SOLAR RADIATION GEOMETRY USING_R Example_2.4.2_Page No_59

Q) Calculate the angle made by beam radiation with the normal to a flat collector on December 1, at 9:00 A.M., solar time for a location at 28⁰ 35’ N. The collector is tilted at an angle of latitude plus 10⁰, with the horizontal and is pointing due south.
Solution :
Given Data:
- Day of the year: December 1^st
- Solar Time : 09:00 A.M(hh:mm)
- Location Latitude, \(\phi\) = 28⁰ 35’ N= 28.(35/60)⁰= 28.58⁰
- Tilt angle(slope, s) = latitude + 10⁰ = 10⁰ + 28⁰ 35’N= 10⁰+28.(35/60)⁰= 38.58⁰
- Collector is pointing due south, \(\gamma\)=0

When the collector is pointing due south,
- we have the equation : \[cos\theta_T = cos(\phi-s)cos\delta cos\omega + sin(\phi-s)sin\delta\]

we need to calculate first declination angle(\(\delta\)) and hour angle(\(\omega\)) we know, \[ Declination angle (\delta)=23.45sin(\frac{360}{365})(284+n)\]

#R program for calculation of Declination angle
library(lubridate)
date<-("2019-12-01")
n<-yday(date)
sprintf("The number of the day is %s",n) 

## [1] "The number of the day is 335"

declinationangle_delta<-round(23.45* sin((360/365)*(284+n)*(pi/180)),digits=2)
sprintf("The declination angle is %s deg respectively", declinationangle_delta)

## [1] "The declination angle is -22.11 deg respectively"

Now, we also have, \[\omega = 15*(12-LST)\]

#R program for calculation of Declination angle
LST<-9
omega_deg<-15*(12-LST)

sprintf("The hour anlge is %s deg", omega_deg)

## [1] "The hour anlge is 45 deg"

Now substitute all values in the equation \(cos\theta_T\)

#R program for calculation of Angle made by the beam radiation to a flat collector theta_T
phi<-28.58
s<-phi+10
d<-declinationangle_delta
w<-omega_deg
theta_T<-acos(cos(phi-s)*cos(d)*cos(w)+sin(phi-s)*sin(d))*(180/pi)
sprintf("Angle made by the beam radiation to a flat collector on 335 day at 9:00AM is %02.05s deg", theta_T)

## [1] "Angle made by the beam radiation to a flat collector on 335 day at 9:00AM is 59.85 deg"