If the solar constant is 1.2kW per meter squared roughly, what would the total energy received per meter squared be at 50 degrees north latitude and 400 feet above ocean level.
Answers:
same
Solar constant is the get-up-and-go received per unit area per second on the surface of the earth ( neglect absorption in earth atmosphere and when the sun is at its aim distance from earth).
It is the same 1.2kW/m^2 by definition of solar constant. It will not change with latitude. A transfer of 400 feet is nothing when compared to earth sun distance of 1.5x10^11m
The height above the ground isn't really a major factor surrounded by the calculation. It's much more important to know the time of the day and the season.
Example: The maximum amount of irradiance would be received at the summer solstice (June 21), when the angle between the sun rays and the vertical would lone be (50-23.5)° = 26.5°, therefore the irradiance would there be sin(26.5°) / tan(26.5°) * 1.2 kW/m^2 = 1.1 kW/m^2
That's the maximum value. On the winter solstice (December 21) though, the best angle you seize is (50+23.5)° = 73.5°
Therefore, the maximum irradiance at that time would only be sin(73.5°) / tan(73.5°) * 1.2 kW/m^2 = 0.34 kW/m^2.
Of course, this applies only to surfaces that are lying flat on the ground. If you would e.g. build a solar array that moves with the sun's direction, you'd be capable of get much higher irradiance values. Source(s): own deduction
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Answers:
same
Solar constant is the get-up-and-go received per unit area per second on the surface of the earth ( neglect absorption in earth atmosphere and when the sun is at its aim distance from earth).
It is the same 1.2kW/m^2 by definition of solar constant. It will not change with latitude. A transfer of 400 feet is nothing when compared to earth sun distance of 1.5x10^11m
The height above the ground isn't really a major factor surrounded by the calculation. It's much more important to know the time of the day and the season.
Example: The maximum amount of irradiance would be received at the summer solstice (June 21), when the angle between the sun rays and the vertical would lone be (50-23.5)° = 26.5°, therefore the irradiance would there be sin(26.5°) / tan(26.5°) * 1.2 kW/m^2 = 1.1 kW/m^2
That's the maximum value. On the winter solstice (December 21) though, the best angle you seize is (50+23.5)° = 73.5°
Therefore, the maximum irradiance at that time would only be sin(73.5°) / tan(73.5°) * 1.2 kW/m^2 = 0.34 kW/m^2.
Of course, this applies only to surfaces that are lying flat on the ground. If you would e.g. build a solar array that moves with the sun's direction, you'd be capable of get much higher irradiance values. Source(s): own deduction
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I need to know how long, if even possible, would the solar panel charge the battery. If you need any other information ask. Battery: 3.7 V, 1219mAh Solar Panel: 4.2 V, 22 mA to be simply, it's need at least : 1219/22=55 hours to full...
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I have in mind a small solar panel that could be plugged into a wall socket. Do I need a converter or some generous of interference between the two power sources so there aren't any unpleasant explosions? I don't know much about this sort of thing so please...