the pipe at the top of the U is below atmospheric, by 1m. So if you drill a hole in the top, air will enter, and break the siphon. The water coming over the U then free-falls down inside the pipe. That is sometimes desirable (siphon breakers aren’t unusual), but then the pump head rises to 3m. Of course, if the end of the pipe is below the tank water level, you can get flow with no pump, as you discovered all those years ago. BTW, I only put a lot of detail in to make sure we each understand the other’s viewpoint, not trying to be patronising! Happy to discuss further if you want.