Posted by: @jamespa

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Taking the example I reproduce above, your standing energy out (which must in the long term equal your house loss) is about 4.3kW at a temp diff between IAT-OAT of about 9C, so about 470W/C.

During your six hour setback your IAT falls from 20 to about 17 roughly linearly, so an average 'setback' of 1.5C,.lets generously say 2C to allow for the slight curve.  It then recovers over about 3 hours also roughly linearly.  Thus the 'saved' loss from the house (due to the fact that the house temp is lower) is roughly 2*470*9 Wh=8.4kWh. 

Your COP is consistently over 4, so that would correspond to saving of energy in of about 2kWh.  Yet you are claiming, and I agree that the mark one eyeball integrator would suggest that you are correct, that the saved energy in is about 6kWh.  Unless my math or reading of your graphs is all wrong, something doesn't stack up.

I certainly welcome you questioning my data results and conclusions, as I have no divine way of getting things right, and am just as fallible as the next person. That's why we publish methods and results so others can scrutinise them, and is also coincidentally why I don't trust conclusions where that process hasn't been followed.  

For ease of reference, here is the relevant chart:

My 'logic' (actually it is a bit of whatiffery...what if the setback had not happened, how much energy would I have used?) is to estimate what would have been used during the setback by eyeballing the period preceding the setback, and extrapolating that into the setback. As it was around 1 kWh per hour with an OAT of 8 most of the time, and the OAT was in that region during the setback, I estimated the energy in to be 1 kWh x 6 = 6 kWh. I think that is OK, and we agree that is a reasonable figure for the saved energy in.

Now, we can do the same thing for the energy out. This will probably become a bit self-fulfilling, but my Mk 1 Eyeball Integrator suggests the energy out  saved is of the order of just over 4 kWh per hour, say 4.2 kWh per hour, or 25.2 kWh for the six hour period. This 4.2 kWh per hour figure is close to your 4.3 kW figure in your first paragraph quoted above (and 4.3kW of power for one hour = 4.3 kWh of energy in that hour), meaning so far we are still in agreement. I also agree that this can be presented as around 4.2 / 9 = 460W per degree C. 

It is your middle paragraph above that I think we need to look at closely, as that is where I think your logic may go awry, in particular the bit that reads "Thus the 'saved' loss from the house (due to the fact that the house temp is lower) is roughly 2*470*9 Wh=8.4kWh." But I am not sure this is the 'saved loss' or even that 'saved loss' is a meaningful concept. The house is still losing heat at much the same rate as before the setback, because the IAT/OAT difference is pretty much the same throughout, and therefore the house is still losing heat at the rate of (on my figures) 460W per degree C, or 4.2 kWh per hour for the 9 degree IAT/OAT difference, which at a COP of around 4 corresponds with an energy in of just over 1 kWh per hour. The same applies to the whole 6 hours period, 25.6 kWh loss, divided by a COP 4 of 4 to get an energy in value of 6.3 (and the universe is still spinning on its proper axis).

I think the problem is the saved loss, which, as I say, I am not sure even exists as a concept. The loss is marginally less, because the IAT/OAT is marginally less, but that's it. We haven't 'saved a loss', we have just experienced a slightly lower loss, and that happens solely because of the marginal change in IAT/OAT difference. Put another way, given zero energy in during the setback, there is nothing to be saved (can't reduce zero below zero, and have negative energy), and, in effect, the heat in and the heat out become decoupled, as we are not in as steady state, and each does its own thing, heat in drops to zero, and heat out declines marginally.

A final thought: do we even need to look at this concept of 'saved loss'? Does it perhaps over-complicate something that doesn't need to be made complicated? If we can clearly see the heat pump uses 1.2 kWh per hour preceding the setback, and nothing during the 6 hour setback which happens with similar IAT/OAT actuals and differences, is it not enough to say we have clearly saved 1.2 x 6 = 7.2 kWh during the setback, and be done with it? 

If we are going to introduce something called 'saved loss', then we may need to think hard about definitions.

Not an 100% perfect answer, but at least some thoughts.              

Perhaps another way of putting it is this: yes, the heat loss will be marginally lower during a setback, and we could even at a stretch call that 'saved heat loss' but it is only related to the IAT and OAT actuals and differences, and has nothing to do with current energy in, which during a setback is zero, which in turn means the concept of a COP is meaningless - can't divide by zero. If the COP during the setback is zero, or meaningless, then it doesn't do to divide the saved loss by COP that applies when the system is running... 

Posted by: @derek-m

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Posted by: @kev-m

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@derek-m 

yes I would be interested in seeing how my data compares with the model.

 

Kev 

If you would care to provide some data from your system covering particular periods I can run it through the model to see how the results match.

Data if possible should include LWT, RWT, IAT, OAT, setpoints, any setback periods, flowrates if you know them, energy in and energy out, and anything else you think may be useful.

Also detail any results that may be of use.

 

I'll see what I can do but it will be a few days as I'm away at the moment.

K

@cathoderay For clarity Im not disputing the figures or your eyeballed numbers, I'm disputing the interpretation which, for some while, I believed but now doubt.

Saved loss (an ugly term but I cant currently think of a better one) is absolutely key to understanding whats going on here.  The important point is that, over the long term, or between any two points in time when the state of the house is the same, the energy put into the house by the heat pump must be exactly equal to the energy lost from the house.  During setback the house, as you correctly say, continues to lose energy, at a rate which is only a bit lower than when not in setback.  This lost energy must be replaced in order to restore the house to its original condition and if it is not restored the house will end up in a different condition, so the next cycle begins from a different position.  Thus the saving in energy which must be put into the house as a result of setback is only the difference between what is lost during setback and what would have been lost had it not been setback (the 'saved loss'). 

Perhaps an analogy would help.  Consider a lake  (house) with an inflow (heat pump) and an outflow (walls etc).  The amount of water passing through the outflow (heat loss) will depend on the height of the water in the lake (indoor temperature) above whatever is the threshold over which the water must pass (bar at the end of the lake, etc - the outdoor temperature).

Now suppose that, during the daytime, I feed water in at a rate which just balances the rate at which water is lost, then the height of the lake (temperature) stays the same.  Lets suppose this is 4 cubic metres per hour.  Now further suppose that I stop feeding water in for six hours.  I have apparently 'saved' 24 cubic metres of water during setback.  Lets also suppose, like your house, that the lake is sufficiently large that its level falls by only a small fraction.  So during the 6 hours the lake will lose just a bit less than 4 cu m per hour, lets suppose the average loss is 3.5cu m per hour.  Thus during the 6 hours it will lose 21cu m instead of 24cu m.

Although the lake has only dropped a bit, it has dropped, and to return the lake to its original position I have to put back 21cu m in addition to whatever it continues to lose during the recovery time.  If I dont then I've depleted the lake, and I cant go on doing this.  So rather than a saving of 24cu m the saving is only in fact 3 cu m.

Exactly the same applies to your house.  If you dont restore all of the energy lost from the house during setback (in addition to the energy lost during the recovery period) you will end up in a different position, basically you will have cooled the fabric.  You cant continue doing this.

Hence my argument and why I (sadly) don't yet have confidence in the interpretation.  

Hopefully that makes the argument clearer and exposes the apparent paradox. 

(Incidentally its valid to divide the saved loss by the COP as a way to work out the saved energy in, because what this is doing is the same as dividing by COP the actual loss during setback, dividing by COP the loss which would have occurred had setback not occurred, and subtracting one from the other: (4-3.5)/5=4/5-3.5/5)

Don't feel bad about this, intellectual rigour is essential, and I would rather be corrected than wrong!

I think I get your reasoning, especially with the helpful water analogy. It seems to me that if you are correct, then the suspect part of my charts is not the setback (zero energy in and out is correct), but the recovery period, when we should see much more energy going in (much more water to restore the lake to its former level).

As you may recall, both the energy in and out are calculated from the raw data. I do however also record the Midea total lifetime energy in and out data, and, depending on how Midea calculate these total numbers, they should be more or less an independent check on my calculated values - in fact, I actually did a check a while back, and they were close enough not to worry me at that time. I will have a look at that data, and see what I find there. 

@kev-m

Thanks Kev, no problem.

Posted by: @cathoderay

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It seems to me that if you are correct, then the suspect part of my charts is not the setback (zero energy in and out is correct), but the recovery period, when we should see much more energy going in (much more water to restore the lake to its former level)

Thanks for the comments, Im still bouncing around in my mind whether I have missed something, but I cant yet see it. This stuff is hard, however you look at it, and one of the problems with experiments (which I fully accept are necessary) is that you cant actually do the 'control'. 

I think you are correct (if I am correct) but it may be quite subtle.  It could well be that full recovery takes a lot longer than is apparent, and a few hundred watts extra is going in over a period of 12 hours or more (bear in mind that the air temperature will recover faster than the fabric temperature).  This could easily get lost in the noise and the uncertainty due to the varying OAT.  The recovery time is likely ultimately to be limited by the output from the emitters (at any given FT) if you are running pure WC, although it may be that your auto adaption script overrides sufficiently that this is not the case.  

@derek-m does account for the 'lake' in his model, which may explain differences of scale (I think) between the qualitatively similar results at these fairly mild OATs.  I am, however, a bit surprised that you don't see a larger peak immediately after setback ends.  Currently my gas boiler is running at about 4.5kW during the day.  I turn it off at night and, for the first hour in the morning, it runs at 8.5kW (which is its peak output - its a 28kW machine that I have throttled back experimentally in prep for switching to a heat pump).  I think my house temp drop is not dissimilar to yours.  The difference may be that I'm not running at a weather compensated flow temp (boiler not capable), so the radiators can emit much more during recovery (because the FT is higher than is strictly necessary) and then settle back down to a time-modulated pattern 

If the house fabric is at certain temp at say 9pm that is satisfactory, you switch off heating and slowly lose that heat, its heat that is essentially not needed overnight if it falls to satisfactory IAT. Following day its easier then getting house fabric back up to temp with help of usually warmer then night time oat. Obviously the house wont be at a constant temp as it is with continuous running but to the person in most houses the savings would outweigh the small bit colder IAT in the mornings. The thread title is that question, not whether more heat is needed to re heat the lost heat in the fabric of the house . Essentially cathodesrays experiments have been showing what the thread title queries.

@jamespa - OK, here are my calculated energy values and the Midea energy values for the same time frame as in the earlier chart plotted one above the other. My calculated values use raw data (amps, volts, LWT/RWT delta t, flow rate, specific heat), the Midea values are the current (for the plot point) total lifetime value minus the value one hour before, ie what has added in the last hour, and are integers, because the lifetime values are integers, with possible/likely rounding errors (one of the reasons I preferred the calculated values). They should be independent of each other, apart from my earlier caveat about it depending on how Midea calculate the lifetime values: 

By and large, they are similar. In the area of interest, the 0300 to 0900 recovery period, my energy in values are 2 or just under, the Midea energy in values all 2. The discrepancy may be the 80% under-reading in the calculated value I mentioned earlier. During the recovery period, the OAT fell compared to the OAT before and during the setback, meaning we expect the energy in to increase. 

Here's the same chart, showing more of the recovery period: 

Using the Mk 1 Integrator, I suggest the saving during the setback was around 1.5 kWh per hour, at total of 9 kWh during the setback. If we take that as the baseline (what would have been used anyway, in reality it might have been a bit more as the OAT was lower), then we end up with what looks like a 6 hour recovery period (though in fact most of the recovery had happened by 0700, see the pop up on the earlier chart) that uses 2 kWh per hour. But 1.5 kWh per hour of that energy would have been used anyway, meaning the extra was 0.5 kWh per hour, or 3 kWh over the full recovery.

We still have the net saving of around 6 kWh over the entire period.

I don't like it either when the theory and observations don't agree! The fact that two so far as we know largely independent methods get the same results increases the likelihood they are both right (or both are wrong for the same reason).

I think the only way we will get a better answer is to look at 24 energy use, based on readings from the heat pump dedicated external kWh meter, in my house (which keeps a lot of variables constant, same building. same system) over periods with similar OATs (and variable nailed) both with and without a setback. Doing this is tedious in the extreme, I have to clamber up a step ladder to read the damn thing, not to mention manually finding periods of similar OATs in the data, but I am nothing if not dedicated to getting at the truth, and so will persevere.          

Posted by: @newhouse87

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If the house fabric is at certain temp at say 9pm that is satisfactory, you switch off heating and slowly lose that heat, its heat that is essentially not needed overnight if it falls to satisfactory IAT. Following day its easier then getting house fabric back up to temp with help of usually warmer then night time oat. Obviously the house wont be at a constant temp as it is with continuous running but to the person in most houses the savings would outweigh the small bit colder IAT in the mornings. The thread title is that question, not whether more heat is needed to re heat the lost heat in the fabric of the house . Essentially cathodesrays experiments have been showing what the thread title queries.

Sorry to disappoint you but I think you have failed to understand the point I am trying to make.  What I am saying is that the figures @cathoderay has extracted from the graphs appear to be inconsistent with the (implied) assumption that the house is returned to its original state following setback, and that therefore either the figures/interpretations are wrong or in fact the house is not returned to its original state and thus the assumption that the process can be repeated/the assumption that the claimed amount of energy has been saved is itself false.  Its absolutely necessary, for the saving to be assessed correctly, that the house starts and ends in the same condition (fabric temperature).  Or put another way, if the house fabric has not returned to the temperature it was the previous day, then you aren't back where you started, and cant do the same for a second night because you have depleted the reservoir of energy held in the fabric. 

Now I'm not saying that energy is not saved at all by set back, but what I am saying is that the claimed energy saving appears to be way to big to be consistent with conservation of energy, and thus it is not yet proven that energy is saved.  Until the conclusions can be reconciled with the first law of thermodynamics (conservation of energy), then the conclusions cannot be relied upon.

I have invited people to point out a fault in the logic of my argument, and that invitation remains open, firstly because I want to be sure I (and others) correctly understand what we are seeing, and secondly because I actually hope I'm wrong , whilst fearing that I am right.

You may find that my lake analogy above helps you to understand the point and its importance to the title of the thread.

I still have Home Assistant on my system, and here is its version of the last month, showing OAT ('byte_09'), IAT ('MD02') and hourly trailing 24 energy in/out (based on Midea's total lifetime values). These plots are not ideal, they confuse my Mk 1 Integrator somewhat, but I haven't found a way to get say noon values only, with the past 24 hours value. Ignore the numbers in brackets after the subplot titles, they are the current value. 

There isn't really enough data here, meaning I am being somewhat speculative, but I think there are one or two periods where comparisons can be tentatively made. The introduction of setback can be seen in the MD02 trace, as the start of the saw tooth pattern. At the same time, we should note the average IAT was 1-2 degrees higher before the introduction of setback compared to the period after the introduction of setback, but also note this was not because of the introduction of setback, it was because I lowered the the WCC a fraction as the house was running hot. You can just the the IAT starting to fall immediately before the first setback.

What we want to do is compare noon energy in values for periods of similar OATs before and after the introduction of setback. Very roughly, the noon to noon period 22-23 Oct is similar to the noon to noon 6-7 Nov period (similar OAT peaks and troughs), with the former being pre setback use, the latter being during setback use. I can then hover of the last 24 hour energy in trace, and get the nearest to noon value for the 23 Oct (29 kWh, no setback) and ditto for the 7th November (25 kWh, with setback) - but the average IAT was also a bit lower...all in all, non-conclusive.

I do have data from the latter part of the 2022/23 heating season, and will have at look at that to see if there any pre-setback periods that can used for comparisons with current setback periods, as the accumulate over time. If there is a forecast of sustained stable weather, I might also turn the auto-adaption off for part of the time, and see what happens.      

Posted by: @cathoderay

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There isn't really enough data here, meaning I am being somewhat speculative, but I think there are one or two periods where comparisons can be tentatively made...

I think it might take a bit of thought about how to analyse data to minimise the amount needed.  Part of the problem is that the signal is quite small relative to the noise.  For example if setback averages 2C for 9hr per day, at a typical iat oat delta of 10C, then the saving in daily energy lost from the house is 9/24 x 2/10 ie 7.5%.  Unless cop variations work in your favour this means that the maximum saving inenergy in is also 7.5 %.  That's not a lot to dig out of the noise.

There, as you may be aware, signal processing techniques to extract signal from noise, but I can't immediately think of one easily applied to heat pumps.