woodgeek wrote: "Thanks for the explanation Dick, if I understand, you avoid lifting power losses because your system works like a siphon....when 'off' the loop is an inverted U, with both ends under the ground water level. I gather this only works because the HP is less than 40' above the ground water level at all times--the point where cavitation would start. Is that right?
That said, have you computed/estimated your COP with and without the (well) pump power included??"
I like the "inverted U siphon" analogy, and that makes it easy to understand. Someone can imagine a 75 foot hose submerged in a pond and full of water. You could grab the middle of the hose and hoist the progressively heavier weight skyward, and water would stay in the hose for perhaps 30 ft or so. Atmospheric pressure pushes up at 14.696 psig (nominal sea level), while the pressure at the top of the loop is the vapor pressure of water at whatever temperature it's at. For well water at 50 F, that VP is 0.178 psi, so the net pressure difference to support a column of water is 14.518 psi. The density of water at 50 F is 62.38 lb/cuft. Dividing by 144 gives 0.433 lb of weight on a square inch for each foot of height. Dividing 14.518 by that gives a theoretical maximum height of 33.5 ft. You'd get something less than that, due to the dissolved air that would start to come out of solution. Pulling the center of the hose higher than the maximum would result in vaporization of water to fill the space above the 33.5 ft level. That vapor actually is steam, but at 50 F and 0.178 psi absolute, or 14.518 of vacuum relative to atmospheric pressure. However, as vaporization occurs, the heat required (around 1000 BTU/lb) comes from the water itself (and a negligible amount from the hose wall), resulting in autorefrigeration, and the temperature would drop. Theoretically, if you lifted high enough (and slow enough so as to maintain equilibrium, and with an insulated hose), the water at the top level would drop to the triple point of water, 32 F and 0.088 psi absolute, at which you have in equilibrium liquid water, vapor water (steam), and ice. Don't worry; you'll never see this happen in your GSHP water loop.
Trying to calculate an actual COP requires knowing actual power to the heat pump compressor and electronics, to the well pump, and to the blower motor. It's my understanding that the mfg tables for COP vs water flow and temperature and air flow and temperature do not include the blower motor, since that is a function of ductwork. If I assume the power and COP from the tables for my water and air info, I need pump and blower power numbers. I don't have any instrumentation on those. The blower is ECM, and the pump is part of the Franklin Subdrive 75 package. For just the HP, the COP is supposed to be around 4.8. I took a stab at estimating the pump power, assuming a pump efficiency typical of what a small pump gives, and I ballparked the overall COP at around 4, but this doesn't include blower power. With proper instrumentation to get total power, and assuming the small rotameter showing water flow is accurate, I could get overall COP from the entering and leaving water temperatures. For the blower power, I'd correct the COP value ignoring it by adding those equivalent BTUs to both the delivered heat (numerator) and power used, since the unit is within conditioned space and that's where the blower power goes. The pump power is assumed lost to the outside surroundings. I suspect I'd have trouble getting a better COP by trying to measure air flow in/out and the temperature rise. The air flow across the rectangular duct won't be uniform.
