Posted by: @bob77

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Heat loss Q depends on the thermal transmission value U, the area A and the differential in temperatures Δt:

Q = U × A × Δt

U and A will remain constant, so the only variable is Δt.

This gives you the heat requirement of the property -- that is, the amount of energy that the heat pump is required to produce. But we're really interested in the amount of energy consumed by the heat pump, and this will be dependent on COP, and ultimately the flow temperature. So we have to add another equation:

Q_produced = E_consumed * COP = U.A.Δt

COP being Coefficient of Performance. The COP is a measure of efficiency, and thus is simply the ratio of Heat Produced per unit of electricity consumed. It is dependent on flow temperature and outdoor temperature: the lower the flow temperature, the higher the COP; the higher the outdoor temperature, the higher the COP. So we have COP = f(T_flow, T_outdoor), thus:

E_consumed = U.A.Δt / f(T_flow, T_outdoor)

We don't have a neat linear formula for the COP at different flow temperatures or outdoor temperatures, but heat pump manuals will have curves for your specific heat pump that contain the relevant information (and there are spreadsheets produced by forum members that contain this as well). Going by my own Mistubishi unit, at 2 degrees C outdoor temperature, the COP drops by about 0.25 for each 5 degree C increase in flow temperature. So, let's say you increase your flow temperature by 10 degrees in order to raise the temperature of your house back up to the thermostat's set-point for the evening. This means that the COP drops by 0.5, so the Energy Consumed during this period will rise:

% change in E_consumed = 0.5/COP_orig

According to my heat pump's manual, the COP at an outdoor temp of 2 degrees C and a flow temp of 35 degrees C is 3.5. This means that the % change in Energy Consumed is 14%; i.e. increasing the flow temperature in order to raise the indoor temperature back up to the thermostat's set point will increase the energy consumption of the heat pump by 14%.

This is before the Δt is considered, however, which I understand is the main thrust of your argument: the average temperature of the house if you turn it off during the day will fall, and therefore the house will be subject to a lower heat loss. This in turn means that the heat pump does not need to produce as much energy to raise the temperature from cold to warm as it does to maintain the house at a warm temperature. (Please correct me if this summary of your argument is incorrect.) There is a nuance to this as well, because while the formula you posted is linear, the rate at which the temperature falls is not linear (the temperature will fall faster when the house is warmer, and slow down as the indoor temperature drops). This means the house spends slightly longer at the cooler temperatures than you'd expect based on a straight average (again, another minor point in favour of turning off your heat pump during the day). But let's just make the simplifying assumption that the temperature decrease is linear (in my experience, it's approximately linear, unless you're turning it off for a few days, so I think this is a fair assumption).

Let's say we have our heating on for 8 hours, and off for 16 hours (to make the maths easier), and let's say that during the time we have it off, the house's temperature drops from 21 degrees C to 18 degrees C. Let's also assume that the house instantly warms to our set temperature (again, another simplifying assumption). On average, during the "off" period, the house's temperature 19.5 degrees C, and during the "on" period the average temperature is 21 degrees C. So our overall average house temperature is 20 degrees C. Our Δt for the above formula is 1 degree C, which means the amount of heat the heat pump is required to produce is reduced by about 5%.

I've made a few assumptions here, which are easy to relax or change if you want to do the proper calculation for your property (the biggest one is probably that the flow temp would need to be 10 degrees higher for the entire 8 hour heating period - likely it will only need to be higher for a fraction of that time, before reverting to the normal flow temperature, but what that fraction is is up for grabs: 2 hours of 10 degree flow temps vs 3 hours of 10 degree flow temps tips the balance in favour of turning the heating off, for example). But the point is, we've made reasonable assumptions, and we've found that although the heating requirement falls when you turn the thermostat off, this is outweighed by the increased power consumption of the heat pump when running at the higher flow temperatures required for a more responsive heating strategy. It's not obvious that turning the heating off and on will actually reduce your power consumption: the increased energy consumption and reduce heating requirement are in the same order of magnitude (i.e. one is not a small fraction of the other), so they both ought to be considered.

This is all theory though, and only half the theory at that (there are loads of things missed off here). The best way of figuring it out is just trial and error, but using the theory as a guide for which direction you should trial next.

To be clear, my point isn't that it's cheaper overall to keep the heating on and not turn it off during the day -- just that it's not obvious that one strategy is better than the other, and you'd need to really do some maths or experiments to figure out what's best for your house, your preferences, and the way you live.