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Diffstat (limited to 'Materials_Science')
| -rwxr-xr-x | Materials_Science/Chapter02.ipynb | 328 | ||||
| -rwxr-xr-x | Materials_Science/Chapter03.ipynb | 1338 | ||||
| -rwxr-xr-x | Materials_Science/Chapter05.ipynb | 666 | ||||
| -rwxr-xr-x | Materials_Science/Chapter07.ipynb | 724 | ||||
| -rwxr-xr-x | Materials_Science/Chapter08.ipynb | 179 | ||||
| -rwxr-xr-x | Materials_Science/Chapter09.ipynb | 200 | ||||
| -rwxr-xr-x | Materials_Science/Chapter15.ipynb | 61 | ||||
| -rwxr-xr-x | Materials_Science/Chapter16.ipynb | 275 | ||||
| -rwxr-xr-x | Materials_Science/Chapter17.ipynb | 60 | ||||
| -rwxr-xr-x | Materials_Science/Chapter18.ipynb | 251 |
10 files changed, 0 insertions, 4082 deletions
diff --git a/Materials_Science/Chapter02.ipynb b/Materials_Science/Chapter02.ipynb deleted file mode 100755 index e709048c..00000000 --- a/Materials_Science/Chapter02.ipynb +++ /dev/null @@ -1,328 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:dd88e5202fe8cb4f62cee212896e4620a6a485517d90fb3c0293ab4baa871eef"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter02: Structure of atoms"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.1:pg-12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 2.1 : radius of the first bohr\"s orbit \n",
- " \n",
- "#given data :\n",
- "\n",
- "ep=8.854*10**-12;#\n",
- "h=6.626*10**-34;#\n",
- "m=9.1*10**-31;#in Kg\n",
- "e=1.602*10**-19;#\n",
- "r1=((ep*(h**2))/((math.pi*m*(e**2))));#\n",
- "print round(r1*10**10,2),\"is radius,r1(in angstrom) \"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.53 is radius,r1(in angstrom) \n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.2:pg-12"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Example 2.2 : radius of the second bohr\"s orbit \n",
- " \n",
- "#given data :\n",
- "\n",
- "r1_h=0.529; # radius for hydrozen atom in Angstrum\n",
- "n1=1;# for the first bohr's orbit of electron in hydrozen atom\n",
- "Z1=1; # for the first bohr's orbit of electron in hydrozen atom\n",
- "k=(r1_h*Z1)/n1**2; # where k is constant\n",
- "n2=2; # for the second bohr orbit\n",
- "Z2=2; #for the second bohr orbit\n",
- "r2_he=k*(n2**2/Z2);\n",
- "print r2_he,\" is radius of the second bohr orbit,r2 in (Angstrom) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.058 is radius of the second bohr orbit,r2 in (Angstrom) \n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.3:pg-13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 2.3: to prove\n",
- " \n",
- "Z=1;#assume\n",
- "n1=1;#orbit 1\n",
- "n2=2;#orbit 2\n",
- "n3=3;#orbit 3\n",
- "e1=((-13.6*Z)/(n1**2));#energy for the first orbit\n",
- "e2=((-13.6*Z)/(n2**2));#energy for the second orbit\n",
- "e3=((-13.6*Z)/(n3**2));#energy for the third orbit\n",
- "e31=e3-e1;#energy emitted by an electron jumping from orbit nuber 3 to orbit nimber 1\n",
- "e21=e2-e1;#energy emitted by an electron jumping from orbit nuber 2 to orbit nimber 1\n",
- "re=e31/e21;#ratio of energy\n",
- "print re,\" is equal to ratio of energy for an electron to jump from orbit 3 to orbit 1 and from orbit 2 to orbit 1 is 32/27 \\n hence proved\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.18518518519 is equal to ratio of energy for an electron to jump from orbit 3 to orbit 1 and from orbit 2 to orbit 1 is 32/27 \n",
- " hence proved\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.4:pg-13"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 2.4 : velocity\n",
- " \n",
- "#given data :\n",
- "\n",
- "h=6.626*10**-34;\n",
- "e=1.6*10**-19;\n",
- "epsilon_o=8.825*10**-12;\n",
- "n=1;\n",
- "Z=1;\n",
- "vn=(Z*e**2)/(2*epsilon_o*n*h);\n",
- "print vn,\" is velocity,vn in (m/s) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "2188990.2342 is velocity,vn in (m/s) \n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.5:pg-14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 2.5 : velocity\n",
- " \n",
- "#given data :\n",
- "n=1;\n",
- "Z=1;\n",
- "k=6.56*10**15; # k is constant\n",
- "fn=k*(Z**2/n**3);\n",
- "print fn,\" is orbital frequency,fn in (Hz) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "6.56e+15 is orbital frequency,fn in (Hz) \n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.6.a:pg-14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 2.6.a : the energy of the photon emitted\n",
- " \n",
- "#given data :\n",
- "Z=1;#for hydrozen\n",
- "n1=3;\n",
- "n2=2;\n",
- "E3=-(13.6*Z**2)/n1**2;\n",
- "E2=-(13.6*Z**2)/n2**2;\n",
- "del_E=E3-E2;\n",
- "print round(del_E,2),\" is the energy of photon emitted, del_E in (eV) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.89 is the energy of photon emitted, del_E in (eV) \n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.6.b:pg-14"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 2.6.b : frequency\n",
- " \n",
- "#given data :\n",
- "\n",
- "Z=1;#for hydrozen\n",
- "n1=3;\n",
- "n2=2;\n",
- "m=6.626*10**-34;# mass of electron in kg\n",
- "E3=-(13.6*Z**2)/n1**2;\n",
- "E2=-(13.6*Z**2)/n2**2;\n",
- "del_E=E3-E2;\n",
- "E=del_E*1.6*10**-19;# in joules\n",
- "v=(E/m);\n",
- "print round(v,2),\"frequency of the photon emitted,v(Hz) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "4.5611563873e+14 frequency of the photon emitted,v(Hz) \n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex2.6.c:pg-15"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 2.6.c : wave length of the photon emitted\n",
- " \n",
- "#given data :\n",
- "\n",
- "Z=1;#for hydrozen\n",
- "n1=3;\n",
- "n2=2;\n",
- "m=6.626*10**-34;# mass of electron in kg\n",
- "C=3*10**8;\n",
- "E3=-(13.6*Z**2)/n1**2;\n",
- "E2=-(13.6*Z**2)/n2**2;\n",
- "del_E=E3-E2;\n",
- "E=del_E*1.6*10**-19;\n",
- "v=E/m;\n",
- "lamda=C/v;\n",
- "print round(lamda,9),\" is wavelength of the photon emitted,(m) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "6.58e-07 is wavelength of the photon emitted,(m) \n"
- ]
- }
- ],
- "prompt_number": 4
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Materials_Science/Chapter03.ipynb b/Materials_Science/Chapter03.ipynb deleted file mode 100755 index f10c658e..00000000 --- a/Materials_Science/Chapter03.ipynb +++ /dev/null @@ -1,1338 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:296c4bd8d9302a92dd3772adca6817eea05f8e0c9e58e45daf4ced8630943a9e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter03:Crystal Structure"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.1:pg-50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.1: miller indices\n",
- "import math \n",
- "#given data \n",
- "x1=1.0;#\n",
- "x2=1.0;#\n",
- "x3=2.0;#\n",
- "h1=1/x1;#\n",
- "h2=1/x2;#\n",
- "h3=1/x3;#\n",
- "print \"Miller indices of the plane (112) are: \",h1,\",\",h2,\",\",h3\n",
- "x11=0.0;#\n",
- "x21=0.0;#\n",
- "x31=1.0;#\n",
- "h11=inf;#\n",
- "h21=inf;#\n",
- "h31=1/x31;#\n",
- "print \"Miller indices of the plane (001) are : \",h11,\",\",h21,\",\",h31\n",
- "x111=1.0;#\n",
- "x211=0.0;#\n",
- "x311=1.0;#\n",
- "h111=1/x111;#\n",
- "h211=inf;#\n",
- "h311=1/x311;#\n",
- "print \"Miller indices of the plane (101) are : \",h111,\",\",h211,\",\",h311\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Miller indices of the plane (112) are: 1.0 , 1.0 , 0.5\n",
- "Miller indices of the plane (001) are : inf , inf , 1.0\n",
- "Miller indices of the plane (101) are : 1.0 , inf , 1.0\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.2:pg-51"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.2: miller indices\n",
- " \n",
- "#given data \n",
- "x1=0.0;#\n",
- "x2=2.0;#\n",
- "x3=0.0;#\n",
- "h1=inf;#\n",
- "h2=1/x2;#\n",
- "h3=inf;#\n",
- "print\"Miller indices of the plane (020) are: \",h1,\",\",h2,\",\",h3\n",
- "x11=1.0;#\n",
- "x21=2.0;#\n",
- "x31=0;#\n",
- "h11=1/x11;#\n",
- "h21=1/x21;#\n",
- "h31=inf;#\n",
- "print\"Miller indices of the plane (120) are : \",h11,\",\",h21,\",\",h31\n",
- "x111=2.0;#\n",
- "x211=2.0;#\n",
- "x311=0.0;#\n",
- "h111=1/x111;#\n",
- "h211=1/x211;#\n",
- "h311=inf;#\n",
- "print\"Miller indices of the plane (220) are : \",h111,\",\",h211,\",\",h311\n",
- "#miller indices for plane (120) is calculated wrong in the book\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Miller indices of the plane (020) are: inf , 0.5 , inf\n",
- "Miller indices of the plane (120) are : 1.0 , 0.5 , inf\n",
- "Miller indices of the plane (220) are : 0.5 , 0.5 , inf\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.3:pg-52"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 3.3: miller indices\n",
- " \n",
- "x=1/2.0;#\n",
- "x1=1/x;#\n",
- "r2=0;#\n",
- "r3=0;#\n",
- "x10=-1;#\n",
- "x2=1.0/x10;#\n",
- "r4=0;#\n",
- "r5=0;#\n",
- "print\"miller indices (Case 1) of the given plane are \",x1,\" : \",r2,\" : \",r3\n",
- "print\"miller indices (Case 2) of the given plane are \",x2,\" : \",r3,\" : \",r4 \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "miller indices (Case 1) of the given plane are 2.0 : 0 : 0\n",
- "miller indices (Case 2) of the given plane are -1.0 : 0 : 0\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.4:pg-52"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 3.4: miller indices\n",
- " \n",
- "a=0.529;#\n",
- "b=1;#\n",
- "c=0.477;#\n",
- "a1=0.264;#\n",
- "b1=1;#\n",
- "c1=0.238;#\n",
- "r1=round(a/a1);#\n",
- "r2=b/b1;#\n",
- "r3=round(c/c1);#\n",
- "print\"miller indices of the given plane are \",r1,\" : \",r2,\" : \",r3\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "miller indices of the given plane are 2.0 : 1 : 2.0\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.5:pg-53"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.5: miller indices\n",
- " \n",
- "#given data \n",
- "x1=1;#\n",
- "x2=1;#\n",
- "x3=0;#\n",
- "h1=1/x1#\n",
- "h2=1/x2;#\n",
- "h3=inf;#\n",
- "print\"Miller indices of the plane (110) are: \",h1,\",\",h2,\",\",h3\n",
- "x11=1;#\n",
- "x21=1;#\n",
- "x31=1;#\n",
- "h11=1/x11;#\n",
- "h21=1/x21;#\n",
- "h31=1/x31;#\n",
- "print\"Miller indices of the plane (111) are : \",h11,\",\",h21,\",\",h31\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Miller indices of the plane (110) are: 1 , 1 , inf\n",
- "Miller indices of the plane (111) are : 1 , 1 , 1\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.9:pg-58"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 3.9: atoms per unit cell\n",
- " \n",
- "c=8;#corners\n",
- "f=6;#faces\n",
- "nf=(1/2.0)*f;#no. of atoms in all six faces\n",
- "nc=(1/8.0)*c;#no. of atoms in all corners\n",
- "ta=nf+nc;#\n",
- "print ta,\"are total number of atoms \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "4.0 are total number of atoms \n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.10:pg-61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.10 : largest diameter\n",
- "import math \n",
- "#given data :\n",
- "\n",
- "a=3.61; # edge length in angstrum\n",
- "r=(a*math.sqrt(2))/4;\n",
- "d=2*r;\n",
- "print round(d,4),\"= largest diameter,d(angstrom) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "2.5527 = largest diameter,d(angstrom) \n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.11:pg-62"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.11 : volume change in percentage\n",
- "import math\n",
- "#given data :\n",
- "r_bcc=0.1258; # in nm\n",
- "r_fcc=0.1292;# in nm\n",
- "a_bcc=(r_bcc*4)/math.sqrt(3);\n",
- "a_fcc=(r_fcc*4)/math.sqrt(2);\n",
- "v_fcc=(a_fcc)**3;# in nmn**3\n",
- "v_bcc=(a_bcc)**3; # in nm**3\n",
- "V=((v_fcc-v_bcc)/v_bcc)*100;\n",
- "print round(V,2),\"=volume change in percentage,V(%) \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "99.01 =volume change in percentage,V(%) \n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.12:pg-64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Example 3.12 : number of atom/mm**2\n",
- " \n",
- "\n",
- "#given data :\n",
- "a=3.03*10**-7; # lattice constant in mm\n",
- "A=1/a**2;# for 100 planes \n",
- "B=0.707/a**2;#for(110) planes\n",
- "C=0.58/a**2;# for(111) planes\n",
- "print round(A,-11),\"=number of atoms for (100) plane \"\n",
- "print round(B,-10),\"=number of atoms for (110) plane \"\n",
- "print round(C,-11),\"=number of atoms for (111) plane \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.09e+13 =number of atoms for (100) plane \n",
- "7.7e+12 =number of atoms for (110) plane \n",
- "6.3e+12 =number of atoms for (111) plane \n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.13:pg-66"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Example 3.13 : number of atom/mm**2 of planes\n",
- " \n",
- "#given data :\n",
- "\n",
- "a=2.87*10**-7; # lattice constant in mm\n",
- "A=1/a**2;# for 100 planes \n",
- "B=1.414/a**2;#for(110) planes\n",
- "C=1.732/a**2;# for(111) planes\n",
- "print round(A,-11),\"=number of atoms for (100) plane \"\n",
- "print round(B,-11),\"=number of atoms for (110) plane \"\n",
- "print round(C,-11),\"=number of atoms for (111) plane \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.21e+13 =number of atoms for (100) plane \n",
- "1.72e+13 =number of atoms for (110) plane \n",
- "2.1e+13 =number of atoms for (111) plane \n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.14:pg-69"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Example 3.14 : number of atom/mm**2 surface area\n",
- " \n",
- "#given data :\n",
- "a=4.93*10**-7; # lattice constant in mm\n",
- "A=2/a**2;# for 100 planes \n",
- "B=1.414/a**2;#for(110) planes\n",
- "C=2.31/a**2;# for(111) planes\n",
- "print round(A,-11),\"=number of atoms for (100) plane \"\n",
- "print round(B,-11),\"=number of atoms for (110) plane \"\n",
- "print round(C,-11),\"=number of atoms for (111) plane \"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "8.2e+12 =number of atoms for (100) plane \n",
- "5.8e+12 =number of atoms for (110) plane \n",
- "9.5e+12 =number of atoms for (111) plane \n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.15:pg-69"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.15 : planar density\n",
- " \n",
- "#given data :\n",
- "\n",
- "a=0.143*10**-6; # atomic radius in mm\n",
- "A=2.31/(a**2);# for(111) planes\n",
- "print round(A,-10),\"= atom,A(atoms/mm**2) \"\n",
- "# answer is wrong in book\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.1296e+14 = atom,A(atoms/mm**2) \n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.16:pg-71"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.16 : volume\n",
- " \n",
- "import math\n",
- "#given data :\n",
- "a=0.2665; # in mm\n",
- "c=0.4947;# in mm\n",
- "V=(3*math.sqrt(3)*a**2*c)/2.0;\n",
- "print round(V,4),\"=volume,V(mm**3) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.0913 =volume,V(mm**3) \n"
- ]
- }
- ],
- "prompt_number": 51
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.17:pg-72"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Example 3.17 : find the packing efficiency and lattice parameter\n",
- " \n",
- "\n",
- "#given data :\n",
- "r=1.22# in angstrum\n",
- "a=(4*r)/math.sqrt(3);\n",
- "efficiency=(math.pi*math.sqrt(3))/8;\n",
- "print round(efficiency,2),\"=efficiency \"\n",
- "print round(a,2),\"= lattice parameter,a(angstrom) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.68 =efficiency \n",
- "2.82 = lattice parameter,a(angstrom) \n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.18:pg-73"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.18 : interplanar distance\n",
- "import math \n",
- "#given data :\n",
- "h=1;\n",
- "k=1;\n",
- "l=1;\n",
- "#d=a/math.sqrt(h**2+k**2+l**2)\n",
- "dBYa=1/math.sqrt(h**2+k**2+l**2);\n",
- "print \"Interplanor distance in (Angstrom) is a*\",round(dBYa,3)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Interplanor distance in (Angstrom) is a* 0.577\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.19:pg-74"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.19 : spacing\n",
- "import math\n",
- "#given data :\n",
- "h1=2;\n",
- "k1=0;\n",
- "l1=0;\n",
- "h2=2;\n",
- "k2=2;\n",
- "l2=0;\n",
- "h3=1;\n",
- "k3=1;\n",
- "l3=1;\n",
- "r=1.246;\n",
- "a=(4*r)/math.sqrt(2);# in angstrum\n",
- "#d=a/math.sqrt(h**2+k**2+l**2)\n",
- "d1=a/math.sqrt(h1**2+k1**2+l1**2);\n",
- "d2=a/math.sqrt(h2**2+k2**2+l2**2);\n",
- "d3=a/math.sqrt(h3**2+k3**2+l3**2);\n",
- "print round(d1,2),\"=d_200 spacind,d1(angstrom) \"\n",
- "print round(d2,2),\"=d_220 spacind,d2(angstrom) \"\n",
- "print round(d3,2),\"=d_111 spacind,d3(angstrom) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.76 =d_200 spacind,d1(angstrom) \n",
- "1.25 =d_220 spacind,d2(angstrom) \n",
- "2.03 =d_111 spacind,d3(angstrom) \n"
- ]
- }
- ],
- "prompt_number": 57
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.20:pg-74"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.20 : interplaner spacing d_220\n",
- "import math \n",
- "\n",
- "#given data :\n",
- "a=0.316;# in nm\n",
- "h=2;\n",
- "k=2;\n",
- "l=0;\n",
- "d=a/math.sqrt(h**2+k**2+l**2);\n",
- "print round(d,3),\"= inter planer spacing d_220,d(nm) \"\n",
- "# answer is wrong in book\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.112 = inter planer spacing d_220,d(nm) \n"
- ]
- }
- ],
- "prompt_number": 59
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.21:pg-74"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 3.21: interplanar spacing d220\n",
- " \n",
- "import math\n",
- "a=1;#constant assume\n",
- "a1=[1,0,0];#lattice planes\n",
- "a2=[1,1,0];#lattice planes\n",
- "a3=[1,1,1];#lattice planes\n",
- "d100=a/(math.sqrt(a1[0]+a1[1]**2+a1[2]**2));#interplanar distance between (100)planes\n",
- "d110=a/(math.sqrt(a2[0]**2+a2[1]**2+a2[2]**2));#interplanar distance between (110)planes\n",
- "d111=a/(math.sqrt(a3[0]**2+a3[1]**2+a3[2]**2));#interplanar distance between (111)planes\n",
- "print \"ratio of interplanar distances is \",d100,\":\",round(d110,2),\":\",round(d111,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "ratio of interplanar distances is 1.0 : 0.71 : 0.58\n"
- ]
- }
- ],
- "prompt_number": 63
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.22:pg-75"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 3.22: perpendicular distance\n",
- "import math \n",
- "a=1;#constant assume\n",
- "a1=[1,1,1];#lattice planes\n",
- "a2=[2,2,2];#lattice planes\n",
- "d1=a/(math.sqrt(a1[0]**2+a1[1]**2+a1[2]**2));#perpendicular distance between origin and (111)planes\n",
- "d2=a/(math.sqrt(a2[0]**2+a2[1]**2+a2[2]**2));#perpendicular distance between origin and (222)planes\n",
- "d22 = d1-d2;#perpendicular distance between the planes (111) and (222)\n",
- "print round(d22,2),\"= perpendicular distance between the planes (111) and (222)\"\n",
- "\n",
- "# a is assumed to be 1\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.29 = perpendicular distance between the planes (111) and (222)\n"
- ]
- }
- ],
- "prompt_number": 65
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.23:pg-76"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 3.23: angle between planes (122) and (111)\n",
- "import math\n",
- "a=1;# assume\n",
- "a1=[1,2,2];#lattice planes\n",
- "a2=[1,1,1];#lattice planes\n",
- "d1=a/(math.sqrt(a1[0]**2+a1[1]**2+a1[2]**2));#perpendicular distance between origin and (111)planes\n",
- "d2=a/(math.sqrt(a2[0]**2+a2[1]**2+a2[2]**2));#perpendicular distance between origin and (222)planes\n",
- "cphi= ((a1[0]*a2[0])+(a1[1]*a2[1])+(a1[2]*a2[2]))*(d1*d2);#\n",
- "d=math.degrees(math.acos((cphi)));# in degree\n",
- "d1=math.floor(d);#\n",
- "d2=d-d1;#\n",
- "print\"angle between planes (122) and (111) is \",d1,\" degree \",round(60*d2),\" minutes\"\n",
- "\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "angle between planes (122) and (111) is 15.0 degree 48.0 minutes\n"
- ]
- }
- ],
- "prompt_number": 72
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.24:pg-77"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.24 : concentration of iron\n",
- " \n",
- "\n",
- "#given data :\n",
- "d=7.87;\n",
- "N=6.023*10**23; # avogadro's number\n",
- "A=55.85;# atomic weight\n",
- "I=A/N;# mass of iron atom\n",
- "atom=d/I;\n",
- "print round(atom,-20),\"= number of atoms(atoms/cm**3) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "8.49e+22 = number of atoms(atoms/cm**3) \n"
- ]
- }
- ],
- "prompt_number": 83
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.25:pg-77"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.25 : lattice constant\n",
- " \n",
- "#given data :\n",
- "\n",
- "n=2;\n",
- "A=55.8;\n",
- "N=6.023*10**26; # avogadro's number in /kg-mole\n",
- "b=7.87*10**3;# in kg/m**3\n",
- "a=((A*n)/(N*b))**(1/3.0);\n",
- "print round(a*10**10,3),\"= lattice constant,a(angstrom)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "2.866 = lattice constant,a(angstrom)\n"
- ]
- }
- ],
- "prompt_number": 85
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.26:pg-77"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.26 : density\n",
- "import math\n",
- "#given data :\n",
- "\n",
- "n=4;\n",
- "N=6.023*10**23; # avogadro's number\n",
- "r=1.278*10**-8;# in cm\n",
- "A=63.5;\n",
- "a=(r*4)/math.sqrt(2);# in cm\n",
- "b=(A*n)/(a**3*N);\n",
- "print round(b,2),\"= density of copper,b(g/cc) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "8.93 = density of copper,b(g/cc) \n"
- ]
- }
- ],
- "prompt_number": 87
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.27:pg-77"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.27 : number of atoms\n",
- " \n",
- "#given data :\n",
- "n=4;\n",
- "N=6.023*10**23; # avogadro's number\n",
- "A=55.85;\n",
- "a=2.9*10**-8;\n",
- "b=7.87;#density in g/cc\n",
- "#a**3=(A*n)/(N*b)\n",
- "n=round((a**3*N*b)/A);\n",
- "print n,\"= number of atoms,n \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "2.0 = number of atoms,n \n"
- ]
- }
- ],
- "prompt_number": 88
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.28:pg-78"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.28 : lattice constant\n",
- " \n",
- "#given data :\n",
- "d=6250;#density\n",
- "N=6.02*10**23;#avogadro's number\n",
- "n=4;\n",
- "m=60.2*10**-3;# atomic mass\n",
- "M=(n*m)/N;\n",
- "V=M/d;\n",
- "a=V**(1/3.0)*10**9;\n",
- "print a,\"= the lattice constant,a(nm) \"\n",
- "#ANSWER IS WRONG IN THE TEXT BOOK\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.4 = the lattice constant,a(nm) \n"
- ]
- }
- ],
- "prompt_number": 89
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.29:pg-78"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.29 : the number of atoms\n",
- " \n",
- "#given data :\n",
- "d=7.87;#in g/cm**3\n",
- "A=55.85;\n",
- "a=2.9*10**-8;# in cm\n",
- "N=6.02*10**23;#avogadro's number\n",
- "n=(d*a**3*N)/A;\n",
- "print round(n),\"= the number of atom,n \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "2.0 = the number of atom,n \n"
- ]
- }
- ],
- "prompt_number": 90
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.30:pg-83"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 3.30: calculate the number of vacancies in the copper\n",
- "import math \n",
- "B=1.38*10**-23;#boltzman constant in J/atom-K\n",
- "B1=8.62*10**-5;# bolzman constant in ev/atom-K\n",
- "Qv=0.9;# eV/atom\n",
- "t=27;# room temperatyre in degree celsius\n",
- "pcu=8.4;#in g/cm**3\n",
- "Acv=63.5;# in g/mol\n",
- "T=t+273;#temperture in kelvin\n",
- "Nv=6.023*10**23;#\n",
- "P=8.4;#\n",
- "Ns=(Nv*P)/Acv;# number of regular lattice sites\n",
- "Nv1=Ns*math.exp(-Qv/(B1*T));#\n",
- "print Nv1,\"is number of vacancies in copper in vacancies/cm**3\"\n",
- "#answer is wrong in the textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "61187298.8086 is number of vacancies in copper in vacancies/cm**3\n"
- ]
- }
- ],
- "prompt_number": 91
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.31:pg-86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.31 : interplanar spacing\n",
- "import math \n",
- "#given data :\n",
- "\n",
- "theta=20.3;#in degree\n",
- "lamda=1.54;# in angstrum\n",
- "n=1.0;\n",
- "a=math.sin(math.radians(theta))\n",
- "d=lamda/(2*a);\n",
- "print round(d,2),\"= interplanar spacing,d(angstrom) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "2.22 = interplanar spacing,d(angstrom) \n"
- ]
- }
- ],
- "prompt_number": 101
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.32:pg-86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.32 : interatomic spacing\n",
- "import math \n",
- "#given data :\n",
- "\n",
- "theta=30;#in degree\n",
- "lamda=1.54;# in angstrum\n",
- "n=1;\n",
- "a=math.sin(math.radians(theta))\n",
- "d=lamda/(2*a);\n",
- "print d,\"=interatomic spacing,d(angstrom) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.54 =interatomic spacing,d(angstrom) \n"
- ]
- }
- ],
- "prompt_number": 102
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.33:pg-87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 3.33 : number of per order\n",
- "import math \n",
- "#given data :\n",
- "\n",
- "theta=90;#in degree\n",
- "lamda=1.54;# in angstrum\n",
- "a=math.sin(math.radians(theta))\n",
- "d=1.181;\n",
- "n=(2*d*a)/lamda;\n",
- "print round(n,2),\"= number of order,n \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.53 = number of order,n \n"
- ]
- }
- ],
- "prompt_number": 106
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.34:pg-87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 3.34: size of unit cell\n",
- "import math \n",
- "n=1.0;#\n",
- "a=1.0;#assume\n",
- "h=0.58;#wavelnegth in armstrong\n",
- "th=9.5;#reflection angle in degree\n",
- "a1=[2.0,0,0];#miller indices\n",
- "d200=a/(math.sqrt(a1[0]**2+a1[1]**2+a1[2]**2));#interplanar distance between (200)planes\n",
- "a=((n*h)/(2*d200*math.sin(math.radians(th))));#zsize of unit cell\n",
- "print round(a,3),\"= size of unit cell in \u00c4\"\n",
- "#amswer is wrong in the textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "3.514 = size of unit cell in \u00c4\n"
- ]
- }
- ],
- "prompt_number": 111
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.35:pg-87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 3.35: bragg angle\n",
- "import math\n",
- "n=1;#\n",
- "a=3.57;#in \u00c4\n",
- "h=0.54;#wavelnegth in \u00c4 \n",
- "a1=[1,1,1];#miller indices\n",
- "d111=a/(math.sqrt(a1[0]**2+a1[1]**2+a1[2]**2));#interplanar distance between (111)planes\n",
- "snd=((n*h)/(2*d111));#\n",
- "th=math.degrees(math.asin(snd));# bragg angle in degree\n",
- "d1=math.floor(th);#\n",
- "d2=th-math.floor(d1);#\n",
- "print\"angle between planes (122) and (111) is \",d1,\" degree \",round(60*d2),\" minutes\"\n",
- "#wavelength is given wrong in example it is 0.54\u00c4 and it is taken as 1.54\u00c4\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "angle between planes (122) and (111) is 7.0 degree 32.0 minutes\n"
- ]
- }
- ],
- "prompt_number": 113
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.36:pg-88"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 3.36: interplanner spacing and miller indices\n",
- " \n",
- "a=3.16;# in \u00c4\n",
- "h=1.54;# in \u00c4\n",
- "n=1;#\n",
- "th=20.3;# in degree\n",
- "d=((n*h)/(2*math.sin(math.radians(th))));# interplanner spacing in \u00c4\n",
- "x=a/d;#\n",
- "y=x**2;#\n",
- "print round(d,2),\"= interplanner spacing in \u00c4 \"\n",
- "print \"miller indices are (110) , (011) or (101)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "2.22 = interplanner spacing in \u00c4 \n",
- "miller indices are (110) , (011) or (101)\n"
- ]
- }
- ],
- "prompt_number": 115
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex3.37:pg-88"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 3.36: interplanner spacing and diffraction angle\n",
- "import math \n",
- "a=.2866;# in \u00c4\n",
- "h=0.1542;# in nm\n",
- "n=1.0;#\n",
- "a1=[2.0,1.0,1.0];#miller indices\n",
- "d211=a/(math.sqrt(a1[0]**2+a1[1]**2+a1[2]**2));#interplanar distance between (211)planes\n",
- "snd=((n*h)/(2*d211));#\n",
- "th=math.degrees(math.asin(snd));# bragg angle in degree\n",
- "d1=math.floor(th);#\n",
- "d2=th-math.floor(d1);#\n",
- "print\"angle between planes (122) and (111) is \",d1,\" degree \",round(60*d2),\" minutes\"\n",
- "print round(d211,2),\"=interplanner spacing in \u00c4 \"\n",
- "#answer is wrong in the textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "angle between planes (122) and (111) is 41.0 degree 13.0 minutes\n",
- "0.12 =interplanner spacing in \u00c4 \n"
- ]
- }
- ],
- "prompt_number": 121
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Materials_Science/Chapter05.ipynb b/Materials_Science/Chapter05.ipynb deleted file mode 100755 index 39b44fee..00000000 --- a/Materials_Science/Chapter05.ipynb +++ /dev/null @@ -1,666 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:ef6d9a0d23fffdb9f7c392e83ea5acb91eccb75cc2e53a11277239fd4fc34966"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter05:Electron Theory of Metals"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.1.i:pg-110"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 5.1.i: probability for diamond\n",
- " \n",
- "# given :\n",
- "\n",
- "Eg=5.6; # in eV\n",
- "k=86.2*10**-6; # in eVk**-1\n",
- "T=273+25.0; # in K\n",
- "E_Ef=Eg/2;\n",
- "f_E=1/(1+math.exp(E_Ef/(k*T)));\n",
- "print f_E,\" is probability for diamond\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "4.58172567644e-48 is probability for diamond\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.1.ii:pg-110"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 5.1.ii: probability for silicon\n",
- " \n",
- "# given :\n",
- "Eg=1.07; # in eV\n",
- "k=86.2*10**-6; # in eVk**-1\n",
- "T=273+25.0; # in K\n",
- "E_Ef=Eg/2;\n",
- "f_E=1/(1+math.exp(E_Ef/(k*T)));\n",
- "print f_E,\"is probability for diamond \"\n",
- "# answer is wrong in book\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "9.01312095705e-10 is probability for diamond \n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.2:pg-119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 5.2: resistance\n",
- " \n",
- "# given :\n",
- "l=1; # length in m\n",
- "A=4*10**-4; # area of cross section in m**2\n",
- "p=0.01*10**-2; # resistivity in ohm-m\n",
- "R=p*(l/A);\n",
- "print R,\"is resistance of wire,R(ohm) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.25 is resistance of wire,R(ohm) \n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.3:pg-120"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 5.3: resistance\n",
- " \n",
- "# given :\n",
- "\n",
- "p=1.7*10**-8; # resistivity i ohm-m\n",
- "d=0.0005; # diameter of the wire in m\n",
- "l=31.4; # length in m\n",
- "A=(math.pi*d**2)/4;\n",
- "R=p*(l/A);\n",
- "print round(R,2),\"is resistance of wire,R(ohm) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "2.72 is resistance of wire,R(ohm) \n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.4:pg-120"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 5.4: conductivity\n",
- " \n",
- "# given :\n",
- "\n",
- "V=.432; # voltage drop across the wire in volts\n",
- "I=10; # current through the wire in A\n",
- "l=1; # length in m\n",
- "d=1*10**-3; # diameter in m\n",
- "R=V/I;\n",
- "A=(math.pi*d**2)/4;\n",
- "p=(R*A)/l;\n",
- "b=1/p;\n",
- "print round(b,2),\" is conductivitty,b(ohm**-1.m**-1) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "29473137.61 is conductivitty,b(ohm**-1.m**-1) \n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.5:pg-124"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 5.5: drift velocity\n",
- " \n",
- "# given :\n",
- "\n",
- "n=10**19; # in m**3\n",
- "b=0.01; # conductivity in ohm**-1. m**-1\n",
- "V=0.17; # in volts\n",
- "d=.27*10**-3; # in m\n",
- "e=1.602*10**-19; # in C\n",
- "m=9.1*10**-31; # in kg\n",
- "E=V/d; # in volt/m\n",
- "v=((b*E)/(n*e));\n",
- "print round(v,2),\"is drift velocity of electron,v (m/sec) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "3.93 is drift velocity of electron,v (m/sec) \n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.6:pg-124"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 5.6: conductivity\n",
- " \n",
- "# given :\n",
- "e=1.6*10**-19; # in C\n",
- "T=300; # temerature in K\n",
- "t=2*10**-14; # time in sec\n",
- "c=63.54; # atomic weight of copper in a.m.u\n",
- "m=9.1*10**-31; # mass in kg\n",
- "# we know that 63.45 grams of copper contains 6.023*10**23 free electrons since one atom contributes one electron.the volume of 63.54 gram of copper is 8.9 cubic centimetre(c.c).\n",
- "n=6.023*10**23/(c/8.9); #number of electrons per unit volume(c.c)\n",
- "n1=n*10**6; # the number of electrons per m**3\n",
- "b=(e**2*n1*t)/m;\n",
- "print round(b,2),\"is conductivity,b(mho/m) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "47466174.12 is conductivity,b(mho/m) \n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.7:pg-125"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 5.7: mobility of electrons\n",
- " \n",
- "# given :\n",
- "\n",
- "e=1.602*10**-19; # in C\n",
- "m=9.1*10**-31; # in kg\n",
- "t=10**-14; # time in sec\n",
- "mu=(e*t)/m;\n",
- "print mu,\"is mobility of electrons,mu(m**2/volts.sec) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.00176043956044 is mobility of electrons,mu(m**2/volts.sec) \n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.8:pg-125"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 5.8: mobility \n",
- " \n",
- "# given :\n",
- "\n",
- "d=10.5; # density of silver in gm/c.c\n",
- "w=107.9; # atomic weight\n",
- "b=6.8*10**5; # conductivity in mhos/cm\n",
- "e=1.602*10**-19; # in C\n",
- "N=6.023*10**23;\n",
- "n=(N*d)/w;\n",
- "mu=b/(e*n);\n",
- "print round(mu,2),\"is mobility of electron,mu(m**2/volt-sec) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "72.42 is mobility of electron,mu(m**2/volt-sec) \n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.9:pg-126"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 5.9: mobility and drift velocity\n",
- " \n",
- "# given :\n",
- "b=6.5*10**7; # conductivity in ohm**-1.m**-1\n",
- "e=1.602*10**-19; # in C\n",
- "n=6*10**23; #\n",
- "E=1; # in V/m\n",
- "mu=b/(e*n);\n",
- "v=mu*E;\n",
- "print round(mu,2),\"is mobility ,mu(m**2/volt-sec) \"\n",
- "print round(v,2),\"is drift velocity,v(m/sec) \"\n",
- "# mobility and drift is calculated wrong in book\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "676.24 is mobility ,mu(m**2/volt-sec) \n",
- "676.24 is drift velocity,v(m/sec) \n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.10:pg-126"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 5.10 : density and drift velocity \n",
- " \n",
- "#given data :\n",
- "\n",
- "e=1.602 *10**-19;\n",
- "b=58*10**6;# in ohm**-1 m**-1\n",
- "mu_n=3.5*10**-3;# in m**2/V s\n",
- "E=0.5; # in V/m\n",
- "n=b/(e*mu_n);\n",
- "print n,\"is density,n(m**-3) \"\n",
- "v=mu_n*E;\n",
- "print v,\"is drift velocity,v(m/s) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.03442125914e+29 is density,n(m**-3) \n",
- "0.00175 is drift velocity,v(m/s) \n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.11:pg-127"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 5.11 : velocity\n",
- "import math\n",
- "#given data :\n",
- "m=9.109*10**-31; # in kg\n",
- "e=1.602 *10**-19;\n",
- "Ef=2.1# in ev\n",
- "Wf=e*Ef;# in J\n",
- "vf=math.sqrt((2*Wf)/m);\n",
- "print vf,\"is velocity,vf(m/s) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "859449.869617 is velocity,vf(m/s) \n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.12.a:pg-127"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 5.12.a : velocity\n",
- " \n",
- "#given data :\n",
- "m=9.1*10**-31; # in kg\n",
- "e=1.602 *10**-19;\n",
- "Ef=3.75;# in ev\n",
- "Wf=(e*Ef);# in J\n",
- "vf=math.sqrt(((2*Wf)/m));\n",
- "print vf,\" is velocity,vf(m/s) \"\n",
- "# answer is wrong in book\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1149055.99095 is velocity,vf(m/s) \n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.12.b:pg-127"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Example 5.12.b : mobility of electron\n",
- " \n",
- "#given data :\n",
- "m=9.1*10**-31; # in kg\n",
- "e=1.602 *10**-19;\n",
- "Ef=3.75;# in ev\n",
- "t=10**-14;# in sec\n",
- "mu=(e*t)/m;\n",
- "print mu,\"is mobility,mu(m**2/V-sec) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.00176043956044 is mobility,mu(m**2/V-sec) \n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.13:pg-127"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 5.13 : the mean free path\n",
- "import math\n",
- "#given data :\n",
- "\n",
- "t=10**-9; # in sec\n",
- "m=9.109*10**-31; # in kg\n",
- "e=1.602 *10**-19;\n",
- "Ef=7# in ev\n",
- "Wf=e*Ef;# in J\n",
- "vf=math.sqrt((2*Wf)/m);\n",
- "lamda=vf*t*10**3;\n",
- "print round(lamda,2),\"is the mean free path,lamda(mm) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.57 is the mean free path,lamda(mm) \n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.14:pg-128"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 5.14 : mobility and average time\n",
- " \n",
- "#given data :\n",
- "\n",
- "m=9.109*10**-31; # in kg\n",
- "e=1.602 *10**-19;\n",
- "d=8.92*10**3;# in kg/m**3\n",
- "p=1.73*10**-8;# ohm-m\n",
- "A=63.5;#atomic weight\n",
- "N=6.023*10**22; # avogadro's number\n",
- "n=(N*d)/A;\n",
- "b=1/p;# conductivity\n",
- "mu=b/(n*e);\n",
- "print round(mu,1),\"= mobility,mu(m**2/V-s) \"\n",
- "t=(mu*m)/e;\n",
- "print round(t*10**9,3),\"= average time,t(ns) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "42.6 = mobility,mu(m**2/V-s) \n",
- "0.242 = average time,t(ns) \n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex5.15:pg-129"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 5.15 : electrical resistivity\n",
- "import math\n",
- "#given data :\n",
- "\n",
- "r=1.86*10**-10;# in m\n",
- "t=3*10**-14;# in sec\n",
- "a=2;\n",
- "m=9.1*10**-31; # in kg\n",
- "e=1.602 *10**-9;\n",
- "A=23*10**-3;#in kg/m\n",
- "N=6.023*10**23; # avogadro's number\n",
- "M=(a*A)/N;\n",
- "V=((4/math.sqrt(3))*r)**3;\n",
- "d=M/V;\n",
- "mu=((e*t)/m);\n",
- "n=(N*d)/A;\n",
- "b=1.602 *10**-19*n*mu;\n",
- "p=(1/b);\n",
- "print p,\"= resistivity,p(ohm-m) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "4.68383991207e-18 = resistivity,p(ohm-m) \n"
- ]
- }
- ],
- "prompt_number": 43
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Materials_Science/Chapter07.ipynb b/Materials_Science/Chapter07.ipynb deleted file mode 100755 index 82a2c634..00000000 --- a/Materials_Science/Chapter07.ipynb +++ /dev/null @@ -1,724 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c09a61f5f13016fbda07546fd7ab3e30960d55a34ee491901311869fc5da9a32"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter07:Mechanical Tests of Metals"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.1:pg-146"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 7.1 : shear modulus of the material\n",
- " \n",
- "#given data :\n",
- "E=210 # youngs's modulus in GN/m**2\n",
- "v=0.3 # poisson ratio\n",
- "G=E/(2*(1+v)) # shear modulus\n",
- "\n",
- "print \"shear modulus,G(GN/m**2) = \",round(G,2)\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "shear modulus,G(GN/m**2) = 80.77\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.2:pg-152"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 7.2 : young's modulus of elasticity,yield point stress, ultimate stress and percentage elongation\n",
- " \n",
- "#given data :\n",
- "d=40.0*10**-3 #in m\n",
- "W=40.0*10**3 # load in N\n",
- "del_l=3.04*10**-5 # in m\n",
- "L=200.0*10**-3 # in m\n",
- "load_max=242.0*10**3 #in N\n",
- "l=249*10.0**-3 # length of specimen in m\n",
- "l0=(d+L) # in m\n",
- "A=(math.pi*d**2)/4.0\n",
- "\n",
- "b=W/A\n",
- "\n",
- "epsilon=del_l/L\n",
- "\n",
- "E=(b/epsilon)\n",
- "\n",
- "print\"young modulus,E(N/m**2) = \",E\n",
- "\n",
- "Y_load=161*10**3\n",
- "\n",
- "Y_stress=Y_load/A\n",
- "\n",
- "print \"yield point stress,Y_stress(N/m**2) = \",Y_stress\n",
- "\n",
- "U_stress=load_max/A\n",
- "\n",
- "print \"ultimate stress,U_stress(N/m**2) = \",U_stress\n",
- "\n",
- "p_elongation=((l-l0)/l0)*100\n",
- "\n",
- "print \"percentage elongation,p_elongation(%) = \",p_elongation\n",
- "#percentage elongation is calculated wrong in textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "young modulus,E(N/m**2) = 2.09414398805e+11\n",
- "yield point stress,Y_stress(N/m**2) = 128119729.189\n",
- "ultimate stress,U_stress(N/m**2) = 192577481.141\n",
- "percentage elongation,p_elongation(%) = 3.75\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.3.a:pg-153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 7.3.a: yield point stress\n",
- " \n",
- "\n",
- "yl=40.0 #yeild load in kN\n",
- "ml=71.5 #maximum load in kN\n",
- "fl=50.5 #fracture load in kN\n",
- "glf=79.5 #gauge length of fratture in mm\n",
- "st=7.75*10**-4 #strain at load of 20kN\n",
- "d=12.5 #specimen diamtere in mm\n",
- "sl=62.5 #specimen length in mm\n",
- "A=(math.pi*(d*10**-3)**2)/4.0 # in meter square\n",
- "ylp=((yl*10.0**3)/(A)) #yeild point stress in N/m**2\n",
- "print \"yeild point stress in N/m**2 is \",ylp \n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "yeild point stress in N/m**2 is 325949323.452\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.3.b:pg-153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 7.3.b: ultimate tensile strength\n",
- " \n",
- "yl=40.0 #yeild load in kN\n",
- "ml=71.5 #maximum load in kN\n",
- "fl=50.5 #fracture load in kN\n",
- "glf=79.5 #gauge length of fratture in mm\n",
- "st=7.75*10**-4 #strain at load of 20kN\n",
- "d=12.5 #specimen diamtere in mm\n",
- "sl=62.5 #specimen length in mm\n",
- "A=(math.pi*(d*10**-3)**2)/4.0 # in meter square\n",
- "ylp=((yl*10.0**3)/(A)) #yeild point stress in N/m**2\n",
- "uts=((ml*10.0**3)/(A)) #ultimate tensile strangth in N/m**2\n",
- "print uts,\"is ultimate tensile strangth in N/m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "582634415.671 is ultimate tensile strangth in N/m**2\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.3.c:pg153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 7.3.c: percentage elongation\n",
- " \n",
- "yl=40 #yeild load in kN\n",
- "ml=71.5 #maximum load in kN\n",
- "fl=50.5 #fracture load in kN\n",
- "glf=79.5 #gauge length of fratture in mm\n",
- "st=7.75*10**-4 #strain at load of 20kN\n",
- "d=12.5 #specimen diamtere in mm\n",
- "sl=62.5 #specimen length in mm\n",
- "a=(math.pi*d*10**-3)**2/4 # in meter square\n",
- "pel=((glf-sl)/sl)*100 #percentage elongation\n",
- "print pel,\"% is percentage elongation\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "27.2 % is percentage elongation\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.3.d:pg-153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 7.3.d:modulus of elasticity\n",
- "import math\n",
- "yl=40 #yeild load in kN\n",
- "ml=71.5 #maximum load in kN\n",
- "fl=50.5 #fracture load in kN\n",
- "glf=79.5 #gauge length of fratture in mm\n",
- "st=7.75*10**-4 #strain at load of 20kN\n",
- "d=12.5 #specimen diamtere in mm\n",
- "sl=62.5 #specimen length in mm\n",
- "A=(math.pi*(d*10**-3)**2)/4.0 # in meter square\n",
- "ylp=((yl*10**3)/(A)) #yeild point stress in N/m**2\n",
- "uts=((ml*10**3)/(A)) #ultimate tensile strangth in N/m**2\n",
- "pel=((glf-sl)/sl)*100 #percentage elongation\n",
- "strss=((20*10**3)/A) #stress at 20kN in N/m**2\n",
- "mel=strss/st #modulus of elasticity in N/m**2\n",
- "print round(mel,2),\"is modulus of elasticity in N/m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "2.10289886098e+11 is modulus of elasticity in N/m**2\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.3.e:pg153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 7.3.e: yield point stress\n",
- "import math\n",
- "yl=40.0 #yeild load in kN\n",
- "ml=71.5 #maximum load in kN\n",
- "fl=50.5 #fracture load in kN\n",
- "glf=79.5 #gauge length of fratture in mm\n",
- "st=7.75*10**-4.0 #strain at load of 20kN\n",
- "d=12.5 #specimen diamtere in mm\n",
- "sl=62.5 #specimen length in mm\n",
- "A=(math.pi*(d*10**-3)**2)/4.0 # in meter square\n",
- "ylp=((yl*10**3)/(A)) #yeild point stress in N/m**2\n",
- "uts=((ml*10**3)/(A)) #ultimate tensile strangth in N/m**2\n",
- "pel=((glf-sl)/sl)*100 #percentage elongation\n",
- "strss=((20*10**3)/A) #stress at 20kN in N/m**2\n",
- "mel=strss/st #modulus of elasticity in N/m**2\n",
- "mrs=((ylp*10**-3)**2/(2*mel)) #modulus of resilience \n",
- "print mrs,\" is modulus of resilience\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.252610725675 is modulus of resilience\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.3.f:pg-153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 7.3.f: fracture stress\n",
- " \n",
- "yl=40 #yeild load in kN\n",
- "ml=71.5 #maximum load in kN\n",
- "fl=50.5 #fracture load in kN\n",
- "glf=79.5 #gauge length of fratture in mm\n",
- "st=7.75*10**-4 #strain at load of 20kN\n",
- "d=12.5#specimen diamter in mm\n",
- "sl=62.5 #specimen length in mm\n",
- "A=(math.pi*(d*10**-3)**2.0)/4 # in meter square\n",
- "ylp=((yl*10**3)/(A)) #yeild point stress in N/m**2\n",
- "uts=((ml*10**3)/(A)) #ultimate tensile strangth in N/m**2\n",
- "pel=((glf-sl)/sl)*100 #percentage elongation\n",
- "strss=((20*10.0**3)/A) #stress at 20kN in N/m**2\n",
- "mel=strss/st #modulus of elasticity in N/m**2\n",
- "mrs=((ylp*10**-3)**2.0/(2*mel)) #modulus of resilience \n",
- "fs=((fl*10.0**3)/(A)) #fracture stress in N/m**2\n",
- "print fs,\"is fracture stress in N/m**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "411511020.858 is fracture stress in N/m**2\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.3.g:pg153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "# Example 7.3.g: modulus of toughness\n",
- " \n",
- "yl=40.0 #yeild load in kN\n",
- "ml=71.5 #maximum load in kN\n",
- "fl=50.5 #fracture load in kN\n",
- "glf=79.5 #gauge length of fratture in mm\n",
- "st=7.75*10**-4 #strain at load of 20kN\n",
- "d=12.5 #specimen diamtere in mm\n",
- "sl=62.5 #specimen length in mm\n",
- "A=(math.pi*(d*10**-3)**2)/4 # in meter square\n",
- "ylp=((yl*10**3)/(A)) #yeild point stress in N/m**2\n",
- "uts=((ml*10**3)/(A)) #ultimate tensile strangth in N/m**2\n",
- "pel=((glf-sl)/sl)*100 #percentage elongation\n",
- "strss=((20*10**3)/A) #stress at 20kN in N/m**2\n",
- "mel=strss/st #modulus of elasticity in N/m**2\n",
- "mrs=((ylp*10**-3)**2/(2*mel)) #modulus of resilience \n",
- "fs=((fl*10**3)/(A)) #fracture stress in N/m**2\n",
- "mth=((ylp+uts)*(pel/100))/2 #modulus of toughness in N/m**2\n",
- "print mth,\" is modulus of toughness in N/m**2\"\n",
- "#percentage reduction in area is not calulated in the textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "123567388.521 is modulus of toughness in N/m**2\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.4:pg-155"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "#Example 7.4 : true breaking stress and nominal breaking stress \n",
- " \n",
- "\n",
- "#given data :\n",
- "d1=12.7; # in mm\n",
- "B_load=14;# in K-N\n",
- "A1=(math.pi*d1**2)/4;# original cross section area\n",
- "d2=7.87; # in mm\n",
- "A2=(math.pi*d2**2)/4;# final cross sction area\n",
- "T_stress=B_load/A2;\n",
- "print T_stress*1000,\" is true breaking stress,T_stress in (N/mm**2) \"\n",
- "N_stress=B_load/A1;\n",
- "print N_stress*1000,\" is nominal breaking stress,N_stress in (N/mm**2) \"\n",
- "#true breaking stress unit is wrong in the textbook\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "287.798608363 is true breaking stress,T_stress in (N/mm**2) \n",
- "110.517413518 is nominal breaking stress,N_stress in (N/mm**2) \n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.5.a:pg-155"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 7.5.a: yield point stress\n",
- " \n",
- "\n",
- "yl=34.0;#yeild load in kN\n",
- "ul=61.0;#ultimate load in kN\n",
- "fl=78.0;#final length in mm\n",
- "glf=60.0;#gauge length of fratture in mm\n",
- "fd=7.0;#final diamtere in mm\n",
- "d=12.0;#specimen diamtere in mm\n",
- "sl=62.5;#specimen length in mm\n",
- "A=(math.pi*(d)**2)/4;# in meter square\n",
- "ylp=((yl*10**3)/(A));# yeild point stress in N/mm**2\n",
- "print floor(ylp),\" is yeild point stress in N/mm**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "300.0 is yeild point stress in N/mm**2\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.5.b:pg-155"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 7.5.b: ultimate tensile stress\n",
- " \n",
- "\n",
- "yl=34.0;#yeild load in kN\n",
- "ul=61.0;#ultimate load in kN\n",
- "fl=78.0;#final length in mm\n",
- "glf=60.0;#gauge length of fratture in mm\n",
- "fd=7.0;#final diamtere in mm\n",
- "d=12.0;#specimen diamtere in mm\n",
- "sl=62.5;#specimen length in mm\n",
- "A=(math.pi*(d)**2)/4;# in meter square\n",
- "uts=((ul*10**3)/(A));#ultimate tensile strangth in N/mm**2\n",
- "print uts,\" is ultimate tensile strangth in N/mm**2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "539.358418256 is ultimate tensile strangth in N/mm**2\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.5.c:pg-155"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 7.5.c: percentage reduction\n",
- " \n",
- " \n",
- "yl=34;#yeild load in kN\n",
- "ul=61;#ultimate load in kN\n",
- "fl=78;#final length in mm\n",
- "glf=60;#gauge length of fratture in mm\n",
- "fd=7;#final diamtere in mm\n",
- "d=12;#specimen diamtere in mm\n",
- "sl=62.5;#specimen length in mm\n",
- "A=(math.pi*(d)**2)/4;# in mm square\n",
- "A1=(math.pi*(fd)**2)/4;# in mm square\n",
- "pr=(A-A1)/A;# reduction\n",
- "print round(pr*100,2),\"% is percentage reduction\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "65.97 % is percentage reduction\n"
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.5.d:pg-155"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 7.5.d: percentage elonagtion\n",
- " \n",
- "\n",
- "yl=34.0;#yeild load in kN\n",
- "ul=61.0;#ultimate load in kN\n",
- "fl=78.0;#final length in mm\n",
- "glf=60.0;#gauge length of fratture in mm\n",
- "fd=7.0;#final diamtere in mm\n",
- "d=12.0;#specimen diamtere in mm\n",
- "sl=62.5;#specimen length in mm\n",
- "A=(math.pi*(d)**2)/4;# in mm square\n",
- "A1=(math.pi*(fd)**2)/4;# in mm square\n",
- "pr=(fl-glf)/glf;# elongation\n",
- "print round(pr*100,2),\"% is percentage elongtion \"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "30.0 % is percentage elongtion \n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.6:pg-156"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 7.6 : strain \n",
- " \n",
- "\n",
- "#given data :\n",
- "b=44.5*10**3;#force\n",
- "E=1.1*10**5;# in N/mm**2\n",
- "A=15.2*19.1# in mm**2\n",
- "epsilon=b/(A*E);\n",
- "print \"strain,epsilon (mm) = \",epsilon"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "strain,epsilon (mm) = 0.00139344672963\n"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.7:pg-156"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 7.7 :stress and strain \n",
- " \n",
- "\n",
- "#given data :\n",
- "sigma=450;#in MPa\n",
- "epsilon=0.63;\n",
- "sigma_t=sigma*(1+epsilon);\n",
- "print \"true stress,sigma_t(MPa) = \",sigma_t\n",
- "epsilon_t=math.log(1+epsilon);\n",
- "print \"true strain,epsilon_t(MPa) = \",epsilon_t\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "true stress,sigma_t(MPa) = 733.5\n",
- "true strain,epsilon_t(MPa) = 0.488580014819\n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex7.8:pg-157"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 7.8: which part has a greater stress\n",
- " \n",
- "l=24.0;#length in mm\n",
- "b=30;#breadth in mm\n",
- "ld=7000;#load in kg\n",
- "sd=10;#steel bar diamtere in mm\n",
- "sl=5000.0;#load in kg\n",
- "al=ld/(l*b);#stress on aluminium bar in kg/mm**2\n",
- "a=((math.pi*sd**2)/4.0);#area in mm**2\n",
- "slb=sl/a;#stress on steel bar in kg/mm**2\n",
- "print\"stress on aluminium bar is \",round(al,2),\" kg/mm**2 is less than stress on steel bar \",round(slb,2),\" kg/mm**2 \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "stress on aluminium bar is 9.72 kg/mm**2 is less than stress on steel bar 63.66 kg/mm**2 \n"
- ]
- }
- ],
- "prompt_number": 29
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Materials_Science/Chapter08.ipynb b/Materials_Science/Chapter08.ipynb deleted file mode 100755 index 7ab3ee40..00000000 --- a/Materials_Science/Chapter08.ipynb +++ /dev/null @@ -1,179 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:34ae042b330defbf3a9166ffcdabea3a9847c3ed545aa16f3851917a2138bfe6"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter08:Deformation of Metals"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.1:pg-175"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 8.1: critical resolved shear stress of silver\n",
- " \n",
- "\n",
- "Ts=15;#tensile stress in Mpa\n",
- "d=[1,1,0];\n",
- "d1=[1,1,1];\n",
- "csda=((d[0]*d1[0])+(d[1]*d1[1])+(d[2]*d1[2]))/((math.sqrt(d[0]**2+d[1]**2+d[2]**2))*math.sqrt(d1[0]**2+d1[1]**2+d1[2]**2));#angle degree\n",
- "d2=[0,1,1];\n",
- "csdb=((d[0]*d2[0])+(d[1]*d2[1])+(d[2]*d2[2]))/((math.sqrt(d[0]**2+d[1]**2+d[2]**2))*math.sqrt(d2[0]**2+d2[1]**2+d2[2]**2));#angle degree\n",
- "t=Ts*csda*csdb;#critical resolved shear stress in MPa\n",
- "print round(t,2),\"= critical resolved shear stress in MPa\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "6.12 = critical resolved shear stress in MPa\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.2:pg-186"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 8.2: yield strength of material\n",
- " \n",
- "import numpy.linalg as lin\n",
- "import math\n",
- "ys1=115;# yeild strength in MN/mm**2\n",
- "ys2=215;# yeild strength in MN/mm**2\n",
- "d1=0.04;#diamtere in mm\n",
- "d2=0.01;#diamtere in mm\n",
- "A=numpy.array([[2 ,10], [1 ,10]]);\n",
- "B=numpy.array([230,215]);\n",
- "x=lin.solve(A,B)\n",
- "si=x[0];# in MN/mm**2\n",
- "k=x[1];#\n",
- "d3=0.016;#in mm\n",
- "sy= si +(k/math.sqrt(d3));#yeild strength for a grain size in MN/mm**2\n",
- "print round(sy,2),\"=yeild strength for a grain size in MN/mm**2\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "173.11 =yeild strength for a grain size in MN/mm**2\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex8.3:pg-186"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 8.3: yield strength of material\n",
- "import numpy.linalg as lin\n",
- "import math\n",
- "ys1=120;# yeild strength in MN/mm**2\n",
- "ys2=220;# yeild strength in MN/mm**2\n",
- "d1=0.04;#diamtere in mm\n",
- "d2=0.01;#diamtere in mm\n",
- "A=numpy.array([[2 ,10], [1 ,10]]);\n",
- "B=numpy.array([240,220]);\n",
- "x=lin.solve(A,B)\n",
- "si=x[0];# in MN/mm**2\n",
- "k=x[1];#\n",
- "d3=0.025;#in mm\n",
- "sy= si +(k/math.sqrt(d3));#yeild strength for a grain size in MN/mm**2\n",
- "print round(sy,2),\"= yeild strength for a grain size in MN/mm**2\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "146.49 = yeild strength for a grain size in MN/mm**2\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex18.4:pg-193"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 8.4 : grain diameter\n",
- "import math \n",
- "\n",
- "#given data :\n",
- "N=9; # ASTM number\n",
- "m=8*2**N; # no. of grains [er square millimetre\n",
- "grain=1/math.sqrt(m);\n",
- "print round(grain,4),\"=the grain diameter(mm) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.0156 =the grain diameter(mm) \n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Materials_Science/Chapter09.ipynb b/Materials_Science/Chapter09.ipynb deleted file mode 100755 index f9450a23..00000000 --- a/Materials_Science/Chapter09.ipynb +++ /dev/null @@ -1,200 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:8681b8cb87a35e316321fbab8aecdb38b5b347586e7dea8acd0d04d98e829c88"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter09:Fracture of Metals"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex9.1:pg-200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 9.1 : difference\n",
- "import math \n",
- "#given data :\n",
- "E=200*10**9; # in N/m**2\n",
- "C=(4*10**-6)/2;# in m\n",
- "gama=1.48; # in J/m**2\n",
- "sigma=math.sqrt((2*E*gama)/(math.pi*C));\n",
- "print round(sigma*10**-6),\"= fracture strength,sigma(MN/m**2) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "307.0 = fracture strength,sigma(MN/m**2) \n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex9.2:pg-200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 9.2 : the fracture strength and compare\n",
- " \n",
- "import math\n",
- "#given data :\n",
- "E=70*10**9; # in N/m**2\n",
- "C=(4.2*10**-6)/2;# in m\n",
- "gama=1.1; # in J/m**2\n",
- "sigma=math.sqrt((2*E*gama)/(math.pi*C));\n",
- "print sigma,\"= fracture strength,sigma(N/m**2) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "152783261.475 = fracture strength,sigma(N/m**2) \n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex9.3:pg-200"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 9.3 : maximum length of surface\n",
- "import math\n",
- "\n",
- "#given data :\n",
- "sigma=36;#in MN/m**2\n",
- "gama=0.27;# in J/m**2\n",
- "E=70*10**9;#in N/m**2\n",
- "C=((2*E*gama)/(sigma**2*math.pi))*10**-6;\n",
- "C2=2*C;\n",
- "print round(C2,3),\"= maximum length of surface flow,C2(micro-m) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "18.568 = maximum length of surface flow,C2(micro-m) \n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex9.4a:pg-203"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 9.4.a: Temperature\n",
- " \n",
- "import math\n",
- "E=350;# in GN/m**2\n",
- "Y=2;# in J/m**2\n",
- "C=2;# in micro meter\n",
- "sg=math.sqrt((2*E*10**9*Y)/(math.pi*C*10**-6));# IN mn/M**2\n",
- "e=10**-2;# per second\n",
- "T=173600/(round(sg*10**-6)-20.6-61.3*(math.log10(e)));# in kelvin\n",
- "print round(T,1),\"= temperature in kelvin for ductile to brittle transition at a strain rate of 10**-2 per second\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "302.4 = temperature in kelvin for ductile to brittle transition at a strain rate of 10**-2 per second\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex9.4b:pg-203"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 9.4.b: Temperature\n",
- "import math\n",
- "\n",
- "E=350;# in GN/m**2\n",
- "Y=2;# in J/m**2\n",
- "C=2;# in micro meter\n",
- "sg=math.sqrt((2*E*10**9*Y)/(math.pi*C*10**-6));# IN mn/M**2\n",
- "e=10**-5;# per second\n",
- "T=173600/(round(sg*10**-6)-20.6-61.3*(math.log10(e)));# in kelvin\n",
- "print round(T),\"= temperature in kelvin for ductile to brittle transition at a strain rate of 10**-5 per second\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "229.0 = temperature in kelvin for ductile to brittle transition at a strain rate of 10**-5 per second\n"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Materials_Science/Chapter15.ipynb b/Materials_Science/Chapter15.ipynb deleted file mode 100755 index f927d51b..00000000 --- a/Materials_Science/Chapter15.ipynb +++ /dev/null @@ -1,61 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:7315f4cd0dbe1ea8c121564011c9ce9945618f308024d617157c5ed6076bc7d9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter15:Composite Materials and Ceramics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex15.1:pg-299"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 15.1 : colume ratio of aluminium and boron\n",
- " \n",
- "import numpy.linalg as lin\n",
- "yal=715;# in GN/,**2\n",
- "yfe=210;# in GN/,**2\n",
- "yb=440;# in GN/,**2\n",
- "A=numpy.array([[71, 71],[71, 440]]);#\n",
- "B=numpy.array([71,210]);#\n",
- "X=lin.solve(A,B)\n",
- "print round(X[0],3),\"is volume ratio of aluminium \"\n",
- "print round(X[1],3),\"is volume ratio of boron \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.623 is volume ratio of aluminium \n",
- "0.377 is volume ratio of boron \n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Materials_Science/Chapter16.ipynb b/Materials_Science/Chapter16.ipynb deleted file mode 100755 index 3ff4c3db..00000000 --- a/Materials_Science/Chapter16.ipynb +++ /dev/null @@ -1,275 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:267ebfe4747dc282d89f7efa1dce2f2f4b1f305bd03bf7fa58c0d8bd7749c9c1"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter16:Semiconductors"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex16.1:pg-315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 16.1 : concentration\n",
- "\n",
- "#given data :\n",
- "e=1.602*10**-19;# Coulomb\n",
- "sigma_i=5*10**-4;# in ohm/m\n",
- "mu_n=0.14;# in m**2/V-sec\n",
- "mu_p=0.05;# in m**2/V-sec\n",
- "n_i=sigma_i/(e*(mu_n+mu_p));\n",
- "print round(n_i*10**6,-20),\"= the concentration,n_i(/cm**3) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.64e+22 = the concentration,n_i(/cm**3) \n"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex16.2:pg-315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 16.2 : intrinsic carrier\n",
- " \n",
- "\n",
- "#given data :\n",
- "e=1.602*10**-19; # Coulomb\n",
- "p_i=2*10**-4;# in ohm-m\n",
- "mu_n=6;# in m**2/V-sec\n",
- "mu_p=0.2;# in m**2/V-sec\n",
- "n_i=1/(e*(mu_n+mu_p)*p_i);\n",
- "print round(n_i,-19),\"= the intrinsic carrier,n_i(/m**3) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "5.03e+21 = the intrinsic carrier,n_i(/m**3) \n"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex16.3:pg-315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 16.3 : neglect the intrinsic conductivity\n",
- " \n",
- "\n",
- "#given data :\n",
- "e=1.6*10**-19; # Coulomb\n",
- "sigma=10**-12;# in mhos/m\n",
- "mu_n=0.18;# in m**2/V-sec\n",
- "n=sigma/(e*mu_n);\n",
- "N=n; # amount of n type impurity\n",
- "print round(N),\"in(/m**3) \"\n",
- "# The answer is slightly different in textbook due to approximation"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "34722222.0 in(/m**3) \n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex16.4:pg-315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 16.4 : number of electron carriers\n",
- " \n",
- "\n",
- "#given data :\n",
- "e=1.6*10**-19; # Coulomb\n",
- "p=20*10**-2;# in ohm-m\n",
- "mu_n=100*10**-4;# in m**2/V-sec\n",
- "n=1/(e*mu_n*p);\n",
- "print round(n,-19),\"= number of electrons carrier,n(/m**3) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "3.12e+21 = number of electrons carrier,n(/m**3) \n"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex16.5:pg-316"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 16.5 : concentration of impurity\n",
- "import math\n",
- "e=1.6*10**-19;# Coulomb\n",
- "l=10;#in mm\n",
- "d=1;#in mm\n",
- "r=100;#in ohms\n",
- "up=0.19;#mobilty of electrons in V-sec\n",
- "a=(math.pi*((d*10**-3)**2))/4;#area in m**2\n",
- "p=((r*a))/(l*10**-3);#resistivity in Ohm-cm\n",
- "n=((1/(p*e*up)));#concentration in per m**3\n",
- "print round(n,-19),\"is impurity concentration is in per m**3\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "4.19e+21 is impurity concentration is in per m**3\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex16.6:pg-316"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 16.6 : intrinsic carrier density\n",
- " \n",
- "#given data :\n",
- "\n",
- "e=1.602*10**-19; # in coulomb\n",
- "p=3000.0;# in ohm/m\n",
- "sigma=1/p;# in ohm/m\n",
- "mu_n=0.14;# in m**2/V-sec\n",
- "mu_p=0.05;# in m**2/V-sec\n",
- "n_i=sigma/(e*(mu_n+mu_p));\n",
- "print round(n_i,-13),\"is the concentration,n_i(/m**3) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "1.095e+16 is the concentration,n_i(/m**3) \n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex16.7:pg-317"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 16.7 : conductivity\n",
- " \n",
- "#given data :\n",
- "e=1.602*10**-19; # in coulomb\n",
- "n_i=5.021*10**15; # in m**-3\n",
- "mu_n=0.48;# in m**2/V-sec\n",
- "mu_p=0.013;# in m**2/V-sec\n",
- "sigma=n_i*(e*(mu_n+mu_p));\n",
- "print round(sigma,9),\"= the conductivity,sigma(ohm**-1 m**-1) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.000396552 = the conductivity,sigma(ohm**-1 m**-1) \n"
- ]
- }
- ],
- "prompt_number": 42
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Materials_Science/Chapter17.ipynb b/Materials_Science/Chapter17.ipynb deleted file mode 100755 index 9a4bd8d7..00000000 --- a/Materials_Science/Chapter17.ipynb +++ /dev/null @@ -1,60 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:03648571a8c8e1b4e8f8e6dad5bff5a090472aa9ccf6236105c75aa2a594a7be"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter17:Insulating Materials"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex17.1:pg-333"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# Example 17.1: greater chanrge\n",
- " \n",
- "\n",
- "er1=6;#\n",
- "d1=0.25;# in mm\n",
- "a=1;# assume\n",
- "er2=2.6;#\n",
- "d2=0.1;# in mm\n",
- "c1=(er1/d1);# in ampere\n",
- "c2=(er2/d2);# in amperes\n",
- "print\" C1 \",c1,\"A will hold the more charge than C2 \",c2,\"A \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " C1 24.0 A will hold the more charge than C2 26.0 A \n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Materials_Science/Chapter18.ipynb b/Materials_Science/Chapter18.ipynb deleted file mode 100755 index 287d3fc7..00000000 --- a/Materials_Science/Chapter18.ipynb +++ /dev/null @@ -1,251 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:6d6e15c1e7376b50229c166646322c80b0a7d8183966dd1596145061fc9a1c4d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter18:Magnetic Materials"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex18.1:pg-346"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 18.1 : magnetization and flux density\n",
- "import math\n",
- "#given data :\n",
- "mu0=4*math.pi*10**-7;\n",
- "H=10**4;# in A/m\n",
- "Xm=3.7*10**-3;# room temperature\n",
- "mu_r=1+Xm;\n",
- "B=mu0*mu_r*H;\n",
- "M=Xm*H;\n",
- "print B,\"= the flux density,B(Wb/m**2) \"\n",
- "print M,\"= magnetization,M(A/m) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.0126128661856 = the flux density,B(Wb/m**2) \n",
- "37.0 = magnetization,M(A/m) \n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex18.2a:pg-350"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 18.2.a : saturation magnetization\n",
- " \n",
- "#given data :\n",
- "mu_b=9.27*10**-24;# A.m**2\n",
- "p=8.9; # in g/cm**3\n",
- "Na=6.023*10**23;# avogadro's number\n",
- "A=58.71; # in g/mol\n",
- "n=((p*Na)/A)*10**6;\n",
- "Ms=0.60*mu_b*n;\n",
- "print Ms,\"= saturation magnetization,Ms(A/m) \"\n",
- "# the answe ris slightly different in textbook due to approximation"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "507834.0 = saturation magnetization,Ms(A/m) \n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex18.2b:pg-350"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 18.2.b : saturation flux density\n",
- "import math\n",
- "#given data :\n",
- "\n",
- "mu0=4*math.pi*10**-7;\n",
- "mu_b=9.27*10**-24;# A.m**2\n",
- "p=8.9; # in g/cm**3\n",
- "Na=6.023*10**23;# avogadro's number\n",
- "A=58.71; # in g/mol\n",
- "n=((p*Na)/A)*10**6;\n",
- "Ms=0.60*mu_b*n;\n",
- "Bs=mu0*Ms;\n",
- "print round(Bs,2),\"= saturation flux density,Bs(tesla) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "0.64 = saturation flux density,Bs(tesla) \n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex18.3:pg-351"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 18.3 : magnetic moment\n",
- "import math \n",
- "#given data :\n",
- "\n",
- "mu0=4*math.pi*10**-7;\n",
- "mu_b=9.27*10**-24;# A.m**2\n",
- "p=8.9; # in g/cm**3\n",
- "Na=6.023*10**23;# avogadro's number\n",
- "A=58.71; # in g/mol\n",
- "n=((p*Na)/A)*10**6;\n",
- "Bs=0.65;#in Wb/m**2\n",
- "Ms=Bs/mu0;\n",
- "m_mu_b=Ms/n;\n",
- "print round(m_mu_b,26),\" is saturation magnetisation,m_mu_b(A.m**2) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "5.67e-24 is saturation magnetisation,m_mu_b(A.m**2) \n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex18.4:pg-355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 18.4 : power loss\n",
- " \n",
- "#given data :\n",
- "V=0.01;#in m**3\n",
- "f=50;# in Hz\n",
- "area=600;#in jm**-1\n",
- "Wh=area*V*f;\n",
- "print Wh,\"= power loss,Wh(watts) \"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "300.0 = power loss,Wh(watts) \n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Ex18.5:pg-355"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Example 18.4 : los of energy\n",
- " \n",
- "#given data :\n",
- "mass=10.0;# in kg\n",
- "energy_loss=250.0;# in J/m**2\n",
- "#energy loss at the rate of 50 cycles/s\n",
- "E=energy_loss*50.0;# in J/m**3\n",
- "E_loss=E*3600.0;#in J/m**3\n",
- "D=7500.0;#density in kg/m**3\n",
- "Volume=mass/D;\n",
- "energy_loss_per_hour=E_loss/Volume;\n",
- "print energy_loss_per_hour,\"= energy_loss_per_hour(J/hour) \"\n",
- "\n",
- "# answer is incorrect in textbook"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "33750000000.0 = energy_loss_per_hour(J/hour) \n"
- ]
- }
- ],
- "prompt_number": 22
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |