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| -rwxr-xr-x | sample_notebooks/amit kumarsaini/Chapter1.ipynb | 179 | ||||
| -rwxr-xr-x | sample_notebooks/amit kumarsaini/Chapter2.ipynb | 202 |
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diff --git a/sample_notebooks/amit kumarsaini/Chapter1.ipynb b/sample_notebooks/amit kumarsaini/Chapter1.ipynb new file mode 100755 index 00000000..a1c4f794 --- /dev/null +++ b/sample_notebooks/amit kumarsaini/Chapter1.ipynb @@ -0,0 +1,179 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e330c51bf6e610e7419d5a8924484e7b299e4592ed588610e8ac90d2d1956ef9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1-Fundamental Concepts"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Chapter 1, Example 1.1, Page 21\n",
+ "import math\n",
+ "##Find Atomic weight of Boron\n",
+ "I10 = 0.199 ## Isotopic abundance of B10 (Value used in question is wrong)\n",
+ "A10 = 10.012937 ##Atomic weight of B10\n",
+ "I11 = 0.801 ## Isotopic abundance of B11\n",
+ "A11 = 11.009306 ##Atomic weight of B11\n",
+ "##Calculation\n",
+ "W = (I10*A10)+(I11*A11)\n",
+ "print'%s %.2f %s'%(\"The atomic weight of Boron = \",W,\"\");\n",
+ "\n",
+ "##Answers may vary due to round off error\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The atomic weight of Boron = 10.81 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Chapter 1, Example 1.2, Page 22\n",
+ "import math\n",
+ "##Find number of 10B molecules in 5g of Boron\n",
+ "m = 5. ##g\n",
+ "Na = 0.6022*10**24 ##atoms/mol\n",
+ "AB = 10.811 ##Atomic weight of 10B , g/mol\n",
+ "NB = (m*Na)/(AB)\n",
+ "print'%s %.2e %s'%(\"The number of Boron atoms = \",NB,\" atoms\");\n",
+ "\n",
+ "##Answers may vary due to round off error\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number of Boron atoms = 2.79e+23 atoms\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Chapter 1, Example 1.3, Page 22\n",
+ "import math\n",
+ "##Estimate the mass on an atom of U 238. From Eq. (1.3)\n",
+ "##Calculating the approximate weight\n",
+ "Mapprox = 238./(6.022*10**23)\n",
+ "##Calculating the precise weight\n",
+ "M = 238.050782/(6.022142*10**23)\n",
+ "print'%s %.2e %s'%(\"The approximate mass on an atom of U 238 = \",Mapprox,\" g/atom\");\n",
+ "print'%s %.2e %s'%(\"\\n The precise mass on an atom of U 238 = \",M,\" g/atom\")\n",
+ "print(\"Varies by a negligible error\")\n",
+ "##Answers may vary due to round off error\n",
+ "\n",
+ " \n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The approximate mass on an atom of U 238 = 3.95e-22 g/atom\n",
+ "\n",
+ " The precise mass on an atom of U 238 = 3.95e-22 g/atom\n",
+ "Varies by a negligible error\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Chapter 1, Example 1.4, Page 23\n",
+ "import math\n",
+ "##Density of Hydrogen atom in water\n",
+ "p = 1. ## density of water in g cm^-3\n",
+ "Na = 6.022*10**23 ## molucules/mol\n",
+ "A = 18. ## atomic weight of water in g/mol\n",
+ "N = (p*Na)/A\n",
+ "NH = 2.*N\n",
+ "print'%s %.2e %s'%(\"The density of water = \",N,\" molecules/cm3\");\n",
+ "print'%s %.2e %s'%(\"\\n The density of hydrogen atoms = \",NH,\" atoms/cm3\");\n",
+ "##Answers may vary due to round off error\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The density of water = 3.35e+22 molecules/cm3\n",
+ "\n",
+ " The density of hydrogen atoms = 6.69e+22 atoms/cm3\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/sample_notebooks/amit kumarsaini/Chapter2.ipynb b/sample_notebooks/amit kumarsaini/Chapter2.ipynb new file mode 100755 index 00000000..9a5cce62 --- /dev/null +++ b/sample_notebooks/amit kumarsaini/Chapter2.ipynb @@ -0,0 +1,202 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ddaa68fa50b83699f8a6f35deb059bff01c78a46802bd57850513a730bade48f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Solar Radiation and its Measurement"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 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 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 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": {}
+ }
+ ]
+}
\ No newline at end of file |