# Chapter2-Solar Radiation and its Measurement¶

## Example 2.4.1-pg 59¶

In [3]:
##Ex2.4.1.;Detremine local solar time and declination
import math
##The local solar time=IST-4(standard time longitude-longitude of location)+Equation of time correstion
##IST=12h 30min;for the purpose of calculation we are writing it as a=12h,b=29 min 60sec;
a=12.;
b=29.60;
##(standard time longitude-longitude of location)=82 degree 30min - 77 degree 30min;
##for the purpose of calculation we are writing it as
STL3=82.5-72.5;
##Equation of time correstion: 1 min 01 sec
##for the purpose of calculation we are writing it as
c=1.01;
##The local solar time=IST-4(standard time longitude-longitude of location)+Equation of time correstion
LST=b-STL3-c;
print'%s %.2f %s %.2f %s'%(" The local solar time=",a,""and "",LST," in hr.min.sec");
##Declination delta can be obtain by cooper's eqn : delta=23.45*sin((360/365)*(284+n))
n=170.;##(on June 19)
##let
a=(360./365.)*(284.+n)
aa=(a*math.pi)/180.
##therefore
delta=23.45*math.sin(aa);
print'%s %.2f %s'%("\n delta=",delta," degree");

 The local solar time= 12.00  18.59  in hr.min.sec

delta= 23.43  degree


## Example 2.4.2-pg 59¶

In [2]:
import math
##Ex2.4.2.;Calculate anglr made by beam radiation with the normal to a flat collector.
gama=0.;##since collector is pointing due south.
##For this case we have equation : cos_(theta_t)=cos(fie-s)*cos(delta)*cos(w)+sin(fie-s)*sin(delta)
##with the help of cooper eqn on december 1,
n=335.;
##let
a=(360./365.)*(284.+n);
aa=(a*math.pi)/180;
##therefore
delta=23.45*math.sin(aa);
print'%s %.2f %s'%(" delta=",delta," degree");
##Hour angle w corresponding to 9.00 hour=45 Degreew
w=45.;##degree
##let
a=math.cos(((28.58*math.pi)/180.)-((38.58*math.pi)/180.))*math.cos(delta*math.pi*180**-1)*math.cos(w*math.pi*180**-1);
b=math.sin(delta*math.pi*180**-1)*math.sin(((28.58*math.pi)/180.)-((38.58*math.pi)/180.));
##therefore
cos_of_theta_t=a+b;
theta_t=math.acos(cos_of_theta_t)*57.3;
print'%s %.2f %s'%("\n theta_t=",theta_t," Degree");

 delta= -22.11  degree

theta_t= 44.73  Degree


## Example 2.7.1-pg 68¶

In [1]:
##Ex2.7.1.;Determine the average values of radiation on a horizontal surface
import math
##Declination delta for June 22=23.5 degree, sunrice hour angle ws
##Sunrice hour angle ws=acosd(-tan(fie)*tan(delta))
ws=math.acos(-math.tan(fie)*math.tan(delta));
print'%s %.2f %s'%(" Sunrice hour angle ws=",ws," Degree");
n=172.;##=days of the year (for June 22)
##We have the relation for Average insolation at the top of the atmosphere
##Ho=(24/%pi)*Isc*[{1+0.033*(360*n/365)}*((cos (fie)*cos(delta)*sin(ws))+(2*%pi*ws/360)*sin(fie)*sin(delta))]
Isc=1353.;##SI unit=W/m^2
ISC=1165.;##MKS unit=kcal/hr m^2
##let
a=24./math.pi;
aa=(360.*172.)/365.;
aaa=(aa*math.pi)/180.;
b=math.cos(aaa);
bb=0.033*b;
bbb=1+bb;
c=(10*math.pi)/180.;
c1=math.cos(c);
cc=(23.5*math.pi)/180;
cc1=math.cos(cc);
ccc=(94.39*math.pi)/180;
ccc1=math.sin(ccc);
c=c1*cc1*ccc1;
d=(2*math.pi*ws)/360.;
e=(10*math.pi)/180;
e1=math.sin(e);
ee=(23.5*math.pi)/180;
ee1=math.sin(ee);
e=e1*ee1;
##therefoe Ho in SI unit
Ho=a*Isc*(bbb*(c+(d*e)));
print'%s %.2f %s'%("\n SI UNIT->Ho=:",Ho," W/m^2");
Hac=Ho*(0.3+(0.51*0.55))
print'%s %.2f %s'%("\n SI UNIT->Hac=",Hac," W/m^2 day");
ho=a*ISC*(bbb*(c+(d*e)));
print'%s %.2f %s'%("\n MKS UNIT->Ho=",ho," kcal/m^2");
hac=ho*0.58;
print'%s %.2f %s'%("\n MKS UNIT->Hac=",hac," kcal/m^2 day");

##The values are approximately same as in textbook

 Sunrice hour angle ws= 1.65  Degree

SI UNIT->Ho=: 9025.25  W/m^2

SI UNIT->Hac= 5239.16  W/m^2 day

MKS UNIT->Ho= 7771.19  kcal/m^2

MKS UNIT->Hac= 4507.29  kcal/m^2 day