Hi All, HVAC contractor here with lots of experience in closed loop design and install. Probably counting in 1000's of tons and 100's of miles of pipe over the past 30 years. Never once installed an open loop until now. Sizing circulator pumps is easy on closed loop. Just friction loss from pipe and dynamic loss from fittings and unit. Pressure or change in elevation doesn't enter into the equation. Spent the last few days researching open loop concepts and have lots of questions. I would like to start off with just the basic head the pump will see (ignoring piping losses) would it be correct per my diagram that the total static head (not sure if that is the correct term) on the pump is 130' or would it be from the water level to the uppermost point in the piping?
Thinking a little more about this it seems like at the moment of the pump starting up and the solenoid on the GSHP opens the pump has to overcome the highest lift in the piping to start flow. Once flow is established throughout the entire piping system is when the head will decrease to what I show in the diagram.. Any thoughts??
Looks like this forum is a ghost town. Maybe someone will come along and give some input. In the meantime I am going to keep beating on it until I understand it completely. Looks like my initial diagram isn't right. Common sense tells me the total static head would be the difference between the two water levels. I can't ignore the head on the discharge pipe because it is submerged some distance from the surface of the water.
So moving on to the next step in this journey of learning. I calculated the pressure loss of the system due to friction loss through the pipe and losses from the fittings and equipment. Adding the piping components to the GWHP gives me a total of 10.3 ft of head at my design flow of 9 GPM. Compared to the static head on the pump from the elevation change this is a very small part. It can probably be ignored completely and would not affect pump selection that much. But in the interest of accuracy we add that to the static of 135' from above and total dynamic head on the pump comes to 145.3 ft. I found some generic pump curves online to play with. Later I will research pumps and do a proper selection. I digitized the curves and overlaid my system curve to get the intersects. So at this point the best choice is the smaller 3/4 HP pump. Next up is I don't have a feel for how much pressure I will have at the inlet of my unit. The unit doesn't really care about pressure and just wants a solid column of water without any air bubbles. Good practice is to run some positive pressure at the inlet of the unit. This is easy enough to do with a small expansion tank or even a ball valve on the outlet side to create a pressure drop. But for now I want to explore the pressure in this system without the addition of those items. I suck at fluid dynamics and will probably jack up the equations but lets see where that takes me.
So it appears I made a mistake on my system curve calculation. The only head that is a function of flow is the piping and unit. The system will always have the static head and can never go to zero. Sorry, did I mention I was a closed loop guy? This makes more sense.
I know my posts are kind of weird. I am just using this to document my steps as I work through this problem. I decided to beat on pressure difference as a function of height difference. I know the head on the riser pipe in a well with a water depth of 140' is... well it is 140'. But what is the actual math behind that. This is going to be a long post.... came across this nice equation: So this equation looks nice, it even give the result in psi. So what is the specific weight of water? Using these equations it is: I had to look up the density of water in a chart and plugging that number in I get this: The unit of slugs is not very intuitive to me; when I hear that word I think of those slimy creatures I would see on my back steps in the morning. I am used to density being found by (m= ρv) Mass = Density x Volume. One slug is defined as a mass that is equal to 32.174 lbs. Beating on it even more using Newton’s second law (F=ma). From high school physics we are taught that weight is just a force due to gravity. Knowing that one cubic foot of water weighs 62.428 lb we divide that by acceleration due to gravity (32.174) and we get a mass of 1.94 slugs. Cool. Now plugging those numbers into my very first equation using 140' as the height I get this: Hmm, we get over 10,000 psi... That is obviously not right. I don't know what I am overlooking in this equation but I can't seem to get it to work. But this one does: So the difference in pressure of water from 0' to a depth of 140' down is 8739.92 lb/square feet. Converting the area from square feet to square inches by dividing by 144 and we get 60.694 psi. Divide 140' by the 60.694 psi we get 2.307 which is the conversion factor used to convert ft of head to psi. So all that math works. Another way to look at it is the weight of 1 cubic foot of water (12"x12"x12") with another 139 cubic feet stacked on top. One cubic foot of water weighs 62.428 lbs. Stack another 139 on top or 8,677.492 lbs and the total weight is 8739.92 lbs like I show above. This also shows that only the height of the pipe matters and not the size. A pipe that is 36" in diameter will have the same head as a pipe with a 3" diameter. So after all this math I understand how ft of head is just a measure of pressure and magically equals the height or change in elevation of the fluid. Fun stuff!
I think this post is turning into a diary of a madman. They say you can't wear out a CD but I am pretty sure I may have wore out that Ozzy disk back in high school. That could explain a few things. Moving on though. I want to analyze the properties of this system at different points. This is unrealistic but I am going to start by ignoring pressure due to elevation change and friction drop through the pipes. So I will only be looking at pressures associated with velocity of the fluid as it travels through the different size pipes. This is just a straight forward rate problem. I do that all the time with airflow through ducts. CFM = FPM x Area(sqrft) It is a simple calculation as long as you take care of converting the units: minutes to seconds, gallons to cubic feet and such. As you would expect velocity increases as the pipe diameter decreases: And Bernoulli's principle tells us energy is conserved which means the total energy (pressure) is constant and potential energy (static pressure) is converted to kinetic energy (velocity pressure) without any losses. Using this equation for velocity pressure: I can solve for static and velocity pressure at each point: The pressure results I show in this chart are converted to feet of head. It is obvious on this system the velocity has little impact in the overall picture. But it has always fascinated me how pressure energy is conserved while converting from potential to kinetic. Velocity pressure is not a linear function because of the square in the equation. I need to think about this a lil before I move on to adding elevation and friction to my calculations.
So happy Valentines day everyone. To be honest this is a depressing holiday for me. My wife has been in poor health for many years now with the last year or so being confined to bed and wheelchair. I don't know; it is just a hard thing for everyone involved to go through. I thought I would have a better understanding of the math behind converting potential to kinetic energy by now. I have been using an object at rest, some height above the ground as my model and looking at how the height energy is changed to velocity energy as it accelerates towards the ground due to gravity. I have thrown spreadsheets at it, created charts that show the velocity increases as a linear function over time. And of course the slope of that line is acceleration due to gravity. I can sum of the feet travel each second as the velocity increases and that gives the total change in height (potential energy). When you draw a sloped line between two points you can create a triangle, and knowing the area of a triangle is equal to 1/2 the base time the height then maybe that is where the 1/2 comes from in the kinetic energy equation. When I curve fit the distance traveled versus time it is a nice 2nd degree polynomial which probably explains where the velocity squared comes from. So far I have refrained from asking the internet for the answer, if I can solve it myself I will have a better understanding. But so far I don't have enough geek in me figure it out. So I may have to put this curiosity on hold for now. I have this nice geothermal heat pump unit I will be setting next week and without water flowing through it.... well, it just becomes a pretty room ornament.
So getting back to my calculations after letting Mr. Newton sidetrack me. I have calculated the total pressure at each of these points in the system. Using the general energy equation which is an extension of Bernoulli’s equation that accounts for other losses: And ignoring energy added/removed by the pump I created this table that sums up all the pressures at each specific point in the system: For instance at point 'A' the pump, it sees pressure from elevation and the friction loss from the entire piping system. And as we move on to each consecutive point the friction head from the piping decreases and the elevation pressure changes. So at point 'D', the inlet to the GWHP the total pressure is 18.31' of head. Convert that to psi and we get 7.936 total pressure which the unit should be happy with. As a double check of all the figures. The total head the pump sees is the difference between point 'A' and point 'G' or 147.617' of head. Which equals my initial calculation: I am probably overlooking some minor losses but I doubt they are anything that would have a significant impact on my numbers. I do want to explore how a small bladder tank might affect things and decide if one is really required. They aren't expensive and it probably is a wise addition. More math to come.
As I was thinking about bladder tanks and how pressures will vary on this system I thought I better look at variations in the well water level first. Common sense tells me there will some change due to pump drawdown and maybe even seasonal changes. This is something I don't have a good feel for at all and I will need to reach out to someone who call help with that. For now I just threw number at it. Using 20' as a maximum difference between best and worse case scenarios this is how the system plots: So even with a difference of 20' the flow volume only changes (increases) by a about 1.5 gpm. Looking at the unit performance data this small change in flow has little impact on the heating efficiency (COP) of the unit. The same also hold true in the cooling mode.: Efficiency does increase with flow rate but very little. That is a whole different topic comparing pump watts to unit watts. But common sense would tell me getting the required water through the unit as cheaply as possible is the primary goal here. When I run my pressure calculation at the inlet of the unit using the higher flow rate it increases by about 1 psi. And that is only because of the higher downstream friction loss associated with the higher flow rate. So to summarize things at this point; if all my calculations and assumptions are correct (that is a still a big if) then I don't see any reason why this system wouldn't work as is without the addition of a bladder tank or flow controls. But as we all know, going from design on paper to real world operation can lead to severe issues if we miss something. That being said I will be incorporating a flow control of some type. First thoughts are a VFD on the pump would be ideal to dial things in. But that is additional cost and complexity. A Hays balancing valve might be a nice, inexpensive choice. Or I have come across cycle stop valves in my research which sound very interesting. Have to do more research on these different methods of flow control to choose the best one. A small bladder tank is probably a wise addition to keep pressure from bleeding down during the off cycle which helps minimize mineral deposits inside the piping and and heat exchanger. I still need to educate myself on selecting the correct size and air pressure setting. There is no reason to add additional head to be pump that is not needed.
Looks I am taking my wife to hospital today. Probably be there for a few days and will have to time to ponder on all of this as I stay with her. It will help pass the time.
Good news. Luck is on our side and so far the wife hasn't had to go into the hospital. Now that I am wrapping my head around all of this I see why this equation doesn't work: They have the units wrong on the specific weight: pounds per cubic feet. When I convert it to cubic inches it works. I should have caught that sooner. I feel like I have beat up on this problem pretty good now. Enough that I have complete confidence in designing a solid open well system that will work without issue. There are a few lingering questions but those will sort themselves out as I move forward with things. It really has been a fun journey of learning for me and I hope haven't embarrassed myself too much. Stay safe everyone!
I thought this forum was taken down. Clicked my bookmark for it just for kicks and was surprised to find it wasn’t. I can’t contribute anything to your posts, but good luck with your project.
