BTU Dissipation in 24 Inch Well

Are you talking about using the water source for a heat pump? In what way? If you're thinking about using it for cooling a condenser in a water source heat pump you will need some way to cool the water, such as a fan/coil or evaporative cooling, it won't work for long without it, and to work properly you need a steady temperature water supply to set the charge properly...

..."As a general rule of thumb, you can assume a minimum flow rate of about 2.5 to 3 GPM is required for every 12,000 BTU/hour (1 ton) of heating and cooling using a water source heat pump (though some units specify flows as low as 1.5 GPM/ton for open-loop systems and 3 GPM/ton for closed-loop systems)."...Feb 19, 2018

Here is some good information to start with....

https://waterwelljournal.com/geothermal-system-design-2/

With a 2' diameter and 37' depth that would be 116 ft³= 867 US gal

This is a large well and I am figuring on the inside surface dissipating a lot of heat.It has a large surface area that is surrounded by wet gravel and clay,which will also act as a heat sink/source..This is where the complications set in,the specific heat of concrete,wet gravel and water in series with each other,with the water filling the gaps in the gravel.With 6 inches of gravel all around the tiles,and the surrounding earth made of wet clay.There would be more things at work here than just the water thermal capacity itself.There is a large heat heat sink area available,I just don't know how much it can dissipate.I realize it is a sliding scale as the delta T changes.

This is generally done with two wells, one for supply and one for discharge...assuming the wells can provide enough water flow...at 3 gpm per ton for a 4 ton system that's 12 gpm...generally you want a couple hundred feet between wells...

Typically you get a 10-12°F temperature differential from inlet to discharge water...

I have seen the bore holes,about 8 inch diameter with a U Tube in them and they are filled with mud grout above the water level.

They do not have much surface area to dissipate heat.

My well has over 200 square ft of surface area under water to act as a radiator.

So basically,I need to determine how many BTU's the well casing can dissipate into the earth,without any flow,just using the volume and area of the well and casing.

I would suspect it would be more than an air heat exchanger of the same area?

I have a second well if required,but I don't want to use it if I don't have to.

I could probably even make it siphon from one well to the other due to height difference.I would inject the return water into the bottom of the well to aid in circulation and prevent stratification of the water.

This would be an open system,with the water being put back into the same well.

The well has a pass through flow rate of about 2 gallons/min.when the level is stable at the 37 feet depth level.The total well depth is 51 feet.

How confident are you about the flow rate of your well? How did you determine this rate?

The well flow rate will be the biggest factor in the amount of heat you can transfer. Water and thus clay and wet gravel have large specific heat but generally less than stellar flow and generally poor conductivity. Without flow, convection doesn't occur (outside casing walls). Radiation is not significant with low dT.

You should reevaluate your plan on pumping from the top and returning to the bottom of the well. Why would you want to mix mess up your usable cool temps? Heat transfer is driven by a difference in temperature. Differences in temperature are things to be utilized on thrown away with mixing.

Really you should pump from low in your well and return to a drain field. That way you can sink heat to much more earth before the water returns to your well, thus keeping your low heat sink temperatures.

Because of the low delta T you won't see much benefit trying to cool your inside air directly. Perhaps consider further cooling your refrigerant after the condenser in your current hvac system. Alternately, it could be a good heat sink for an indirect adsorption chiller or dehumidifier.

It would depend on whether there was a drive to boil the water in the well.

If so, then it would be quite a spectacular show for the neighbours...

Heat pump technology has dramatically improved in the past few years. They now have outside air heat sink heat pumps for heating and cooling that are rated to handle winter conditions here on Long Island NY. I'm certain the heat capacity of your well is much greater than air. However, everything depends on the type and quality of the heat pump you will be using. Consult with the heat pump vendor what their requirements for the heat pump and relevant local zoning codes.

A few engineering concerns come to my mind that may preclude using this existing well with a certain heat pump:

• Is the well inner diameter large enough for the two pipes and U-turn the heat pump needs?
• Is the aquifer this well reaches a potable water layer for you or others?
• What is the frost level for your location?

Once a suitable heat pump for your heating and cooling loads has been identified then you should be on your way to seeing if this well is suitable.

My intention is to use the well water as a source of heating/cooling.I will pump water from the well,through a fin-type radiator and return it to the well.

The water will be a heat sink as well as a radiator.

The diameter of the well is 24 inches.The water depth is 37 feet,approximately 850 gallons of water.

The well casing will provide over 200 square feet of surface area to dissipate or collect heat.That is a lot of surface area,and water is more efficient than air.

My main concern is how much heat can I get rid of into the surrounding earth from this much area.I may not need a water flow into/out of the well if I can sink/source enough with the well casing.

If I went with a closed loop system,I would put a slinky-type tubing coil in the well for maximum effect,and perhaps a bubble tube to prevent stratification of the hot/cold water.

I guess I'm not getting this, maybe if you could draw a diagram we might get someplace....

Can you mark temperatures expected ....I still think you need active circulation in the water source....but it depends on the amount of gpm flow and temperature differential...

That's a "ground-source heat pump", removing heat from the well and pumping it round the house. Many homes in Switzerland do this, for example.

What the original post is after is the reverse: dissipate heat in the well, if read correctly.

So basically you are talking about pulling the water up from the well and feeding it through a heat exchanger before sending it back down the well. Which leads me to ask why not store the water in a large tank after it has done the cooling.

You have done the work and used the energy pumping the water up, why not use it for other purposes or if you have to return the water, store it in the tank and drain the bottom water off back down to the well. Maybe use the returned water in an evaporative spray cooling heat exchanger to cool the well water even more before use.

Of course I looked up you Mercator and saw where you lived but have no real idea of your temperature extremes but we are having 44c days and in winter we get down to below zero at night tempered by 20c days while the bore water here seems to be a constant 15c in summer and 10c in winter though we are pulling it from 90m.

Just my 2.2c with VAT/GST applied

Quite. Once the water is on the surface one might as well do something useful with it, especially if the area classes as being in "water stress".

"Of course I looked up you Mercator and saw where you lived"

I don't believe that HiTek really lives in the Nikola Tesla memorial centre in Croatia.

What I am considering is much simpler,although less efficient than a conventional ground source heat pump with a condensable-gas heat exchanger.

Cooling is the most expensive part in my area,heating is relatively less expensive by comparison.

I am considering pumping the water from the well into a heat exchanger/radiator,located in the existing air return of the heat pump.

The water will be returned to the well,but not confined to a coil or exchanger within the well,it will simply be returned to the bottom of the well.The hot water will rise to the top,causing a natural circulation.The bottom return pipe will be directed parallel and at a slight upward angle to the bottom tile to encourage circulation.A tee halfway down the return will be directed in the opposite direction to encourage mixing of the water.

Pumping costs will be minimal because it will be returned to a lower height than it's source in the well.The round trip distance of the water will be approximately 150 feet,at 10 gpm.With 1 1/4 " pvc,I "guesstimate" about a 2 to 3 psi pressure loss due to internal pipe friction and elbows,and the fin type coil I will be using.

The intake pipe will be near the center near top half of the well.

With a thorough mixing of the water,an approximately even temperature could be achieved,utilizing as much of the internal tile area as possible.

The average water temperature around here is 12.7C,much cooler than the ambient in summer,and much warmer than the ambient in winter.

A Freon based system could recover the heat from the water in the winter,but that is not my objective.

I do not want to incur the expense of an evaporative cooler with the attendant chemical and maintenance costs; Any solids will concentrate, the water will become more acidic, requiring frequent flushing,and exposure to light will allow algae to accumulate on the evap. coils.

The way I am proposing requires no breaking of the existing sealed Freon system,no danger of a cooling fluid leaking into the ground water and no evaporative loss of water.

The pipe will be buried deep and insulated and heat traced where required and drained in the winter.

The system will have all winter to disperse any accumulated heat from the summer.

It will simply cool the return air going into the existing system.

A thermostat will control the volume of the water,and hence the cooling, via bypass valve back to the well.I don't want to get the return air too cool and cause icing of the evap coil in the heat pump.

I will may have to adjust the Freon volume to get the desired super-heat and sub cooling.

The automatic expansion valve should handle some of the variation of the heat load.

I will incorporate a drain to handle condensation from my additional coil.

Normally,you want a 25F degree drop in temperature from intake to exhaust on a very good heat pump system.If I can achieve a 10-15F degree pre-cooling it will be fine.

That is what I am considering,but the plans are not chiseled in stone,and I have not begun work on it yet.

The ability of the well to get rid of the heat in the well is an unknown at this time,but my gut feeling is it will be more than adequate.

I am still open for suggestions on this matter.

Thanks to everyone for their valuable feedback and ideas!

The bacteriological implications of the concept need to be addressed.

Groundwaters are relatively sterile, as there is nothing much in the way of biology at depth to make them otherwise.

Injecting warm water from a non-disinfected process into deep strata seems to be asking for trouble, with both the process equipment in question and also the neighbours', which process might be abstraction for drinking; there are health issues looming.

Which is sort-of-why it isn't really done.

This won't be worth the effort or expense, it will only have a minimal effect (maybe 1-2°) for a very short time...

You are right..not worth the trouble.

Thanks to everyone!

You are working with 7,000 lbs of water, and a nearly constant temperature of 50 degrees F outside of your well casing. I think your original idea to calculate the insulation resistance of the wet casing still bears looking into. I’d assume the constant 50F temp outside the casing, based on the low heat loads you’re looking at and your well flow capacity of several GPM. For cyclical cooling, peaking during the day, I’d be surprised if you couldn’t run a 2000 square foot house through the summer months for 80% of the time, with no supplemental refrigeration at all.

Perhaps coming up with a target heat load will put the required capacity in focus. A man-cooler with a constant 55F to 60F air flow sounds pretty good to me. 900 lb/hr water flow x 10DegF rise is 9,000 BTU/h.

A regular air conditioner has an evaporator coil temperature below freezing, and with an indoor temperature of 80°F might hope to get a 16°F drop across the coil...that would be a supply air temperature of 64°...With a 50^°F evaporator coil you would not have enough cooling to overcome the heat load...so you would just be spinning your wheels....Likely that you would cool the air slightly and raise the relative humidity making the space uncomfortable and clammy, at least until the well water heated up enough to do nothing at all....

Now if you lived in a low humidity area <20%RH you could use the well water in a swamp cooler and probably get good results...or you could use the well water as a cold shower to cool down periodically....

How would you raise the humidity? This is a coil, not evaporative cooler, like a chilled water commercial system. 50F is not ideal, but not running a compressor either.

Cooling the air raises the relative humidity, the air molecules shrink....this is why air conditioning systems remove humidity as they cool the air, and why air conditioning systems need to be sized correctly, an oversized unit can cool the house down too quickly and raise the RH to an uncomfortable level...Some air conditioning systems now have humidity control built in to the air handlers which control the air flow volume making it slower in a high humidity situation and faster in a low humidity condition...Part of an air conditioning system is humidity control for comfort...

None of this makes much sense. Dropping moisture out of the air is a result of chilling the air, it will drop out on your ‘chiller’ coil. The less moisture in your air, the better it will feel.

Only if the coil is well below dew point temperature....

All,

I've been reading this for a couple days and think that SE and some others are completely missing the basics of a ground source heat pump's operation.

An air source heat pump, in cooling mode, is pumping the heat from the cooler interior of the house to the warmer exterior--this takes energy, because the temperature gradient can be 36-deg F (20-deg C). The refrigerant flows through a closed loop with the outdoor condensing coil being fairly hot.

A ground source heat pump, in cooling mode, is surrounding the "outdoor" condensing coil in a water-filled heat exchanger whose "outdoor" temperature is the temperature of the water being pumped from the well, and generally will be cooler than the air inside the house, so the temperature gradient can be negative and thus the efficiency goes way up. The water is either in a closed loop with a pipe in the well having a U-bend at the bottom and the heat transfer is through the walls of the buried pipe to the surrounding water or wet soil/gravel/clay. If not a closed loop, the somewhat warmer water is returned to the well at a different level than the one from which it was pumped. Either way, the cost of pumping the water is far less than the savings in compressor energy consumption, so there is a very good reduction in costs of operation.

This is the system HTRN has been talking about. Not an evaporative (swamp) cooler, not a system with a cooling tower, and not anything else. Questions about indoor humidity don't enter in either. His question simply was one of how much energy could be dissipated through the walls of his 24-inch diameter well that is 30 feet deep. Nobody has answered this simple question.

The heat input to the system is the heat that needs to be removed from the house during the hottest period of the summertime day. This depends on the heat flow through the house's insulation and the heat requirement for conditioning replacement air. Assuming the heat pump is designed to handle 5-tons of cooling, or 60,000 Btu per hour. This would heat approximately 60,000 lb of water 1-deg F in one hour. The well's water content is about 94 ft³ and weighs about 5,900 pounds. Thus the heat output from the house would raise the well water's temperature by about 10-deg F per hour at the hottest times of the day. I expect his home to have a much lower cooling load, such as 1.5 or 2 tons, so the well heating would be much less.

I'm not good at calculating the heat flow from the well, but I expect someone on this forum can look it and graph the well water's temperature rise over a 24-hour period of time. I strongly doubt that the water temperature will rise too much, but I invite another guru to do this.

Thanks--JMM

He doesn't want to use the water to cool the condenser of a heat pump....he wants to add a coil in the return duct and circulate water from the well as supplementary cooling/heating...

No, he says right in the original posting that this is for a heat pump. I think later on he was just politely appeasing your misunderstanding to reduce disagreements.

#8 is a reply to your post #7....

"My intention is to use the well water as a source of heating/cooling.I will pump water from the well,through a fin-type radiator and return it to the well."...

My mistake on when the fin type radiator was added to this thread. It was a clarification after I recommended contacting the heat pump vendor. Placing a secondary heat exchanger in the air stream of the heat sink part of another heat pump system might aid, even often aid but I can also see times where it might cause damage (e.g. icing in condenser coils) or make no difference at all to a climate system if not properly sized.

I indicated on one of my prior posts that I would provide a drain for any condensation on the coil.
This is intended to pre-cool the return air to the heat pump,thereby reducing the load and power consumption.
The system might run on water cooled at at certain times with the fan in manual,such as at night.
A properly designed and sized heat pump will run only about 10-15 minutes every hour.
The specific heat of water is 1,which is 1 btu.lb/degree F.
850 gallons is 7000 btu per degree F.
With a 10 degree exchange rate,that is 70,000 btu without considering the flow through of the well,and the heat absorbing capacity of the surrounding saturated clay and water outside of the tile.
So the key here is determining how much heat will be absorbed from the water by the surrounding media.
The thermal conductivity of concrete is 1.7
The thermal conductivity of saturated clay is 2.5
Water is .606
Saturated Gravel is .7
Sandstone is .58
The units are in BTU/hour X degree F X square feet of area.
The tile is concrete 2 inches thick,surrounded by 6 inches of saturated gravel,surrounded by an almost infinite amount of saturated clay(mother Earth).
The bottom of the well is sandstone,many feet thick.
This combination of materials makes it hard for me to calculate the heat exchange rate.
I understand the negative effects of pre cooling the return air too much,and reducing load below a certain point.

If the Freon liquid entering the evaporator is not almost totally vaporized, the compressor can be hit by liquid which will damage the compressor,unless it is a screw compressor.
A receiver ahead of the compressor can hold only a certain amount of liquid.
This is why I am now considering running the water portion at night,or in high demand situations.
A need for a 3 mode (PID) control seems appropriate here.
I have some left over controllers and sensors from previous jobs many years ago.
This will require me to monitor the super heat/subcooling temperatures and adjust accordingly.
The reason for returning the heat to the bottom of the well,and sourcing from the top is to encourage mixing of the water to increase the total heat exchange area,but this has an offsetting factor of reducing the differential temperature of the water vs the surrounding media,and hence the exchange rate with the surrounding media.
Hard to determine the most effective method.
Considering the complexity of all of the interactions,the best way might be to try to obtain some empirical date by constructing the system outside of the heat pump to get some real world figures to work with.
I am still in the information gathering mode,and I am still open for suggestions,and appreciate all contributions,past and future.

How long an air conditioner will run depends on heat load, temperature differential, and size of the unit....Since the heat load is variable, there is no set run time...there is a design run time that is calculated at max design ambient temperature vs interior design temperature....here in Florida the design temperatures are 95°F ambient and 75°F interior, and a 50/50 run time, 20 min on, 20 min off....

....but the heat load is variable depending on time of year, traffic in and out of the house, window types and treatments, roofing type and shade, and any number of other variables that come into play....Even the air filter can play a role, proper air balancing, return and t-stat location...

...and certainly the humidity plays a major role....the humidity should be kept above 35% and below 65%, 50% relative humidity is ideal...Below 35% RH tends to dry out the mucous membranes and makes susceptibility to colds and flu more prevalent, over 65% tends to increase fungal growth, also unhealthy...

Those are pretty high values for conductivity for concrete and clay.

.

".... The units are in BTU/hour X degree F X square feet of area. ..."

This is not a standard measure of thermal conductivity. Thermal conductivity is measured as:

BTU/(hr × deg F × sqr ft area ÷ ft thickness)

....which simplifies to BTU/ (hr °F ft)

The heat transfer through the walls will not be that significant compared to the flow of the water and the overall thermal mass of the well and first inch of well casing.

Circulating the well water and destroying the thermocline by returning the hot water to the bottom will reduce the already low delta T available between inside air and well water as well as between well water and well casing.

You are right.I am still exploring different scenarios at this time,nothing final yet.

Here is a pertinent image of a similar situation:

The first one is what am currently exploring.

Note the large diameter pipe extending deep into the well for the intake water.

Here is an example showing the calculations for a standing column GSHP.

I may end up using the water to cool the Freon to source/sink more heat from the water.

Still researching.

My pumping power requirements will be small because of the nearly balanced positive/negative head pressures.