Artesian Well as Heat source in middle of Minnesota?

Discussion in 'Surface Water Loops' started by BrianP, Oct 24, 2021.

  1. BrianP

    BrianP New Member

    My family is building a small house in central MN on a property with a 30 gallon/minute Artesian well at 47 °F.
    These calculations, if correct, show that there should be more than enough thermal availability in the well water to supply a 1.5 - 2 Ton heat pump.

    I am trying to size the heat exchanger for the Artesian well.

    Proposal for HEAT EXCHANGER:
    • 3 Ton spiral-coil Turbotec heat exchanger, 36 kBTU/hr
    • 0.833" ID outer spring water pipe.
    • 0.625" ID Copper, twisted heat pump fluid pipe inside outer pipe.
    • OR possibly a 1.125 ID outer pipe and 0.875 ID Inner pipe for less flow restriction.

    Model Number Nominal Size A B C D E F G H J K L M Unit Weight
    C-5984-01 36K 0.883 0.883 0.625 0.625 14.00 15.75 5.19 10.88 2.50 2.50 4.44 4.00 26.00 lbs.
    C-5924-01 60K 1.125 1.125 0.875 0.875 17.25 18.00 7.63 14.00 2.50 2.50 6.75 5.38 56.00 lbs.
    turbotecproducts d@t com /products/spiral-coils


    Water Temperatures on coldest Winter night with 47 °F Artesian spring water temperature.
    • 30 °F Cold pipe at house, cooled by heat pump
    • 31 °F Cold pipe at Artesian Well
    • 42 °F Hot pipe at Artesian Well, 11 °F temperature rise with 16 °F differential.
    • 41 °F Hot pipe at house; 11 °F temperature differential for heat pump to work with.
    For each Gallon/min flow rate and 1 °F temperature, there are 147W of available thermal energy.
    At 1/4 spring flow rate, 30/4= 7.5 GPM -> 1104 W per °F
    2.5 kW would need 2500/1104 = 2.26 °F (2.5 kW is calculated actual heat load, < 1 Ton)

    Given the heat exchanger Diameter "E" of 15.75, 3 full revs (from drawing), count the 4 smaller, inner revs as 2 outer -> 5, 15" revs.
    5 * PI * 15 = 235" = 19.6' heat exchanger linear length.

    The Volume of spring water in the heat exchanger would be the differences in the circular areas of the outer and inner tubes times the length. I estimate 0.311 gallons.

    At 7 gal/min -> 2.67 seconds trip time from A to B through exchanger.
    The inner tube is twisted to create turbulence and increase surface area of heat transfer.
    (5.42 seconds running the Artesian water through the inner, 0.875" with the larger heat exchanger)

    With almost 20 linear feet of contact length and 2.67 seconds contact time and 47 °F spring water contacting 31 °F antifreeze for 16 °F temperature difference, 2.26 °F final temperature rise should be quite easily done.

    I am not sure at which temperature difference Turbotech sizes its devices.
    Is this analysis reasonable?
    Last edited: Oct 25, 2021
  2. geoxne

    geoxne Active Member Forum Leader

    I'm not quite sure what you are trying to do here. Are you trying to isolate artesian well water from the heat pump with the heat exchanger? If yes, why? Does the water quality from the well require isolation from the heat pump coil?

    The point being the heat pump will run a lot more efficiently with direct flow with a source EWT (Entering Water Temperature) of 47F.

    Also, the TurboTec coil you are specifying is a fluid to refrigerant heat exchanger. The refrigerant side outer tube is steel and is not appropriate for water flow. If you need to isolate source water flow from the geo heat pump than you should use a heat exchanger appropriate for water flow (Hydronic) on both circuits source and load. A swimming pool heat exchanger would be more appropriate especially if you are dealing with poor water quality. They are more commonly available in different sizes and materials.
  3. BrianP

    BrianP New Member

    A closed loop system with distilled water and propylene glycol and some anti-corrosion additives would protect the heat exchanger in the heat pump coil as well as the copper and steel at the well.

    In the event of an extended power outage, pure water in the 80 foot pipes between the house and the Artesian well could freeze. The recommended "burial depth for water supply lines" or Frost Line in Minnesota is 80 inches.

    Turbotec has both Copper and CopperNickel inner tube options. The steel outer tube should be fine with distilled water, glycol and corrosion inhibitors (RhoGard, ThermalStar, DowFrost ?).

    Would the steel in the Turbotec react with the copper in the heat pump? The Artesian well water quality is quite good and almost drinkable.

    There would be 4 thermocouples on the inputs and the outputs for each channel. If the heat exchange efficiency goes down significantly, the heat exchanger could be treated with citric or some other acid to reduce the oxidized copper.

    With the 60 kBTU/hr model, 11+ degrees of temperature difference, ~20 linear feet and 5+ seconds of contact time, extracting 2.5 kW should be doable.

    How much heat transfer is possible per foot per degree F for 0.875 OD L type CU tubing?
    length=12", ID = 0.785", wall = 0.045" = 0.001143 m
    Q = K * Area * ΔT / thickness
    Area = π * 0.785" * 12" * (1m/39.37in)^2 = 0.01909 m^2

    Q = 401 W/m/K * 0.01909 m^2 * 1 °F * 1 K/1.8 °F / 0.001143m
    Q = 401 * 0.01909 / 1.8 / 0.001143
    Q = 3720 = 3.72 kW per linear foot per °F

    Even with a reduced K value due to the cold work of twisting, there looks to be vastly more thermal power than needed.

    I have read horror stories about open loop systems and see great advantages in a closed loop.

    How much temperature difference does the Heat Pump produce in the working fluid? I have seen measurements of 7 °F online.

    2% nickel would cut conductivity in half.
  4. geoxne

    geoxne Active Member Forum Leader

    I stand by my assertion that the TurboTec coax coil is not appropriate for your application. Sizing a hydronic heat exchanger requires calculating the fluid flow rate and head on both source and load circuits. You will not find any of published data for fluid flow on a circuit designed for refrigerant. Again you would be better off with a pool heat exchanger. Plus calculators are commonly available to size them with your design input ie. source and load GPM, EWT, BTU's required and resulting Delta T's to size for the proper heat transfer required.
    Quite frankly, I seriously doubt you will be able to pump 6GPM through the refrigerant side of that Turbotec coil.

    At a flow rate of 3gpm/ton you would calculate a 8F DeltaT.
    BTU's=GPMs x DeltaT x 500

    At that rate, in reality, the DeltaT in heating would be around 6F and DeltaT in cooling would be around 10F because the heat of extraction is less while heating than the heat of rejection is while cooling given the nominal capacity of the heat pump. You can apply the actual HE and HR from manufacturers specification tables for a specific heat pump under your conditions.
  5. SShaw

    SShaw Active Member Forum Leader

    As geoxne wrote, that's not the type of HX normally used to isolate a GSHP from ground water. Normally, a plate heat exchanger would be used.

    Why is a secondary heat exchanger necessary?

    Have your well water tested. If the chemistry is suitable, just install a conventional GSHP with a cupronickel heat exchanger. You would need to pump well water through the heat pump at about 1.5 GPM/ton of capacity.

    Even if you use a secondary HX, you'd still have an "open loop" system. If you want a closed loop system, then install vertical or horizontal ground loops.

    It it were my house I wouldn't want open loop, and I wouldn't want to waste all that well water. Where are you going to dump it?
  6. geoxne

    geoxne Active Member Forum Leader

    I respectfully disagree. A flatplate HX has micro channels that can easily foul with open loop water flow. A flateplate HX can be destroyed quickly in freezing conditions. It doesn't take much for ice to distort the plates enough to break seal between primary and secondary. Again, a tube and shell pool HX is more appropriate.
    Manufacturers recommend 2gpm/ton with EWT below 50F.
    If you want to introduce a HX with antifreeze to avoid trenching below the frost line, please consider the following.
    -With a primary source EWT of 47F, even with a close approach HX (really large) you should NOT expect your secondary EWT to be any more than 42F. It takes a DeltaT to drive the HX process.
    -You don't know how much heat you might loose on the 160 foot of pipe run shallow buried in frozen ground. Your efforts should be concentrated on providing the highest possible EWT to the Heat Pump. A 10F increase in EWT will net you an approximate increase of 1/2 of a COP in efficiency.
    -Even though you propose antifreeze in the secondary loop from the heat pump the, HX has the potential to freeze the open loop source side when your antifreeze loop drops below freezing.
  7. SShaw

    SShaw Active Member Forum Leader

    I don't mean to say which is better. It's just that every reference I've seen to an open loop system with a secondary HX, including by northern installers on this site, has used plate heat exchangers. They have also been installed inside the building for freeze protection. I don't know the technical reasons for this. Could be compact size, better HX with similar temperature fluids, ease of maintenance, etc. Maybe you've got direct experience that contradicts this practice.

    Good point. 2 GPM/ton would certainly be safer with 47 degree EWT.
  8. BrianP

    BrianP New Member

    First of all, I calculate that there is 2-4 times the thermal availability than needed.
    I want to have a closed loop system to keep the heat pump pristine and to prevent freezing.

    I would like to be able to unscrew 2 (4?) clamps, remove the COPPER heat exchanger, remove oxidation/gunk with acetic/citric/hydrochloric/??? acid and a mechanical abrasion and be good to go for another decade.

    A small heat exchanger submerged in a pool by the creek with a waterfall out the back side and perhaps a gazebo around it it is a possibility.
    The same place it is dumping now, the creek.
    This property, two adjacent properties and a neighbor across the river all have constantly flowing Artesian wells.
  9. BrianP

    BrianP New Member

    How do you calculate how much length you need to get to a finite temperature delta?
    If the temperature of the antifreeze is 32°F and the water is 47°F and flow rates are equal, a very, very long HX would approach (32+47)/2 = 39.5°F
    But, it depends on the heat capacity of the antifreeze mixture; 50/50 mix = 0.81485 BTU/(lb °F) vs. 1.0038 for water.

    Knowing the internal pipe volume, you can calculate the transit time through the pipe.

    For some reason, I am thinking of 2 revolutions of 3 foot diameter, 1.25 Copper tube twisted inside 1.5" HDPE ->
    2 * 3.14 * 3 = 18.8 feet.

    How would one calculate the exit temperatures on the spring water and heat pump antifreeze flows?

    Thermal design between the House and the Spring:
    • 5 foot deep trench, 1-2 feet wide.
    • 8" between hot and cold pipes.
    • 1.5" PE 4710, DR 9.0, 1.453" ID, 255 PSI HDPE, ~80 feet in each direction.
    • 4" thermal "Durox" type "AirCrete" bricks with 4" pipe to brick gap, K=0.11 W/mK.
    • Fill with Pumice concrete with 4X the R value of "sand and gravel" concrete.
    I would expect (hope?) to see not more than 2°F maximum heat loss between the house and the spring in January.
    Is there a better way to insulate the pipes and keep the ground water off of them?

    [1] ASTM B-88
  10. BrianP

    BrianP New Member

    I am a mechanical engineer, not an HVAC expert.
    I was not sure where the 500 figure came from so I calculated it:

    Power = heat capacity * mass flow rate * DeltaT
    = 3 gal/min * 3.78 kg/gal * 8 °F * 1°C/1.8°F * 4184 J/(K °C) * 1min/60sec * 1W/(J · Sec^-1) * 3.41214 BTU·hr^-1/W
    = 11,920 BTU/hr ( == 3.515 kW)

    I calculated 496.7. Just using a 500 conversion factor is a lot easier!
  11. BrianP

    BrianP New Member

    What is the suitability of using pure, annealed Copper of type K or L with Artesian Spring water with a pH of 7.8 and "alkaline" = 240?
    I would like to explore the use of pure copper instead of CopperNickel or Titanium which are commonly used in swimming pools and corrosive fluids like salts or Chlorine.

    90% Copper w/ 10% Nickel reduces the conductivity by 88%, 401 -> 50 W/m·K. Even 5% Ni reduces the conductivity from 400 to 100.

    A twisted Copper tube inside another tube like the Turbotech design (above) is quite appealing for the efficiency and compactness.

    The 2 questions:
    • How many how many years between cleanings (loss of 20% conductivity) using spring water?
    • How many years between replacements due to corrosion, pressure loss (K->0.072" wall, L->0.060 wall)?
    Any recommendations on Vendors?
  12. geoxne

    geoxne Active Member Forum Leader

    Wow! Good on your calculation. Now I know where the 500 comes from.

    Universal Hydronic Formula
    BTUs/hr = GPM x DeltaT x 500 for pure water (or 485 for typical antifreeze solutions)
  13. SShaw

    SShaw Active Member Forum Leader

    The calculation is almost exactly 500 and is very easy to understand if you look at it this simple way...

    A BTU is defined as the amount of heat required to raise the temperature of one pound of water by one degree Fahrenheit. At one GPM, you are pumping 500 pounds of water per hour, so the BTU/hr exchanged is 500 times the delta-t.

    The math:

    BTU/Hr = Heat Capacity * Mass Flow Rate * delta-t

    BTU/Hr = (1.001 BTU/lb/F) * (8.33 lb/gal * X gal/min * 60 min/hr) * delta-t F

    BTU/Hr = 500.3 BTU/hr * X * delta-t

    where X is flow rate in GPM and delta-t is degrees F.
  14. BrianP

    BrianP New Member

    I figured out where the difference in our estimations were.
    Looks like both the Specific heat and the density:
    11,920/12,000 * 500 = 496.666667 => 496.7 original estimate

    Convert water's Specific Heat Capacity from SI to English:
    4184 J/(kg · K) * 1BTU / 1055.06J * 1K/1.8°F * 1kg/2.204623lb = 0.999327 BTU/(lb · °F)

    0.999327 BTU/(lb · °F) * (8.33 lb/gal * X gal/min * 60 min/hr) * delta-t F

    BTU/Hr = 499.46 gal/min * delta-t @ 69.5 °F

    The constants are different:
    1.001 BTU/lb °F = 4190.99 joule/kg °C
    4184 joule/kg °C = 999.33123 BTU/lb °F

    And the water density varies with temperature:
    temp g/cm^3 lb/ft^3
    40°F/4.4°C 0.99999 62.423
    50°F/10°C 0.99975 62.408

    Find water density at 47°F:
    xyztl2m 40 0.99999 50 0.99975 47 -> 0.999822 g/cm^3 @ 47 °F

    0.999822 g/cm^3 * 1kg/1000g * 3785.4 cc/gal * 2.204623 lb/kg = 8.34388308 lb/gal @ 47 °F ( = 3.7847262 kg / gal)

    0.999327 BTU/(lb · °F) * 8.34388308 lb/gal * 60 min/hr = 500.296 << DEAD NUTS @ 47 °F!
  15. SShaw

    SShaw Active Member Forum Leader

    The two physical constants I used (1.001 BTU/lb/F and 8.33 lbs/gallon) are based on 60 degree F water.

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