Closed loop pump and dump

Discussion in 'Hybrid Systems' started by Argofanatic, Nov 1, 2010.

  1. Argofanatic

    Argofanatic New Member

    Hello everyone, nice forum you folks have here. I'm looking to share some information as well as learn from everybody here.

    I am getting ready to put the finishing touches on my DIY geothermal installation.

    House location: Just north of Ottawa, Ontario.
    House size: 3400 sq. ft.
    Heat loss per Manual J: 90,000 BTU
    Heat gain per Manual J: 22,000 BTU

    Basement has 2300 sq. ft of in-floor radiant.
    Kitchen/Entry area has 750 sq. ft in ceiling radiant using aluminum plates.
    3 Hydronic fan coils supply the rest of the rooms in the house.
    All ducting over-sized to supply large volumes of low temperature air.

    This is a water to water system.

    After shopping around for an appropriate heat pump I think I will probably go with a 6 Ton Geostar (Made by WaterFurnace) GSW075R.

    My house is built next to a large deep lake.
    I was going to drop some piping in the lake but found out that getting a permit to do this is a complicated and next to impossible feat in Quebec.

    Here is the solution I came up with.
    If I can't put the pipes in the lake, I'll bring the lake to the pipes !

    I know I could have just done a simple pump and dump, but closed loop seems more reliable and less problematic.

    I buried two 1200 gallon concrete holding tanks.
    Inside the tanks I have placed 300 feet of 1" copper tubing.
    I drilled two 4" holes in the tanks. 1 to drain the tank, and 1 to feed the tank and supply the copper coil. 4" PVC made a nice conduit for 3 x 1" plastic pipe.

    My drain hole is at roughly 3/4 of the tank, so my capacity in effect becomes 900 gallons.

    The bottom lake water temperature in the dead of winter is 40 deg. F

    During the coldest days, the geo unit is capable of roughly 67,000BTU with a COP of around 3 with a 40 deg EWT.

    I figure each tank needs to supply roughly 23k BTU for every hour of run-time with a temperature drop of 1.8 degrees per tank. So, I need to replenish the tanks with 900 gallons every hour so I don't freeze them up solid if running full time.

    Here's where I need a bit of help.

    I have 150 ft of head from the bottom of the lake to the tank(s). Since I don't have a need for pressure, can I run the submersibles without a pressure tank ?

    Will the (Myers Predator or Rustler) 3/4hp 20GPM pumps survive 24/7 pumping during the coldest days ?

    Here are a couple of pictures of my current setup.
    Marc.
     

    Attached Files:

    Last edited: Jan 4, 2011
  2. Argofanatic

    Argofanatic New Member

    In business !

    In case anyone is interested, I'm updating this thread.

    I finally finished plumbing everything together and now have an operational geothermal system.

    Whoohoo.

    I purchased and installed a WaterFurnace Envision 6 Ton Water-Water unit.

    The lake pumps are operational while the geo unit is active.
    Lake water temperature is currently 42F and the temperature in my tanks is a constant 42F, I have two probes in each of the tanks, one in the bottom and one just below the surface, every sensor returns 42F. It's currently unusually warm for our area, the outdoor temperature is a balmy 31F.

    I'm running the system in outdoor reset with Tekmar controls. So far the unit appears to be running for about 1 hour raising the 140 Gallons of storage tanks to about 110 degrees. My primary/secondary loop then runs for about another hour on the stored energy (geo unit off).

    Here are a couple of numbers.
    Source EWT = 42F
    Source LWT = 27F
    Delta T = 15F degrees

    Isn't that a little on the high side ???

    After nearly an hour of running
    Load EWT = 100F
    Load LWT = 108F
    Delta T = 8F degrees

    Isn't that a little on the low side ???

    I just tried a little experiment and isolated the geo/buffer tanks from the primary loop and measured the following ...

    Starting tank temperature = 89F
    Geo run time = 15 minutes
    Ending tank temperature = 98F

    A quick calculation to determine BTU output gives me the following ...
    143 Gallons of water
    Delta T of 9 degrees (98-89)
    BTU = 143*8.33*9 = 10.7kBTU

    If I now multiply this by 4 (4x15 minutes) I end up with roughly 43kBTU

    Does this make any sense ?
    The unit should be able to produce roughly 75kBTU with the current operating parameters.

    Any insight would be appreciated.
    Cheers.
     
  3. urthbuoy

    urthbuoy Well-Known Member Industry Professional Forum Leader

    It's heat of extraction.

    You need to run the numbers on the ewt/lwt source side differential.
     
  4. Palace GeoThermal

    Palace GeoThermal Well-Known Member Industry Professional Forum Leader

    Yes it is on the high side, can you up the flow?

    BTW, congrats on getting your system going. Are the water bosses OK with you pumping water out of the lake and back?
     
  5. Argofanatic

    Argofanatic New Member

    Thanks for replies guys.

    urthbuoy, I don't quite follow what you mean. I understand the HE side of things but don't quite get how I can work backwards on the source side differential. Any clarification would be appreciated.

    palacegeo, I figured I would need to bump up the flow, I have two UPS26-99 in parallel (one per tank) but I guess the flow is just marginal, my calcs showed adequate for the loop lengths, but the coax drop might have just put me over the edge. The only way I could up the flow would be add a third UPS25-99 just ahead of my parallels to boost the overall flow.
     
  6. urthbuoy

    urthbuoy Well-Known Member Industry Professional Forum Leader

    As i type on a mobile device, I'm usually a little too brief.

    You need your source side flows from pressure drops and heat pump charts. Your source side temperature differential. And your number 500 for pure water.

    That's what you put in to calculate HE.

    I think you have all except source side flows.
     
  7. Palace GeoThermal

    Palace GeoThermal Well-Known Member Industry Professional Forum Leader

    As long as you have antifreeze in your closed loop, there is no real danger.

    You would probably see some increase in efficiency with more flow, not sure how much.
     
  8. Argofanatic

    Argofanatic New Member

    Ah, a true addict :)

    My flows are a bit of a mystery right now. That's why I was trying to estimate my BTU output over a 15 minute period. I installed tri-gauges which give me pretty useless delta - p's (100 psi in a 1" span). Buying some nice 4" 60 PSI gauges on fleebay to remedy that situation.

    Does a higher delta on the source side cause me a performance penalty ? Intuitively, I would think that a higher delta would be a good thing.
     
  9. Argofanatic

    Argofanatic New Member

    palacegeo, thanks for the reply once again. I was typing my reply while you were posting.

    I wasn't going to put any anti-freeze in, but my vendor "suggested" it would be wise.
    Glad I followed his advice and added 20% methanol.
     
  10. Argofanatic

    Argofanatic New Member

    Technically I'm not putting the water back.
    I have 150' of drainage pipe over a gravel bed. Eventually the water will make it back to the lake, indirectly.
     
  11. AMI Contracting

    AMI Contracting A nice Van Morrison song Industry Professional Forum Leader

    Yes I confess we may have an open loop system or two that opts for "ground dispersal" vs the improper discharge of water into a governed body.

    :eek: I hope none of it accidently spills into Lake Erie :eek:
    joe
     
  12. Argofanatic

    Argofanatic New Member

    I could use a little more help ...

    My system has been running for several days now.

    Yesterday I ran the unit continuously for about 18 hrs raising the temperature of my 2300 sq. ft. 5" slab, my source EWT and LWT were rock steady: EWT = 42 and LWT = 27, they never fluctuate.

    Looking at the WF Envision performance tables (NSW075), source delta T's of around 15 are achieved with flow rates of 10GPM. So if I use 10GPM I calculate a HE of roughly 72MBTU therefore my load HC should be even higher.

    I ran another test yesterday to determine my HC and this was the result of my experiment:

    With all heat emitter pumps turned off, the buffer tank starting temperature was 86F, after 18 minutes the temperature was 97F (delta 11F).

    My load delta-t is a consistent 8F.

    Tank capacity: roughly 143 Gallons. Energy supplied = 8.33x143x11 = 13.1MBTU over 18 minutes gives me roughly 43.7MBTU/h where is all my extracted heat going ???
    I appear to be getting only half of the heat I'm expecting.

    The only potential issue I can see is my load pump setup.

    I'm using a Taco 011 pump pumping into the geo unit, the buffer tanks a piped in parallel reverse-return using 1-1/4" pipe, pressure drop "should" be minimal so I would expect the flow to be pretty close to the recommended 19GPM (once again, accurate gauges are on the way). Is it possible that my flow is actually much lower ? i.e.: 43700/8/500 = 10.9 GPM which would explain why my output is nearly half of what I expect ?

    Here are a couple of pictures of my setup.
     

    Attached Files:

  13. Argofanatic

    Argofanatic New Member

    Confused.

    Anyone ?
    Suggestions ?
     
  14. AMI Contracting

    AMI Contracting A nice Van Morrison song Industry Professional Forum Leader

    Hang tough busy week for a math problem.....
     
  15. Palace GeoThermal

    Palace GeoThermal Well-Known Member Industry Professional Forum Leader

    A couple of thoughts:

    1)
    This is not how you calculate GPM. You need to measure the pressure drop through the heat pump and then use the attached table to calculate GPM.

    2) In looking at the performance data for your unit, I see there is data for EWT of 30 and 50. Yours is 42 so we have to interpolate between the two. There is also data for entering load temps of 80 and 100. Yours is 87 so we have to interpolate again.

    3) If I use 10 gpm and interpolate as above, I see HE of ~ 48,000 BTUH. Not the 72,000 you arrived at. This is closer to the 43,000 of actual HE that you measured in your buffer tanks.

    May be I missed something in all of this....if so let me know.

    4) According the charts, you would need an EWT of 70 to produce HE in the 70,000 range.

    I have attached charts that show the above numbers.

    I suspect that your GPM might be lower than 10 which would explain why your actual HE is lower than what the table shows. If you can measure the pressure drop through the unit, then we will know.
     

    Attached Files:

    Last edited: Jan 6, 2011
  16. Palace GeoThermal

    Palace GeoThermal Well-Known Member Industry Professional Forum Leader

    HE = Brine x GPM x (EWT - LWT)

    If we use the above equation and solve for GPM, we get

    GPM= HE/ (Brine*Delta T)

    You measured your HE in your buffer tanks to be 44,000 so a good guess at your GPM is

    44,000/ 7500 = 5.9.

    Do the pressure drop readings and see what GPM you come up with.
     
  17. Argofanatic

    Argofanatic New Member

    Another piece of the puzzle ...

    Thanks again Dewayne and others, I appreciate you guys taking time out to help both myself and other die hard tinkerers.

    The interpolation all makes sense as you point out. I re-evaluated my friction losses today to make sure I didn't mess up on the pump sizing. Measured all pipes, counted fittings, looked up tables, I calculate that my parallel UPS26-99's are pushing about 8GPM each into the tank loops, and recombining into my 1-1/4 plastic pipe at roughly 16-17 GPM.

    The head losses I used include the heat exchanger losses. If I guesstimate the pressure drop on my POS pressure gauges, I figure I have about a 7-8 PSI drop (better gauges on the way). Correcting for methanol that works out to between 15 and 17 GPM.

    I guess my outdoor tank temperatures have stabilized now after several days because my EST and LST are now 36 and 26. (Outdoor temps are averaging 18F). Doing the interpolation thing gives me an HE = 42,790, COP = 3.04, LLT = 107(measured & matched), Power KW = 6.11, HC should be 68,610 BTU. with a 19GPM load flow.

    This is where I'm getting hung up, HE and HC.
    From what I understand HE is the extraction of heat from the source, and the HC is the combined heating capacity of HE AND the electrical conversion of heat energy. Is this correct ?

    Experimentally I know that my HC (load side output ?) is around 47000BTUs, which would work out to a loop flow of 6 GPM. This would agree with the low indistinguishable pressure drop I see on my POS gauges.

    I think I'm getting there !
    Marc.
     
  18. Palace GeoThermal

    Palace GeoThermal Well-Known Member Industry Professional Forum Leader


    That is the way I understand it as well
     
  19. Palace GeoThermal

    Palace GeoThermal Well-Known Member Industry Professional Forum Leader

    Abbreviations and Definitions
    ELT = entering load fluid temperature to heat pump
    kW = kilowatts
    SWPD = source coax water pressure drop
    EST = entering source fluid temperature to heat pump
    LLT = leaving load fluid temperature from heat pump
    HE = heat extracted in MBTUH
    PSI = pressure drop in pounds per square inch
    LST = leaving source fluid temperature from heat pump
    LGPM = load flow in gallons per minute
    HC = total heating capacity in MBTUH
    FT HD = pressure drop in feet of head
    COP = coefficient of performance, heating [HC/kW x 3.413]
    LWPD = load coax water pressure drop
    EER = energy efficiency ratio, cooling
    LWT = leaving water temperature
    TC = total cooling capacity in MBTUH
    EWT = entering water temperature
    HR = heat rejected in MBTUH
    Brine = water with a freeze inhibiting solution
     

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