Question

A flat plate solar collector located at Bhopal (23° 15' N, 77° 42' E) is tilted at an angle of 30° with the horizontal and facing due South. Considering hour angle as 45°, the angle made by the beam radiation with normal to the collector plate on June 30, 2023 at 3:00 PM (Local Apparent Time) is _________ degree. (Rounded off to 2 decimal places)

Hint:

To calculate the angle of beam radiation, use the solar angle formula, considering the latitude, solar declination, and hour angle for the time of the day.

Answer 1

The angle between the beam radiation and the normal to the collector plate is given by the formula: \[ \gamma = \cos^{-1} \left( \sin(\phi) \sin(\delta) + \cos(\phi) \cos(\delta) \cos(H) \right) \] where:
\( \phi \) is the latitude of the location (23° 15' N = 23.25°),
\( \delta \) is the solar declination angle,
\( H \) is the hour angle, which is 45° at 3:00 PM (Local Apparent Time).
The solar declination angle on June 30, 2023, is approximately \( \delta = 23.44^\circ \) (since June 21 is around the summer solstice and the declination angle is near its maximum value). Now, substitute the values into the formula: \[ \gamma = \cos^{-1} \left( \sin(23.25^\circ) \sin(23.44^\circ) + \cos(23.25^\circ) \cos(23.44^\circ) \cos(45^\circ) \right) \] First, calculate the terms:
\( \sin(23.25^\circ) \approx 0.396 \)
\( \sin(23.44^\circ) \approx 0.399 \)
\( \cos(23.25^\circ) \approx 0.917 \)
\( \cos(23.44^\circ) \approx 0.917 \)
\( \cos(45^\circ) \approx 0.707 \)
Now, substitute these values into the equation: \[ \gamma = \cos^{-1} \left( (0.396)(0.399) + (0.917)(0.917)(0.707) \right) \] \[ \gamma = \cos^{-1} \left( 0.158 + 0.588 \right) \] \[ \gamma = \cos^{-1} \left( 0.746 \right) \] \[ \gamma \approx 53^\circ \] Thus, the angle made by the beam radiation with the normal to the collector plate is approximately 53°.