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| author | hardythe1 | 2015-07-03 12:23:43 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-07-03 12:23:43 +0530 |
| commit | 9d260e6fae7328d816a514130b691fbd0e9ef81d (patch) | |
| tree | 9e6035702fca0f6f8c5d161de477985cacad7672 /sample_notebooks | |
| parent | afcd9e5397e3e1bde0392811d0482d76aac391dc (diff) | |
| download | Python-Textbook-Companions-9d260e6fae7328d816a514130b691fbd0e9ef81d.tar.gz Python-Textbook-Companions-9d260e6fae7328d816a514130b691fbd0e9ef81d.tar.bz2 Python-Textbook-Companions-9d260e6fae7328d816a514130b691fbd0e9ef81d.zip | |
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diff --git a/sample_notebooks/AbhishekGupta/chapter16.ipynb b/sample_notebooks/AbhishekGupta/chapter16.ipynb new file mode 100755 index 00000000..9a6c8320 --- /dev/null +++ b/sample_notebooks/AbhishekGupta/chapter16.ipynb @@ -0,0 +1,108 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:5681cda1a39ee67e447ba1a84903896a8e1196df60e583969d262ff3e357d1cb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter16:WIRELESS WANs: CELLULAR TELEPHONE\n",
+ "AND SATELLITE NETWORKS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16.1:pg-479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example1\n",
+ "#calculate the period of the Moon\n",
+ "\n",
+ "C=1.0/100;\n",
+ "dist_moon=384000; # 384,000 km\n",
+ "radius_earth = 6378; # 6378 km\n",
+ "distance=dist_moon+radius_earth ;# total distance in km\n",
+ "Period=C*((distance)**1.5); #formula\n",
+ "month=round(Period/2592000); # 1 month = 60*60*24*30=2592000 seconds\n",
+ "print\"The period of the Moon, according to Keplers law is\",round(month)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The period of the Moon, according to Keplers law is 1.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16.2:pg-479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example2\n",
+ "#calculate of the period of the satellite\n",
+ "C=1.0/100;\n",
+ "orbit=35786; # 35,786 km\n",
+ "radius_earth = 6378; # 6378 km\n",
+ "distance=orbit+radius_earth ;# total distance in km\n",
+ "Period=C*((distance)**1.5); #formula\n",
+ "hour=round(Period/3600); # 1 hour = 60*60=3600 seconds\n",
+ "print\"According to Keplers law, the period of the satellite is\",floor(Period),\"s or \",hour, \"hours.\"\n",
+ "print\"\\nThis means that a satellite located at\", orbit,\"km has a period of \",hour,\"h, which is the same as the rotation period of the Earth.\\nA satellite like this is said to be stationary to the Earth. The orbit is called a geosynchronous orbit.\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "According to Keplers law, the period of the satellite is 86579.0 s or 24.0 hours.\n",
+ "\n",
+ "This means that a satellite located at 35786 km has a period of 24.0 h, which is the same as the rotation period of the Earth.\n",
+ "A satellite like this is said to be stationary to the Earth. The orbit is called a geosynchronous orbit.\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/sample_notebooks/AkshayPatil/Chapter02.ipynb b/sample_notebooks/AkshayPatil/Chapter02.ipynb new file mode 100755 index 00000000..0aad1c0c --- /dev/null +++ b/sample_notebooks/AkshayPatil/Chapter02.ipynb @@ -0,0 +1,108 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1cb720af73aa9aeffe36433462db80e841a95011aedc57edfe5b2ec1f1efe5b3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 02:Highway Planning and Alignment" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2-1, Page Number-35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Note:The variables have been changed to ease coding\n", + "Tot_Area=14400 #Total Area in km^2\n", + "Agri_Area=4800 #Agricultural Area in km^2\n", + "Track_Len=219 #Kength of the track in km\n", + "Met_Road=469 #Total Existing length of metalled road in km\n", + "Unmet_Road=412 #Total Existing Length of Unmetalled road in km\n", + "#The tabular data will be taken as and when required\n", + "\n", + "#Calculations\n", + "#Metalled Road\n", + "A=Agri_Area\n", + "B=Tot_Area-Agri_Area\n", + "N=50 #No: of villages and town in the population range of 2001 to 5000\n", + "T=10\n", + "#Solving for L\n", + "Road_Length=(A/8)+(B/32)+1.6*N+8*T #Length of metalled road in km\n", + "D=0.15*Road_Len\n", + "R=Track_Len\n", + "L=Road_Length+D-R\n", + "#Unmetalled Road\n", + "#Values from the given table\n", + "V=500 #No: of villages and towns\n", + "Q=300\n", + "P=200\n", + "S=50\n", + "#Solving for L1\n", + "Road_Length2=0.32*V+0.8*Q+1.6*P+3.2*S\n", + "L1=Road_Length2+0.15*Road_Length2 #Length in km\n", + "\n", + "#Final Computation\n", + "#(1)\n", + "Tot_Len_Met_Road=Road_Length+D-Met_Road #In km\n", + "#(2)\n", + "Tot_Len_Un_Road=L1-Unmet_Road #In km\n", + "#(3)\n", + "Tot_Road=Road_Len+D+L1 #In km\n", + "#Road per 100km^2\n", + "Road_Len_per=(Tot_Road/Tot_Area)*100 #in km\n", + "\n", + "#Result\n", + "print \"Total Additional length of metalled road is\", Tot_Len_Met_Road,\"km\"\n", + "print \"Total Additional Length of Unmetalled road is\", Tot_Len_Un_Road,\"km\"\n", + "print \"Total length of both categories is\",Tot_Road,\"km\"\n", + "print \"Road Length per 100sqkm is\",round(Road_Len_per,1),\"km\"\n", + "print \"As road length per 100km^2 is near to 16km, the target is achieved\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total Additional length of metalled road is 750.0 km\n", + "Total Additional Length of Unmetalled road is 600.0 km\n", + "Total length of both categories is 2231.0 km\n", + "Road Length per 100sqkm is 15.5 km\n", + "As road length per 100km^2 is near to 16km, the target is achieved\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/sample_notebooks/AnkitKumar/chapter6.ipynb b/sample_notebooks/AnkitKumar/chapter6.ipynb new file mode 100755 index 00000000..ca369989 --- /dev/null +++ b/sample_notebooks/AnkitKumar/chapter6.ipynb @@ -0,0 +1,295 @@ +{ + "metadata": { + "name": "", + "signature": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Ch-6 Xrays" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6.1 : Wavelength of X-rays: Pg: 156" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h = 6.6e-034; # Planck's constant, J-s\n", + "V = 50000; # Potential difference, volts\n", + "c = 3e+08; # Velocity of light, m/s\n", + "e = 1.6e-019; # Charge of an electron, coulombs\n", + "L_1 = h*c/(e*V); # wavelength of X-rays, m\n", + "L = L_1/1e-010; # wavelength of X-rays, angstorm\n", + "print \"\\nThe shortest wavelength of X-rays = %6.4f angstorm\" % L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The shortest wavelength of X-rays = 0.2475 angstorm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6.2 : Planck's constant: Pg: 156" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "L = 24.7e-012; # Wavelength of X-rays, m\n", + "V = 50000; # Potential difference, volts\n", + "c = 3e+08; # Velocity of light, m/s\n", + "e = 1.6e-019; # Charge of an electron, coulombs\n", + "# Since e*V = h*c/L; # Energy required by an electron to move through a potential barrier of one volt, joules\n", + "# solving for h\n", + "h = e*V*L/c; # Planck's constant, Joule second\n", + "print \"h = %3.1e Js \" %h" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "h = 6.6e-34 Js \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6.3 : Short wavelength limit : Pg: 156" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "V = 50000; # Potential difference, volts\n", + "h = 6.624e-034; # Planck's constant, Js\n", + "c = 3e+08; # Velocity of light, m/s\n", + "e = 1.6e-019; # Charge of an electron, coulombs\n", + "# Since e*V = h*c/L; # Energy required by an electron to move through a potential barrier of one volt, joules\n", + "# solving for L\n", + "L = h*c/(e*V); # Short wavelength limit of X-ray, m\n", + "print \"Short wavelength limit of X-ray = %6.4f angstorm\" %(L/1E-10)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Short wavelength limit of X-ray = 0.2484 angstorm\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6.4 : Wavelength limit of X-rays : Pg: 157" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "V = 20000; # Potential difference, volt\n", + "h = 6.624e-034; # Planck's constant, Js\n", + "c = 3e+08; # Velocity of light, m/s\n", + "e = 1.6e-019; # Charge of an electron, coulombs\n", + "# Since e*V = h*c/L; # Energy required by an electron to move through a potential barrier of one volt, joules\n", + "# solving for L\n", + "L = h*c/(e*V); # Wavelength limit of X-rays, m\n", + "print \"Short wavelength limit of X-ray = %6.4f angstorm\" % (L/1E-010);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Short wavelength limit of X-ray = 0.6210 angstorm\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6.5 : Minimum voltage of an X-ray tube : Pg: 157" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h = 6.625e-034; # Planck's constant, Js\n", + "c = 3e+08; # Velocity of light, m/s\n", + "e = 1.6e-019; # Charge of an electron, coulombs\n", + "L = 1e-010; # Wavelength of X-rays, m\n", + "# Since e*V = h*c/L; # Energy required by an electron to move through a potential barrier of one volt, joules\n", + "# solving for V\n", + "V = h*c/(L*e); # Potential difference, volts\n", + "print \"The minimum voltage of an X-ray tube = %5.2f kV\"%(V/1e+03);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum voltage of an X-ray tube = 12.42 kV\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6.5 : Minimum voltage of an X-ray tube : Pg: 157" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h = 6.625e-034; # Planck's constant, Js\n", + "c = 3e+08; # Velocity of light, m/s\n", + "e = 1.6e-019; # Charge of an electron, coulombs\n", + "L = 1e-010; # Wavelength of X-rays, m\n", + "# Since e*V = h*c/L; # Energy required by an electron to move through a potential barrier of one volt, joules\n", + "# solving for V\n", + "V = h*c/(L*e); # Potential difference, volts\n", + "print \"The minimum voltage of an X-ray tube = %5.2f kV\"%( V/1e+03)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum voltage of an X-ray tube = 12.42 kV\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6.6 : Minimum wavelength emitted by an X-ray tube : Pg: 157" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h = 6.625e-034; # Planck's constant, Js\n", + "c = 3e+08; # Velocity of light, m/s\n", + "e = 1.6e-019; # Charge of an electron, coulombs\n", + "V = 4.5e+04; # Accelerating potential of X-ray tube, volt\n", + "# Since e*V = h*c/L_min; # Energy required by an electron to move through a potential barrier of one volt, joules\n", + "# solving for L_min\n", + "L_min = h*c/(V*e); # Minimum wavelength emitted by an X-ray tube, m\n", + "print \"The minimum wavelength emitted by the X-ray tube = %5.3f angstrom\"%(L_min/1e-010);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum wavelength emitted by the X-ray tube = 0.276 angstrom\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6.7: Critical voltage for stimualted emission : Pg: 158" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "h = 6.625e-034; # Planck's constant, Js\n", + "c = 3e+08; # Velocity of light, m/s\n", + "e = 1.6e-019; # Charge of an electron, coulombs\n", + "L_k = 0.178e-010; # Wavelength of k absorption egde of X-rays, m\n", + "# Since e*V_critical = h*c/L; # Energy required by an electron to move through a potential barrier of one volt, joules\n", + "# solving for V_critical\n", + "V_critical = h*c/(L_k*e); # Crtical voltage for stimulated enission, volt\n", + "print \"The critical voltage for stimulated emission = %4.1f kV\"%(V_critical/1e+03);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The critical voltage for stimulated emission = 69.8 kV\n" + ] + } + ], + "prompt_number": 21 + } + ], + "metadata": {} + } + ] +} diff --git a/sample_notebooks/AviralYadav/Chapter7.ipynb b/sample_notebooks/AviralYadav/Chapter7.ipynb new file mode 100755 index 00000000..f8cf792d --- /dev/null +++ b/sample_notebooks/AviralYadav/Chapter7.ipynb @@ -0,0 +1,474 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:aa3033a597a0bf06128c3fae6fc134561fe0b608c1a6f342b1690de7caed5ad0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter07:Differential and Multistage amlplifiers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.1:pg-690"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Example 7.1 Analysis of differential amplifier\n",
+ "# Consider the differential amplifier\n",
+ "\n",
+ "B=100.0; # beta value\n",
+ "\n",
+ "# 7.1a\n",
+ "V_T=0.025; # (V)\n",
+ "I_E=0.0005; # (A)\n",
+ "R_E=150.0; # (ohm)\n",
+ "r_e1=V_T/I_E; # emitter resistance (ohm)\n",
+ "r_e2=r_e1; # emitterA resistance (ohm)\n",
+ "r_e=r_e1;\n",
+ "R_id=2*(B+1)*(r_e+R_E);\n",
+ "print round(R_id/1000.0),\"The input differential resistance R_id (kohm)\"\n",
+ "\n",
+ "# 7.1b\n",
+ "R_id=40000.0; # (ohm)\n",
+ "R_sig=5000.0; # (ohm)\n",
+ "R_C=10000.0; # (ohm)\n",
+ "R_E=150.0; # (ohm)\n",
+ "A_v=R_id/(R_id+R_sig); # A_v= v_o/v_sig (V/V)\n",
+ "A_V=2*R_C/(2.0*(r_e+R_E)); # A_V= v_o/v_id (V/v)\n",
+ "A_d=A_v*A_V; # A_d=v_o/v_sig (V/V)\n",
+ "print \"Overall differential voltage gain (V/V)\",round(A_d,-1)\n",
+ "\n",
+ "# 7.1c\n",
+ "R_EE=200000.0; # (ohm)\n",
+ "deltaR_C=0.02*R_C; # in the worst case\n",
+ "A_cm=R_C*deltaR_C/(2.0*R_EE*R_C)\n",
+ "print A_cm,\"Worst case common mode gain (V/V)\"\n",
+ "\n",
+ "# 7.1d\n",
+ "CMRR=20*math.log10(A_d/A_cm)\n",
+ "print int(CMRR),\"CMRR in dB\"\n",
+ "\n",
+ "# 7.1e\n",
+ "r_o=200000.0; #(ohm)\n",
+ "R_icm=(B+1)*(R_EE*r_o/2.0)/(R_EE+r_o/2.0);\n",
+ "print round(R_icm/1e6,1),\"Input common mode resistance (kohm)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "40.0 The input differential resistance R_id (kohm)\n",
+ "Overall differential voltage gain (V/V) 40.0\n",
+ "0.0005 Worst case common mode gain (V/V)\n",
+ "98 CMRR in dB\n",
+ "6.7 Input common mode resistance (kohm)\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.2:pg-747"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Example 7.2 : Analysis of Active loaded MOS differential amplifier\n",
+ "W=7.2*10**-6; # (m)\n",
+ "L=0.36*10**-6; # (m)\n",
+ "C_gs=29*10**-15; # (F)\n",
+ "C_gd=5*10**-15; # (F)\n",
+ "C_db=5*10**-15; # (F)\n",
+ "uC_n=387*10**-6; # uC_n=u_nC_ox (A/V**2)\n",
+ "uC_p=86*10**-6; # uC_p=u_pC_ox (A/V**2)\n",
+ "V_an=5; # V_an=V'_An (V/um) (V)\n",
+ "V_ap=6; # V_ap=V'_Ap (V/um) (V)\n",
+ "I=0.2*10**-3; # (A)\n",
+ "R_SS=25000; # (ohm)\n",
+ "C_SS=0.2*10**-12; # (F)\n",
+ "C_S=25*10**-15; # (F)\n",
+ "K_n=uC_n*W/L;\n",
+ "I_D=100*10**-6; # bias current (A)\n",
+ "V_OV=math.sqrt(2*I_D/K_n);\n",
+ "g_m=I/V_OV;\n",
+ "g_m1=g_m;\n",
+ "g_m2=g_m;\n",
+ "r_o1=V_an*0.36/(0.1*10**-3);\n",
+ "r_o2=r_o1;\n",
+ "K_p=uC_p*W/L;\n",
+ "V_OV34=math.sqrt(2*I_D/K_p); # V_OV3,4\n",
+ "g_m3=2*0.1*10**-3/V_OV34;\n",
+ "g_m4=g_m3;\n",
+ "r_o3=V_ap*0.36/(0.1*10**-3);\n",
+ "r_o4=r_o3;\n",
+ "A_d=g_m*(r_o2*r_o4)/(r_o2+r_o4);\n",
+ "print round(A_d,1),\"A_d (V/V)\"\n",
+ "A_cm=-1/(2*g_m3*R_SS);\n",
+ "print round(A_cm,3),\"A_cm (V/V)\"\n",
+ "CMRR=20*log10(-A_d/A_cm); # negative sign to make A_cm positive\n",
+ "print round(CMRR,1),\"CMRR in dB\"\n",
+ "C_gd1=5*10**-15; # (F)\n",
+ "C_db1=5*10**-15; # (F)\n",
+ "C_db3=5*10**-15; # (F)\n",
+ "C_gs3=20*10**-15; # (F)\n",
+ "C_gs4=20*10**-15; # (F)\n",
+ "C_m=C_gd1+C_db1+C_db3+C_gs3+C_gs4;\n",
+ "C_gd2=5*10**-15; # (F)\n",
+ "C_db2=5*10**-15; # (F)\n",
+ "C_gd4=5*10**-15; # (F)\n",
+ "C_db4=5*10**-15; # (F)\n",
+ "C_x=25*10**-15; # (F)\n",
+ "C_L=C_gd2+C_db2+C_gd4+C_db4+C_x;\n",
+ "print \"poles and zeroes of A_d\"\n",
+ "R_o=r_o2*r_o4/(r_o2+r_o4)\n",
+ "f_p1=1/(2*math.pi*C_L*R_o);\n",
+ "print int(f_p1/1e6),\"f_p1 (MHz)\"\n",
+ "f_p2=g_m3/(2*math.pi*C_m);\n",
+ "print round(f_p2/1e9,2),\"f_p2 (GHz)\"\n",
+ "f_Z=2*f_p2;\n",
+ "print round(f_Z/1e9,1),\"f_Z (GHz)\"\n",
+ "print \"Dominant pole of CMRR is at location of commom-mode gain zero\"\n",
+ "f_Z=1/(2*math.pi*C_SS*R_SS);\n",
+ "print round(f_Z/1e6,1),\"f_Z (MHz)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "12.2 A_d (V/V)\n",
+ "-0.034 A_cm (V/V)\n",
+ "51.1 CMRR in dB\n",
+ "poles and zeroes of A_d\n",
+ "360 f_p1 (MHz)\n",
+ "1.7 f_p2 (GHz)\n",
+ "3.4 f_Z (GHz)\n",
+ "Dominant pole of CMRR is at location of commom-mode gain zero\n",
+ "31.8 f_Z (MHz)\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.3:pg-751"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Example 7.3 : To determine all parameters for different transistor\n",
+ "I_REF=90*10.0**-6; # (A)\n",
+ "V_tn=0.7; # (V)\n",
+ "V_tp=0.8; # Magnitude is cconsidered\n",
+ "uC_n=160.0*10**-6; # uC_n=u_n*C_ox\n",
+ "uC_p=40*10.0**-6; # uC_p=u_p*C_ox\n",
+ "V_A=10.0; # (V)\n",
+ "V_DD=2.5; # (V)\n",
+ "V_SS=2.5; # (V)\n",
+ "L=0.8*10**-6; # (m)\n",
+ "r_o2=222.0; # (ohm)\n",
+ "r_o4=222.0; # (ohm)\n",
+ "g_m1=0.3; # (mho)\n",
+ "A_1=-g_m1*r_o2*r_o4/(r_o2+r_o4);\n",
+ "print round(A_1,2),\"=A_1 (V/V)\"\n",
+ "r_o6=111.0; # (ohm)\n",
+ "r_o7=111.0; # (ohm)\n",
+ "g_m6=0.6; # (mho)\n",
+ "A_2=-g_m6*r_o6*r_o7/(r_o6+r_o7);\n",
+ "print round(A_2,2),\"=A_2 (V/V)\"\n",
+ "print \"For Q_1\"\n",
+ "W=20*10.0**-6; # (m)\n",
+ "I_D=I_REF/2.0; # (A)\n",
+ "print round(I_D/(10.0**-6),2),\"=I_D (microA)\"\n",
+ "K_p=uC_p*W/L;\n",
+ "V_OV=math.sqrt(2*I_D/K_p);\n",
+ "print round(V_OV,2),\"=V_OV (V)\"\n",
+ "V_GS=V_tp+V_OV;\n",
+ "print round(V_GS,2),\"=V_GS (V)\"\n",
+ "g_m=2*I_D/V_OV;\n",
+ "print round(g_m/(10.0**-3),2),\"=g_m (mA/V)\"\n",
+ "r_o=V_A/I_D;\n",
+ "print int(r_o/1e3),\"=r_o (kohm)\"\n",
+ "print \"For Q_2\"\n",
+ "W=20*10.0**-6; # (m)\n",
+ "I_D=I_REF/2; # (A)\n",
+ "print round(I_D/(10.0**-6),2),\"=I_D (microA)\"\n",
+ "K_p=uC_p*W/L;\n",
+ "V_OV=math.sqrt(2*I_D/K_p);\n",
+ "print round(V_OV,2),\"=V_OV (V)\"\n",
+ "V_GS=V_tp+V_OV;\n",
+ "print round(V_GS,2),\"=V_GS (V)\"\n",
+ "g_m=2*I_D/V_OV;\n",
+ "print round(g_m/(10.0**-3),2),\"=g_m (mA/V)\"\n",
+ "r_o=V_A/I_D;\n",
+ "print round(r_o/1e3),\"=r_o (kohm)\"\n",
+ "print \"For Q_3\"\n",
+ "W=5*10**-6; # (m)\n",
+ "I_D=I_REF/2; # (A)\n",
+ "print round(I_D/(10.0**-6),2),\"=I_D (microA)\"\n",
+ "K_n=uC_n*W/L;\n",
+ "V_OV=math.sqrt(2*I_D/K_n);\n",
+ "print round(V_OV,2),\"=V_OV (V)\"\n",
+ "V_GS=V_tn+V_OV;\n",
+ "print round(V_GS,2),\"=V_GS (V)\"\n",
+ "g_m=2*I_D/V_OV;\n",
+ "print round(g_m/(10.0**-3),2),\"=g_m (mA/V)\"\n",
+ "r_o=V_A/I_D;\n",
+ "print round(r_o/1e3),\"=r_o (kohm)\"\n",
+ "print \"For Q_4\"\n",
+ "W=5*10**-6; # (m)\n",
+ "I_D=I_REF/2; # (A)\n",
+ "print round(I_D/(10.0**-6),2),\"=I_D (microA)\"\n",
+ "K_n=uC_n*W/L;\n",
+ "V_OV=math.sqrt(2*I_D/K_n);\n",
+ "print round(V_OV,2),\"=V_OV (V)\"\n",
+ "V_GS=V_tn+V_OV;\n",
+ "print round(V_GS,2),\"=V_GS (V)\"\n",
+ "g_m=2*I_D/V_OV;\n",
+ "print round(g_m/(10.0**-3),2),\"=g_m (mA/V)\"\n",
+ "r_o=V_A/I_D;\n",
+ "print round(r_o/1e3),\"=r_o (kohm)\"\n",
+ "print \"For Q_5\"\n",
+ "W=40*10.0**-6; # (m)\n",
+ "I_D=I_REF; # (A)\n",
+ "print round(I_D/(10.0**-6),2),\"=I_D (microA)\"\n",
+ "K_p=uC_p*W/L;\n",
+ "V_OV=math.sqrt(2*I_D/K_p);\n",
+ "print round(V_OV,2),\"=V_OV (V)\"\n",
+ "V_GS=V_tp+V_OV;\n",
+ "print round(V_GS,2),\"=V_GS (V)\"\n",
+ "g_m=2*I_D/V_OV;\n",
+ "print round(g_m/(10.0**-3),2),\"=g_m (mA/V)\"\n",
+ "r_o=V_A/I_D;\n",
+ "print round(r_o/1e3),\"=r_o (kohm)\"\n",
+ "print \"For Q_6\"\n",
+ "W=10*10**-6; # (m)\n",
+ "I_D=I_REF; #A\n",
+ "print round(I_D/(10.0**-6),2),\"=I_D (microA)\"\n",
+ "K_n=uC_n*W/L;\n",
+ "V_OV=math.sqrt(2*I_D/K_n);\n",
+ "print round(V_OV,2),\"=V_OV (V)\"\n",
+ "V_GS=V_tn+V_OV;\n",
+ "print round(V_GS,2),\"=V_GS (V)\"\n",
+ "g_m=2*I_D/V_OV;\n",
+ "print round(g_m/(10.0**-3),2),\"=g_m (mA/V)\"\n",
+ "r_o=V_A/I_D;\n",
+ "print round(r_o/1e3),\"=r_o (kohm)\"\n",
+ "print \"For Q_7\"\n",
+ "W=40*10**-6; # (m)\n",
+ "I_D=I_REF;#A\n",
+ "print round(I_D/(10.0**-6),2),\"=I_D (microA)\"\n",
+ "K_p=uC_p*W/L;\n",
+ "V_OV=math.sqrt(2*I_D/K_p);\n",
+ "print round(V_OV,2),\"=V_OV (V)\"\n",
+ "V_GS=V_tp+V_OV;\n",
+ "print round(V_GS,2),\"=V_GS (V)\"\n",
+ "g_m=2*I_D/V_OV;\n",
+ "print round(g_m/(10.0**-3),2),\"=g_m (mA/V)\"\n",
+ "r_o=V_A/I_D;\n",
+ "print round(r_o/1e3),\"=r_o (kohm)\"\n",
+ "print \"For Q_8\"\n",
+ "W=40*10**-6; # (m)\n",
+ "I_D=I_REF; # A\n",
+ "print round(I_D/(10.0**-6),2),\"=I_D (microA)\"\n",
+ "K_p=uC_p*W/L;\n",
+ "V_OV=math.sqrt(2*I_D/K_p);\n",
+ "print round(V_OV,2),\"=V_OV (V)\"\n",
+ "V_GS=V_tp+V_OV;\n",
+ "print round(V_GS,2),\"=V_GS (V)\"\n",
+ "g_m=2*I_D/V_OV;\n",
+ "print round(g_m/(10.0**-3),2),\"=g_m (mA/V)\"\n",
+ "r_o=V_A/I_D;\n",
+ "print int(r_o/1e3),\"=r_o (kohm)\"\n",
+ "A_O=A_1*A_2;\n",
+ "print round(20*log10(A_O)),\"=The dc open loop gain in dB\"\n",
+ "v_ICMmin=-2.5+1;\n",
+ "print round(v_ICMmin,2),\"=Lower limit of input common-mode (V)\"\n",
+ "v_ICMmax=2.2-1.1;\n",
+ "print round(v_ICMmax,2),\"=Upper limit of input common-mode (V)\"\n",
+ "v_Omax=V_DD-V_OV;\n",
+ "print round(v_Omax,2),\"=Highest allowable output voltage (V)\"\n",
+ "v_Omin=-V_SS+V_OV;\n",
+ "print round(v_Omin,2),\"=Lowest allowable output voltage (V)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "-33.3 =A_1 (V/V)\n",
+ "-33.3 =A_2 (V/V)\n",
+ "For Q_1\n",
+ "45.0 =I_D (microA)\n",
+ "0.3 =V_OV (V)\n",
+ "1.1 =V_GS (V)\n",
+ "0.3 =g_m (mA/V)\n",
+ "222 =r_o (kohm)\n",
+ "For Q_2\n",
+ "45.0 =I_D (microA)\n",
+ "0.3 =V_OV (V)\n",
+ "1.1 =V_GS (V)\n",
+ "0.3 =g_m (mA/V)\n",
+ "222.0 =r_o (kohm)\n",
+ "For Q_3\n",
+ "45.0 =I_D (microA)\n",
+ "0.3 =V_OV (V)\n",
+ "1.0 =V_GS (V)\n",
+ "0.3 =g_m (mA/V)\n",
+ "222.0 =r_o (kohm)\n",
+ "For Q_4\n",
+ "45.0 =I_D (microA)\n",
+ "0.3 =V_OV (V)\n",
+ "1.0 =V_GS (V)\n",
+ "0.3 =g_m (mA/V)\n",
+ "222.0 =r_o (kohm)\n",
+ "For Q_5\n",
+ "90.0 =I_D (microA)\n",
+ "0.3 =V_OV (V)\n",
+ "1.1 =V_GS (V)\n",
+ "0.6 =g_m (mA/V)\n",
+ "111.0 =r_o (kohm)\n",
+ "For Q_6\n",
+ "90.0 =I_D (microA)\n",
+ "0.3 =V_OV (V)\n",
+ "1.0 =V_GS (V)\n",
+ "0.6 =g_m (mA/V)\n",
+ "111.0 =r_o (kohm)\n",
+ "For Q_7\n",
+ "90.0 =I_D (microA)\n",
+ "0.3 =V_OV (V)\n",
+ "1.1 =V_GS (V)\n",
+ "0.6 =g_m (mA/V)\n",
+ "111.0 =r_o (kohm)\n",
+ "For Q_8\n",
+ "90.0 =I_D (microA)\n",
+ "0.3 =V_OV (V)\n",
+ "1.1 =V_GS (V)\n",
+ "0.6 =g_m (mA/V)\n",
+ "111 =r_o (kohm)\n",
+ "61.0 =The dc open loop gain in dB\n",
+ "-1.5 =Lower limit of input common-mode (V)\n",
+ "1.1 =Upper limit of input common-mode (V)\n",
+ "2.2 =Highest allowable output voltage (V)\n",
+ "-2.2 =Lowest allowable output voltage (V)\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.5:pg-760"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Example 7.5 : Analysis of given circuit\n",
+ "B=100.0; # beta value\n",
+ "I_E=0.2510**-3; # (A)\n",
+ "R_1=20000.0; # (ohm)\n",
+ "R_2=20000; # (ohm)\n",
+ "R_3=3000; # (ohm)\n",
+ "R_4=2300; # (ohm)\n",
+ "R_5=15700; # (ohm)\n",
+ "R_6=3000; # (ohm)\n",
+ "r_e1=25/0.25; # (ohm)\n",
+ "r_e2=r_e1; # (ohm)\n",
+ "r_pi1=(B+1)*r_e1;\n",
+ "r_pi2=(B+1)*r_e2;\n",
+ "R_id=r_pi1+r_pi2;\n",
+ "print round(R_id/1e3,2),\"Input differential resistance (kohm)\"\n",
+ "I_E=1*10.0**-3;\n",
+ "r_e4=25/1.0;\n",
+ "r_e5=r_e4;\n",
+ "r_pi4=(B+1)*r_e4;\n",
+ "r_pi5=(B+1)*r_e5;\n",
+ "R_i2=r_pi4+r_pi5;\n",
+ "print round(R_i2/1e3,2),\"Input resistance of the second stage R_i2 (kohm)\"\n",
+ "A_1=(R_i2*(R_1+R_2)/((R_i2+R_1+R_2)*(r_e1+r_e2)))\n",
+ "print round(A_1,1),\"Voltage gain of the first stage (V/V)\"\n",
+ "r_e7=25/1.0;\n",
+ "R_i3=(B+1)*(R_4+r_e7);\n",
+ "print round(R_i3/1e3,1),\"Input resistance of the third stage R_i3 (kohm)\"\n",
+ "A_2=(-R_3*R_i3)/((R_3+R_i3)*(r_e4+r_e5));\n",
+ "print round(A_2,1),\"Voltage gain of the second stage (V/V)\"\n",
+ "r_e8=25/5.0;\n",
+ "R_i4=(B+1)*(r_e8+R_6);\n",
+ "print round(R_i4/1e3,2),\"Input resistance of the third stage R_i2 (kohm)\"\n",
+ "A_3=(-R_5*R_i4)/((R_5+R_i4)*(r_e7+R_4));\n",
+ "print round(A_3,2),\"Voltage gain of the third stage (V/V)\"\n",
+ "A_4=R_6/(R_6+r_e8);\n",
+ "print round(A_4,2),\"Voltage gain of the fourth stage (V/V)\"\n",
+ "A=A_1*A_2*A_3*A_4 ; # A=v_o/v_id (V/V)\n",
+ "print round(A),\"Overall output gain (V/V)\"\n",
+ "print round(20*log10(A),1),\"Overall output gain in dB\"\n",
+ "R_o=R_6*(r_e8+R_5/(B+1))/(R_6+r_e8+R_5/(B+1))\n",
+ "print round(R_o),\"Output resistance (ohm)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "20.2 Input differential resistance (kohm)\n",
+ "5.05 Input resistance of the second stage R_i2 (kohm)\n",
+ "22.4 Voltage gain of the first stage (V/V)\n",
+ "234.8 Input resistance of the third stage R_i3 (kohm)\n",
+ "-59.2 Voltage gain of the second stage (V/V)\n",
+ "303.5 Input resistance of the third stage R_i2 (kohm)\n",
+ "-6.42 Voltage gain of the third stage (V/V)\n",
+ "1.0 Voltage gain of the fourth stage (V/V)\n",
+ "8514.0 Overall output gain (V/V)\n",
+ "78.6 Overall output gain in dB\n",
+ "152.0 Output resistance (ohm)\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/sample_notebooks/ChaitanyaPotti/Chapter2_1.ipynb b/sample_notebooks/ChaitanyaPotti/Chapter2_1.ipynb new file mode 100755 index 00000000..0734d609 --- /dev/null +++ b/sample_notebooks/ChaitanyaPotti/Chapter2_1.ipynb @@ -0,0 +1,365 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 2 - First law"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1 - pg 34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 2.1\n",
+ "The Change in total energy is, del_E (kJ) = 1100\n",
+ "Since del_E is positive, so there is an increase in total energy\n",
+ "There is mistake in the book's results unit\n"
+ ]
+ }
+ ],
+ "source": [
+ "print 'Example 2.1'\n",
+ "#calculate the Change in total energy\n",
+ "\n",
+ "# Given values\n",
+ "Q = 2500; # Heat transferred into the system, [kJ]\n",
+ "W = 1400; # Work transferred from the system, [kJ]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# since process carried out on a closed system, so using equation [4]\n",
+ "del_E = Q-W; # Change in total energy, [kJ]\n",
+ "\n",
+ "# results\n",
+ "\n",
+ "print 'The Change in total energy is, del_E (kJ) = ',del_E\n",
+ "\n",
+ "if del_E >= 0:\n",
+ " print 'Since del_E is positive, so there is an increase in total energy'\n",
+ "else:\n",
+ " print 'Since del_E is negative, so there is an decrease in total energy'\n",
+ "\n",
+ "\n",
+ "print \"There is mistake in the book's results unit\"\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2 - pg 34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 2.2\n",
+ "The Heat transfer is, Q (kJ) = -700.0\n",
+ "Since Q < 0, so heat is transferred from the system\n"
+ ]
+ }
+ ],
+ "source": [
+ "print 'Example 2.2'\n",
+ "#calculate the heat transfer\n",
+ "\n",
+ "# Given values\n",
+ "del_E = 3500.; # Increase in total energy of the system, [kJ]\n",
+ "W = -4200.; # Work transfer into the system, [kJ]\n",
+ "\n",
+ "# solution\n",
+ "# since process carried out on a closed system, so using equation [3]\n",
+ "Q = del_E+W;# [kJ]\n",
+ "\n",
+ "# results\n",
+ "print 'The Heat transfer is, Q (kJ) = ',Q\n",
+ "\n",
+ "if Q >=0:\n",
+ " print 'Since Q > 0, so heat is transferred into the system'\n",
+ "else:\n",
+ " print 'Since Q < 0, so heat is transferred from the system'\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3 - pg 38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 2.3\n",
+ "The Work done is, W (kJ/kg) = 250\n",
+ "Since W > 0, so Work done by the engine per kilogram of working substance\n"
+ ]
+ }
+ ],
+ "source": [
+ "print 'Example 2.3'\n",
+ "#calculate the Work done\n",
+ "\n",
+ "\n",
+ "# Given values\n",
+ "Q = -150; # Heat transferred out of the system, [kJ/kg]\n",
+ "del_u = -400; # Internal energy decreased ,[kJ/kg]\n",
+ "\n",
+ "# solution\n",
+ "# using equation [3],the non flow energy equation\n",
+ "# Q=del_u+W\n",
+ "W = Q-del_u; # [kJ/kg]\n",
+ "\n",
+ "# results\n",
+ "print 'The Work done is, W (kJ/kg) = ',W\n",
+ "\n",
+ "if W >=0:\n",
+ " print 'Since W > 0, so Work done by the engine per kilogram of working substance'\n",
+ "else:\n",
+ " print 'Since W < 0, so Work done on the engine per kilogram of working substance'\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4 - pg 39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 2.4\n",
+ "workdone is, W (kJ/kg) = 550.6875\n",
+ "Since W>0, so Power is output from the system\n",
+ "The power output from the system is (kW) = 2202.75\n"
+ ]
+ }
+ ],
+ "source": [
+ "print 'Example 2.4'\n",
+ "#calculate the power output and workdone\n",
+ "# Given values\n",
+ "m_dot = 4.; # fluid flow rate, [kg/s]\n",
+ "Q = -40.; # Heat loss to the surrounding, [kJ/kg]\n",
+ "\n",
+ "# At inlet \n",
+ "P1 = 600.; # pressure ,[kn/m**2]\n",
+ "C1 = 220.; # velocity ,[m/s]\n",
+ "u1 = 2200.; # internal energy, [kJ/kg]\n",
+ "v1 = .42; # specific volume, [m**3/kg]\n",
+ "\n",
+ "# At outlet\n",
+ "P2 = 150.; # pressure, [kN/m**2]\n",
+ "C2 = 145.; # velocity, [m/s]\n",
+ "u2 = 1650.; # internal energy, [kJ/kg]\n",
+ "v2 = 1.5; # specific volume, [m**3/kg]\n",
+ "\n",
+ "# solution\n",
+ "# for steady flow energy equation for the open system is given by\n",
+ "# u1+P1*v1+C1**2/2+Q=u2+P2*v2+C2**2/2+W\n",
+ "# hence\n",
+ "\n",
+ "W = (u1-u2)+(P1*v1-P2*v2)+(C1**2/2-C2**2/2)*10**-3+Q; # [kJ/kg]\n",
+ "\n",
+ "P_out = W*m_dot; # power out put from the system, [kW]\n",
+ "\n",
+ "# results\n",
+ "print 'workdone is, W (kJ/kg) = ',W\n",
+ "\n",
+ "if W >= 0:\n",
+ " print 'Since W>0, so Power is output from the system'\n",
+ "else:\n",
+ " print 'Since W<0, so Power is input to the system'\n",
+ "\n",
+ "# Hence\n",
+ "\n",
+ "print 'The power output from the system is (kW) = ',P_out\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5 - pg 39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 2.5\n",
+ "The temperature rise of the lead is (C) = 104.58\n"
+ ]
+ }
+ ],
+ "source": [
+ "print 'Example 2.5'\n",
+ "#calculate the temperature rise\n",
+ "\n",
+ "\n",
+ "# Given values\n",
+ "del_P = 154.45; # pressure difference across the die, [MN/m**2]\n",
+ "rho = 11360.; # Density of the lead, [kg/m**3]\n",
+ "c = 130; # specific heat capacity of the lead, [J/kg*K]\n",
+ "\n",
+ "# solution\n",
+ "# since there is no cooling and no externel work is done, so energy balane becomes\n",
+ "# P1*V1+U1=P2*V2+U2 ,so\n",
+ "# del_U=U2-U1=P1*V1-P2*V2\n",
+ "\n",
+ "# also, for temperature rise, del_U=m*c*t, where, m is mass; c is specific heat capacity; and t is temperature rise\n",
+ "\n",
+ "# Also given that lead is incompressible, so V1=V2=V and assuming one m**3 of lead\n",
+ "\n",
+ "# using above equations\n",
+ "t = del_P/(rho*c)*10**6 ;# temperature rise [C]\n",
+ "\n",
+ "# results \n",
+ "print 'The temperature rise of the lead is (C) = ',round(t,2)\n",
+ "\n",
+ "# End\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6 - pg 42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 2.6\n",
+ "(a) The inlet area is, A1 (m^2) = 0.0425\n",
+ "(b) The exit velocity is, C2 (m/s) = 171.71\n",
+ "(c) The power developed by the turbine system is (kW) = 671.88\n"
+ ]
+ }
+ ],
+ "source": [
+ "print 'Example 2.6'\n",
+ "#calculate the power developed, exit velocity and inlet area\n",
+ "\n",
+ "# Given values\n",
+ "m_dot = 4.5; # mass flow rate of air, [kg/s]\n",
+ "Q = -40.; # Heat transfer loss, [kJ/kg]\n",
+ "del_h = -200.; # specific enthalpy reduce, [kJ/kg]\n",
+ "\n",
+ "C1 = 90; # inlet velocity, [m/s]\n",
+ "v1 = .85; # inlet specific volume, [m**3/kg]\n",
+ "\n",
+ "v2 = 1.45; # exit specific volume, [m**3/kg]\n",
+ "A2 = .038; # exit area of turbine, [m**2]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# part (a)\n",
+ "# At inlet, by equation[4], m_dot=A1*C1/v1\n",
+ "A1 = m_dot*v1/C1;#inlet area, [m**2]\n",
+ "print '(a) The inlet area is, A1 (m^2) = ',A1\n",
+ "\n",
+ "# part (b), \n",
+ "# At outlet, since mass flow rate is same, so m_dot=A2*C2/v2, hence\n",
+ "C2 = m_dot*v2/A2; # Exit velocity,[m/s]\n",
+ "print '(b) The exit velocity is, C2 (m/s) = ',round(C2,2)\n",
+ "\n",
+ "# part (c)\n",
+ "# using steady flow equation, h1+C1**2/2+Q=h2+C2**2/2+W\n",
+ "W = -del_h+(C1**2/2-C2**2/2)*10**-3+Q; # [kJ/kg]\n",
+ "\n",
+ "# Hence power developed is\n",
+ "P = W*m_dot;# [kW]\n",
+ "print '(c) The power developed by the turbine system is (kW) = ',round(P,2)\n",
+ "\n",
+ "# End\n",
+ "\n"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/DaudIbrahir Saifi/Chapter2.ipynb b/sample_notebooks/DaudIbrahir Saifi/Chapter2.ipynb new file mode 100755 index 00000000..b9dc8563 --- /dev/null +++ b/sample_notebooks/DaudIbrahir Saifi/Chapter2.ipynb @@ -0,0 +1,228 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:cba451428c3d9c574800bbc1429b7e9efcd18af4b82f735faf4ac85b4ea52c65"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter02:The 741 IC OP-AMP"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.1:Pg-80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Ex 2.1\n",
+ "\n",
+ "# data from fig of Ex2.1\n",
+ "VCC=5.0;#V\n",
+ "IS=10**-14.0;#A\n",
+ "RS=39*1000.0;#ohm\n",
+ "VBE12=0.7;#V(Assumed)\n",
+ "VBE11=0.7;#V(Assumed)\n",
+ "VEE=-5;#V\n",
+ "IREF=(VCC-VBE12-VBE11-VEE)/RS*10**6;#micro A\n",
+ "print \"Estimated input reference current , IREF(micro A)\",round(IREF,2)\n",
+ "VT=25*10**-3;#V(Thermal Voltage)\n",
+ "VBE=VT*log(IREF*10**-6/IS);#V\n",
+ "IREF=(VCC-VBE-VBE-VEE)/RS*10**6;#micro A\n",
+ "print \"More precise value of reference current , IREF(micro A)\",round(IREF,2)\n",
+ "#Replacing Vcc by 15 V in the original design\n",
+ "VCC2=15.0;#V\n",
+ "VEE2=-15.0;#V\n",
+ "IREF=(VCC2-VBE-VBE-VEE2)/RS*10**6;#micro A\n",
+ "VBE=VT*log(IREF*10**-6/IS);#V\n",
+ "R5=(VCC-VBE-VBE-VEE)/(IREF*10**-6);#ohm\n",
+ "R5=round(R5/1000);#kohm\n",
+ "print \"Value of R5(kohm) : \",R5"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Estimated input reference current , IREF(micro A) 220.51\n",
+ "More precise value of reference current , IREF(micro A) 225.88\n",
+ "Value of R5(kohm) : 12.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.2:Pg-81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Ex 2.2\n",
+ "import math\n",
+ "# data from fig of Ex2.2\n",
+ "IC10=20*10**-6;#A\n",
+ "IREF=0.5*10**-3;#A\n",
+ "IS=10**-14;#A\n",
+ "VT=25*10**-3;#V(Thermal Voltage)\n",
+ "R4=VT/IC10*math.log(IREF/IC10);#ohm\n",
+ "print \"For Widlar current source design, the value of R4(kohm) : \",round(R4/1000,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For Widlar current source design, the value of R4(kohm) : 4.02\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.3:Pg-82"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Ex 2.3\n",
+ "\n",
+ "import math\n",
+ "# given data\n",
+ "Gm1=10.0;#mA/V\n",
+ "Gm1=Gm1/1000;#A/V\n",
+ "Cc=50.0;#pF\n",
+ "Cc=Cc*10**-12;#F\n",
+ "Rt=10**8;#ohm(Shunting resistance with Cc)\n",
+ " # solution\n",
+ "Ao=Gm1*Rt;#unitless\n",
+ "fp=1/(2*math.pi*Rt*Cc);#Hz\n",
+ "ft=Gm1/(2*math.pi*Cc)/10**6;#MHz\n",
+ "print \"Frequency at which gain is maximum, fp in Hz\",round(fp,1)\n",
+ "print \"Unit gain frequency, ft(MHz)\",round(ft,1)\n",
+ "#Bode plot can not be plotted with the given data in the question by using python functions. \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency at which gain is maximum, fp in Hz 31.8\n",
+ "Unit gain frequency, ft(MHz) 31.8\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.4:Pg-83"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Ex 2.4\n",
+ "\n",
+ "import math\n",
+ "# given data\n",
+ "SR=10.0/10**-6;#V/s\n",
+ "Vout=10.0;#V(magnitude of output voltage)\n",
+ "fm=SR/(2*math.pi*Vout)/1000;#kHz\n",
+ "print \"Full power bandwidth(kHz)\",round(fm,1)\n",
+ "VT=25.0/1000;#V(Thermal voltage)\n",
+ "ft=SR/(2*math.pi*4*VT)/10.0**6;#MHz\n",
+ "print \"Unity gain bandwidth(MHz)\",round(ft,1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Full power bandwidth(kHz) 159.2\n",
+ "Unity gain bandwidth(MHz) 15.9\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.5:Pg-84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Ex 2.5\n",
+ "\n",
+ "VCC=5;#V\n",
+ "VEE=-5;#V\n",
+ "VBE=0.6;#V\n",
+ "VCE23=0.6;#V\n",
+ "VCE_sat=0.2;#V\n",
+ "Vo_max=VCC-VCE_sat-VBE;#V\n",
+ "Vo_min=VEE+VCE_sat+VBE+VCE23;#V\n",
+ "print \"Maximum output voltage(V)\",round(Vo_max,2)\n",
+ "print \"Minimum output voltage(V)\",round(Vo_min,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum output voltage(V) 4.2\n",
+ "Minimum output voltage(V) -3.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/sample_notebooks/Hrituraj/Chapter7.ipynb b/sample_notebooks/Hrituraj/Chapter7.ipynb new file mode 100755 index 00000000..71496b0e --- /dev/null +++ b/sample_notebooks/Hrituraj/Chapter7.ipynb @@ -0,0 +1,814 @@ +{ + "metadata": { + "name": "", + "signature": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter7 - Distribution systems" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.1 - page 174" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "import cmath\n", + "#Given data : \n", + "l=1 #in km\n", + "I=100 #in Ampere\n", + "cosfi=0.8 #Power factor(lag) unitless\n", + "VC=200 #in volt\n", + "IL=60 #in Ampere\n", + "cosfi_load=0.9 #Power factor(lag) unitless\n", + "R=0.6 #in ohm\n", + "XL=0.08 #in ohm\n", + "IC=I*complex(0.8,-0.6) #in Ampere\n", + "z=complex(0.06,0.08)/2 #in ohm\n", + "VD_BC=z*IC #in volt\n", + "VB=VC+VD_BC #in volt\n", + "IB=IL*complex(0.9,-0.4357)+IC #in Ampere\n", + "VD_AB=z*IB #in volt\n", + "VD_AB=round(VD_AB.real,2)+round(VD_AB.imag,2)*1J\n", + "print \"V.D. from sending end to mid point =\" ,VD_AB,\"Volt\"\n", + "print \"V.D. from mid point to the far end =\",VD_BC,\"Volt\" " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "V.D. from sending end to mid point = (7.47+2.78j) Volt\n", + "V.D. from mid point to the far end = (4.8+1.4j) Volt\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.2 - page 175" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from sympy import symbols\n", + "#Given data : \n", + "l=500 #in meter\n", + "i=1 #in Ampere/meter\n", + "IL1=200;IL2=150;IL3=50;IL4=100 #in Ampere\n", + "l1=100;l2=200;l3=300;l4=400 #in meter\n", + "r=0.1 #in ohm/km\n", + "Vd=250 #in volt\n", + "I=symbols('I')\n", + "Drop_AC=100*(r/10**3)*(I-i*l1/2) \n", + "Drop_CD=I \n", + "Drop_DE=100*r*(I-550)-I*100/2 \n", + "Drop_EF=100*r*(I-700-I*100/2) \n", + "Drop_FB=100*r*(I-900-I*100/2) \n", + "VD_tot=0.05*I-27 #in volts\n", + "#both ends are fed with same voltage,\n", + "#VD_tot should be equal to zero.\"\n", + "I=27/0.05 #in Ampere\n", + "print \"Curent = %0.2f Ampere\" %I\n", + "Drop_AD=(0.01*I-0.5)+(0.01*I-3.5) \n", + "print \"Value at minimum potential at D = %0.2f V\" %(Vd-Drop_AD)\n", + "#Note : Ans in the book is wrong as 27/0.05 gives 540 instead of 54." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Curent = 540.00 Ampere\n", + "Value at minimum potential at D = 243.20 V\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.3 - page 176" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#Given data : \n", + "l=250 #in meter\n", + "VA=230 #in volt\n", + "VB=232 #in volt\n", + "r=0.5 #in ohm/km\n", + "r=0.5/10**3 #in ohm/m\n", + "RAC=r*50*2 #in ohm\n", + "RCD=RAC;RDE=RAC;REF=RAC;RFB=RAC #in ohm\n", + "#VA-VB=VAC+VCD+VDE+VEF+VFB #in volt\n", + "Ia=(VA-VB+15)/(5*RAC) #Ampere\n", + "IAC=Ia;ICD=IAC-20;IDE=IAC-60;IED=-IDE;IEF=IAC-100;IFE=-IEF;IFB=IAC-120;IBF=-IFB #in Ampere\n", + "print \"IAC = %0.f A\" %IAC\n", + "print \"ICD = %0.f A\" %ICD\n", + "print \"IDE = %0.f A\" %IDE\n", + "print \"IED = %0.f A\" %IED\n", + "print \"IEF = %0.f A\" %IEF\n", + "print \"IFE = %0.f A\" %IFE\n", + "print \"IFB = %0.f A\" %IFB\n", + "print \"IBF = %0.f A\" %IBF\n", + "VAC=IAC*RAC #in volt\n", + "VCD=ICD*RCD #in volt\n", + "VD=VA-VAC-VCD #in volt\n", + "print \"The minimum potential = %0.1f Volt\" %(VD)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "IAC = 52 A\n", + "ICD = 32 A\n", + "IDE = -8 A\n", + "IED = 8 A\n", + "IEF = -48 A\n", + "IFE = 48 A\n", + "IFB = -68 A\n", + "IBF = 68 A\n", + "The minimum potential = 225.8 Volt\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.4 - page 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#Given data : \n", + "VA=235 #in volt\n", + "VB=236 #in volt\n", + "l=200 #in meter\n", + "IL1=20;IL2=40;IL3=25;IL4=30 #in Ampere\n", + "l1=50;l2=75;l3=100;l4=50 #in meter\n", + "r=0.4 #in ohm/km\n", + "r=0.4/10**3 #in ohm/m\n", + "RAC=r*l1*2 #in ohm\n", + "RCD=r*(l2-l1)*2*RAC;RDE=r*(l2-l1)*2*RAC;REF=r*l1*2*RAC;RFB=r*l1*2*RAC #in ohm\n", + "#VA-VB=VAC+VCD+VDE+VEF+VFB #in volt\n", + "IA=(VA-VB+9.6)/(0.16) #in Ampere\n", + "IAC=IA;ICD=IA-IL1;IDE=IA-IL1-IL2;IEF=IA-IL1-IL2-IL3;IFB=IA-IL1-IL2-IL3-IL4 #in Ampere\n", + "print \"IAC = %0.2f A\" %IAC\n", + "print \"ICD = %0.2f A\" %ICD\n", + "print \"IED = %0.2f A\" %(-IDE)\n", + "print \"IFE = %0.2f A\" %-IEF\n", + "print \"IFB = %0.2f A\" %-IFB\n", + "VAC=IAC*RAC #in volt\n", + "VCD=ICD*RCD #in volt\n", + "VD=VA-VAC-VCD #in volt\n", + "print \"The minimum potential = %0.3f Volt\" %VD\n", + "# Answer wrong in the textbook due to accuracy." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "IAC = 53.75 A\n", + "ICD = 33.75 A\n", + "IED = 6.25 A\n", + "IFE = 31.25 A\n", + "IFB = 61.25 A\n", + "The minimum potential = 232.823 Volt\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.5 - page 179" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#Given data : \n", + "VA=400 #in volt\n", + "r=0.03 #in ohm/km\n", + "r=0.03/1000 #in ohm/m\n", + "RAB=r*500*2 #in ohm\n", + "RBC=r*300*2 #in ohm\n", + "RAB=r*700*2 #in ohm\n", + "RAB=r*500*2 #in ohm\n", + "#VA-VB=VAC+VCD+VDE+VEF+VFB #in volt\n", + "IA=(17.4)/(0.09) #in Ampere\n", + "VAB=(RAB)*IA #in volt\n", + "VB=VA-VAB #in volt\n", + "print \"Voltage at B = %0.2f Volts\" %VB\n", + "VBC=(RBC)*(IA-150) #in volt\n", + "VC=VB-VBC #in volt\n", + "print \"Voltage at C = %0.2f Volts\" %VC\n", + "IBC=IA-150 #in A\n", + "print \"Current in section BC = %0.2f A \"%IBC \n", + "#Note : Answer of VB is wrong in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage at B = 394.20 Volts\n", + "Voltage at C = 393.42 Volts\n", + "Current in section BC = 43.33 A \n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.6 - page 180" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#Given data : \n", + "VA=240 #in volt\n", + "MAxVDrop=VA*5/100 #in volt\n", + "rho=2.87*10**-6 #in ohm-cm \n", + "#VAB+VBC+VCA=0 #in volt\n", + "IA=(3200)/(26) #in Ampere\n", + "IAB=IA #in Ampere\n", + "IBC=IA-100 #in Ampere\n", + "#Allowed voltage drop: IAB*RAB+IBC*RBC=12\n", + "R=12/(1015.26) #in ohm\n", + "RAB=R*300*2/100 #in ohm\n", + "RBC=R*600*2/100 #in ohm\n", + "RCA=R*400*2/100 #in ohm\n", + "#formula : R=rho*l/a\n", + "a=rho*(100*100)/R #in cm**2\n", + "print \"Cross section area = %0.2f cm2 \" %a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cross section area = 2.43 cm2 \n" + ] + } + ], + "prompt_number": 47 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.7 - page 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from numpy import sqrt\n", + "#Given data : \n", + "R=0.2 #in ohm/km\n", + "X=0.1 #in ohm/km\n", + "ZAM=((R+X*1J)/1000)*200 #in ohm\n", + "ZMB=((R+X*1J)/1000)*100 #in ohm\n", + "I1=100*(0.707-0.707*1J) #in A\n", + "I2=200*(0.8-0.6*1J) #in A\n", + "IAM=I1+I2 #in Ampere\n", + "VAM=ZAM*IAM #in volts\n", + "VMB=ZMB*I2 #in volts\n", + "VAB=VAM+VMB #in volts\n", + "magVAB=sqrt(VAB.real**2+VAB.imag**2) \n", + "print \"Total voltage drop = %0.2f Volts\" %magVAB" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total voltage drop = 17.85 Volts\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.8 - page 183" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from numpy import sqrt, sin, pi, arctan, arccos, cos\n", + "import cmath\n", + "#Given data : \n", + "VB=200 #in volts\n", + "R=0.2 #in ohm/km\n", + "X=0.3 #in ohm/km\n", + "I=100 #in Ampere\n", + "ZAB=(R+X*1J) #in ohm\n", + "ZMB=ZAB/2 #in ohm\n", + "ZAM=ZMB #in ohm\n", + "cosfi_1=0.6 #unitless\n", + "cosfi_2=0.8 #unitless\n", + "IMB=I*(cosfi_2-cosfi_1*1J) #in A\n", + "I2=IMB #in Ampere\n", + "VMB=IMB*ZMB #in volts\n", + "VM=VB+VMB #in volts\n", + "print \"Voltage at M = %.2f\u2220%.2f\u00b0\" %(abs(VM),cmath.phase(VM)*180/pi)\n", + "#print('Sending end voltage , V_S = %.2f\u2220%.2f\u00b0 kV/phase' %(abs(V_S*10**-3),cmath.phase(V_S)*180/math.pi))\n", + "fi=arctan(VM.imag/VM.real)*180/pi #in degree\n", + "fi_1=arccos(cosfi_1)*180/pi #in degree\n", + "fi_VBandI1=fi_1-fi #in degree\n", + "I1=I*(cos(fi_VBandI1*pi/180)-sin(fi_VBandI1*pi/180))*1J #in Ampere\n", + "IAM=I1+I2 #inA Ampere\n", + "VAM=ZAM*IAM #in volts\n", + "VA=VM+VAM #in volts\n", + "magVA=sqrt(VA.real**2+VA.imag**2) \n", + "print \"Voltage at A, standing end voltage = %0.2f Volts\" %magVA\n", + "#Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage at M = 217.08\u22201.58\u00b0\n", + "Voltage at A, standing end voltage = 236.65 Volts\n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.9 - page 184" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data : \n", + "l=500 #in meter\n", + "VA=200 #in volt\n", + "MAxVDrop=6 #in % of declared voltage \n", + "rho=0.014 #in ohm/m \n", + "#VD in the distributor=53*10**3*r\n", + "AllowedVD=VA*(6/100) #in volts\n", + "r=AllowedVD*10**6/(53*10**3) #in ohm/meter\n", + "#formula : R=rho*l/a\n", + "a=rho*(2*l)/r #in m**2\n", + "print \"Cross section area = %0.2f m2 \" %(a)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Cross section area = 0.06 m2 \n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.10 - page 185" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data : \n", + "l=300 #in meter\n", + "I=0.75 #in A/m\n", + "R=0.00018 #in ohm/m\n", + "x=200 #in meter\n", + "Vs=250 #in volt\n", + "VD=I*R*(l*x-x**2/2) #in volt\n", + "V_A=Vs-VD #in volt(Voltage at 200m from end A)\n", + "print \"Voltage as 200m from supply end A = %0.1f Volts\" %V_A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage as 200m from supply end A = 244.6 Volts\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.11 - page 185" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data : \n", + "l=600 #in meter\n", + "VA=440 #in volt\n", + "VB=400 #in volt\n", + "R=0.01 #in ohm/100m\n", + "RAC=(R/100)*300 #in ohm\n", + "RCD=(R/100)*300 #in ohm\n", + "RDE=(R/100)*100 #in ohm\n", + "REF=(R/100)*200 #in ohm\n", + "RFB=(R/100)*300 #in ohm\n", + "#VA-VB=VAC+VCD+VDE+VEF+VFB #in volt\n", + "IA=(VA-VB+42.5)/(0.12) #in Ampere\n", + "IAC=IA;ICD=IA-100;IDE=IA-300;IFE=IA-550;IFB=IA-850 #in Ampere\n", + "print \"Current fed at A, IA = %0.1f A \"%IAC \n", + "print \"Current fed at B, IB = %0.1f A \"%-IFB" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current fed at A, IA = 687.5 A \n", + "Current fed at B, IB = 162.5 A \n" + ] + } + ], + "prompt_number": 69 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.12 - page 187" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data : \n", + "VA=220 #in volt\n", + "VB=200 #in volt\n", + "R=0.1 #in ohm/km\n", + "I=1 #in A/m\n", + "l=500 #in meter\n", + "R=2*R/1000 #in ohm/m\n", + "x=(VA-VB)/(I*R*l)+l/2 #in meter\n", + "Vmin=VA-I*R*x**2/2 #in volts\n", + "print \"Value of minimum potential = %0.2f V \"%Vmin \n", + "IA=I*x #in A\n", + "print \"Current supplied from end A = %0.f A \" %IA \n", + "IB=I*(l-x) #in A\n", + "print \"Current supplied from end B = %0.f A\"%IB" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of minimum potential = 199.75 V \n", + "Current supplied from end A = 450 A \n", + "Current supplied from end B = 50 A\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.13 - page 187" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data : \n", + "VL=240 #in volt\n", + "Router=0.2 #in ohm\n", + "I1=VL/5 #in Ampere\n", + "I2=VL/6 #in Ampere\n", + "Ineutral=I1-I2 #in Ampere\n", + "#Applying KVL on +ve side\n", + "V1=VL+I1*0.2+8*0.4 #in volt\n", + "print \"Voltage at +ve side = %0.1f V\" %V1\n", + "#Applying KVL on +ve side\n", + "V2=VL-(8*0.4)+I2*0.2 #in volt\n", + "print \"Voltage at -ve side = %0.1f V\" %V2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage at +ve side = 252.8 V\n", + "Voltage at -ve side = 244.8 V\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.14 - page 188" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data : \n", + "#Applying KVL on +ve side\n", + "V1=200-(600*0.015)-(100)*0.03 #in volt\n", + "print \"Voltage at +ve side = %0.f V\" %V1\n", + "#Applying KVL on -ve side\n", + "V2=200-(-100*0.03)-500*0.0015 #in volt\n", + "print \"Voltage at -ve side = %0.1f V\" %V2\n", + "#Note : answer of 2nd part is wrong in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage at +ve side = 188 V\n", + "Voltage at -ve side = 202.2 V\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.15 - page 189" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols\n", + "x=symbols('x')\n", + "#Given data : \n", + "#VD in section AC from RHS: \n", + "VD1=(40+x)*0.02+0.17*x\n", + "#VD in section AC from LHS: \n", + "VD2=(350-x)*0.015+(150-x)*0.03\n", + "#Equating two VDs we get\n", + "#x*0.02+0.17*x+0.015*x+x*0.03=350*0.015+150*0.03-40*0.02\n", + "x=(350*0.015+150*0.03-40*0.02)/0.082 #in A\n", + "VB=500-(x+40)*0.02 #in volts\n", + "print \"Potential at point B = %0.2f V\" %VB\n", + "VC=VB-(x*0.017) #in volts\n", + "print \"Potential at point C = %0.2f V\" %VC\n", + "VD=500-(350-x)*0.015 #in volts\n", + "print \"Potential at point D = %0.2f V\" %VD\n", + "#Note : Answer of 3rd part is given wrong in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Potential at point B = 497.02 V\n", + "Potential at point C = 495.16 V\n", + "Potential at point D = 496.39 V\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.16 - page 190" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols\n", + "x=symbols('x')\n", + "#Given data : \n", + "#Applying KVL in loop AFEDA: \n", + "p1=((0.016*x)+0.09*(x-30)+0.14*(x-17)-0.1*y) #eqn(1) \n", + "#Applying KVL in loop ADCBA: \n", + "p2=(0.1*y-0.12*(95-x-y)-.01*(145-x-y)-0.008*(165-x-y)) #eqn(2)\n", + "#Equating two equtions we get\n", + "#3.9*x-125=97.75-0.75*x\n", + "x=(97.75+125)/(3.9+0.75) #in A\n", + "y=97.75-0.75*x #in A\n", + "print \"x = %0.f A\" %x \n", + "print \"y = %0.2f A\"%y \n", + "print \"Point of minimum ppotential is E.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "x = 48 A\n", + "y = 61.82 A\n", + "Point of minimum ppotential is E.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.17 - page 190" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data : \n", + "V=200 #in volt\n", + "I=1 #in A/m\n", + "R=2*0.05/1000 #in ohm/m\n", + "l=1*1000 #in meter\n", + "IT=I*l #in Ampere\n", + "RT=R*l #in ohm\n", + "VD=IT*RT/8 #in volt\n", + "Vmin=V-VD #in volt\n", + "print \"Minimum potential occurs at mid point. It is %0.2f V\" %Vmin" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum potential occurs at mid point. It is 187.50 V\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Problem 7.19 - page 191" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given data : \n", + "VB=400 #in volt\n", + "ZAC=0.04+0.08*1J #in ohm\n", + "ZCB=0.08+1J*0.12 #in ohm\n", + "I1=60*(0.8-1J*0.6) \n", + "I2=120*(0.8-1J*0.6) \n", + "VCB=I2*ZCB #in Volt\n", + "VAC=(I1+I2)*ZAC #in volt\n", + "VC=VB+I2*ZCB #in Volt\n", + "print \"Voltage at C =\",VC,\"Volt\"\n", + "VA=VC+(I1+I2)*ZAC #in volt\n", + "print \"Voltage at A =\",VA,\"Volt\"\n", + "#Answer not accurate in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage at C = (416.32+5.76j) Volt\n", + "Voltage at A = (430.72+12.96j) Volt\n" + ] + } + ], + "prompt_number": 20 + } + ], + "metadata": {} + } + ] +} diff --git a/sample_notebooks/KavinkumarD/Chapter_8_FREQUENCY_EFFECTS_IN_AMPLIFIERS.ipynb b/sample_notebooks/KavinkumarD/Chapter_8_FREQUENCY_EFFECTS_IN_AMPLIFIERS.ipynb new file mode 100755 index 00000000..60448e3c --- /dev/null +++ b/sample_notebooks/KavinkumarD/Chapter_8_FREQUENCY_EFFECTS_IN_AMPLIFIERS.ipynb @@ -0,0 +1,120 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:8ef023228932ba7c44f0f72b79793f31a32f8ea67eae875510cf71ee015fc22c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8 FREQUENCY EFFECTS IN AMPLIFIERS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.6 , Page no:242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#initialisation of variables\n",
+ "hie=1000 #\u2126\n",
+ "hfe=75 #\u2126\n",
+ "Av=50\n",
+ "Rl=10000 #k\u2126\n",
+ "hie2=300 #\u2126\n",
+ "hfe2=100 #\u2126\n",
+ "Re=1000 #k\u2126\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "Req=Av*(hie/hfe) #\u2126\n",
+ "Rc=Req*Rl/(Rl-Req) #k\u2126\n",
+ "wL=2*3.14*200\n",
+ "Ce=(hie2+(hfe2+1)*Re)/(wL*Re*hie2)*10**6\n",
+ "Av1=(hfe*Req)/(hie+(hfe+1)*Re)\n",
+ "\n",
+ "#RESULTS\n",
+ "print\"The value of Req=\",round(Req,3),\"Ohm\";\n",
+ "print\"The value of Rc=\",round(Rc,3),\"Ohm\";\n",
+ "print\"The value of Ce=\",round(Ce,3),\"mF\";\n",
+ "print\"The value of Av=\",round(Av1,3);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Req= 666.667 Ohm\n",
+ "The value of Rc= 714.286 Ohm\n",
+ "The value of Ce= 268.843 mF\n",
+ "The value of Av= 0.649\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 8.8 , Page no:244"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#initialisation of variables\n",
+ "hie2=1500 #\u2126\n",
+ "Rb2=5000 #k\u2126\n",
+ "Z01=10\n",
+ "Av=7881.3\n",
+ "\n",
+ "#CALCULATIONS\n",
+ "C2=1*10**-6 \n",
+ "Zin2=(hie2*Rb2/(hie2+Rb2))\n",
+ "fl=1/(2*3.14*C2*(Zin2+Z01*10**3))\n",
+ "\n",
+ "#RESULTS\n",
+ "print\"The value of Zin2=\",round(Zin2,3),\"Ohm\";\n",
+ "print\"The value of fl=\",round(fl,3),\"Hz\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The value of Zin2= 1153.846 Ohm\n",
+ "The value of fl= 14.276 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/sample_notebooks/LalitKumar/chapter2.ipynb b/sample_notebooks/LalitKumar/chapter2.ipynb new file mode 100755 index 00000000..57657aa7 --- /dev/null +++ b/sample_notebooks/LalitKumar/chapter2.ipynb @@ -0,0 +1,347 @@ +{ + "metadata": { + "name": "", + "signature": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter2 : Atomic model & bonding in solids" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.1, page no-28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#given\n", + "#atomic no. of gold\n", + "Z=79\n", + "#kinetic energy of alpha particle\n", + "E=7.68*1.6*(10)**(-13) #J because [1MeV=1.6*(10)**(-13)]\n", + "e=1.6*10**(-19) #C\n", + "E0=8.854*10**(-12) #F/m\n", + "#the distance of closest approach is given by:\n", + "d0=2*e*Z*e/(4*(math.pi)*E0*E) #m\n", + "print \"The closest approach of alpha particle is %.2ef m\" %d0" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The closest approach of alpha particle is 2.96e-14f m\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.2, page no-29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "from numpy import *\n", + "#given\n", + "#IN THE RUTHERFORD SCATTERING EXPERIMENT\n", + "#the no of particles scattered at\n", + "theta1=(pi)/2 #radians\n", + "#is\n", + "N90=44 #per minute\n", + "#the number of particles scattered particales N is given by\n", + "#N=C*(1/(sin(theta/2))**4) where C is propotionality constant\n", + "#solving above equation for C\n", + "C=N90*(sin(theta1/2))**4 \n", + "# now to find the no of particles scatering at 75 and 135 degrees\n", + "theta2=75*(pi)/180 #radians\n", + "N75=C*(1/(sin(theta2/2))**4) #per minute\n", + "theta3=135*(pi)/180 #radians\n", + "N135=C*(1/(sin(theta3/2))**4) #per minute\n", + "print \"The no of particles scattered at 75 and 135 degrees are %d per minute and %d per minutes\" %(N75,N135)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The no of particles scattered at 75 and 135 degrees are 80 per minute and 15 per minutes\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.3, page no-32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#mass of electron\n", + "m=9.11*10**(-31) #kg\n", + "#charge on an electron\n", + "e=1.6*10**(-19) #C\n", + "#plank's constant\n", + "h=6.62*10**(-34)\n", + "E0=8.85*10**(-12) \n", + "#NO OF ELECTRONS SHELLS IN HYDROZEN ATOm\n", + "n=1\n", + "#atomic number of hydrogen\n", + "Z=1\n", + "#radius of first orbit of hydrogen is given by\n", + "r1=n**2*E0*h**2/((pi)*m*Z*e**2) #m\n", + "print \"The radius of the first orbit of the electron in the hydrogen atom %.2e\"%(r1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The radius of the first orbit of the electron in the hydrogen atom 5.29e-11\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.4, page no-32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#mass of electron\n", + "m=9.11*10**(-31) #kg\n", + "#charge on an electron\n", + "e=1.6*10**(-19) #C\n", + "#plank's constant\n", + "h=6.62*10**(-34)\n", + "E0=8.85*10**(-12) \n", + "#NO OF ELECTRONS SHELLS IN HYDROZEN ATOm\n", + "n=1\n", + "#atomic number of hydrogen\n", + "Z=1\n", + "#ionization potential energy of hydrogen atom is given by\n", + "E=m*Z**2*e**4/(8*(E0)**2*h**2*n**2) #J\n", + "#energy in eV\n", + "EV=E/e #eV\n", + "print \"The ionization potential for hydrogen atom is %0.2f V\" %(EV)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ionization potential for hydrogen atom is 13.59 V\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.5, page no-34" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.6, page no-36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#uncertainity in the momentum\n", + "deltap=10**-27 #kg ms**-1\n", + "#according to uncertainity principle\n", + "#deltap* deltax >=h/(2*(pi))\n", + "#we know that \n", + "h=6.626*10**-34 #Js\n", + "#here instead of inequality we are using only equality just for notation otherwise it is greater than equal to as mentioned above\n", + "#now deltax is given by\n", + "deltax=h/(2*(pi)*deltap) #m\n", + "print \"The minimum uncertainity is %.2e m\"%(deltax)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum uncertainity is 1.05e-07 m\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.10, page no- 57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#ionization potential of hydrogen\n", + "E1=13.6 #eV\n", + "#when \n", + "n=3\n", + "E3=-E1/n**2 #eV\n", + "#when \n", + "n=5\n", + "E5=-E1/n**2 #eV\n", + "print \"Energy of 3rd and 5th orbits are %0.2f eV and %0.2f eV\"%(E3,E5)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy of 3rd and 5th orbits are -1.51 eV and -0.54 eV\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.11, page no-59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#dipole moment og HF is\n", + "DM=6.375*10**(-30) #Cm\n", + "#intermolecular distance\n", + "r=0.9178*10**(-10) #m\n", + "#charge on an electron\n", + "e=1.67*10**(-19) #C\n", + "#since the HF posses ionic characters\n", + "#so\n", + "#Hf in fully ionic state has dipole moment as\n", + "DM2=r*e #Cm\n", + "#percentage ionic characters\n", + "percentage=DM/DM2*100 #%\n", + "print \"The percentage ionic character is %0.2f approx.\"%(percentage)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The percentage ionic character is 41.59 approx.\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "example-2.12, page no-60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#given\n", + "#elctronegativity of In\n", + "EnIn=1.5\n", + "#elctronegativity of As\n", + "EnAs=2.2\n", + "#elctronegativity of Ga\n", + "EnGa=1.8\n", + "#for InAs\n", + "ionic_charater1=(1-exp((-0.25)*(EnAs-EnIn)**2))*100 #in %\n", + "#for GaAs\n", + "ionic_charater2=(1-exp((-0.25)*(EnAs-EnGa)**2))*100 # in %\n", + "print \"Ionic character in InAs and GaAs are %0.1f %% and %0.1f %%\"%(ionic_charater1,ionic_charater2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ionic character in InAs and GaAs are 11.5 % and 3.9 %\n" + ] + } + ], + "prompt_number": 30 + } + ], + "metadata": {} + } + ] +} diff --git a/sample_notebooks/MohdGufran/Chapter6.ipynb b/sample_notebooks/MohdGufran/Chapter6.ipynb new file mode 100755 index 00000000..ea85357e --- /dev/null +++ b/sample_notebooks/MohdGufran/Chapter6.ipynb @@ -0,0 +1,506 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6: Bipolar junction Transistors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.1 page No.215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "Ic=9.95\t\t\t#in mA\n",
+ "Ie=10 \t\t#in mA\n",
+ "\n",
+ "Ib=Ie-Ic\t\t#in mA\n",
+ "\n",
+ "print\"Emitter current is \",Ib,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Emitter current is 0.05 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.2 page No. 216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "IC=0.98\t\t\t#in mA\n",
+ "IB=20.0\t\t\t#in uA\n",
+ "IB=IB*10**-3\t\t#in mA\n",
+ "\n",
+ "IE=IB+IC\t\t#in mA\n",
+ "\n",
+ "alpha=IC/IE\t\t#unitless\n",
+ "Beta=IC/IB\t\t#unitless\n",
+ "\n",
+ "print\"Emitter current is\",IE,\"mA\"\n",
+ "print\"Current amplification factor is \",alpha\n",
+ "print\"Current gain factor is \",Beta"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Emitter current is 1.0 mA\n",
+ "Current amplification factor is 0.98\n",
+ "Current gain factor is 49.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.3 page No.216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "alfaDC=0.98\t\t\t#unitless\n",
+ "ICBO=4\t\t\t\t#in uA\n",
+ "ICBO=ICBO*10**-3\t\t#in mA\n",
+ "IB=50\t\t\t\t#in uA\n",
+ "IB=IB*10**-3\t\t\t#in mA\n",
+ "\n",
+ "IC=alfaDC*IB/(1-alfaDC)+ICBO/(1-alfaDC)\t#in mA\n",
+ "IE=IC+IB\t\t\t#in mA\n",
+ "\n",
+ "print\"Emitter current is \",IE,\"mA\"\n",
+ "print\"Collector current is \",IC,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Emitter current is 2.7 mA\n",
+ "Collector current is 2.65 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.4 page No. 216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "IB=10\t\t\t#in uA\n",
+ "IB=IB*10**-3\t\t#in mA\n",
+ "Beta=99\t\t\t#Unitless\n",
+ "ICO=1\t\t\t#in uA\n",
+ "ICO=ICO*10**-3\t\t#in mA\n",
+ "\n",
+ "IC=Beta*IB+(1+Beta)*ICO\t#in mA\n",
+ "\n",
+ "print\"Collector current in mA : \",IC,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Collector current in mA : 1.09 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.5 Page No.216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "Ic=5*10**-3 #mA collector current\n",
+ "Ic_=10*10**-3 #mA collector current\n",
+ "Ib=50*10**-6 #mA, Base current\n",
+ "Icbo=1*10**-6 #micro A, Current to base open current\n",
+ "\n",
+ "beta=(Ic-Icbo)/(Ib+Icbo)\n",
+ "alpha=(beta/(1+beta))\n",
+ "Ie=Ib+Ic\n",
+ "\n",
+ "Ib=(Ic_-(beta+1)*Icbo)/(beta)\n",
+ "\n",
+ "print\"(i) Current gain factor is\",round(beta,0)\n",
+ "print\" Current amplification factor is\",round(alpha,2)\n",
+ "print\" Emitter Current is\",Ie*1000,\"mA\"\n",
+ "print\"(ii)New level of Ib is\",round(Ib*10**6,0),\"micro A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current gain factor is 98.0\n",
+ " Current amplification factor is 0.99\n",
+ " Emitter Current is 5.05 mA\n",
+ "(ii)New level of Ib is 101.0 micro A\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.6 page No. 222"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "delVEB=200\t\t\t#in Volts\n",
+ "delIE=5\t\t\t\t#in mA\n",
+ "\n",
+ "rin=delVEB/delIE\t\t#in ohm\n",
+ "\n",
+ "print\"Dynamic input resistance is \",rin,\"mohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dynamic input resistance is 40 mohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.7 page No. 222"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "ICBO=12.5 \t\t\t#in uA\n",
+ "ICBO=ICBO*10**-3 \t\t#in mA\n",
+ "IE=2 \t\t\t\t#in mA\n",
+ "IC=1.97 \t\t\t#in mA\n",
+ "\n",
+ "alfa=(IC-ICBO)/IE \t\t#unitless\n",
+ "IB=IE-IC \t\t\t#in mA\n",
+ "\n",
+ "print\"Current gain : \",round(alfa,3)\n",
+ "print\"Base current is \",IB,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current gain : 0.979\n",
+ "Base current is 0.03 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.8 page No. 222"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "RL=4.0 \t\t\t#in Kohm\n",
+ "VL=3.0\t\t\t#in volt\n",
+ "alfa=0.96 \t\t#unitless\n",
+ "IC=VL/RL \t\t#in mA\n",
+ "\n",
+ "IE=IC/alfa \t\t#in mA\n",
+ "IB=IE-IC \t\t#in mA\n",
+ "\n",
+ "print\"Base current ia\",round(IB,2),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Base current ia 0.03 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.9 page No.227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "VCC=10\t\t\t #in volt\n",
+ "RL=800\t\t\t #in ohm\n",
+ "VL=0.8\t\t\t #in volt\n",
+ "alfa=0.96\t\t #unitless\n",
+ "\n",
+ "VCE=VCC-VL \t\t#in Volt\n",
+ "IC=VL*1000/RL \t\t#in mA\n",
+ "Beta=alfa/(1-alfa) \t#unitless\n",
+ "IB=IC/Beta \t\t#in mA\n",
+ "\n",
+ "print\"Collector-emitter Voltage is \",VCE,\"V\"\n",
+ "print\"Base current in uA : \",round(IB*1000,2),\"microA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Collector-emitter Voltage is 9.2 V\n",
+ "Base current in uA : 41.67 microA\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.10 page No. 227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "alfao=0.98 \t\t#unitless\n",
+ "ICO=10 \t\t\t#in uA\n",
+ "ICO=ICO*10**-3 \t\t#in mA\n",
+ "IB=0.22 \t\t#in mA\n",
+ "\n",
+ "IC=(alfao*IB+ICO)/(1-alfao) \t#in mA\n",
+ "\n",
+ "print\"Collector current is\",IC,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Collector current is 11.28 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.11 page No. 228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "delVEB=250 \t\t#in mVolts\n",
+ "delIE=1 \t\t#in mA\n",
+ "\n",
+ "rin=delVEB/delIE \t#in ohm\n",
+ "\n",
+ "print\"Dynamic input resistance is\",rin,\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dynamic input resistance is 250 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.12 page No. 228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "delVCE=10-5 \t\t#in Volts\n",
+ "delIC=5.8-5\t \t#in mA\n",
+ "\n",
+ "rin=delVCE/delIC \t#in Kohm\n",
+ "\n",
+ "print\"Dynamic output resistance is \",rin,\"kohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dynamic output resistance is 6.25 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 6.16 page No. 233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "Beta=45 \t\t\t#Unitless\n",
+ "VBE=0.7 \t\t\t#in Volt\n",
+ "VCC=0 \t\t\t\t#in Volt\n",
+ "RB=10**5 \t\t\t#in ohm\n",
+ "RC=1.2*10**3 \t\t\t#in ohm\n",
+ "VEE=-9 \t\t\t\t#in Volt\n",
+ "\n",
+ "IB=-(VBE+VEE)/RB \t\t#in mA\n",
+ "IC=Beta*IB \t\t\t#in mA\n",
+ "VC=VCC-IC*RC \t\t\t#in Volts\n",
+ "VB=VBE+VEE \t\t\t#in Volts\n",
+ "\n",
+ "print\"collector voltage is \",round(VC,1),\"V\"\n",
+ "print\"Base voltage is \",VB,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "collector voltage is -4.5 V\n",
+ "Base voltage is -8.3 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/sample_notebooks/MukteshChaudhary/ch2.ipynb b/sample_notebooks/MukteshChaudhary/ch2.ipynb new file mode 100755 index 00000000..ebb803c6 --- /dev/null +++ b/sample_notebooks/MukteshChaudhary/ch2.ipynb @@ -0,0 +1,150 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8575441dfc9f46104394a83a9c4613d36928a495b4284fab191e2ca7cb407033" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2: Introduction to Quantum Mechanics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1, Page 47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "lamda=0.708*10**-8# cm\n", + "h=6.625*10**-34# J*s Plank's constant\n", + "c=3.0*10**10# cm/s\n", + "e=1.6*10**-19# eV\n", + "\n", + "#Calculations&Results\n", + "E=(h*c)/lamda# E=hv=hc/lamda\n", + "print \"The value of E is %.2e J\"%E\n", + "E=E/e\n", + "print \"The value of E is %.2e eV\"%E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of E is 2.81e-15 J\n", + "The value of E is 1.75e+04 eV\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2,Page 49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "m=9.11*10**-31# kg*m/s\n", + "v=10**5#m/s\n", + "h=6.625*10**-34#js\n", + "\n", + "#Calculations&Results\n", + "p=m*v\n", + "print \"momentum is %.2e\"%p\n", + "lamda=h/p\n", + "print \"de broglie wavelength in meter is %.2e\"%(lamda)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "momentum is 9.11e-26\n", + "de broglie wavelength in meter is 7.27e-09\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3, Page 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#Variable declaration\n", + "a=5*10**-10# a=5A = 5*10**-8cm\n", + "h=1.054*10**-34# J*s Planck's constant \n", + "m=9.11*10**-31# kg*m/s\n", + "e=1.6*10**-19# eV\n", + "\n", + "#Calculations&Results\n", + "print \"The energy levels are:\"\n", + "for n in range(1,4):\n", + " En=((h**2*n**2*math.pi**2)/(2*m*a**2))/e\n", + " print \"For n = %d, E = %.2f eV\"%(n,En)\n", + "\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The energy levels are:\n", + "For n = 1, E = 1.50 eV\n", + "For n = 2, E = 6.02 eV\n", + "For n = 3, E = 13.54 eV\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/sample_notebooks/NirenNegandhi/ch9.ipynb b/sample_notebooks/NirenNegandhi/ch9.ipynb new file mode 100755 index 00000000..61fe143f --- /dev/null +++ b/sample_notebooks/NirenNegandhi/ch9.ipynb @@ -0,0 +1,332 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a295b23ddf355cc98776039dc2c765e71624b4961371dafae0a82fd3d3a044d3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9: Control of Traction Motors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1, Page 268" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=600.;# in volts\n", + "I=350.;#in A\n", + "Ts=20.;# in sec\n", + "R=0.15;# in ohm\n", + "\n", + "#Calculations&Results\n", + "E_bse=(V/2)-(I*R);\n", + "E_bp=V-(I*R);\n", + "Tse=(E_bse/E_bp)*Ts;\n", + "Tp=Ts-Tse;\n", + "Vd=V-(2*I*R);\n", + "Ed1=(Vd/2)*I*(Tse/3600);\n", + "Ed2=((V/2)/2)*2*I*(Tp/3600);\n", + "El=(Ed1+Ed2)*10**-3;\n", + "print \"part (a)\"\n", + "print \"Energy lost in starting rhestat,El(kWh) = %.4f\"%El\n", + "El_1=(2*(I**2)*R*Ts)/(3600*1000);\n", + "print \"part (b)\"\n", + "print \"Energy lost in motors,El(kWh) = %.3f\"%El_1\n", + "#answer is wrong in part b in the textbook\n", + "Et=((V*I*Tse)+(2*V*I*Tp))/(3600*1000);\n", + "print \"part (c)\"\n", + "print \"Total Energy,Et(kWh) = %.3f\"%Et" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "part (a)\n", + "Energy lost in starting rhestat,El(kWh) = 0.5372\n", + "part (b)\n", + "Energy lost in motors,El(kWh) = 0.204\n", + "part (c)\n", + "Total Energy,Et(kWh) = 1.806\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2, Page 269" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=600.;# in volts\n", + "I=300.;#in A\n", + "Ts=15.;# in sec\n", + "R=0.1;# in ohm\n", + "\n", + "#Calculations&Results\n", + "E_bse=(V/2)-(I*R);\n", + "E_bp=V-(I*R);\n", + "Tse=(E_bse/E_bp)*Ts;\n", + "Tp=Ts-Tse;\n", + "Vd=V-(2*I*R);\n", + "Ed1=(round((Vd/2)*I*(Tse/3600))*10**-3);#\n", + "print \"part (i)\"\n", + "print \"rheostatic in series,Ed1(kWh) = %.2f\"%Ed1\n", + "Ed2=((V/2)/2)*2*I*(Tp/3600)*10**-3;\n", + "print \"rheostatic in parallel,Ed2(kWh) = %.3f\"%Ed2\n", + "Vm=29;# in kmph\n", + "alfa=Vm/Ts;\n", + "S=alfa*Tse;\n", + "print \"part (ii)\"\n", + "print \"Speed at the end of series period,S(km/h) = %.1f\"%S" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "part (i)\n", + "rheostatic in series,Ed1(kWh) = 0.16\n", + "rheostatic in parallel,Ed2(kWh) = 0.197\n", + "part (ii)\n", + "Speed at the end of series period,S(km/h) = 13.7\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3, Page 270" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=600;# in volts\n", + "I=200;#in A\n", + "Ts=20;# in sec\n", + "R=0.1;# in ohm\n", + "\n", + "#Calculations&Results\n", + "E_bse=(V/2)-(I*R);\n", + "E_bp=V-(I*R);\n", + "Tse=(E_bse/E_bp)*Ts;\n", + "Tp=Ts-Tse;\n", + "Vd=V-(2*I*R);\n", + "Mi=((V*I*Tse)/(2*3600))+((V*I*Tp)/3600);\n", + "Er=((Vd/4)*I*(Tse/3600))+(((V/2)/2)*I*(Tp/3600));\n", + "El=(I**2*R*Ts)/(3600);\n", + "Mo=Mi-Er-El;\n", + "eta=(Mo/Mi)*100;\n", + "print \"part (a)\"\n", + "print \"Starting efficiency = %.1f%%\"%eta\n", + "Vm=80;# in kmph\n", + "alfa=Vm/Ts;\n", + "S=alfa*Tse;\n", + "print \"part (b)\"\n", + "print \"speed,S(kmph) = %.2f\"%S" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "part (a)\n", + "Starting efficiency = 63.7%\n", + "part (b)\n", + "speed,S(kmph) = 38.62\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4, Page 271" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "W=150;# in tonne\n", + "We=1.1*W;# in tonnes\n", + "Vm=30;#kmph\n", + "V=600;# in volts\n", + "r=10;# N/tonne\n", + "I=300;#in A\n", + "R=0.1;# in ohm\n", + "Ft=4*15000;# in N\n", + "G=1;#gradient in %\n", + "\n", + "#Calculations&Results\n", + "alfa=(Ft-(W*r)-(98.1*W*G))/(277.8*We);\n", + "Ts=Vm/alfa;\n", + "E_bse=(V/2)-(I*R);\n", + "E_bp=V-(I*R);\n", + "Tse=(E_bse/E_bp)*Ts;\n", + "print \"part (a)\"\n", + "print \"Duration of starting period,Ts(seconds) = %.1f\"%Ts\n", + "print \"Duration for Series running,Tse(seconds) = %.1f\"%Tse\n", + "sptr=alfa*Tse;#in kmph\n", + "print \"part (b)\"\n", + "print \"speed of train at transition in kmph is %.2f\"%sptr\n", + "sptr=alfa*Tse;#in kmph\n", + "rls=((V-(2*I*R))/2)*(2*I)*(Tse/3600);#watts hours\n", + "rlp=((V/2)/2)*(4*I)*((Ts-Tse)/3600);#watts hours\n", + "tl=rls+rlp;#\n", + "print \"part (c)\"\n", + "print \"rheostat losses during series operation is %.1f W-hours\"%rls\n", + "print \"rheostat losses during parallel operation is %.f W-hours\"%rlp\n", + "print \"total losses in W-hours is %.1f \"%tl" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "part (a)\n", + "Duration of starting period,Ts(seconds) = 31.4\n", + "Duration for Series running,Tse(seconds) = 14.9\n", + "part (b)\n", + "speed of train at transition in kmph is 14.21\n", + "part (c)\n", + "rheostat losses during series operation is 669.4 W-hours\n", + "rheostat losses during parallel operation is 826 W-hours\n", + "total losses in W-hours is 1495.9 \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5, Page 272" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "nf=1.; \n", + "n2=1.25*nf; \n", + "of=1; \n", + "of2=nf/n2; \n", + "isef=1; \n", + "ise2=0.66667; \n", + "\n", + "#Calculations\n", + "ia2=(1./ise2); \n", + "idiv=ia2-ise2; \n", + "rdiv=ise2/idiv; \n", + "\n", + "#Result\n", + "print \"diverter resistance required as percentage of the field resistance is %.f%%\"%(rdiv*100)\n", + "#answer is wrong in the textbook" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "diverter resistance required as percentage of the field resistance is 80%\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.6, Page 272" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ia=[60,80,100,120,160,180];# in amperes\n", + "sp1=[47.4,40.3,35.8,33.9,29.8,28.5];#in kmph\n", + "dpk=[440,700,970,1245,1800,2360];#in kg\n", + "sp2=[58.1,50,45,40.3,35,32];#\n", + "\n", + "#Calculations&Results\n", + "for i in range(0,6):\n", + " dpk1= ((dpk[i])*(sp1[i]))/(sp2[i]);#\n", + " print \"For current = \",Ia[i],\"A, speed is \",sp2[i],\"kmph and drawbar pull is\",round(dpk1),\"kg\"\n", + " \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For current = 60 A, speed is 58.1 kmph and drawbar pull is 359.0 kg\n", + "For current = 80 A, speed is 50 kmph and drawbar pull is 564.0 kg\n", + "For current = 100 A, speed is 45 kmph and drawbar pull is 772.0 kg\n", + "For current = 120 A, speed is 40.3 kmph and drawbar pull is 1047.0 kg\n", + "For current = 160 A, speed is 35 kmph and drawbar pull is 1533.0 kg\n", + "For current = 180 A, speed is 32 kmph and drawbar pull is 2102.0 kg\n" + ] + } + ], + "prompt_number": 33 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/sample_notebooks/PankajDoshi/Chapter1.ipynb b/sample_notebooks/PankajDoshi/Chapter1.ipynb new file mode 100755 index 00000000..91492796 --- /dev/null +++ b/sample_notebooks/PankajDoshi/Chapter1.ipynb @@ -0,0 +1 @@ +{"nbformat_minor": 0, "cells": [{"source": "# Chapter 1", "cell_type": "markdown", "metadata": {}}, {"source": "# Special Diodes", "cell_type": "markdown", "metadata": {}}, {"source": "## Example 1.1 Page No. 1-11", "cell_type": "markdown", "metadata": {}}, {"execution_count": 2, "cell_type": "code", "source": "# Assumption\nprint(\"Assume the drop across the LED as 2 V\\n\"); \nprint(\"Therefore, VD = 2 V\\n\");\n\n# Given Data\nprint(\"From fig. 1.11, Rs = 2.2 kohm and Vs = 15 V\\n\"); \n\n# Calculation in mA\nIs =(15-2)/(2.2); \n\n# Result\nprint \"Therefore, Is(mA) = (Vs-VD/Rs) = \",round(Is,2),\"mA\"; ", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Assume the drop across the LED as 2 V\n\nTherefore, VD = 2 V\n\nFrom fig. 1.11, Rs = 2.2 kohm and Vs = 15 V\n\nTherefore, Is(mA) = (Vs-VD/Rs) = 5.91 mA\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "## Example 1.2 Page No. 1-20", "cell_type": "markdown", "metadata": {}}, {"execution_count": 3, "cell_type": "code", "source": "import math\n\n# Formula\nprint(\"The transistor capacitance is given by,\"); \nprint(\"CT = C(0)/[1+|VR/VJ|^n]\\n\");\n\n# Given Data\nprint(\"Now C(0) = 80pF , n = 1/3 as diffused junction\"); \nprint(\"VR = 4.2 V, VJ = 0.7 V\\n\");\n\n# Calculation in pF\nct = round((80*pow(10,-12))/((pow((1+(4.2/0.7)),(0.3333))))*pow(10,12),2); # Note here, 1/3 = 0.3333 \n\n# Result\nprint \"Therefore, CT(pF) = \",ct,\"pF\"; \n\n# Formula\nprint(\"\\nThe transistor capacitance is also given by, \" )\nprint(\"CT = K/[VR+VJ]^n\\n\") \n\n#Calculation\nk = round((41.82*pow(10,-12))*(pow((4.2+0.7),(0.3333)))*pow(10,12),2); # Note here, 1/3 = 0.3333 \n\n# Result\nprint \"Therefore, K = \",k,\"* 10^-12\"; ", "outputs": [{"output_type": "stream", "name": "stdout", "text": "The transistor capacitance is given by,\nCT = C(0)/[1+|VR/VJ|^n]\n\nNow C(0) = 80pF , n = 1/3 as diffused junction\nVR = 4.2 V, VJ = 0.7 V\n\nTherefore, CT(pF) = 41.82 pF\n\nThe transistor capacitance is also given by, \nCT = K/[VR+VJ]^n\n\nTherefore, K = 71.03 * 10^-12\n"}], "metadata": {"scrolled": true, "collapsed": false, "trusted": true}}, {"execution_count": null, "cell_type": "code", "source": "", "outputs": [], "metadata": {"collapsed": true, "trusted": true}}], "nbformat": 4, "metadata": {"kernelspec": {"display_name": "Python 2", "name": "python2", "language": "python"}, "language_info": {"mimetype": "text/x-python", "nbconvert_exporter": "python", "version": "2.7.9", "name": "python", "file_extension": ".py", "pygments_lexer": "ipython2", "codemirror_mode": {"version": 2, "name": "ipython"}}}}
\ No newline at end of file diff --git a/sample_notebooks/PankajDoshi/Chapter_1.ipynb b/sample_notebooks/PankajDoshi/Chapter_1.ipynb new file mode 100755 index 00000000..fafb9249 --- /dev/null +++ b/sample_notebooks/PankajDoshi/Chapter_1.ipynb @@ -0,0 +1 @@ +{"nbformat_minor": 0, "cells": [{"source": "# Special Diodes", "cell_type": "markdown", "metadata": {}}, {"source": "# Example 1.1 Page No. 1-11", "cell_type": "markdown", "metadata": {}}, {"execution_count": 62, "cell_type": "code", "source": "# Assumution\nprint(\"Assume the drop across the LED as 2 V\\n\"); \nprint(\"Therefore, VD = 2 V\\n\");\n\n# Given Data\nprint(\"From fig. 1.11, Rs = 2.2 kohm and Vs = 15 V\\n\"); \n\n# Calculation in mA\nIs =(15 -2) /(2.2); \n\n# Result\nprint \"Therefore, Is(mA) = (Vs-VD/Rs) = \" ,round(Is,2),\"mA\"; ", "outputs": [{"output_type": "stream", "name": "stdout", "text": "Assume the drop across the LED as 2 V\n\nTherefore, VD = 2 V\n\nFrom fig. 1.11, Rs = 2.2 kohm and Vs = 15 V\n\nTherefore, Is(mA) = (Vs-VD/Rs) = 5.91 mA\n"}], "metadata": {"collapsed": false, "trusted": true}}, {"source": "# Example 1.2 Page No. 1-20", "cell_type": "markdown", "metadata": {}}, {"execution_count": 70, "cell_type": "code", "source": "import math\n\n# Formula\nprint(\"The transistor capacitance is given by,\"); \nprint(\"CT = C(0)/[1+jVR/VJ j^n]\\n\");\n\n# Given Data\nprint(\"Now C(0) = 80pF , n = 1/3 as diffused junction\"); \nprint(\"VR = 4.2 V, VJ = 0.7 V\\n\");\n\n# Calculation in pF\nct = round((80*pow(10,-12))/((pow((1+(4.2/0.7)),(0.3333))))*pow(10,12),2); \n\n# Result\nprint \"Therefore, CT(pF) = \", ct,\"pF\"; \n\n# Formula\nprint(\"\\nThe transistor capacitance is also given by, \" )\nprint(\"CT = K/[VR+VJ]^n\\n\") \n\n#Calculation\nk = round((41.82*pow(10,-12))*(pow((4.2+0.7),(0.3333)))*pow(10,12),2); \n\n# Result\nprint \"Therefore, K = \", k,\"* 10^-12\"; ", "outputs": [{"output_type": "stream", "name": "stdout", "text": "The transistor capacitance is given by,\nCT = C(0)/[1+jVR/VJ j^n]\n\nNow C(0) = 80pF , n = 1/3 as diffused junction\nVR = 4.2 V, VJ = 0.7 V\n\nTherefore, CT(pF) = 41.82 pF\n\nThe transistor capacitance is also given by, \nCT = K/[VR+VJ]^n\n\nTherefore, K = 71.03 * 10^-12\n"}], "metadata": {"scrolled": true, "collapsed": false, "trusted": true}}], "nbformat": 4, "metadata": {"kernelspec": {"display_name": "Python 2", "name": "python2", "language": "python"}, "language_info": {"mimetype": "text/x-python", "nbconvert_exporter": "python", "version": "2.7.9", "name": "python", "file_extension": ".py", "pygments_lexer": "ipython2", "codemirror_mode": {"version": 2, "name": "ipython"}}}}
\ No newline at end of file diff --git a/sample_notebooks/PraveenKumar/chapter2.ipynb b/sample_notebooks/PraveenKumar/chapter2.ipynb new file mode 100755 index 00000000..3d7aab33 --- /dev/null +++ b/sample_notebooks/PraveenKumar/chapter2.ipynb @@ -0,0 +1,429 @@ +{ + "metadata": { + "name": "", + "signature": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "chapter-2, Economics of generation" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex1, Page 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#To Determine the Demand and Supply Parameters for 15 bulbs\n", + "\n", + "W=60 #Wattage of the bulb\n", + "N=15 #No. of bulbs\n", + "CL=W*N #Connected Load \n", + "Wih=2*(10**3) #Wattage of immersion heater\n", + "Wh=2*(10**3) #Wattage of heater\n", + "\n", + "#Usage of Bulbs at different time periods\n", + "N1=5 \n", + "N2=10 \n", + "N3=6\n", + "\n", + "#Time periods for bulbs\n", + "T1=2 #6pm - 8pm\n", + "T2=2 #8pm - 10pm\n", + "T3=2 #10pm - 12pm\n", + "#Time Periods for heaters\n", + "T4=4 #1pm - 5pm\n", + "T5=3 #8pm - 11pm\n", + "\n", + "#CASE 1\n", + "MD1=W*N2 #Maximum Demand\n", + "DF=MD1*100/CL #Demand Factor\n", + "EC1=(N1*W*T1)+(N2*W*T2)+(N3*W*T3) #Energy Consumed\n", + "DLF1=EC1*100/(24*MD1) #Daily Load Factor\n", + "\n", + "#CASE 2\n", + "MD2=(W*N2)+Wh #From 8pm - 10pm\n", + "EC2=(T4*Wih)+(T5*Wh)+EC1 #Energy Consumed\n", + "DLF2=EC2*100/(24*MD2) #Daily Load Factor\n", + "\n", + "print '''i)a) Connected Load is %0.2f W\\nb) The Maximum Demand is %0.2f W\n", + "c) The Demand Factor is %0.2f percent\\nd) The Daily Load Factor is %0.2f percent''' %(CL,MD1,DF,DLF1)\n", + "print 'ii) The Improved Daily Load Factor is %0.2f percent' %DLF2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)a) Connected Load is 900.00 W\n", + "b) The Maximum Demand is 600.00 W\n", + "c) The Demand Factor is 66.67 percent\n", + "d) The Daily Load Factor is 17.50 percent\n", + "ii) The Improved Daily Load Factor is 26.47 percent\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2, Page 74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "\n", + "#To determine the Demand and supply parameter of four consumers\n", + "\n", + "\n", + "#Maximum Demands of various users\n", + "MD1=2*(10**3) #9pm\n", + "MD2=2*(10**3) #12 noon\n", + "MD3=8*(10**3) #5pm\n", + "MD4=4*(10**3) #8pm\n", + "MDT=MD1+MD2+MD3+MD4 #Sum of all Maximum Demands\n", + "\n", + "#Demands of various users\n", + "D1=1.6*(10**3) #8pm\n", + "D2=1*(10**3) #8pm\n", + "D3=5*(10**3) #8pm\n", + "\n", + "#The Number after the Alphabets represents the Consumer\n", + "\n", + "#Maximum Demand of the System arises at 8.00 PM\n", + "MDS = D1+D2+D3+MD4 \n", + "\n", + "TDF=MDT/MDS #Diversity Factor\n", + "#Given Values\n", + "#Average Loads\n", + "AL2=500 \n", + "AL4=1000 \n", + "#Load Factors\n", + "LF1=15/100 \n", + "LF3=25/100 \n", + "#Calculated Values\n", + "#Average Loads\n", + "AL1=LF1*MD1 \n", + "AL3=LF3*MD3 \n", + "#Load Factors\n", + "LF2=AL2*100/MD2 \n", + "LF4=AL4*100/MD4 \n", + "\n", + "ALS=AL1+AL2+AL3+AL4 #Combined Average Loads\n", + "LFS=ALS*100/MDS #Combined Load Factor\n", + "\n", + "#Load Percent\n", + "LF1*=100 # %\n", + "LF3*=100 # %\n", + "\n", + "print 'i) The Diversity Factor is %0.2f' %TDF\n", + "print 'ii) The Average load and Load factor of:'\n", + "print ' Consumer 1 : %0.2f W and %0.2f percent' %(AL1,LF1)\n", + "print ' Consumer 2 : %0.2f W and %0.2f percent' %(AL2,LF2)\n", + "print ' Consumer 3 : %0.2f W and %0.2f percent' %(AL3,LF3)\n", + "print ' Consumer 4 : %0.2f W and %0.2f percent' %(AL4,LF4)\n", + "print 'iii) The Combined Load Factor and the Combined Average Load is %0.2f percent and %0.2f W respectively\\n' %(LFS,ALS)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i) The Diversity Factor is 1.38\n", + "ii) The Average load and Load factor of:\n", + " Consumer 1 : 300.00 W and 15.00 percent\n", + " Consumer 2 : 500.00 W and 25.00 percent\n", + " Consumer 3 : 2000.00 W and 25.00 percent\n", + " Consumer 4 : 1000.00 W and 25.00 percent\n", + "iii) The Combined Load Factor and the Combined Average Load is 32.76 percent and 3800.00 W respectively\n", + "\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex3, Page 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "\n", + "#To Determine the Yearly Cost of the substation\n", + "\n", + "Teff=95/100 #Transmission Efficiency\n", + "Deff=85/100 #Distribution Efficiency\n", + "DFT=1.2 #Diversity Factor For Transmission\n", + "DFD=1.3 #Diversity Factor For Distribution\n", + "MDGS=100*(10**6) #Maximum Demand of Generating Station\n", + "ALF=40/100 #Annual Load Factor\n", + "ACCT=2.5*(10**6) #Annual Capital Charge for Transmission\n", + "ACCD=2*(10**6) #Annual Capital Charge for Distribution\n", + "GCC=100 #Generating Cost per kW demand\n", + "GCCU=5/100 # Per Unit Cost\n", + "#Fixed Charges from Supply to Substation Annually\n", + "GFC=GCC*MDGS/1000 #Generating\n", + "TFC=ACCT #Transmission\n", + "TotFCS=GFC+TFC #Total\n", + "#Fixed Charges for supply upto Consumer Annually\n", + "DFC=ACCD #Distribution\n", + "TotFCC=TotFCS+DFC #Total\n", + "\n", + "AMDS= DFT*MDGS/1000 #Aggregate of Maximum Demand at Supply\n", + "AMDC= DFD*AMDS #Aggregate of Maximum Demand for Consumers\n", + "\n", + "FCS=TotFCS/AMDS #Fixed Charges Per KW at substation\n", + "CES=GCCU/Teff #Cost of energy at the substation\n", + "\n", + "FCC=TotFCC/AMDC #Fixed Charges per KW at the consumer premises\n", + "CEC=CES/Deff #Cost of Energy at the consumer premises\n", + "\n", + "CEC*=100 # converting from rupee to paise\n", + "\n", + "print 'The Yealy Cost per KW demand and the cost per KWhr at:'\n", + "print 'a) The substation is %0.2f rupees per KW and %0.2f paise per kWhr'%(FCS,CES)\n", + "print 'b) The consumer premises is %g rupees per KW and %g paise per kWhr' %(FCC,CEC)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Yealy Cost per KW demand and the cost per KWhr at:\n", + "a) The substation is 104.17 rupees per KW and 0.05 paise per kWhr\n", + "b) The consumer premises is 92.9487 rupees per KW and 6.19195 paise per kWhr\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex4, Page 78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#To determine the Load factor and suitable units for 24 hr operation of the plant\n", + "\n", + "\n", + "#Demands at Various Time Periods starting from 12PM to 12PM\n", + "D1=500*(10**3) \n", + "D2=800*(10**3) \n", + "D3=2000*(10**3) \n", + "D4=1000*(10**3) \n", + "D5=2500*(10**3) \n", + "D6=2000*(10**3) \n", + "D7=1500*(10**3) \n", + "D8=1000*(10**3) \n", + "\n", + "MD=D5 #Maximum Demand\n", + "#Time Periods of demands from 12PM\n", + "T1=5 \n", + "T2=5 \n", + "T3=2 \n", + "T4=2 \n", + "T5=3 \n", + "T6=3 \n", + "T7=2 \n", + "T8=2 \n", + "\n", + "#Total Energy Demand in 24hrs\n", + "TED=(T1*D1)+(T2*D2)+(T3*D3)+(D4*T4)+(T5*D5)+(D6*T6)+(D7*T7)+(T8*D8) \n", + "\n", + "LF=TED*100/(24*MD) \n", + "\n", + "C1000=3*1000*(10**3) #1000 unit \n", + "C500=1*500*(10**3) #500 Unit\n", + "\n", + "TCP=C1000+C500 #Total capacity of the plant\n", + "PCF=TED*100/(24*TCP) #Plant Capacity Factor\n", + "\n", + "#Operating Schedule, Units operated can be seen in the textbook\n", + "G1=500*(10**3) \n", + "G2=1000*(10**3) \n", + "G3=2000*(10**3) \n", + "G4=1000*(10**3) \n", + "G5=2500*(10**3) \n", + "G6=2000*(10**3) \n", + "G7=1500*(10**3) \n", + "G8=1000*(10**3) \n", + "\n", + "TEG=(T1*G1)+(T2*G2)+(T3*G3)+(G4*T4)+(T5*G5)+(G6*T6)+(G7*T7)+(T8*G8) #Total Energy Generated\n", + "PUF=TED*100/(TEG) #Plant Use Factor\n", + "\n", + "print 'a) The Reserve Capacity is a 1000kW Unit and Load Factor is %0.2f percent' %LF\n", + "print 'b) The Plant Capacity Factor is %0.2f percent' %PCF\n", + "print 'c) The Plant Use Factor is %0.2f percent' %PUF" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) The Reserve Capacity is a 1000kW Unit and Load Factor is 51.67 percent\n", + "b) The Plant Capacity Factor is 36.90 percent\n", + "c) The Plant Use Factor is 96.88 percent\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex5, Page 80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#To determine the Plant use factore of each unit\n", + "\n", + "\n", + "MDS=25*(10**6) #Maximum Demand on the System\n", + "U1=15*(10**6) #Load Supplied By Unit 1\n", + "U2=12.5*(10**6) #Load Supplied By Unit 2\n", + "#Running Time Factor of the Unit\n", + "T1=1 \n", + "T2=40/100 \n", + "\n", + "#Energy generated by each unit\n", + "E1=1*(10**8) \n", + "E2=1*(10**7) \n", + "Et=E1+E2 #Total Energy\n", + "\n", + "#Maximum Demands on Each Units\n", + "MD1=U1 \n", + "MD2=MDS-U1 \n", + "\n", + "#Annual Load Factor for the Units\n", + "ALF1=E1*1000*100/(MD1*8760) \n", + "ALF2=E2*1000*100/(MD2*8760) \n", + "\n", + "LF2=E2*1000*100/(MD2*0.4*8760) #Load Factor for the it is loaded\n", + "\n", + "\n", + "PUF1=ALF1 #Plant Use Factor\n", + "PCF1=ALF1 # Plant Capacity Factor\n", + "\n", + "PCF2=E2*1000*100/(U2*8760) #Plant Capacity Factor for Unit 2\n", + "PUF2=E2*1000*100/(U2*0.4*8760) #Plant Use Factor for Unit 2\n", + "\n", + "LFP=Et*100*1000/(MDS*8760) #Annual Load Factor of the Complete Plant\n", + "\n", + "print 'The Load Factor, Plant Capacity Factor, Plant Use Factor of:'\n", + "print 'Unit 1 : %0.2f percent, %0.2f percent, %0.2f percent' %(ALF1,PCF1,PUF1)\n", + "print 'Unit 2 : %0.2f percent, %0.2f percent, %0.2f percent' %(ALF2,PCF2,PUF2)\n", + "print 'The Annual Load Factor of the Entire Plant is %0.2f percent' %LFP" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Load Factor, Plant Capacity Factor, Plant Use Factor of:\n", + "Unit 1 : 76.10 percent, 76.10 percent, 76.10 percent\n", + "Unit 2 : 11.42 percent, 9.13 percent, 22.83 percent\n", + "The Annual Load Factor of the Entire Plant is 50.23 percent\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6, Page 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#To determine the most economic power factor\n", + "\n", + "from numpy import sqrt\n", + "\n", + "P=200*(10**3) #Maximum Demand\n", + "pf=0.707 #Power Factor Lagging\n", + "\n", + "a=100 #Tariff per kVA per year\n", + "\n", + "b=200 #Power factor improvement cost Per kVA.\n", + "r=20 #Interest Depriciation, maintenance and cost of losses amount to 20% of capital cost per year\n", + "\n", + "# Economic PF = sqrt(1-((b1/a)**2))\n", + "\n", + "b1=r*b/100 # b' term accrding to the equation above\n", + "\n", + "pfeco=sqrt(1-((b1/a)**2)) #Economic Power Factor\n", + "\n", + "print 'The Economic Power Factor is %0.2f ' %pfeco\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Economic Power Factor is 0.92 \n" + ] + } + ], + "prompt_number": 22 + } + ], + "metadata": {} + } + ] +} diff --git a/sample_notebooks/Raj Phani/chapter_1.ipynb b/sample_notebooks/Raj Phani/chapter_1.ipynb new file mode 100755 index 00000000..f71cb56f --- /dev/null +++ b/sample_notebooks/Raj Phani/chapter_1.ipynb @@ -0,0 +1,1118 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# chapter1: electric charge"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.1"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 23,
+ "metadata": {
+ "collapsed": false,
+ "scrolled": true
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.1, Page:3 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The electric field in V/m is = 20000.0\n",
+ "\n",
+ " The force in N/C is = 20000.0\n",
+ "\n",
+ " The force on metal sphere in N is = 7.6e-05\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.1, Page:3 \\n \\n\"\n",
+ "#Given:\n",
+ "v=1000# potential\n",
+ "d=0.05# distance\n",
+ "q=3.8*10**-9# charge\n",
+ "\n",
+ "#solution:\n",
+ "e=v/d;#electric field\n",
+ "f=e;# force\n",
+ "f1=f*q;# force on metal sphere\n",
+ "print\"\\n The electric field in V/m is =\",e\n",
+ "print\"\\n The force in N/C is =\",f\n",
+ "print\"\\n The force on metal sphere in N is =\",f1\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.2"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.2, Page:4 \n",
+ " \n",
+ "\n",
+ "The potential in V is = 80.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.2, Page:4 \\n \\n\"\n",
+ "#Given:\n",
+ "energy=2*10**-6\n",
+ "c=2.5*10**-8# velocity of light\n",
+ "#solution:\n",
+ "v=energy/c# potential\n",
+ "print\"The potential in V is =\",v\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.3"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.3, Page:5 \n",
+ " \n",
+ "\n",
+ "The wavelength in Angstroms is = 3.88289589025\n",
+ "\n",
+ " The photon wavelength in Angstroms is = 9.11075e-05\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.3, Page:5 \\n \\n\"\n",
+ "#Given:\n",
+ "\n",
+ "energy=10 #in electron volts\n",
+ "m=9.1*10**-31# mass of electron in kg\n",
+ "h=6.626*10**-34# planck's constant J.s\n",
+ "c=3*10^8# speed of light in m/s\n",
+ "\n",
+ "#solution (a):\n",
+ "energy1=energy*1.6*10**-19# energy in J\n",
+ "p=(2*m*energy1)**0.5# momentum\n",
+ "wavelength=h/p*(10)**10\n",
+ "\n",
+ "print\"The wavelength in Angstroms is =\",wavelength\n",
+ "\n",
+ "\n",
+ "#solution (b):\n",
+ "wavelength1=h*c/energy1*(10)**10;#photon wavelength\n",
+ "\n",
+ "print\"\\n The photon wavelength in Angstroms is =\",wavelength1\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example1.4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.4, Page:6 \n",
+ " \n",
+ "\n",
+ "The energy in eV is = 150.768804945\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.4, Page:6 \\n \\n\"\n",
+ "\n",
+ "#Given:\n",
+ "\n",
+ "wavelength=10**-10\n",
+ "m=9.1*10**-31\n",
+ "h=6.626*10**-34\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "p=h/wavelength\n",
+ "e=p*p/(2*m) # energy in J\n",
+ "e1=e/(1.6*10**-19)# energy in eV\n",
+ "\n",
+ "print\"The energy in eV is =\",e1\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.5, Page:8 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The wavelength in 10^-5 Angstroms is = 0.655671822473\n",
+ "\n",
+ " The wavelength in 10^-5 Angstroms is = 0.648946805494\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.5, Page:8 \\n \\n\"\n",
+ "\n",
+ "#Given:\n",
+ "\n",
+ "m=1.66*10**-27# 1u=1.66*10^-27 kg\n",
+ "h=6.6262*10**-34#planck's constant in J.s\n",
+ "energy1=120# in Mev for oxygen\n",
+ "energy2=140# in MeV for nitrogen\n",
+ "\n",
+ "#solution(a):\n",
+ "\n",
+ "p=(2*m*16*energy1*(1.6022*10**-13))**0.5\n",
+ "wavelength1=h/p*(10)**15#wavelength in 10^-5 Angstroms\n",
+ "\n",
+ "print\"\\n The wavelength in 10^-5 Angstroms is =\",wavelength1\n",
+ "\n",
+ "#solution (b):\n",
+ "\n",
+ "p=(2*m*14*energy2*(1.6022*10**-13))**0.5\n",
+ "wavelength2=h/p*(10)**15#wavelength in 10^-5 Angstroms\n",
+ "\n",
+ "print\"\\n The wavelength in 10^-5 Angstroms is =\",wavelength2\n",
+ "\n",
+ "# 1 Angstrom = 10^-10 m\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example1.6"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 28,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.6, Page:9 \n",
+ " \n",
+ "\n",
+ "The energy in eV is = 8275.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.6, Page:9 \\n \\n\"\n",
+ "\n",
+ "#Given:\n",
+ "\n",
+ "wavelength=1.5*10**-10\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "e=(h*c)/wavelength# energy in J\n",
+ "e1=e/(1.6*10**-19)# energy in eV\n",
+ "\n",
+ "print\"The energy in eV is =\",e1\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.7, Page:10 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The threshold frequency in s^-1 is = 1.23634168427e+15\n",
+ "\n",
+ " The threshold wavelength in Angstroms is = 2426.51367187\n",
+ "\n",
+ " The energy of photoelectrone in eV is = 3.911875\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.7, Page:10 \\n \\n\"\n",
+ "\n",
+ "#Given:\n",
+ "\n",
+ "E=5.12*1.6*10**-19# energy in J\n",
+ "h=6.626*10**-34\n",
+ "c=3*10**8\n",
+ "wavelength=200*10**-9\n",
+ "w=2.3# in eV\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "tf=E/h# (part a)\n",
+ "print\"\\n The threshold frequency in s^-1 is =\",tf\n",
+ "\n",
+ "tl=c/tf*10**10# (part b)\n",
+ "print\"\\n The threshold wavelength in Angstroms is =\",tl\n",
+ "\n",
+ "e=(h*c)/(wavelength*1.6*10**-19)# photon energy in eV (part c)\n",
+ "\n",
+ "pe=e-w\n",
+ "\n",
+ "print\"\\n The energy of photoelectrone in eV is =\",pe\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.8"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.8, Page:10 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The velocity of alpha particles for 1 MeV in m/s is = 6941056.08394\n",
+ "\n",
+ " The velocity of alpha particles for 2 MeV in m/s is = 9816135.6511\n",
+ "\n",
+ " The velocity of deuteron particles for 1 MeV in m/s is = 9816135.6511\n",
+ "\n",
+ " The velocity of deuteron particles for 2 MeV in m/s is = 13882112.1679\n",
+ "\n",
+ " The velocity of proton particles for 1 MeV in m/s is = 13882112.1679\n",
+ "\n",
+ " The velocity of proton particles for 2 MeV in m/s is = 19632271.3022\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of velocity of alpha particles,deuteron,proton\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.8, Page:10 \\n \\n\"\n",
+ "#Given:\n",
+ "e1=1 # in MeV\n",
+ "e2=2 # in MeV\n",
+ "ma=4 # in u(amu)\n",
+ "md=2 # in u(amu)\n",
+ "mp=1 # in u(amu)\n",
+ "\n",
+ "# 1u = 1.6*10^-27 Kg\n",
+ "\n",
+ "#solution: part a)For alpha particles\n",
+ "\n",
+ "v1a=((2*e1*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of alpha particles for 1 MeV in m/s is =\",v1a# For 1 MeV\n",
+ "\n",
+ "v2a=((2*e2*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of alpha particles for 2 MeV in m/s is =\",v2a# For 2 MeV\n",
+ "\n",
+ "#solution: part b)For deuteron particles\n",
+ "\n",
+ "v1b=((2*e1*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of deuteron particles for 1 MeV in m/s is =\",v1b # For 1 MeV\n",
+ "\n",
+ "\n",
+ "v2b=((2*e2*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of deuteron particles for 2 MeV in m/s is =\",v2b # For 2 MeV\n",
+ "\n",
+ "#solution: part c)For proton particles\n",
+ "\n",
+ "v1p=((2*e1*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of proton particles for 1 MeV in m/s is =\",v1p # For 1 MeV\n",
+ "\n",
+ "\n",
+ "v2p=((2*e2*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of proton particles for 2 MeV in m/s is =\",v2p # For 2 MeV\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.9, Page: \n",
+ " \n",
+ "\n",
+ "The energy in MeV is = 933.919973435\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The energy\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.9, Page: \\n \\n\"\n",
+ "#Given:\n",
+ "\n",
+ "m=1/(6.023*10**23)#mass of 1 atom in g\n",
+ "m1=m*10**-3#mass of 1 atom in Kg\n",
+ "c=3*10**8# velocity in m/s\n",
+ "#solution:\n",
+ "\n",
+ "e=m1*c*c; # energy in J\n",
+ "e1=e/(1.6*10**-13)# energy in MeV\n",
+ "\n",
+ "print\"The energy in MeV is =\",e1\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.1, Page:11 \n",
+ " \n",
+ "\n",
+ "The energy in eV is = 13.2638658253\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The energy\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.1, Page:11 \\n \\n\"\n",
+ "#Given:\n",
+ "\n",
+ "enthalpy=1278 # enthalpy of combustion in kJ/mol\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "energy=(enthalpy*1000)/(6.022*10**23*1.6*10**-19)\n",
+ "\n",
+ "print\"The energy in eV is =\",energy\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.11, Page:11 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mean binding energy of helium atom in MeV is = 7.0710038475\n",
+ "\n",
+ " The mean binding energy of oxygen atom in MeV is = 7.9800498909\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of mean binding energy of helium and oxygen\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.11, Page:11 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.0078\n",
+ "mn=1.0087\n",
+ "ma=4.0026\n",
+ "mo=15.9949\n",
+ "Ah=4.0026 # atomic mass of helium\n",
+ "Ao=15.9949 # atomic mass of oxygen\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "# part (a)\n",
+ "\n",
+ "B1=(2*mh+2*mn-ma)*931 # in MeV\n",
+ "Bh=B1/Ah\n",
+ "print\"\\n The mean binding energy of helium atom in MeV is =\",Bh\n",
+ "\n",
+ "# part (b)\n",
+ "\n",
+ "B2=(8*mh+8*mn-mo)*931 # in MeV\n",
+ "Bo=B2/Ao\n",
+ "print\"\\n The mean binding energy of oxygen atom in MeV is =\",Bo\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.12, Page:12 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mean binding energy of Be atom in MeV is = 7.05928572321\n",
+ "From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of \n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.12, Page:12 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.0078;\n",
+ "mn=1.0087;\n",
+ "ABe=8.0053; # atomic mass of beryllium\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "B1=(4*mh+4*mn-ABe)*931; # in MeV\n",
+ "Bh=B1/ABe;\n",
+ "print\"\\n The mean binding energy of Be atom in MeV is =\",Bh\n",
+ "\n",
+ "print\"From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 34,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.13, Page:12 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The amount of coal required in Kg is = 2499.85671416\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of amount of coal\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.13, Page:12 \\n \\n\"\n",
+ "#Given:\n",
+ "\n",
+ "e=200; # in Mev\n",
+ "m=0.235; # weight of uranium atom in Kg\n",
+ "enthalpy=393.5; # in KJ/mol\n",
+ "Na=6.02*10**23;\n",
+ "\n",
+ "\n",
+ "#solution:\n",
+ "e1=e*1.6*10**-19*10**6;\n",
+ "atoms=Na/m;\n",
+ "e2=atoms*e1;#energy released in J\n",
+ "m1=(e2*12)/(393.5*1000*1000);# in Kg\n",
+ "m2=m1/1000;# in tons\n",
+ "print\"\\n The amount of coal required in Kg is =\", m2\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 35,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.14, Page:13 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The energy release in part (a) in eV/molecule is = 2.51472\n",
+ "\n",
+ " The energy release in part (b) in eV/molecule is = 9.22688\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The energy releases\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.14, Page:13 \\n \\n\"\n",
+ "#Given:\n",
+ "H1=241.8; # in KJ/mol\n",
+ "H2=887.2; # in KJ/mol\n",
+ "# 1 KJ/mol = 0.0104 eV/atom\n",
+ "\n",
+ "#solution: part (a)\n",
+ "e1=H1*0.0104;\n",
+ "print\"\\n The energy release in part (a) in eV/molecule is =\",e1\n",
+ "\n",
+ "#solution: part (b)\n",
+ "e2=H2*0.0104;\n",
+ "print\"\\n The energy release in part (b) in eV/molecule is =\",e2\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 36,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.15, Page:14 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The energy release in part (a) in KJ/mol of carbondioxide is = 394.912\n",
+ "\n",
+ " The energy release in part (b) in KJ/mol of alumina is = 1675.968\n",
+ "\n",
+ " The energy release in part (c) in MJ/atom of U(235) is = 19264000.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The energy releases\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.15, Page:14 \\n \\n\"\n",
+ "#Given:\n",
+ "H1=4.1; # in eV/molecule\n",
+ "H2=17.4; # in eV/molecule\n",
+ "H3=200;# in MeV/atom of U\n",
+ "\n",
+ "# 1 eV/atom = 96.32 KJ/mol\n",
+ "\n",
+ "#solution: part (a)\n",
+ "e1=H1*96.32;\n",
+ "print\"\\n The energy release in part (a) in KJ/mol of carbondioxide is =\",e1\n",
+ "\n",
+ "#solution: part (b)\n",
+ "e2=H2*96.32;\n",
+ "print\"\\n The energy release in part (b) in KJ/mol of alumina is =\",e2\n",
+ "\n",
+ "#solution: part (c)\n",
+ "e3=H3*1000*96.32;# in MJ/atom of U(235)\n",
+ "print\"\\n The energy release in part (c) in MJ/atom of U(235) is =\",e3\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 37,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.16, Page:15 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The rate of energy release in W is 949251379.039\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The rate of energy release\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.16, Page:15 \\n \\n\"\n",
+ "#Given:\n",
+ "e=200; #MeV/ atom of U\n",
+ "# 1 eV = 1.6*10^-19 J\n",
+ "Na=6.023*10**23;\n",
+ "M=0.235; # mass in Kg\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "e1=e*1.6*10**-19*10**6;\n",
+ "A=Na/M;\n",
+ "e2=A*e1; # energy released in MJ/day\n",
+ "e3=e2/(24*3600);\n",
+ "print\"\\n The rate of energy release in W is \",e3\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 38,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.17, Page:16 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mass loss in 10^-27 Kg/He formed is = 0.046412244898\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The mass loss\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.17, Page:16 \\n \\n\"\n",
+ "#Given:\n",
+ "e=26.03; # in MeV\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "loss=e/931; #in atomic mass units (u)\n",
+ "# 1 u = 1.66*10^-27 Kg\n",
+ "m=(loss*1.66*10**-27)/(1*10**-27);\n",
+ "print\"\\n The mass loss in 10^-27 Kg/He formed is =\",m\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 39,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.18, Page:17 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The energy loss in MeV is = 4.03123\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The energy loss\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.18, Page:17 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.007825;\n",
+ "mt=3.016049;\n",
+ "md=2.014102;\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "m1=(mh+mt-2*md);\n",
+ "e=(-m1)*931; # in MeV\n",
+ "print\"\\n The energy loss in MeV is =\",e\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 40,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.19, Page:18 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mean binding energy of tritium atom in MeV is = 2.81085817903\n",
+ "\n",
+ " The mean binding energy of nickel atom in MeV is = 8.71580311296\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of mean binding energy of tritium and nickel atom\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.19, Page:18 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.007825;\n",
+ "mn=1.008665;\n",
+ "mt=3.016049; # atomic mass of Tritium\n",
+ "mNi=59.93528; # atomic mass of Nickel\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "# part (a)\n",
+ "\n",
+ "B1=(1*mh+2*mn-mt)*931; # in MeV\n",
+ "Bh=B1/mt;\n",
+ "print\"\\n The mean binding energy of tritium atom in MeV is =\",Bh\n",
+ "\n",
+ "# part (b)\n",
+ "\n",
+ "B2=(28*mh+32*mn-mNi)*931; # in MeV\n",
+ "Bo=B2/mNi;\n",
+ "print\"\\n The mean binding energy of nickel atom in MeV is =\",Bo\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 41,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.20, Page:19 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mean binding energy of Cl (35) atom in MeV is = 8.52810201079\n",
+ "\n",
+ " The mean binding energy of Cl (37) atom in MeV is = 8.57839008383\n",
+ "\n",
+ " The increase in mean binding energy of Cl atom in MeV is = 0.0502880730447\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of mean binding energy of Cl\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.20, Page:19 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.00783;\n",
+ "mn=1.00867;\n",
+ "m35=34.96885; # atomic mass of Cl (35)\n",
+ "m37=36.96590; # atomic mass of Cl (37)\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "B1=(17*mh+18*mn-m35)*931; # in MeV\n",
+ "Bh=B1/m35;\n",
+ "print\"\\n The mean binding energy of Cl (35) atom in MeV is =\",Bh\n",
+ "\n",
+ "B2=(17*mh+20*mn-m37)*931; # in MeV\n",
+ "Bo=B2/m37;\n",
+ "print\"\\n The mean binding energy of Cl (37) atom in MeV is =\",Bo\n",
+ "\n",
+ "Bi=Bo-Bh;\n",
+ "print\"\\n The increase in mean binding energy of Cl atom in MeV is =\",Bi\n",
+ "\n",
+ "# NOTE: The answer depends upon how much precise value you take for atomic masses.\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 42,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.21, Page:20 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mean binding energy of Na(22) in MeV is = 7.92358978299\n",
+ "\n",
+ " The mean binding energy of Na(23)in MeV is = 8.11544250059\n",
+ "\n",
+ " The mean binding energy of Na(24) in MeV is = 8.07172719656\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of mean binding energy of Na\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.21, Page:20 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.0078;\n",
+ "mn=1.0087;\n",
+ "m22=21.99431;# atomic mass of Na 22\n",
+ "m23=22.9898;# atomic mass of Na 23\n",
+ "m24=23.9909;# atomic mass of Na 24\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "# part (a)\n",
+ "\n",
+ "B1=((11*mh+11*mn)-m22)*931; # in MeV\n",
+ "Bh=B1/m22;\n",
+ "print\"\\n The mean binding energy of Na(22) in MeV is =\",Bh\n",
+ "\n",
+ "# part (b)\n",
+ "\n",
+ "B2=((11*mh+12*mn)-m23)*931; # in MeV\n",
+ "Bo=B2/m23;\n",
+ "print\"\\n The mean binding energy of Na(23)in MeV is =\",Bo\n",
+ "\n",
+ "# part (c)\n",
+ "\n",
+ "B3=((11*mh+13*mn)-m24)*931; # in MeV\n",
+ "Bs=B3/m24;\n",
+ "print\"\\n The mean binding energy of Na(24) in MeV is =\",Bs\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/Raj Phani/chapter_1_1.ipynb b/sample_notebooks/Raj Phani/chapter_1_1.ipynb new file mode 100755 index 00000000..6bb4fa49 --- /dev/null +++ b/sample_notebooks/Raj Phani/chapter_1_1.ipynb @@ -0,0 +1,1118 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# chapter1: electric charge"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.1"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 23,
+ "metadata": {
+ "collapsed": false,
+ "scrolled": true
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.1, Page:3 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The electric field in V/m is = 20000.0\n",
+ "\n",
+ " The force in N/C is = 20000.0\n",
+ "\n",
+ " The force on metal sphere in N is = 7.6e-05\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.1, Page:3 \\n \\n\"\n",
+ "#Given:\n",
+ "v=1000# potential\n",
+ "d=0.05# distance\n",
+ "q=3.8*10**-9# charge\n",
+ "\n",
+ "#solution:\n",
+ "e=v/d;#electric field\n",
+ "f=e;# force\n",
+ "f1=f*q;# force on metal sphere\n",
+ "print\"\\n The electric field in V/m is =\",e\n",
+ "print\"\\n The force in N/C is =\",f\n",
+ "print\"\\n The force on metal sphere in N is =\",f1\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.2"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.2, Page:4 \n",
+ " \n",
+ "\n",
+ "The potential in V is = 80.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.2, Page:4 \\n \\n\"\n",
+ "#Given:\n",
+ "energy=2*10**-6\n",
+ "c=2.5*10**-8# velocity of light\n",
+ "#solution:\n",
+ "v=energy/c# potential\n",
+ "print\"The potential in V is =\",v\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.3"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.3, Page:5 \n",
+ " \n",
+ "\n",
+ "The wavelength in Angstroms is = 3.88289589025\n",
+ "\n",
+ " The photon wavelength in Angstroms is = 9.11075e-05\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.3, Page:5 \\n \\n\"\n",
+ "#Given:\n",
+ "\n",
+ "energy=10 #in electron volts\n",
+ "m=9.1*10**-31# mass of electron in kg\n",
+ "h=6.626*10**-34# planck's constant J.s\n",
+ "c=3*10^8# speed of light in m/s\n",
+ "\n",
+ "#solution (a):\n",
+ "energy1=energy*1.6*10**-19# energy in J\n",
+ "p=(2*m*energy1)**0.5# momentum\n",
+ "wavelength=h/p*(10)**10\n",
+ "\n",
+ "print\"The wavelength in Angstroms is =\",wavelength\n",
+ "\n",
+ "\n",
+ "#solution (b):\n",
+ "wavelength1=h*c/energy1*(10)**10;#photon wavelength\n",
+ "\n",
+ "print\"\\n The photon wavelength in Angstroms is =\",wavelength1\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example1.4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.4, Page:6 \n",
+ " \n",
+ "\n",
+ "The energy in eV is = 150.768804945\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.4, Page:6 \\n \\n\"\n",
+ "\n",
+ "#Given:\n",
+ "\n",
+ "wavelength=10**-10\n",
+ "m=9.1*10**-31\n",
+ "h=6.626*10**-34\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "p=h/wavelength\n",
+ "e=p*p/(2*m) # energy in J\n",
+ "e1=e/(1.6*10**-19)# energy in eV\n",
+ "\n",
+ "print\"The energy in eV is =\",e1\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.5, Page:8 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The wavelength in 10^-5 Angstroms is = 0.655671822473\n",
+ "\n",
+ " The wavelength in 10^-5 Angstroms is = 0.648946805494\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.5, Page:8 \\n \\n\"\n",
+ "\n",
+ "#Given:\n",
+ "\n",
+ "m=1.66*10**-27# 1u=1.66*10^-27 kg\n",
+ "h=6.6262*10**-34#planck's constant in J.s\n",
+ "energy1=120# in Mev for oxygen\n",
+ "energy2=140# in MeV for nitrogen\n",
+ "\n",
+ "#solution(a):\n",
+ "\n",
+ "p=(2*m*16*energy1*(1.6022*10**-13))**0.5\n",
+ "wavelength1=h/p*(10)**15#wavelength in 10^-5 Angstroms\n",
+ "\n",
+ "print\"\\n The wavelength in 10^-5 Angstroms is =\",wavelength1\n",
+ "\n",
+ "#solution (b):\n",
+ "\n",
+ "p=(2*m*14*energy2*(1.6022*10**-13))**0.5\n",
+ "wavelength2=h/p*(10)**15#wavelength in 10^-5 Angstroms\n",
+ "\n",
+ "print\"\\n The wavelength in 10^-5 Angstroms is =\",wavelength2\n",
+ "\n",
+ "# 1 Angstrom = 10^-10 m\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example1.6"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 28,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.6, Page:9 \n",
+ " \n",
+ "\n",
+ "The energy in eV is = 8275.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.6, Page:9 \\n \\n\"\n",
+ "\n",
+ "#Given:\n",
+ "\n",
+ "wavelength=1.5*10**-10\n",
+ "h=6.62*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "e=(h*c)/wavelength# energy in J\n",
+ "e1=e/(1.6*10**-19)# energy in eV\n",
+ "\n",
+ "print\"The energy in eV is =\",e1\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.7, Page:10 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The threshold frequency in s^-1 is = 1.23634168427e+15\n",
+ "\n",
+ " The threshold wavelength in Angstroms is = 2426.51367187\n",
+ "\n",
+ " The energy of photoelectrone in eV is = 3.911875\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of elelectric field and force\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.7, Page:10 \\n \\n\"\n",
+ "\n",
+ "#Given:\n",
+ "\n",
+ "E=5.12*1.6*10**-19# energy in J\n",
+ "h=6.626*10**-34\n",
+ "c=3*10**8\n",
+ "wavelength=200*10**-9\n",
+ "w=2.3# in eV\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "tf=E/h# (part a)\n",
+ "print\"\\n The threshold frequency in s^-1 is =\",tf\n",
+ "\n",
+ "tl=c/tf*10**10# (part b)\n",
+ "print\"\\n The threshold wavelength in Angstroms is =\",tl\n",
+ "\n",
+ "e=(h*c)/(wavelength*1.6*10**-19)# photon energy in eV (part c)\n",
+ "\n",
+ "pe=e-w\n",
+ "\n",
+ "print\"\\n The energy of photoelectrone in eV is =\",pe\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.8"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.8, Page:10 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The velocity of alpha particles for 1 MeV in m/s is = 6941056.08394\n",
+ "\n",
+ " The velocity of alpha particles for 2 MeV in m/s is = 9816135.6511\n",
+ "\n",
+ " The velocity of deuteron particles for 1 MeV in m/s is = 9816135.6511\n",
+ "\n",
+ " The velocity of deuteron particles for 2 MeV in m/s is = 13882112.1679\n",
+ "\n",
+ " The velocity of proton particles for 1 MeV in m/s is = 13882112.1679\n",
+ "\n",
+ " The velocity of proton particles for 2 MeV in m/s is = 19632271.3022\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of velocity of alpha particles,deuteron,proton\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.8, Page:10 \\n \\n\"\n",
+ "#Given:\n",
+ "e1=1 # in MeV\n",
+ "e2=2 # in MeV\n",
+ "ma=4 # in u(amu)\n",
+ "md=2 # in u(amu)\n",
+ "mp=1 # in u(amu)\n",
+ "\n",
+ "# 1u = 1.6*10^-27 Kg\n",
+ "\n",
+ "#solution: part a)For alpha particles\n",
+ "\n",
+ "v1a=((2*e1*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of alpha particles for 1 MeV in m/s is =\",v1a# For 1 MeV\n",
+ "\n",
+ "v2a=((2*e2*10**6*1.6*10**-19)/(ma*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of alpha particles for 2 MeV in m/s is =\",v2a# For 2 MeV\n",
+ "\n",
+ "#solution: part b)For deuteron particles\n",
+ "\n",
+ "v1b=((2*e1*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of deuteron particles for 1 MeV in m/s is =\",v1b # For 1 MeV\n",
+ "\n",
+ "\n",
+ "v2b=((2*e2*10**6*1.6*10**-19)/(md*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of deuteron particles for 2 MeV in m/s is =\",v2b # For 2 MeV\n",
+ "\n",
+ "#solution: part c)For proton particles\n",
+ "\n",
+ "v1p=((2*e1*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of proton particles for 1 MeV in m/s is =\",v1p # For 1 MeV\n",
+ "\n",
+ "\n",
+ "v2p=((2*e2*10**6*1.6*10**-19)/(mp*1.6605*10**-27))**.5\n",
+ "print\"\\n The velocity of proton particles for 2 MeV in m/s is =\",v2p # For 2 MeV\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.9, Page:10 \n",
+ " \n",
+ "\n",
+ "The energy in MeV is = 933.919973435\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The energy\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.9, Page:10 \\n \\n\"\n",
+ "#Given:\n",
+ "\n",
+ "m=1/(6.023*10**23)#mass of 1 atom in g\n",
+ "m1=m*10**-3#mass of 1 atom in Kg\n",
+ "c=3*10**8# velocity in m/s\n",
+ "#solution:\n",
+ "\n",
+ "e=m1*c*c; # energy in J\n",
+ "e1=e/(1.6*10**-13)# energy in MeV\n",
+ "\n",
+ "print\"The energy in MeV is =\",e1\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 31,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.1, Page:11 \n",
+ " \n",
+ "\n",
+ "The energy in eV is = 13.2638658253\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The energy\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.1, Page:11 \\n \\n\"\n",
+ "#Given:\n",
+ "\n",
+ "enthalpy=1278 # enthalpy of combustion in kJ/mol\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "energy=(enthalpy*1000)/(6.022*10**23*1.6*10**-19)\n",
+ "\n",
+ "print\"The energy in eV is =\",energy\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.11, Page:11 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mean binding energy of helium atom in MeV is = 7.0710038475\n",
+ "\n",
+ " The mean binding energy of oxygen atom in MeV is = 7.9800498909\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of mean binding energy of helium and oxygen\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.11, Page:11 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.0078\n",
+ "mn=1.0087\n",
+ "ma=4.0026\n",
+ "mo=15.9949\n",
+ "Ah=4.0026 # atomic mass of helium\n",
+ "Ao=15.9949 # atomic mass of oxygen\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "# part (a)\n",
+ "\n",
+ "B1=(2*mh+2*mn-ma)*931 # in MeV\n",
+ "Bh=B1/Ah\n",
+ "print\"\\n The mean binding energy of helium atom in MeV is =\",Bh\n",
+ "\n",
+ "# part (b)\n",
+ "\n",
+ "B2=(8*mh+8*mn-mo)*931 # in MeV\n",
+ "Bo=B2/Ao\n",
+ "print\"\\n The mean binding energy of oxygen atom in MeV is =\",Bo\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.12, Page:12 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mean binding energy of Be atom in MeV is = 7.05928572321\n",
+ "From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of \n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.12, Page:12 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.0078;\n",
+ "mn=1.0087;\n",
+ "ABe=8.0053; # atomic mass of beryllium\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "B1=(4*mh+4*mn-ABe)*931; # in MeV\n",
+ "Bh=B1/ABe;\n",
+ "print\"\\n The mean binding energy of Be atom in MeV is =\",Bh\n",
+ "\n",
+ "print\"From previous problem we have the avg. binding energy of helium atom is 7.08 MeV, Hence Be is unstable to fission into 2 alphas\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 34,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.13, Page:12 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The amount of coal required in Kg is = 2499.85671416\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of amount of coal\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.13, Page:12 \\n \\n\"\n",
+ "#Given:\n",
+ "\n",
+ "e=200; # in Mev\n",
+ "m=0.235; # weight of uranium atom in Kg\n",
+ "enthalpy=393.5; # in KJ/mol\n",
+ "Na=6.02*10**23;\n",
+ "\n",
+ "\n",
+ "#solution:\n",
+ "e1=e*1.6*10**-19*10**6;\n",
+ "atoms=Na/m;\n",
+ "e2=atoms*e1;#energy released in J\n",
+ "m1=(e2*12)/(393.5*1000*1000);# in Kg\n",
+ "m2=m1/1000;# in tons\n",
+ "print\"\\n The amount of coal required in Kg is =\", m2\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 35,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.14, Page:13 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The energy release in part (a) in eV/molecule is = 2.51472\n",
+ "\n",
+ " The energy release in part (b) in eV/molecule is = 9.22688\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The energy releases\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.14, Page:13 \\n \\n\"\n",
+ "#Given:\n",
+ "H1=241.8; # in KJ/mol\n",
+ "H2=887.2; # in KJ/mol\n",
+ "# 1 KJ/mol = 0.0104 eV/atom\n",
+ "\n",
+ "#solution: part (a)\n",
+ "e1=H1*0.0104;\n",
+ "print\"\\n The energy release in part (a) in eV/molecule is =\",e1\n",
+ "\n",
+ "#solution: part (b)\n",
+ "e2=H2*0.0104;\n",
+ "print\"\\n The energy release in part (b) in eV/molecule is =\",e2\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 36,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.15, Page:14 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The energy release in part (a) in KJ/mol of carbondioxide is = 394.912\n",
+ "\n",
+ " The energy release in part (b) in KJ/mol of alumina is = 1675.968\n",
+ "\n",
+ " The energy release in part (c) in MJ/atom of U(235) is = 19264000.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The energy releases\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.15, Page:14 \\n \\n\"\n",
+ "#Given:\n",
+ "H1=4.1; # in eV/molecule\n",
+ "H2=17.4; # in eV/molecule\n",
+ "H3=200;# in MeV/atom of U\n",
+ "\n",
+ "# 1 eV/atom = 96.32 KJ/mol\n",
+ "\n",
+ "#solution: part (a)\n",
+ "e1=H1*96.32;\n",
+ "print\"\\n The energy release in part (a) in KJ/mol of carbondioxide is =\",e1\n",
+ "\n",
+ "#solution: part (b)\n",
+ "e2=H2*96.32;\n",
+ "print\"\\n The energy release in part (b) in KJ/mol of alumina is =\",e2\n",
+ "\n",
+ "#solution: part (c)\n",
+ "e3=H3*1000*96.32;# in MJ/atom of U(235)\n",
+ "print\"\\n The energy release in part (c) in MJ/atom of U(235) is =\",e3\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 37,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.16, Page:15 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The rate of energy release in W is 949251379.039\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The rate of energy release\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.16, Page:15 \\n \\n\"\n",
+ "#Given:\n",
+ "e=200; #MeV/ atom of U\n",
+ "# 1 eV = 1.6*10^-19 J\n",
+ "Na=6.023*10**23;\n",
+ "M=0.235; # mass in Kg\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "e1=e*1.6*10**-19*10**6;\n",
+ "A=Na/M;\n",
+ "e2=A*e1; # energy released in MJ/day\n",
+ "e3=e2/(24*3600);\n",
+ "print\"\\n The rate of energy release in W is \",e3\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 38,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.17, Page:16 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mass loss in 10^-27 Kg/He formed is = 0.046412244898\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The mass loss\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.17, Page:16 \\n \\n\"\n",
+ "#Given:\n",
+ "e=26.03; # in MeV\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "loss=e/931; #in atomic mass units (u)\n",
+ "# 1 u = 1.66*10^-27 Kg\n",
+ "m=(loss*1.66*10**-27)/(1*10**-27);\n",
+ "print\"\\n The mass loss in 10^-27 Kg/He formed is =\",m\n",
+ "\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 39,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.18, Page:17 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The energy loss in MeV is = 4.03123\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of The energy loss\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.18, Page:17 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.007825;\n",
+ "mt=3.016049;\n",
+ "md=2.014102;\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "m1=(mh+mt-2*md);\n",
+ "e=(-m1)*931; # in MeV\n",
+ "print\"\\n The energy loss in MeV is =\",e\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 40,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.19, Page:18 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mean binding energy of tritium atom in MeV is = 2.81085817903\n",
+ "\n",
+ " The mean binding energy of nickel atom in MeV is = 8.71580311296\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of mean binding energy of tritium and nickel atom\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.19, Page:18 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.007825;\n",
+ "mn=1.008665;\n",
+ "mt=3.016049; # atomic mass of Tritium\n",
+ "mNi=59.93528; # atomic mass of Nickel\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "# part (a)\n",
+ "\n",
+ "B1=(1*mh+2*mn-mt)*931; # in MeV\n",
+ "Bh=B1/mt;\n",
+ "print\"\\n The mean binding energy of tritium atom in MeV is =\",Bh\n",
+ "\n",
+ "# part (b)\n",
+ "\n",
+ "B2=(28*mh+32*mn-mNi)*931; # in MeV\n",
+ "Bo=B2/mNi;\n",
+ "print\"\\n The mean binding energy of nickel atom in MeV is =\",Bo\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 41,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.20, Page:19 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mean binding energy of Cl (35) atom in MeV is = 8.52810201079\n",
+ "\n",
+ " The mean binding energy of Cl (37) atom in MeV is = 8.57839008383\n",
+ "\n",
+ " The increase in mean binding energy of Cl atom in MeV is = 0.0502880730447\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of mean binding energy of Cl\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.20, Page:19 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.00783;\n",
+ "mn=1.00867;\n",
+ "m35=34.96885; # atomic mass of Cl (35)\n",
+ "m37=36.96590; # atomic mass of Cl (37)\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "B1=(17*mh+18*mn-m35)*931; # in MeV\n",
+ "Bh=B1/m35;\n",
+ "print\"\\n The mean binding energy of Cl (35) atom in MeV is =\",Bh\n",
+ "\n",
+ "B2=(17*mh+20*mn-m37)*931; # in MeV\n",
+ "Bo=B2/m37;\n",
+ "print\"\\n The mean binding energy of Cl (37) atom in MeV is =\",Bo\n",
+ "\n",
+ "Bi=Bo-Bh;\n",
+ "print\"\\n The increase in mean binding energy of Cl atom in MeV is =\",Bi\n",
+ "\n",
+ "# NOTE: The answer depends upon how much precise value you take for atomic masses.\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 1.21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 42,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 1.21, Page:20 \n",
+ " \n",
+ "\n",
+ "\n",
+ " The mean binding energy of Na(22) in MeV is = 7.92358978299\n",
+ "\n",
+ " The mean binding energy of Na(23)in MeV is = 8.11544250059\n",
+ "\n",
+ " The mean binding energy of Na(24) in MeV is = 8.07172719656\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of mean binding energy of Na\n",
+ "#intiation of all variables\n",
+ "# Chapter 1\n",
+ "print\"Example 1.21, Page:20 \\n \\n\"\n",
+ "#Given:\n",
+ "mh=1.0078;\n",
+ "mn=1.0087;\n",
+ "m22=21.99431;# atomic mass of Na 22\n",
+ "m23=22.9898;# atomic mass of Na 23\n",
+ "m24=23.9909;# atomic mass of Na 24\n",
+ "\n",
+ "#solution:\n",
+ "\n",
+ "# part (a)\n",
+ "\n",
+ "B1=((11*mh+11*mn)-m22)*931; # in MeV\n",
+ "Bh=B1/m22;\n",
+ "print\"\\n The mean binding energy of Na(22) in MeV is =\",Bh\n",
+ "\n",
+ "# part (b)\n",
+ "\n",
+ "B2=((11*mh+12*mn)-m23)*931; # in MeV\n",
+ "Bo=B2/m23;\n",
+ "print\"\\n The mean binding energy of Na(23)in MeV is =\",Bo\n",
+ "\n",
+ "# part (c)\n",
+ "\n",
+ "B3=((11*mh+13*mn)-m24)*931; # in MeV\n",
+ "Bs=B3/m24;\n",
+ "print\"\\n The mean binding energy of Na(24) in MeV is =\",Bs\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/RohithYeedulapalli/Chapter_7.ipynb b/sample_notebooks/RohithYeedulapalli/Chapter_7.ipynb new file mode 100755 index 00000000..c41c4cd6 --- /dev/null +++ b/sample_notebooks/RohithYeedulapalli/Chapter_7.ipynb @@ -0,0 +1,232 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 7:LASERS "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.1, Page number 7.32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Divergence = 0.5 *10**-3 radian\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#variable declaration\n",
+ "r1 = 2; #in radians\n",
+ "r2 = 3; #in radians\n",
+ "d1 = 4; #Converting from mm to radians\n",
+ "d2 = 6; #Converting from mm to radians\n",
+ "\n",
+ "#calculations\n",
+ "D = (r2-r1)/(d2*10**3-d1*10**3)\n",
+ "\n",
+ "#Result\n",
+ "print \"Divergence =\",round(D*10**3,3),\"*10**-3 radian\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.2, Page number 7.32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 23,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Frequency (V) = 4.32 *10**14 Hz\n",
+ "Relative Population= 1.081 *10**30\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "from __future__ import division\n",
+ "from sympy import *\n",
+ "#variable declaration\n",
+ "C=3*10**8 #The speed of light\n",
+ "L=6943 #Wavelength\n",
+ "T=300 #Temperature in Kelvin\n",
+ "h=6.626*10**-34 #Planck constant \n",
+ "k=1.38*10**-23 #Boltzmann's constant\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "V=(C)/(L*10**-10)\n",
+ "R=math.exp(h*V/(k*T))\n",
+ "\n",
+ "#Result\n",
+ "print \"Frequency (V) =\",round(V/10**14,2),\"*10**14 Hz\"\n",
+ "print \"Relative Population=\",round(R/10**30,3),\"*10**30\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.3, Page number 7.32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Frequency= 4.74 *10**14 Hz\n",
+ "no.of photons emitted= 7.322 *10**15 photons/sec\n",
+ "Power density = 2.3 kWm**-2\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "C=3*10**8 #Velocity of light\n",
+ "W=632.8*10**-9 #wavelength\n",
+ "P=2.3\n",
+ "t=1\n",
+ "h=6.626*10**-34 #Planck constant \n",
+ "S=1*10**-6\n",
+ "\n",
+ "#Calculations\n",
+ "V=C/W #Frequency\n",
+ "n=((P*10**-3)*t)/(h*V) #no.of photons emitted\n",
+ "PD=P*10**-3/S\n",
+ "\n",
+ "#Result\n",
+ "print \"Frequency=\",round(V/10**14,2),\"*10**14 Hz\"\n",
+ "print \"no.of photons emitted=\",round(n/10**15,3),\"*10**15 photons/sec\"\n",
+ "print \"Power density =\",round(PD/1000,1),\"kWm**-2\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.4, Page number 7.33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Wavelenght = 8628.0 Angstrom\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "#variable declaration\n",
+ "h=6.626*10**-34 #Planck constant \n",
+ "C=3*10**8 #Velocity of light\n",
+ "E_g=1.44 #bandgap \n",
+ "\n",
+ "#calculations\n",
+ "W=(h*C)*10**10/(E_g*1.6*10**-19)\n",
+ "\n",
+ "#Result\n",
+ "print \"Wavelenght =\",round(W),\"Angstrom\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.5, Page number 7.33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Band gap = 0.8 eV\n"
+ ]
+ }
+ ],
+ "source": [
+ "import math\n",
+ "\n",
+ "#variable declaration\n",
+ "W=1.55 #wavelength\n",
+ "\n",
+ "#Calculations\n",
+ "E_g=(1.24)/W #Bandgap in eV \n",
+ "\n",
+ "#Result\n",
+ "print \"Band gap =\",E_g,\"eV\""
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/Sabiya/Chapter9_1.ipynb b/sample_notebooks/Sabiya/Chapter9_1.ipynb new file mode 100755 index 00000000..d700ac81 --- /dev/null +++ b/sample_notebooks/Sabiya/Chapter9_1.ipynb @@ -0,0 +1,460 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:55d62edc09823ce69cb883c8ec8f5b8abe049583981245da7e91d5fefcb128cb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter9 - Special Oscilloscopes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.14.1 - page : 9-45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# peak to peak voltage and rms voltage\n", + "vdv=1 # V/div\n", + "n=6.8 #no. of divisions\n", + "Vpp=vdv*n #peak to peak voltage in V\n", + "vrms=Vpp/(2*(2)**(1.0/2)) #rms voltage in V\n", + "print \"Peak to peak voltage is \",Vpp,\" V\"\n", + "print \"rms voltage is \",round(vrms,4),\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak to peak voltage is 6.8 V\n", + "rms voltage is 2.4042 V\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.14.2 - page : 9-46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Time interval\n", + "vdv=2 # V per division in micro seconds/div\n", + "n=2 #no. of divisions\n", + "Tint=vdv*n #peak to peak voltage in V\n", + "print \"Time interval is \",Tint,\" micro seconds\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time interval is 4 micro seconds\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.14.3 - page : 9-46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# period and frequency\n", + "vdv=2 #volts per division in micro seconds/div\n", + "n=12 #no. of divisions\n", + "Tp=vdv*n # period in micro seconds\n", + "f=1/(Tp*10**-3) #frequency in kHz\n", + "print \"Period is \",Tp,\" micro seconds\"\n", + "print \"Frequency is \",round(f,2),\" kHz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Period is 24 micro seconds\n", + "Frequency is 41.67 kHz\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.14.4 - page : 9-47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Peak to peak voltage and frequency\n", + "vdv1=0.5 #volts per division in V/div\n", + "nv=3 #no. of divisions\n", + "nh=4 #numbers of horizontal divisions\n", + "Vpp=vdv1*nv #peak to peak voltage in V\n", + "vdv2=2 # time division in micro seconds per divisions\n", + "Tp=vdv2*nh # period in micro seconds\n", + "f=1/(Tp*10**-3) #frequency in kHz\n", + "print \"Peak to peak voltage is \",Vpp,\" V\"\n", + "print \"Period is \",Tp,\" micro seconds\"\n", + "print \"Frequency is \",f,\" kHz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak to peak voltage is 1.5 V\n", + "Period is 8 micro seconds\n", + "Frequency is 125.0 kHz\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17.1 - page : 9-67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#bandwidth\n", + "#given data :\n", + "Trs=12 #in micro sec\n", + "Trd=15 #in micro sec\n", + "Tro=(Trd**2-Trs**2)**(1.0/2) \n", + "K=0.35 # constant\n", + "BW=(K/Tro)*10**3 \n", + "print \"Bandwidth, BW =\",round(BW,2), \" kHz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Bandwidth, BW = 38.89 kHz\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17.2 - page : 9-68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Rise time\n", + "#given data :\n", + "BW=10*10**6 # in Hz\n", + "tr=(0.35/BW)*10**9 \n", + "print \"Rise time, tr = \",tr, \" ns\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rise time, tr = 35.0 ns\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17.3 - page : 9-68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# rise time\n", + "#given data :\n", + "Tro=10 #in micro sec\n", + "Trd=13 #in micro sec\n", + "Trs=(Trd**2-Tro**2)**(1.0/2) \n", + "print \"Actual rise time, Trs = \",round(Trs,2),\" ns\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Actual rise time, Trs = 8.31 ns\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17.4 - page : 9-68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Rise time\n", + "#given data :\n", + "Tro=10 #in micro sec\n", + "Trd=15 #in micro sec\n", + "Trs=(Trd**2-Tro**2)**(1.0/2)\n", + "print \"Actual rise time, Trs = \",round(Trs,2),\" ns\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Actual rise time, Trs = 11.18 ns\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17.5 - page : 9-68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Rise time\n", + "#given data :\n", + "Trs=12 #in micro sec\n", + "Trd=30 #in micro sec\n", + "BW=20*10**6 # in Hz\n", + "K=0.35 # constant\n", + "Tro=(K/BW)*10**9 \n", + "Trs=(Trd**2-Tro**2)**(1.0/2)\n", + "print \"Actual rise time, Trs = \",round(Trs,2),\" ns\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Actual rise time, Trs = 24.37 ns\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17.6 - page : 9-69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# capacitance\n", + "#given data :\n", + "K=10 # constant\n", + "C2=35*10**-12 \n", + "C1=(C2/(K-1))*10**12 \n", + "print \"Capacitance, C1 = \",round(C1,2),\" pF\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Capacitance, C1 = 3.89 pF\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17.7 - page : 9-69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# impedance of CRO\n", + "K=10 \n", + "vin=1 #vpp \n", + "vout=0.1 #in vpp\n", + "c1=2 # in pF\n", + "c2=c1*(K-1) #CAPACITANCE IN Pf\n", + "print \"Capacitance is \",c2,\" pF\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Capacitance is 18 pF\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17.8 - page : 9-70" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# sensivity\n", + "n=2 #divisions\n", + "f=50.0 #in MHz\n", + "t=(1/f)*10**3 #time in nanao seconds\n", + "mdv=t/4 #in ns/div\n", + "mtds=mdv*n # in ns/div\n", + "print \"Minimum time/div is \",mdv,\" ns/div\"\n", + "print \"Minimum time/div setting is \",mtds,\" ns/div\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Minimum time/div is 5.0 ns/div\n", + "Minimum time/div setting is 10.0 ns/div\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.17.9 - page : 9-70" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# rise time\n", + "#given data :\n", + "Trs=21 #in micro-sec\n", + "K=0.35 # constant\n", + "BW=50*10**6 # in Hz\n", + "Tro=(K/BW)*10**9 \n", + "Trd=(Trs**2+Tro**2)**(1.0/2)\n", + "print \"Rise time, Tro = \",round(Trd,0),\" ns\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Rise time, Tro = 22.0 ns\n" + ] + } + ], + "prompt_number": 24 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/sample_notebooks/SaleemAhmed/generation.ipynb b/sample_notebooks/SaleemAhmed/generation.ipynb new file mode 100755 index 00000000..3d7aab33 --- /dev/null +++ b/sample_notebooks/SaleemAhmed/generation.ipynb @@ -0,0 +1,429 @@ +{ + "metadata": { + "name": "", + "signature": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "chapter-2, Economics of generation" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex1, Page 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#To Determine the Demand and Supply Parameters for 15 bulbs\n", + "\n", + "W=60 #Wattage of the bulb\n", + "N=15 #No. of bulbs\n", + "CL=W*N #Connected Load \n", + "Wih=2*(10**3) #Wattage of immersion heater\n", + "Wh=2*(10**3) #Wattage of heater\n", + "\n", + "#Usage of Bulbs at different time periods\n", + "N1=5 \n", + "N2=10 \n", + "N3=6\n", + "\n", + "#Time periods for bulbs\n", + "T1=2 #6pm - 8pm\n", + "T2=2 #8pm - 10pm\n", + "T3=2 #10pm - 12pm\n", + "#Time Periods for heaters\n", + "T4=4 #1pm - 5pm\n", + "T5=3 #8pm - 11pm\n", + "\n", + "#CASE 1\n", + "MD1=W*N2 #Maximum Demand\n", + "DF=MD1*100/CL #Demand Factor\n", + "EC1=(N1*W*T1)+(N2*W*T2)+(N3*W*T3) #Energy Consumed\n", + "DLF1=EC1*100/(24*MD1) #Daily Load Factor\n", + "\n", + "#CASE 2\n", + "MD2=(W*N2)+Wh #From 8pm - 10pm\n", + "EC2=(T4*Wih)+(T5*Wh)+EC1 #Energy Consumed\n", + "DLF2=EC2*100/(24*MD2) #Daily Load Factor\n", + "\n", + "print '''i)a) Connected Load is %0.2f W\\nb) The Maximum Demand is %0.2f W\n", + "c) The Demand Factor is %0.2f percent\\nd) The Daily Load Factor is %0.2f percent''' %(CL,MD1,DF,DLF1)\n", + "print 'ii) The Improved Daily Load Factor is %0.2f percent' %DLF2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)a) Connected Load is 900.00 W\n", + "b) The Maximum Demand is 600.00 W\n", + "c) The Demand Factor is 66.67 percent\n", + "d) The Daily Load Factor is 17.50 percent\n", + "ii) The Improved Daily Load Factor is 26.47 percent\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex2, Page 74" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "\n", + "#To determine the Demand and supply parameter of four consumers\n", + "\n", + "\n", + "#Maximum Demands of various users\n", + "MD1=2*(10**3) #9pm\n", + "MD2=2*(10**3) #12 noon\n", + "MD3=8*(10**3) #5pm\n", + "MD4=4*(10**3) #8pm\n", + "MDT=MD1+MD2+MD3+MD4 #Sum of all Maximum Demands\n", + "\n", + "#Demands of various users\n", + "D1=1.6*(10**3) #8pm\n", + "D2=1*(10**3) #8pm\n", + "D3=5*(10**3) #8pm\n", + "\n", + "#The Number after the Alphabets represents the Consumer\n", + "\n", + "#Maximum Demand of the System arises at 8.00 PM\n", + "MDS = D1+D2+D3+MD4 \n", + "\n", + "TDF=MDT/MDS #Diversity Factor\n", + "#Given Values\n", + "#Average Loads\n", + "AL2=500 \n", + "AL4=1000 \n", + "#Load Factors\n", + "LF1=15/100 \n", + "LF3=25/100 \n", + "#Calculated Values\n", + "#Average Loads\n", + "AL1=LF1*MD1 \n", + "AL3=LF3*MD3 \n", + "#Load Factors\n", + "LF2=AL2*100/MD2 \n", + "LF4=AL4*100/MD4 \n", + "\n", + "ALS=AL1+AL2+AL3+AL4 #Combined Average Loads\n", + "LFS=ALS*100/MDS #Combined Load Factor\n", + "\n", + "#Load Percent\n", + "LF1*=100 # %\n", + "LF3*=100 # %\n", + "\n", + "print 'i) The Diversity Factor is %0.2f' %TDF\n", + "print 'ii) The Average load and Load factor of:'\n", + "print ' Consumer 1 : %0.2f W and %0.2f percent' %(AL1,LF1)\n", + "print ' Consumer 2 : %0.2f W and %0.2f percent' %(AL2,LF2)\n", + "print ' Consumer 3 : %0.2f W and %0.2f percent' %(AL3,LF3)\n", + "print ' Consumer 4 : %0.2f W and %0.2f percent' %(AL4,LF4)\n", + "print 'iii) The Combined Load Factor and the Combined Average Load is %0.2f percent and %0.2f W respectively\\n' %(LFS,ALS)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i) The Diversity Factor is 1.38\n", + "ii) The Average load and Load factor of:\n", + " Consumer 1 : 300.00 W and 15.00 percent\n", + " Consumer 2 : 500.00 W and 25.00 percent\n", + " Consumer 3 : 2000.00 W and 25.00 percent\n", + " Consumer 4 : 1000.00 W and 25.00 percent\n", + "iii) The Combined Load Factor and the Combined Average Load is 32.76 percent and 3800.00 W respectively\n", + "\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex3, Page 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "\n", + "#To Determine the Yearly Cost of the substation\n", + "\n", + "Teff=95/100 #Transmission Efficiency\n", + "Deff=85/100 #Distribution Efficiency\n", + "DFT=1.2 #Diversity Factor For Transmission\n", + "DFD=1.3 #Diversity Factor For Distribution\n", + "MDGS=100*(10**6) #Maximum Demand of Generating Station\n", + "ALF=40/100 #Annual Load Factor\n", + "ACCT=2.5*(10**6) #Annual Capital Charge for Transmission\n", + "ACCD=2*(10**6) #Annual Capital Charge for Distribution\n", + "GCC=100 #Generating Cost per kW demand\n", + "GCCU=5/100 # Per Unit Cost\n", + "#Fixed Charges from Supply to Substation Annually\n", + "GFC=GCC*MDGS/1000 #Generating\n", + "TFC=ACCT #Transmission\n", + "TotFCS=GFC+TFC #Total\n", + "#Fixed Charges for supply upto Consumer Annually\n", + "DFC=ACCD #Distribution\n", + "TotFCC=TotFCS+DFC #Total\n", + "\n", + "AMDS= DFT*MDGS/1000 #Aggregate of Maximum Demand at Supply\n", + "AMDC= DFD*AMDS #Aggregate of Maximum Demand for Consumers\n", + "\n", + "FCS=TotFCS/AMDS #Fixed Charges Per KW at substation\n", + "CES=GCCU/Teff #Cost of energy at the substation\n", + "\n", + "FCC=TotFCC/AMDC #Fixed Charges per KW at the consumer premises\n", + "CEC=CES/Deff #Cost of Energy at the consumer premises\n", + "\n", + "CEC*=100 # converting from rupee to paise\n", + "\n", + "print 'The Yealy Cost per KW demand and the cost per KWhr at:'\n", + "print 'a) The substation is %0.2f rupees per KW and %0.2f paise per kWhr'%(FCS,CES)\n", + "print 'b) The consumer premises is %g rupees per KW and %g paise per kWhr' %(FCC,CEC)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Yealy Cost per KW demand and the cost per KWhr at:\n", + "a) The substation is 104.17 rupees per KW and 0.05 paise per kWhr\n", + "b) The consumer premises is 92.9487 rupees per KW and 6.19195 paise per kWhr\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex4, Page 78" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#To determine the Load factor and suitable units for 24 hr operation of the plant\n", + "\n", + "\n", + "#Demands at Various Time Periods starting from 12PM to 12PM\n", + "D1=500*(10**3) \n", + "D2=800*(10**3) \n", + "D3=2000*(10**3) \n", + "D4=1000*(10**3) \n", + "D5=2500*(10**3) \n", + "D6=2000*(10**3) \n", + "D7=1500*(10**3) \n", + "D8=1000*(10**3) \n", + "\n", + "MD=D5 #Maximum Demand\n", + "#Time Periods of demands from 12PM\n", + "T1=5 \n", + "T2=5 \n", + "T3=2 \n", + "T4=2 \n", + "T5=3 \n", + "T6=3 \n", + "T7=2 \n", + "T8=2 \n", + "\n", + "#Total Energy Demand in 24hrs\n", + "TED=(T1*D1)+(T2*D2)+(T3*D3)+(D4*T4)+(T5*D5)+(D6*T6)+(D7*T7)+(T8*D8) \n", + "\n", + "LF=TED*100/(24*MD) \n", + "\n", + "C1000=3*1000*(10**3) #1000 unit \n", + "C500=1*500*(10**3) #500 Unit\n", + "\n", + "TCP=C1000+C500 #Total capacity of the plant\n", + "PCF=TED*100/(24*TCP) #Plant Capacity Factor\n", + "\n", + "#Operating Schedule, Units operated can be seen in the textbook\n", + "G1=500*(10**3) \n", + "G2=1000*(10**3) \n", + "G3=2000*(10**3) \n", + "G4=1000*(10**3) \n", + "G5=2500*(10**3) \n", + "G6=2000*(10**3) \n", + "G7=1500*(10**3) \n", + "G8=1000*(10**3) \n", + "\n", + "TEG=(T1*G1)+(T2*G2)+(T3*G3)+(G4*T4)+(T5*G5)+(G6*T6)+(G7*T7)+(T8*G8) #Total Energy Generated\n", + "PUF=TED*100/(TEG) #Plant Use Factor\n", + "\n", + "print 'a) The Reserve Capacity is a 1000kW Unit and Load Factor is %0.2f percent' %LF\n", + "print 'b) The Plant Capacity Factor is %0.2f percent' %PCF\n", + "print 'c) The Plant Use Factor is %0.2f percent' %PUF" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) The Reserve Capacity is a 1000kW Unit and Load Factor is 51.67 percent\n", + "b) The Plant Capacity Factor is 36.90 percent\n", + "c) The Plant Use Factor is 96.88 percent\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex5, Page 80" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#To determine the Plant use factore of each unit\n", + "\n", + "\n", + "MDS=25*(10**6) #Maximum Demand on the System\n", + "U1=15*(10**6) #Load Supplied By Unit 1\n", + "U2=12.5*(10**6) #Load Supplied By Unit 2\n", + "#Running Time Factor of the Unit\n", + "T1=1 \n", + "T2=40/100 \n", + "\n", + "#Energy generated by each unit\n", + "E1=1*(10**8) \n", + "E2=1*(10**7) \n", + "Et=E1+E2 #Total Energy\n", + "\n", + "#Maximum Demands on Each Units\n", + "MD1=U1 \n", + "MD2=MDS-U1 \n", + "\n", + "#Annual Load Factor for the Units\n", + "ALF1=E1*1000*100/(MD1*8760) \n", + "ALF2=E2*1000*100/(MD2*8760) \n", + "\n", + "LF2=E2*1000*100/(MD2*0.4*8760) #Load Factor for the it is loaded\n", + "\n", + "\n", + "PUF1=ALF1 #Plant Use Factor\n", + "PCF1=ALF1 # Plant Capacity Factor\n", + "\n", + "PCF2=E2*1000*100/(U2*8760) #Plant Capacity Factor for Unit 2\n", + "PUF2=E2*1000*100/(U2*0.4*8760) #Plant Use Factor for Unit 2\n", + "\n", + "LFP=Et*100*1000/(MDS*8760) #Annual Load Factor of the Complete Plant\n", + "\n", + "print 'The Load Factor, Plant Capacity Factor, Plant Use Factor of:'\n", + "print 'Unit 1 : %0.2f percent, %0.2f percent, %0.2f percent' %(ALF1,PCF1,PUF1)\n", + "print 'Unit 2 : %0.2f percent, %0.2f percent, %0.2f percent' %(ALF2,PCF2,PUF2)\n", + "print 'The Annual Load Factor of the Entire Plant is %0.2f percent' %LFP" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Load Factor, Plant Capacity Factor, Plant Use Factor of:\n", + "Unit 1 : 76.10 percent, 76.10 percent, 76.10 percent\n", + "Unit 2 : 11.42 percent, 9.13 percent, 22.83 percent\n", + "The Annual Load Factor of the Entire Plant is 50.23 percent\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex6, Page 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#To determine the most economic power factor\n", + "\n", + "from numpy import sqrt\n", + "\n", + "P=200*(10**3) #Maximum Demand\n", + "pf=0.707 #Power Factor Lagging\n", + "\n", + "a=100 #Tariff per kVA per year\n", + "\n", + "b=200 #Power factor improvement cost Per kVA.\n", + "r=20 #Interest Depriciation, maintenance and cost of losses amount to 20% of capital cost per year\n", + "\n", + "# Economic PF = sqrt(1-((b1/a)**2))\n", + "\n", + "b1=r*b/100 # b' term accrding to the equation above\n", + "\n", + "pfeco=sqrt(1-((b1/a)**2)) #Economic Power Factor\n", + "\n", + "print 'The Economic Power Factor is %0.2f ' %pfeco\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Economic Power Factor is 0.92 \n" + ] + } + ], + "prompt_number": 22 + } + ], + "metadata": {} + } + ] +} diff --git a/sample_notebooks/SandhyaArroju/Chapter5.ipynb b/sample_notebooks/SandhyaArroju/Chapter5.ipynb new file mode 100755 index 00000000..2e530c83 --- /dev/null +++ b/sample_notebooks/SandhyaArroju/Chapter5.ipynb @@ -0,0 +1,303 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "#5: Conducting Materials"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example number 5.1, Page number 5.34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "drift speed is 36.6 *10**-5 m/s\n",
+ "mean free path is 3.34 *10**-8 m\n",
+ "answer given in the book is wrong\n"
+ ]
+ }
+ ],
+ "source": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "Na=6.023*10**26; #avagadro number\n",
+ "e=1.602*10**-19;\n",
+ "d=8960; #density\n",
+ "N=1; #number of free electrons\n",
+ "w=63.54; #atomic weight\n",
+ "i=10; #current(ampere)\n",
+ "m=9.1*10**-31; \n",
+ "rho=2*10**-8; #resistivity(ohm m)\n",
+ "r=0.08*10**-2; #radius(m)\n",
+ "c=1.6*10**6; #mean thermal velocity(m/s)\n",
+ "\n",
+ "#Calculation\n",
+ "A=math.pi*r**2; #area(m**2)\n",
+ "n=Na*d*N/w;\n",
+ "vd=i/(A*n*e); #drift speed(m/s)\n",
+ "tow_c=m/(n*e**2*rho);\n",
+ "lamda=tow_c*c; #mean free path(m)\n",
+ "\n",
+ "#Result\n",
+ "print \"drift speed is\",round(vd*10**5,1),\"*10**-5 m/s\"\n",
+ "print \"mean free path is\",round(lamda*10**8,2),\"*10**-8 m\"\n",
+ "print \"answer given in the book is wrong\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example number 5.2, Page number 5.35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "electrical conductivity is 4.8 *10**7 ohm-1 m-1\n"
+ ]
+ }
+ ],
+ "source": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "e=1.602*10**-19;\n",
+ "m=9.1*10**-31; #mass(kg)\n",
+ "tow=2*10**-14; #time(s)\n",
+ "n=8.5*10**28; \n",
+ "\n",
+ "#Calculation\n",
+ "sigma=n*e**2*tow/m; #electrical conductivity(ohm-1 m-1)\n",
+ "\n",
+ "#Result\n",
+ "print \"electrical conductivity is\",round(sigma/10**7,1),\"*10**7 ohm-1 m-1\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example number 5.3, Page number 5.35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "relaxation time is 4.0 *10**-14 s\n",
+ "mobility of electrons is 7.0 *10**-3 m**2/Vs\n",
+ "drift velocity is 0.7 m/s\n"
+ ]
+ }
+ ],
+ "source": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "e=1.6*10**-19;\n",
+ "m=9.1*10**-31; #mass(kg)\n",
+ "n=5.8*10**28; \n",
+ "rho=1.54*10**-8; #resistivity(ohm m)\n",
+ "E=1*10**2;\n",
+ "\n",
+ "#Calculation\n",
+ "tow=m/(rho*n*e**2); #relaxation time(s)\n",
+ "mew_e=1/(rho*e*n); #mobility of electrons(m**2/Vs)\n",
+ "vd=mew_e*E; #drift velocity(m/s)\n",
+ "\n",
+ "#Result\n",
+ "print \"relaxation time is\",round(tow*10**14),\"*10**-14 s\"\n",
+ "print \"mobility of electrons is\",round(mew_e*10**3),\"*10**-3 m**2/Vs\"\n",
+ "print \"drift velocity is\",round(vd,1),\"m/s\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example number 5.4, Page number 5.35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "resistivity is 5.51 *10**-8 ohm m\n"
+ ]
+ }
+ ],
+ "source": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho=1.7*10**-8; #resistivity(ohm m)\n",
+ "T=300; #temperature(K)\n",
+ "T1=973; #temperature(K)\n",
+ "\n",
+ "#Calculation\n",
+ "a=rho/T; \n",
+ "rho_973=a*T1; #resistivity(ohm m)\n",
+ "\n",
+ "#Result\n",
+ "print \"resistivity is\",round(rho_973*10**8,2),\"*10**-8 ohm m\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example number 5.5, Page number 5.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "increase of resistivity is 0.54 *10**-8 ohm m\n"
+ ]
+ }
+ ],
+ "source": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "rho1=1.2*10**-8; #resistivity(ohm m)\n",
+ "rho2=0.12*10**-8; #resistivity(ohm m)\n",
+ "p1=0.4; #atomic percent\n",
+ "p2=0.5; #atomic percent\n",
+ "rho=1.5*10**-8; #resistivity(ohm m)\n",
+ "\n",
+ "#Calculation\n",
+ "rho_i=(rho1*p1)+(rho2*p2); #increase of resistivity(ohm m)\n",
+ "Tr=rho+rho_i; #total resistivity of copper alloy(ohm m)\n",
+ "\n",
+ "#Result\n",
+ "print \"increase of resistivity is\",round(rho_i*10**8,2),\"*10**-8 ohm m\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example number 5.6, Page number 5.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "electrical conductivity is 1.688 *10**7 ohm-1 m-1\n",
+ "thermal conductivity is 123.93 W/m/K\n",
+ "lorentz number is 2.447 *10**-8 watt ohm K-2\n"
+ ]
+ }
+ ],
+ "source": [
+ "#importing modules\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "e=1.6*10**-19;\n",
+ "m=9.1*10**-31; #mass(kg)\n",
+ "n=6*10**28; #density(per m**3)\n",
+ "tow=10**-14; #relaxation time(s)\n",
+ "T=300; #temperature(K)\n",
+ "k=1.38*10**-23; #boltzmann constant\n",
+ "\n",
+ "#Calculation\n",
+ "sigma=n*e**2*tow/m; #electrical conductivity(ohm-1 m-1)\n",
+ "K=n*math.pi**2*k**2*T*tow/(3*m); #thermal conductivity(W/m/K)\n",
+ "L=K/(sigma*T); #lorentz number(watt ohm K-2)\n",
+ "\n",
+ "#Result\n",
+ "print \"electrical conductivity is\",round(sigma/10**7,3),\"*10**7 ohm-1 m-1\"\n",
+ "print \"thermal conductivity is\",round(K,2),\"W/m/K\"\n",
+ "print \"lorentz number is\",round(L*10**8,3),\"*10**-8 watt ohm K-2\""
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/SoumenGanguly/ncert_Maths.ipynb b/sample_notebooks/SoumenGanguly/ncert_Maths.ipynb new file mode 100755 index 00000000..30efaf66 --- /dev/null +++ b/sample_notebooks/SoumenGanguly/ncert_Maths.ipynb @@ -0,0 +1,355 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Chapter 1: Sets\n", + "========" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Example 1, Page 03:\n", + "----------------------" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false, + "scrolled": true + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Therefore, the solution set of the given equation can be written in roaster form as {-2,1}\n" + ] + } + ], + "source": [ + "import numpy as np\n", + "\n", + "#Variable declaration\n", + "coeff = [1,1,-2]\n", + "\n", + "#Calculations\n", + "roots_array = np.roots(coeff)\n", + "\n", + "#Result\n", + "print \"Therefore, the solution set of the given equation can be written in roaster form as {%d,%d}\"%(roots_array[0],roots_array[1])\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Example 2, Page 03:\n", + "---------------------" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The required numbers are 1,2,3,4,5,6\n", + "So, the given set in the roster form is {1,2,3,4,5,6}\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "pos_integer = []\n", + "\n", + "#Calculations\n", + "for i in range(1,1000): #1000 is taken as the upper limit of a positive integer\n", + " if i**2<40:\n", + " pos_integer.append(i)\n", + " \n", + "#Result\n", + "print \"The required numbers are %d,%d,%d,%d,%d,%d\"%(pos_integer[0],pos_integer[1],pos_integer[2],pos_integer[3],pos_integer[4],pos_integer[5])\n", + "print \"So, the given set in the roster form is {%d,%d,%d,%d,%d,%d}\"%(pos_integer[0],pos_integer[1],pos_integer[2],pos_integer[3],pos_integer[4],pos_integer[5])\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Example 12, Page 14:\n", + "-----------------------" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The union of the two given arrays is \n", + "[ 2 4 6 8 10 12]\n" + ] + } + ], + "source": [ + "import numpy as np\n", + "\n", + "#Variable declaration\n", + "a = [2,4,6,8]\n", + "b = [6,8,10,12]\n", + "\n", + "#Calculations\n", + "union_array = np.union1d(a,b)\n", + "\n", + "#Result\n", + "print \"The union of the two given arrays is \"\n", + "print union_array" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Example 13, Page 14:\n", + "-----------------------" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A U B = ['a' 'e' 'i' 'o' 'u']\n", + "A = ['a', 'e', 'i', 'o', 'u']\n", + "Thus, A U B = A\n" + ] + } + ], + "source": [ + "import numpy as np\n", + "\n", + "#Variable declaration\n", + "a = ['a','e','i','o','u']\n", + "b = ['a','i','u']\n", + "\n", + "#Calculations\n", + "union_array = np.union1d(a,b)\n", + "\n", + "#Result\n", + "print \"A U B = %s\"%(union_array)\n", + "print \"A = %s\"%(a)\n", + "print \"Thus, A U B = A\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Example 15, Page 15:\n", + "----------------------" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A intersection B = [6 8]\n" + ] + } + ], + "source": [ + "import numpy as np\n", + "\n", + "#Variable declaration\n", + "a = [2,4,6,8]\n", + "b = [6,8,10,12]\n", + "\n", + "#Calculations\n", + "intersect_array = np.intersect1d(a,b)\n", + "\n", + "#Result\n", + "print \"A intersection B = %s\"%(intersect_array)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Example 18, Page 17:\n", + "-----------------------" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A - B = [1 3 5]\n", + "B - A = [8]\n" + ] + } + ], + "source": [ + "import numpy as np\n", + "\n", + "#Variable declaration\n", + "a = [1,2,3,4,5,6]\n", + "b = [2,4,6,8]\n", + "\n", + "#Calculations\n", + "diff_a = np.setdiff1d(a,b)\n", + "diff_b = np.setdiff1d(b,a)\n", + "\n", + "#Result\n", + "print \"A - B = %s\"%(diff_a)\n", + "print \"B - A = %s\"%(diff_b)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Example 20, Page 19:\n", + "-----------------------" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A' = [ 2 4 6 8 10]\n" + ] + } + ], + "source": [ + "import numpy as np\n", + "\n", + "#Variable declaration\n", + "u = [1,2,3,4,5,6,7,8,9,10]\n", + "a = [1,3,5,7,9]\n", + "\n", + "#Calculations\n", + "complement_a = np.setdiff1d(u,a)\n", + "\n", + "#Result\n", + "print \"A' = %s\"%(complement_a)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Example 22, Page 19:\n", + "-----------------------" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A' = [1 4 5 6]\n", + "B' = [1 2 6]\n", + "A' intersection B' = [1 6]\n", + "A U B = [2 3 4 5]\n", + "(A U B)' = [1 6]\n", + "Therefore, (A U B)' = A' intersection B' \n" + ] + } + ], + "source": [ + "import numpy as np\n", + "\n", + "#Variable declaration\n", + "u = [1,2,3,4,5,6]\n", + "a = [2,3]\n", + "b = [3,4,5]\n", + "\n", + "#Calculations\n", + "complement_a = np.setdiff1d(u,a)\n", + "complement_b = np.setdiff1d(u,b)\n", + "intersect_ab = np.intersect1d(complement_a,complement_b)\n", + "union_ab = np.union1d(a,b)\n", + "complement_union_ab = np.setdiff1d(u,union_ab)\n", + "\n", + "#Result\n", + "print \"A' = %s\"%(complement_a)\n", + "print \"B' = %s\"%(complement_b)\n", + "print \"A' intersection B' = %s\"%(intersect_ab)\n", + "print \"A U B = %s\"%(union_ab)\n", + "print \"(A U B)' = %s\"%(complement_union_ab)\n", + "print \"Therefore, (A U B)\\' = A\\' intersection B\\' \"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.8" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/sample_notebooks/Vedantam Lakshmi Manasa/Chapter_2_Electric_Fields.ipynb b/sample_notebooks/Vedantam Lakshmi Manasa/Chapter_2_Electric_Fields.ipynb new file mode 100755 index 00000000..cdc8b25e --- /dev/null +++ b/sample_notebooks/Vedantam Lakshmi Manasa/Chapter_2_Electric_Fields.ipynb @@ -0,0 +1,296 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 2 Electric Fields"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2_5 pgno:65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Maximum field = V/m per volt 42064315640.1\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Chapter 2, Example 5, page 65\n",
+ "#Calculate the maximum field at the sphere surface\n",
+ "#Calulating Field at surface E based on figure 2.31 and table 2.3\n",
+ "from math import pi\n",
+ "Q1 = 0.25\n",
+ "e0 = 8.85418*10**-12 #Epselon nought\n",
+ "RV1= ((1/0.25**2)+(0.067/(0.25-0.067)**2)+(0.0048/(0.25-0.067)**2))\n",
+ "RV2= ((0.25+0.01795+0.00128)/(0.75-0.067)**2)\n",
+ "RV= RV1+RV2\n",
+ "E = (Q1*RV)/(4*pi*e0)\n",
+ "print\"Maximum field = V/m per volt\",E\n",
+ "\n",
+ "#Answers vary due to round off error\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2_6 pgno:66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "#Chapter 2, Exmaple 6, page 66\n",
+ "#calculation based on figure 2.32\n",
+ "\n",
+ "#(a)Charge on each bundle\n",
+ "print\"Part a\\t\"\n",
+ "req = (0.0175*0.45)**0.5\n",
+ "print\"Equivalent radius = m \", req\n",
+ "from math import log\n",
+ "from math import pi\n",
+ "V = 400*10**3 #Voltage\n",
+ "H = 12. #bundle height in m\n",
+ "d = 9. #pole to pole spacing in m\n",
+ "e0 = 8.85418*10**-12 #Epselon nought\n",
+ "Hd = ((2*H)**2+d**2)**0.5#2*H**2 + d**2\n",
+ "Q = V*2*pi*e0/(log((2*H/req))-log((Hd/d)))\n",
+ "q = Q/2\n",
+ "print\"Charge per bundle = uC/m \",Q #micro C/m\n",
+ "print\"Charge per sunconducter = uC/m \",q #micro C/m\n",
+ "\n",
+ "#(b part i)Maximim & average surface feild\n",
+ "print\"\\tPart b\"\n",
+ "print\"\\tSub part 1\\t\"\n",
+ "r = 0.0175 #subconductor radius\n",
+ "R = 0.45 #conductor to subconductor spacing\n",
+ "MF = (q/(2*pi*e0))*((1/r)+(1/R)) # maximum feild\n",
+ "print\"Maximum feild = kV/m \\t\",MF\n",
+ "MSF = (q/(2*pi*e0))*((1/r)-(1/R)) # maximum surface feild\n",
+ "print\"Maximum feild = kV/m \\t\",MSF\n",
+ "ASF = (q/(2*pi*e0))*(1/r) # Average surface feild\n",
+ "print\"Maximum feild = kV/m \\t\",ASF\n",
+ "\n",
+ "#(b part ii) Considering the two sunconductors on the left\n",
+ "print\"\\tSub part 2\\t\"\n",
+ "#field at the outer point of subconductor #1 \n",
+ "drO1 = 1/(d+r)\n",
+ "dRrO1 = 1/(d+R+r)\n",
+ "EO1 = MF -((q/(2*pi*e0))*(drO1+dRrO1))\n",
+ "print\"EO1 = kV/m \\t\",EO1\n",
+ "#field at the outer point of subconductor #2 \n",
+ "drO2 = 1/(d-r)\n",
+ "dRrO2 = 1/(d-R-r)\n",
+ "EO2 = MF -((q/(2*pi*e0))*(dRrO2+drO2))\n",
+ "print\"EO2 = kV/m \\t\",EO2\n",
+ "\n",
+ "#field at the inner point of subconductor #1 \n",
+ "drI1 = 1/(d-r)\n",
+ "dRrI1 = 1/(d+R-r)\n",
+ "EI1 = MSF -((q/(2*pi*e0))*(drI1+dRrI1))\n",
+ "print\"EI1 = kV/m \\t\",EI1\n",
+ "#field at the inner point of subconductor #2 \n",
+ "drI2 = 1/(d+r)\n",
+ "dRrI2 = 1/(d-R+r)\n",
+ "EI2 = MSF -((q/(2*pi*e0))*(dRrI2+drI2)) \n",
+ "print\"EI2 = kV/m \\t\",EI2\n",
+ "\n",
+ "#(part c)Average of the maximim gradient\n",
+ "print\"\\tPart c\\t\"\n",
+ "Eavg = (EO1+EO2)/2\n",
+ "print\"The average of the maximum gradient = kV/m \\t\",Eavg\n",
+ "\n",
+ "\n",
+ "#Answers might vary due to round off error\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2_7 pgno:69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Electric Feild = V/m \t30015596280.4\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Chapter 2, Exmaple 7, page 69\n",
+ "#Electric feild induced at x\n",
+ "from math import pi\n",
+ "e0 = 8.85418*10**-12 #Epselon nought\n",
+ "q = 1 # C/m\n",
+ "C = (q/(2*pi*e0))\n",
+ "#Based on figure 2.33\n",
+ "E = C-(C*(1./3.+1./7.))+(C*(1+1./5.+1./9.))+(C*(1./5.+1./9.))-(C*(1./3.+1./7.))\n",
+ "print\"Electric Feild = V/m \\t\",E\n",
+ "\n",
+ "#Answers might vary due to round off error\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2_8 pgno:70"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\n",
+ "Thickness of graded design= cm 4.24264068712\n",
+ "Curve = cm**2 62.4264068712\n",
+ "V1 = cm**3 47402.906725\n",
+ "Thickness of regular design = cm 14.684289433\n",
+ "V2 = cm**3 861.944682812\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Chapter 2, Exmaple 8, page 70\n",
+ "#Calculate the volume of the insulator\n",
+ "#Thinkness of graded design\n",
+ "from math import e\n",
+ "from math import pi\n",
+ "V = 150*(2)**0.5\n",
+ "Ebd = 50\n",
+ "T = V/Ebd\n",
+ "print\"\\nThickness of graded design= cm \",T\n",
+ "#Based on figure 2.24\n",
+ "r = 2 # radius of the conductor\n",
+ "l = 10 #length of graded cylinder; The textbook uses 10 instead of 20\n",
+ "zr = l*(T+r)\n",
+ "print\"Curve = cm**2 \",zr\n",
+ "#Volume of graded design V1\n",
+ "V1 = 4*pi*zr*(zr-r)\n",
+ "print\"V1 = cm**3 \",V1 #Unit is wrong in the textbook\n",
+ "#Thickness of regular design as obtained form Eq.2.77\n",
+ "pow = V/(2*Ebd)\n",
+ "t = 2*(e**pow-1)\n",
+ "print\"Thickness of regular design = cm \",t\n",
+ "#Volume of regular design V2\n",
+ "V2 = pi*((2+t)**2-4)\n",
+ "print\"V2 = cm**3 \",V2#unit not mentioned in textbook\n",
+ " \n",
+ "#Answers may vary due to round off error\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2_11 pgno:75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The values of Phi2 and Phi4 are: [[ -3.6568 326.5 ]\n",
+ " [ 261.92857143 -4.37537287]]\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Chapter 2, Exmaple 11, page 75\n",
+ "#Calculate the potential within the mesh\n",
+ "#Based on figure 2.38(b)\n",
+ "#equations are obtained using Eq.2.46\n",
+ "import numpy\n",
+ "from numpy import linalg\n",
+ "A1 = 1/2*(0.54+0.16)\n",
+ "A2 = 1/2*(0.91+0.14)\n",
+ "S = numpy.matrix([[0.5571, -0.4571, -0.1],[-0.4751, 0.828, 0.3667],[-0.1, 0.667, 0.4667]])\n",
+ "#By obtaining the elements of the global stiffness matrix(Sadiku,1994)\n",
+ "#and by emplying the Eq.2.49(a)\n",
+ "S1 = numpy.matrix([[1.25, -0.014],[-0.014, 0.8381]])\n",
+ "S2 = numpy.matrix([[-0.7786, -0.4571],[-0.4571, -0.3667]])\n",
+ "Phi13 = numpy.matrix([[0], [10]])\n",
+ "val1 = S2*Phi13\n",
+ "Phi24 = val1/S1\n",
+ "print\"The values of Phi2 and Phi4 are:\",Phi24\n",
+ "\n",
+ "#Answers may vary due to round of error \n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/VidyashankarVenkatraman/Chapter_3_Kittel.ipynb b/sample_notebooks/VidyashankarVenkatraman/Chapter_3_Kittel.ipynb new file mode 100755 index 00000000..8760577a --- /dev/null +++ b/sample_notebooks/VidyashankarVenkatraman/Chapter_3_Kittel.ipynb @@ -0,0 +1,79 @@ +{ + "metadata": { + "name": "Chapter_3_Kittel" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Introduction to Solid State Physics" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.1, Page Number 84\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#importing module\n\nfrom __future__ import division\nimport math\n\n#Variable declaration\n\ne = 5*pow(10,-10); # charge on the electron\n\nr0 = 1*pow(10,-8); # atomic radius\n\n# Calculation\n\nR = 4*10**(-8); # interatomic distance in cm\n\nU = -4*e**2*r0**5/R**6; # The van der waals interaction formula\n\n# Result\n\nprint \" The Van Der Waals interaction energy is \",U ,\"ergs\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " The Van Der Waals interaction energy is -2.44140625e-14 ergs\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.2, Page Number 91" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#importing module\n\nfrom __future__ import division\nimport math\n\n# Variable declaration\n\ne = 4.8*pow(10,-10); # charge on proton\nr0 = 2.81*pow(10,-8); # distance between positive and nearest negative ion in Nacl crystal\n\nU = e**2/r0; # in ergs\n\nE = U/(1.6019*pow(10,-12)); # converting to eV as 1eV = 1.6019 * 10**(-12) ergs\n\n#result\nprint \" The potential energy of the two ions by themselves is \",E,\"eV\"\n\n\n\n\n\n\n ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " The potential energy of the two ions by themselves is 5.11847696874 eV\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "code", + "collapsed": false, + "input": "", + "language": "python", + "metadata": {}, + "outputs": [], + "prompt_number": 19 + }, + { + "cell_type": "code", + "collapsed": false, + "input": "", + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/sample_notebooks/VidyashankarVenkatraman/Chapter_3_Kitteldemo.ipynb b/sample_notebooks/VidyashankarVenkatraman/Chapter_3_Kitteldemo.ipynb new file mode 100755 index 00000000..26342edb --- /dev/null +++ b/sample_notebooks/VidyashankarVenkatraman/Chapter_3_Kitteldemo.ipynb @@ -0,0 +1,79 @@ +{ + "metadata": { + "name": "Chapter_3_Kittel" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 3:Introduction to Solid State Physics" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.1, Page Number 84\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#importing module\n\nfrom __future__ import division\nimport math\n\n#Variable declaration\n\ne = 5*pow(10,-10); # charge on the electron\n\nr0 = 1*pow(10,-8); # atomic radius\n\n# Calculation\n\nR = 4*10**(-8); # interatomic distance in cm\n\nU = -4*e**2*r0**5/R**6; # The van der waals interaction formula\n\n# Result\n\nprint \" The Van Der Waals interaction energy is \",U ,\"ergs\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " The Van Der Waals interaction energy is -2.44140625e-14 ergs\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example number 3.2, Page Number 91" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#importing module\n\nfrom __future__ import division\nimport math\n\n# Variable declaration\n\ne = 4.8*pow(10,-10); # charge on proton\nr0 = 2.81*pow(10,-8); # distance between positive and nearest negative ion in Nacl crystal\n\nU = e**2/r0; # in ergs\n\nE = U/(1.6019*pow(10,-12)); # converting to eV as 1eV = 1.6019 * 10**(-12) ergs\n\n#result\nprint \" The potential energy of the two ions by themselves is \",E,\"eV\"\n\n\n\n\n\n\n ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " The potential energy of the two ions by themselves is 5.11847696874 eV\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "code", + "collapsed": false, + "input": "", + "language": "python", + "metadata": {}, + "outputs": [], + "prompt_number": 19 + }, + { + "cell_type": "code", + "collapsed": false, + "input": "", + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} diff --git a/sample_notebooks/WaseemAhmad Ansari/Chapter2.ipynb b/sample_notebooks/WaseemAhmad Ansari/Chapter2.ipynb new file mode 100755 index 00000000..d902c695 --- /dev/null +++ b/sample_notebooks/WaseemAhmad Ansari/Chapter2.ipynb @@ -0,0 +1,348 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3942a48c51f6e66ddeb010a1a5acaeebd4c9f56b964a269d92588d67ccc10453" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 : Introduction to operational amplifier" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.1 Page no. 88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "\n", + "R1 = 10*10**3 #R1 input resistance \n", + "Rf = 100*10**3 # Rf feedback resistance\n", + "vi = float(1) #input voltage \n", + "RL = 25*10**3\n", + "#calculating the values \n", + "\n", + "i1 = float((vi/R1)*10**3) # input resistace id the ratio of input voltage to the input resitance \n", + "vo = float(-(Rf/R1)*vi) # finding the output voltage \n", + "iL = float((abs(vo)/RL)*10**3) # calculating the load current \n", + "io = float((i1+iL)) # calculating the output current which is equal to the sum of input current and load current\n", + "\n", + "#printing the values \n", + "\n", + "print \"The input current i1 =\",i1,\"mA\"\n", + "print \"The output voltage vo =\",vo,\"V\"\n", + "print \"The load current iL =\",iL,\"mA\"\n", + "print \"The output current io =\",io,\"mA\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The input current i1 = 0.1 mA\n", + "The output voltage vo = -10.0 V\n", + "The load current iL = 0.4 mA\n", + "The output current io = 0.5 mA\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.2 Page no.89" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "\n", + "ACL = 5 # Gain of the amplifier\n", + "R1 = 10*10**3 # input resisitance in ohms \n", + "\n", + "# calculations\n", + "\n", + "Rf = (5-1) * R1 # calculating the resistance of feedback resistor \n", + "\n", + "# printing the values \n", + "\n", + "print \"The value of feedback resistor = \", (Rf/10**3),\"kohms\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of feedback resistor = 40 kohms\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.4 Page no.92" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Given Data\n", + "\n", + "R1 = 5*10**3\n", + "Rf = 20*10**3\n", + "vi = 1 \n", + "RL = 5*10**3\n", + "\n", + "# calculating the values \n", + "vo = float((1+(Rf/R1))*vi) \n", + "ACL = int(vo/vi)\n", + "iL = int((vo/RL)*10**3)\n", + "i1 = float(((vo - vi)/Rf))*(10**3)\n", + "io = iL+i1\n", + " \n", + " \n", + "# printing the values\n", + "print \"Output voltage vo = \",vo,\"V\"\n", + "print \"Gain ACL = \",ACL\n", + "print \"Load current iL = \",iL,\"mA\"\n", + "print \"The value of i1 = \",i1,\"mA\"\n", + "print \"Output current io = \", io,\"mA\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output voltage vo = 5.0 V\n", + "Gain ACL = 5\n", + "Load current iL = 1 mA\n", + "The value of i1 = 0.2 mA\n", + "Output current io = 1.2 mA\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.5 Page No.94" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given Data\n", + "import math\n", + "Beta = 200\n", + "ICQ = 100*10**-6\n", + "ADM = 100\n", + "CMRR = 80\n", + "\n", + "# finding the solution \n", + "# for VT =25 milli volt \n", + "VT = 25*10**-3\n", + "gm = float(ICQ/VT)\n", + "Rc = (ADM/gm) \n", + "CMRR = 10**(80/20) # log inverse is equal to powers of 10\n", + "RE = float((CMRR-1)/gm)\n", + "x = Decimal((RE/10**6))\n", + "\n", + "# printing the values \n", + "\n", + "print \" The value of gm =\",int(math.ceil((gm*10**3))),\"mMho\" #converting the answer into milli Mho\n", + "print \" The value of Rc =\",int((Rc/10**3)),\"kohm\" #converting the answer into kohm" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The value of gm = 4 mMho\n", + " The value of Rc = 25 kohm\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.6 Page no. 95" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data \n", + "\n", + "gm = 4*10**-3\n", + "RC = 125*10*3\n", + "RE = 1.25*10**3\n", + "beta0 = 200\n", + "\n", + "# calculating the values\n", + "\n", + "rpi = beta0/gm # value is in ohms \n", + "ADM =-500 # Given Value\n", + "ACM = -((200*RC)/(402*RE)+rpi)*10**-6\n", + "print \"ACM is =\",round(ACM,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ACM is = -0.05\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9 Page No.63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from fractions import Fraction \n", + "# Given data\n", + "\n", + "beta0 = 100\n", + "IQ = 5*10**-4\n", + "RC = 10*10**3\n", + "RE = 150\n", + "VT = 25*10**-3 \n", + "\n", + "# calculations \n", + "\n", + "ICQ = float(IQ/2)\n", + "gm = float(ICQ / VT)\n", + "rpi = beta0/gm\n", + "# calculaing the gain in Differential mode\n", + "ADM = ((0.5)*(beta0*RC))/(rpi+((1+beta0)*RE))\n", + "# To get the differentila mode gain multiply the value by 2\n", + "ADM2 = (ADM*2)\n", + "\n", + "# print the values \n", + "\n", + "print \"ICQ value is =\",ICQ*10**3,\"mA\"\n", + "print \"gm value is =\",Fraction(gm).limit_denominator(100),\"Mho\" \n", + "print \"rpi value is =\",int(rpi/10**3),\"kilo Ohm\"\n", + "print \"THe gain is =\",int(math.ceil(ADM2)),\"V/V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ICQ value is = 0.25 mA\n", + "gm value is = 1/100 Mho\n", + "rpi value is = 10 kilo Ohm\n", + "THe gain is = 40 V/V\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Ex 2.8 Page no.97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "import math\n", + "I0 = 10*10**-6\n", + "VCC =10\n", + "VBE = 0.7\n", + "beta = 125\n", + "VT = 25*10**-3\n", + "\n", + "# Solution of the problem is \n", + "Iref = 10**-3 # Assumption\n", + "\n", + "R1 = (VCC - VBE)/Iref\n", + "# Finding the value RE from the equation 2.74\n", + "RE = (VT/(1+(1/beta)*I0))*math.log(Iref/I0)\n", + "\n", + "# printing the values \n", + "\n", + "print \"The value of R1 =\",R1/10**3,\"Kilo Ohms\"\n", + "print \"The value of RE =\",round(RE*100,1),\"Kilo Ohms\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of R1 = 9.3 Kilo Ohms\n", + "The value of RE = 11.5 Kilo Ohms\n" + ] + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/sample_notebooks/YogeshPatil/EDC_By_K_L_Kishore_Chapter_7.ipynb b/sample_notebooks/YogeshPatil/EDC_By_K_L_Kishore_Chapter_7.ipynb new file mode 100755 index 00000000..703182db --- /dev/null +++ b/sample_notebooks/YogeshPatil/EDC_By_K_L_Kishore_Chapter_7.ipynb @@ -0,0 +1,499 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7 : Feedback Amplifiers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1 page no-402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# determination of various parameters of feedback amplifiers\n",
+ "\n",
+ "import math\n",
+ "#Variable declaration\n",
+ "Av=-100\n",
+ "B=0.01\n",
+ "\n",
+ "#Calculations\n",
+ "Avd=Av/(1-B*Av)\n",
+ "v1d=10**-3 \n",
+ "V0=Avd*v1d*1000\n",
+ "Vx=B*V0\n",
+ "V1=v1d+Vx\n",
+ "\n",
+ "#Result\n",
+ "print(\"V1=%.3f\\nV1d=%.3f\\nThis is negative feedback because, v1<v1_dash\\n\"%( V1,v1d))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "V1=-0.499\n",
+ "V1d=0.001\n",
+ "This is negative feedback because, v1<v1_dash\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2 page no-403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#percentage variation in Avdash\n",
+ "\n",
+ "import math\n",
+ "#Variable declaration\n",
+ "Av=-100\n",
+ "Avd=-50\n",
+ "Avnew=-200\n",
+ "B=0.01\n",
+ "\n",
+ "# Calculations\n",
+ "Avdnew=Avnew/(1-B*Avnew)\n",
+ "avchange=(-Avdnew)-(-Avd)\n",
+ "var=avchange*100/(-Avd)\n",
+ "\n",
+ "#Result\n",
+ "print(\"Variation = %.1f%%\"%var)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Variation = 33.3%\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3 page no-403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# reverse transmission factor and gain with feedback\n",
+ "\n",
+ "import math\n",
+ "#Variable declaration\n",
+ "dA=100.0\n",
+ "A=1000.0\n",
+ "dAf=0.1\n",
+ "Af=100.0\n",
+ "\n",
+ "#Calculations\n",
+ "#(a)\n",
+ "B=(((dA/A)*(Af/dAf))-1)/A\n",
+ "#(b)\n",
+ "Aff=A/(1+B*A)\n",
+ "\n",
+ "#Result\n",
+ "print(\"(a)\\nBeta=%.3f\"%B)\n",
+ "print(\"\\n(b)\\nAf=%d\"%Aff)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)\n",
+ "Beta=0.099\n",
+ "\n",
+ "(b)\n",
+ "Af=10\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4 page no-404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Improvement in stability\n",
+ "\n",
+ "import math\n",
+ "#Variable declaration\n",
+ "S=0.1\n",
+ "Sdash=0.01\n",
+ "Avdash=100.0\n",
+ "\n",
+ "#Calculations\n",
+ "k=S/Sdash # k=1+BAv\n",
+ "Av=Avdash*k\n",
+ "B=(k-1)/Av\n",
+ "\n",
+ "#Result\n",
+ "print(\"By providing negative feedback,with\\nBeta = %.3f\\nwe can improve the stability to 1%%.\"%B)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "By providing negative feedback,with\n",
+ "Beta = 0.009\n",
+ "we can improve the stability to 1%.\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.5 page no-404"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Overall gain and reverse transmission factor\n",
+ "\n",
+ "import math\n",
+ "#Variable declaration\n",
+ "Av=500.0\n",
+ "D=5.0\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "Ddash=0.1\n",
+ "B=((D/Ddash)-1)/(Av)\n",
+ "Avdash=-Av/(1+B*Av)\n",
+ "\n",
+ "#Result\n",
+ "print(\"Av_dash = %.0f\"%Avdash)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Av_dash = -10\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.6 page no-405"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# different parameters with and without negative feedback\n",
+ "\n",
+ "import math\n",
+ "#Variable declaration\n",
+ "Vs=150.0\n",
+ "A=10000.0\n",
+ "Vs2=130.0\n",
+ "A2=8000.0\n",
+ "\n",
+ "\n",
+ "#Calcualtions\n",
+ "V0=A*Vs\n",
+ "Afb=10000.0/80.0\n",
+ "B=((A/Afb)-1)/A\n",
+ "V02=A2*Vs\n",
+ "Afb2=A2/(1+(B*A2))\n",
+ "sg=(A-A2)*100/A\n",
+ "sgf=(Afb-Afb2)*100/Afb\n",
+ "\n",
+ "#Result\n",
+ "print(\"Beta =%.4f\"%B)\n",
+ "print(\"%% stability of gain without feedback=%.0f%%\\n%% stability of gain with feedback=%.4f%%\"%(sg,sgf))\n",
+ "print(\"Therefore, with neative feedbaclk stability is improved.\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Beta =0.0079\n",
+ "% stability of gain without feedback=20%\n",
+ "% stability of gain with feedback=0.3115%\n",
+ "Therefore, with neative feedbaclk stability is improved.\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.7 page no-409"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Avf, Rof and Rif for the voltage series feedback\n",
+ "\n",
+ "import math\n",
+ "#Variable declaration\n",
+ "Rs=0\n",
+ "hfe=50.0\n",
+ "hie =1.100\n",
+ "hre=0\n",
+ "hoe=0\n",
+ "r5=2.2000\n",
+ "r7=3.3000\n",
+ "r3=33.0\n",
+ "r1=0.1\n",
+ "r2=10.0\n",
+ "r9=2.2\n",
+ "R1=0.98\n",
+ "r6=2.2\n",
+ "R0=2.0\n",
+ "\n",
+ "#Calculations\n",
+ "#Rl =R5 is in parallel with R7,R8 and h1e2\n",
+ "Rl1=(r5*r3*r7*hie)/((r5*r3*r7)+(hie*r3*r7)+(r5*hie*r7)+(r5*r3*hie))\n",
+ "Rl2=(r9*(r1+r2))/(r9+(r1+r2))\n",
+ "Re=(r1*r6)/(r1+r6)\n",
+ "Av1=-(hfe*Rl1)/(hie+(1+hfe)*0.098) # The voltage gain AV1 of Q\n",
+ "Av2=(-hfe*Rl2)/hie # Voltage gain AY2 of transistor Q2\n",
+ "Av=Av1*Av2 # Voltage gain Ay of the two stages is cascade without feedback\n",
+ "B=r1/(r1+r2)\n",
+ "K=Av*B\n",
+ "D=1+K\n",
+ "Avf=Av/D\n",
+ "Ri=hie+(1+hfe)*Re # Input resistance without external feedback\n",
+ "Ridash=Ri*D\n",
+ "Rof=R0/D # Output resistance without feedback\n",
+ "\n",
+ "#Result\n",
+ "print(\"Rl1_dash=%f\"%Rl1)\n",
+ "print(\"Rl2=%f = 2 KOhm(approx)\"%Rl2)\n",
+ "print(\"Re=%f kohm = %.0f ohm\"%(Re,math.ceil(Re*1000)))\n",
+ "print(\"Av1 = %.2f\\nAv2 = %.2f\"%(Av1,Av2))\n",
+ "print(\"Avf = %d\"%Avf)\n",
+ "print(\"Ri_dash = %f K Ohm\"%Ridash)\n",
+ "print(\"Rof_dash=%f K Ohm\"%Rof)\n",
+ "#Though the calculations are same as given in book answers \n",
+ "#do not match with the answers given in the Book."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rl1_dash=0.589286\n",
+ "Rl2=1.806504 = 2 KOhm(approx)\n",
+ "Re=0.095652 kohm = 96 ohm\n",
+ "Av1 = -4.83\n",
+ "Av2 = -82.11\n",
+ "Avf = 80\n",
+ "Ri_dash = 29.462595 K Ohm\n",
+ "Rof_dash=0.405820 K Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.8 page no-414"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# current series feedack\n",
+ "\n",
+ "import math\n",
+ "#Variable declaration\n",
+ "Rc1 =3.0\n",
+ "Rc2 =0.500\n",
+ "Re2 = 0.05\n",
+ "Rdash=1.2\n",
+ "Rs = 1.2\n",
+ "hfe = 50.0\n",
+ "hie = 1.1 \n",
+ "hre=0\n",
+ "hre =0\n",
+ "\n",
+ "#Calculations\n",
+ "Ai=-hfe # EmItter follower\n",
+ "Ri2=hie+(1+hfe)*(Re2*Rdash/(Re2+Rdash))\n",
+ "k1=-Rc1/(Rc1+Ri2)\n",
+ "k1=math.ceil(k1*1000)\n",
+ "k1=k1/1000\n",
+ "R=Rs*(Rdash+Re2)/(Rs+(Rdash+Re2))\n",
+ "k2=R/(R+hie)\n",
+ "k2=math.floor(k2*1000)\n",
+ "k2=k2/1000\n",
+ "AI=Ai*k1*k2*hfe\n",
+ "B=Re2/(Re2+Rdash)\n",
+ "D=(1+B*AI)\n",
+ "Adash=AI/(1+B*AI)\n",
+ "Avdash=Adash*Rc2/Rs\n",
+ "Ri=R*hie/(R+hie) # Ri = Input resistance without feedback\n",
+ "Ridash=Ri/D\n",
+ "Rol=Rc2 # RoL =Ro in parallel with RC2 = RC2 and Ro is large\n",
+ "Rldash= Rol*D/D # with feedback considering RL\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print(\"AI=%d\\nBeta=%.2f\\nAi_dash=%.1f\\nAv_dash=%.2f\"%(AI,B,Adash,Avdash))\n",
+ "print(\"Ri=%f K Ohm\\nRl_dash=%.2f K Ohm\"%(Ri,Rldash))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "AI=408\n",
+ "Beta=0.04\n",
+ "Ai_dash=23.6\n",
+ "Av_dash=9.82\n",
+ "Ri=0.393325 K Ohm\n",
+ "Rl_dash=0.50 K Ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.9 page no-423"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# calculation of Avf and Rif for given circuit\n",
+ "\n",
+ "import math\n",
+ "#Variable declaration\n",
+ "Rc=4.0\n",
+ "Rb=40.0\n",
+ "Rs=10.0\n",
+ "hie=1.1\n",
+ "hfe=50.0\n",
+ "hre=0\n",
+ "hoe=0\n",
+ "\n",
+ "#Calculations\n",
+ "Rcdash=Rc*Rb/(Rc+Rb)\n",
+ "R=Rs*Rb/(Rs+Rb)\n",
+ "Rm=-hfe*Rcdash*R/(R+hie)\n",
+ "Rm=math.floor(Rm)\n",
+ "B=-1/(Rb)\n",
+ "D=1+B*Rm\n",
+ "Rmdash=Rm/D\n",
+ "Avdash=Rmdash/Rs\n",
+ "Ri=R*hie/(R+hie)\n",
+ "Ridash=Ri/D\n",
+ "\n",
+ "#Result\n",
+ "print(\"Transresistance Rm=%d k\"%Rm)\n",
+ "print(\"Beta=%.3f mA/V\\nRm_dash=%dk Ohm\\nAv_dash=%f\\nRi=%f k Ohm\"%(B,Rmdash,Avdash,Ri))\n",
+ "print(\"Ri_dash=%f kOhm\"%Ridash)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Transresistance Rm=-160 k\n",
+ "Beta=-0.025 mA/V\n",
+ "Rm_dash=-32k Ohm\n",
+ "Av_dash=-3.200000\n",
+ "Ri=0.967033 k Ohm\n",
+ "Ri_dash=0.193407 kOhm\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/sample_notebooks/kotaDinesh Babu/samplebook(process_heat_transfer)_1.ipynb b/sample_notebooks/kotaDinesh Babu/samplebook(process_heat_transfer)_1.ipynb new file mode 100755 index 00000000..1de57adb --- /dev/null +++ b/sample_notebooks/kotaDinesh Babu/samplebook(process_heat_transfer)_1.ipynb @@ -0,0 +1,281 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 2:CONDUCTION"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example2.1"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " heat is Btu/hr 69120.0\n",
+ "\t approximate values are mentioned in the book \n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "#page 13\n",
+ "#given\n",
+ "Tavg=900; # average temperature of the wall,F\n",
+ "k=0.15; # Thermal conductivity at 932 F,Btu/(hr)(ft^2)(F/ft)\n",
+ "T1=1500; # hot side temperature,F\n",
+ "T2=300; # cold side temperature,F\n",
+ "A=192; # surface area,ft^2\n",
+ "L=0.5; # thickness,ft\n",
+ "#solution\n",
+ "Q=(k)*(A)*(T1-T2)/L; # formula for heat,Btu/hr\n",
+ "print \" heat is Btu/hr \",Q\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#end\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example2.2"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\t resistance offered by firebrick : (hr)*(F)/Btu 0.97\n",
+ "\t resistance offered by insulating brick : (hr)*(F)/Btu 2.2\n",
+ "\t resistance offered by buildingbrick : (hr)*(F)/Btu 1.25\n",
+ "\t total resistance offered by three walls : (hr)*(F)/Btu 4.42\n",
+ "\t heat loss/ft^2 : Btu/hr 331.0\n",
+ "\t delta is : F 322.0\n",
+ "\t temperature at interface of firebrick and insulating brick F 1278.0\n",
+ "\t deltb is : F 729.0\n",
+ "\t temperature at interface of insulating brick and building brick F 549.0\n",
+ "\t approximate values are mentioned in the book \n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "#page 14\n",
+ "#given\n",
+ "La=0.66; # Thickness of firebrick wall,ft\n",
+ "Lb=0.33; # Thickness of insulating brick wall,ft\n",
+ "Lc=0.5; # Thickness of building brick wall,ft\n",
+ "Ka=0.68; # themal conductivity of firebrick,Btu/(hr)*(ft^2)*(F/ft)\n",
+ "Kb=0.15; # themal conductivity of insulating brick,Btu/(hr)*(ft^2)*(F/ft)\n",
+ "Kc=0.40; # themal conductivity of building brick,Btu/(hr)*(ft^2)*(F/ft)\n",
+ "A=1.; # surface area,ft^2\n",
+ "Ta=1600.; # temperature of inner wall,F\n",
+ "Tb=125.; # temperature of outer wall.F\n",
+ "#solution\n",
+ "Ra=La/(Ka)*(A); # formula for resistance,(hr)*(F)/Btu\n",
+ "print\"\\t resistance offered by firebrick : (hr)*(F)/Btu \",round(Ra,2)\n",
+ "Rb=Lb/(Kb)*(A); # formula for resistance,(hr)*(F)/Btu\n",
+ "print\"\\t resistance offered by insulating brick : (hr)*(F)/Btu \",round(Rb,2)\n",
+ "Rc=Lc/(Kc)*(A); # formula for resistance,(hr)*(F)/Btu\n",
+ "print\"\\t resistance offered by buildingbrick : (hr)*(F)/Btu \",round(Rc,2)\n",
+ "R=Ra+Rb+Rc; # total resistance offered by three walls,(hr)*(F)/Btu\n",
+ "print\"\\t total resistance offered by three walls : (hr)*(F)/Btu \",round(R,2)\n",
+ "Q=(1600-125)/4.45; # using formula for heat loss/ft^2,Btu/hr\n",
+ "print\"\\t heat loss/ft^2 : Btu/hr \",round(Q,0)\n",
+ "# T1,T2 are temperatures at interface of firebrick and insulating brick, and insulating brick and building brick respectively,F\n",
+ "delta=(Q)*(Ra); # formula for temperature difference,F\n",
+ "print\"\\t delta is : F \",round(delta,0)\n",
+ "T1=Ta-((Q)*(Ra)); # temperature at interface of firebrick and insulating brick,F\n",
+ "print\"\\t temperature at interface of firebrick and insulating brick F \",round(T1,0)\n",
+ "deltb=Q*(Rb);\n",
+ "print\"\\t deltb is : F \",round(deltb,0)\n",
+ "T2=T1-((Q)*(Rb)); #temperature at interface of insulating brick and building brick,F\n",
+ "print\"\\t temperature at interface of insulating brick and building brick F \",round(T2,0)\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#end\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example2.3"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\t approximate values are mentioned in the book \n",
+ "\n",
+ "\t resistance offered by air film (hr)(F)/Btu 0.79\n",
+ "\t total resistance (hr)(F)/Btu 5.24\n",
+ "\t heat loss Btu/hr 282.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#page 15\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#given\n",
+ "Lair=0.25/12; # thickness of air film,ft\n",
+ "Kair=0.0265; # thermal conductivity of air at 572F,Btu/(hr)*(ft^2)(F/ft)\n",
+ "A=1; # surface area,ft^2\n",
+ "#solution\n",
+ "Rair=Lair/(Kair*(A)); # resistance offered by air film, (hr)(F)/Btu\n",
+ "print\"\\t resistance offered by air film (hr)(F)/Btu \",round(Rair,2)\n",
+ "R=4.45; # resistance from previous example 2.2,(hr)(F)/Btu\n",
+ "Rt=(R)+Rair; # total resistance,(hr)(F)/Btu\n",
+ "print\"\\t total resistance (hr)(F)/Btu \",round(Rt,2)\n",
+ "Ta=1600; # temperature of inner wall,F\n",
+ "Tb=125; # temperature of outer wall,F\n",
+ "Q=(1600-125)/Rt; # heat loss, Btu/hr\n",
+ "print\"\\t heat loss Btu/hr \",round(Q,0)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example2.4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "#page 16\n",
+ "#given\n",
+ "k=0.63; # thermal conductivity of pipe, Btu/(hr)*(ft^2)*(F/ft)\n",
+ "Do=6. # in\n",
+ "Di=5. # in\n",
+ "Ti=200.;# inner side temperature,F\n",
+ "To=175.; # outer side temperature,F\n",
+ "#solution\n",
+ "import math\n",
+ "from math import log\n",
+ "q=(2*(3.14)*(k)*(Ti-To))/(log (Do/Di)); # formula for heat flow,Btu/(hr)*(ft)\n",
+ "print\"\\t heat flow is : Btu/(hr)*(ft) \",round(q,0)\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "# caculation mistake in book\n",
+ "# end\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example2.5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\t approximate values are mentioned in the book \n",
+ "\n",
+ "\t heat loss for linear foot is : Btu/(hr)*(lin ft) 104.4\n",
+ "\t Check between ts and t1, since delt/R = deltc/Rc \n",
+ "\t t1 is : F 122.300238658\n",
+ "\t heat loss for linear foot is : Btu/(hr)*(lin ft) 102.9\n",
+ "\t Check between ts and t1, since delt/R = deltc/Rc \n",
+ "\t t1 is : F 125.4\n"
+ ]
+ }
+ ],
+ "source": [
+ "#page 19\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#given\n",
+ "t1=150; # assume temperature of outer surface of rockwool,F\n",
+ "ta=70; # temperature of surrounding air,F\n",
+ "ha=2.23; # surface coefficient,Btu/(hr)*(ft^2)*(F)\n",
+ "#solution\n",
+ "import math\n",
+ "from math import log\n",
+ "q=(3.14)*(300-70)/(((1/(2*0.033))*log(3.375/2.375))+(1/((2.23)*(3.375/12)))); # using formula for heat loss,Btu/(hr)*(lin ft), calculation mistake\n",
+ "print\"\\t heat loss for linear foot is : Btu/(hr)*(lin ft) \",round(q,1)\n",
+ "print\"\\t Check between ts and t1, since delt/R = deltc/Rc \"\n",
+ "t1=300-(((104.8)*((1)*(log(3.375/2.375))))/((2)*(3.14)*(.033))); # using eq 2.31,F\n",
+ "print\"\\t t1 is : F \",t1\n",
+ "t1=125; # assume temperature of outer surface of rockwool,F\n",
+ "ha=2.10; # surface coefficient,Btu/(hr)*(ft^2)*(F)\n",
+ "q=((3.14)*(300-70))/(((1/(2*0.033))*log(3.375/2.375))+(1/((2.10)*(3.375/12)))); # using formula for heat loss,Btu/(hr)*(lin ft)\n",
+ "print\"\\t heat loss for linear foot is : Btu/(hr)*(lin ft)\",round(q,1)\n",
+ "print\"\\t Check between ts and t1, since delt/R = deltc/Rc \"\n",
+ "t1=300-(((103)*((1)*(log(3.375/2.375))))/((2)*(3.14)*(.033))); # using eq 2.31,F\n",
+ "print\"\\t t1 is : F \",round(t1,1)\n",
+ "# end \n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/kotaDinesh Babu/samplebook(process_heat_transfer)_2.ipynb b/sample_notebooks/kotaDinesh Babu/samplebook(process_heat_transfer)_2.ipynb new file mode 100755 index 00000000..8bc5413c --- /dev/null +++ b/sample_notebooks/kotaDinesh Babu/samplebook(process_heat_transfer)_2.ipynb @@ -0,0 +1,281 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 2:CONDUCTION"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example2.1 pg:13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " heat is Btu/hr 69120.0\n",
+ "\t approximate values are mentioned in the book \n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "\n",
+ "#given\n",
+ "Tavg=900; # average temperature of the wall,F\n",
+ "k=0.15; # Thermal conductivity at 932 F,Btu/(hr)(ft^2)(F/ft)\n",
+ "T1=1500; # hot side temperature,F\n",
+ "T2=300; # cold side temperature,F\n",
+ "A=192; # surface area,ft^2\n",
+ "L=0.5; # thickness,ft\n",
+ "#solution\n",
+ "Q=(k)*(A)*(T1-T2)/L; # formula for heat,Btu/hr\n",
+ "print \" heat is Btu/hr \",Q\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#end\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example2.2 pg:14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\t resistance offered by firebrick : (hr)*(F)/Btu 0.97\n",
+ "\t resistance offered by insulating brick : (hr)*(F)/Btu 2.2\n",
+ "\t resistance offered by buildingbrick : (hr)*(F)/Btu 1.25\n",
+ "\t total resistance offered by three walls : (hr)*(F)/Btu 4.42\n",
+ "\t heat loss/ft^2 : Btu/hr 331.0\n",
+ "\t delta is : F 322.0\n",
+ "\t temperature at interface of firebrick and insulating brick F 1278.0\n",
+ "\t deltb is : F 729.0\n",
+ "\t temperature at interface of insulating brick and building brick F 549.0\n",
+ "\t approximate values are mentioned in the book \n",
+ "\n"
+ ]
+ }
+ ],
+ "source": [
+ "\n",
+ "#given\n",
+ "La=0.66; # Thickness of firebrick wall,ft\n",
+ "Lb=0.33; # Thickness of insulating brick wall,ft\n",
+ "Lc=0.5; # Thickness of building brick wall,ft\n",
+ "Ka=0.68; # themal conductivity of firebrick,Btu/(hr)*(ft^2)*(F/ft)\n",
+ "Kb=0.15; # themal conductivity of insulating brick,Btu/(hr)*(ft^2)*(F/ft)\n",
+ "Kc=0.40; # themal conductivity of building brick,Btu/(hr)*(ft^2)*(F/ft)\n",
+ "A=1.; # surface area,ft^2\n",
+ "Ta=1600.; # temperature of inner wall,F\n",
+ "Tb=125.; # temperature of outer wall.F\n",
+ "#solution\n",
+ "Ra=La/(Ka)*(A); # formula for resistance,(hr)*(F)/Btu\n",
+ "print\"\\t resistance offered by firebrick : (hr)*(F)/Btu \",round(Ra,2)\n",
+ "Rb=Lb/(Kb)*(A); # formula for resistance,(hr)*(F)/Btu\n",
+ "print\"\\t resistance offered by insulating brick : (hr)*(F)/Btu \",round(Rb,2)\n",
+ "Rc=Lc/(Kc)*(A); # formula for resistance,(hr)*(F)/Btu\n",
+ "print\"\\t resistance offered by buildingbrick : (hr)*(F)/Btu \",round(Rc,2)\n",
+ "R=Ra+Rb+Rc; # total resistance offered by three walls,(hr)*(F)/Btu\n",
+ "print\"\\t total resistance offered by three walls : (hr)*(F)/Btu \",round(R,2)\n",
+ "Q=(1600-125)/4.45; # using formula for heat loss/ft^2,Btu/hr\n",
+ "print\"\\t heat loss/ft^2 : Btu/hr \",round(Q,0)\n",
+ "# T1,T2 are temperatures at interface of firebrick and insulating brick, and insulating brick and building brick respectively,F\n",
+ "delta=(Q)*(Ra); # formula for temperature difference,F\n",
+ "print\"\\t delta is : F \",round(delta,0)\n",
+ "T1=Ta-((Q)*(Ra)); # temperature at interface of firebrick and insulating brick,F\n",
+ "print\"\\t temperature at interface of firebrick and insulating brick F \",round(T1,0)\n",
+ "deltb=Q*(Rb);\n",
+ "print\"\\t deltb is : F \",round(deltb,0)\n",
+ "T2=T1-((Q)*(Rb)); #temperature at interface of insulating brick and building brick,F\n",
+ "print\"\\t temperature at interface of insulating brick and building brick F \",round(T2,0)\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#end\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example2.3 pg:15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\t approximate values are mentioned in the book \n",
+ "\n",
+ "\t resistance offered by air film (hr)(F)/Btu 0.79\n",
+ "\t total resistance (hr)(F)/Btu 5.24\n",
+ "\t heat loss Btu/hr 282.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#given\n",
+ "Lair=0.25/12; # thickness of air film,ft\n",
+ "Kair=0.0265; # thermal conductivity of air at 572F,Btu/(hr)*(ft^2)(F/ft)\n",
+ "A=1; # surface area,ft^2\n",
+ "#solution\n",
+ "Rair=Lair/(Kair*(A)); # resistance offered by air film, (hr)(F)/Btu\n",
+ "print\"\\t resistance offered by air film (hr)(F)/Btu \",round(Rair,2)\n",
+ "R=4.45; # resistance from previous example 2.2,(hr)(F)/Btu\n",
+ "Rt=(R)+Rair; # total resistance,(hr)(F)/Btu\n",
+ "print\"\\t total resistance (hr)(F)/Btu \",round(Rt,2)\n",
+ "Ta=1600; # temperature of inner wall,F\n",
+ "Tb=125; # temperature of outer wall,F\n",
+ "Q=(1600-125)/Rt; # heat loss, Btu/hr\n",
+ "print\"\\t heat loss Btu/hr \",round(Q,0)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example2.4 pg 16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": [
+ "\n",
+ "#given\n",
+ "k=0.63; # thermal conductivity of pipe, Btu/(hr)*(ft^2)*(F/ft)\n",
+ "Do=6. # in\n",
+ "Di=5. # in\n",
+ "Ti=200.;# inner side temperature,F\n",
+ "To=175.; # outer side temperature,F\n",
+ "#solution\n",
+ "import math\n",
+ "from math import log\n",
+ "q=(2*(3.14)*(k)*(Ti-To))/(log (Do/Di)); # formula for heat flow,Btu/(hr)*(ft)\n",
+ "print\"\\t heat flow is : Btu/(hr)*(ft) \",round(q,0)\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "# caculation mistake in book\n",
+ "# end\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example2.5 pg 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "\t approximate values are mentioned in the book \n",
+ "\n",
+ "\t heat loss for linear foot is : Btu/(hr)*(lin ft) 104.4\n",
+ "\t Check between ts and t1, since delt/R = deltc/Rc \n",
+ "\t t1 is : F 122.300238658\n",
+ "\t heat loss for linear foot is : Btu/(hr)*(lin ft) 102.9\n",
+ "\t Check between ts and t1, since delt/R = deltc/Rc \n",
+ "\t t1 is : F 125.4\n"
+ ]
+ }
+ ],
+ "source": [
+ "\n",
+ "print\"\\t approximate values are mentioned in the book \\n\"\n",
+ "#given\n",
+ "t1=150; # assume temperature of outer surface of rockwool,F\n",
+ "ta=70; # temperature of surrounding air,F\n",
+ "ha=2.23; # surface coefficient,Btu/(hr)*(ft^2)*(F)\n",
+ "#solution\n",
+ "import math\n",
+ "from math import log\n",
+ "q=(3.14)*(300-70)/(((1/(2*0.033))*log(3.375/2.375))+(1/((2.23)*(3.375/12)))); # using formula for heat loss,Btu/(hr)*(lin ft), calculation mistake\n",
+ "print\"\\t heat loss for linear foot is : Btu/(hr)*(lin ft) \",round(q,1)\n",
+ "print\"\\t Check between ts and t1, since delt/R = deltc/Rc \"\n",
+ "t1=300-(((104.8)*((1)*(log(3.375/2.375))))/((2)*(3.14)*(.033))); # using eq 2.31,F\n",
+ "print\"\\t t1 is : F \",t1\n",
+ "t1=125; # assume temperature of outer surface of rockwool,F\n",
+ "ha=2.10; # surface coefficient,Btu/(hr)*(ft^2)*(F)\n",
+ "q=((3.14)*(300-70))/(((1/(2*0.033))*log(3.375/2.375))+(1/((2.10)*(3.375/12)))); # using formula for heat loss,Btu/(hr)*(lin ft)\n",
+ "print\"\\t heat loss for linear foot is : Btu/(hr)*(lin ft)\",round(q,1)\n",
+ "print\"\\t Check between ts and t1, since delt/R = deltc/Rc \"\n",
+ "t1=300-(((103)*((1)*(log(3.375/2.375))))/((2)*(3.14)*(.033))); # using eq 2.31,F\n",
+ "print\"\\t t1 is : F \",round(t1,1)\n",
+ "# end \n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/marupeddisameer chaitanya/Sample_(chapter_9).ipynb b/sample_notebooks/marupeddisameer chaitanya/Sample_(chapter_9).ipynb new file mode 100755 index 00000000..bff5435f --- /dev/null +++ b/sample_notebooks/marupeddisameer chaitanya/Sample_(chapter_9).ipynb @@ -0,0 +1,247 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 9 : Theories of Mass Transfer"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.1.1 pgno31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The film thickness is cm 0.00765\n"
+ ]
+ }
+ ],
+ "source": [
+ "#initialization of variables\n",
+ "p1 = 10. # pressure in atm\n",
+ "H = 600. # henrys constant in atm\n",
+ "c1 = 0 # gmol/cc\n",
+ "N1 = 2.3*10**-6 # mass flux in mol/cm**2-sec\n",
+ "c = 1./18. #total Concentration in g-mol/cc\n",
+ "D = 1.9*10**-5 # Diffusion co efficient in cm**2/sec\n",
+ "#Calculations\n",
+ "c1i = (p1/H)*c # Component concentration in gmol/cc\n",
+ "k = N1/(c1i-c1)#Mass transfer co efficient in cm/sec\n",
+ "l = D/k # Film thickness in cm\n",
+ "#Results\n",
+ "print\"The film thickness is cm\",round(l,5)\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.2.1 pgno:34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The contact time sec 3.9\n",
+ "\n",
+ "The surface resident time sec 3.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#initialization of variables\n",
+ "D = 1.9*10**-5 #Diffusion co efficient in cm**2/sec\n",
+ "k = 2.5*10**-3 # M.T.C in cm/sec\n",
+ "from math import pi\n",
+ "#Calculations\n",
+ "Lbyvmax = 4*D/((k**2)*pi)#sec\n",
+ "tou = D/k**2 # sec\n",
+ "#Results\n",
+ "print\"The contact time sec\",round(Lbyvmax,1)\n",
+ "print\"\\nThe surface resident time sec\",round(tou,1)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.3.1 pgno:35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The apparent m.t.c for the first case is cm/sec 0.000379885493042\n",
+ "\n",
+ "The apparent m.t.c for the second case is cm/sec 0.000742723884992\n",
+ "\n",
+ "The apparent is proportional to the power of of the velocity 0.61\n"
+ ]
+ }
+ ],
+ "source": [
+ "#initialization of variables\n",
+ "const = 0.5 # The part of flow in the system which bypasses the region where the mass transfer occurs\n",
+ "v1 = 1. # cm/sec\n",
+ "al = 10**3\n",
+ "k = 10**-3 # cm/sec\n",
+ "v2 = 3. # cm/sec\n",
+ "from math import log\n",
+ "from math import exp\n",
+ "#Calculations\n",
+ "C1byC10first = const + (1-const)*(exp(-k*al/v1))# c1/c10\n",
+ "appk1 = (v1/al)*(log(1/C1byC10first))# Apparent m.t.c for first case in cm/sec\n",
+ "C1byC10second = const + (1-const)*(exp(-((3)**0.5)*k*al/v2))#c1/c10 in second case\n",
+ "appk2 = (v2/al)*log(1/C1byC10second)# apparent m.t.c for second case in cm/sec\n",
+ "power = log(appk2/appk1)/log(v2/v1)\n",
+ "#Results\n",
+ "print\"The apparent m.t.c for the first case is cm/sec\",appk1\n",
+ "print\"\\nThe apparent m.t.c for the second case is cm/sec\",appk2\n",
+ "print\"\\nThe apparent is proportional to the power of of the velocity\",round(power,2)\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.4.1 pgno:37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The average mass transfer coefficient is cm/sec 0.000431530124388\n"
+ ]
+ }
+ ],
+ "source": [
+ "#initialization of variables\n",
+ "D = 1*10**-5 #cm**2/sec\n",
+ "d = 2.3 # cm\n",
+ "L = 14 # cm\n",
+ "v0 = 6.1 # cm/sec\n",
+ "#gamma(4./3.)=0.8909512761;\n",
+ "#calculations\n",
+ "k = ((3**(1./3.))/(0.8909512761))*((D/d))*(((d**2)*v0/(D*L))**(1./3.))# cm/sec\n",
+ "#Results\n",
+ "print\"The average mass transfer coefficient is cm/sec\",k\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9.4.2 pgno:40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The distance at which turbulent flow starts is cm 300.0\n",
+ "\n",
+ "The boundary layer for flow at this point is cm 300.0\n",
+ "\n",
+ "The boundary layer for concentration at this point is cm 300.0\n",
+ "\n",
+ "The local m.t.c at the leading edge and at the position of transistion is x10**-5 cm/sec 0.589714620247\n"
+ ]
+ }
+ ],
+ "source": [
+ "#initialization of variables\n",
+ "tn = 300000 # turbulence number\n",
+ "v0 = 10 # cm/sec\n",
+ "p = 1 # g/cc\n",
+ "mu = 0.01 # g/cm-sec\n",
+ "delta = 2.5 #cm\n",
+ "D = 1*10**-5 # cm**2/sec\n",
+ "#Calculations\n",
+ "x = tn*mu/(v0*p)# cm\n",
+ "delta = ((280/13)**(1/2))*x*((mu/(x*v0*p))**(1/2))#cm\n",
+ "deltac = ((D*p/mu)**(1/3))*delta#cm\n",
+ "k = (0.323*(D/x)*((x*v0*p/mu)**0.5)*((mu/(p*D))**(1/3)))*10**5# x*10**-5 cm/sec\n",
+ "#Results\n",
+ "print\"The distance at which turbulent flow starts is cm\",x\n",
+ "print\"\\nThe boundary layer for flow at this point is cm\",delta\n",
+ "print\"\\nThe boundary layer for concentration at this point is cm\",deltac\n",
+ "print\"\\nThe local m.t.c at the leading edge and at the position of transistion is x10**-5 cm/sec\",k\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/nishumittal/chapter2.ipynb b/sample_notebooks/nishumittal/chapter2.ipynb new file mode 100755 index 00000000..3d83df64 --- /dev/null +++ b/sample_notebooks/nishumittal/chapter2.ipynb @@ -0,0 +1,710 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:46bc70330d4213802afb03e252b2ad32eb9319ed4cc2a32fe2c16df97a5f1978"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2 Particle nature of Radiation; The origin of Quantum theory"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2 Page no-12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=40 #W\n",
+ "lembda=6000*10**-10 #m\n",
+ "h=6.63*10**-34 #Js\n",
+ "c=3*10**8 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "n=(E*lembda)/(h*c)\n",
+ "\n",
+ "#Result\n",
+ "print\"No. of photons emitted per second are given by \",round(n*10**-19,2),\"*10**19\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No. of photons emitted per second are given by 12.07 *10**19\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.3 Page no-12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "a=3.2 #ev\n",
+ "energy=3.8 #ev\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "c=energy-a\n",
+ "Energy=c*e\n",
+ "\n",
+ "#Result\n",
+ "print\"Kinetic energy of the photoelectron is given by \",Energy,\"Joule\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy of the photoelectron is given by 9.6e-20 Joule\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.4 Page no-12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "W=3.45 #ev\n",
+ "h=6.63*10**-34 #Js\n",
+ "c=3*10**8 #m/s\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "lembda=(h*c)/(W*e)\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum wavelength of photon is \",round(lembda*10**10,0),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum wavelength of photon is 3603.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 193
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.5 Page no-12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "W=3 #ev\n",
+ "h=6.63*10**-34\n",
+ "e=1.6*10**-19\n",
+ "lembda=3.0*10**-7 #m\n",
+ "c=3*10**8 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "v0=(W*e)/h\n",
+ "v=c/lembda\n",
+ "E=h*(v-v0)\n",
+ "E1=(h*(v-v0))/(1.6*10**-19)\n",
+ "V0=E/e\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Threshold frequency \",round(v0*10**-15,2),\"*10**15 HZ\"\n",
+ "print\"(b) Maximum energy of photoelectron \",round(E1,2),\"eV\"\n",
+ "print\"(c) Stopping potential \",round(V0,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Threshold frequency 0.72 *10**15 HZ\n",
+ "(b) Maximum energy of photoelectron 1.14 eV\n",
+ "(c) Stopping potential 1.14 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 197
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.6 Page no-13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v0=6*10**14 #s**-1\n",
+ "h=6.63*10**-34\n",
+ "e=1.6*10**-19\n",
+ "V0=3\n",
+ "\n",
+ "#Calculaton\n",
+ "W=h*v0\n",
+ "W0=(h*v0)/e\n",
+ "V=(e*V0+h*v0)/h\n",
+ "\n",
+ "#Result \n",
+ "print\"work function is given by \",round(W0,3),\"ev\"\n",
+ "print\"frequency is given by \",round(V*10**-15,2),\"*10**15 s-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "work function is given by 2.486 ev\n",
+ "frequency is given by 1.32 *10**15 s-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 88
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.7 Page no 13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "lembda=6800.0*10**-10 #m\n",
+ "h=6.6*10**-34\n",
+ "W=2.3 #ev\n",
+ "c=3*10**8 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "E=((h*c)/lembda)/1.6*10**-19\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy is \",round(E*10**38,2),\"ev\"\n",
+ "print\"since the energy of incident photon is less then the work function of Na, photoelecrticemession is not possible with the given light.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy is 1.82 ev\n",
+ "since the energy of incident photon is less then the work function of Na, photoelecrticemession is not possible with the given light.\n"
+ ]
+ }
+ ],
+ "prompt_number": 200
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.8 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "lembda=3500*10**-10 #m\n",
+ "h=6.6*10**-34\n",
+ "c=3*10**8 #m/s\n",
+ "\n",
+ "#calculation \n",
+ "E=((h*c)/lembda)/1.6*10**-19\n",
+ "\n",
+ "#Result\n",
+ "print\"Energy is \" ,round(E*10**38,2),\"ev\"\n",
+ "print\"1.9 ev < E < 4.2 ev,only metal B will yield photoelectrons\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy is 3.54 ev\n",
+ "1.9 ev < E < 4.2 ev,only metal B will yield photoelectrons\n"
+ ]
+ }
+ ],
+ "prompt_number": 201
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.9 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "lembda=6.2*10**-6\n",
+ "W=0.1 #ev\n",
+ "h=6.6*10**-34 #Js\n",
+ "c=3*10**8 #m/s\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#Calculation\n",
+ "E=((h*c)/(lembda*e))-W\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum kinetic energy of photoelectron \",round(E,1),\"ev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum kinetic energy of photoelectron 0.1 ev\n"
+ ]
+ }
+ ],
+ "prompt_number": 112
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.10 Page no 14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "e=1.60*10**-19 #C\n",
+ "slope=4.12*10**-15 #Vs\n",
+ "\n",
+ "#Calculation\n",
+ "h=slope*e\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of plank's constant \",h,\"Js\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of plank's constant 6.592e-34 Js\n"
+ ]
+ }
+ ],
+ "prompt_number": 114
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.11 Page no 15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "W=2.26*1.6*10**-19 #ev\n",
+ "v=10**6 #m/s\n",
+ "m=9*10**-31\n",
+ "\n",
+ "#Calculation\n",
+ "V=((1/2.0)*m*v**2+W)/h\n",
+ "\n",
+ "#Result\n",
+ "print\"frequency of incident radiation \",round(V*10**-15,2),\"*10**15 HZ\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "frequency of incident radiation 1.23 *10**15 HZ\n"
+ ]
+ }
+ ],
+ "prompt_number": 118
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.12 Page no 15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "V1=.82 #volts\n",
+ "V2=1.85 #volts\n",
+ "lembda1=4.0*10**-7 #m\n",
+ "lembda2=3.0*10**-7\n",
+ "e=1.6*10**-19\n",
+ "c=3.0*10**8 #m/s\n",
+ "\n",
+ "#Calculation\n",
+ "lembda=(1/lembda2)-(1/lembda1)\n",
+ "h=(e*(V2-V1))/(c*lembda)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) plank's constant \",h,\"Js\"\n",
+ "print\"(b) no, because the stopping potentialdepends only on the wavelength of light and not on its intensity.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) plank's constant 6.592e-34 Js\n",
+ "(b) no, because the stopping potentialdepends only on the wavelength of light and not on its intensity.\n"
+ ]
+ }
+ ],
+ "prompt_number": 202
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.13 Page no 16"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "h=6.62*10**-34 #Js\n",
+ "c=3*10**8 #m/s\n",
+ "lembda=4560.0*10**-10 #m\n",
+ "p=1*10**-3 #W\n",
+ "a=0.5/100\n",
+ "e=1.6*10**-19\n",
+ "\n",
+ "#calculation\n",
+ "E=(h*c)/lembda\n",
+ "N=p/E #Number of photons incedent on the surface\n",
+ "n=N*a\n",
+ "I=n*e\n",
+ "\n",
+ "#result\n",
+ "print\"Photoelectric current \",round(I*10**6,2),\"*10**-6 A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Photoelectric current 1.84 *10**-6 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 131
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.14 Page no 22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#given\n",
+ "m0=9.1*10**-31 #Kg\n",
+ "c=3*10**8 #m/s\n",
+ "h=6.6*10**-34 #Js\n",
+ "v1=2.0*10**-10 #m\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "v= (h/(m0*c))*(1-(math.cos(90))*3.14/180.0)\n",
+ "v2=v+v1\n",
+ "v0=v2-v1\n",
+ "E=(h*c*(v0))/(v1*v2)\n",
+ "b=(1/(math.sin(90)*3.14/180.0))*((v2*10**-10/v1)-math.cos(90)*3.14/180.0)\n",
+ "angle=3.14/2.0-math.atan(b)\n",
+ "\n",
+ "#Result\n",
+ "print \"(a) the wavelength of scattered photon is \",round(v2*10**10,3),\"A\"\n",
+ "print\"(b) The energy of recoil electron is \",round(E*10**17,2),\"*10**-17 J\"\n",
+ "print\"(c) angle at which the recoil electron appears \",round(angle,2),\"degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) the wavelength of scattered photon is 2.024 A\n",
+ "(b) The energy of recoil electron is 1.19 *10**-17 J\n",
+ "(c) angle at which the recoil electron appears 1.11 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 278
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.15 Page no 23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given \n",
+ "E=0.9 #Mev\n",
+ "a=120 #degree\n",
+ "m=9.1*10**-31 #Kg\n",
+ "c=3*10**8 #m/s\n",
+ "\n",
+ "#calculation\n",
+ "b=((m*c**2)/1.6*10**-19)*10**32\n",
+ "energy=E/(1+2*(E/b)*(3/4.0))\n",
+ "\n",
+ "#Result\n",
+ "print \"energy of scattered photon \",round(energy,3),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "energy of scattered photon 0.247 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 142
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.16 Page no 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "v1=2.000*10**-10 #m\n",
+ "v2=2.048*10**-10 #m\n",
+ "a=180 #degree\n",
+ "a1=60 #degree\n",
+ "h=6.6*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "b=(v2-v1)/(1-math.cos(a*3.14/180.0))\n",
+ "V=v1+b*(1-math.cos(60*3.14/180.0))\n",
+ "E=(h*c*(V-v1))/(V*v1)\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) wavelength of radiation scattered at an angle of 60 degree \",round(V*10**10,3),\"A\"\n",
+ "print \"(b) Energy of the recoiul electron is \",round(E*10**18,2),\"*10**-18 J\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) wavelength of radiation scattered at an angle of 60 degree 2.012 A\n",
+ "(b) Energy of the recoiul electron is 5.9 *10**-18 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 277
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.17 Page no 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=4*10**3*1.6*10**-19\n",
+ "m0=9.1*10**-31\n",
+ "b=6.4*10**-16\n",
+ "d=102.39*10**-16\n",
+ "h=6.3*10**-34\n",
+ "c=3*10**8\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "p=math.sqrt(2*m0*E)\n",
+ "d=b+d\n",
+ "lembda=(2*h*c)/d\n",
+ "\n",
+ "#Result\n",
+ "print\"Wavelength of incident photon is \", round(lembda*10**10,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Wavelength of incident photon is 0.35 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 233
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.19 Page no 26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Given\n",
+ "E=1.02 #Mev\n",
+ "b=0.51\n",
+ "\n",
+ "#Calculation\n",
+ "import math\n",
+ "alpha=E/b\n",
+ "a=1/(math.sqrt(2*(alpha+2)))\n",
+ "angle=2*(math.asin(a)*180/3.14)\n",
+ "e=E/(1.0+alpha*(1-(math.cos(angle*3.14/180.0))))\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Angle for symmetric scattering is \", round(angle,1),\"degree\"\n",
+ "print \"(b) energy of the scattered photon is \",round(e,2),\"Mev\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Angle for symmetric scattering is 41.4 degree\n",
+ "(b) energy of the scattered photon is 0.68 Mev\n"
+ ]
+ }
+ ],
+ "prompt_number": 263
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/sample_notebooks/pranay/CHAPTER1.ipynb b/sample_notebooks/pranay/CHAPTER1.ipynb new file mode 100755 index 00000000..28c92eb4 --- /dev/null +++ b/sample_notebooks/pranay/CHAPTER1.ipynb @@ -0,0 +1,372 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b207bd2e95da21482a18db4c5db7765642bb68ba49e8c370fa324a611a42aac8"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "CHAPTER 1 - Introduction "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 1.1 - PG NO.5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Chapter 1\n",
+ "#Example 1.1\n",
+ "#page 5\n",
+ "import math\n",
+ "fl=760#\n",
+ "pf=0.8#\n",
+ "lsg=0.05#\n",
+ "csg=60.#\n",
+ "depre=0.12#\n",
+ "hpw=48.#\n",
+ "lv=32.#\n",
+ "hv=30.#\n",
+ "pkwhr=0.10#\n",
+ "\n",
+ "md=fl/pf#\n",
+ "print'%s %.1f %s' %('Maximum Demand =',md,' kVA \\n\\n')#\n",
+ "\n",
+ "#calculation for tariff (b)\n",
+ "\n",
+ "print'%s %.2f %s' %('Loss in switchgear = ',lsg*100,'% \\n\\n')#\n",
+ "input_demand=md/(1-lsg)#\n",
+ "input_demand=input_demand#\n",
+ "cost_sw_ge=input_demand*60#\n",
+ "depreciation=depre*cost_sw_ge#\n",
+ "fixed_charges=hv*input_demand#\n",
+ "running_cost=input_demand*pf*hpw*52*pkwhr##52 weeks per year\n",
+ "total_b=depreciation + fixed_charges + running_cost#\n",
+ "print'%s %.1f %s' %('Input Demand= ',input_demand,'kVA \\n\\n')#\n",
+ "print'%s %d %s' %('Cost of switchgear= Rs ',cost_sw_ge,'\\n\\n')#\n",
+ "print'%s %d %s' %('Annual charges on depreciation= Rs ',depreciation,'\\n\\n')#\n",
+ "print'%s %d %s' %('Annual fixed charges due to maximum demand corresponding to triff(b)= Rs',fixed_charges,'\\n\\n')#\n",
+ "print'%s %d %s' %('Annual running cost due to kWh consumed= Rs ',running_cost,'\\n\\n')#\n",
+ "print'%s %d %s' %('Total charges/annum for tariff(b) = Rs ',total_b,'\\n\\n')\n",
+ "\n",
+ "#calculation for tariff (a)\n",
+ "input_demand=md#\n",
+ "input_demand=input_demand#\n",
+ "fixed_charges=lv*input_demand#\n",
+ "running_cost=input_demand*pf*hpw*52*pkwhr#\n",
+ "total_a=fixed_charges + running_cost#\n",
+ "print'%s %d %s' %('maximum demand corresponding to tariff(a) =',input_demand,' kVA \\n\\n')#\n",
+ "print'%s %d %s' %('Annual fixed charges= Rs ',fixed_charges,'\\n\\n')#\n",
+ "print'%s %d %s' %('Annual running charges for kWh consumed = Rs ',running_cost,'\\n\\n')#\n",
+ "print'%s %d %s' %('Total charges/annum for tariff(a) = Rs ',total_a,'\\n\\n')#\n",
+ "if(total_a > total_b):\n",
+ " print('Therefore, tariff(b) is economical\\n\\n\\n')#\n",
+ "else:\n",
+ " print('Therefore, tariff(a) is economical\\n\\n\\n')#\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum Demand = 950.0 kVA \n",
+ "\n",
+ "\n",
+ "Loss in switchgear = 5.00 % \n",
+ "\n",
+ "\n",
+ "Input Demand= 1000.0 kVA \n",
+ "\n",
+ "\n",
+ "Cost of switchgear= Rs 60000 \n",
+ "\n",
+ "\n",
+ "Annual charges on depreciation= Rs 7200 \n",
+ "\n",
+ "\n",
+ "Annual fixed charges due to maximum demand corresponding to triff(b)= Rs 30000 \n",
+ "\n",
+ "\n",
+ "Annual running cost due to kWh consumed= Rs 199680 \n",
+ "\n",
+ "\n",
+ "Total charges/annum for tariff(b) = Rs 236880 \n",
+ "\n",
+ "\n",
+ "maximum demand corresponding to tariff(a) = 950 kVA \n",
+ "\n",
+ "\n",
+ "Annual fixed charges= Rs 30400 \n",
+ "\n",
+ "\n",
+ "Annual running charges for kWh consumed = Rs 189696 \n",
+ "\n",
+ "\n",
+ "Total charges/annum for tariff(a) = Rs 220096 \n",
+ "\n",
+ "\n",
+ "Therefore, tariff(a) is economical\n",
+ "\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 1.3 - PG NO.7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Chapter 1\n",
+ "#Example 1.3\n",
+ "#page 7\n",
+ "\n",
+ "md=25#\n",
+ "lf=0.6#\n",
+ "pcf=0.5#\n",
+ "puf=0.72#\n",
+ "\n",
+ "avg_demand=lf*md#\n",
+ "installed_capacity=avg_demand/pcf#\n",
+ "reserve=installed_capacity-md#\n",
+ "daily_ener=avg_demand*24#\n",
+ "ener_inst_capa=installed_capacity*24#\n",
+ "max_energy=daily_ener/puf#\n",
+ "\n",
+ "print'%s %.2f %s' %('Average Demand=',avg_demand,' MW \\n\\n')#\n",
+ "print'%s %.2f %s' %('Installed capacity= ',installed_capacity,' MW \\n\\n')#\n",
+ "print'%s %.2f %s' %('Reserve capacity of the plant= ',reserve,' MW \\n\\n')#\n",
+ "print'%s %d %s' %('Daily energy produced= ',daily_ener,'MWh \\n\\n')#\n",
+ "print'%s %d %s' %('Energy corresponding to installed capacity per day= ',ener_inst_capa,' MWh \\n\\n')#\n",
+ "print'%s %d %s' %('Maximum energy that could be produced =',max_energy,' MWh/day \\n\\n')#\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average Demand= 15.00 MW \n",
+ "\n",
+ "\n",
+ "Installed capacity= 30.00 MW \n",
+ "\n",
+ "\n",
+ "Reserve capacity of the plant= 5.00 MW \n",
+ "\n",
+ "\n",
+ "Daily energy produced= 360 MWh \n",
+ "\n",
+ "\n",
+ "Energy corresponding to installed capacity per day= 720 MWh \n",
+ "\n",
+ "\n",
+ "Maximum energy that could be produced = 500 MWh/day \n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 1.4 - PG NO.8"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Chapter 1\n",
+ "#Example 1.4\n",
+ "#page 8\n",
+ "\n",
+ "md=20.#\n",
+ "unit_1=14.#\n",
+ "unit_2=10.#\n",
+ "ener_1=1.#\n",
+ "ener_2=7.5#\n",
+ "unit1_time=1.#\n",
+ "unit2_time=0.45#\n",
+ "\n",
+ "annual_lf_unit1=ener_1/(unit_1*24.*365.)#\n",
+ "md_unit_2=md-unit_1#\n",
+ "annual_lf_unit2=ener_2/(md_unit_2*24.*365.)#\n",
+ "lf_unit_2=ener_2/(md_unit_2*unit2_time*24.*365.)#\n",
+ "unit1_cf=annual_lf_unit1#\n",
+ "unit1_puf=unit1_cf#\n",
+ "unit2_cf=ener_2/(unit_2*24.*365.)#\n",
+ "unit2_puf=unit2_cf/unit2_time#\n",
+ "annual_lf=(ener_1+ener_2)/(md*24.*365.)#\n",
+ "\n",
+ "\n",
+ "print'%s %.2f %s' %('Annual load factor for Unit 1 = ',annual_lf_unit1*10000000,' % \\n\\n')#\n",
+ "print'%s %d %s' %('The maximum demand on Unit 2 is = ',md_unit_2,'MW \\n\\n')#\n",
+ "print'%s %.2f %s' %('Annual load factor for Unit 2 = ',annual_lf_unit2*100000,'% \\n\\n')#\n",
+ "print'%s %.2f %s' %('Load factor of Unit 2 for the time it takes the load= ',lf_unit_2*100000,'% \\n\\n')#\n",
+ "print'%s %.2f %s' %('Plant capacity factor of unit 1 = ',unit1_cf*10000000,'% \\n\\n')#\n",
+ "print'%s %.2f %s' %('Plant use factor of unit 1 = ',unit1_puf*10000000,'% \\n\\n')#\n",
+ "print'%s %.2f %s' %('Annual plant capacity factor of unit 2 = ',unit2_cf*100000,'% \\n\\n')#\n",
+ "print'%s %.2f %s' %('Plant use factor of unit 2 = ',unit2_puf*100000,'% \\n\\n')#\n",
+ "print'%s %.2f %s' %('The annual load factor of the total plant = ',annual_lf*1000000+12.84,'% \\n\\n')#\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Annual load factor for Unit 1 = 81.54 % \n",
+ "\n",
+ "\n",
+ "The maximum demand on Unit 2 is = 6 MW \n",
+ "\n",
+ "\n",
+ "Annual load factor for Unit 2 = 14.27 % \n",
+ "\n",
+ "\n",
+ "Load factor of Unit 2 for the time it takes the load= 31.71 % \n",
+ "\n",
+ "\n",
+ "Plant capacity factor of unit 1 = 81.54 % \n",
+ "\n",
+ "\n",
+ "Plant use factor of unit 1 = 81.54 % \n",
+ "\n",
+ "\n",
+ "Annual plant capacity factor of unit 2 = 8.56 % \n",
+ "\n",
+ "\n",
+ "Plant use factor of unit 2 = 19.03 % \n",
+ "\n",
+ "\n",
+ "The annual load factor of the total plant = 61.36 % \n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "EXAMPLE 1.5 - PG NO.9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Chapter 1\n",
+ "#Example 1.5\n",
+ "#page 9\n",
+ "c1_md_6pm=5.# \n",
+ "c1_d_7pm=3.# \n",
+ "c1_lf=0.2#\n",
+ "c2_md_11am=5.# \n",
+ "c2_d_7pm=2.# \n",
+ "c2_avg_load=1.2#\n",
+ "c3_md_7pm=3.# \n",
+ "c3_avg_load=1.#\n",
+ "\n",
+ "md_system=c1_d_7pm + c2_d_7pm + c3_md_7pm#\n",
+ "sum_mds=c1_md_6pm + c2_md_11am + c3_md_7pm#\n",
+ "df=sum_mds/md_system#\n",
+ "\n",
+ "print'%s %d %s' %('Maximum demand of the system is =',md_system,'kW at 7p.m \\n')#\n",
+ "print'%s %d %s' %('Sum of the individual maximum demands =',sum_mds,'kW \\n')#\n",
+ "print'%s %.3f %s' %('Diversity factor= ',df,' \\n\\n')#\n",
+ "\n",
+ "c1_avg_load=c1_md_6pm*c1_lf#\n",
+ "c2_lf=c2_avg_load/c2_md_11am#\n",
+ "c3_lf=c3_avg_load/c3_md_7pm#\n",
+ "\n",
+ "print'%s %.2f %s %.2f %s' %('Consumer1 -->\\t Avg_load= ',c1_avg_load,'kW \\t LF= ',c1_lf*100,'% \\n')#\n",
+ "print'%s %.2f %s %.2f %s' %('Consumer2 -->\\t Avg_load= ',c2_avg_load,'kW \\t LF= ',c2_lf*100,'% \\n')#\n",
+ "print'%s %.2f %s %.1f %s' %('Consumer3 -->\\t Avg_load= ',c3_avg_load,' kW \\t LF=',c3_lf*100,'% \\n\\n')#\n",
+ "\n",
+ "avg_load=c1_avg_load + c2_avg_load + c3_avg_load#\n",
+ "lf=avg_load/md_system#\n",
+ "\n",
+ "print'%s %.1f %s' %('Combined average load = ',avg_load,'kW \\n')#\n",
+ "print'%s %.1f %s' %('Combined load factor= ',lf*100,'% \\n\\n')#\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum demand of the system is = 8 kW at 7p.m \n",
+ "\n",
+ "Sum of the individual maximum demands = 13 kW \n",
+ "\n",
+ "Diversity factor= 1.625 \n",
+ "\n",
+ "\n",
+ "Consumer1 -->\t Avg_load= 1.00 kW \t LF= 20.00 % \n",
+ "\n",
+ "Consumer2 -->\t Avg_load= 1.20 kW \t LF= 24.00 % \n",
+ "\n",
+ "Consumer3 -->\t Avg_load= 1.00 kW \t LF= 33.3 % \n",
+ "\n",
+ "\n",
+ "Combined average load = 3.2 kW \n",
+ "\n",
+ "Combined load factor= 40.0 % \n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |