# Chapter3-Solar Energy Collectors¶

## Example 3.6.1-pg100¶

In [1]:
##Ex3.6.1.;calculate: solar altitude anglr,Incident angle,Collector efficiency
import math
##Solar declination :delta
n=1
delta=23.45*math.sin((360./365.)*(284.+n));
print'%s %.2f %s'%(" Solar declination delta=",delta," degree");
fie=22.;##degree
##solar hour angle ws=0,(at mean of 11:30 and 12:30)
ws=0.;
##Solar altitude anglr alpha is given by

##alpha=asind(((cos(fie)*cos(delta)*cos(ws))+(sin(fie)*sin(delta)))
##let
a=math.cos((22*math.pi)/180.)*math.cos((-23*math.pi)/180.)*math.cos(0);
b=math.sin((22*math.pi)/180.)*math.sin((-23*math.pi)/180.);
##therefore
sin_alpha=a+b;
print'%s %.2f %s'%("\n sin_aplha=",sin_alpha,"");
alpha=math.asin(sin_alpha);
print'%s %.2f %s'%("\n aplha=",alpha,"degree");
##Incident angle
theta=(180./2.)-alpha;
print'%s %.2f %s'%("\n Incident angle=",theta,"degree");
##Rb is given by
Rb=((math.cos(((22*math.pi)/180.)-(37*math.pi)/180.)*math.cos((-23*math.pi)/180.)*math.cos(0))+(math.sin(((22*math.pi)/180.)-(37*math.pi)/180)* math.sin((-23*math.pi)/180)))/sin_alpha;
print'%s %.2f %s'%("\n Rb=",Rb,"");
##Effective absorptance product is <t.alpha>=t.alpha/ 1-(1-alpha)*pd
pd=0.24;##Diffuse reflectance for two glass covers
##let TA=<t.alpha>
TA=(0.88*0.90)/(1-(1-0.90)*pd);
print'%s %.2f %s'%("\n Effective absorptance product is <t.alpha>=",TA,"");
##Hb=0.5 ly/mm = 0.5 cal/cm^2 * min
Hb=((0.5*10**4)/10**3)*60;##unit=kcal/m^2 hr
print'%s %.2f %s'%("\n Hb=",Hb," kcal/m^2 hr");
Hb=Hb*1.163;##unit=W/m^2 hr;   [since  1 kcal = 1.163 watt]
print'%s %.2f %s'%("\n Hb=",Hb," W/m^2 hr");
##S=Hb*Rb*<t.alpha>
S=Hb*Rb*TA;
print'%s %.2f %s'%("\n S=",S," W/m^2 hr");
s=S/1.163;
print'%s %.2f %s'%("\n S=",s," kcal/m^2 hr");
##Useful gain
##qu=FR(S-UL*(Tfi-Ta))
qu=0.810*(s-(6.80*(60-15)))
print'%s %.2f %s'%("\n qu=",qu," kcal/m^2 hr");
##Qu=FR(S-UL*(Tfi-Ta))
Qu=0.810*(S-(7.88*(60-15)))
print'%s %.2f %s'%("\n qu=",Qu," W/m^2 hr");
##Collection Efficiency  : nc=(qu/(Hb*Rb))*100;
nc=(28.07/(300*Rb))*100.;
print'%s %.2f %s'%("\n Collection Efficiency=",nc," persent");

##values of "sine alpha" in the textbook is taken approximate to the real values

 Solar declination delta= -23.38  degree

sin_aplha= 0.71

aplha= 0.79 degree

Incident angle= 89.21 degree

Rb= 1.40

Effective absorptance product is <t.alpha>= 0.81

Hb= 300.00  kcal/m^2 hr

Hb= 348.90  W/m^2 hr

S= 396.50  W/m^2 hr

S= 340.93  kcal/m^2 hr

qu= 28.29  kcal/m^2 hr

qu= 33.94  W/m^2 hr

Collection Efficiency= 6.68  persent


## Example 3.9.1-pg 119¶

In [2]:
##calculate the useful gain,exit fluid temperature and collection efficiency
##Optical properties are estimated as
p=0.85;
import math
##(T. alpha)=0.77;let A=(T. alpha)
A=0.77
gama=0.94;
Do=0.06;
L=8;##unit=meter,##L=length of concentrator
W=2;##W=width of concentrator in meter
dco=0.09;##dco=diameter of transpaarent cover
Ar= math.pi*Do*L;##Ar=area of the receiver pipe
A_alpha=(W-dco)*L;##aperture area of the concentration
Cp=0.30;##unit=kcal/kg degree calcius
m=400;##unit=kg/hr,m=flow rate
HbRb=600;##unit=kcal/hr m^2
Tfi=150;##degree calcius
T_alpha=25;##degree calcius
##Heat transfer coefficient from fluid inside to surroundings,
Uo=5.2;##unit=kcal/hr-m^2
##Heat transfer coefficient from absorber cover surface to surroundings,
UL=6;##unit=kcal/hr-m^2
F=(Uo/UL);
##Heat removed factor FR is
##FR=((m*Cp)/(Ar*UL))*(1-(%e^-((Ar*UL*F)/(m*Cp))))
##let X=(m*Cp)/(Ar*UL);Y=(%e^-((Ar*UL*F)/(m*Cp)))
X=(m*Cp)/(1.51*UL*0.86);
Y=math.e**(-1/X);
FR=X*0.86*(1-Y);
##Absorbed solar energy is
S=HbRb*p*gama*A;
print'%s %.2f %s %.2f %s'%(" Area of the receiver pipe Ar= ",Ar,"=1.51 m^2"and" \n A_aplha= ",A_alpha," m^2=collection efficiency factor ");
print'%s %.2f %s'%("\n value of F= ",F,"");
print'%s %.2f %s %.2f %s '%("\n Heat removed factor FR=",FR,""and" \n Absorbed solar energy is \n S=",S," kcal/Hr m^2 .....(MKS) ");
##for unit in S.I.   , 1 kcal/Hr m^2 = 1.16298 W/m^2
s= S*1.16298; ##in W/m^2
print'%s %.2f %s'%("\n S=",s," W/m^2.....(SI)");
##the values of F,FR will be same in any unit,since they are factors(dimensionless)
##Useful Gain=Qu=A_alpha*FR*(S-((Ar*UL)/A_alpha)*(Tfi-T_alpha))
##In MKS unit
Qu=A_alpha*FR*(S-((1.51*UL)/A_alpha)*(Tfi-T_alpha))
print'%s %.2f %s'%("\n useful gain in (MKS) Qu=",Qu," kcal/hr");
##IN SI unit
qu=A_alpha*FR*(s-((1.51*6.98)/A_alpha)*(Tfi-T_alpha))##UL=6.98 W/m^2 degree celcius
print'%s %.2f %s'%("\n useful gain in (SI) Qu=",qu," Watt");
##the exit fluid temperature can be obtained from
tci=150;##degree celcius
tco=tci+(Qu/(m*Cp));##from Qu=mCp(tco-tc);  where, tco=collector fluid temp. at outlet,tci=Fluid inlet temp.
n=(Qu/(16*HbRb))*100;##ncollector=Qu/(A_alpha*HbRb)*100;
print'%s %.2f %s %.2f %s'%("\n collector fluid temp. at outlet  tco=",tco," degree celcius"and " \n ncollector = ",n," percent ");

##The values/results/answers is approximate in the text book to the real calculated value

 Area of the receiver pipe Ar=  1.51
A_aplha=  15.28  m^2=collection efficiency factor

value of F=  0.87

Heat removed factor FR= 0.83  369.14  kcal/Hr m^2 .....(MKS)

S= 429.30  W/m^2.....(SI)

useful gain in (MKS) Qu= 3753.64  kcal/hr

useful gain in (SI) Qu= 4365.07  Watt

collector fluid temp. at outlet  tco= 181.28
ncollector =  39.10  percent