How Can I Calculate The Heat Gained By A Sunspace (with Given Dimensions And Location Info.)?
How Can I Calculate The Heat Gained By A Sunspace (with Given Dimensions And Location Info.)?
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Passive solar is a method of using the sun's energy to heat a home. It is very popular because the process is free once you create a passive solar house.
Passive solar can be used to heat a home in colder areas, but you must enter it with reasonable expectations. While cold weather is an obstacle, The real problem will be the length of time the sun hits his property. If your home receives only four or five hours of direct sunlight a day, forget it. Never occur enough energy to keep the house warm for sufficient periods of time.
Passive solar design is very popular in warm to moderate, because is more or less a free method to heat a home. Manipulation of the house position and placement of large windows in the south against the wall is typical of strategies to address the issue. Obviously, large windows in cold weather will result in significant heat loss, regardless of the quality of with which they are built. So what can you do?
There are two primary approaches to creating a passive solar design that works in the winter. One is the use of a large Trombe wall and the other is the greenhouse approach. Leta's take a look.
Trombe walls are popular in passive solar design, as effectively convert sunlight to heat and are interesting from an aesthetic standpoint. Typically, a Trombe wall is 8 to 12 feet long, south, facing the wall of a house. In colder areas significantly, the wall will need to be much higher, perhaps the full length of the house, depending on energy analysis and cold weather is expected. A home energy audit is the only to reach a definitive answer.
You are also going to have to incorporate back strategy for heat. When the sun enters the glass plate and heated masonry wall, you risk losing large amounts of it through the surface glass. This means you need to create an air circulation method by which hot air is drawn into a secondary space behind the wall. This simply may be a closed room or space intended for that purpose. The circulation must be done in a timer like solar thermostats used in solar panels hot water. The point is to keep the accumulated heat from escaping back into the environment.
The greenhouse approach simplifies things. The essential idea is to build isolated a greenhouse to collect and store the sun's heat during the day. Often called SunSpace, the greenhouse effect is similar to those used for plants. Even in climates cold, the sun will produce a great amount of heat. Once again, the problem is to keep heat from escaping once they have accumulated. Since the sun has to come through a transparent surface, which inevitably have the problem of heat to escape through it. The best option is to use a timer-controlled to blow air through of the house once certain temperatures are reached. Not very efficient, but has few options.
An alternative to passive solar heating in very cold areas is biomass. Corn burning furnaces are popular. This is a much cheaper solution, such as corn. This biomass energy is also much more reliable and, personally, is the way I would go.
Rick Chapo is with SolarCompanies.com, a directory of solar energy companies. Visit us to read more articles on solar power.
How do I calculate the heat gained by a SunSpace (with dimensions given and the location.)?
I do not need software because they are too complicated. I have no time to learn it. If you can say also to calculate heat gain Trombe walls and any other passive solar'll very happy.
The solar radiation incident on a plane perpendicular to the direction of the Sun is 1366 W / m ^ 2 on a clear day. Some use a constant of 1.251 W / m ^ 2 to take into account the atmospheric attenuation and reflection. This amount varies with latitude, season and time of day. When you begin to work out solar heating for a building, you should also have the orientation of the building into account. The calculations for the roof and every wall will different. You can avoid dealing with the time of day by multiplying the constant changes of 2 / π (≈ 0.6366) and multiplying by 12 hours to obtain a figure of 9.5569 kWh / day / m ^ 2. You can also get an average of seasonal variations of integration for the average incidence angle of your latitude. I I have not made the referral, so can not give a factor. Two things remain to be aware of: atmospheric attenuation for non-peak conditions, and overcast. I "botched" the attenuation factor by using 2 / π again, and you may be able to obtain a "clear day rate" of your local weather office.