It makes no sense to use the noon position of the sun to predict the full-day solar energy received by a collector. The middle of June is very close to summer solstice time in the northern hemisphere. The latitude of Gothenburg is 58 degrees N. At noon on the solstice, the sun will be 90-58+23 = 55 degrees above the horizon. For optimum collection at that time, the solar panels should be tilted to the south by about 35 degrees. Then the sun's rays will be perpendicular to the panels at noon.
That will not be the optimum panel tilt angle for all times of day and days of the year, however.
There is a good discussion of this subject at
http://www.macslab.com/optsolar.html
For a single tilt that will not be changed during the year, the best angle is the latitude, times 0.75, plus 3 degrees. That would be about 46 degrees in Gothenburg
This refers to passive solar collector, where solar radiation is used to heat water. It is therefore not on the solar cells.
What characteristics should solar collectors have, to function in an optimal way?
Why should these properties be fulfilled?
If there is a water flow of 500 g / minute to a solar panel on one m2, how much water is heated up under optimal conditions during the passage of the collector? Expect that the collector is in Gothenburg and the calculation is done for one day in mid-June (noon)
3 answers
how should i consider the flow of water
500 g / minute and the area of the collector
500 g / minute and the area of the collector
500 g/minute is a very high flow rate for a home solar water heater. How much do you need to heat the water in a single pass?
For a single pass through the solar collector,
(mass flow rate)*(specific heat of water)*(delta T) = (Solar heating per area)*(collector area) - convection losses.
Convetion losses become high when you try to heat the water more than 10 degrees C above the ambient temperature.
The solar heating per area will be about 1000 W/m^2 times a cosine factor, if the sun is out.
For a single pass through the solar collector,
(mass flow rate)*(specific heat of water)*(delta T) = (Solar heating per area)*(collector area) - convection losses.
Convetion losses become high when you try to heat the water more than 10 degrees C above the ambient temperature.
The solar heating per area will be about 1000 W/m^2 times a cosine factor, if the sun is out.