Posts Tagged ‘Thermal Model’

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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