Based on the low change in water temp, it looks like you are getting about 1 ton out of the well - or an excessive flow rate.

We return to the OP Question...So, now, after 2.5 hrs:

Outside Air temp = 16f

Inside Room Temp = 64f

GEO inlet = 54f

GEO out = 52f

Inside (Floor) Supply = 83

Inside (Floor) Return = 73



To get the calc btu, what is it, lb water*flow*delta T* conversion factor btu to KW?

Currently the Pump was last measured at 17.6 gpm, slightly throttled from 20gpm.

What is reasonable change? 4f?
just a note: I am surprised at the EWT as I expected 40f; summer must have warmed us more than expected.

To get the calc btu, what is it, lb water*flow*delta T* conversion factor btu to KW?

8.35 * GPM * delta_T * 60 = btu/hr from the well

4F is about right. I didn't follow your head calculations - you should account for the fact that you are returning water to the same level.

8.35 * GPM * delta_T * 60 = btu/hr from the well

After 14 hours run time : 8.35*18*1.75*60=15781 BTU/Hr from Well. 1 ton = 12,000BTU. Thanks, I didn't know about how to get this... Now what? Is 1 ton from the pump and dump wells a bad thing? Is it lowering or raising the COP? Not being dumb here (trying not), but looking to know. Throttle the pump? Change out the pump.
Is the fact that the flow through the Outside HX is reversed from manufacturer design an issue?


> Depth to water = zero

> Depth to submersible pump from water level = 14'

> Depth of well - 14'
> Pumping height from pump discharge to top of loop (Heat Exchanger (HX) is at
top of loop) = 29'

> Vertical height from HX to water = 15'

From Jacques

> > > If you are dumping lower that the pump discharge, there will be a gain
> > > in energy, the static head will be negative instead of positive. The
> > > total head of the pump is the sum of the friction head and static head.
> > > The static head will subtract from the friction head in this case. Since
> > > normally the friction head in such a system will be fairly high there
> > > will be a net positive head required from the pump but much less than
> > > otherwise.
> > >
> > >
The static head has to calculated based on the level of the discharge lake or tank fluid surface. It does not matter how deep the pipe discharge end is.
Jacques

So, what my head total is a number I do not know how to get, but the GPM is @20 from the pump.
20gpm is at 30' head on the manuf. pump curve.


Current Numbers after 14 hours: Outside Air temp = 8f

Inside Room Temp = 69f

GEO inlet = 55.75f

GEO out = 54f

Inside (Floor) Supply = 94.5

Inside (Floor) Return = 83
3.3KW/hr energy usage.

How/where did you measure your GPM? Are you sure that the temperatures are perfectly accurate?

Who said: A man with a thermometer knows what temperature is; a man with two thermometers is never sure?


Temperature and GPM error measurement could make a difference...

size the pump for and set the pressure switch to the pressure needed to get the desired GPM with the throttle valve almost fully open (to leave a little bit of pressure and margin

GPM measured with a bucket and stop watch at the discharge well end of GEO pipe. I was able to throttle the 20gpm back to 18 with 2" ball valves.
What was done on this setup is the GSHP turns on the high flow GEO pump. A solenoid and a check valve and a pressure tank keep pressure on the system when the unit is not calling for GEO pump....Pressure at the HX is quite low, but flow is high. No pressure switch controls the pump.


Temps are measured with cheap digital thermometers that are taped on to the four pipes (1. geo supply 2. geo return 3. inside supply 4. inside return)


The geo side thermometers, with unit not running for 1.5 hours, are within .1c of each other. Seems rt, eh? But the numbers are higher than what was measured at the two wells with a 'Hydrolab QUANTA', last August, whose accuracy I trust over the cheapy ones. (47f, now I get 55.6f)


Any ideas for ensuring good temperature #'s? Swap supply and return temperature probes?

After writing this, my plan is swapping probes and insulating probes better, then posting run numbers again.

GPM with a bucket needs to be measured at the same discharge level as the system will use when running. Ie, well water level. Or just close the valve until you get the correct temp drop.

> Pressure at the HX is quite low

Probably below atmospheric which you may want to address if you find that you get fouling.

Posted By nooboo on 23 Oct 2012 05:50 PM
Who said: A man with a thermometer knows what temperature is;
a man with two thermometers is never sure?

That's Segal's Law -- except it was watches, not thermometers.

http://web.mit.edu/newsoffice/1998/segal.html


Cole's Law: chopped cabbage

Where is it 8*F outside this time of year?

Posted By engineer on 24 Oct 2012 09:24 PM
Where is it 8*F outside this time of year?

Yikes!!

Does geo actually work that far north?

I would think ground temps would be so low as to render geo unfeasible...the only reliable heat in mid winter would be to set some petroleum product on fire...

Does geo actually work that far north?

I would think ground temps would be so low as to render geo unfeasible...the only reliable heat in mid winter would be to set some petroleum product on fire...

Don't forget bio-mass:) There is more than one GSHP in my area of the woods.
How does one calculate COP on a running system accurately?
What I have figured just now is that the HP made 300,000 BTUs in 14 hours at a cost of 80kw. Doing some math I get 21k/11k or about COP 2
Disclaimer: My estimated BTU's made is from my spreadsheet of water and concrete thermal storage. I may have greater thermal storage. The cost of energy was from the electric meter.
My input water is expected to cool down to 40f.

Reworking my approach, I figured 5.3 COP
Inside flow rate of 10 gpm, delta T is 12f

Using the formulas:

COP = hh / hw

3.3 kilowatts of the appliance = 11 260.0674 btu / hour

Assume a flow rate of 10gpm based upon my earlier (inside) pump calcs with found flow delta T of 12f

Q (Btu/hr) = GPM x 500 x delta T

So, 10*500*12 = 60,000

60,000/11260 = COP of 5.3

Not encumbered by the facts and certainly with errors in GPM and energy consumption, I say that GSHP does work in this freezer we call home. 5.3 would be pretty good, eh?

It would be a miracle if your source water temps and flows are correct. Energy from nowhere.

I may be the luckiest guy on this forum!

An evening with sunshine, but losing daylight at 5.5. minutes a day, we are just grateful the heat is on! And seem to be working!!

Sure, it is a cold climate, but we love it, as long as we have a way to come in from the cold.
We are ice skating on the lake this saturday night.

It would be a miracle if your source water temps and flows are correct. Energy from nowhere.

I take the advice seriously, wanting to understand, and will check the temps and flows again.

With the temp probes are better insulated now and with GEO side unchanged from 18 gpm, and all four thermometers checked against each other, all reading within +- .05°
On the geo side, the delta t is 3.35, 18 gpm.
Doing the math
8.35*gpm*delta_t*60=
8.35*18*3.35*60=
30,210BTU/hr. (GEO side)
Is that more reasonable comparing it with the inside flow rate of 10 gpm and 12.1 delta_t
8.35*10*12.1*60=60,000btu/hr (Inside)? Another miracle, rt?
I have a Grundfos 26-64 with about 10' total head. Maybe my flow is slower on the inside pump? Our Woodstove is currently firing.

The input btu from the well plus the electricity btu must equal the output btu (except for small losses/gains from the internal air). It may not matter, but something isn't reading quite right.

Posted By jonr on 28 Oct 2012 07:57 PM
The input btu from the well plus the electricity btu must equal the output btu (except for small losses/gains from the internal air). It may not matter, but something isn't reading quite right.

30k from geo, 12k from electricity (3.3kw=12k+-)=42k...
60-42=18k btu/hr...Could part of that 18k be woodstove BTU input? And the other part measurement error.
or perhaps ALL measurement error:laugh:
Thanks for all the help!!

After five months of use, one problem has shown up in this self-install of mine: the 2" long sweep 90 used on the injection well pipe, at the GSHP, has collapsed on itself from the negative static head; resulting in a probable low ground source flow rate. It was rubber, like in a no-hub, and it is logical that the 90 could not withstand the negative pressure.

Glad to read you survived the winter up there.