Posts Tagged ‘Setback’

Setback? Should you lower heating overnight?

December 19, 2022

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!

 

Estimating the heat capacity of my house

December 19, 2022

Friends, the spell of cold weather at the start of December 2022 has led to me breathlessly examining data on the thermal performance of the heat pump and the house.

During this period, outside temperatures fell as low as -5 °C and average daily temperatures were below 0 °C. In order to try to keep the internal temperature constant, I studied measurements of internal temperature taken every 2 minutes. The data were pretty stable, only rarely falling outside the bound of 19.5 °C ± 0.5 °C.

But looking in detail, I noticed a curious pattern.

Click on image for a larger version. Two graphs from the period 6th to 18th December 2022. The upper graph shows the air temperature in the middle of the house. At around 01:30 each night the temperature fell sharply. The lower graph shows the rate of change of the air temperature versus time (°C/hour). From this graph it is clear that the rate at which the temperature fell was approximately -0.95 °C/hour.

The upper graph shows sharp falls in temperature at 01:30 each night. These were caused by the heat pump switching to its hot water heating cycle. Prior to this, the heat flowing into the house from the heat pump was more-or-less balanced by the heat flowing out. But when the heat pump switches to heating the domestic hot water, there was no heating from the heat pump and the internal temperature fell.

The lower graph shows the rate of change of the air temperature (°C/hour) versus time over the same period. From this graph it is clear that the rate at which the air temperature fell during the domestic hot water cycles was approximately 0.95 °C/hour.

With a little mathematical analysis (which you can read here if you care) this cooling rate can be combined with knowledge of the heat transfer coefficient (which I estimated a couple of weeks ago) to give estimates of (a) the time constant for the house to cool and (b) the effective heat capacity of the house.

Analysis: Time Constant 

The time constant for the house, is the time for the temperature difference between the inside and outside of the house to fall to ~37% of its initial value after the heating is removed.

The time constant is estimated as (the initial temperature difference) divided by (the initial cooling rate). In this case  the initial temperature difference was typically ~20 °C and the initial cooling rate was 0.95 °C/hour, so the time constant of the house is roughly 21 hours. Sometimes it’s useful to express this in seconds: i.e. 21 x 3,600 = 75,600 seconds.

This suggests that if we switched off all the heating when the house was at 20 °C and the external temperature was 0 °C, the house would cool to roughly 7.4 °C after 21 hours. Intuitively this seems right, but for obvious reasons, I don’t want to actually do this experiment!

Note that this time constant is a characteristic of the house and does not vary with internal or external temperature.

Analysis: Thermal Resistance  

A couple of weeks ago I posted an analysis of the heating power required to heat our house as the ‘temperature demand’  increased as the external temperature fell. The summary graph is shown below.

Click on graph for a larger version. Graph of average heating power (in kW) versus temperature demand (°C) for the first 10 days of December 2022.

From this I concluded that Heat Transfer Coefficient (HTC) for the house was around 165 W/°C.

The inverse of the HTC is known as the thermal resistance that connects the inside of the house to the external environment. So the thermal resistance for the house is ~ 1/165 = 0.00606 °C/W.

Analysis: Heat Capacity  

A general feature of simple thermal analyses is that the time constant, thermal resistance and heat capacity are connected by the formula:

Time constant = Thermal resistance x Heat Capacity

Since we have estimates for the time constant (75,600 s) and the thermal resistance (0.00606 °C/W) we can this estimate the heat capacity of the house as 12,474,000 joules per °C.

This extremely large number is difficult to comprehend, but if we change to units more appropriate for building physics we can express the heat capacity as 3.5 kWh/°C. In other words, if the house were perfectly insulated, it would take 3.5 kWh of heat to raise its temperature by 1 °C.

We can check whether the number makes sense by noticing that the main mass of the house is the bricks from which it is built. A single brick weighs ~3 kg and has a heat capacity of ~2,400 J/°C. So thermally it looks like my house consists of 12,474,000/2,400 ~ 5,200 bricks.

However this estimate is too small. Even considering just the 133 square metres of external walls, if these have the equivalent of 120 bricks per square metre that would come to ~16,000 bricks.

So I think this heat capacity estimate just applies the heat capacity of the first internal parts of the house to cool. This refers to all the surfaces in contact with the air. So I think this is the effective heat capacity for cooling just a degree or two below ambient.

Why did I bother with this? The ‘Setback’ Problem

Friends, sometimes I go upstairs and forget why I went. And sometimes I start analysing things and can’t remember why I started! Fortunately, in this case, I had a really good reason for wanting to know the effective heat capacity of my house.

When I analysed the heat flows previously, I have had to assume that the temperature of the house was stable i.e. that there was a balance between the heat flowing in and the heat flowing out. As long as the temperature of the fabric of the house is stable, then it is neither storing or releasing heat.

However this isn’t enough if we want to understand some very common problems in the thermal physics of houses, such as “the setback problem”: this is the question of whether it’s smart to reduce the temperature of a dwelling (say) overnight and then to re-heat it once again in the morning. To answer this question we need to know about the rate at which a house cools down (it’s time constant) which is equivalent to knowing its heat capacity.

And that is why I have done this prolonged and tedious analysis. The next article will be an analysis of ‘The Setback Problem’. And it will be much more exciting!


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