Allen G Myerson wrote: The VRV prevents vacuum. Therefore, when there is a VRV, you will never get a head gain. There is no head gain to lose.
My point was that when the VRV opens, you effectively lost some of the pressure gain from the non-VRV case because air entered the return pipe. However, this situation is not necessarily binary and can have some gradation of static head. You will only lose all of the pressure gain if all of the water is removed out of the return pipe and there is no elevation to the water level in the return pipe. Even a small amount of water will cause a pressure differential from the input to output of the pipe.
When the VRV opens, air enters the plumbing system but it also encounters head loss due to the friction loss through the VRV itself so the pressure on the inside of the plumbing at the location of the VRV does not immediately go to 0 PSI. This is much like a balloon deflating through a small hole. The entering air mixes with the flow from the pump and creates an air/water mixture which is lighter than water alone so the pressure differential across the return pipe starts to decrease as more air enters the pipe and the percentage of air increases. However, there will always remain a pressure differential between the ends of the return pipe as long as there is at least some water in the mixture creating weight at the bottom of the pipe. If the flow rate of the pump decreases, due to the extra head, enough such that water cannot collect at the bottom of the return pipe, then and only then will the pressure differential across the return pipe go to zero. For example, if the pump can provide enough flow rate so the air/water mixture remains at 50%, then half of the pressure differential between the ends of the pipe remain.
Also, because a pump is dynamic and the flow rates are changing with the changing total head, you can get into a situation where the solar system will go through cycles of the VRV closing and opening so it oscillates back and forth between two states. This is often described as periodic bursts of bubbles out of the returns rather than a constant stream.
In the case of a constant stream of bubbles, it is likely that the there is an equilibrium of water and air in the return pipe and it is still creating at least some pressure differential in the return pipe so not all of the return gain is lost.
But the bottom line is that only under the situation where the pump cannot deliver enough flow rate to at least partially fill the return pipe, does the pressure differential in the return pipe go to zero and all of the static head disappears.
Allen G Myerson wrote: This is just not accurate. When there is not a VRV, the siphon effect does cause the static head to cancel. The weight of the water falling creates a lower pressure at the top, which equals the static head pressure. This siphon effect works whether there is a pump or not. In this case, the siphon effect is equal, which cancels the height difference.
The classical definition of a siphon is a device that moves fluid from one height to a lower height using ONLY gravity. In the case that you have outlined, the two pipes are of equal length, contain equal amounts of water and are at identical elevation so a siphon is not possible under those conditions. If you turn off the pump, water will remain in the pipes and there will be no siphoning of water from one pipe to other. That requires a pressure differential which can only be caused by difference in water height that each pipe is connected to. Also, plumbing systems with pumps are not usually referred to as siphons at least not in text books I have read. Also, I don’t think the term “siphon effect” does not accurately describe what is actually going on and can be misleading. The only force which creates static head is the force of gravity pulling downward on each of the columns of water. There are no other forces involved with static head. However, a difference in elevation from two bodies of water can create pressure differences which then force water movement like in the case of the siphon. But that doesn’t really apply in this case.
The water in each of the two pipes is exerting a downward force or pressure due only to the effect of gravity on the water. The common node of the two pipes has the same pressure and is lower than the pressure at the bottom of each of the pipes. It is the higher pressure on the bottom of the pipe which holds up the water in the pipes. If you separated and capped each pipe at the top, the pressure at the top and the bottom of the pipe would remain the same and water would still remain in the pipe so the net force at the common node between the two pipes is zero and the pipes are in equilibrium which means that neither side of the pipe is exerting force on the other. Therefore, without an external force, no water will flow. Now if a pump is attached to the setup, then the pump creates an imbalance in pressure which then causes water to flow.
Also, the concept of static head being “cancelled” is somewhat misleading. One side doesn’t really cancel the other it is just a case where the weight of the water is the same above both pipes so there is no difference in static head and it doesn’t add to the total head of the pump. In other words, the pump does not need to lift the water to a higher elevation. One way to make this distinction is to imagine two locations at the bottom of a lake and at the same depth, do the locations have the same pressure because one side cancels the other or is it simply because they have the same amount of water overhead? I believe it is later.
Perhaps I am being too picky about this but I think if the wrong terminology is used, there can a misunderstanding of exactly what is going on.