How much heat can I get from the sun?

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To estimate the heat absorbed through a south-facing window in Boulder, Colorado, on a sunny day in early April, one can start with the average solar radiation intensity of about 1 kW/m². Factors such as the angle of the sun, the window's glass properties, and the roof's overhang will significantly affect the actual heat gain. Calculating the total heat involves considering the window area, the duration of sunlight, and the specific absorption and reflection characteristics of the materials used. For a rough estimate, one can convert the solar intensity to BTUs over a few hours of full sunlight. This project is complex and may require simplifications or statistical methods for a high school level assignment.
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Hi,
I'm a student doing a project that involves building a model house with passive solar heating system. I'm curious if there is a way to approximate the amount of heat I will absorb through a south facing window on a sunny day, in early April. I realize this depends on many variables, but I'm just looking for a ball park figure (10 BTUs, a billion BTUs?)

I'm in Boulder Colorado (North Latitude 40 degrees). The average high is about 55F.

Anyone have any thoughts? Thanks!
 
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zzinfinity said:
Hi,
I'm a student doing a project that involves building a model house with passive solar heating system. I'm curious if there is a way to approximate the amount of heat I will absorb through a south facing window on a sunny day, in early April. I realize this depends on many variables, but I'm just looking for a ball park figure (10 BTUs, a billion BTUs?)

I'm in Boulder Colorado (North Latitude 40 degrees). The average high is about 55F.

Anyone have any thoughts? Thanks!

Here's a start.
http://en.wikipedia.org/wiki/Insolation
 
A airconditioning heat load program would be great otherwise you have to get charts for your Latitude showing the radiation intensities on a south wall, Note the amount of window in direct sunlight is affected by the overhang of the roof (which you would have to calculate, yes for each hour) The radiation intensities would be given in per square meter so the final result depends on how many square meters the wall is assuming nothing shades it. You would have to do the calculation for each hour for the differing intensities and work out the heat for each hour (Watts multiplied by seconds (3600) If I remember right there is a certain amount reflected by the glass and some absorbed and lost back to outside. (I think it depends on what type of glass)
 
Bright sunlight has an intensity of roughly 1kW/m^2. Clouds (->weather data), other sources of shadows (->architecture) and surfaces not perpendicular to the sunlight (->architecture and latitude) will reduce this number for your house. In addition, the window will reflect some fraction of the sunlight (->glass properties).

For a ballpark figure, take a few hours of full sunlight and convert 1kW/m^2 to all sorts of strange unit systems if you like.
 
You need to know a few things:
1 - The relative angle of the sun and roof. This is insolation.

2 - The surface properties of the roof. Specifically the absorbtivity and reflectivity. This defines how the radiation behaves at the surface.

3 - The roof wall properties. This includes thickness and the conduction coefficient which defines how the heat propagates through the roof.

4 - Air properties of the house interior. This includes initial temperature and depending how complicated you want to get, conduction and convection coefficients.

You picked a complicated project. If this is for high school I recommend something simpler.


If possible use statistical methods.
 
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