11
Solar Heat Gain
Windows can bring in significant energy when the sun shines through them. That energy warms the surfaces the light falls upon and eventually, by convection, warms the whole living space. (Convection is the transfer of heat energy by motion of a fluid like air.) We can see these effects from a sample for the living room in my condo. The figure is for a Saturday, so the programmable thermostat kept the heat off all night, then turned it on all day, moving the setpoint from 12.5 up to 21C. The heaters came on in the early morning before sunrise, and by mid-morning the sun shining in the south facing windows took over, raising the temperature to a peak in the early afternoon, well beyond the 21C setpoint. Although I’m not sure the application is accurate for quantity, the energy histogram shows that the heat was only on from about 6:30 to 10:00.
Non-Equilibrium Models
If the solar heat gain is enough to increase the temperature in the living space there must be more total energy coming in $(q_i)$ than is going out $(q_o)$ and we are no longer in equilibrium with our surroundings. In concept this can be expressed simply:
\begin{equation}
m C_p\frac{dT_i}{dt} = q_i – q_o
\end{equation}
where the mass $m$ and specific heat $C_p$ include everything in the living space: air, furniture, construction materials, etc. The derivative is the rate of change of the indoor temperature, which will go up slowly for a large thermal mass $mC_p$ or more quickly if the living space is empty except for the air.
In practice the system is much more complicated because different parts of the living space will have different temperatures, different specific heats, and will warm and cool at different rates. You can design with massive elements sitting in the sunshine to absorb the energy and release most of it much later to smooth out the daily variation, but that subtlety is for more complex models.
Cooling at Night with the Heat Off
We could make a guess at an overall $mC_p$ by watching how quickly the space cools.
\begin{align}
m C_p&\approx (q_i – q_o)/\frac{dT_i}{dt}\\
&\approx \frac{\Delta t}{\Delta T_i}(q_i – q_o)
\end{align}
The design load $q_o$ for our condo was 3255W at a temperature differential of 45C, so we would expect it to be about 2000W in the early morning hours of January 26, when the temperature difference between indoors and outdoors was only about 30C. (All of our losses were proportional to $(T_i-T_o)$ totalling about $72\rm\;W$ for every degree Celcius of difference.) By eye, the temperature drop $\Delta T_i$ is about $1\rm\;C$ over $\Delta t$ of six hours ($21600\rm\;s$). The energy input $q_i$ would be close to zero if the condo was empty, with the power off, but with two people and the usual collection of computers, appliances, etc. it was probably about 500W, which is close to our average summer electrical consumption.
\begin{align}
m C_p&\approx \frac{\Delta t}{\Delta T_i}(q_i – q_o)\\
&= \frac{21600}{-1}(500 – 2000)\\
&\approx