Calculate the angle of incident of beam radiation or surface located at Abuja at 130solar time on march 16,1996, if the surface is tilted 45 degree Celsius from the horizontal and pointed 30 degree west of south. Take one degree =60mins. 

To calculate the angle of incidence of beam radiation on a surface located in Abuja at a specific time (13:00 solar time on March 16, 1996), we need to account for several factors:

1. **Latitude of the location** (Abuja):  
   Abuja, Nigeria, is located at approximately 9.08° N latitude.

2. **Surface tilt angle**:  
   The surface is tilted at \(45^\circ\) from the horizontal.

3. **Surface azimuth angle**:  
   The surface is pointed \(30^\circ\) west of south, meaning the azimuth angle is \(180^\circ + 30^\circ = 210^\circ\).

4. **Solar declination angle (\(\delta\))**:  
   The declination angle can be calculated using the following formula for March 16, 1996:
   \[
   \delta = 23.45^\circ \times \sin\left( \frac{360^\circ}{365} \times (n - 81) \right)
   \]
   where \(n\) is the day of the year (March 16 is the 76th day of the year).  
   Substituting \(n = 76\):
   \[
   \delta = 23.45^\circ \times \sin\left( \frac{360^\circ}{365} \times (76 - 81) \right) = 23.45^\circ \times \sin\left( \frac{-1800^\circ}{365} \right)
   \]

5. **Solar hour angle (H)**:  
   Solar time is 13:00 (1:00 PM), so the solar hour angle is calculated as:
   \[
   H = (Solar\ time - 12) \times 15^\circ = (13 - 12) \times 15^\circ = 15^\circ
   \]

6. **Solar zenith angle (\(\theta_z\))**:  
   The solar zenith angle is given by:
   \[
   \cos(\theta_z) = \sin(\phi) \sin(\delta) + \cos(\phi) \cos(\delta) \cos(H)
   \]
   where \(\phi\) is the latitude (9.08° N), \(\delta\) is the declination, and \(H\) is the solar hour angle. Substituting the values will give the zenith angle.

7. **Angle of incidence (\(\theta\))**:  
   The angle of incidence is given by:
   \[
   \cos(\theta) = \cos(\theta_z) \cos(\beta) + \sin(\theta_z) \sin(\beta) \cos(\gamma)
   \]
   where:
   - \(\beta\) is the surface tilt angle (\(45^\circ\)),
   - \(\gamma\) is the angle between the surface normal and the sun's direction (\(30^\circ\)).

Let me calculate these values step by step for the final angle of incidence.

Here are the results of the calculations:

1. **Solar declination angle (\(\delta\))**: \(-2.02^\circ\)  
   This indicates that the sun is slightly south of the celestial equator on March 16.

2. **Solar zenith angle (\(\theta_z\))**: \(18.62^\circ\)  
   This is the angle between the sun's rays and the vertical direction at the given time and location.

3. **Angle of incidence (\(\theta\))**: \(30.05^\circ\)  
   This is the angle at which the solar radiation hits the tilted surface.

If you'd like more details on any part of the process, feel free to ask!

Here are five related questions you might find helpful:
1. How does solar declination vary throughout the year?
2. What is the effect of tilt angle on solar energy absorption?
3. How does the solar hour angle change throughout the day?
4. What is the relationship between solar zenith and azimuth angles?
5. How do location and latitude affect solar radiation?

**Tip**: Adjusting the tilt angle of solar panels to match the latitude can optimize energy absorption throughout the year.

Calculate the angle of incident of beam radiation or surface located at Abuja at 130 solar time on March 16, 1996, if the surface is tilted 45 degrees from the horizontal and pointed 30 degrees west of south. Take one degree = 60 mins.

Learn how to calculate the angle of incidence of beam radiation on a surface in Abuja, Nigeria, on March 16, 1996. This problem involves solar geometry, trigonometric formulas for solar zenith and hour angles, and geometric methods to find the angle of incidence on a tilted surface. Suitable for advanced high school students (Grades 11-12).

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