|
Earth, our small planet, gets some of this energy produced by the sun. We receive the equivalent of some 1000 watts/square meter. A one square meter surface placed at an angle of 90 degree to the sun will get two kinds of energy: Direct solar radiation and diffused radiation. The former comes from the sun where as the latter is the reflexion from various objects. The addition of both types of radiation is called global radiation. This global radiation may reach some 1400 watts per square meter in some instances. |
|
Solar captors
There are two types of solar captors: The one that transforms solar energy into electricity and the one that transforms it into heat. The former is very expensive and not very efficient. The solar captors presented in the DVD transform solar radiation into heat. This heat will then be used to heat up your swimming pool.
Like others, you have noticed that any black object exposed to the sun's rays becomes warm. The explanation is quite simple: Any dark object transforms solar energy into heat which is then dissipated into the atmosphere. A black surface is therefore a simple solar captor.
A completely black surface exposed to the sun's rays will transform this solar radiation into calorific energy, in other words heating energy. As already mentioned, the energy is about 1000 watts/m² during a nice sunny day.
|
|
The idea is to recuperate this enormous amount of energy coming from the sun and use it to heat up the water of a swimming pool using a series of solar panels which you will build yourself following the instructions on the DVD.
|
|
 |
|
|
|
Thermodynamic
All matter is made of atoms and molecules in constant movement. The movement of these particles generates an internal energy called thermal energy. Temperature measurement informs us of the intensity of these movements. The higher the temperature, the higher the level of agitation.
Heat is defined as being a transfer of thermal energy. When
something is heated, a transfer of energy occurs between the heating body and that which is heated. Both thermal energy and heat are expressed in joules and the symbol is Q.
All bodies do not absorb heat in the same manner. Depending on their nature and weight, they heat up and cool down at different speeds. For that reason, the notion of thermal mass has been introduced. The thermal mass of any substance is the amount of heat needed to bring an increase of 1 Kelvin to a kilogram of this substance. This thermal mass is represented by the letter C. For instance, the thermal heat of water is 4186.8. This means that 4186.8 joules of energy must be supplied to 1 kg of water to increase the temperature of 1 Celsius or 1 Kelvin.
|
|

Thermal mass of a few substances
|
|
Notice that water as such has a very high thermal mass. This is the reason why it is so difficult to heat up the water of a swimming pool. We see how important it is to have an efficient heating system.
When a hot material is put in contact with a cold one, there is a spontaneous transfer of heat from the hot to the cold material until both materials reach the same temperature. We then say that the system is in thermal equilibrium. This is the basic principle surrounding the solar heating system presented in this document. The correlation between heat and temperature is currently expressed by the following mathematical equation: |
|
 |
|
Let us calculate the heat needed to bring 1 liter of water to the boiling point (100°C), assuming that the atmospheric pressure is constant:
|
|
Q = mcDt,
Q= Amount of heat in Joules
m= 1 kg (because 1 liter of water = 1kg)
c= 4186j/kg•C
Δt= Difference in temperature = 100°C - 20°C = 80°C
Q= 1kg X 4186j/kg.C X 80°C
Q= 334944 joules or 335 Kj
|
|
|
We could have solved this problem differently if the notion of Calorie had been used.
A calorie is defined as the amount of energy needed to bring a 1 degree-Celsius increase in temperature to 1cm³ (1 ml) of water. Therefore:
|
1 calorie = 4.1868 joules |
We could have solved this problem in a different way:
Since it takes 1 calorie to increase by 1 degree 1 cm3 of water, 1000 calories are needed to increase by 1 degree 1 kg (1000 cm³) of water. Therefore 80,000 calories (80 x 1000) are required to increase by 80°C 1 kg of water. Since 1 calorie = 4.1868 joules, the result is 80,000 calories X 4.1868= 334,944 joules or 335 kj.
|
|
Miscellaneous technical information

|
|
Here is another example: A coffee machine contains 1 liter of water. Knowing that the electrical resistance is 1500 watts, how long will it take to bring the water from 15° C to 80° C?
-
1500 watts = 1500 j/s = 358 calories/second (1500/4.186)
-
Difference in temperature (Δt) = 80° - 15° = 65° C
-
1 Liter of water = 1000ml = 1000 cm3
-
To increase 1000 ml by 1° C requires 1000 calories.
-
To increase 1000 ml by 65° C requires 65000 calories.
-
65,000 calories/358 calories per second = 181 seconds = 3 minutes
|
To summarize, we must remember that:
· On the earth surface, the power of solar radiations is a little over 1000 watts per square meter;
· A black object converts the solar radiation into heat;
· Water is difficult to heat up;
· Finally, to heat up the water of a swimming pool, a large quantity of heat is needed.
|
|
 |
|
|