This week I am going to run the numbers on a hypothetical active solar heating system. Active solar differs from passive solar in that it requires mechanical means for moving the heat energy (pumps, fans, etc.). Passive solar can be as simple as opening your curtains to let some sunlight in.
My room gets pretty cold at night and heating is expensive. I could leave the blinds open during the day but then the room gets excessively warm in the afternoon and, since my room has little thermal mass (ability to store heat energy), the warmth quickly dissipates when the sun goes down. To solve this problem I could increase the thermal mass of my room by stacking bricks under my bed or installing a Trombe wall in front of a window. Both of these options are feasible but don’t sound like much fun.
I think I want to build a solar collector outside my window. My solar collector, insulated water tank, would capture the sun’s energy during the day and at night I could pipe that heat into my room with the flip of a switch. For simplicity’s sake my collector consists of a 1x1m double-pane glass window, mounted at a 38-degree angle (equal to my latitude for maximum solar exposure), a water tank that measures 1x1m and is 25cm deep (with a capacity of 250 liters = 250 kg), and walls insulated using three inch thick polyisocyanurate foil-faced panels.
First I want to see how much energy I can get from the sun during the darkest part of the year. In my area the lowest amount of daily solar insolation occurs in December and measures 174 W/m². Since a Watt is defined as Joules per second (J/s), and there are 86,400 seconds in a day, we know that we can expect 15,033,600 Joules per square meter every day (J/m²/day). I chose water as my energy storage material since it has excellent characteristics for use as thermal mass; it has a very high specific heat (4.1813 J/gK), has high density (1g/cm³) and it conducts heat well (but not too well).
By plugging our known values into the specific heat equation (c=J/gK) we can solve for K (the temperature change in Kelvin, 1K=1°C), which is 14.36K per day. So, if we had a hyper-insulated box, the water temperature would rise by 14.36 degrees every day. Our box is not hyper-insulated though, so we need to factor in some heat-loss as well. Last week in AskPablo I explored home insulation and purposefully made the results of my example applicable to this week’s column.
In last week’s column I determined that our insulated box would lose heat at a rate of 18.7 J/s for every 10F difference between the inside and outside of the box. With our 250kg of water this works out to a 1.55°C loss for every 10°F (5.6°C) temperature difference every day. If my spreadsheet simulation is correct, and there is no other loss of heat energy, this closed system could reach water’s boiling point of 100°C (212°F) within 20 days.
Of course we want to use the heat stored in the water for heating my room, so the actual temperature would remain a bit lower. At 60°C, reached in 5 days, we can use roughly 625kJ of energy. This equates to running a 15W heater all night (8 hours). Not bad considering that it’s free but also not that much. This system could be scaled up and made more efficient with more insulation and/or different materials. We can also decrease the amount of thermal mass (water) and increase the amount of energy captured by using parabolic mirrors to focus more energy on our thermal mass. It is really complicated, but terribly interesting to find the exact balance of thermal mass, insulation, and insolation in order to maximize your heat energy output. Systems such as this are installed on rooftops near you and provide radiant floor heating, or hot water, for free year-round. Isn’t thermodynamics fun?!?
Pablo Päster, MBA