Introduction

Comfort, in terms of hot/cold, is basically related to the energy balance of ones body. If we loose to much, we feel cold. A simple and well known strategy to overcome this uncomfort is to wear clothes. They provide an insulation barrier that resists heat flow. This phenomena is decribed by Fourier’s law:

\[\phi_{h}= -\lambda \frac{dT}{dx}\]

The law basically states that the heat flux is proportional to thermal conductivity of a material and a function of the temperature gradient with respect to distance. In simple words, weiring thicker (bigger dx) clothes with better insulating properties (smaller \(\lambda\)) restricts heat flow from your body to the surrounding and thus will make you feel warmer. Of course there is other heat transport mechansims, such as radiation, at play as well. To ease my life, I ignore these.

It might sound very silly, but wearing (thicker) clothes is a very cost effective strategy to fight your bodies heat loss. A winter jacket allows us to sustain temperatures below 0 C while still feeling comfortable. As it is typically a weird habbit to wear a winter jacket inside a building, the civilized world has come up with the strategy to heat ones surroundings. When the temperature differential (dT) between the body an its surrounding is smaller, heat flux is decreased.

Generally, we surround ourselfs by air. Air is a gas and therefore has pretty low volumetric heat capacity. The energy required to heat up one cubic metre of air with 1 degree only requires 1.2 kJ, see below.

\[Q = V \times \rho \cdot C_p \cdot \Delta T = 1 \ m^3 \ \cdot 1.2 \ \frac{kg}{m^3} \cdot 1 \frac{kJ}{kg \ K} \cdot 1 \ K = 1.2 \ kJ\]

This demonstrates that the energy required to heat up a reasonably large room (say 100 m\(^3\)) with several degrees is fairly small. However, it turned out that my gas consumption is much much larger than what is suggested by the calculation. For example, at certain day within 40 minutes the temperature rose from 15 to 17 C (dT=2C), while 0.5 m\(^3\) of gas (~15MJ) was consumed. This means that the air loses heat to furnature and its enclosure, most importantly, the walls of my house. And it seems to do that fairly quickly. In other words, heat tranfer is probably limited within the wall, not from the air to the wall. The wall thus act as a thermal buffer. Whether we like or dislike this is a matter of debate.

To take one step back. I am interested in minimizing my energy consumption. One important strategy that I maintain is to only turn on the heating when the house is occupied. A lower average temperature differential (dT) will lead to less heat loss. However, it takes a long time to reach a lower delta T if the walls have a very large buffering capacity. This somewhat long intro led me to simple question of “how quickly does a stone wall suck up heat”?

Let us consider a scenario where the wall has attained a stable temperature of 15C. We then apply a heat source at such that the inner side of the wall is kept at 20C. On the other side off the wall we have a layer of insulation material, see figure 1. To keep things simple, I assume the insulation layer does not transfer any heat (i.e. a perfect insulator). The brickwall typically has a depth of 10 cm (i.e. typicall fire-brick thicknesss is 10cm).

Figure 1: Schematic representation of scenario.

Some boring Math…

In other words, we assume that heat transport is limited within the stone wall. This instationary heat conduction in the wall can be described by:

\[\frac{dT}{dt} = a \frac{d^2T}{dx^2}\]

In which a, the thermal diffusivity (m\(^2\) / s), for a “fire-brick” is equal to:

\[a = \frac{\lambda}{C_p \cdot \rho} = \frac{0.47}{1000 \cdot 2000} = 2.35 \cdot 10^{-7}\]

To activate my brain a bit I have decide to discretize this equation in order to solve it. I used a finite difference method and solved the following equation:

\[T^{t+1}_x = T^{t}_x + \frac{a \Delta t}{ \Delta x^2} (T^{t}_{x-1} - 2T^{t}_{x} + T^{t}_{x+1})\]

In this equation t represents the timestep and x the spatial step. Timesteps of 1 second and positional steps of 1 mm were taken.

Temperature profile

I have run this simulation which is displayed in the video below. It almost takes one hour before the heat front has penetrated through the stone and reached the other side. Until this first hour the insulation layer “does nothing”. Yes, of course it restricts heat flow. The insulation layer restricts the flow for dT of 15C - T\(_{outside}\), but it does not restrict heat flow for >15C within the stone for the first hour.

Temperature profile development in brick wall as function of time, for 1 hour

To save you from watching a very long video I have created a figure with lines of the temperature profile for several hours after the heating has started, see figure 2 below. Once the temperature profile has reached “the other end” of the stone (distance = 100mm), delta T over the stone starts to decrease which further slows down heat transfer. This can be seen from the smaller jumps in temperature from hour to hour (at distance = 100 mm).

Figure 2: Development of temperature profile in a fire-brick.

Energy demand

I am actually more interested in the energy consumption than the development of the temperature profile. Let us first consider the maximum amount of energy required to heat up 1 m\(^2\) of wall from 15 to 20C. The area that needs to be heated is 1 m\(^2\) times the thickness of the stone wall which is 0.1 m (100mm), which gives a volume of 0.1 m\(^3\).

\[Q = \rho \cdot V \cdot C_p \cdot \Delta T = 2000\ kg \cdot 0.1 m^3 \cdot 1000 \frac{J}{kg \ K} \cdot 5 K = 10^6 J = 1 MJ\]

So to completely compensate the temperature differential of 5C, we need 1MJ per m\(^2\) of stone wall. The time to heat up the stone is calculated by dividing the required energy (1MJ per m\(^2\)) divided by the heat flux (W per m\(^2\)). I have calculated the local heat flux as function of time, see Figure 3.

Figure 3: Local heat flux in a fire-brick wall as function of time.

Obviously, initial the heat flux initially is “infinite” and than rapidly decreases. After an hour the heat flux is around 50 W per m\(^2\), still about 2 times the heat flux in case of stationary conduction (24 W / m\(^2\)). The integral (area under the curve) of Figure 3 provides the total amount of energy put in. I have created such graph and displayed it in Figure 4.

After 1 hour some 0.3 MJ per m2 wall is absorbed wich is roughly 30% of the total energy required to heat up the wall (1MJ per m2), but it takes almost 10 hours to “Fill” the wall for 90%.

Figure 4: Energy accumulated in stone wall as function of time. Dotted line represents a fill level of 90%.

To save energy, I am aiming for a lower average delta T. If I want to cycle my heat demand, for example, having a temperature of 20C between 7 and 21h and 15C during the night, the buffering effect of the wall is an important element to consider. Note that “Filling” the wall with energy and “Emptying” it will proceed differently. I will come back to that in the future. Nevertheless, the simulations here show that the walls of an house have an important buffering effect.

Realistic Fill level

The insulation layer in Figure 1 is not a perfect insulator and thus does conduct some heat. The heat conducted by such insulation is much less than conducted by the brick layer. Consequently, the temperature gradient in the insulation is much bigger than in the brick layer, see Figure 5.

Figure 5: Schematic representation of steady state temperature profile in an insulated wall.

I consider an insulation layer of 5cm composed of polystyrene (\(\lambda\) = 0.03 W / m K). The heat flow through such “composite” wall can be calculated according to:

\[\phi = \frac{\Delta T}{\left( \frac{dx_{Wall}}{\lambda_{Wall}} + \frac{dx_{Insulation}}{\lambda_{Insulation}} \right)} = \frac{\Delta T}{\left( \frac{0.1}{0.47} + \frac{0.05}{0.03} \right)} = \frac{\Delta T}{1.9}\]

Almost 90% of the heat transfer resistence can be ascribed to the insulation layer. In steady state, when Tw = 20C, for an outside temperature of 15 or 0C, the middle temperature (Tm) is 19.4C or 17.7C respectively. To find the corresponding “Fill” levels, I have plotted the fill level versus the right-side temperature of the brick wall (Tm).

Figure 6: Fill level versus right hand side temeprature of brick wall (i.e. Tm in Figure 5)

To reach a temperature of 17.7C a fill level of 70% is required, and to reach a temperature of 19.4 a fill level of 91% is required. This corresponds to almost 5 and 10h respectively. A pretty slow process. Once the brick wall is “filled”, heat transfer is fully dominated by the insulation layer.

Conclusion

A lower average delta T will lead to reduced energy consumption. Operating a building/room at a lower temperature when unoccupied is an attractive option. When a building/room is in a “cooled” state (Tsetpoint = low) it first needs te be heated up “prior to use”. This heat input is equal to 1) the energy required to compensate heat losses and 2) the energy needed to heat up the thermal mass to the desired setpoint. I have here investigated the latter, how important is instationary conduction in a brick wall? It turns out that it takes almost an hour before heat has fully penetrated through a stone wall. So, within this first hour, the insulation layer does not restrict heat flow for the “instationary part”. It then takes several more hours to raise the temperature significantly on the right-side of the brick wall (i.e. transistion point of brick wall to insulation layer), only until then the insulation layer starts to “work” for the “instationary” part.

In my personal opinion, one should try to reduce the (“accesible”) thermal mass of a building as much as possible. Especially for buildings/rooms that are not occupied all the time. Or alternatively stated: require the same temperature all the time. This will reduce the heat buffering capacity of a building and thus respond quicker to the setpoint. Removing accesible thermal mass can be done by placing a (thin) insulation layer in front of the brick wall. This will make cycling of the building/room temperature setpoint more efficient.

On the contrary, a potential advantage of a large thermal mass could be to “store” solar energy during the day and subsequent “slow” release during the night.