Solar Panel System

[Allen G Myerson]

chem geek
Though there is no vacuum in the panels, I don't count the static head since it does cancel each other out up and down (it's just that one shouldn't go negative in the panels).

If there is no vacuum, they can't cancel each other out. You're still adding the static head; you're just doing it on the return side.

Think about this:

There is a pipe that rises 23 feet vertically out of the water, turns 90 degrees and runs 10 feet horizontally, and then turns 90 degrees back down 10 feet to the water. No vacuum relief. Once the pipe is full of water, the pressure in the top horizontal pipe is -10 psi (gauge) and the pressure at water level is zero psi (gauge) (static before moving the water). In this case, the rise and fall cancel each other out because the falling water "pulls" the rising water by the suction created from the water falling.

However, if you put a vacuum relief at the top, there cannot be a vacuum, and the pressure at water level has to be +10 psi static when the pump pushes the water to the roof. Since the water falling back down cannot "pull" the rising water, its potential energy can only be used for the return trip to the pool. If the return trip generates only 3 psi of head loss, you lose the other 7 psi.

You only get back the potential energy if the return trip generates at least 10 psi, or more. In this case, you have to add an extra 7 psi to close the vacuum relief valve. Therefore, you do get to 10 psi, but only by adding 7 psi of extra restriction. You're still losing the 7 psi. Further, you don't need to add the 7 psi restriction. You can just allow the vacuum valve to stay open and suck in air.

To calculate the head loss for an open loop vs. a closed loop, you have to add the height of the panels, but you can subtract the head loss due to the dynamic head created from the panels to the pool (up to the amount of static head).

In your case, the rise and fall do cancel, but only due to the creation of additional head loss in the return from the panels to the pool due to the addition of some type of restriction. The math comes out the same, but I think that it's an important conceptual point to understand that the rise and fall do not cancel each other out in the same way that they do in a closed loop system.

chem geek
The height of the top of the highest solar panel may be more than 15 feet, possibly closer to 20 feet, though I'll have to measure that. So that's 6.5 to 8.6 PSI required in losses from the top of the panel to the returns, probably with a flow restrictor in there somewhere.

This is where you are adding the height of the panels.

chem geek
If I add that to what it takes to get to that highest point I get around (2.7 + 2.2/2 + .9 + either 6.5 or 8.6) = 11.2 to 13.3

You are adding the height (6.5 or 8.6 psi) to the dynamic head caused by the water going to the panels + the dynamic head caused by the water going through the panels

Allen G Myerson
Your total head would be 6.5 psi static to get up to the panels + the dynamic head caused by the water going to the panels + the dynamic head caused by the water going through the panels + the pressure at the vacuum breaker (which should be close to zero).

You can add the height going up or coming down, either way, it's still adding the same thing.

You do get some of the potential energy back from the water falling, but it is limited to the dynamic head loss from the panels to the pool, which should be much less than the height, which is why you need to add extra restriction if you want to close the vacuum relief valve, which you don't necessarily need to do.

[mas985]

If there is no vacuum, they can't cancel each other out. You're still adding the static head; you're just doing it on the return side.

Forgive me for butting in here but I would like to carify this statement, it seems contradictory and not quite correct. First of all pressure is relative and what a vacuum is defined as depends on what you define as 0 PSIR. At sea level, 14.7 PSIA is defined as 0 PSIR. So what is a vacuum at sea level may not be a vacuum on top of a mountain. The atmospheric pressure only determines the pressure that the vacuum release valve opens.

Second, not to be picky but techically ALL pool systems with or without a vacuum release valve are open loop systems since the pool itself is exposed to the air.

Next is the concept of “pulling” water. In the past, there has ben some debate on if water is pulled over a siphon or is pushed over a siphon. Most physicists now agree water is pushed over a siphon and Bernoulli’s equation fully explains the flow rate of water with pressure differentials and changes in height. Also, a siphon will still siphon if both tanks are under the same positive pressure and the top of the siphon is under positive pressure albeit less than the tanks. The pressure differentials are relative and it does not depend on a vacuum.

Next, absolute pressure is irrelavent for static head gain or loss, only the weight of the water matters which is dependent on the water column height. For a fully primed pipe, and it has to be fully primed, there is static head loss going up the pipe and static head gain going down the pipe both are equal but sum to zero and do not contribute to the total head. It doesn't matter if there is a VRV or not. Whenever there is an elevation change in a fully primed pipe, there is a change in static head. The static head comes from the weight of the water in the pipe so water pressure drops with elevation rise and visa versa. It doesn't matter if it is a closed loop system or open loop. As long as the vacuum release valve remains closed air is kept out of the pipe, both the open loop and close loop systems behave identically. If you were to measure the pressure along the two plumbing systems, they would have identical pressures.

Where things get tricky is when the pressure is insufficient to keep the vacuum release valve closed and air gets sucked into the return pipe. At this point the system effectively breaks into two pieces that are somewhat independent. The first part is the pump overcoming the the static head loss going up the pipe as well as the dynamic head loss in the pipe to the vacuum release valve. At this point the pressure is 0 PSIR and the pump only needs to over come the pressure to this point. After this point, the system behaves as a gravity fed plumbing system. Water falls down the return pipe through the air and the return pipe may still be partially filled creating some head gain. This water height then determines the flow rate through the remaining part of the plumbing.

So in summary, static head gain and loss can occur with or without a vacuum release valve and the plumbing system behaves identically as long as the vacuum release valve stays closed.

[chem geek]

When I wrote I wasn't counting static head (because it netted to 0 up and down), I was specifically talking about the calculation I was doing to account for the incremental head loss (pressure difference seen at the filter's pressure gauge with vs. without solar) going from the filter to the panels and back to the pad (I corrected my earlier post where I underestimated the flow restrictor amount). I did not use any static head in that analysis (the 2nd paragraph in this post) nor should I unless, as Mark points out, the vacuum relief valve was open and I assumed it was not.

The later calculations (in the third paragraph) where I did count static head were only to figure out the minimum pressure needed (again, as measured at the filter gauge) so that I would not have less than zero net pressure (relative to air pressure -- I called this a vacuum, but technically it's just a pressure lower than atmospheric pressure). Essentially, I was just trying to figure out the implied loss from a flow restrictor. If I calculated the overall head loss with that assumption, then I don't use any static head calculation: 2.7 + 2.2/2 + 0.9 + 2.2/2 + 2.7 + ( 2.7 to 4.8 )) = 11.2 to 13.3 PSI which is reasonably close to the 13.5 PSI increment I measured for the solar given my underestimation (not counting the numerous 90 degree elbows, T-junctions and associated extra pipe length on the roof, etc.). There is no static head in this calculation (i.e. it cancels out up and down so I did not include it), but that's because I assume the vacuum relief valve is closed due to sufficient pressure helped by a flow restrictor.

The interesting thing is that my unusually high head loss even with the solar off may imply that the flow restrictor was put in the general piping line to the pool, not just in the loop that included the solar. I'll see if I can find out where the flow restrictor is in my system. Maybe it was in the main line for the solar off situation to give the 1 HP pump more resistance to limit the flow rates somewhat. If that is the case, then now with my variable speed pump I can have that flow restrictor moved to return side of the solar loop instead so that my solar off energy costs can be further reduced.

[Allen G Myerson]

If there is no vacuum, they can't cancel each other out. You're still adding the static head; you're just doing it on the return side.

Forgive me for butting in here but I would like to clarify this statement, it seems contradictory and not quite correct. First of all pressure is relative and what a vacuum is defined as depends on what you define as 0 PSIR. At sea level, 14.7 PSIA is defined as 0 PSIR. So what is a vacuum at sea level may not be a vacuum on top of a mountain. The atmospheric pressure only determines the pressure that the vacuum release valve opens.

I don't think that the statement is at all contradictory, or incorrect. I think that it is obvious that I was referring to local gauge pressure, where vacuum means a pressure less than 0 gauge, which is 14.7 psi absolute. In a closed loop system that is less than 32 feet high, the static height up and down cancel due to the vacuum created by the falling water and the siphon effect..

mas985
Second, not to be picky but technically ALL pool systems with or without a vacuum release valve are open loop systems since the pool itself is exposed to the air.

This is not correct. As long as the loop is airtight, it is closed. There is definitely a difference between a closed loop and one that has a vacuum relief.

mas985
Next is the concept of “pulling” water. In the past, there has been some debate on if water is pulled over a siphon or is pushed over a siphon. Most physicists now agree water is pushed over a siphon and Bernoulli’s equation fully explains the flow rate of water with pressure differentials and changes in height. Also, a siphon will still siphon if both tanks are under the same positive pressure and the top of the siphon is under positive pressure albeit less than the tanks. The pressure differentials are relative and it does not depend on a vacuum.

Of course water cannot be "pulled". It is just a conceptual device that can be used to make things easier to explain. When someone sucks water through a straw, they are just lowering the pressure at the top of the straw enough for the ambient atmospheric pressure to push the water up the straw. If there is a vacuum relief valve, then a siphon cannot work. If there is no vacuum relief, then a siphon can work. This is the difference between an open loop and a closed loop.

mas985
As long as the vacuum release valve remains closed air is kept out of the pipe, both the open loop and close loop systems behave identically.

This is the key point. If the vacuum relief valve remains closed, then the return head has to be greater than the static head and the static head will cancel out. However, if one decides not to worry about closing the vacuum relief, then the static head does not cancel. The potential energy of the water coming from the roof can only be used for the trip from the panels to the pool. If the head loss of the plumbing from the panels to the pool is less than the height of the top of the panels, then the difference is lost and the static head does not cancel.

mas985
Where things get tricky is when the pressure is insufficient to keep the vacuum release valve closed and air gets sucked into the return pipe. At this point the system effectively breaks into two pieces that are somewhat independent. The first part is the pump overcoming the static head loss going up the pipe as well as the dynamic head loss in the pipe to the vacuum release valve. At this point the pressure is 0 PSIR and the pump only needs to overcome the pressure to this point. After this point, the system behaves as a gravity fed plumbing system. Water falls down the return pipe through the air and the return pipe may still be partially filled creating some head gain.

This agrees with what I have been saying.

mas985
This water height then determines the flow rate through the remaining part of the plumbing.

The flow rate, in gallons of water per minute, remains the same. It has to be the same as what is supplied. If the vacuum relief valve is open, you will get a turbulent mix of water and air. The water height will determine the velocity of the water air mix, but the flow rate remains the same.

mas985
So in summary, static head gain and loss can occur with or without a vacuum release valve and the plumbing system behaves identically as long as the vacuum release valve stays closed.

This agrees with what I have been saying. My point is what happens when the vacuum relief opens due to the return trip having less head loss than the height of the panels.

For example:

Solar off: Going from out of the filter to the pool, you have the equivalent of 100 feet of straight 2-inch PVC. Flow rate 48 gpm. The head loss is 4.24 feet.

Solar on, with a vacuum relief valve, and allowing the vacuum relief to remain open:
Going from out of the filter to the panels, you have the equivalent of 100 feet of straight 2-inch PVC. Flow rate 48 gpm. Head loss through the panels = 2 feet. Height to the top of the panels = 20 feet. From the panels back to the filter pad you have the equivalent of 100 feet of straight 2-inch PVC. And then, going to the pool, you have the equivalent of 100 feet of straight 2-inch PVC.

With the solar on, you have 4.24 feet of dynamic head to the panels + 2 feet of dynamic head through the panels + 20 feet of static head, which equals 26.24 feet of total head pressure that the pump has to push against.

From the panels back to the pool, you have the equivalent of 4.24 feet + 4.24 feet = 8.48 feet of head. Since this is less than the static head, you don't have to count it. Therefore, of the 20 feet of static head potential energy, you only use 8.48 feet and the rest (11.52 feet) is lost. This is a case where the static head going up was only partially offset by the static head coming down.

[Allen G Myerson]

The situation can be made clearer if we use larger diameter pipes to reduce the total dynamic head loss, and take it out of the equation. Of course, it would take an infinitely large pipe to reduce the total dynamic head loss to zero.

However, if we decide that any total dynamic head loss of less than 1 foot can be considered to be negligible, then we could use all 4-inch PVC and the total dynamic head loss would be 0.5 feet. (48 gpm going through 300 foot of 4-inch diameter PVC.

In the case of an open loop, you are left with the static head due to the height of the panels and the dynamic head loss due to the water going through the panels. In my earlier example, this would be 20 feet + 2 feet for a total head loss of 22 feet. This is the theoretical design limit for an open loop.

In the case of a closed loop (less than 32 feet high), the static head automatically cancels due to the siphon effect, and you are left with only the dynamic head loss due to the water going through the panels. This would be 2 feet. This is the theoretical design limit for a closed loop.

The difference in theoretical design limits is the height of the panels. The height of the panels has to be added to an open loop and it does not automatically cancel, whereas for a closed loop, it does automatically cancel.

[mas985]

I have only a couple of disagreements with what you posted last.

The classical definition of an open loop plumbing system in both the plumbing and hydraulic industries is a plumbing system that has ANY point where the water contacts the air and is at atmospheric pressure. This happens at the pool surface so by definition any pool plumbing with or without a VRV is an open loop plumbing system. A closed loop plumbing system is defined as being isolated from the environment where the water has no contact with the air which is not the case for a pool even one without a VRV. A section of pipe may be open or closed to the air but the entire pool system is always open. But I think we have wasted enough time on that point so let’s move on.

The flow rate, in gallons of water per minute, remains the same. It has to be the same as what is supplied. If the vacuum relief valve is open, you will get a turbulent mix of water and air. The water height will determine the velocity of the water air mix, but the flow rate remains the same.

I disagree with this statement as written. A pump’s flow rate is dependent upon the head loss (static and dynamic) that the plumbing system creates. So if a plumbing system has more head loss, the flow rate from the pump decreases. When the VRV opens and total head loss increases because of the loss of head gain on the other side, the pump flow rate will slow down. In fact, depending on the pump’s head curve it can slow down so much that the pump can become dead headed where the water will stop flowing at a certain height in the pipe. The pump will raise the water to maximum head of the pump curve and simply maintain that level. However, if the pump head curve is high enough, then water will continue to flow albeit at a slower rate but not the same rate as when the VRV was closed. Not only does an open VRV have more head loss, it will also have much less flow rate.

Third, static head is not dependent on the siphon effect but siphoning is dependent on static head differences. Static head can exist without siphoning but siphoning cannot exist without static head differences. So the canceling effect of static head has absolutely nothing to do with a siphon. A siphon cannot exist without a differnce in static head. Static head only has to do with the change in elevation, siphoning is not even involved since the two pipes are at the same height plus the system is on a pump which does not meet the criteria for the definition of a siphon where it must be gravity fed.

Other then that, I think we agree.

[Allen G Myerson]

mas985
When the VRV opens and total head loss increases because of the loss of head gain on the other side, the pump flow rate will slow down.

The VRV prevents vacuum. Therefore, when there is a VRV, you will never get a head gain. There is no head gain to lose.

I was responding to this statement that you made:

mas985
This water height then determines the flow rate through the remaining part of the plumbing.

It seems like you were saying that the flow rate of the water returning to the pool would be determined by the height of the panels. If that is what you were saying, my point is that the flow rate can't be anything more than the flow rate being supplied. It seems like you were saying that the flow rate from the panels back to the pool could be higher than the flow rate from the pump to the panels.

mas985
So the cancelling effect of static head has absolutely nothing to do with a siphon. A siphon cannot exist without a difference in static head. Static head only has to do with the change in elevation, siphoning is not even involved since the two pipes are at the same height plus the system is on a pump which does not meet the criteria for the definition of a siphon where it must be gravity fed.

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.

[mas985]

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.

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.

[Allen G Myerson]

Example:

You have a 2-inch PVC pipe rise out of the water 23 feet straight up, turn 90 degrees and run 10 feet horizontally, turn 90 degrees back down 23 feet into the water. Assume that the pump is at the water level and has a check valve. Sea level ambient atmospheric pressure =14.7 psi absolute or 0 psi gauge. Flow rate = 48 gpm.

With no vacuum relief valve:
If the pump turns on, fills the entire pipe with water, and then turns off, the pressure at the check valve would be 14.7 psi absolute or 0 psi gauge. The pressure at the top would be 4.7 psi absolute or -10 gauge. When the pump runs, the only head loss would be due to the dynamic head loss. The total length of pipe would be 56 + 11.4 for the 90 degree fittings = 67.4 feet. The dynamic head loss would be 2.86 feet.

With a vacuum relief valve:
If the pump turns on, fills the entire pipe with water, and then turns off, the water would run out of the downward pipe and the pressure at the check valve would be 24.7 psi absolute or 10 psi gauge. The pressure at the top would be 14.7 psi absolute or 0 psi gauge. When the pump runs, the head loss would be due to the dynamic head loss up to the downward pipe + 23 feet for the static height. The total head loss would be 1.88 feet + 23 feet = 24.88 feet. Without a vacuum relief valve, the pump has to push against an additional 23 feet of head. The additional head is only partially offset by the subtracting dynamic head loss due to the water's flow through the downward pipe. The water's potential energy due to its height at the panels can only be used by the return trip.

As you can see, the difference is quite significant. The difference is 24.88 - 2.86 = 22.02 feet of head, which is almost equal to the height. You could increase the resistance of the downward pipe by 22.02 psi to close the vacuum relief valve, but it wouldn't make any difference to the total head loss that the pump has to push against

My primary point is that, when there is a VRV, the static head does not necessarily cancel. It only partially cancels unless the return head is unusually large. It seemed to me that some people were automatically assuming that the static head up and down automatically cancelled.

[Allen G Myerson]

The simplest and most obvious way that I can put this in non technical terms is that when there is not a vacuum relief valve, the falling water creates a suction in the top pipe, which allows it to "reach back" all of the way to the suction inlet and "pull" on the entire column of water with a force that exactly cancels the static head loss.

When there is a VRV, the falling water cannot create a suction in the top pipe. Therefore, it cannot "reach back" and "pull" on the preceding water column. However, the water has still gained a certain potential energy due to its height. This potential energy can only be used to overcome the dynamic head loss due to the water's flow from the top pipe back to the pool. This means that the static head loss is only partially recovered. The static head loss recovery is limited to the dynamic head loss due to the water's flow from the top pipe back to the pool.

Yes, I know that water cannot be "pulled". However, it is a perfectly valid conceptual device just as when someone says that water is sucked through a straw or pipe.

[Allen G Myerson]

We don't try to make a closed loop where the water falling back down pulls the feed water up creating a syphon effect meaning the pump doesn't have to do any work. No No No. That would create negative pressure in the solar collectors and a negative pressure (a vacuum) can collapse pvc plumbing when it gets hot and hot it will get. http://www.h2otsun.com/pools/index.html

[mas985]

Yes, I know that water cannot be "pulled". However, it is a perfectly valid conceptual device just as when someone says that water is sucked through a straw or pipe.

My primary objection is in your use of the concept that suction is pulling on the water. Suction does not and cannot pull on water, period. You keep saying you know that and yet you continue to do so which I really don't understand. Read through any physics book or for that matter any elementary science book and the act of sucking on a straw is explained as simply as the higher pressure pushing on lower pressure which moves the fluid. The same forces are at work for the sealed vertical pipe except the water weight is creating the pressure differential instead of the mouth sucking on the straw. I really think you would better serve the forum by explaining the physics in proper terms rather than propagating a myth.

BTW, I would stick to educational text books when looking for the proper definition of a siphon rather then a commercial web site. They are using the term incorrectly as well.

In reality, it is the solar system WITH a VRV at an elevation lower than the top most point of the plumbing which can siphon water out of the panels. The case without a VRV or even with a VRV at the top most point of the panels, does not qualify as a siphon because gravity can not move the fluid up and over a high point. Only the case were a VRV is installed at a point lower than the solar high point can a siphon exist. in this case when the VRV opens, water will be siphoned away from the VRV up and over the top of the solar panels. The act of siphoning is the higher air pressure pushing up on the lower water pressure thereby moving the water up and over the high point.

[Allen G Myerson]

mas985
My primary objection is in your use of the concept that suction is pulling on the water. Suction does not and cannot pull on water, period. You keep saying you know that and yet you continue to do so which I really don't understand.

OK, you are right about this. It's easier than trying to explain it the correct way, but that's no excuse. I always get aggravated when I see science principles being explained improperly, especially on TV shows that should know better. I will make the effort to explain things correctly, and not take shortcuts that are not scientifically accurate. Thanks for helping so many people with your advice.

[chem geek]

Likewise Allen. Thank you for your contributions. I hadn't considered what was going on with my solar panels with regard to looking at static head due to the vacuum relief valve and not having too little pressure (i.e. too much less than atmospheric) near the top of the panels so you helped me out there. I was hoping you and Mark would see eye-to-eye since it seemed to me you were both saying pretty much the same thing, though not necessarily in the same (technically accurate) way.

[mas985]

Allen,

Thank you for understanding my point of view. You have been both open minded and polite in the discussion which is sometimes a rarity on these forums. I didn't expect this to go this far but I am glad that we came to an agreement. To a fault, I often find debate stimulating, engaging and I usually come away with even deeper understanding of the scientific principles involved.


Also, Richard,

There have been several threads about this subject over at TFP mostly around solar and 2 speed/variable speed pumps but about also about air in the returns caused by the VRV opening although you may have missed some of the postings. Some have have experienced an open VRV due to a dirty filter, so too much head loss on the supply side can also be a problem which leads me to think that may be your issue with the check valve.

Also, some have asked if they can run on low speed with solar and in the past, someone would chime in and say that the static head loss is canceled so it should work ok. But in fact, static head still exists and affects the pressure at the VRV but simply does not add to total head with respect to the pump. This is why I don’t like to use the word cancelled for static head because some have interpreted that to mean the static head has disappeared all together when it really hasn’t. This slight nuance is often missed and can cause misinterpretations in what is really happening.

I could go on and on ad nauseum about this topic but I think it is time to get off my soap box now.