Answer :
Answer:
z=60.32°, i=0.32°, Beam Radiation = 1097.2 W/m², Id = 94.2 W/m², Ir=14.1W/m², total radiation = 1205.4 W/m², Local time=1:21PM
Explanation:
A. Zenith Angle:
As we know that,
Zenith angle=z=90⁰-α=L(latitude)=29.68⁰
Another way to do it is to find α first,
At solar time hour angle is 0⁰. So, solar altitude becomes equal to latitude which could be written as
sinα=cosL
α=sin⁻¹(cosL)=sin⁻¹(cos29.68⁰)=60.32°
B. Angle of incidence:
angle of incidence= cosi=sin(α+β)=sin(60.32°+32°)=sin92.32°
i=cos⁻¹(sin92.32°)=0.32°
C. Beam Radiation:
First we need to calculate extra terrestrial radiations
Iext.=1353[1+0.034cos(360n/365)]
where n=264
=1345 W/m²
Now,
Beam Radiation=CIext⁻ⁿ
where n=0.1/sin60.32°
Beam Radiation = 1097.2 W/m²
D. Diffude Radiation:
difuse radiation = Id = 0.0921ₙcos²(β/2)
where β=30°
Id = 94.2 W/m²
E. Reflected Radiations:
Ir=pIn(sinα+0.092)sin²(β/2)
= (0.2)(1097.1)(sin60.32+0.092)sin²(30/2)
= 14.1W/m²
F. Total Radiation:
total radiation = beam radiation + diffuse radiation + reflected raddiation
= 1205.4 W/m²
G. Local Time:
LST= ST-ET-(lₓ-l(local))4min/₀
= 12:00-7.9min-(75°-82.27°)4min/₀
=12:21PM
Local time
LDT=LST+=12:21+1:00=1:21PM