Q) Determine the average value of solar radiation on a horizontal surface for June 22, at the latitude of 100N, if constants a and b are given as equal to 0.30 and 0.51 respectively, and the ratio (\(\frac{\bar{n}}{N}\))=0.55.
Solution: Given data:
- Date = June 22nd,2020(year assumed)
- Latitude of location, \(\phi\) = 100N
- Constants a = 0.30 and b = 0.51
- Ratio of avg daily hours to max daily hours, (\(\frac{\bar{n}}{N}\))=0.55
- Average value of solar radiaton, Hav= ? (to be calculated)
We know as suggested by Page(1964),
\[ Hav = H_o *(a+b*(\frac{\bar{n}}{N})) ------> (1)\]
Where Ho = Monthly average insolartion at the top of the atmosphere
\[Ho = \frac{24}{\pi}*Isc*[{1+0.033*cos(\frac{360n}{365})}(cos\phi cos\delta sin\omega_s + \frac{2\pi\omega_S}{360}sin\phi sin\delta] ------> (2)\]
which is not given in the data and needs to be calculated. In the data given, declination angle \(\delta\) and \(\omega_s\) are not available and requires further estimation of the same. ### Calculation Declination angle \(\delta\) \[ Declination angle (\delta)=23.45sin(\frac{360}{365})(284+n) ------> (3)\] where n is the numbe of the day in that year
#R program for calculation of Declination angle (delta)
library(lubridate)
date<-("2020-06-22")
n<-yday(date)
sprintf("The number of the day is %s",n) ## [1] "The number of the day is 174"declinationangle_delta<-round(23.45* sin((360/365)*(284+n)*(pi/180)),digits=2)
d<-declinationangle_delta
sprintf("The declination angle is %s deg", d)## [1] "The declination angle is 23.44 deg"\[ Sunrise angle(\omega_s) = acos(-tan\phi tan\delta) ------> (4)\] “acos means cosinverse”
#R program for calculation of sunrise angle(omega_s)
phi<-10
d<-declinationangle_delta
omega_s<-acos(-1*tan(phi*pi/180)*tan(d*pi/180))
w_s<-omega_s*(180/pi)
sprintf("The sunrise angle is is %02.05s deg",w_s) ## [1] "The sunrise angle is is 94.38 deg"Substituting the values of given data and the values obtained from (3) and (4) in equation (2), we get the average insolation at the top of the atmosphere “Ho”.
#R program for calculation average insolation at the top of the atmosphere (Ho)
Isc<-1353 #Solar constant in watts per sq.mtr
x<-cos(phi*pi/180)*cos(d*pi/180)*sin(w_s*pi/180)
y<-(2*pi*(w_s*pi/180)/360)*(sin(d*pi/180)*sin(phi*pi/180))
z<-(1+(0.003*cos((360*n*pi)/(360*180))))*(x+y)
Ho<-(24/pi)*(Isc)*(z) #Ho in w/sq.m per day in SI Units
sprintf("The average insolation at the top of the atmosphere Ho is %02.05s w/sq.m/day",Ho) ## [1] "The average insolation at the top of the atmosphere Ho is 9304. w/sq.m/day"Substitute the value of Ho in equation(1) to get the average solar radiation on a horizontal surface,Hav:
#R program for calculation average insolation on the horizontal surface (Hav)
a<-0.30 #given
b<-0.51 #given
ratio<-0.55 #given ratio:(n/N)
Hav<-Ho*(a+(b*ratio))
sprintf("The average insolation on the horizontal surface Havg is %02.05s w/sq.m/day",Hav)## [1] "The average insolation on the horizontal surface Havg is 5401. w/sq.m/day"Result: The average insolation on the horizontal surface Havg for June 22nd of the assumed year 2020 is 5401 watts/sq.m/day.