OT - Geothermal Heat issue...?

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On Mon, 26 Nov 2007 18:32:56 -0600, "Greg O"

Hi Greg,
Your explanation makes sense to me (and is essentially what I offered to someone else in this thread) but...
Though, indeed, the efficiency of the system decreases as the temperature of the well goes down, could that loss of efficiency compensate for the significant savings we would have if we were to drop the temp of our house by, say, 10 degrees for 8 or 10 hours each day?
My intuition tells me that it would not.
What do you think?
Thanks,
--
Kenneth

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I don't think you will see significant savings setting back the temps. Also consider the time it takes to recover after a 10 degree setback. It is very possible that equipment may run longer to recover the temps than it would to just maintain a "normal" occupied temp, more possible when you consider the stress it may put on the well. Again, I would set the temps and leave them at one set point, unless the home is not occupied for days. Can I ask where you may be located? Greg
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"Kenneth" wrote:

In a nutshell, thermal inertia.
Once the system is balanced, it requires minimum energy to maintain the balance.
Change the set point to a lower level, remain there for a while, then return to the higher level requires a lot of thermal work.
Heat intensive industries such as steel, refineries, etc, run 24/7 for just this reason.
Lew
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On Mon, 26 Nov 2007 16:44:48 -0800, "Lew Hodgett"

Hi Lew,
I don't deny it... I just don't understand it:
(Though there may be parallels to industrial applications, I'll stick with home heating for my example.)
For a given outside and inside temperature, the house loses a constant amount of heat per hour, and that amount must be replaced if we are to keep the internal temperature constant.
If the internal temperature of the house is allowed to drop, two things happen. First, there is the direct energy savings because it takes fewer BTUs to keep the house at the lower temp; but perhaps less obviously, the rate of heat loss to the outside environment is decreased. (Because the greater the temperature differential, the more rapid the rate of equalization.)
So, for the eight hours or so that the interior temperature was lowered, there are savings for two reasons: We are providing less heat to the house, and we are losing less per hour of what heat we do supply.
When we decide to go back to the original interior temperature, at every stage (prior to reaching that temp) the hourly rate of heat loss is something less than it would be when we reach the desired internal temperature.
Now, of course, heating up the house those 10 degrees will take a bushel of BTUs, but (unless I am way off here) that would have to be fewer than those saved.
I well understand that the efficiency of the system goes down as the well cools, but it seems to me that the diminished efficiency, though regrettable, is more than balanced by the savings at the lower temperatures.
With all of this, I may be completely out to lunch, but I'd love to understand where I am going astray.
Thanks for any further thoughts,
--
Kenneth

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On Mon, 26 Nov 2007 21:44:01 -0500, Kenneth

I have a closed loop slinky coils in three trenches as deep as the backhoe could dig. Covered back with the clay that came out of the trenches. I think it is a 3 ton unit conditioning 2700 sf, a little east of Atlanta Ga. No booster heat. Hot water heater option. When it is extremely cold, ie 10F or extremely hot 100F the unit runs a lot. It makes lots of hot water when the delta T is enough for the unit to run more than a few minutes. I think the thermostat is at 74 in the winter and 78 in the summer. My wife my cycles the thermostat a degree when she is too cold or too hot. No 68F in the winter that I grew up with. Warm blooded woman I married.
I have no idea if setback works as she is awake when I sleep. Rolling the thermostat when you have people living different shifts does not work so swell.
The neighbor down the road has a couple of geothermal units that they zoned for the main part of the house they lived in and another zone and unit for the extra bedrooms. They claimed power bills less than my house even though their house was much bigger. Makes me mad enough to finish insulating the concrete walls in my conditioned basement.
I have not thought about how dry the earth might be and how that might affect the efficiency of the unit. Not a lot of rain over the summer. I do know I have added dirt to the trenches once in 10 years. I need to add a few more inches to some of the trenches close to the house as they have settled. Settled is good suggesting better heat transfer, maybe.
If you can get the same night to happen back to back read your power meter. My house seems to do about 1000 kwh a month or about 38-45 kwh per day average in the coldest of winter days. You could read the meter before you go to bed on a normal night and read it at 8 am. Next night assuming same wind conditions, cloud cover and night temps do the setback and rollup. You might get your answer. Or put an hour/minute meter on the air handler and get similar results. The hour/minute meter might be more accurate. You could also attach thermometers to the two water lines to see the delta T and what if any measurable influence the setback does to the well temps.
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On Mon, 26 Nov 2007 22:59:57 -0500, Jim Behning

Hey, don't feel bad about that...
Someone I work with used exactly the same equipment to heat a home about the size of ours, and her costs are about half of ours...
Though she is only a few miles away, she is served by a different power company.
Her's has a dual-rate policy. Mine, does not.
As a result, she pays slightly more than half of what I pay.
'Feels great...
All the best,
--
Kenneth

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"Kenneth" wrote:

<snip>
OK, let's try a different way.
Moving thermal energy from place to place is at best a very inefficient process.
You pay to move energy(heat) into the building, but not all the energy is used due to efficiency losses.
Then you turn down the thermostat and move energy from the building to a colder source, but again, efficiency losses come into play.
Then you turn the thermostat back up, and you need to move energy back into the building, again with efficiency losses.
System efficiencies are the problem.
Other examples where efficiency bites you.
1) Wet cell batteries.
For every 100 ampere-hours you consume out of a wet cell battery, you must replace 125 ampere-hours.
Batteries are convenient, not efficient.
2) A big ball.
It takes a lot of energy to get a big ball rolling.
Once it is up to speed, to keep it rolling, all that is needed is to replace the frictional losses.
If you let the ball slow down, it takes a bunch of energy to bring the ball back up to speed.
Maybe these are poor analogies, but it's late, and I'm lazy tonight.
I'd refer you to one of my old thermo text books, but it's easier to run a field test than wade thru one of those books.
I'd keep my sticky fingers off the thermostat.
Lew
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_THOSE_ BTUs are just 'deferred spending'. you spend exactly that amount to raise the temp back to the original setting.

That is the -totality- of the energy savings -- the lowered losses. from the reduced temperature.

FALSE. you are double-counting the same saving there.
to maintain any system 'at equilibrium', all you do is replace the losses. if you are maintaining a lower equilibrium point, the 'savings' are exactly equal to the difference in the losses at the two equilibrium points.

authoritative answer: "it depends".
1) _how_much_ lower are the building thermal losses for the temperature reduction employed? 2) _how_much_ less efficient is the heat plant as the -rate- of draw increases?
Depending on the _quantitative_ answers to those two questions the 'savings' can 'net' to either a positive or negative result.
The exact answers to both questions will be specific to a particular installation.
Getting an answer by 'science' is -very- messy. It's much simpler to use the 'experimentalist' approach and simply 'measure' what actually happens.
The building loss rates are relatively easy -- measure the required heat input at both equilibrium points. It _is_ reasonable to assume that the delta on the loss rates is the same for both temperature rising and falling, so the cool-down, and warm-up phases effectively cancel each other.
The changing 'efficiency' of the heat plant is harder. You really need to have a running monitor on the well-water temperature for that. (with that you can tell 'when' things have 'recovered' from the excessive consumption to raise the building back to the higher level.
Failing instrumentation on the water temperature, one can use outside air temperatures as a -rough- basis for comparison. (it helps greatly if you have historical power usage data [at stable inside temperature operation] that you can correlate with 'heating degree days' for various periods)
If you have the above-mentioned historical data, you'll see that 'cost of operation' goes up as the heat demand increases. both in absolute terms and on a per unit basis.
Now, run the system for a while in 'set-back' mode. Total the 'heating degree days', and the cost. See where that 'per unit' cost falls relative to the same degree-days for stable temperature operation.
NOTE: this is all figuring 'cost' on the basis of "how cold it is outside" -not- on a "per BTU of heat added" basis, so you have a direct comparison of the 'efficiency' of the methods, and can reasonably predict what, if any, the overall savings will be.
Heat pumps are, by their nature, less efficient, the larger the temperature differential between the 'external' and 'internal' sides. And that efficiency does degrade significantly with relatively small increases in that differential.
Things will depend 'a whole lot' on the thermal conductivity of the external heat reservoir, and how fast stuff in the vicinity of the 'radiator' there recovers to equilibrium after a draw-down.
W/o extensive geological testing, that's hard to quantify.
I _would_ tend to believe that the designers/installers *DO* know what they're talking about when they recommend stable (and not 'set back') operation, counter-intuitive though it may seem.
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On Tue, 27 Nov 2007 18:07:39 -0000, snipped-for-privacy@host122.r-bonomi.com (Robert Bonomi) wrote:
Hi Robert,
I thank you for your detailed response...
Perhaps I am not understanding what you have written, but allow me to ask something further.
Please see my comments inline below:
I wrote:

You responded:

I wrote:

You responded:

In my attempt to understand this...
Suppose I lowered the temperature of the house 10 degrees, but not merely overnight. Instead, I left them lower for a month.
Would I not have very significant savings for that month?
If so, would not the reasons for those savings apply as well to my overnight lowering of the house's internal temperature (though with decreased benefit because of the diminished duration)?
Thanks again,
--
Kenneth

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On Tue, 27 Nov 2007 13:43:34 -0500, Kenneth

Why do you need a second electric meter to excrement with? You can read your own meter every day at the same time. As long as you record when you are drying clothes or other significant electricity burning events you should be able to test for no hardware costs.
Are you torturing the group by not doing your own meter reading and reporting back?
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On Tue, 27 Nov 2007 18:56:19 -0500, Jim Behning

Hi Jim,
I am lost...
How would reading my meter at the same time every day tell me my heat-related energy consumption?
Thanks,
--
Kenneth

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On Wed, 28 Nov 2007 09:46:22 -0500, Kenneth

Day zero meter reads 1480 Day one with no thermostat changes reads 1520 Day two no thermostat change 1570 day 3 no thermostat change 1620 Day 4 after thermostat rollback 1680 day 5 with thermostat rollback 1740 day6 with thermostat rollback 1800 day 7 no thermostat change 1850 day 8 thermostat rollback 1910 day 9 thermostat rollback 1970 day 10 no thermostat change 2020
and so on. Just read the meter every day at the same time. Do the math. Do it over enough days to factor out clothes dryers and baking festivals. If there is indeed any significant energy saving to be had by thermostat rollback it will show up with a month of measuring. Especially if you do rollback every other night.
That said I think when I feel rich I will buy http://www.theenergydetective.com/store or one of these with the split core http://eyomenergy.com/Merchant2/merchant.mvc?Screen=CTGY&Store_Code S&Category_Code=ES
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On Wed, 28 Nov 2007 18:56:46 -0500, Jim Behning

You are not measuring how much you spend on heating, you are measuring much less or more you are spending by trying different thermostat strategies. You also need to note wind, sunshine and outside temp. My heat does not run on a 50 degree day with sun and little to no wind.
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Jim Behning wrote:

Just for some perspective, last week the temperature here was 60 degrees, which melted the snow that had come down a couple of days previously.
--
--
--John
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Relative to what you would have 'spent' at the 10degree higher temperature yes.

Yes and no. <grin>
There are three intervals to consider. 1) while the temperature is falling from 'X' to 'X-10', 2) while the temperature is stable at 'X-10', 3) while the temperature is rising from 'X-10' to 'X'.
As the house cools from 'X' to 'X-10', you aren't providing any heat input at that time. *THAT* 'savings', is cancelled by the 'extra' energy you have to put back into the building the next day, to raise the temperature from 'X-10' back to 'X'. For complicated reasons, it usually takes a little more energy to go from 'X-10' to 'X' than was 'saved' by letting things fall from 'X' to 'X-10'. This differential is usually fairly minor, however it can be magnified if the -rates- at which the temperature falls and rises are different.
The heat input required to maintain the house at a constant "X" is exactly the heat losses being radiated by the house to the exterior.
The heat input required to maintain the house at a constant "X-10" is exactly the heat losses being radiated by the house to the exterior.
In both cases the rate of loss is a function of (a) the temperature differential, _and_ the quality of the insulation.
The point is, however, that the difference in heat input is exactly the difference in thermal losses, at a constant temperature. You cannot count a savings for less heat input, -and- a savings for lower thermal losses.
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No, the reason is that it costs them money to keep it warm and it also costs them money to warm it up , but they can produce a product while it is warm and not while it is warming up.
It is not how much they are spending on energy, it is their return on that investment--less than zero (due to other operating costs) while warming up, and greater than zero (hopefully) when at operating temperature.
There are other considerations such as thermal stresses during warm-up and cooling down.
--
FF





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Because you have no control over the systems ability to "produce" more or less. It's out put is a relative constant. If you request more out of the system you pay more through an external source.

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Leon wrote:

Water Furnace uses avariable speed (or at least two-speed) units so there is some control.

Beyond high-speed, correct, but that can also be controlled as to whether it is used or not in a couple of different ways.
It's possible (probable?) the initial installation didn't not take advantage of any of those options and the installer isn't clever enough to recognize/implement them, but there are alternatives for most of the issues.
When/if the unit does "max out" w/ the ground source, then the only choice is an aux heat output, but OP has indicated they chose not to use one anyway owing to having sized the unit(s) at a quite high output.
So, my conclusion is still that it would be very unusual set of circumstances in this case if the setback would not reduce overall usage.

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wrote:

Hi again,
Of course, the experiment is a very simple one, but right now, we have only one electric meter.
We will soon have two, and with that, I should know.
All the best,
--
Kenneth

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On Tue, 27 Nov 2007 14:09:49 GMT, "Leon"

Hi Leon,
Yes, its output is constant, but does that lead to the conclusion that we would, or would not save with a setback?
Thanks,
--
Kenneth

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