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.
Only if you decide to keep the building's mass at that temperature. If you want to restore the temperature of the model, you also have to re-heat the container.
If the structure had no thermal mass then that would be the case. But it does have thermal mass and changing the temperature of that thermal mass requires the addition or removal of heat.
But the time at lower setpoint more than compensates for the differential loss. It's well-established in general that a setback lowers overall heating costs in general. It would take unusual circumstances for that to not be so.
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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.
"Robatoy" wrote
I had an old volvo that had a thermostat that would die on a regular basis. So I just ran without the thermostat. The problem with this particular configuration is that it took over twenty minutes for it to heat up. And until it heated up, you had no heat, defrosters and the engine did not run well.
But volvo engineers had a unique solution. They had a window shade type device located in front of the radiator. You pulled a chain under the dash to pull the shade up over the radiator and it heated up quick! This model also had a baby bottle warmer under the dash as well.
But the time integral is still less...
As I described just above your comment, we do not have auxiliary heat at all.
All the best,
Hi again, OP here...
That is just what I am trying to sort out:
For reasons that I have not been able to understand, the Geothermal folks say that for their systems, it does not work that way (and they seem to be consistent on this.)
All the best,
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,
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,
It's not something one should just assume. Especially with alternative energy.
Or not as the case may be.
Hi John,
Might you know of some reasons that the general principle of savings through setbacks would not apply to my geothermal source...?
As I have said before here, I certainly do not (know enough to) disagree, but I have no understanding of why that should be true.
Sincere thanks,
_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.
I was responding to the guy who does, not you...
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I don't think it is, in general, any different unless one of the conditions outlined previously were to be true for a given installation.
I've not looked recently, but when we were looking into it initially, Oklahoma State was the leading research university on geothermal. I'm sure there are many others with useful information that a google would uncover as well. The other place that had a wealth of geothermal information online back then was also TVA (tva.gov). I would also expect the EIA and DOE energy conservation web sites to be potentially useful as well. You might try Water Furnace directly rather than the local distributor/installer to see what they say -- I found them to be quite knowledgeable with their evaluation/sizing software packages. It wouldn't surprise me at all but what they could actually make that a part of their analysis for your particular system -- of course, being as it is already installed, they might not want to run another gratis...
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Are you sure it is constant--at least I was unaware that W-F used single-speed units--they weren't when we did ours anyway, but that's been quite a while ago.
Even if so, it does not lead to the conclusion. The savings of a setback depend on the integral of the the demand over the time period--if the average demand is lower, then the input required is lower for a similar set of external conditions.
The only kicker in the mix is whether there really would be such a significant loss in efficiency owing to the heat source "drawdown" that the overall system efficiency drops sufficiently to cause more energy to be used than is saved. I have an extremely difficult time believing that to be at all likely.
See my other response for some suggested places to look for some more definitive research and sources.
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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:
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,
I don't know enough details of your setup or the structure to be able to run the numbers. The main objection I'm seeing is that it's likely to cause the resistance elements in the heat pump to kick in but you say that you don't have any, but there might be something else unique to your situation.
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