Question

Determine the following for a south facing surface at 30� slope in

Answer 1

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