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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter2-Solar Radiation and its Measurement"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.4.1-pg 59"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Ex2.4.1.;Detremine local solar time and declination\n",
"import math\n",
"##The local solar time=IST-4(standard time longitude-longitude of location)+Equation of time correstion\n",
"##IST=12h 30min;for the purpose of calculation we are writing it as a=12h,b=29 min 60sec;\n",
"a=12.;\n",
"b=29.60;\n",
"##(standard time longitude-longitude of location)=82 degree 30min - 77 degree 30min;\n",
"##for the purpose of calculation we are writing it as\n",
"STL3=82.5-72.5;\n",
"##Equation of time correstion: 1 min 01 sec\n",
"##for the purpose of calculation we are writing it as\n",
"c=1.01;\n",
"##The local solar time=IST-4(standard time longitude-longitude of location)+Equation of time correstion\n",
"LST=b-STL3-c;\n",
"print'%s %.2f %s %.2f %s'%(\" The local solar time=\",a,\"\"and \"\",LST,\" in hr.min.sec\");\n",
"##Declination delta can be obtain by cooper's eqn : delta=23.45*sin((360/365)*(284+n))\n",
"n=170.;##(on June 19)\n",
"##let\n",
"a=(360./365.)*(284.+n)\n",
"aa=(a*math.pi)/180.\n",
"##therefore\n",
"delta=23.45*math.sin(aa);\n",
"print'%s %.2f %s'%(\"\\n delta=\",delta,\" degree\");\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" The local solar time= 12.00 18.59 in hr.min.sec\n",
"\n",
" delta= 23.43 degree\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.4.2-pg 59"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"##Ex2.4.2.;Calculate anglr made by beam radiation with the normal to a flat collector.\n",
"gama=0.;##since collector is pointing due south.\n",
"##For this case we have equation : cos_(theta_t)=cos(fie-s)*cos(delta)*cos(w)+sin(fie-s)*sin(delta)\n",
"##with the help of cooper eqn on december 1,\n",
"n=335.;\n",
"##let\n",
"a=(360./365.)*(284.+n);\n",
"aa=(a*math.pi)/180;\n",
"##therefore\n",
"delta=23.45*math.sin(aa);\n",
"print'%s %.2f %s'%(\" delta=\",delta,\" degree\");\n",
"##Hour angle w corresponding to 9.00 hour=45 Degreew\n",
"w=45.;##degree\n",
"##let\n",
"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);\n",
"b=math.sin(delta*math.pi*180**-1)*math.sin(((28.58*math.pi)/180.)-((38.58*math.pi)/180.));\n",
"##therefore\n",
"cos_of_theta_t=a+b;\n",
"theta_t=math.acos(cos_of_theta_t)*57.3;\n",
"print'%s %.2f %s'%(\"\\n theta_t=\",theta_t,\" Degree\");\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" delta= -22.11 degree\n",
"\n",
" theta_t= 44.73 Degree\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 2.7.1-pg 68"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"##Ex2.7.1.;Determine the average values of radiation on a horizontal surface\n",
"import math\n",
"##Declination delta for June 22=23.5 degree, sunrice hour angle ws\n",
"delta=(23.5*math.pi)/180;##unit=radians\n",
"fie=(10*math.pi)/180;##unit=radians\n",
"##Sunrice hour angle ws=acosd(-tan(fie)*tan(delta))\n",
"ws=math.acos(-math.tan(fie)*math.tan(delta));\n",
"print'%s %.2f %s'%(\" Sunrice hour angle ws=\",ws,\" Degree\");\n",
"n=172.;##=days of the year (for June 22)\n",
"##We have the relation for Average insolation at the top of the atmosphere\n",
"##Ho=(24/%pi)*Isc*[{1+0.033*(360*n/365)}*((cos (fie)*cos(delta)*sin(ws))+(2*%pi*ws/360)*sin(fie)*sin(delta))]\n",
"Isc=1353.;##SI unit=W/m^2\n",
"ISC=1165.;##MKS unit=kcal/hr m^2\n",
"##let\n",
"a=24./math.pi;\n",
"aa=(360.*172.)/365.;\n",
"aaa=(aa*math.pi)/180.;\n",
"b=math.cos(aaa);\n",
"bb=0.033*b;\n",
"bbb=1+bb;\n",
"c=(10*math.pi)/180.;\n",
"c1=math.cos(c);\n",
"cc=(23.5*math.pi)/180;\n",
"cc1=math.cos(cc);\n",
"ccc=(94.39*math.pi)/180;\n",
"ccc1=math.sin(ccc);\n",
"c=c1*cc1*ccc1;\n",
"d=(2*math.pi*ws)/360.;\n",
"e=(10*math.pi)/180;\n",
"e1=math.sin(e);\n",
"ee=(23.5*math.pi)/180;\n",
"ee1=math.sin(ee);\n",
"e=e1*ee1;\n",
"##therefoe Ho in SI unit\n",
"Ho=a*Isc*(bbb*(c+(d*e)));\n",
"print'%s %.2f %s'%(\"\\n SI UNIT->Ho=:\",Ho,\" W/m^2\");\n",
"Hac=Ho*(0.3+(0.51*0.55))\n",
"print'%s %.2f %s'%(\"\\n SI UNIT->Hac=\",Hac,\" W/m^2 day\");\n",
"ho=a*ISC*(bbb*(c+(d*e)));\n",
"print'%s %.2f %s'%(\"\\n MKS UNIT->Ho=\",ho,\" kcal/m^2\");\n",
"hac=ho*0.58;\n",
"print'%s %.2f %s'%(\"\\n MKS UNIT->Hac=\",hac,\" kcal/m^2 day\");\n",
"\n",
"##The values are approximately same as in textbook\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" Sunrice hour angle ws= 1.65 Degree\n",
"\n",
" SI UNIT->Ho=: 9025.25 W/m^2\n",
"\n",
" SI UNIT->Hac= 5239.16 W/m^2 day\n",
"\n",
" MKS UNIT->Ho= 7771.19 kcal/m^2\n",
"\n",
" MKS UNIT->Hac= 4507.29 kcal/m^2 day\n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}