Setback? Should you lower heating overnight?

Friends, it has become a fact of my conversational life that people ask me questions about heating their homes. And one of the most common questions I am asked is whether or not people should turn down their heating overnight? This is referred to by heating engineers as overnight ‘setback’.

Up until today I have only had a generic and unsatisfactory answer:

Well, it depends…”.

But last night I finally saw how the question could be answered and then I wrote a spreadsheet to test out my idea. And now I can I answer more fulsomely. My answer will now be

Well, it depends, but it’s marginal, either way”.

Before I get into the gory details, let me just outline that this article is about working out which strategy uses less energy. In contrast to this dull, but understandable, utilitarian perspective, because any gains are marginal, I would urge you to keep doing whatever makes your home a place of joy.

This is a long and technical article, and follows on from a previous dull post in which I estimated the heat capacity of house. Sorry. If you are looking for something lighter, please allow me to recommend this article about candles! But if you really want to know the details, read on.

The Question

The question people are really asking is this:

  • If I set back the temperature from (say) 20 °C during the day to (say) 16 °C at night I will save energy.
  • But then if I want the house to be (say) 20 °C when I get up, I need to apply additional or boost heating for (say) a couple of hours before I get up and this requires extra energy.
  • On balance, will applying this ‘setback’ save energy? Use extra energy? Or will it make no difference?

The Answer

The answer is as follows,

  • If the heating is 100% efficient, then a setback period will always save energy. It’s generally not a big saving, but it is always a saving.
  • If the efficiency of heating varies with power, then a setback period may save energy, or may not.

Specifically

  • Direct electrical heating with fan heaters, storage heaters or infrared heaters.
    • A setback period will always save energy.
    • The longer the setback period and the lower the temperature, the greater the saving.
  • Gas heating with a gas boiler
    • The efficiency of a modern condensing gas boiler is generally in the range 85% ± 5%.
    • If the efficiency falls with increased power – which can happen – then the small saving in energy can be offset by the decreased efficiency of the boiler at high power.
  • Heat Pump
    • The efficiency of an Air Source Heat Pump (ASHP) is generally around 300%.
    • If the efficiency falls with increased power – which can happen – then the small saving in energy can be offset by the decreased efficiency of the ASHP at high power.

Spreadsheet Calculation

To calculate the energy savings I wrote a spreadsheet that simulates the way energy flows into and out of a house over a 24 hour period.

I modelled 4 separate ‘modes’ of heating the house.

  • The steady state: is where the temperature is stable and the input heating power immediately flows out of the house.
  • Off: is the condition on entering the setback period where there is initially no heating and the temperature falls as the house loses heat ‘naturally’. The rate at which the house cools depends upon the heat transfer coefficient and the heat capacity of the house.
  • Setback: is where the temperature is stable at a lower temperature than in the steady state and the input heating power immediately flows out of the house.
  • Boost: is where the heating power is increased to rapidly raise the temperature of the house.  The rapidity with which the house cools depends upon the heat transfer coefficient and the heat capacity of the house.

Click on image for larger version. Illustration of the temperature variation during a setback period. The dotted orange line shows the data for the situation where the house is maintained at the same temperature 24/7 and the red line shows data for the modelled ‘setback’.

I then divided the day into 0.1 hour periods and calculated (a) the heating energy and (b) the fuel consumed, in each of these periods. I then summed up the heating power and fuel consumed taking account of the possibility that the efficiency might be different in each phase.

Click on image for larger version. Illustration of the cumulative use of heat energy through a setback period. The dotted orange line shows the data for the situation where the house is maintained at the same temperature 24/7 and the red line shows data for the modelled ‘setback’.

Unfortunately there are a large number of variables that can be – well – varied: specifically

  • Internal Steady State Temperature
  • External Temperature
  • Setback Temperature
  • Length of setback period
  • Length of boost period/Boost power
  • Heat Transfer Coefficient/Thermal Resistance
  • House Heat Capacity/Time constant
  • Heating Efficiency in each stage

If you want to play with the spreadsheet you can download the Excel™  spreadsheet here:

I must warn you it is an experimental spreadsheet and I can give no guarantees that it is error-free. In fact I can almost guarantee it is error-strewn!

It’s all about the boost!

The reason that the balance of benefits in a ‘setback’ is nuanced is because of the use of ‘boost’ power.

Without boost power, or with only a low power boost, the internal temperature will only return to its set temperature slowly.

For example, in the illustration below, the boost power is 4 kW while the steady state power is 3 kW. Even with this extra 1 kW, the boost takes several hours to bring the temperature back to the set point.

Click on image for larger version. When the boost power is low, it takes a long time to return the house to its set temperature. The model settings are shown at the top. The upper graph shows temperature versus time and the lower graph shows the cumulative energy used during the day. On each of the graphs the dotted orange line shows the data for the situation where the house is maintained at the same temperature 24/7 and the red line shows data for the modelled ‘setback’.

Obviously this is isn’t satisfactory because the internal temperature remains well below the set point for hours after it should have stabilised. However the savings (12%) compared with maintaining the steady state continuously are substantial!

If we increase the boost power to 6 kW, then the internal temperature returns relatively rapidly to the set point, and the savings are still a reasonable 9% compared with maintaining the steady state.

Click on image for larger version. Increasing the boost power reduces the time to return the house to its set temperature. The model settings are shown at the top. The upper graph shows temperature versus time and the lower graph shows the cumulative energy used during the day. On each of the graphs the dotted orange line shows the data for the situation where the house is maintained at the same temperature 24/7 and the red line shows data for the modelled ‘setback’.

However, if the heating in the boost phase is less efficient than the heating in the steady state, then these energy savings can easily disappear. For example, in the situation below, the boost heating efficiency is reduced from 90% to 80%. This has exactly the same thermal behaviour as the example above, but would now use 3% more fuel.

Click on image for larger version. If the boost heating is 10% less efficient than the steady state heating, then the energy savings can be wiped out. The model settings are shown at the top. The upper graph shows temperature versus time and the lower graph shows the cumulative energy used during the day. On each of the graphs the dotted orange line shows the data for the situation where the house is maintained at the same temperature 24/7 and the red line shows data for the modelled ‘setback’.

The examples above might be appropriate to the behaviour of gas boilers which sometimes condense water vapour in their exhaust fumes less effectively when operating at higher power.

But exactly the same principle could also apply with a heat pump. To pump higher thermal power it might be necessary to raise the temperature of the water flowing in the radiators.

The example below has the same power settings as the examples above, but now assumes an efficiency of 300% (i.e. COP = 3) for all heating phases. You can see that overall energy consumed is 3 times lower than in the examples above.

Click on image for larger version. If the heating efficiency is the same for all heating phases, then the energy savings are the same whether that efficiency is 90% or 300%. The model settings are shown at the top. The upper graph shows temperature versus time and the lower graph shows the cumulative energy used during the day. On each of the graphs the dotted orange line shows the data for the situation where the house is maintained at the same temperature 24/7 and the red line shows data for the modelled ‘setback’.

But if the efficiency in the boost phase falls to 250% from 300%, then once again, the savings are reversed and the setback strategy actually costs more than the steady state strategy.

Click on image for larger version. If the heating efficiency in the boost phase is 250% rather than 300%, then the setback strategy costs more energy. The model settings are shown at the top. The upper graph shows temperature versus time and the lower graph shows the cumulative energy used during the day. On each of the graphs the dotted orange line shows the data for the situation where the house is maintained at the same temperature 24/7 and the red line shows data for the modelled ‘setback’.

Conclusions

Developing this simulation has helped me understand some of the basic physics behind the use of setback strategies. And so my advice has developed from “Well, it depends…” to “Well, it depends, but it’s marginal, either way”.

The critical factor is the relative efficiency of heating at higher power (boost) compared with heating at lower power (steady state). If the boost heating is even marginally less efficient than the steady state heating then any energy savings are reduced, and may even be reversed.

This begs the question of whether there is any way to know what is the situation in a particular household. And my first thought is, “No“. Without detailed measurements, there is no way to tell!

 

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12 Responses to “Setback? Should you lower heating overnight?”

  1. Simon Says:

    I wonder if it’s easier to start from the rating of the heating system, whether heat pump or gas boiler, that detail doesn’t matter so much as comfort, and the idea that, whatever the indoor temperature might fall to after midnight, say, it needs to be back at the right level by 8am, say.
    As you’ve noted, when the heating comes back on, it has to be at a higher power than if the house hadn’t cooled down, and being able to do this relies on there being some headroom available. If that headroom were a factor of two, then the heating could be off for 4 hours and came back on, at about double the usual power, for 4 hours. If the house cools down enough during those first 4 hours for the rate of heat loss (the loss to outdoors is what matters, and this rate may not be reliably indicated by the measured indoor temperature, because things are not in equilibrium) to reduce, that reduction is where any saving in energy has to come from.
    I think the slowest mode *is* relevant here. If its time constant were, say, 24 hours, then the rate of energy loss to the outdoors might might reduce to about 5/6 of its normal rate after 4 hours, and the average over those 4 hours would be about 11/12 of its normal rate. There would be a similar reduction during the 4 hours it takes for the house to warm up again. This would suggest that the scope for saving is about 8/24 of 1/12 of the energy normally used in a day, i.e. 1/36, or a little less than 3%. That’s not much, if running the heating system at double the normal output risks reducing its efficiency.
    This reasoning seems simple, but I honestly think it might be valid.
    BW

    • protonsforbreakfast Says:

      Simon, Good Morning,

      I think your estimation may well be about right. And I did think about trying to express the savings as a closed form formula. If the exponential temperature decays and rises are linearised it’s not that complicated (as you point out) but it is a clumsy formula. So I thought a spreadsheet was best.

      But your main point is that there is a limit to how much can be saved. I agree.

      I should also have run a comparison of just turning down the thermostat by 1 °C!

      Anyway: all the best

      Michael

      • Simon Duane Says:

        What interested me about the other way of thinking was that, of all of the variables in your calculation that can be varied, it’s only the time constant and the lengths of time that the heating is below/above the equilibrium level that matter, and working out the linear approximation (which is plenty good enough for me) is hardly more than an exercise in mental arithmetic. But, if you like spreadsheets, … 😉

  2. Edmond Hui Says:

    Thank you as always for the learning. I’ve always thought of it like vehicles. If I ride a bike and want to average a cruising speed, I feel like it’s more efficient just to cycle at that speed. Otherwise I’d have to cycle hard to get above that speed, with its extra air resistance etc, and then rest my muscles (insufficiently) until I coast to a low speed when I’d have to do it all over again. So setback on a bicycle is intuitively a bad idea.

    On the other hand, in an internal combustion engine car, the engine works best at a certain load, and as a result I understand that for ultra high efficiency one has to accelerate through the gears at about 70% throttle, changing gear at the lowest comfortable revs on the engine until you get to top gear, and then coast back down to low speed. So setback on an internal combustion engined car is a good idea. I assume in an electric car, it would be best to go at a steady speed.

    I guess the answer for a house, as you so clearly explained, is that you don’t know until you measure it whether it’s a bicycle or an internal combustion car.

    • protonsforbreakfast Says:

      Ed, Good Morning,

      Yes! both very good points. In fact I was speaking with my son (a physicist) this morning, and he raised exactly the issue of a car. But I think the issue of a bicycle is better because the extra effort is more apparent.

      I trust you are warm and well and that the School can still keep the heating on!

      Michael

  3. Simon Duane Says:

    As a follow-up, I would say that your bricks estimate of the house heat capacity is actually more reliable than air temperature measurements over an hour or two. Those bricks will surely dominate the heat capacity that is relevant to the slowest mode. On that basis, I think the relevant time constant (for your very well-insulated!) house is likely to be three times larger, perhaps of the order 60+ hours. The scope for you to make savings by setback is surely negligible.
    Although those bricks store a lot of heat energy (being inside your external wall insulation, they give you a gigantic thermal buffer – as someone remarked on your previous post, equivalent to a tank containing 3000 litres of water), they don’t lose it quickly. Those of us living in leaky houses know all too well how the the relatively small heat capacity of the indoor air doesn’t stop the heating of that air being a significant cost.

    • protonsforbreakfast Says:

      Simon, Good morning yet again!

      Yes! I agree. As well as there being several different ‘heat reservoirs’ there are also several different ‘thermal resistances’ appropriate to heat transfer through insulation and heat transfer through air leaks.

      It really would be a great experiment to switch off the power for several days when it was cold outside.

      But I have gone soft and no longer have a hard-core experimentalist’s mettle for such challenges!

      All the best

      Michael

  4. Dan Grey Says:

    “I want the house to be (say) 20 °C when I get up” – get a Nest or Netatmo smart thermostat which both calculate how early the boiler needs to fire to bring a house to the set temperature at a set time 🙂. It lets you set a much lower temperature as you don’t need to then worry about the house being warm on time.

    (Hive can do it to it too but it’s limited to 1 hour early, which isn’t enough. My Netatmo was starting my heating 3 hours early to bring the house from 16° to 18° ontime during the cold weather!)

  5. protonsforbreakfast Says:

    It’s always reassuring to agree with Nicola!

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