Ground Source Heat Pump

Don't have any theoretical way but, What do you think about making a small tank with same material but with dimensions 1foot= 1mm and doing it empirically?

If the scheme goes ahead, bang goes the opportunity of using the structure as a well; it becomes a sump. The reason is that increasing its temperature causes all sorts of biological activity to increase, rendering the water practically useless for drinking.

The well will not be used for potable water supply,only for occasional lawn watering.

OK!

I doubt it is a good storage of heat. Concrete and gravel is not good conductor, but they are neither good insulator. Because of the surface is so large, you will have very high loss, hardly a storage for heat.

A closed tank surrounded on six sides by air gap is a better choice. Aslo, make the internal surface very shiny. it is a themo flask.

Ground source heat pumps extract heat from the ground because it is much more efficient than extracting it from air.The ground temperature at below 6 feet is around 55 F year round.I do not expect it to act as a thermos for storing and holding heat,but merely to act as a heat sink for summer use, and a heat source for winter use.The input heat will be absorbed by the surrounding earth.I just need to know the amount of heat the surrounding earth can absorb; basically the heat throughput of the well.

I have read of heat pumps using 2 or more wells, with a coil of pipe embedded in Bentonite, with water circulating thru them and through a isolation heat exchanger for the freon-to water heat transfer.

When calculating thermal conductivity and thermal mass of materials in series, is it a simple additive function?;Water+concrete+gravel+clay?

The ground temperature remaining a constant 55° F is a myth that people maintain. You can find the approximate temperature for your location on the net.

Calculating the ability of the well to disperse/collect the heat is quite complex for what I have seen.

Really the constant high (relatively) ground temperature (in the winter) and cool (in the summer) provide the savings.

The University of Oklahoma and the Oregon Technical Institute have lots of information available on the web.

Unless you live in a cold climate where an air source heat pump is not cost effective you will never manage to pay off the difference in costs - GSHP systems are expensive.

The 55F number I quoted is the AVERAGE temperature for my location.I have a good grasp of heat pump theory and hands on experience, and I intend to install this unit myself,so costs will be a lot lower than otherwise.The thermal conductivity of the various materials complicated the calculation.Thermal mass calculation is no problem, rather how fast the heat can travel into the surrounding earth is the crux of the problem.There are several materials in series between the water and the wet clay,which I consider to be an infinite heat source,for all practical purposes of the calculation, and I realize the absorbsion of heat will decrease as the differential temperature between the elements decrease.

My real question is how to calculate the thermal transfer of the media between the water and the clay.Do I simply take the lowest value as the limiting factor, or are there more factors involved?

The concrete tile has holes allowing water to pass thru to the gravel layer surrounding it,so the gravel is totally flooded at all times, and the water acts as a heat sink to average the temperature of the gravel, which has a higher thermal mass than the water.The flooded gravel is in intimate contact with the clay,being coupled by the water surrounding it, and it is also coupled with the concrete tile in the same manner.

I intend to install the exchanger coils in a spiral, to create a convection flow from the bottom to the top of the well, with no 2 coils directly over another.This should tend to mix the water and create a relative motion between the sides and center, enhancing heat transfer,and preventing stagnant thermal boundaries(thermoclines).

Any problems with this plan if the calculations of heat transfer are worked out?

I think because the well has large surface area and a large amount of water (hugh heat capacity), for average home, I do not think , you need to worry about the well not able to cope. Not only that, it is tied to big earth whiich is a big reservoir .

The make up of the concrete is unknown or the type of gravel. Heat conduction rate for those are an unknown with out their make up. Either way the thermal conduction of those materials are higher then water. So base it on the waters conduction rate. There are other factors that would make difference as is the ground water moving.

Also by your dimensions the water volume is 16% larger then 750 gals. That would make big difference in your calculations.

Hi

I to have a well some 4 feet in diameter but with only 12feet depth of water and I also keep thinking abut whether or not it would be a good source for heat energy so I am interested in any replies. I have no thermal lining but a red sand stone wall some 12" thick.

Many thanks

Oliver Dunthorne

I believe the sink and storage capacity of your well would not depend on the materials surrounding your well but in the amount of bleed you are going to use to keep the well at ground water temperature. As long as the well has good flow, water can be removed or bled off to allow new water into the well to maintain groundwater temperature. If you don't bleed the well the standing water column will gradually heat or cool depending on whether you are using the well for heating or cooling. Here is some information for standing wells in use today.

http://www.hvac.okstate.edu/research/Documents/Orio_Johnson_Rees_Chiasson_Deng_Spitler_05.pdf

Thank you for the attached link it makes interesting reading

Oliver Dunthorne

Thanks for the link.All of the standing column wells I saw were 6inch diameter, and they rely on through put of water to move heat into and out of the well, because of the relatively small surface area of the well bore.

If I add the thickness of the gravel to the diameter of the tile, I get around 4 ft. diameter, which equals about 465 square feet of radiator area at 37 feet of depth of water.

This is a very large heat sink area in contact with clay.Water is a much better conductor of heat than air,and so is wet clay, so for comparison, imagine an air heat exchanger that is 10 feet high by 46 feet long, and it can shed a lot of heat.

Using 1300 feet of 1 inch UHMW tubing in this well,I am thinking the well could handle much more.

Perhaps I am wrong, but a seat- of- the-pants calculation looks feasible to me.