SOLAR RADIATION GEOMETRY_PAGE 59_Example 2.4.1

Q) Determine the Local Solar time and declination at a location latitude 23 deg 15 minutes,longitude 77 deg 30 minutes at 12:30 IST on June 19. Equation of time correction is given from standart table or chart = -(1’ 01").
Solution:
Given data:

• Latitude of location \(\phi\) = 23⁰ 15’
• Standard time longitude(STL) = 82⁰ 30’
• Longitude of location (LOL) = 77⁰ 30’
• IST = 12:30(h:m)
• Equation of time of correction(C)= -1’01"
• Day of the year = June 19^th

A) Calculation of Local Solar Time
- we know that Local Solar Time
\[LST=IST-[4*(STL-LOL)]+C\]

#R program for calculation of Local Solar Time(LST)
library(lubridate)

#Conversion of required given data to seconds for ease of calculatons 
IST<-as.numeric(hms("12:30:00"))
STL<-as.numeric(hms("00:82:30"))
LOL<-as.numeric(hms("00:77:30"))
C<-(-as.numeric(hms("00:01:01")))

#Substituting in the data in the formula
LST<-IST-(4*(STL-LOL))+(C)

#Converting the answer from seconds to h:m:s format
td <- seconds_to_period(LST)

#Displayin the output
sprintf('Local Solar Time : %02dhrs :%02dmins :%02dsecs',td@hour, minute(td), second(td))

## [1] "Local Solar Time : 12hrs :08mins :59secs"

B) Calculation of Declinatio angle \(\delta\):
- We know that, \[ Declination angle (\delta)=23.45sin(\frac{360}{365})(284+n)\]

#R program for calculation of Local Solar Time(LST)

date<-("2020-06-19")
n<-yday(date)

declinationangle_delta<-round(23.45* sin((360/365)*(284+n)*(pi/180)),digits=3)
declinationangle_delta

## [1] 23.445