diff options
27 files changed, 11379 insertions, 0 deletions
diff --git a/Electronic_Circuits_by_M._H._Tooley/Chapter12_2.ipynb b/Electronic_Circuits_by_M._H._Tooley/Chapter12_2.ipynb new file mode 100644 index 00000000..2b19d126 --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/Chapter12_2.ipynb @@ -0,0 +1,152 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:99e910301f8d94e773f6d9f81e6e674a29d03eeb4decca35ea0a1abf9f727037"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter12-The 555 timer"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:12.1\n",
+ "import math\n",
+ "C=100.*10**-9;\n",
+ "t_on=10.*10**-3;\n",
+ "R=(t_on/(1.1*C))/1000.;\n",
+ "print'%s %.2f %s'%(\"R= \",R,\" kohm\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "R= 90.91 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:12.2\n",
+ "import math\n",
+ "C=100.*10**-6;\n",
+ "t_on=60.;\n",
+ "R=(t_on/(1.1*C))/1000.;\n",
+ "print'%s %.2f %s'%(\"R= \",R,\" kohm\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "R= 545.45 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg223"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:12.3\n",
+ "import math\n",
+ "##R1=R2=R\n",
+ "prf=10.;\n",
+ "C=1.*10**-6;\n",
+ "R=0.48/(prf*C);\n",
+ "print'%s %.2f %s'%(\"R= \",R,\" ohm\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "R= 48000.00 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg223"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:12.4\n",
+ "import math\n",
+ "prf=50.;\n",
+ "C=100*10**-9;\n",
+ "R=0.72/(prf*C);##in ohms\n",
+ "print'%s %.2f %s'%(\"R= \",R/1000,\" kohm\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "R= 144.00 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuits_by_M._H._Tooley/Chapter13_2.ipynb b/Electronic_Circuits_by_M._H._Tooley/Chapter13_2.ipynb new file mode 100644 index 00000000..8f1f8a80 --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/Chapter13_2.ipynb @@ -0,0 +1,289 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d56d12390f39b518475a37bfc6bbb499a3e335e6c5fa1b48bf6416cc300a4e24"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter13-Radio"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:13.1\n",
+ "import math\n",
+ "c=3.*10**8;\n",
+ "wl=15.;\n",
+ "f=c/wl;\n",
+ "print'%s %.2f %s'%(\"The frequency = \",f,\" Hz\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The frequency = 20000000.00 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:13.2\n",
+ "import math\n",
+ "c=3.*10**8;\n",
+ "f=150.*10**6;\n",
+ "wl=c/f;\n",
+ "print'%s %.2f %s'%(\"The wavelength =\",wl,\" m\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The wavelength = 2.00 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:13.3\n",
+ "import math\n",
+ "wl=8.;\n",
+ "f=30.*10**6;\n",
+ "v=f*wl;\n",
+ "print'%s %.2f %s'%(\"The veocity of propagation = \",v,\" m/s\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The veocity of propagation = 240000000.00 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:13.4\n",
+ "import math\n",
+ "f_rf=162.5;##in kHz\n",
+ "f_af=1.25;##in kHz\n",
+ "f_bfo_max=f_rf+f_af;\n",
+ "f_bfo_min=f_rf-f_af;\n",
+ "print'%s %.2f %s %.2f %s '%(\"The two possible BFO freq. = \",f_bfo_max,\" kHz\" and \"\",f_bfo_min,\"kHz\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The two possible BFO freq. = 163.75 161.25 kHz \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:13.5\n",
+ "import math\n",
+ "f_rf_1=88.;##in MHz\n",
+ "f_rf_2=108.;##in MHz\n",
+ "f_if=10.7;##in MHz\n",
+ "f_lo_1=f_rf_1+f_if;\n",
+ "f_lo_2=f_rf_2+f_if;\n",
+ "print'%s %.2f %s %.2f %s '%(\"The range local oscillator be tuned = \",f_lo_1,\" MHz\" and \"\",f_lo_2,\" MHz\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The range local oscillator be tuned = 98.70 118.70 MHz \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg238"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:13.6\n",
+ "import math\n",
+ "f_rf_1=88.;##in MHz\n",
+ "f_rf_2=108.;##in MHz\n",
+ "f_if=10.7;##in MHz\n",
+ "f_lo_1=f_rf_1+f_if;\n",
+ "f_lo_2=f_rf_2+f_if;\n",
+ "print'%s %.2f %s %.2f %s '%(\"The range local oscillator be tuned = \",f_lo_1,\" MHz\" and \"\",f_lo_2,\"MHz\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The range local oscillator be tuned = 98.70 118.70 MHz \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:13.7\n",
+ "import math\n",
+ "r=12.;##in ohms\n",
+ "i=0.5;##in amps\n",
+ "P_r=i*i*r;##in W\n",
+ "print'%s %.2f %s'%(\"Power radiated = \",P_r,\" W\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power radiated = 3.00 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:13.8\n",
+ "import math\n",
+ "r=2.;##in ohms\n",
+ "i=0.5;##in amps\n",
+ "P_r=4.;##in W\n",
+ "P_loss=i*i*r;\n",
+ "P_eff=(P_r/(P_r+P_loss))*100.;\n",
+ "print'%s %.2f %s'%(\"The power loss = \",P_loss,\" W\");\n",
+ "print'%s %.2f %s'%(\"\\n The power loss = \",P_eff,\"\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The power loss = 0.50 W\n",
+ "\n",
+ " The power loss = 88.89 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuits_by_M._H._Tooley/Chapter1_2.ipynb b/Electronic_Circuits_by_M._H._Tooley/Chapter1_2.ipynb new file mode 100644 index 00000000..5c47a9e8 --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/Chapter1_2.ipynb @@ -0,0 +1,770 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b9799d17ce9c5e8acfde2941052198091704f3207bfb93e90879b4a048464a60"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1-Electrical Fundamentals"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg3"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa:1.4\n",
+ "import math\n",
+ "ang_d=215.;##given\n",
+ "ang_r=ang_d*math.pi/180.;\n",
+ "print'%s %.2f %s %.2f %s '%(\"%f degree angle is \",ang_d,\" radians\" and \"\",ang_r,\"\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "%f degree angle is 215.00 3.75 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg3"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa:1.5\n",
+ "import math\n",
+ "ang_r=2.5;##given\n",
+ "ang_d=2.5*180./math.pi;##angle in degrees\n",
+ "print'%s %.2f %s %.2f %s '%(\"%f degree angle is \",ang_d,\" radians\" and \"\",ang_r,\"\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "%f degree angle is 143.24 2.50 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa:1.6\n",
+ "import math\n",
+ "i_amp=0.075;##given\n",
+ "i_milamp=i_amp*1000.;##current in milliamp.\n",
+ "print'%s %.2f %s'%(\"%f amp current is \",i_milamp,\" mA\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "%f amp current is 75.00 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa:1.7\n",
+ "import math\n",
+ "fq_khz=1495.;##given\n",
+ "fq_Mhz=fq_khz/1000.;\n",
+ "print'%s %.2f %s'%(\" kHz frequency is \",fq_Mhz,\" MHz\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " kHz frequency is 1.50 MHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa:1.8\n",
+ "import math\n",
+ "c_pF=27000.;##given\n",
+ "c_uF=c_pF/1000.;\n",
+ "print'%s %.2f %s'%(\"picofarad capacitance is \",c_uF,\" microfarad\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "picofarad capacitance is 27.00 microfarad\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa:1.9\n",
+ "import math\n",
+ "c_mA=7.25;##given\n",
+ "c_A=c_mA*1000.;\n",
+ "print'%s %.2f %s'%(\" milliampere current is \",c_A,\" ampere\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " milliampere current is 7250.00 ampere\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa:1.10\n",
+ "import math\n",
+ "vg_v=3.75*10**-6;##given\n",
+ "vg_mv=vg_v*1000.;\n",
+ "print'%s %.2e %s'%(\" volt voltage is \",vg_mv,\" mV\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " volt voltage is 3.75e-03 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.11\n",
+ "import math\n",
+ "r=33000.;##in ohms\n",
+ "i=0.003;##in amp\n",
+ "v=i*r;\n",
+ "print'%s %.2f %s'%(\"Voltage dropped = \",v,\" volts\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage dropped = 99.00 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.12\n",
+ "import math\n",
+ "t=20.*10**-3;##in sec\n",
+ "i=45.*10**-6;##in amp\n",
+ "q=i*t*10**9;\n",
+ "print'%s %.2f %s'%(\"Charge transferred = \",q,\" nC\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge transferred = 900.00 nC\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.13\n",
+ "import math\n",
+ "p=0.3;##in watts\n",
+ "v=1500.;##in volts\n",
+ "i=(p/v)*10**6;\n",
+ "print'%s %.2f %s'%(\"Current supplied = \",i,\" microamp\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current supplied = 200.00 microamp\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.14\n",
+ "import math\n",
+ "r=12.;##in ohms\n",
+ "v=6.;##in volts\n",
+ "i=(v/r);\n",
+ "print'%s %.2f %s'%(\"Current = \",i,\" Amp\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current = 0.50 Amp\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.15\n",
+ "import math\n",
+ "r=56.;##in ohms\n",
+ "i=0.1;##in amp\n",
+ "v=i*r;\n",
+ "print'%s %.2f %s'%(\"Voltage dropped = \",v,\" volts\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage dropped = 5.60 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.16\n",
+ "import math\n",
+ "v=15.;##in volts\n",
+ "i=0.001;##in amp\n",
+ "r=v/i;\n",
+ "print'%s %.2f %s'%(\"Resistance = \",r,\" ohms\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance = 15000.00 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.17\n",
+ "import math\n",
+ "p=1.724*10**-8;##in ohm-meter\n",
+ "l=8.;##in meters\n",
+ "a=1.*10**-6;##in sq. meter\n",
+ "r=(p*l)/a;\n",
+ "print'%s %.2f %s'%(\"Resistance = \",r,\" ohms\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance = 0.14 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex18-pg8"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.18\n",
+ "import math\n",
+ "p=1.724*10**-8;##in ohm-meter\n",
+ "l=20.;##in meters\n",
+ "a=1.*10**-6;##in sq. meter\n",
+ "i=5.;##in amperes\n",
+ "r=(p*l)/a;\n",
+ "v=i*r;\n",
+ "print'%s %.2f %s'%(\"Voltage dropped = \",v,\" volts\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage dropped = 1.72 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex19-pg9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.19\n",
+ "import math\n",
+ "v=3.;##in volts\n",
+ "i=1.5;##in amperes\n",
+ "p=v*i;\n",
+ "print'%s %.2f %s'%(\"Power supplied = \",p,\" watts\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power supplied = 4.50 watts\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex20-pg9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.20\n",
+ "import math\n",
+ "v=4.;##in volts\n",
+ "r=100.;##in ohms\n",
+ "p=(v**2)/r;\n",
+ "print'%s %.2f %s'%(\"Power dissipated = \",p,\" watts\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power dissipated = 0.16 watts\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex21-pg9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.21\n",
+ "import math\n",
+ "i=20.*10**-3;##in amps\n",
+ "r=1000.;##in ohms\n",
+ "p=(i**2)*r;\n",
+ "print'%s %.2f %s'%(\"Power dissipated = \",p,\" watts\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power dissipated = 0.40 watts\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex22-pg10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.22\n",
+ "import math\n",
+ "v=600;##in volts\n",
+ "d=25*10^-3;##in meters\n",
+ "E=(v)/d;\n",
+ "print'%s %.2f %s'%(\"Electric Field Strength = \",E/10000,\" kV/m\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electric Field Strength = -1.00 kV/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex23-pg13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.23\n",
+ "import math\n",
+ "u=4.*math.pi*10**-7;##in H/m\n",
+ "i=20.;##in amps\n",
+ "d=50.*10**-3;##in meters\n",
+ "B=(u*i)/(2.*math.pi*d);\n",
+ "print'%s %.2e %s'%(\"Flux Density = \",B,\" Tesla\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Flux Density = 8.00e-05 Tesla\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex24-pg13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.24\n",
+ "import math\n",
+ "B=(2.5*10**-3);##in Tesla\n",
+ "a=(20.*10**-4);##in sq. meter\n",
+ "flux=B*a;\n",
+ "print'%s %.2e %s'%(\"Flux = \",flux,\" webers\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Flux = 5.00e-06 webers\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex25-pg15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.25\n",
+ "import math\n",
+ "B1=0.6;##in Tesla\n",
+ "u1=B1/800.;\n",
+ "u_r1=u1/(4.*math.pi*10**-7);\n",
+ "print'%s %.2f %s'%(\"reltive permitivity at 0.6T = \",u_r1,\"\");\n",
+ "B2=1.6;##in Tesla\n",
+ "u2=0.2/4000.;\n",
+ "u_r2=u2 /(4.*math.pi*10**-7);\n",
+ "print'%s %.2f %s'%(\"\\n reltive permitivity at 1.6T = \",u_r2,\"\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "reltive permitivity at 0.6T = 596.83 \n",
+ "\n",
+ " reltive permitivity at 1.6T = 39.79 \n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex26-pg16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:1.26\n",
+ "import math\n",
+ "flux=0.8*10**-3;\n",
+ "a=(500.*10**-6);##in sq. meter\n",
+ "l=0.6;##in meter\n",
+ "N=800.;\n",
+ "B=flux/a;\n",
+ "print'%s %.2e %s'%(\"Flux Density = \",B,\" Tesla\");\n",
+ "H=3500.;##in A/m\n",
+ "i=(H*l)/N;\n",
+ "print'%s %.2f %s'%(\"\\n Current required = \",i,\" amp.s\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Flux Density = 1.60e+00 Tesla\n",
+ "\n",
+ " Current required = 2.62 amp.s\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuits_by_M._H._Tooley/Chapter2_2.ipynb b/Electronic_Circuits_by_M._H._Tooley/Chapter2_2.ipynb new file mode 100644 index 00000000..20419576 --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/Chapter2_2.ipynb @@ -0,0 +1,1045 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b5194708e2447220722c4dfa1402fc797342ce58013c4fb6e50908b3fd6172f1"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Passive Components "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.1\n",
+ "import math\n",
+ "marked=220.;##in ohms\n",
+ "measured=207.;##in ohms\n",
+ "err=marked-measured;\n",
+ "tol=(err/marked)*100.;\n",
+ "print'%s %.2f %s'%(\"Tolerance = \",tol,\"\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Tolerance = 5.91 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.2\n",
+ "import math\n",
+ "r=39.;##in ohms\n",
+ "v=9.;##in volts\n",
+ "i=(v/r);##in Amps\n",
+ "print'%s %.2f %s'%(\"Current = \",i*1000,\" mA\");\n",
+ "tol=0.1;##i.e, 10%\n",
+ "r_min=r-(tol*r);\n",
+ "i_max=v/r_min;\n",
+ "r_max=r+(tol*r);\n",
+ "i_min=v/r_max;\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\n Max.Current = \",i_max*1000,\" mA \" and \"\",i_min*1000,\" Min Current\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current = 230.77 mA\n",
+ "\n",
+ " Max.Current = 256.41 209.79 Min Current \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.3\n",
+ "import math\n",
+ "v=28.;##in volts\n",
+ "i=0.1;##in A\n",
+ "r=v/i;\n",
+ "p=v*i;\n",
+ "print'%s %.2f %s %.2f %s '%(\"Resistance Value = \",r,\" ohms & Power dissipated\"\" = \",p,\" W\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance Value = 280.00 ohms & Power dissipated = 2.80 W \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.4\n",
+ "import math\n",
+ "r=10*(10**2);\n",
+ "print'%s %.2f %s'%(\"Resistor value = \",r,\" ohm\");\n",
+ "print(\"\\nTolerance = 10 \");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistor value = 1000.00 ohm\n",
+ "\n",
+ "Tolerance = 10 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.5\n",
+ "import math\n",
+ "r=27.*(10**3);\n",
+ "print'%s %.2f %s'%(\"Resistor value = \",r,\" ohm\");\n",
+ "print(\"\\nTolerance = 5 \");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistor value = 27000.00 ohm\n",
+ "\n",
+ "Tolerance = 5 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.6\n",
+ "import math\n",
+ "r=56*(10);\n",
+ "print'%s %.2f %s'%(\"Resistor value = \",r,\" ohm\");\n",
+ "print(\"\\nTolerance = 5 \");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistor value = 560.00 ohm\n",
+ "\n",
+ "Tolerance = 5 \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.7\n",
+ "import math\n",
+ "r=25.*(10**0);\n",
+ "print'%s %.2f %s'%(\"Resistor value = \",r,\" ohm\");\n",
+ "print(\"\\nTolerance = 20 \");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistor value = 25.00 ohm\n",
+ "\n",
+ "Tolerance = 20 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.8\n",
+ "import math\n",
+ "r=22.*(10**3);\n",
+ "print(\"Bands are Red, Red, Red, Red\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bands are Red, Red, Red, Red\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.9\n",
+ "import math\n",
+ "print(\"Resistance = 4.7 ohm with 10%% tolerance\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance = 4.7 ohm with 10%% tolerance\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.10\n",
+ "import math\n",
+ "print(\"Resistance = 330 ohms with 2%% tolerance\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance = 330 ohms with 2%% tolerance\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.11\n",
+ "import math\n",
+ "print(\"Resistance = 0.22 ohm with 20%% tolerance\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance = 0.22 ohm with 20%% tolerance\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.12\n",
+ "import math\n",
+ "r1=22.;##in ohms\n",
+ "r2=47.;##in ohms\n",
+ "r3=33.;##in ohms\n",
+ "r_ser=r1+r2+r3;\n",
+ "print'%s %.2f %s'%(\"Effective resistance in series = \",r_ser,\" ohms\");\n",
+ "r_parel=((1./r1)+(1./r2)+(1./r3))**-1;\n",
+ "print'%s %.2f %s'%(\"\\n Effective resistance in parallel = \",r_parel,\" ohms\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Effective resistance in series = 102.00 ohms\n",
+ "\n",
+ " Effective resistance in parallel = 10.31 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.13\n",
+ "import math\n",
+ "r1=4.7;##in ohms\n",
+ "r2=47.;##in ohms\n",
+ "r3=12.;##in ohms\n",
+ "r4=27.;##in ohms\n",
+ "r5=r3+r4;\n",
+ "r_parel=((1./r5)+(1./r2))**-1;\n",
+ "r_eff=r_parel+r1;\n",
+ "print'%s %.2f %s'%(\"Effective resistance = \",r_eff,\" ohms\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Effective resistance = 26.01 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.14\n",
+ "import math\n",
+ "print(\"Two 100 ohm resistor of 1 W\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Two 100 ohm resistor of 1 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.15\n",
+ "import math\n",
+ "temp_coeff=0.001;##in per degree centigrade\n",
+ "r_o=1500.;##in ohm\n",
+ "t=80.;##temperature diff.\n",
+ "r_t=r_o*(1.+(temp_coeff)*t)\n",
+ "print'%s %.2f %s'%(\"Resistance at \",r_t,\" degree = ohms\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance at 1620.00 degree = ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.16\n",
+ "import math\n",
+ "temp_coeff=0.0005;##in per degree centigrade\n",
+ "r_t1=680.;##in ohm\n",
+ "t1=20.;##temperature diff.\n",
+ "t2=90.;\n",
+ "r_o=r_t1/(1.+(temp_coeff)*t1);\n",
+ "r_t2=r_o*(1.+(temp_coeff)*t2);\n",
+ "print'%s %.2f %s %.2f %s '%(\"Resistance at \",t2,\" degree = \",r_t2,\" ohms\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistance at 90.00 degree = 703.56 ohms \n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.17\n",
+ "import math\n",
+ "r_o=40.;##resis at 0 degree\n",
+ "r_t=44.;##at 100 degree\n",
+ "t=100.;##temperature diff.\n",
+ "temp_coeff=(1./t)*((r_t/r_o)-1.);\n",
+ "print'%s %.2e %s'%(\"Temperature Coefficient = \",temp_coeff,\" per degree centigrade\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Temperature Coefficient = 1.00e-03 per degree centigrade\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex18-pg33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.18\n",
+ "import math\n",
+ "V_1=50.;\n",
+ "V_2=10.;\n",
+ "dV=V_1-V_2;##in volts\n",
+ "dt=0.1;##in seconds\n",
+ "C=22.*10**-6;\n",
+ "i=C*(dV/dt)*1000.;##in mA\n",
+ "print'%s %.2f %s'%(\"Current flow = \",i,\" milliAmps\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current flow = 8.80 milliAmps\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex19-pg33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.19\n",
+ "import math\n",
+ "C=10.*10**-6;\n",
+ "V=250.;##in volts\n",
+ "Q=V*C*1000.;##in millicoulomb\n",
+ "print'%s %.2f %s'%(\"Charged stored =\",Q,\" mC\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charged stored = 2.50 mC\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex20-pg33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.20\n",
+ "import math\n",
+ "C=47.*10**-6;##in farads\n",
+ "W=4.;##energy in joules\n",
+ "V=math.sqrt(W/(0.5*C));\n",
+ "print'%s %.2f %s'%(\"Voltage tht be applied = \",V,\" volts\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage tht be applied = 412.57 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex21-pg34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.21\n",
+ "import math\n",
+ "E_o=8.85*10**-12;\n",
+ "E_r=5.4;\n",
+ "C=1*10**-9;\n",
+ "d=0.1*10**-3;\n",
+ "A=(C*d)/(E_o*E_r)*10**4;\n",
+ "print'%s %.2f %s'%(\"Required plate area = \",A,\" sq. cm\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Required plate area = 20.92 sq. cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex22-pg34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.22\n",
+ "import math\n",
+ "E_o=8.85*10**-12;\n",
+ "E_r=4.5;\n",
+ "n=6.;##no. of plates\n",
+ "d=0.2*10**-3;##in meter\n",
+ "A=20.*10**-4;##in sq.meter\n",
+ "C=((E_o*E_r*(n-1.)*A)/d)*10**11;\n",
+ "print'%s %.2f %s'%(\"Capacitance = \",C,\" pF\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance = 199.12 pF\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex23-pg36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.23\n",
+ "import math\n",
+ "print(\"Capacitance = 10000 pF of 10%%\"); "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance = 10000 pF of 10%%\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex24-pg36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.24\n",
+ "import math\n",
+ "print(\"Capacitance = 150 pF of 2%% tolerance at 100 V\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance = 150 pF of 2%% tolerance at 100 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex25-pg37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.25\n",
+ "import math\n",
+ "C1=2.;##in nF\n",
+ "C2=4.;##in nF\n",
+ "C3=2.;\n",
+ "C4=4.;\n",
+ "C_a=C1+C2;\n",
+ "C_b=C_a*C3/(C_a+C3);\n",
+ "C_eff=C4+C_b;\n",
+ "print'%s %.2f %s'%(\"Capacitance = \",C_eff,\" nF\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance = 5.50 nF\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex26-pg37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.26\n",
+ "import math\n",
+ "C=100.;##in uF\n",
+ "C_eff=C*C/(C+C);\n",
+ "print'%s %.2f %s'%(\"Two capacitors of uF be in parallel used to make \",C_eff,\" uF capacitance\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Two capacitors of uF be in parallel used to make 50.00 uF capacitance\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex27-pg40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.27\n",
+ "import math\n",
+ "L=600.*10**-3;##in H\n",
+ "I1=6.;##in A\n",
+ "I2=2.;##in A\n",
+ "dI=I1-I2;\n",
+ "dt=250.*10**-3;##in sec.\n",
+ "E=-L*(dI/dt);\n",
+ "print'%s %.2f %s'%(\"Induced voltage = \",E,\" volts\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Induced voltage = -9.60 volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex28-pg40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.28\n",
+ "import math\n",
+ "E=2.5;##energy in joules\n",
+ "L=20.*10**-3;##in henry\n",
+ "I=math.sqrt(E/(0.5*L));\n",
+ "print'%s %.2f %s'%(\"Current = \",I,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current = 15.81 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex29-pg40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.29\n",
+ "import math\n",
+ "u_o=12.57*10**-7;\n",
+ "u_r=500.;\n",
+ "A=15.*10**-4;##area of cross-section in sq. meters\n",
+ "l=20.*10**-2;##length\n",
+ "L=100.*10**-3;##in henry\n",
+ "n=math.sqrt((L*l)/(u_r*u_o*A));\n",
+ "print'%s %.2f %s'%(\"Inductor requires \",n,\" turns of wire\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inductor requires 145.65 turns of wire\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex30-pg42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.30\n",
+ "import math\n",
+ "##L=(L1*L2)/(L1+L2)\n",
+ "L_eq=5.;##in millihenry\n",
+ "print'%s %.2f %s'%(\"Inductor of 10 mH wired in parallel would provide \",L_eq,\" mH\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inductor of 10 mH wired in parallel would provide 5.00 mH\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex31-pg42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:2.31\n",
+ "import math\n",
+ "L1=60.;##in mH\n",
+ "L2=60.;##in mH\n",
+ "L_a=L1+L2;\n",
+ "L3=120.;##in mH\n",
+ "L_b=L_a*L3/(L_a+L3);\n",
+ "L4=50.;##in mH\n",
+ "L_eq=L4+L_b;\n",
+ "print'%s %.2f %s'%(\"Equivalent Inductance = \",L_eq,\" mH\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Equivalent Inductance = 110.00 mH\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuits_by_M._H._Tooley/Chapter3_2.ipynb b/Electronic_Circuits_by_M._H._Tooley/Chapter3_2.ipynb new file mode 100644 index 00000000..781f46d7 --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/Chapter3_2.ipynb @@ -0,0 +1,586 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f23f1f12c5c13ad8f68288fd358f70344404a6fb4aa236ccdacf571419cfbed9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter3-DC Circuits"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.1\n",
+ "import math\n",
+ "i1=1.5;\n",
+ "i2=2.7;##in amp.s\n",
+ "i5=i1+i2;\n",
+ "i4=3.3;\n",
+ "i3=i4+i5;\n",
+ "print'%s %.2f %s'%(\"Current b/w A & B = \",i5,\" A\");\n",
+ "print'%s %.2f %s'%(\"\\n Current I3 = \",i3,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current b/w A & B = 4.20 A\n",
+ "\n",
+ " Current I3 = 7.50 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.2\n",
+ "import math\n",
+ "E1=6.;\n",
+ "E2=3.;\n",
+ "V2=E1-E2;\n",
+ "V1=4.5;\n",
+ "E3=V1-E2;\n",
+ "print'%s %.2f %s'%(\"Value of V2 = \",V2,\" A\");\n",
+ "print'%s %.2f %s'%(\"\\n Value of E3 = \",E3,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of V2 = 3.00 A\n",
+ "\n",
+ " Value of E3 = 1.50 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.3\n",
+ "import math\n",
+ "V1=7.5;##in volts\n",
+ "V2=4.5;\n",
+ "V3=4.5;\n",
+ "r1=110.;##in ohms\n",
+ "r2=33.;\n",
+ "r3=22.;\n",
+ "i1=V1/r1;\n",
+ "i2=V2/r2;\n",
+ "i3=V3/r3;\n",
+ "print'%s %.2f %s'%(\"Current I1 = \",i1,\" A\");\n",
+ "print'%s %.2f %s'%(\"\\n Current I2 = \",i2,\" A\");\n",
+ "print'%s %.2f %s'%(\"\\n Current I3 = \",i3,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current I1 = 0.07 A\n",
+ "\n",
+ " Current I2 = 0.14 A\n",
+ "\n",
+ " Current I3 = 0.20 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.4\n",
+ "import math\n",
+ "V_in=5.;##in volts\n",
+ "r1=4000.;\n",
+ "r2=1000.;\n",
+ "r_p=r1*r2/(r1+r2);\n",
+ "V_out=V_in*(r2/(r1+r2));\n",
+ "V_out_p=V_in*(r_p/(r_p+r2));\n",
+ "print'%s %.2f %s'%(\"output voltage at no load = \",V_out,\" A\");\n",
+ "print'%s %.2f %s'%(\"\\n output voltage when loaded by 10kohms = \",V_out_p,\"A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "output voltage at no load = 1.00 A\n",
+ "\n",
+ " output voltage when loaded by 10kohms = 2.22 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.5\n",
+ "import math\n",
+ "I_in=5.;##in mA\n",
+ "R_m=100.;\n",
+ "I_m=1.;\n",
+ "R_s=R_m*I_m/(I_in-1.);\n",
+ "print'%s %.2f %s'%(\"Value of parallel shunt resistor = \",R_s,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of parallel shunt resistor = 25.00 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.6\n",
+ "import math\n",
+ "r1=100.;\n",
+ "r2=1000.;\n",
+ "R_x_1=(r2/r1)*10000.;\n",
+ "R_x_2=(r1/r2)*10.;\n",
+ "print'%s %.2f %s %.2f %s '%(\"Range extends from \",R_x_2,\" ohms to\" and \" \",R_x_1,\" ohms\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Range extends from 1.00 100000.00 ohms \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.7\n",
+ "import math\n",
+ "E=10.;\n",
+ "r1=500.;\n",
+ "r2=600.;\n",
+ "r3=500.;\n",
+ "r4=400.;\n",
+ "V_a=E*(r2/(r1+r2));\n",
+ "V_b=E*(r4/(r3+r4));\n",
+ "V_oc=V_a-V_b;\n",
+ "r=((r1*r2)/(r1+r2))+((r3*r4)/(r3+r4));\n",
+ "i=(V_oc/(r+100.))*1000.;\n",
+ "print'%s %.2f %s'%(\"Current flow in 100 ohm resistor = \",i,\" mA\" );"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current flow in 100 ohm resistor = 1.70 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.8\n",
+ "import math\n",
+ "I_sc=19.;##in uA\n",
+ "R=1000.;\n",
+ "R_m=968.;\n",
+ "V_out=I_sc*(R*R_m/(R+R_m));\n",
+ "print'%s %.2f %s'%(\"Voltage produced = \",V_out,\" uV\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage produced = 9345.53 uV\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.9\n",
+ "import math\n",
+ "c=1.*10**-6;##in farads\n",
+ "r=3.3*10**6;##in ohms\n",
+ "t=1.;##in sec.\n",
+ "V_s=9.;##in volts\n",
+ "V_c=V_s*(1.-math.e**(-t/(r*c)));\n",
+ "print'%s %.2f %s'%(\"Voltage produced = \",V_c,\" V\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage produced = 2.35 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.10\n",
+ "import math\n",
+ "c=100.*10**-6;##in farads\n",
+ "r=1.*10**3;##in ohms\n",
+ "t1=50.*10**-3;##in sec.\n",
+ "t2=100.*10**-3;##in sec.\n",
+ "V_s=350;##in volts\n",
+ "i1=(V_s/1000.)*(math.e**(-t1/(r*c)));\n",
+ "i2=(V_s/1000.)*(math.e**(-t2/(r*c)));\n",
+ "print'%s %.2f %s %.2f %s '%(\"Charging current after \",t1,\" sec\" and \"\",i1,\" A\");\n",
+ "\n",
+ "print'%s %.2f %s %.2f %s '%(\"Charging current after \",t2,\" sec\" and \"\",i2,\" A\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charging current after 0.05 0.21 A \n",
+ "Charging current after 0.10 0.13 A \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.11\n",
+ "import math\n",
+ "c=10.*10**-6;##in farads\n",
+ "r=47.*10**3;##in ohms\n",
+ "V_s=20.;##in volts\n",
+ "V_c=10.;\n",
+ "t=-c*r*math.log(V_c/V_s);\n",
+ "print'%s %.2f %s'%(\"time taken = \",t,\" sec.\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "time taken = 0.33 sec.\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.12\n",
+ "import math\n",
+ "c=150.*10**-6;##in farads\n",
+ "r=2.*10**6;##in ohms\n",
+ "V_s=150.;##in volts\n",
+ "V_c=0.8187*V_s;\n",
+ "print'%s %.2f %s'%(\"Capacitor voltage = \",V_c,\" V\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitor voltage = 122.80 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.13\n",
+ "import math\n",
+ "r=10.*10**3;##in ohms\n",
+ "t=1.*10**-3;\n",
+ "c=(0.1*t/r)*10**9;\n",
+ "print'%s %.2f %s'%(\"Capacitor = \",c,\" nF\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitor = 10.00 nF\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.14\n",
+ "import math\n",
+ "r=10.*10**3;##in ohms\n",
+ "t=1.*10**-3;\n",
+ "c=(10.*t/r)*10**6;\n",
+ "print'%s %.2f %s'%(\"Capacitor = \",c,\" uF\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitor = 1.00 uF\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg64"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.15\n",
+ "import math\n",
+ "L=6.;##in henry\n",
+ "r=24.;##in ohms\n",
+ "t=0.1;##in sec.\n",
+ "V_s=12.;##in volts\n",
+ "i=(V_s/r)*(1.-math.e**(-t*r/L));\n",
+ "print'%s %.2f %s'%(\"current = \",i,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current = 0.16 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg64"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:3.16\n",
+ "import math\n",
+ "V_s=5.;##in volts\n",
+ "V_c=0.8647*V_s;\n",
+ "print'%s %.2f %s'%(\"Inductor voltage = \",V_c,\" V\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Inductor voltage = 4.32 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuits_by_M._H._Tooley/Chapter4_2.ipynb b/Electronic_Circuits_by_M._H._Tooley/Chapter4_2.ipynb new file mode 100644 index 00000000..3272c4a7 --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/Chapter4_2.ipynb @@ -0,0 +1,586 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:348e913cbcf303bd30e2a0fca661cdd75a027447d5cfac9117537a9c7cb2c1c0"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter4-Alternating voltage and current"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg70"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.1\n",
+ "import math\n",
+ "V_m=20.;##in volts\n",
+ "f=50.;##in Hz\n",
+ "t1=2.5*10**-3;\n",
+ "t2=15.*10**-3;\n",
+ "V1=V_m*math.sin(2.*math.pi*f*t1);\n",
+ "V2=V_m*math.sin(2.*math.pi*f*t2);\n",
+ "print'%s %.2f %s'%(\"Voltage at 2.5ms = \",V1,\" V\");\n",
+ "print'%s %.2f %s'%(\"\\n Voltage at 15ms = \",V2,\" V\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage at 2.5ms = 14.14 V\n",
+ "\n",
+ " Voltage at 15ms = -20.00 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg70"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.2\n",
+ "import math\n",
+ "f=400.;##in Hz\n",
+ "T=1./f;\n",
+ "print'%s %.2f %s %.4f %s '%(\"Time period of \",f,\" Hz\" and \" waveform = \",T,\" sec\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time period of 400.00 waveform = 0.0025 sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.3\n",
+ "import math\n",
+ "T=40.*10**-3;##in Hz\n",
+ "f=1./T;\n",
+ "print'%s %.2f %s'%(\"Frequency of 40 ms waveform = \",f,\" Hz\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Frequency of 40 ms waveform = 25.00 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.4\n",
+ "import math\n",
+ "V_rms=240.;##in Volts\n",
+ "V_pk=1.414*V_rms;\n",
+ "print'%s %.2f %s %.2f %s' %(\"Peak voltage of \",V_rms,\" V RMS\" and \" voltage = \",V_pk,\" V\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak voltage of 240.00 voltage = 339.36 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.5\n",
+ "import math\n",
+ "I_pk=50.*10**-3;##in Amps\n",
+ "I_rms=0.353*I_pk;\n",
+ "print'%s %.2f %s'%(\"RMS current of 50mA peak current = \",I_rms,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "RMS current of 50mA peak current = 0.02 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.6\n",
+ "import math\n",
+ "V=10.;##pk-pk voltage\n",
+ "r=1000.;##ohms\n",
+ "I_pk=V/r;##in Amps\n",
+ "I_rms=0.353*I_pk*1000.;##milliamps\n",
+ "print'%s %.2f %s'%(\"RMS current of 10V peak-peak voltage = \",I_rms,\" mA\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "RMS current of 10V peak-peak voltage = 3.53 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.7\n",
+ "import math\n",
+ "c=1.*10**-6;\n",
+ "f1=100.;\n",
+ "f2=10000.;\n",
+ "X_c1=1./(2.*math.pi*f1*c);\n",
+ "X_c2=1./(2.*math.pi*f2*c);\n",
+ "print'%s %.2f %s'%(\"Reactance at 100Hz = \",X_c1,\" mA\");\n",
+ "print'%s %.2f %s'%(\"\\n Reactance at 10kHz = \",X_c2,\" mA\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reactance at 100Hz = 1591.55 mA\n",
+ "\n",
+ " Reactance at 10kHz = 15.92 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.8\n",
+ "import math\n",
+ "V=240.;\n",
+ "c=100.*10**-9;\n",
+ "f=50.;\n",
+ "X_c=1./(2.*math.pi*f*c);\n",
+ "I_c=V/X_c;\n",
+ "print'%s %.2f %s'%(\"Current flow = \",I_c,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current flow = 0.01 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.9\n",
+ "import math\n",
+ "L=1.*10**-3;\n",
+ "f1=100.;\n",
+ "f2=10000.;\n",
+ "X_L1=(2.*math.pi*f1*L);\n",
+ "X_L2=(2.*math.pi*f2*L);\n",
+ "print'%s %.2f %s'%(\"Reactance at 100Hz = \",X_L1,\" ohm\");\n",
+ "print'%s %.2f %s'%(\"\\nReactance at 10kHz = \",X_L2,\" ohm\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reactance at 100Hz = 0.63 ohm\n",
+ "\n",
+ "Reactance at 10kHz = 62.83 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.10\n",
+ "import math\n",
+ "L=1.*10**-3;\n",
+ "f1=100.;\n",
+ "f2=10000.;\n",
+ "X_L1=(2.*math.pi*f1*L);\n",
+ "X_L2=(2.*math.pi*f2*L);\n",
+ "print'%s %.2f %s'%(\"Reactance at 100Hz = \",X_L1,\" ohm\");\n",
+ "print'%s %.2f %s'%(\"\\nReactance at 10kHz = \",X_L2,\" ohm\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reactance at 100Hz = 0.63 ohm\n",
+ "\n",
+ "Reactance at 10kHz = 62.83 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.11\n",
+ "import math\n",
+ "C=2.*10**-6;\n",
+ "f=400.;\n",
+ "V=115.;\n",
+ "X_C=1./(2.*math.pi*f*C);\n",
+ "r=199.;\n",
+ "z=math.sqrt(r**2+X_C**2);\n",
+ "I_s=V/z;\n",
+ "print'%s %.2f %s'%(\"Reactance = \",X_C,\" ohm\");\n",
+ "print'%s %.2f %s'%(\"\\n Current = \",I_s,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reactance = 198.94 ohm\n",
+ "\n",
+ " Current = 0.41 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.12\n",
+ "import math\n",
+ "L=150.*10**-3;\n",
+ "f=400.;\n",
+ "V=115.;\n",
+ "X_L=(2.*math.pi*f*L);\n",
+ "r=250.;\n",
+ "z=math.sqrt(r**2+X_L**2);\n",
+ "I_s=V/z;\n",
+ "print'%s %.2f %s'%(\"Reactance = \",X_L,\" ohm\");\n",
+ "print'%s %.2f %s'%(\"\\n Current = \",I_s,\" A\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reactance = 376.99 ohm\n",
+ "\n",
+ " Current = 0.25 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.13\n",
+ "import math\n",
+ "L=100.*10**-3;\n",
+ "f=400.;\n",
+ "C=(1./(4.*math.pi*math.pi*f*f*L))*10**6;\n",
+ "print'%s %.2f %s'%(\"Capacitance required = \",C,\" uF\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacitance required = 1.58 uF\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg78"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.14\n",
+ "import math\n",
+ "L=20.*10**-3;\n",
+ "f=2000.;\n",
+ "V=1.5;\n",
+ "r=100.;\n",
+ "C=10.*10**-9;\n",
+ "X_L=(2.*math.pi*f*L);\n",
+ "X_C=1./(2.*math.pi*f*C);\n",
+ "z=math.sqrt(r**2+(X_L-X_C)**2);\n",
+ "i=V/z;\n",
+ "v=i*r;\n",
+ "print'%s %.4f %s'%(\"Current supplied = \",i,\" mA\");\n",
+ "print'%s %.2f %s'%(\"\\nVoltage developed = \",v,\" V\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current supplied = 0.0002 mA\n",
+ "\n",
+ "Voltage developed = 0.02 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.15\n",
+ "import math\n",
+ "N_s=120.;\n",
+ "V_p=220.;\n",
+ "N_p=2000.;\n",
+ "V_s=N_s*V_p/N_p;\n",
+ "print'%s %.2f %s'%(\"Secondry voltage = \",V_s,\" V\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Secondry voltage = 13.20 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:4.16\n",
+ "import math\n",
+ "V_p=200.;\n",
+ "V_s=10.;\n",
+ "N_p=1200.;\n",
+ "N_s=N_p*V_s/V_p;\n",
+ "i_s=2.5;\n",
+ "i_p=N_s*i_s/N_p;\n",
+ "print'%s %.2f %s'%(\"Secondry turns = \",N_s,\"\");\n",
+ "print'%s %.2f %s'%(\"\\nprimary current = \",i_p,\"A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Secondry turns = 60.00 \n",
+ "\n",
+ "primary current = 0.12 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuits_by_M._H._Tooley/Chapter5_2.ipynb b/Electronic_Circuits_by_M._H._Tooley/Chapter5_2.ipynb new file mode 100644 index 00000000..b0c877d6 --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/Chapter5_2.ipynb @@ -0,0 +1,303 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:55165db38b61e651606f5e47e79ec33dbbcd65dc316fccbc779cf43ddfeefa00"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter5-Semiconductors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:5.1\n",
+ "import math\n",
+ "v1=0.43;##volts\n",
+ "i1=2.5*10**-3;##in Amps.\n",
+ "v2=0.65;##volts\n",
+ "i2=7.4*10**-3;##in Amps.\n",
+ "r1=v1/i1;\n",
+ "r2=v2/i2;\n",
+ "print'%s %.2f %s'%(\"Diode resistance for 2.5A current = \",r1,\" ohms\");\n",
+ "print'%s %.2f %s'%(\"\\n Diode resistance for 0.65V = \",r2,\" ohms\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diode resistance for 2.5A current = 172.00 ohms\n",
+ "\n",
+ " Diode resistance for 0.65V = 87.84 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:5.2\n",
+ "import math\n",
+ "i=15.*10**-3;\n",
+ "R=(21.-2.2)/i;\n",
+ "v=18.8;##in volts\n",
+ "P=i*v*1000.;\n",
+ "print'%s %.2f %s %.2f %s '%(\"Resistor \",R,\" ohms\" and \" of \",P,\" mW\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Resistor 1253.33 of 282.00 mW \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:5.3\n",
+ "import math\n",
+ "I_c=30.;##in mA\n",
+ "I_b=0.6;\n",
+ "I_e=I_c+I_b;\n",
+ "hfe=I_c/I_b;\n",
+ "print'%s %.2f %s %.2f %s '%(\"Emitter current = \",I_e,\" ohms\" and \"hfe = \",hfe,\"\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Emitter current = 30.60 hfe = 50.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:5.4\n",
+ "import math\n",
+ "I_c=30.;##in mA\n",
+ "I_b=0.6;\n",
+ "I_e=I_c+I_b;\n",
+ "hfe=I_c/I_b;\n",
+ "print'%s %.2f %s %.2f %s '%(\"Emitter current = \",I_e,\" ohms\" and \"hfe = \",hfe,\"\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Emitter current = 30.60 hfe = 50.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:5.5\n",
+ "import math\n",
+ "I_e=98.;##in mA\n",
+ "I_c=97.;\n",
+ "I_b=I_e-I_c;\n",
+ "hfe=I_c/I_b;\n",
+ "print'%s %.2f %s %.2f %s '%(\"Emitter current = \",I_b,\" ohms\" and \"hfe = \",hfe,\"\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Emitter current = 1.00 hfe = 97.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:5.6\n",
+ "import math\n",
+ "I_c=1.5;##in A\n",
+ "I_b=50.*10**-3;\n",
+ "V_ce=6.;##volts\n",
+ "hfe=I_c/I_b;\n",
+ "P=I_c*V_ce;\n",
+ "print'%s %.2f %s'%(\"hfe required = \",hfe,\"\");\n",
+ "print'%s %.2f %s'%(\"\\n collector power dissipation = \",P,\" W\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "hfe required = 30.00 \n",
+ "\n",
+ " collector power dissipation = 9.00 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:5.7\n",
+ "import math\n",
+ "hfe=200.\n",
+ "I_c=10.*10**-3;\n",
+ "dI_b=I_c/hfe;\n",
+ "dI_c=hfe*dI_b/100.;\n",
+ "print'%s %.4f %s'%(\"Base current = \",dI_b,\" A \");\n",
+ "print'%s %.4f %s'%(\"\\nChange in collector current = \",dI_c,\" mA\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Base current = 0.0001 A \n",
+ "\n",
+ "Change in collector current = 0.0001 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:5.8\n",
+ "import math\n",
+ "dV_gs=0.025;\n",
+ "g_fs=-0.5;\n",
+ "dI_d=dV_gs*g_fs;##in mA\n",
+ "I_d1=50.*10**-3;##in mA\n",
+ "I_d2=dI_d+I_d1;\n",
+ "print'%s %.2f %s'%(\"Change in drain current = \",dI_d,\" A\");\n",
+ "print'%s %.2f %s'%(\"\\nNew value of drain current = \",I_d2,\" A\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in drain current = -0.01 A\n",
+ "\n",
+ "New value of drain current = 0.04 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuits_by_M._H._Tooley/Chapter7_2.ipynb b/Electronic_Circuits_by_M._H._Tooley/Chapter7_2.ipynb new file mode 100644 index 00000000..e8e27f4f --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/Chapter7_2.ipynb @@ -0,0 +1,343 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2256e43b0ccc4d0255b5cf781a9c914f08adb93f8f0caa254bf09c9ef401f2ba"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter7-Amplifiers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:7.1\n",
+ "import math\n",
+ "I_i=4.;\n",
+ "V_o=2.;\n",
+ "V_i=50.*10**-3;\n",
+ "I_o=200.;\n",
+ "A_v=V_o/V_i;\n",
+ "A_i=I_o/I_i;\n",
+ "print'%s %.2f %s'%(\" Volt gain = \",A_v,\"\"); \n",
+ "print'%s %.2f %s'%(\"\\n Current gain = \",A_i,\"\"); \n",
+ "print'%s %.2f %s'%(\"\\n Power gain = \",A_i*A_v,\"\"); \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Volt gain = 40.00 \n",
+ "\n",
+ " Current gain = 50.00 \n",
+ "\n",
+ " Power gain = 2000.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:7.2\n",
+ "import math\n",
+ "A_v_max=35.;\n",
+ "A_v_cutoff=0.707*A_v_max;\n",
+ "print'%s %.2f %s'%(\" Mid-band Volt gain = \",A_v_cutoff,\"\"); \n",
+ "print(\"\\n upper freq = 590Hz & lower freq = 57Hz\"); \n",
+ " \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Mid-band Volt gain = 24.74 \n",
+ "\n",
+ " upper freq = 590Hz & lower freq = 57Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:7.3\n",
+ "import math\n",
+ "A=50.;\n",
+ "b=0.1;\n",
+ "G=A/(1.+b*A);\n",
+ "print'%s %.2f %s'%(\" overall Volt gain = \",G,\"\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " overall Volt gain = 8.33 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg138"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:7.4\n",
+ "import math\n",
+ "A=50.;\n",
+ "A_new=A+0.2*A;\n",
+ "b=0.1;\n",
+ "G=A_new/(1.+b*A_new);\n",
+ "dG=8.33-G/8.33;\n",
+ "print'%s %.2f %s'%(\" percentage change in overall volt gain = \",dG,\"\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " percentage change in overall volt gain = 7.30 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg138"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:7.5\n",
+ "import math\n",
+ "A=100.;\n",
+ "G=20.;\n",
+ "b=(1./G)-(1./A);\n",
+ "print'%s %.2f %s'%(\"amount of feedback required = \",b,\"\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "amount of feedback required = 0.04 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:7.6\n",
+ "import math\n",
+ "h_oe=80.*10**-6;\n",
+ "R_l=10000.;\n",
+ "I_f=320.*10**-6;\n",
+ "I_c=I_f*(1./h_oe)/((1./h_oe)+R_l);\n",
+ "V_out=I_c*R_l;\n",
+ "print'%s %.2f %s'%(\"Output voltage = \",V_out,\" V\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Output voltage = 1.78 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg144"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:7.7\n",
+ "import math\n",
+ "b=200.;\n",
+ "h_ie=1.5*10**3;##in ohms\n",
+ "h_fe=150.;\n",
+ "R_l=b*h_ie/h_fe;\n",
+ "print'%s %.2f %s'%(\"Load resistance = \",R_l,\" ohms\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Load resistance = 2000.00 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg145"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:7.8\n",
+ "import math\n",
+ "V=9.;\n",
+ "V_e=2.;\n",
+ "R4=1000.;\n",
+ "V_b=2.6;\n",
+ "R2=33.*10**3;\n",
+ "R1=68000.;\n",
+ "I_r1=(V-V_b)/R1;\n",
+ "R3=2.2*10**3;\n",
+ "I_b=15.1*10**-6;\n",
+ "I_c=2.0151*10**-3;\n",
+ "V_r3=I_c*R3;\n",
+ "V_c=V-V_r3;\n",
+ "print'%s %.2f %s'%(\"Collector voltage = \",V_c,\" V\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Collector voltage = 4.57 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:7.9\n",
+ "import math\n",
+ "V_pp=14.8-3.3;\n",
+ "print(\"Collector quiescent voltage = 9.2 V\"); \n",
+ "print(\"\\nCollector quiescent current = 7.3mA\"); \n",
+ "print'%s %.2f %s'%(\"\\nOutput peak-peak voltage = \",V_pp,\" V\"); "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Collector quiescent voltage = 9.2 V\n",
+ "\n",
+ "Collector quiescent current = 7.3mA\n",
+ "\n",
+ "Output peak-peak voltage = 11.50 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuits_by_M._H._Tooley/Chapter8_2.ipynb b/Electronic_Circuits_by_M._H._Tooley/Chapter8_2.ipynb new file mode 100644 index 00000000..b999ba17 --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/Chapter8_2.ipynb @@ -0,0 +1,186 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:69bad2cc1b97411ab194c6989d24ce9e98f48b912e29dd4ff77d5d659094fdfe"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter8-Operational Amplifiers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg158"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:8.1\n",
+ "import math\n",
+ "V_out=2.;\n",
+ "V_in=400.*10**-6;\n",
+ "A_v=V_out/V_in;\n",
+ "A_v_dB=math.ceil (20.*(math.log (A_v)/math.log (10.)));\n",
+ "print'%s %.2f %s'%(\"open loop voltage gain = \",A_v_dB,\" dB\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "open loop voltage gain = 74.00 dB\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg159"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:8.2\n",
+ "import math\n",
+ "V_in=5.*10**-3;\n",
+ "R_in=2.*10**6;\n",
+ "I_in=V_in/R_in;\n",
+ "print'%s %.2e %s'%(\"Input current = \",I_in,\" A\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Input current = 2.50e-09 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:8.3\n",
+ "import math\n",
+ "V_out=10.;\n",
+ "t=4.;\n",
+ "SR=V_out/t;\n",
+ "print'%s %.2f %s'%(\"Slew rate = \",SR,\" V/us\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slew rate = 2.50 V/us\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:8.4\n",
+ "import math\n",
+ "V_out=2.;\n",
+ "SR=15.;##in V/us\n",
+ "t=V_out/SR;\n",
+ "print'%s %.2f %s'%(\"Time taken = \",t,\" us\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time taken = 0.13 us\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:8.6\n",
+ "import math\n",
+ "R_in=10000.;\n",
+ "f1=250.;\n",
+ "f2=15000.;\n",
+ "C_in=0.159/(f1*R_in);\n",
+ "C_f=0.159/(f2*R_in);\n",
+ "print'%s %.2e %s'%(\"C_f = \",C_f,\" F\"); \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "C_f = 1.06e-09 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuits_by_M._H._Tooley/Chapter9_2.ipynb b/Electronic_Circuits_by_M._H._Tooley/Chapter9_2.ipynb new file mode 100644 index 00000000..594ffb15 --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/Chapter9_2.ipynb @@ -0,0 +1,127 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e9fb88f63f2c9c2769603c77ea25313f638342259fa02ac4950a18408033aeb6"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter9-Oscillators"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:9.1\n",
+ "import math\n",
+ "C=10.*10**-9;\n",
+ "R=10000.;\n",
+ "f=(1./(2.*math.pi*math.sqrt (6.)*C*R));\n",
+ "print'%s %.2f %s'%(\"The freq of oscillation = \",f,\" Hz\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The freq of oscillation = 649.75 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:9.2\n",
+ "import math\n",
+ "r1=1000.;\n",
+ "r2=1000.;\n",
+ "c=100.*10**-9;\n",
+ "f=(1./(2.*math.pi*c*r1));\n",
+ "print'%s %.2f %s'%(\"The freq of oscillation at 1 kohm= \",f,\" Hz\");\n",
+ "R1=6000.;\n",
+ "R2=6000.;\n",
+ "F=(1/(2.*math.pi*c*R1));\n",
+ "print'%s %.2f %s'%(\"\\nThe freq of oscillation at 6 kohm= \",F,\" Hz\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The freq of oscillation at 1 kohm= 1591.55 Hz\n",
+ "\n",
+ "The freq of oscillation at 6 kohm= 265.26 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:9.3\n",
+ "import math\n",
+ "f=1000.;\n",
+ "t=1./f;\n",
+ "C=10.*10**-9;\n",
+ "R=t/(1.4*C);\n",
+ "print'%s %.2f %s'%(\"R= \",R/1000,\" kohm\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "R= 71.43 kohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuits_by_M._H._Tooley/chapter6_2.ipynb b/Electronic_Circuits_by_M._H._Tooley/chapter6_2.ipynb new file mode 100644 index 00000000..7b68d56d --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/chapter6_2.ipynb @@ -0,0 +1,195 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:208e4f11f55abacdac1b020238bf181604676d4f63a97a09e00055db476b2f63"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter6-Power Supplies "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg117"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:6.1\n",
+ "import math\n",
+ "V_p=220.;\n",
+ "V_s=V_p/44.;\n",
+ "V_pk=1.414*V_s;##in volts\n",
+ "V_l=V_pk-0.6;\n",
+ "print'%s %.2f %s'%(\"Peak voltage that appear across load = \",V_l,\" V\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak voltage that appear across load = 6.47 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:6.2\n",
+ "import math\n",
+ "X_c=3.18;\n",
+ "R=100.;\n",
+ "V_rip=1.*(X_c/math.sqrt(R**2+X_c**2));\n",
+ "print'%s %.2f %s'%(\"Ripple voltage = \",V_rip,\" V\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ripple voltage = 0.03 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:6.3\n",
+ "import math\n",
+ "f=50.;\n",
+ "L=10.;\n",
+ "X_l=2.*math.pi*f*L;\n",
+ "X_c=3.18;\n",
+ "V_rip=1.*(X_c/math.sqrt(X_l**2+X_c**2));\n",
+ "print'%s %.2e %s'%(\"Ripple voltage = \",V_rip,\" V\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ripple voltage = 1.01e-03 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg122"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:6.4\n",
+ "import math\n",
+ "R_l=400.;\n",
+ "V_in=9.;\n",
+ "V_z=5.;\n",
+ "P_z_max=0.5;\n",
+ "R_s_max=R_l*((V_in/V_z)-1.);\n",
+ "R_s_min=((V_z*V_in)-V_z**2)/P_z_max;\n",
+ "print'%s %.2f %s'%(\"Suitable value of resistor = \",(R_s_max+R_s_min)/2,\" ohm\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Suitable value of resistor = 180.00 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg123"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex:6.5\n",
+ "import math\n",
+ "dI_i=20.;\n",
+ "dV_o=0.5;\n",
+ "dV_o_reg=0.1;\n",
+ "dI_o=2.;\n",
+ "R_out=dV_o/dI_o;\n",
+ "Regulation=(dV_o_reg/dI_i)*100.;\n",
+ "print'%s %.2f %s'%(\" output resis. = \",R_out,\" ohm\"); \n",
+ "print'%s %.2f %s'%(\" \\n regulation. = \",Regulation,\"\"); "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " output resis. = 0.25 ohm\n",
+ " \n",
+ " regulation. = 0.50 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Electronic_Circuits_by_M._H._Tooley/screenshots/Chapter1_1.png b/Electronic_Circuits_by_M._H._Tooley/screenshots/Chapter1_1.png Binary files differnew file mode 100644 index 00000000..bdbc5fb8 --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/screenshots/Chapter1_1.png diff --git a/Electronic_Circuits_by_M._H._Tooley/screenshots/Chapter2_1.png b/Electronic_Circuits_by_M._H._Tooley/screenshots/Chapter2_1.png Binary files differnew file mode 100644 index 00000000..6c5b5ee9 --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/screenshots/Chapter2_1.png diff --git a/Electronic_Circuits_by_M._H._Tooley/screenshots/Chapter3_1.png b/Electronic_Circuits_by_M._H._Tooley/screenshots/Chapter3_1.png Binary files differnew file mode 100644 index 00000000..3d29068d --- /dev/null +++ b/Electronic_Circuits_by_M._H._Tooley/screenshots/Chapter3_1.png diff --git a/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter10_1.ipynb b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter10_1.ipynb new file mode 100644 index 00000000..2fd0ff50 --- /dev/null +++ b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter10_1.ipynb @@ -0,0 +1,735 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1a92a9838b2fe8c9a27b79bf686d1269c5c2194ba826da4aec9e1a105fb2b2ae"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10-Radiation Shielding "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg553"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 10.1\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "E0 = 2.; ## Energy of gamma rays in MeV\n",
+ "a = 10.; ## Thickeness of lead shield in cm\n",
+ "phi0 = 10**6; ## Intensity of gamma rays in gamma-rays/cm^2-sec\n",
+ "\n",
+ "## 1.\n",
+ "## Using the data from Table II.4 for E0 = 2 MeV\n",
+ "mu_rho = 0.0457; ## The ratio of total attenuation coefficient to density in cm^2/g\n",
+ "## From standard data tables for lead\n",
+ "rho = 11.34; ## Density of lead in g/cm^3\n",
+ "## Calculation\n",
+ "phi_u = phi0*math.exp(-(mu_rho*rho*a));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Uncollided flux at the rear side of lead shield = \",phi_u,\" gamma-rays/cm^2-sec \\n\")\n",
+ "\n",
+ "## 2.\n",
+ "## Using the data from Table 10.1 for 2 MeV of lead material\n",
+ "mua = mu_rho*rho*a;\n",
+ "B_4 = 2.41; ## Buildup factor if mu*a = 4\n",
+ "B_7 = 3.36; ## Buildup factor if mu*a = 7\n",
+ "## Using two point method of straight line for calculating buildup factor at mu*a\n",
+ "B_m = B_4+((mua-4.)*((B_7-B_4)/(7.-4.)));\n",
+ "## Calculation\n",
+ "phi_b = phi_u*B_m;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Buildup flux at the rear side of lead shield = \",phi_b,\" gamma-rays/cm^2-sec \\n\");\n",
+ "\n",
+ "## 3.\n",
+ "## Using the data from Table II.5 for 2 MeV \n",
+ "mua_rho_air = 0.0238; ## The ratio of total attenuation coefficient to density of air in cm^2/g\n",
+ "## Calculation\n",
+ "X_dot = 0.0659*E0*mua_rho_air*phi_b;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Exposure rate at the rear side of lead shield = \",X_dot,\" mR/hour \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Uncollided flux at the rear side of lead shield = 5614.63 gamma-rays/cm^2-sec \n",
+ "\n",
+ "\n",
+ " Buildup flux at the rear side of lead shield = 15633.48 gamma-rays/cm^2-sec \n",
+ "\n",
+ "\n",
+ " Exposure rate at the rear side of lead shield = 49.04 mR/hour \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg555"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "## Example 10.2\n",
+ "import math\n",
+ "\n",
+ "import warnings\n",
+ "warnings.filterwarnings('ignore')\n",
+ "import numpy\n",
+ "%matplotlib inline\n",
+ "import matplotlib\n",
+ "from matplotlib import pyplot\n",
+ "\n",
+ "## Given data\n",
+ "E = 1.; ## Energy of gamma rays in MeV\n",
+ "X_dot = 1.; ## Exposure rate in mR/hour\n",
+ "phi0 = 10**8; ## Intensity of gamma rays in gamma-rays/cm^2-sec from isotropic point source\n",
+ "## Using the data from Table II.5 for 1 MeV \n",
+ "mua_rho_air = 0.028; ## The ratio of total attenuation coefficient to density of air in cm^2/g\n",
+ "phi_b = X_dot/(0.0659*E*mua_rho_air); ## Buildup flux in gamma-rays/cm^2-sec\n",
+ "## Using Eq 10.14\n",
+ "print'%s %.2f %s'%(\" \\n The equation to calculate radius is \\n %.2E = %E * Bp*exp(-mu*R)/(4*pi*R^2) \\n\",phi_b,phi0);\n",
+ "## Using the data from Table II.4 for E = 1 MeV for Iron\n",
+ "mu_rho = 0.0595; ## The ratio of total attenuation coefficient to density in cm^2/g\n",
+ "## From standard data tables for iron\n",
+ "rho = 7.864; ## Density of iron in g/cm^3\n",
+ "mu = mu_rho*rho;\n",
+ "## On solving the right hand side of equation\n",
+ "## RHS = 3.22*10^3*Bp*exp(-mu*R)/(mu*R)^2\n",
+ "## Let mu*R = x\n",
+ "## Using the data from Table 10.2 for isotropic point source of 1 MeV incident on iron material\n",
+ "Bp = numpy.array([1.87, 2.89, 5.39, 10.2, 16.2, 28.3, 42.7]);\n",
+ "x = numpy.array([1 ,2 ,4 ,7 ,10, 15, 20]);\n",
+ "leng=len(Bp)\n",
+ "RHS= numpy.zeros(leng);\n",
+ "for i in range(0,7):\n",
+ " RHS[i] = (3.22*10**3*Bp[i]*math.exp(-x[i])/x[i]**2);\n",
+ "\n",
+ "pyplot.plot(x,RHS)\n",
+ "pyplot.legend(\"Conical\",\"2D-CD\")\n",
+ "pyplot.xlabel(\"mu*R\")\n",
+ "pyplot.ylabel(\"RHS\")\n",
+ "pyplot.title(\"Semilog plot of RHS vs mu*R\")\n",
+ "pyplot.show()\n",
+ "\n",
+ "## From the graph\n",
+ "muR = 6.55; ## This is the value when RHS = 1\n",
+ "## Calculation\n",
+ "R = muR/mu;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n The shield radius required = \",math.ceil(R),\" cm \\n\");\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The equation to calculate radius is \n",
+ " %.2E = %E * Bp*exp(-mu*R)/(4*pi*R^2) \n",
+ " 541.95 100000000\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x81ef970>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The shield radius required = 14.00 cm \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg561"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 10.3\n",
+ "import math\n",
+ "import warnings\n",
+ "warnings.filterwarnings('ignore')\n",
+ "import numpy\n",
+ "%matplotlib inline\n",
+ "import matplotlib\n",
+ "from matplotlib import pyplot\n",
+ "\n",
+ "## Given data\n",
+ "E = 2.; ## Energy of gamma rays in MeV\n",
+ "X_dot = 2.5; ## Exposure rate in mR/hour\n",
+ "phi0 = 10**9; ## Intensity of gamma rays in gamma-rays/cm^2-sec from isotropic point source\n",
+ "## Using the data from Table II.5 for 1 MeV \n",
+ "mua_rho_air = 0.0238; ## The ratio of total attenuation coefficient to density of air in cm^2/g\n",
+ "phi_b = X_dot/(0.0659*E*mua_rho_air); ## Buildup flux in gamma-rays/cm^2-sec\n",
+ "\n",
+ "## From standard data tables for concrete\n",
+ "rho = 2.35; ## Density of concrete in g/cm^3\n",
+ "## Using the data from Table 10.3 for concrete at 2 MeV\n",
+ "A1 = 18.089;\n",
+ "A2 = 1-A1;\n",
+ "alpha1 = -0.0425;\n",
+ "alpha2 = 0.00849;\n",
+ "## Using Eq 10.26\n",
+ "#print'%s %.2f %s'%(\" \\n The equation to calculate thickness is \\n %.2E = (%E/2) *(%4.3f*E1(%4.3f*mu*a) %4.3f*E1(%4.3f*mu*a)) \\n\",phi_b,phi0,A1,(1+alpha1),A2,(1+alpha2));\n",
+ "## Using the data from Table II.4 for E = 1 MeV for concrete\n",
+ "mu_rho = 0.0445; ## The ratio of total attenuation coefficient to density in cm^2/g\n",
+ "mu = mu_rho*rho;\n",
+ "## On solving the right hand side of equation\n",
+ "## RHS = 1.13*10^7*(E1(0.9575*mu*a)-0.94*E1(1.00849*mu*a))\n",
+ "## Let mu*a = x\n",
+ "x = numpy.array([1 ,2 ,4 ,7 ,10, 15, 20]);\n",
+ "leng=len(x)\n",
+ "RHS= numpy.zeros(leng);\n",
+ "for i in range(0,leng):\n",
+ " RHS[i] = 1.13*10**7*(math.exp(-0.9575*x[i])*((1./(0.9575*x[i])+(1/(0.9575*x[i])**3))) - math.exp(-1.00849*x[i])*((1./(1.00849*x[i])+(1./(1.00849*x[i])**3))));\n",
+ "\n",
+ "pyplot.plot(x,RHS)\n",
+ "pyplot.legend(\"Conical\",\"2D-CD\")\n",
+ "pyplot.xlabel(\"mu*R\")\n",
+ "pyplot.ylabel(\"RHS\")\n",
+ "pyplot.title(\"Semilog plot of RHS vs mu*R\")\n",
+ "pyplot.show()\n",
+ "## From the graph\n",
+ "mua = 13.6; ## This is the value when RHS = 1\n",
+ "## Calculation\n",
+ "a = mua/mu;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n The concrete thickness = \",a,\" cm \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x30dd6d0>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The concrete thickness = 130.05 cm \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg575"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 10.4\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "E = 6.; ## Energy of gamma rays in MeV\n",
+ "phi0 = 10**2; ## Intensity of gamma rays in gamma-rays/cm^2-sec from mono-directional beam\n",
+ "x_w = 100.; ## Thickness of water in cm\n",
+ "x_Pb = 8.; ## Thickness of lead in cm\n",
+ "## Using data from Table II.4 at 6 MeV\n",
+ "mu_w = 0.0275; ## Total attenuation coefficient of water in cm^(-1)\n",
+ "mu_Pb = 0.4944; ## Total attenuation coefficient of lead in cm^(-1)\n",
+ "\n",
+ "mua_w = x_w*mu_w; ## Attenuation due to thickness of water\n",
+ "mua_Pb = x_Pb*mu_Pb; ## Attenuation due to thickness of lead\n",
+ "\n",
+ "## Case (a) - Water is placed before the lead\n",
+ "print'%s %.2f %s'%(\" \\n In case (a), Buildup factor is only due to lead measured at \",mua_Pb,\"\");\n",
+ "## Using the data from Table 10.1 at 6 MeV\n",
+ "B_Pb = 1.86;\n",
+ "phi_b_a = phi0*B_Pb*math.exp(-(mua_w+mua_Pb));\n",
+ "\n",
+ "## Using the data from Table II.5 for 6 MeV \n",
+ "mua_rho_air = 0.0172; ## The ratio of total attenuation coefficient to density of air in cm^2/g\n",
+ "## Calculation\n",
+ "X_dot_a = 0.0659*E*mua_rho_air*phi_b_a;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Exposure rate if water is placed before lead shield = \",X_dot_a*1000,\" uR/hour \\n\");\n",
+ "\n",
+ "## Case (b) - Lead is placed before water\n",
+ "print'%s %.2f %s %.2f %s'%(\" \\n In case (b), Buildup factor is due to water and lead measured at \",mua_w,\"\" and \"\",mua_Pb,\" respectively\");\n",
+ "## Using the data from Table 10.1 for water at 3.2 MeV,, which is the minimum point of mu_Pb curve\n",
+ "B_w = 2.72;\n",
+ "B_m = B_Pb*B_w;\n",
+ "phi_b_b = phi0*B_m*math.exp(-(mua_w+mua_Pb));\n",
+ "\n",
+ "## Calculation\n",
+ "X_dot_b = 0.0659*E*mua_rho_air*phi_b_b;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Exposure rate if lead is placed before water = \",X_dot_b*1000,\" uR/hour \\n\");\n",
+ "## The answer given in the textbook is wrong. This is because the intensity of gamma rays is wrongly taken for calculation. \n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " In case (a), Buildup factor is only due to lead measured at 3.96 \n",
+ "\n",
+ " Exposure rate if water is placed before lead shield = 1.55 uR/hour \n",
+ "\n",
+ " \n",
+ " In case (b), Buildup factor is due to water and lead measured at 2.75 3.96 respectively\n",
+ "\n",
+ " Exposure rate if lead is placed before water = 4.21 uR/hour \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg582"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 10.5\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "fission_density = 4*10**7; ## Fission density in fissions/cm^2-sec\n",
+ "## 1 inches = 2.54 cm\n",
+ "d = 28*2.54; ## Diamaeter of plate in cm\n",
+ "R = d/2.; ## Radius of plate in cm\n",
+ "v = 2.42; ## Number of fission neutrons emitted per fission\n",
+ "x = 75.; ## Distance of point from center of plate in cm\n",
+ "## Using the data from Table 10.4 for removal macroscopic cross section of water\n",
+ "sigma_RW = 0.103; ## Removal macroscopic cross section of water in cm^-1\n",
+ "S = v*fission_density; ## Strength of neutron source in terms of neutrons/cm^2-sec\n",
+ "A = 0.12; ## A constant\n",
+ "## From Figure 10.19\n",
+ "\n",
+ "## Let the upper limit of integral be 20\n",
+ "x_limit = 20;\n",
+ "E1_x11 = 0.0000512 \n",
+ " \n",
+ " \n",
+ "E1_x21 = 0.0000205 \n",
+ "## Calculation\n",
+ "phi_1 = S*A/2.*(E1_x11-E1_x21);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The fast flux without iron shield = \",phi_1,\" neutrons/cm^2-sec \\n\");\n",
+ "\n",
+ "## 2. Iron slab is inserted in front of the fission plate\n",
+ "## Using the data from Table 10.4 for removal macroscopic cross section of iron\n",
+ "sigma_R = 0.168; ## Removal macroscopic cross section of iron in cm^-1\n",
+ "t = 3*2.54; ## Thickness of iron slab in cm\n",
+ "## Now the analysis is similar to multi layered shielding\n",
+ "\n",
+ "x_limit = 20;\n",
+ "\n",
+ "E1_x12 = 0.0000124 \n",
+ "E1_x22 = 0.0000043 \n",
+ " \n",
+ " \n",
+ "## Calculation\n",
+ "phi_2 = S*A/2*(E1_x12-E1_x22);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The fast flux with iron shield = \",phi_2,\" neutrons/cm^2-sec \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The fast flux without iron shield = 178.31 neutrons/cm^2-sec \n",
+ "\n",
+ " \n",
+ " The fast flux with iron shield = 47.04 neutrons/cm^2-sec \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg587"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 10.6\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "## Assuming average energy produced per fission reaction is 200 MeV \n",
+ "P = 250.*10**3; ## Power of research reactor in Watts\n",
+ "P_fission = 200*10**6*1.6*10**(-19); ## Energy produced in a fission reaction in terms of joule\n",
+ "f = 0.75; ## Metal volume fraction\n",
+ "## In this problem, both reflector and shield act as a composite shield\n",
+ "a = 150.+15.; ## Net shield distance in cm\n",
+ "## 1 litre = 1000 grams\n",
+ "V = 32.*1000.; ## Core volume in gram\n",
+ "\n",
+ "fission_density = (P/P_fission)*(1./V);\n",
+ "v = 2.42; ## Number of fission neutrons emitted per fission\n",
+ "S = fission_density*v; ## Neutron source strength in neutrons/cm^3-sec\n",
+ "## Assuming spherical shape\n",
+ "## Volume of sphere = (4/3)*pi*(radius)^3\n",
+ "R = ((3.*V)/(4.*math.pi))**(1./3.);\n",
+ "## Using the data from Table 10.4 for removal macroscopic cross section \n",
+ "sigma_R = 0.174; ## Removal macroscopic cross section of uranium in cm^-1\n",
+ "sigma_RW = 0.103; ## Removal macroscopic cross section of water in cm^-1\n",
+ "A = 0.12; ## A constant\n",
+ "alpha = ((1.-f)*sigma_RW)+(f*sigma_R); ## A parameter\n",
+ "## Calculation\n",
+ "theta = (S*A/(4*alpha))*(math.ceil(R)/(math.ceil(R)+a))**2*math.exp(-sigma_RW*a)*(1-math.exp(-2*alpha*math.ceil(R)));\n",
+ "## Converting into equivalent dose rate by referring Figure 9.12\n",
+ "## Fast neutron flux of 6.8 neutrons/cm^2-sec is equivalent to 1 mrem/hr\n",
+ "H_dot = theta/6.8;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Fast neutron dose rate near the surface of the shield = \",H_dot,\" mrem/hour \\n \");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Fast neutron dose rate near the surface of the shield = 8.10 mrem/hour \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg593"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 10.7\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "E = 14.; ## Energy of neutrons in MeV\n",
+ "phi0 = 10**9; ## Intensity of neutrons in neutrons/cm^2-sec from isotropic point source\n",
+ "## 1 feet = 30.48 cm\n",
+ "d = 10*30.48; ## Distance of concrete wall from a point source in cm\n",
+ "## As Intensity obeys inverse square law\n",
+ "I = phi0/(4.*math.pi*d**2); ## Intensity of neutron beam in terms of neutrons/cm^2-sec\n",
+ "H_dot = 1.; ## The required dose equivalent rate in mrem/hour\n",
+ "## From Figure 10.23(b)\n",
+ "H0_dot = H_dot/I; ## The dose equivalent rate\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The reduced dose equivalent rate due to concrete wall is = \",H0_dot,\" mrem/hr \\n\");\n",
+ "## By looking into Figure 10.23(b) on thickness axis\n",
+ "print(\" \\n The concrete slab thickness is = 70 cm \\n\");\n",
+ "\t\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The reduced dose equivalent rate due to concrete wall is = 0.00 mrem/hr \n",
+ "\n",
+ " \n",
+ " The concrete slab thickness is = 70 cm \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg598"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 10.8\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "R = 7.*30.48; ## Distance of core from the center of shield in cm\n",
+ "## Assuming average energy produced per fission reaction is 200 MeV \n",
+ "P = 10.; ## Power of teaching reactor in Watts\n",
+ "P_fission = 200*10**6*1.6*10**(-19); ## Energy produced in a fission reaction in terms of joule\n",
+ "fission_rate = P/P_fission; ## Number of fission reactions\n",
+ "\n",
+ "## By assuming that the gammma rays are of equal energy of 1 MeV (Group 1) and looking into Table 10.5\n",
+ "E1 = 1.; ## Energy of gamma rays in MeV (Assumed)\n",
+ "chi_pn1 = 5.2; ## Number of prompt gamma rays emitted per fission with energy 2 MeV\n",
+ "S1 = chi_pn1*fission_rate; ## Source strength in gamma rays/sec\n",
+ "## Using the data from Table II.4 for E = 1 MeV for water\n",
+ "mu1 = 0.0996; ## Mass attenuation coefficient at 1 MeV in cm^-1\n",
+ "print'%s %.2f %s'%(\" \\n Buildup factor is due to water measured at \",mu1*R,\"\");\n",
+ "## Using the data from Table 10.2 at 1 MeV\n",
+ "B_p1 = 373.;\n",
+ "phi_b1 = (S1/(4.*math.pi*R**2))*B_p1*math.exp(-mu1*R); ## Buildup flux\n",
+ "## Using the data from Table II.5 for 1 MeV \n",
+ "mua_rho_air1 = 0.028; ## The ratio of total attenuation coefficient to density of air in cm^2/g\n",
+ "## Calculation\n",
+ "X_dot1 = 0.0659*E1*mua_rho_air1*phi_b1;\n",
+ "print'%s %.2f %s'%(\"\\n Exposure rate due to group 1 = \",X_dot1,\" mR/hour \\n\");\n",
+ "\n",
+ "## By assuming that the gammma rays are of equal energy of 2 MeV (Group 2) and looking into Table 10.5\n",
+ "E2 = 2.; ## Energy of gamma rays in MeV (Assumed)\n",
+ "chi_pn2 = 1.8; ## Number of prompt gamma rays emitted per fission with energy 2 MeV\n",
+ "S2 = chi_pn2*fission_rate; ## Source strength in gamma rays/sec\n",
+ "## Using the data from Table II.4 for E = 2 MeV for water\n",
+ "mu2 = 0.0493; ## Mass attenuation coefficient at 2 MeV in cm^-1\n",
+ "print'%s %.2f %s'%(\" \\n Buildup factor is due to water measured at \",mu2*R,\"\");\n",
+ "## Using the data from Table 10.2 at 2 MeV\n",
+ "B_p2 = 13.1;\n",
+ "phi_b2 = (S2/(4.*math.pi*R**2))*B_p2*math.exp(-mu2*R); ## Buildup flux \n",
+ "## Using the data from Table II.5 for 2 MeV \n",
+ "mua_rho_air2 = 0.0238; ## The ratio of total attenuation coefficient to density of air in cm^2/g\n",
+ "## Calculation \n",
+ "X_dot2 = 0.0659*E2*mua_rho_air2*phi_b2;\n",
+ "print'%s %.2f %s'%(\"\\n Exposure rate due to group 2 = \",X_dot2,\" mR/hour \\n\");\n",
+ "\n",
+ "## By assuming that the gammma rays are of equal energy of 4 MeV (Group 3) and looking into Table 10.5\n",
+ "E3 = 4.; ## Energy of gamma rays in MeV (Assumed)\n",
+ "chi_pn3 = 0.22; ## Number of prompt gamma rays emitted per fission with energy 4 MeV\n",
+ "S3 = chi_pn3*fission_rate; ## Source strength in gamma rays/sec\n",
+ "## Using the data from Table II.4 for E = 4 MeV for water\n",
+ "mu3 = 0.0339; ## Mass attenuation coefficient at 4 MeV in cm^-1\n",
+ "print'%s %.2f %s'%(\" \\n Buildup factor is due to water measured at\",mu3*R,\"\");\n",
+ "## Using the data from Table 10.2 at 4 MeV\n",
+ "B_p3 = 5.27;\n",
+ "phi_b3 = (S3/(4.*math.pi*R**2))*B_p3*math.exp(-mu3*R); ## Buildup flux \n",
+ "## Using the data from Table II.5 for 4 MeV \n",
+ "mua_rho_air3=0.0194; ## The ratio of total attenuation coefficient to density of air in cm^2/g\n",
+ "## Calculation\n",
+ "X_dot3 = 0.0659*E3*mua_rho_air3*phi_b3;\n",
+ "print'%s %.2f %s'%(\"\\n Exposure rate due to group 3 = \",X_dot3,\" mR/hour \\n\");\n",
+ "\n",
+ "## By assuming that the gammma rays are of equal energy of 6 MeV (Group 4) and looking into Table 10.5\n",
+ "E4 = 6; ## Energy of gamma rays in MeV (Assumed)\n",
+ "chi_pn4 = 0.025; ## Number of prompt gamma rays emitted per fission with energy 4 MeV\n",
+ "S4 = chi_pn4*fission_rate; ## Source strength in gamma rays/sec\n",
+ "## Using the data from Table II.4 for E = 6 MeV for water\n",
+ "mu4 = 0.0275; ## Mass attenuation coefficient at 6 MeV in cm^-1\n",
+ "print'%s %.2f %s'%(\" \\n Buildup factor is due to water measured at \",mu4*R,\"\");\n",
+ "## Using the data from Table 10.2 at 6 MeV\n",
+ "B_p4 = 3.53;\n",
+ "phi_b4 = (S4/(4.*math.pi*R**2))*B_p4*math.exp(-mu4*R); ## Buildup flux \n",
+ "## Using the data from Table II.5 for 4 MeV \n",
+ "mua_rho_air4=0.0172; ## The ratio of total attenuation coefficient to density of air in cm^2/g\n",
+ "## Calculation\n",
+ "X_dot4 = 0.0659*E4*mua_rho_air4*phi_b4;\n",
+ "print'%s %.2f %s'%(\"\\n Exposure rate due to group 3 = \",X_dot4,\" mR/hour \\n\");\n",
+ "\n",
+ "##Calculation\n",
+ "X_dot = X_dot1+X_dot2+X_dot3+X_dot4;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n The total exposure rate due to all groups = \",X_dot,\" mR/hour \\n\");\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Buildup factor is due to water measured at 21.25 \n",
+ "\n",
+ " Exposure rate due to group 1 = 0.00 mR/hour \n",
+ "\n",
+ " \n",
+ " Buildup factor is due to water measured at 10.52 \n",
+ "\n",
+ " Exposure rate due to group 2 = 1.09 mR/hour \n",
+ "\n",
+ " \n",
+ " Buildup factor is due to water measured at 7.23 \n",
+ "\n",
+ " Exposure rate due to group 3 = 2.34 mR/hour \n",
+ "\n",
+ " \n",
+ " Buildup factor is due to water measured at 5.87 \n",
+ "\n",
+ " Exposure rate due to group 3 = 0.93 mR/hour \n",
+ "\n",
+ "\n",
+ " The total exposure rate due to all groups = 4.36 mR/hour \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg603"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 10.9\n",
+ "import math\n",
+ "## Given data\n",
+ "## Assuming average energy produced per fission reaction is 200 MeV \n",
+ "P = 55.; ## Power density of reactor in watts/cm^3\n",
+ "rho_eff_U = 0.33; ## Effective density of uranium in g/cm^3\n",
+ "rho_eff_W = 1-rho_eff_U; ## Effective density of water in g/cm^3\n",
+ "t_i = 3.; ## Average time spent by water in the reactor in seconds\n",
+ "t_0 = 2.; ## Average time spent by water in the external coolant circuit in seconds\n",
+ "## 1 eV = 1.6*10^(-19) J\n",
+ "P_fission = 200.*10**6*1.6*10**(-19); ## Energy produced in a fission reaction in terms of joule\n",
+ "fission_density = P/P_fission; ## Number of fission reactions\n",
+ "v = 2.42; ## Number of fission neutrons emitted per fission\n",
+ "S = v*fission_density; ## Strength of neutron source in terms of neutrons/cm^2-sec\n",
+ "## Atom density of oxygen at normal density of 1 g/cm^3 is\n",
+ "rho = 1.; ## Density of water in g/cm^3\n",
+ "N_A = 6.02*10**(23); ## Avogadro's constant\n",
+ "M = 18.; ## Molecular weight of water\n",
+ "atom_density = (rho*N_A)/M;\n",
+ "\n",
+ "## Using the data from Table 10.7\n",
+ "sigma_r = 1.9*10**(-5)*10**(-24); ## Reaction cross section in cm^2\n",
+ "T_12 = 7.1; ## Half life of the given reaction in seconds\n",
+ "lambd = 0.693/T_12; ## Decay constant of the given reaction in seconds^(-1)\n",
+ "sigma_act = rho_eff_W*atom_density*sigma_r; ## Average macroscopic activation cross section\n",
+ "## Using the data from Table 10.4\n",
+ "sigma_RW = 0.103; ## Removal cross section of water in cm^-1\n",
+ "sigma_RU = 0.174; ## Removal cross section of Uranium in cm^-1\n",
+ "sigma_R = (sigma_RU*rho_eff_U)+(sigma_RW*rho_eff_W); ## Removal cross section of mixture\n",
+ "## Let activation rate given by (sigma_act*phi_av) be denoted as AR\n",
+ "AR = (sigma_act*S)/sigma_R;\n",
+ "## Calculation\n",
+ "alpha = AR*(1.-math.exp(-t_i*lambd))/(1.-math.exp(-(t_i+t_0)*lambd));\n",
+ "## 1 curie = 3.7*10^(10) disintegrations/sec\n",
+ "## Result\n",
+ "print'%s %.2e %s %.2f %s '%(\"\\n Equilibrium activity of water due to neutron capture of oxygen = \",alpha,\" disintegrations/cm^3-sec\" and \" \",math.ceil(alpha*10**6/(3.7*10**10)),\" uCi/cm^3 \\n\");\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Equilibrium activity of water due to neutron capture of oxygen = 9.21e+06 249.00 uCi/cm^3 \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter11_1.ipynb b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter11_1.ipynb new file mode 100644 index 00000000..51ac6a10 --- /dev/null +++ b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter11_1.ipynb @@ -0,0 +1,736 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7b65f64c10e68c2da7cb8433ded519718072a74fa1329803744f3f81b62d3e1a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter11-Reactor Licensing Safety and the Environment "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg646"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.1\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "h = 30.; ## Height at which the effluent is relaesed\n",
+ "## Calculation of maxima location \n",
+ "sigma_z = h/math.sqrt(2.); ## Vertical dispersion coefficient\n",
+ "## Using the plot given in Figure 11.12 for Type F condition\n",
+ "## The corresponding value with calculated maximum location is noted. \n",
+ "h_max = 1900.;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The point of maximum concentration of non-radioactive effluent = \",h_max,\" m \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The point of maximum concentration of non-radioactive effluent = 1900.00 m \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg654"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.2\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "A = 2*10**3; ## Amount of radioactivity release due to Xenon-133 in curie\n",
+ "t = 365.*24.*3600.; ## Time in seconds\n",
+ "Q_dash = A/t; ## Average emission rate of Xenon-133\n",
+ "h = 100.; ## Location of radioactivity release through vent\n",
+ "v_bar = 1.; ## Wind speed in m/sec\n",
+ "## Using the plot given in Figure 11.11 and 11.12 for Type F condition at 100 m\n",
+ "sigma_y = 275.; ## Horizontal dispersion coefficient\n",
+ "sigma_z = 46.; ## Vertical dispersion coefficient\n",
+ "chi = (Q_dash*math.exp(-h**2/(2.*sigma_z**2)))/(math.pi*v_bar*sigma_y*sigma_z); ## Radionuclide concentration in terms of Ci/cm^3\n",
+ "## Using data from Table 11.3\n",
+ "Eg_bar = 0.03; ## Average gamma decay energy per disintegration in MeV\n",
+ "## Calculation \n",
+ "H_dot = 0.262*chi*Eg_bar*t*10**3; ## The units are in mrem/year\n",
+ "## Expressing the dose rate in SI units\n",
+ "H_dot_SI = 2.62*chi*Eg_bar*t*10**3;\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n The external gamma dose rate due to xenon release under type F condition = \",H_dot,\" mrem/year\" or \" \",H_dot_SI,\"mSv/year \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The external gamma dose rate due to xenon release under type F condition = 0.04 mrem/year 0.37 mSv/year \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg655"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.3\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Using data from Table 11.3\n",
+ "Eg_bar = 0.00211; ## Average gamma decay energy per disintegration in MeV\n",
+ "## Calculation\n",
+ "C_y = 0.262*Eg_bar;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" \\n The dose rate factor due to krypton exposure = \",C_y,\" rem*m^3/sec-Ci \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The dose rate factor due to krypton exposure = 5.53e-04 rem*m^3/sec-Ci \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg656"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.4\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## The results given in Example 11.2 are to be used in this problem\n",
+ "chi = 1.5*10**(-10); ## Radionuclide concentration in terms of Ci/cm^3\n",
+ "t = 365.*24.*3600.; ## Time in seconds\n",
+ "## Using data from Table 11.3\n",
+ "Ebeta_bar = 0.146; ## Average gamma decay energy per disintegration in MeV\n",
+ "f = 1.; ## Experimentally detemined factor\n",
+ "## Calculation\n",
+ "H_dot = 0.229*Ebeta_bar*chi*f*t;\n",
+ "## Expressing the result in milli-rem\n",
+ "print'%s %.2f %s'%(\" \\n The external beta dose rate due to xenon exposure for a year = \",H_dot*10**3,\" mrem/year \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The external beta dose rate due to xenon exposure for a year = 0.16 mrem/year \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg658"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.5\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "A = 1.23; ## Amount of radioactivity release due to I-131 in curie in one year\n",
+ "h = 30.; ## Location of radioactivity release through vent in meter\n",
+ "v_bar = 1.2; ## Wind speed in m/sec\n",
+ "T_12 = 8.04; ## Half life of Iodine 131 in days\n",
+ "T_12b = 138.; ## Biological half life of Iodine 131 in days\n",
+ "zeta = 0.23; ## Fraction of core inventory in MeV \n",
+ "q = 0.23; ## Fraction of Iodine-131 in thyroid\n",
+ "M = 20.; ## Mass of an adult thyroid in grams\n",
+ "\n",
+ "## 1.\n",
+ "t = 365.*24.*3600.; ## Time in seconds\n",
+ "Q_dash = A/t; ## Average emission rate of Iodine-131\n",
+ "## Converting days into seconds by using 1 day = 86400 seconds\n",
+ "lambd = 0.693/(T_12*86400); ## Decay constant of Iodine-131\n",
+ "lambda_b = 0.693/(T_12b*86400.); ## Biological decay constant of Iodine-131\n",
+ "## Let the quantity chi*v_bar/Q_bar = x\n",
+ "## Using the plot given in Figure 11.13 for Type E condition at 2000 m\n",
+ "x = 6.*10**(-5);\n",
+ "## Solving for chi\n",
+ "chi = (x*Q_dash)/v_bar;\n",
+ "## As per the standards of International Commission on Radiolgical Protection (ICRP) \n",
+ "B = 2.32*10**(-4); ## Normal breathing rate in m^3/sec\n",
+ "## Calculation\n",
+ "H_dot = (592.*B*zeta*q*chi)/(M*(lambd+lambda_b));\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" \\n The equilibrium dose rate to an adult thyroid = \",H_dot,\"sem/sec \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "## Calculation\n",
+ "H = H_dot*t;\n",
+ "## Expressing the result in milli-rem\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The annual dose to an adult thyroid = \",H*10**3,\" mrem \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The equilibrium dose rate to an adult thyroid = 6.71e-10 sem/sec \n",
+ "\n",
+ " \n",
+ " The annual dose to an adult thyroid = 21.16 mrem \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg662"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.6\n",
+ "import math\n",
+ "\n",
+ "## Given datah\n",
+ "E = 0.66; ## Energy of gamma ray emitted by caesium in MeV\n",
+ "x = 100.; ## Height of exposure in cm\n",
+ "## Using the data from Table II.5 for air at E = 0.66 MeV\n",
+ "mua_rho = 0.0294; ## The ratio of absorption coefficient to density of air in cm^2/g\n",
+ "## Using the data from Table II.4 for air at E = 0.66 MeV\n",
+ "mu_rho = 0.0775; ## The ratio of attenuation coefficient to density of air in cm^2/g\n",
+ "## Using standard value for density of air\n",
+ "rho = 1.293*10**(-3);\n",
+ "mu = mu_rho*rho;\n",
+ "mux = mu*x;\n",
+ "gamma = 0.57722; ## Euler's constant\n",
+ "E1 = -gamma+math.log(1./mux)+mux; ## Conversion factor \n",
+ "## Using parameter data from Table 11.16\n",
+ "C = 1.41; ## A constant\n",
+ "beta = 0.0857; ## A constant\n",
+ "## Calculation\n",
+ "H_dot_S = 3.39*10**(-2)*E*mua_rho*(E1+(C*math.exp(-(1.-beta)*mux)/(1.-beta)));\n",
+ "## Converting time in hours by 1 hour = 3600 seconds\n",
+ "## Result\n",
+ "print'%s %.2e %s %.2f %s '%(\" \\n The gamma ray dose rate conversion factor due to caesium-137 = \",H_dot_S,\" rem*m^2/sec-Ci\" or \" \",H_dot_S*3600,\"rem*m^2/hour-Ci\\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The gamma ray dose rate conversion factor due to caesium-137 = 3.66e-03 rem*m^2/sec-Ci 13.18 rem*m^2/hour-Ci\n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg665"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.7\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "C0 = 6.25*10**6; ## Amount of initial radioactivity release due to I-131 in curie\n",
+ "p = 0.1; ## Leakage rate in percent\n",
+ "t0 = 2*3600.; ## Analysis time in seconds\n",
+ "v_bar = 1.; ## Wind speed in m/sec\n",
+ "\n",
+ "## 1.\n",
+ "lambdal = 0.01*p/86400.; ## Decay constant corresponding to leakage rate in seconds (1 day = 86400 seconds)\n",
+ "## Using the data from Example 11.5 for the half life of Iodine-131\n",
+ "T_12 = 8.04; ## Half life of Iodine 131 in days\n",
+ "lambdac = 0.693/(T_12*86400.); ## Decay constant of Iodine-131 (I-131) in seconds\n",
+ "## Using data from Table 11.3\n",
+ "Eg_bar = 0.371; ## Average gamma decay energy per disintegration of I-131 in MeV\n",
+ "## Using the plot given in Figure 11.11 and 11.12 for Type F condition at 100 m\n",
+ "sigma_y = 21.; ## Horizontal dispersion coefficient\n",
+ "sigma_z = 70.; ## Vertical dispersion coefficient\n",
+ "## As lambdac*t << 1, \n",
+ "## Calculation\n",
+ "H = (0.262*Eg_bar*lambdal*C0*t0)/(math.pi*v_bar*sigma_y*sigma_z);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The total external dose due to gamma ray exposure = \",H,\" rem\\n\")\n",
+ "\n",
+ "## 2.\n",
+ "## Using the data given in Example 11.5\n",
+ "B = 2.32*10**(-4); ## Normal breathing rate in m^3/sec\n",
+ "zeta = 0.23; ## Fraction of core inventory in MeV \n",
+ "q = 0.23; ## Fraction of Iodine-131 in thyroid\n",
+ "M = 20.; ## Mass of an adult thyroid in grams\n",
+ "## Calculation\n",
+ "H_dot = (592.*B*zeta*q*lambdal*C0*t0)/(math.pi*v_bar*sigma_y*sigma_z*M);\n",
+ "## Converting the units from rem/sec to milli-rem/hour by multiplying by (1000*3600)\n",
+ "## Result\n",
+ "print'%s %.2e %s %.2f %s '%(\" \\n The dose rate to an adult thyroid after 2 hours = \",H_dot,\" rem/sec\" or\"\",math.ceil(H_dot*(1000*3600)),\" mrem/hour\\n\");\n",
+ "\n",
+ "## 3.\n",
+ "## Using the data given in Example 11.5\n",
+ "T_12 = 8.04; ## Half life of Iodine 131 in days\n",
+ "T_12b = 138.; ## Biological half life of Iodine 131 in days\n",
+ "lambd = 0.693/(T_12*86400.); ## Decay constant of Iodine-131 in sec^(-1)\n",
+ "lambda_b = 0.693/(T_12b*86400.); ## Biological decay constant of Iodine-131 in sec^(-1)\n",
+ "## Calculation\n",
+ "H = H_dot/(lambd+lambda_b);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The dose commitment to thyroid by Iodine-131 exposure after 2 hours = \",H,\" rem \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The total external dose due to gamma ray exposure = 0.01 rem\n",
+ "\n",
+ " \n",
+ " The dose rate to an adult thyroid after 2 hours = 4.10e-05 rem/sec 148.00 mrem/hour\n",
+ " \n",
+ " \n",
+ " The dose commitment to thyroid by Iodine-131 exposure after 2 hours = 38.81 rem \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg667"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.8\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given\n",
+ "E = 2.4; ## Energy of gamma rays in MeV\n",
+ "r = 1000*100; ## Distance from the building where radiation is exposed in cm\n",
+ "t0 = 2*3600; ## Time of exposure in seconds\n",
+ "A = 3*10^7; ## Amount of initial radioactivity release due to Kr-88 in curie\n",
+ "f = 0.4; ## Fraction of disintegrations which release 2.4 MeV gamma rays\n",
+ "C0 = A*f; ## Effective number of curies \n",
+ "T_12 = 2.79; ## Half life of Iodine 131 in hours\n",
+ "\n",
+ "lambd = 0.693/(T_12*3600.); ## Decay constant of Iodine-131 in sec^(-1)\n",
+ "## Using the result given in Example 11.7\n",
+ "lambdal = 1.16*10**(-8); ## Decay constant corresponding to fission prouduct release from building\n",
+ "lambdac = lambd+lambdal; ## Total decay constant in sec^(-1) \n",
+ "## Using the data from Table II.4 for air at E = 2.4 MeV\n",
+ "mu_rho = 0.041; ## The attenuation coefficient in cm^2/g\n",
+ "## Using standard value for density of air in g/cm^3\n",
+ "rho = 1.293*10**(-3);\n",
+ "mu = mu_rho*rho;\n",
+ "## Using the data from Table II.5 for air at E = 2.4 MeV\n",
+ "mua_rho = 0.0227; ## The ratio of absorption coefficient to density of air in cm^2/g\n",
+ "print'%s %.2f %s'%(\" \\n Buildup factor is measured at \",mu*r,\"\");\n",
+ "## Using Berger's form in Problem 11.9 \n",
+ "B_p = 4.7; ## Buildup factor due to a point source\n",
+ "## Calculation\n",
+ "H = (54.*C0*(1.-math.exp(-lambdac*t0))/lambdac)*(E*mua_rho*B_p*math.exp(-mu*r)/r**2);\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" \\n The direct dose due to gamma ray exposure = \",H,\" rem \\n\")\n",
+ "## There is a slight deviation in the answer due to approximation of value in the textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Buildup factor is measured at 5.30 \n",
+ " \n",
+ " The direct dose due to gamma ray exposure = 3.91e-07 rem \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.9\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "gammai = 0.0277; ## Fission yield of Iodine-131 in fraction\n",
+ "P = 3200.; ## Thermal power of the plant in MW\n",
+ "## Calculation\n",
+ "alphai = 8.46*10**5*P*gammai;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The saturation activity of Iodine-131 during reactor operation = \",alphai,\" curie \\n\")\n",
+ "\n",
+ "## Using assumption 2 of Nuclear Regulatory Commission (NRC) in calculation of radii of exclusion zone and Low Population Zone (LPZ)\n",
+ "## Due to core meltdown, 25 percent of iodine inventory is released and out of which 91 percent is in elemental form.\n",
+ "Fp = 0.25*0.91; ## Fraction of Iodine-131 released from the fuel into the reactor containment\n",
+ "## As entire iodine escapes through air\n",
+ "Fb = 1.; ## Fraction of 'Fp' which remains airborne and is capable of escaping from the building\n",
+ "## Calculation\n",
+ "C0 = alphai*Fp*Fb;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The activity of Iodine-131 in elemental form due to core meltdown = \",C0,\" curie \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The saturation activity of Iodine-131 during reactor operation = 74989440.00 curie \n",
+ "\n",
+ " \n",
+ " The activity of Iodine-131 in elemental form due to core meltdown = 17060097.60 curie \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg714"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.10\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "P = 1000.; ## Electrical power of the plant in Mwe\n",
+ "eta = 0.38; ## Plant efficiency\n",
+ "P_th = P/eta; ## Thermal power of the plant in MW\n",
+ "h = 100.; ## Height of stack in metre\n",
+ "t = 24*365.; ## The number of hours in a year\n",
+ "m0 = 13000.; ## Amount of coal in the plant in Btu/lb\n",
+ "m0_ash = 0.09; ## Fraction of ash in the coal\n",
+ "\n",
+ "## 1.\n",
+ "E = P_th*t; ## Energy released in a year in MW-hour\n",
+ "## Converting the units in Btu by using 1 MW-hour = 3.412*10^6 Btu\n",
+ "m = (E*3.412*10**6)/m0;\n",
+ "## Converting the units in g/year by using 1 lb/year = 453.59237 g/year\n",
+ "m = m*453.59237;\n",
+ "## Assuming the fly ash equipment has an efficiency of 97.5% of fly ash removal\n",
+ "eta_flyash = 0.025; ## Only (1-efficiency) is exhausted\n",
+ "m_ash = eta_flyash*m0_ash*m;\n",
+ "## A typical power plant contains 3pCi/g of each nuclide (Radium-226) in one year\n",
+ "A = 3*10**(-12);\n",
+ "## Calculation\n",
+ "A_total = A*m_ash;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Total activity of Radium-226 emitted = \",A_total,\" curie \\n\")\n",
+ "\n",
+ "## 2.\n",
+ "v_bar = 1.; ## Wind speed in m/sec\n",
+ "t = 24.*365.*3600.; ## Analysis time of one year equivalently in seconds\n",
+ "MPC = 3.*10**(-12.); ## Maximum Permissible Concentration in micro-curie/cm^3\n",
+ "Q_bar = A_total/t; ## Emission rate for one year in curie/year\n",
+ "## Let the quantity chi*v_bar/Q_bar = x\n",
+ "## Using the plot for Pasquill F, given in Fig. A.7, Pg no 413 from Slade, D. H., Editor, Meteorology and Atomic Energy-1968. U. S. Atomic Energy Commission Report TID-24190, 1968.\n",
+ "x = 2.5*10**(-6.); ## Maximum value of x at 10^4 m \n",
+ "## Solving for chi\n",
+ "chi = (x*Q_bar)/v_bar;\n",
+ "## Converting the units from Ci/m^3 to micro-curie/cm^3 by multiplying by 10^6/10^6 = 1\n",
+ "print'%s %.2e %s'%(\" \\n Concentration of Radium-226 present = \",chi,\" micro-curie/cm^3 \\n\")\n",
+ "## Let c be the comparison factor\n",
+ "## Calculation\n",
+ "c = MPC/chi;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n On comparison, the total concentration of Radium-226 is \",c,\" times smaller than Maximum Permissible Concentration (MPC) \\n\")\n",
+ "## The comparison factor is slightly different from the value in the textbook. This is because of approximation used in the textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Total activity of Radium-226 emitted = 0.02 curie \n",
+ "\n",
+ " \n",
+ " Concentration of Radium-226 present = 1.47e-15 micro-curie/cm^3 \n",
+ "\n",
+ " \n",
+ " On comparison, the total concentration of Radium-226 is 2042.83 times smaller than Maximum Permissible Concentration (MPC) \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg718"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.11\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "Qy_bar = 1.04*10**(-2); ## Emission rate for one year in curie/year\n",
+ "## Let (chi/Q_bar) = d which is called 'Dilution factor'\n",
+ "d = 4*10**(-8); ## Dilution factor in year/cm^3\n",
+ "vd = 0.01; ## Experimentally determined constant\n",
+ "\n",
+ "## 1.\n",
+ "T_12 = 8.04; ## Half life of Iodine 131 in days\n",
+ "T_12f = 14; ## First order half life of Iodine 131 in days\n",
+ "## Converting days into years by using 1 year = 365 days\n",
+ "lambd = 0.693/(T_12/365.); ## Decay constant of Iodine-131\n",
+ "lambdaf = 0.693/(T_12f/365.); ## First order decay constant of Iodine-131\n",
+ "## Calculation\n",
+ "Cf = (Qy_bar*d*vd)/(lambd+lambdaf);\n",
+ "## Expressing the result in micro-curie \n",
+ "Cf = Cf*10**6;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" \\n The activity of I-131 on the vegetation = \",Cf,\" micro-curie/m^2 \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "## The proportionality factor has a value 9*10^(-5) Ci/cm^3 of milk per Ci/m^2 of grass\n",
+ "## Calculation \n",
+ "Cm = 9*10**(-5)*Cf;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" \\n The concentration of I-131 in the milk = \",Cm,\" micro-curie/m^2 \\n\");\n",
+ "\n",
+ "## 3.\n",
+ "MPC = 3*10**(-7); ## Maximum Permissible Concentration in micro-curie/cm^3\n",
+ "## Calculation\n",
+ "H_dot = (2270.*Cm)/MPC;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" \\n The annual dose rate to an infant thyroid by consuming radiated milk = \",H_dot,\" mrem/year \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The activity of I-131 on the vegetation = 8.40e-08 micro-curie/m^2 \n",
+ "\n",
+ " \n",
+ " The concentration of I-131 in the milk = 7.56e-12 micro-curie/m^2 \n",
+ "\n",
+ " \n",
+ " The annual dose rate to an infant thyroid by consuming radiated milk = 5.72e-02 mrem/year \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg721"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 11.12\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "Qy_bar = 0.197; ## Emission rate for one year in micro-curie/year\n",
+ "## Let (chi/Q_bar) = d which is called 'Dilution factor'\n",
+ "d = 6.29*10**(-16); ## Dilution factor in year/cm^3\n",
+ "MPC_w = 6*10**(-5); ## Maximum Permissible Concentration (MPC) of iron in micro-curie/cm^3\n",
+ "\n",
+ "Cw = Qy_bar*d; ## The concentration of Fe-59 \n",
+ "## For fish\n",
+ "Rs_fish = 110.; ## Consumption rate in g/day\n",
+ "## Using the data from Table 11.15 for saltwater concentration of fish for iron\n",
+ "CF_fish = 1800.; ## Concentration Factor of fish\n",
+ "Cs_fish = CF_fish*Cw; ## Activity of fish\n",
+ "H_dot_fish = (Cs_fish*Rs_fish*500.)/(MPC_w*2200.); ## Dose rate for fish\n",
+ "\n",
+ "## For mollusks\n",
+ "Rs_mollusk = 10.; ## Consumption rate in g/day\n",
+ "## Using the data from Table 11.15 for saltwater concentration of mollusk for iron\n",
+ "CF_mollusk = 7600.; ## Concentration Factor of mollusk\n",
+ "Cs_mollusk = CF_mollusk*Cw; ## Activity of mollusk\n",
+ "H_dot_mollusk = (Cs_mollusk*Rs_mollusk*500.)/(MPC_w*2200.); ## Dose rate for mollusk\n",
+ "\n",
+ "## For crustaceans\n",
+ "Rs_crustacean = 10.; ## Consumption rate in g/day\n",
+ "## Using the data from Table 11.15 for saltwater concentration of crustacean for iron\n",
+ "CF_crustacean = 2000.; ## Concentration Factor of crustacean\n",
+ "Cs_crustacean = CF_crustacean*Cw; ## Activity of crustacean\n",
+ "H_dot_crustacean = (Cs_crustacean*Rs_crustacean*500.)/(MPC_w*2200.); ## Dose rate for crustacean\n",
+ "\n",
+ "## Calculation\n",
+ "H_dot = H_dot_fish+H_dot_mollusk+H_dot_crustacean;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" \\n The annual dose rate to GI tract by consuming fish = \",H_dot_fish,\" mrem/year\");\n",
+ "print'%s %.2e %s'%(\" \\n The annual dose rate to GI tract by consuming mollusk = \",H_dot_mollusk,\" mrem/year\");\n",
+ "print'%s %.2e %s'%(\" \\n The annual dose rate to GI tract by consuming crustaceans = \",H_dot_crustacean,\" mrem/year\");\n",
+ "print'%s %.2e %s'%(\" \\n The annual dose rate to GI tract by consuming seafood = \",H_dot,\" mrem/year \\n\");\n",
+ "## The answer for annual dose rate to GI tract by consuming fish is wrong in the textbook. This is because the value of fish consumption rate is wrongly considered.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The annual dose rate to GI tract by consuming fish = 9.29e-08 mrem/year\n",
+ " \n",
+ " The annual dose rate to GI tract by consuming mollusk = 3.57e-08 mrem/year\n",
+ " \n",
+ " The annual dose rate to GI tract by consuming crustaceans = 9.39e-09 mrem/year\n",
+ " \n",
+ " The annual dose rate to GI tract by consuming seafood = 1.38e-07 mrem/year \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter2_1.ipynb b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter2_1.ipynb new file mode 100644 index 00000000..421dbd88 --- /dev/null +++ b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter2_1.ipynb @@ -0,0 +1,818 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:b88f8253908896ceb538355ebb97029247893ecdea7fc487afd787ce682bc949"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Atomic and Nuclear Physics "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg8"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.1\n",
+ "import math\n",
+ "#determine\n",
+ "## Given data\n",
+ "atom_h = 6.6*10**24; ## Number of atoms in Hydrogen\n",
+ "## Using the data given in Table II.2, Appendix II for isotropic abundance of deuterium\n",
+ "isoab_H2 = 0.015; ## Isotropic abundance of deuterium\n",
+ "## Calculation\n",
+ "totatom_d=(isoab_H2*atom_h)/100.;\n",
+ "## Result\n",
+ "print\"%s %.2e %s \"%('\\n Number of deuterium atoms = ',totatom_d,'');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Number of deuterium atoms = 9.90e+20 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.2\n",
+ "import math\n",
+ "#determine\n",
+ "## Given data \n",
+ "## Using the data given in the example 2.2\n",
+ "atwt_O16 = 15.99492; ## Atomic weight of O-16 isotope\n",
+ "isoab_O16 = 99.759; ## Abundance of O-16 isotope\n",
+ "atwt_O17 = 16.99913; ## Atomic weight of O-17 isotope\n",
+ "isoab_O17 = 0.037; ## Abundance of O-17 isotope\n",
+ "atwt_O18 = 17.99916; ## Atomic weight of O-18 isotope\n",
+ "isoab_O18 = 0.204; ## Abundance of O-18 isotope\n",
+ "## Calculation\n",
+ "atwt_O=(isoab_O16*atwt_O16 + isoab_O17*atwt_O17 + isoab_O18*atwt_O18)/100.;\n",
+ "## Result\n",
+ "print\"%s %.2f %s \"%('\\n Atomic Weight of Oxygen = ',atwt_O,'');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Atomic Weight of Oxygen = 16.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.3\n",
+ "import math\n",
+ "#determine\n",
+ "## Given data\n",
+ "me = 9.1095*10**(-28); ## Mass of electron in grams\n",
+ "c = 2.9979*10**10; ## Speed of light in vacuum in cm/sec\n",
+ "## Calculation\n",
+ "rest_mass = me*c**2;\n",
+ "## Result\n",
+ "print\"%s %.2e %s \"%('\\n Rest mass energy of electron = ',rest_mass,' ergs\\n');\n",
+ "print('Expressing the result in joules')\n",
+ "## 1 Joule = 10^(-7)ergs\n",
+ "rest_mass_j = rest_mass*10**(-7);\n",
+ "print\"%s %.2e %s \"%('\\n Rest mass energy of electron = ',rest_mass_j,' joules\\n');\n",
+ "print('Expressing the result in MeV')\n",
+ "## 1 MeV = 1.6022*10^(-13)joules\n",
+ "rest_mass_mev = rest_mass_j/(1.6022*10**(-13));\n",
+ "print\"%s %.2f %s \"%('\\n Rest mass energy of electron = ',rest_mass_mev,' MeV\\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Rest mass energy of electron = 8.19e-07 ergs\n",
+ " \n",
+ "Expressing the result in joules\n",
+ "\n",
+ " Rest mass energy of electron = 8.19e-14 joules\n",
+ " \n",
+ "Expressing the result in MeV\n",
+ "\n",
+ " Rest mass energy of electron = 0.51 MeV\n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.4\n",
+ "\n",
+ "import math\n",
+ "#Calculate the \n",
+ "## From the result of Example 2.3\n",
+ "## Rest mass energy of electron = 0.5110 MeV\n",
+ "rest_mass_mev = 0.5110;\n",
+ "me = 9.1095*10**(-28); ## Mass of electron in grams\n",
+ "## From standard data table\n",
+ "## 1 amu = 1.6606*10^(-24)g\n",
+ "amu = 1.6606*10**(-24);\n",
+ "## Calculation\n",
+ "en_eq = (amu/me)*rest_mass_mev;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Energy equivalent of one amu = ',en_eq,' MeV\\n');\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Energy equivalent of one amu = 931.52 MeV\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.5\n",
+ "\n",
+ "import math\n",
+ "#Calculate the \n",
+ "## From the standard data table\n",
+ "h = 6.626*10**(-34); ## Planck's constant in J-s\n",
+ "c = 3*10**8; ## Speed of light in vacuum in m/sec\n",
+ "## Given data \n",
+ "print('The ionization energy of K shell electron in Lead atom is 88keV');\n",
+ "E = 88*10**3; ## Ionization energy in keV\n",
+ "## Expressing the result in joules by using 1 eV = 1.6022*10^(-19) J\n",
+ "E = E*1.6022*10**(-19);\n",
+ "print(\"From Planck\\''s law of photoelectric effect \\n Energy = (h*c)/lambda\\n\");\n",
+ "## Calculation \n",
+ "lambd = (h*c)/E; \n",
+ "## Result\n",
+ "print'%s %.3e %s'%('\\n Wavelength of radiation = ',lambd,' m\\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ionization energy of K shell electron in Lead atom is 88keV\n",
+ "From Planck''s law of photoelectric effect \n",
+ " Energy = (h*c)/lambda\n",
+ "\n",
+ "\n",
+ " Wavelength of radiation = 1.410e-11 m\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.6\n",
+ "import math\n",
+ "#Calculate the \n",
+ "## Given data\n",
+ "T12 = 64.8; ## Half life = 64.8 hour\n",
+ "lambd = 0.693/T12; ## Decay constant in hour^(-1)\n",
+ "t = 12.; ## Analysis time of gold sample in hours\n",
+ "alpha = 0.9; ## Activity of gold sample after analysis time\n",
+ "\n",
+ "## 1. \n",
+ "## Calculation\n",
+ "R = alpha/(1-math.exp(-lambd*t));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Theoretical maximum activity = ',R,' curie (Ci) \\n');\n",
+ "\n",
+ "## 2. \n",
+ "## Calculation\n",
+ "## The expression to calculate 80 percent of maximum activity is \\n 0.8R = R*(1-exp(-lambda*t)) \n",
+ "t = -math.log(0.2)/lambd;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Time to reach 80 percent of maximum activity = ',t,' hours \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Theoretical maximum activity = 7.47 curie (Ci) \n",
+ "\n",
+ "\n",
+ " Time to reach 80 percent of maximum activity = 150.49 hours \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.7\n",
+ "import math\n",
+ "#Calculate the \n",
+ "print('The reactants are Nitrogen and neutron')\n",
+ "## The total atomic number of reactants \n",
+ "Z_reactant = 7.+0.;\n",
+ "## The total atomic mass number of reactants\n",
+ "A_reactant = 14.+1.;\n",
+ "print('One of the known product is Hydrogen')\n",
+ "Z_H = 1.; ## The atomic number of Hydrogen\n",
+ "A_H = 1.; ## The atomic mass number of Hydrogen \n",
+ "## The atomic number of unknown element\n",
+ "Z_unknown = Z_reactant-Z_H;\n",
+ "## The atomic mass number of unknown element \n",
+ "A_unknown = A_reactant-A_H;\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n For unknown element the atomic number is \",Z_unknown,\" and atomic mass number is \",A_unknown,\" \\n\");\n",
+ "## From periodic table \n",
+ "print('The element corresponds to Carbon-14');\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reactants are Nitrogen and neutron\n",
+ "One of the known product is Hydrogen\n",
+ " \n",
+ " For unknown element the atomic number is 6.00 and atomic mass number is 14.00 \n",
+ " \n",
+ "The element corresponds to Carbon-14\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.8\n",
+ "import math\n",
+ "#Calculate the \n",
+ "print('The reaction is Tritium(d,n)Helium-4');\n",
+ "## Using standard data table of mass in amu\n",
+ "M_H3 = 3.016049; ## Atomic mass of Tritium\n",
+ "M_He4 = 4.002604; ## Atomic mass of Helium\n",
+ "M_d = 2.014102; ## Atomic mass of Deuterium\n",
+ "M_n = 1.008665; ## Atomic mass of neutron\n",
+ "## Calculation of total mass of reactants\n",
+ "tot_reac = M_H3+M_d;\n",
+ "## Calculation of total mass of products\n",
+ "tot_prod = M_He4+M_n;\n",
+ "## Calculation \n",
+ "Q = tot_reac-tot_prod;\n",
+ "## Expressing in MeV by using 1 amu = 931.5 MeV\n",
+ "Q_mev = Q*931.5;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Q value for the reaction = \",Q_mev,\" MeV\");\n",
+ "if Q_mev > 0:\n",
+ " print(\"\\n The reaction is exothermic. \\n\");\n",
+ "else:\n",
+ " print(\"\\n The reaction is endothermic. \\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reaction is Tritium(d,n)Helium-4\n",
+ " \n",
+ " Q value for the reaction = 17.59 MeV\n",
+ "\n",
+ " The reaction is exothermic. \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.9\n",
+ "import math\n",
+ "#Calculate the \n",
+ "## Using standard data table of mass in amu\n",
+ "M_C12 = 12.; ## Atomic mass of Carbon-12\n",
+ "M_n = 1.00866; ## Atomic mass of neutron\n",
+ "M_C13 = 13.00335; ## Atomic mass of Carbon-13\n",
+ "##If one neutron is removed from carbon-13, carbon-12 is obtained\n",
+ "tot = M_C12+M_n;\n",
+ "dm = tot-M_C13; ## Mass defect\n",
+ "## Converting to energy equivalent from mass by using 1 amu = 931.5 MeV\n",
+ "Es = dm*931.5;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The binding Energy of the last neutron in Carbon-13 atom = \",Es,\" MeV\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The binding Energy of the last neutron in Carbon-13 atom = 4.95 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.10\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Using standard data table of mass and the coefficients of mass equation for Silver-107\n",
+ "N = 60.; ## Number of neutrons\n",
+ "Z = 47.; ## Atomic number\n",
+ "A = 107.; ## Atomic mass number\n",
+ "## The coefficients used in mass equation are \n",
+ "alpha = 15.56;\n",
+ "bet = 17.23;\n",
+ "gam = 0.697;\n",
+ "zeta = 23.285;\n",
+ "mn = 939.573; ## Mass of neutron in terms of energy\n",
+ "mH = 938.791; ## Mass of proton in terms of energy\n",
+ "## Calculation \n",
+ "print('Using mass equation');\n",
+ "M = (N*mn)+(Z*mH)-(alpha*A)+(bet*(A**(2/3.)))+(gam*Z**2/A**(1/3.))+(zeta*(A-2*Z)**2/A);\n",
+ "## Expressing in amu by using 1 amu = 931.5 MeV\n",
+ "M_amu = M/931.5;\n",
+ "print'%s %.2f %s %.2f %s '%(\" Mass = \",M,\" MeV\" and \" = \",M_amu,\"f u\" );\n",
+ "print('Actual mass = 106.905092 u');\n",
+ "## Calculation \n",
+ "BE = (alpha*A)-(bet*(A**(2/3.)))-(gam*Z**2/A**(1/3.))-((zeta*(A-(2.*Z))**2)/A);\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\n Binding Energy = \",BE,\" MeV\" or\"\" ,BE/107,\"MeV/nucleon \\n\");\n",
+ "## The value is different from the answer given in the textbook. The textbook answer is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Using mass equation\n",
+ " Mass = 99582.07 = 106.91 f u \n",
+ "Actual mass = 106.905092 u\n",
+ "\n",
+ " Binding Energy = 915.49 MeV 8.56 MeV/nucleon \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.11\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "T_C = 38.; ## Given temeperature in celsius\n",
+ "##The temperature in Kelvin \n",
+ "T_K = T_C+273.15;\n",
+ "T_0 = 293.61; ## The temperature in kelvin equivalent to 0 deg celsius\n",
+ "kT = 0.0253; ## The term 'kT' in eV at temperature T0\n",
+ "## Calculation\n",
+ "Ep = 0.5*kT*(T_K/T_0);\n",
+ "Ebar = 3*Ep;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" Most probable energy of air molecules = \",Ep,\" eV \\n\");\n",
+ "print'%s %.2f %s'%(\" Average energy of air molecules = \",Ebar,\" eV \\n\");\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Most probable energy of air molecules = 0.01 eV \n",
+ "\n",
+ " Average energy of air molecules = 0.04 eV \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.12\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "rho = 0.97; ## Density of Sodium in gram/cm^3\n",
+ "## From standard data table\n",
+ "NA = 0.6022*10**24; ## Avagodro number\n",
+ "M = 22.99; ## Atomic weight of Sodium\n",
+ "## Calculation\n",
+ "N = rho*NA/M;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\"Atom density of sodium = \",N,\" atoms/cm^3 \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Atom density of sodium = 2.54e+22 atoms/cm^3 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.13\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "rho_NaCl = 2.17; ## Density of Sodium Chloride(NaCl) in gram/cm^3\n",
+ "## From standard data table\n",
+ "NA = 0.6022*10**24; ## Avogodro number\n",
+ "M_Na = 22.99; ## Atomic weight of Sodium(Na)\n",
+ "M_Cl = 35.453; ## Atomic weight of Chlorine(Cl)\n",
+ "M_NaCl = M_Na+M_Cl; ## Molecular weight of Sodium Chloride(NaCl)\n",
+ "## Calculation\n",
+ "N = rho_NaCl*NA/M_NaCl;\n",
+ "## As in NaCl, there is one atom of Na and Cl\n",
+ "N_Na = N;\n",
+ "N_Cl = N;\n",
+ "## Result\n",
+ "print'%s %.4e %s'%(\" Atom density of Sodium and Chlorine = \",N,\" molecules/cm^3 \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Atom density of Sodium and Chlorine = 2.2360e+22 molecules/cm^3 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.14\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "rho = 1.; ## Density of water in gram/cm^3\n",
+ "\n",
+ "## 1.\n",
+ "M_H = 1.00797; ## Atomic weight of Hydrogen(H)\n",
+ "M_O = 15.9994; ## Atomic weight of Oxygen(O)\n",
+ "## As in water, there is two atoms of Hydrogen(H) and one atom of Oxygen(O)\n",
+ "M = (2*M_H)+M_O; ## Molecular weight of water\n",
+ "## From standard data table\n",
+ "NA = 0.6022*10**24; ## Avagodro number\n",
+ "## Calculation \n",
+ "N = rho*NA/M;\n",
+ "## Result\n",
+ "print'%s %.4e %s'%(\"Atom density of water = \",N,\" molecules/cm^3 \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "## As in water, there is two atoms of Hydrogen(H) and one atom of Oxygen(O)\n",
+ "N_H = 2*N; ## Atom density of Hydrogen\n",
+ "N_O = N; ## Atom density of Oxygen\n",
+ "## Result\n",
+ "print'%s %.4e %s'%(\"Atom density of Hydrogen(H) = \",N_H,\" atoms/cm^3 \\n\");\n",
+ "print'%s %.4E %s'%(\"Atom density of Oxygen(O)= \",N_O,\" atoms/cm^3 \\n\");\n",
+ "\n",
+ "## 3.\n",
+ "## Using the data given in Table II.2, Appendix II for isotropic abundance of deuterium\n",
+ "isoab_H2 = 0.015;\n",
+ "## Calculation\n",
+ "N_H2 = isoab_H2*N_H/100.;\n",
+ "## Result\n",
+ "print'%s %.4E %s'%(\"Atom density of Deuterium(H-2)= \",N_H2,\" atoms/cm^3 \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Atom density of water = 3.3427e+22 molecules/cm^3 \n",
+ "\n",
+ "Atom density of Hydrogen(H) = 6.6854e+22 atoms/cm^3 \n",
+ "\n",
+ "Atom density of Oxygen(O)= 3.3427E+22 atoms/cm^3 \n",
+ "\n",
+ "Atom density of Deuterium(H-2)= 1.0028E+19 atoms/cm^3 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.15\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data \n",
+ "rho = 19.1; ## Density of Uranium-235 in gram/cm^3\n",
+ "wt = 1500.; ## Weight of uranium rods in a reactor in kg\n",
+ "nr = 0.2; ## Enrichment(w/o) of Uranium-235\n",
+ "\n",
+ "## 1.\n",
+ "## As Enrichment is 20(w/o)\n",
+ "wt_U235 = nr*wt; ## Amount of Uranium-235\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"Amount of Uranium-235 in the reactor = \",wt_U235,\" kg \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "## From standard data table\n",
+ "NA = 0.6022*10**24; ## Avagodro number\n",
+ "M_U235 = 235.0439; ## Atomic weight of Uranium-235\n",
+ "M_U238 = 238.0508; ## Atomic weight of Uranium-238\n",
+ "## Calculation\n",
+ "N_U235 = nr*rho*NA/M_U235; ## Atom density of Uranium-235\n",
+ "N_U238 = (1.-nr)*rho*NA/M_U238; ## Atom density of Uranium-238\n",
+ "## Result\n",
+ "print'%s %.4e %s'%(\"Atom density of Uranium-235 = \",N_U235,\" atoms/cm^3 \\n\");\n",
+ "print'%s %.4e %s'%(\"Atom density of Uranium-238 = \",N_U238,\" atoms/cm^3 \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Amount of Uranium-235 in the reactor = 300.00 kg \n",
+ "\n",
+ "Atom density of Uranium-235 = 9.7871e+21 atoms/cm^3 \n",
+ "\n",
+ "Atom density of Uranium-238 = 3.8654e+22 atoms/cm^3 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 2.16\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "rho_UO2 = 10.5; ## Density of UO2 pellets in gram/cm^3\n",
+ "nr = 0.3; ## Enrichment(w/o) of Uranium-235\n",
+ "## From standard data table\n",
+ "M_U235 = 235.0439; ## Atomic weight of Uranium-235\n",
+ "M_U238 = 238.0508; ## Atomic weight of Uranium-238\n",
+ "M_O = 15.999; ## Atomic weight of Oxygen\n",
+ "NA = 0.6022*10**24; ## Avogodro number\n",
+ "\n",
+ "M = 1./((nr/M_U235)+((1.-nr)/M_U238));\n",
+ "M_UO2 = M+(2.*M_O); ## Molecular weight of UO2\n",
+ "nr_U = M/M_UO2*100.; ## The percent(w/o) of Uranium in UO2 pellet\n",
+ "rho_U = nr_U*rho_UO2/100. ## Density of Uranium in g/cm^3\n",
+ "rho_U235 = nr*rho_U ## Density of Uranium-235 in g/cm^3\n",
+ "## Calculation\n",
+ "N_U235=rho_U235*NA/M_U235;\n",
+ "## Result\n",
+ "print'%s %.4e %s'%(\"Atom density of Uranium-235 = \",N_U235,\" atoms/cm^3 \\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Atom density of Uranium-235 = 7.1110e+21 atoms/cm^3 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter3_1.ipynb b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter3_1.ipynb new file mode 100644 index 00000000..724ffe07 --- /dev/null +++ b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter3_1.ipynb @@ -0,0 +1,813 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f82ab49623953e3c7cad47eb0e1bf1bfaa9c0013f242573b2cbe63ac1a09a05d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter3-Interaction of Radiation with Matter "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.1\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigma = 2.6*10**(-24); ## Cross section of carbon-12 in cm^2\n",
+ "I = 5*10**8; ## Intensity of neutron beam in neutrons/cm^2-sec\n",
+ "A = 0.1; ## Cross sectional area of the beam in cm^2;\n",
+ "X = 0.05; ## Thickness of the target in cm\n",
+ "\n",
+ "## 1.\n",
+ "## Using the data given in Table I.3, Appendix II for carbon-12\n",
+ "N = 0.08*10**(24); ## Atom density in atoms/cm^3\n",
+ "## Calculation \n",
+ "IR = sigma*I*N*A*X;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%('\\n Total interaction rate = ',IR,' interactions/sec \\n');\n",
+ "\n",
+ "## 2. \n",
+ "no = I*A; ## Neutron rate in neutrons/sec\n",
+ "## Calculation \n",
+ "p = IR/no;\n",
+ "print'%s %.2e %s'%('\\n Probability of collision = ',p,' \\n');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Total interaction rate = 5.20e+05 interactions/sec \n",
+ "\n",
+ "\n",
+ " Probability of collision = 1.04e-02 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.2\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "sigmaf = 582.; ## Fission cross section of U-235 on bombardment of neutron in barn\n",
+ "sigmay = 99.; ## Radiative capture cross section of U-235 on bombardment of neutron in barn\n",
+ "## Calculation\n",
+ "pf = sigmaf/(sigmaf+sigmay);\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%('\\ n Probability of fission = ',pf,''and ' = ',pf*100,' percent\\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\\ n Probability of fission = 0.85 85.46 percent\n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.3\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "## Using the data given in the example 3.1\n",
+ "N = 0.08*10**(24.); ## Atom density of Carbon-12 in atoms/cm^3\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigma = 2.6*10**(-24); ## Cross section of carbon-12 in cm^2\n",
+ "I = 5*10**8; ## Intensity of neutron beam in neutrons/cm^2-sec\n",
+ "\n",
+ "## 1.\n",
+ "## Calculation \n",
+ "SIGMAt = N*sigma;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Macroscopic cross section of carbon-12 = ',SIGMAt,' cm^(-1)\\n');\n",
+ "\n",
+ "##2. \n",
+ "## Calculation \n",
+ "F= I*SIGMAt;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%('\\n Collision density in the carbon-12 target = ',F,' collisions/cm^(3)-sec\\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Macroscopic cross section of carbon-12 = 0.21 cm^(-1)\n",
+ "\n",
+ "\n",
+ " Collision density in the carbon-12 target = 1.04e+08 collisions/cm^(3)-sec\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.4\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "E = 100.; ## Neutron energy in keV\n",
+ "## Using the data given in Table II.3, for E = 100 keV\n",
+ "atom_density = 0.0254*10**(24); ## Atom density of sodium in atoms/cm^3\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigma = 3.4*10**(-24); ## Microscopic cross section of sodium in cm^2\n",
+ "## Calculation\n",
+ "SIGMA = atom_density*sigma;\n",
+ "lambd = 1./SIGMA;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Macroscopic cross section = ',SIGMA,' cm^(-1)\\n');\n",
+ "print'%s %.2f %s'%('\\n Mean Free Path = ',lambd,' cm\\n',);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Macroscopic cross section = 0.09 cm^(-1)\n",
+ "\n",
+ "\n",
+ " Mean Free Path = 11.58 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg60"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.5\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "atom_density_U235 = 3.48*10**(-4)*10**(24); ## Atom density of Uranium-235 in atoms/cm^3\n",
+ "atom_density_U238 = 0.0483*10**(24); ## Atom density of Uranium-238 in atoms/cm^3\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigmaa_U235 = 680.8*10**(-24); ## Absorption cross section of Uranium-235 incm^2\n",
+ "sigmaa_U238 = 2.7*10**(-24); ## Absorption cross section of Uranium-238 incm^2\n",
+ "## Calculation\n",
+ "SIGMAA=(atom_density_U235*sigmaa_U235)+(atom_density_U238*sigmaa_U238);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Macroscopic absorption cross section = ',SIGMAA,' cm^(-1)\\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Macroscopic absorption cross section = 0.37 cm^(-1)\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg60"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.6\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "sigmas_H_1 = 3; ## Scattering cross section of Hydrogen in barn at 1 MeV\n",
+ "sigmas_O_1 = 8; ## Scattering cross section of Oxygen in barn at 1 MeV\n",
+ "sigmas_H_th = 21; ## Scattering cross section of Hydrogen in barn at 0.0253 eV \n",
+ "sigmas_O_th = 4; ## Scattering cross section of Oxygen in barn at 0.0253 eV\n",
+ "## Calculation\n",
+ "sigmas_H20_1 = (2*sigmas_H_1)+(1*sigmas_O_1);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Scattering cross section of Water at 1 MeV = ',sigmas_H20_1,' b \\n');\n",
+ "## The equation used to calculate the scattering cross section at 1 MeV cannot be used at thermal energy. \n",
+ "print'%s %.2f %s'%(' Experimental value of scattering cross section of Water at 0.0253 eV = ',103,' b \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Scattering cross section of Water at 1 MeV = 14.00 b \n",
+ "\n",
+ " Experimental value of scattering cross section of Water at 0.0253 eV = 103.00 b \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.7\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "phi = 1*10**(13); ## Neutron flux in neutrons/cm^3\n",
+ "v = 64000.; ## Volume of research reactor in cm^3\n",
+ "sigmaf = 0.1; ## Macroscopic fission cross section in cm^(-1)\n",
+ "## The energy released per fission reaction is 200 MeV\n",
+ "## 1 MeV = 1.6*10^(-13) joule\n",
+ "E = 200*1.6*10**(-13);\n",
+ "## Calculation \n",
+ "fiss_rate = sigmaf*phi; ## Fission rate in neutrons/cm^2-sec\n",
+ "power_cc = E*fiss_rate/10**6; ## Reactor power/cc\n",
+ "power = power_cc*v;\n",
+ "print'%s %.2f %s'%('\\n Reactor power of a research reactor = ',power,' MW\\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Reactor power of a research reactor = 2.05 MW\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.8\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "## From the Figure 3.4 given in the textbook\n",
+ "sigmae = 4.8*10**(-24); ## Experimental cross section of carbon from 0.02eV to 0.01MeV\n",
+ "## Assuming spherical shape and elstic scattering\n",
+ "R = math.sqrt(sigmae/(4.*math.pi));\n",
+ "## Result\n",
+ "print'%s %.2e %s'%('\\n Radius of carbon nucleus = ',R,' cm\\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Radius of carbon nucleus = 6.18e-13 cm\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.9\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data \n",
+ "E0 = 0.0253; ## Thermal energy in eV\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigmay_E0 = 0.332*10**(-24); ## Radiative capture cross section at 0.0253 eV in cm^2\n",
+ "E = 1.; ## Energy in eV at which radiative cross section is to be found\n",
+ "## Calculation \n",
+ "sigmay_E = sigmay_E0*math.sqrt(E0/E);\n",
+ "## Result\n",
+ "## Expressing the result in barn\n",
+ "print'%s %.2e %s'%('\\n Radiative capture cross section of hydrogen at 1 eV = ',sigmay_E*10*24,' b\\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Radiative capture cross section of hydrogen at 1 eV = 1.27e-23 b\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.10\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "E = 1.; ## Energy of neutron in MeV\n",
+ "A = 2.; ## Atomic mass number of deuterium\n",
+ "v = 45.; ## Scattering angle in degree\n",
+ "\n",
+ "## 1.\n",
+ "## Calculation \n",
+ "E_dash = E/(A+1.)**2 *((math.cos (v/57.3)+math.sqrt(A**2-(math.sin(v/57.3))**2))**2);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Energy of scattered neutron = ',E_dash,' MeV \\n');\n",
+ "\n",
+ "## 2.\n",
+ "## Calculation \n",
+ "E_A = E-E_dash;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Energy of recoil nucleus = ',E_A,' MeV \\n');\n",
+ "\n",
+ "## 3.\n",
+ "## Calculation \n",
+ "deltau = math.log(E/E_dash);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Change in lethargy of neutron on collision = ',deltau,' \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Energy of scattered neutron = 0.74 MeV \n",
+ "\n",
+ "\n",
+ " Energy of recoil nucleus = 0.26 MeV \n",
+ "\n",
+ "\n",
+ " Change in lethargy of neutron on collision = 0.30 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.11\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "phi = 5*10**(12); ## Neutron flux in neutrons/cm^2-sec\n",
+ "T = 600.; ## Temperature of neutron in degree\n",
+ "## Using the data given in Table II.3, Appendix II for indium\n",
+ "N = 0.0383*10**(24); ## Atom density in atoms/cm^3\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigmaa_E0 = 194.*10**(-24); ## Microscopic absorption cross section in cm^2\n",
+ "SIGMA_E0 = N*sigmaa_E0; ## Macroscopic absorption cross section in cm^(-1)\n",
+ "## From Table 3.2 \n",
+ "ga_600 = 1.15; ## Non 1/v factor at 600 degree celsius\n",
+ "## Calculation \n",
+ "F_a = ga_600*SIGMA_E0*phi;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%('\\n Absorption rate of neutrons per cc in indium foil = ',F_a,' neutrons/cm^3-sec \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Absorption rate of neutrons per cc in indium foil = 4.27e+13 neutrons/cm^3-sec \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.12\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "N = 120.; ## Number of fuel rods\n",
+ "P = 100.; ## Reactor power in MW\n",
+ "t = 1.; ## Estimation time of fuel rod after removal in days\n",
+ "T = 365.; ## Time of reactor operation\n",
+ "## Estimation\n",
+ "Activity_total = 1.4*10**6*P*(t**(-0.2)-(t+T)**(-0.2));\n",
+ "Activity_one = Activity_total/N; ## For one fuel rod\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n The activity of a fuel rod = ',Activity_one,' Ci \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The activity of a fuel rod = 808363.56 Ci \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.13\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Using the data given in Table 3.4 and Table II.2 for uranium\n",
+ "v_235 = 2.418; ## Average number of neutrons released per fission\n",
+ "y_235 = 0.72; ## Isotropic abundance of Uranium-235 on the earth\n",
+ "sigmaf_235 = 582.2; ## Fission cross section of Uranium-235\n",
+ "sigmaa_235 = 680.8; ## Absorption cross section of Uranium-235\n",
+ "N_235 = y_235;\n",
+ "y_238 = 99.26; ## Isotropic abundance of Uranium-238 on the earth\n",
+ "sigmaa_238 = 2.7; ## Absorption cross section of Uranium-238\n",
+ "## Calculation\n",
+ "n = (v_235*y_235*sigmaf_235)/((y_235*sigmaa_235)+(y_238*sigmaa_238));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Eta for natural uranium = ',n,' \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Eta for natural uranium = 1.34 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.14\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Fission of 1 g of Uranium-235 releases approximately 1 MW/day of energy. \n",
+ "## 1 MW/day = 8.64*10^(10) J\n",
+ "energy_uranium = 8.64*10**10;\n",
+ "\n",
+ "## 1. Coal\n",
+ "h_coal = 3*10**7; ## Heat contenet of coal in J/kg\n",
+ "## Calculation\n",
+ "amt_coal = energy_uranium/h_coal;\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%('\\n Amount of coal required for energy equivalent of fission = ',amt_coal,' kg ' and '\\n ',amt_coal/10**3,'metric tons' and '',amt_coal*1.10231/10**3,'short tons\\n');\n",
+ "## The result is expressed in all units of commercial importance.\n",
+ "\n",
+ "## 2. Oil\n",
+ "h_oil = 4.3*10**7; ## Heat contenet of oil in J/kg\n",
+ "## Calculation\n",
+ "amt_oil = energy_uranium/h_oil;\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%('\\n Amount of oil required for energy equivalent of fission = ',amt_oil,' kg\\n' or'',amt_oil/10**3,' tons' or '',amt_oil*6.3/10**3,'barrels\\n');\n",
+ "## The result is expressed in all units of commercial importance.\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Amount of coal required for energy equivalent of fission = 2880.00 \n",
+ " 2.88 3.17 short tons\n",
+ " \n",
+ "\n",
+ " Amount of oil required for energy equivalent of fission = 2009.30 kg\n",
+ " 2.01 tons 12.66 barrels\n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.15\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "rho = 10.; ## Density of UO2 in g/cm^3\n",
+ "mol_wt_UO2 = 238.+(16.*2.); ## Molecular weight of UO2\n",
+ "per_U = (238./mol_wt_UO2)*100.; ## Percent by weight of Uranium\n",
+ "per_O = 100.-per_U; ## Percent by weight of Oxygen\n",
+ "\n",
+ "## Calculation \n",
+ "##Using the data given in Table II.4 for uranium and oxygen\n",
+ "mup_U = 0.0757; ## Ratio of mass attenuation coefficient to density of uranium in cm^2/g\n",
+ "mup_O = 0.0636; ## Ratio of mass attenuation coefficient to density of oxygen in cm^2/g\n",
+ "mup = (per_U/100.*mup_U)+(per_O/100.*mup_O); ## The total ratio of mass attenuation coefficient in cm^2/g\n",
+ "mu = mup*rho;\n",
+ "## Calculation \n",
+ "lambd = 1/mu;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Mass attenuation coefficient of Uranium dioxide (UO2) = ',mu,' cm^(-1) \\n');\n",
+ "print'%s %.2f %s'%('\\n Mean free path = ',lambd,' cm \\n');\n",
+ "## The answer is marked wrongly in the textbook. But the solution is correctly evaluated.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Mass attenuation coefficient of Uranium dioxide (UO2) = 0.74 cm^(-1) \n",
+ "\n",
+ "\n",
+ " Mean free path = 1.35 cm \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.16\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "E = 0.8; ## Average gamma ray energy in MeV\n",
+ "I = 3*10**(11); ## Intensity of gamma rays incident on the container in gamma rays/cm^2-sec\n",
+ "## Using the data given in Table II.5 for iron at 0.8 MeV\n",
+ "mup_iron = 0.0274; ## Ratio of mass attenuation coefficient to density of iron in cm^2/g\n",
+ "## Calculation \n",
+ "dep_rate = E*I*mup_iron;\n",
+ "## Expressing the result in SI units\n",
+ "## 1 MeV = 1.6*10^(-13) J\n",
+ "## 1 kg = 1000 g\n",
+ "dep_rate_SI = dep_rate*(1.6*10**(-13)*1000.);\n",
+ "print'%s %.2e %s %.2f %s '%('\\n Rate of energy deposited = ',dep_rate,' MeV/g-sec' and '',dep_rate_SI,' J/kg-sec \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Rate of energy deposited = 6.58e+09 1.05 J/kg-sec \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 3.17\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data \n",
+ "E_max = 1.39; ## Maximum energy of beta rays in MeV\n",
+ "## Calculation \n",
+ "R_max = 0.412*E_max**(1.265-(0.0954*math.log(E_max)));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Maximum distance of beta rays traversed = ',R_max,' cm \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Maximum distance of beta rays traversed = 0.62 cm \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter4_1.ipynb b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter4_1.ipynb new file mode 100644 index 00000000..8212531f --- /dev/null +++ b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter4_1.ipynb @@ -0,0 +1,362 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:880d5b96d7a6ce9a20a4fd5f1bbd8bee89ba44d4461c16f4ef09e18728323ef7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter4-Nuclear Reactors and Nuclear Power"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg121"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 4.1\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "## Number of neutrons absorbed by Uranium-238 in resonances for every neutron absorbed in Uranium-235\n",
+ "n_resonance = 0.254;\n",
+ "## Number of neutrons absorbed by Uranium-238 at thermal energy for every neutron absorbed in Uranium-235\n",
+ "n_th = 0.64;\n",
+ "m = 1; ## Amount of Uranium-235 consumed in kg\n",
+ "A_U = 235; ## Atomic mass number of Uranium-235\n",
+ "A_Pu = 239; ## Atomic mass number of Plutonium-239\n",
+ "\n",
+ "## 1.\n",
+ "## Calculation \n",
+ "C = n_resonance+n_th;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Conversion ratio of the reactor = ',C,' \\n');\n",
+ "\n",
+ "## 2. \n",
+ "## Calculation \n",
+ "amt_Pu = m*C*A_Pu/A_U;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Amount of Plutonium-239 produced in the reactor = ',amt_Pu,' kg \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Conversion ratio of the reactor = 0.89 \n",
+ "\n",
+ "\n",
+ " Amount of Plutonium-239 produced in the reactor = 0.91 kg \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg127"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 4.2\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "wP0 = 1.; ## Total fuel consumption rate in terms of kg/day\n",
+ "M = 500.; ## Amount of Plutonium-239 in kg at startup of the reactor\n",
+ "breeding_gain = 0.15; ## Breeding gain of the reactor\n",
+ "\n",
+ "## 1.\n",
+ "print'%s %.2f %s'%(\" The Fast breeder reactor produces \",breeding_gain,\"kg of plutonium-239 more for every kilogram consumed \\n\");\n",
+ "## Calculation\n",
+ "## 1 year = 365 days\n",
+ "production_rate = math.ceil(breeding_gain*365);\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\n Production rate of plutonium-239 = \",breeding_gain,\" kg/day\"and \" = \",production_rate,\" kg/year\");\n",
+ "\n",
+ "## 2.\n",
+ "## Calculation \n",
+ "t_Dl = M/production_rate;\n",
+ "t_De = math.log(2)*t_Dl;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Linear doubling time of plutonium fuel in the reactor = \",t_Dl,\" years \\n\");\n",
+ "print'%s %.2f %s'%(\" \\n Exponential doubling time of plutonium fuel in the reactor = \",t_De,\" years \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The Fast breeder reactor produces 0.15 kg of plutonium-239 more for every kilogram consumed \n",
+ "\n",
+ "\n",
+ " Production rate of plutonium-239 = 0.15 = 55.00 kg/year \n",
+ " \n",
+ " Linear doubling time of plutonium fuel in the reactor = 9.09 years \n",
+ "\n",
+ " \n",
+ " Exponential doubling time of plutonium fuel in the reactor = 6.30 years \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg128"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 4.3\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "power = 3300.; ## Reactor power in MW\n",
+ "time = 750.; ## Reactor operation time in days\n",
+ "amt_UO2 = 98.; ## Amount of Uranium dioxide (UO2) in metric tons\n",
+ "atwt_U = 238.; ## As the enrichment of Uranium-235 is 3 w/o the majority portion is Uranium-238\n",
+ "molwt_O = 16.; ## Molecular weight of Oxygen\n",
+ "\n",
+ "\n",
+ "## 1.\n",
+ "amt_U = amt_UO2*atwt_U/(atwt_U+2*molwt_O); ## Amount of uranium in tonne\n",
+ "total_burnup = power*time; ## Total burnup in MWd\n",
+ "## Calculation \n",
+ "specific_burnup = total_burnup/amt_U;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Specific burnup = \",specific_burnup,\" MWd/tonne \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "## Due to fission of 1.05 g of Uranium-235, 1 MWd of energy is released.\n",
+ "m = 1.05;\n",
+ "P = 10**6;\n",
+ "maxth_burnup = P/m; ## Theoritical maximum burnup \n",
+ "## Calculation of Fractional burnup\n",
+ "bet = specific_burnup/maxth_burnup;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Fractional burnup = \",bet*100,\" percent \\n\");\n",
+ "## Due to approximation of specific burnup value, there is a slight change in fractional burnup value as compared to the textbook value.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Specific burnup = 28650.75 MWd/tonne \n",
+ "\n",
+ " \n",
+ " Fractional burnup = 3.01 percent \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 4.4\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "ratpower = 1075.; ## Output rated electrical power in MWe of the reactor\n",
+ "delpower_yr = 255000.; ## Net output power delivered in one year in terms of MWd\n",
+ "time_refuel = 28.; ## Number of days the plant was shutdown for refuelling\n",
+ "time_repairs = 45.; ## Number of days the plant was shutdown for repairs\n",
+ "time_convrepairs = 18.; ## Number of days the plant was shutdown for conventional repairs\n",
+ "\n",
+ "## 1.\n",
+ "## 1 year = 365 days\n",
+ "ratpower_yr = ratpower*365.; ## Net output rated power in one year in terms of MWd\n",
+ "## Calculation\n",
+ "cap_factor = delpower_yr/ratpower_yr;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Plant capacity factor = \",math.ceil(cap_factor*100),\" percent\\n\");\n",
+ "\n",
+ "## 2.\n",
+ "## Number of days the plant was shutdown in one year \n",
+ "total_shutdown = time_refuel+time_repairs+time_convrepairs;\n",
+ "## Number of days the plant was operable in one year\n",
+ "total_operation = 365.-total_shutdown;\n",
+ "## Calculation\n",
+ "ava_factor = total_operation/365.;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Plant availability factor = \",ava_factor*100,\" percent\\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Plant capacity factor = 65.00 percent\n",
+ "\n",
+ " \n",
+ " Plant availability factor = 75.07 percent\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 4.5\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data \n",
+ "t = 30.; ## Time of uranium sufficiency in years \n",
+ "## Assuming once through Light Water Reactor (LWR)fuel cycle\n",
+ "U_LWR = 0.0055; ## Uranium Utilization factor for LWR\n",
+ "## Assuming once through Liquid Metal cooled Fast Breeder Reactor (LMFBR) fuel cycle\n",
+ "U_LMFBR = 0.67; ## Uranium Utilization factor for LMFBR\n",
+ "## Estimation \n",
+ "est_time = 30.*U_LMFBR/U_LWR;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"The time for which Uranium would fuel LMFBR = \",math.ceil(est_time),\" years \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The time for which Uranium would fuel LMFBR = 3655.00 years \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 4.6\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "A_U = 238.; ## Atomic Mass number of Uranium\n",
+ "A_O = 16.; ## Atomic Mass number of Oxygen\n",
+ "amt_UO2 = 33000.; ## Amount of Uranium dioxide (UO2) present in kilogram(kg)\n",
+ "x_P = 0.032; ## Enrichment of 3.2 w/o uranium product\n",
+ "x_T = 0.002; ## Enrichemnt of 0.2 w/o residual tails\n",
+ "## From Figure 4.45\n",
+ "x_F = 0.00711; ## Enrichemnt of 0.711 w/o feed\n",
+ "\n",
+ "## 1.\n",
+ "## Estimation of enriched uranium in kg\n",
+ "M_P = A_U*amt_UO2/(A_U+2*A_O);\n",
+ "## Estimation of amount of Uranium feed in kg\n",
+ "M_F = ((x_P-x_T)/(x_F-x_T))*M_P;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The amount of uranium feed required per reload = \",math.ceil(M_F),\" kg \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "V_x_P = (1.-2.*x_P)*math.log((1.-x_P)/x_P); ## Value function of uranium product with enrichemnt of 3.2 w/o\n",
+ "V_x_F = (1.-2.*x_F)*math.log((1.-x_F)/x_F); ## Value function of feed with enrichemnt of 0.711 w/o\n",
+ "V_x_T = (1.-2.*x_T)*math.log((1.-x_T)/x_T); ## Value function of tallings with enrichemnt of 0.2 w/o\n",
+ "rate_SWU = 130.75; ## Enrichment cost in dollars per SWU\n",
+ "## Calculation \n",
+ "SWU = M_P*(V_x_P-V_x_T)-M_F*(V_x_F-V_x_T); ## Separative Work (SWU) in kg\n",
+ "enrich_cost = math.ceil(SWU)*rate_SWU; ## Enrichment cost in dollars\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n The enrichment cost = $ \",math.ceil(enrich_cost),\" \\n\");\n",
+ "## Due to approximation of Separative Work Unit(SWU), there is a difference in the value of enrichment cost on comparison with the textbook value.\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The amount of uranium feed required per reload = 170777.00 kg \n",
+ "\n",
+ "\n",
+ " The enrichment cost = $ 18052522.00 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter5_1.ipynb b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter5_1.ipynb new file mode 100644 index 00000000..3e370377 --- /dev/null +++ b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter5_1.ipynb @@ -0,0 +1,230 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:2bf96dc6c6d5f49eacd7b677501192a040f4a59aa7d269c591cd7699891db60d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter5-Neutron Diffusion and Moderation "
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg234"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 5.2\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigma_s = 4.8*10**(-24) ## Scattering cross section of carbon in cm^2\n",
+ "A_C = 12.; ## Atomic Mass number for carbon-12\n",
+ "E = 1.; ## Energy of carbon-12 atom in eV\n",
+ "## Using the data given in Table II.3, for carbon (graphite) at energy 1 eV\n",
+ "N = 0.08023*10**(24); ## Atom density in terms of atom/cm^3\n",
+ "mu_bar = 2./(3.*A_C); ## Average value of the cosine of the angle at which neutrons are scattered in the med/ium\n",
+ "SIGMA_s = N*sigma_s ## Macroscopic scattering cross section of carbon-12\n",
+ "## Calculation\n",
+ "D = 1/(3.*SIGMA_s*(1.-mu_bar));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Diffusion coefficeint of graphite at 1 eV = ',D,' cm \\n');\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Diffusion coefficeint of graphite at 1 eV = 0.92 cm \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 5.5\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "phi1 = 6*10**(14); ## Neutron flux of Group 1\n",
+ "phi2 = 1*10**(15); ## Neutron flux of Group 2\n",
+ "phi3 = 3*10**(15); ## Neutron flux of Group 3\n",
+ "\n",
+ "## 1.\n",
+ "## Using the data given in Table II.3, for atom density of sodium\n",
+ "N = 0.02541*10**(24); ## Atom density in terms of atom/cm^3\n",
+ "## Using the data given for sigmay (Microscopic radiative capture cross section) in Table II.3,\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigmay1 = 0.0005*10**(-24); ## Microscopic gamma cross section of Group 1\n",
+ "sigmay2 = 0.001*10**(-24); ## Microscopic gamma cross section of Group 2\n",
+ "sigmay3 = 0.001*10**(-24); ## Microscopic gamma cross section of Group 3\n",
+ "## Calculation \n",
+ "F_a = N*((sigmay1*phi1)+(sigmay2*phi2)+(sigmay3*phi3));\n",
+ "## Result\n",
+ "print'%s %.2e %s'%('\\n Total absorption rate for three groups = ',F_a,' neutrons/cm^3-sec \\n');\n",
+ "\n",
+ "## 2.\n",
+ "## Calculation\n",
+ "sigmag_12 = 0.24*10**(-24); ## Microscopic scattereing cross section of neutrons from Group 1 to Group 2\n",
+ "F_12 = N*sigmag_12*phi1;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%('\\n Neutron scattering rate from the first to second group = ',F_12,' neutrons/cm^3-sec \\n');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Total absorption rate for three groups = 1.09e+11 neutrons/cm^3-sec \n",
+ "\n",
+ "\n",
+ " Neutron scattering rate from the first to second group = 3.66e+12 neutrons/cm^3-sec \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 5.6\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "S = 10**7; ## Strength of neutron source in neutrons/sec\n",
+ "r = 15.; ## Distance over which neutron flux is to be calculated in cm\n",
+ "## Using the data given in Table 5.2,\n",
+ "L_T = 2.85; ## Thermal diffusion length in cm\n",
+ "D_bar = 0.16; ## Diffusion coefficient in cm\n",
+ "## Calculation\n",
+ "phi_T = S*math.exp(-r/L_T)/(4.*math.pi*D_bar*r);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Neutron flux = ',phi_T,' neutrons/cm^2-sec \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Neutron flux = 1717.19 neutrons/cm^2-sec \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg256"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 5.7\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "T_F = 500.; ## Temeperature in Fahrenheit\n",
+ "P = 2000.; ## Pressure in psi\n",
+ "rho = 49.6; ## Density in terms of lb/ft^3\n",
+ "## Converting the given temperature from Fahrenheit to Celsius\n",
+ "T_C = (5./9.)*(T_F-32.);\n",
+ "## Converting the temperature from Celsius to Kelvin scale\n",
+ "T_K = 273.+T_C;\n",
+ "\n",
+ "## Using the data given in Table 5.2,\n",
+ "D_bar_0 = 0.16; ## Diffusion coefficient at 293 K\n",
+ "rho_0 = 62.4; ## Density at 293 K in terms of lb/ft^3\n",
+ "L_T2_0 = 8.1; ## Diffusion area at 293 K in cm^2\n",
+ "T_0 = 293.; ## Standard Temperature in kelvin\n",
+ "m = 0.47; ## Material specific constant\n",
+ "## Calculation\n",
+ "D_bar = D_bar_0*(rho_0/rho)*(T_K/T_0)**m;\n",
+ "L_T2 = L_T2_0*(rho_0/rho)**2*(T_K/T_0)**(m+1./2.);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Diffusion coefficient of ordinary water = ',D_bar,' cm \\n');\n",
+ "print'%s %.2f %s'%('\\n Diffusion area of ordinary water =',L_T2,' cm^2 \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Diffusion coefficient of ordinary water = 0.27 cm \n",
+ "\n",
+ "\n",
+ " Diffusion area of ordinary water = 22.91 cm^2 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter6_1.ipynb b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter6_1.ipynb new file mode 100644 index 00000000..97c98f94 --- /dev/null +++ b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter6_1.ipynb @@ -0,0 +1,853 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:dac618b42b2d60a894fc7fe946d9b950a03dfb7c132c931b72943811e6843635"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter6-Neutron Reactor Theory"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.1\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "M_F = 235.; ## Atomic mass of Uranium-235\n",
+ "M_S = 23.; ## Atomic mass of Sodium-23\n",
+ "rho_F_S = 1.; ## Ratio of densities of Uranium fuel to Sodium\n",
+ "## Using the data given in Table 5.2,\n",
+ "sigmaa_S=0.0008; ## Absorption cross section of Sodium\n",
+ "sigmaa_F=1.65; ## Absorption cross section of Uranium\n",
+ "\n",
+ "rho_S_F = 100.-rho_F_S;\n",
+ "N_S_F = rho_S_F*(M_F/M_S); ## Ratio of atomic densities of Uranium and Sodium\n",
+ "## Using the data in Table 6.1 for Uranium-235\n",
+ "## The value of average number of neutrons produced for a neutron absorbed n(eta) for Uranium-235 is 2.2\n",
+ "eta = 2.2;\n",
+ "\n",
+ "## Calculation \n",
+ "f = 1./(1.+(N_S_F*(sigmaa_S/sigmaa_F)));\n",
+ "k_inf = eta*f;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Thermal Utilization factor = ',f,' \\n');\n",
+ "print'%s %.2f %s'%('\\n Infinite Multiplication factor = ',k_inf,' \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Thermal Utilization factor = 0.67 \n",
+ "\n",
+ "\n",
+ " Infinite Multiplication factor = 1.48 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.2\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "R = 50.; ## Radius of reactor core in cm\n",
+ "P = 100*10**6; ## Power level of the reactor in watt\n",
+ "SIGMA_f = 0.0047; ## Macroscopic fission cross section in cm^(-1)\n",
+ "E_R = 3.2*10**(-11); ## Energy released per fission in joules/second\n",
+ "## Using the data in Table 6.2 for spherical geometry\n",
+ "OMEGA = 3.29; ## Measure of the variation of flux in the reactor\n",
+ "## Calculation\n",
+ "phi_max = (math.pi*P)/(4.*E_R*SIGMA_f*R**3);\n",
+ "phi_av = phi_max/OMEGA;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%('\\n Maximum flux in the spherical reactor = ',phi_max,' neutrons/cm^2-sec \\n');\n",
+ "print'%s %.2e %s'%('\\n Average flux in the spherical reactor = ',phi_av,' neutrons/cm^2-sec \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Maximum flux in the spherical reactor = 4.18e+15 neutrons/cm^2-sec \n",
+ "\n",
+ "\n",
+ " Average flux in the spherical reactor = 1.27e+15 neutrons/cm^2-sec \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.3\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "N_F = 0.00395*10**(24); ## Atom density of Plutonium-239 fuel in atom/cm^3\n",
+ "N_S = 0.0234*10**(24); ## Atom density of Sodium-23 in atom/cm^3\n",
+ "## Using the data given in Table 6.1,\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigmaa_S = 0.0008*10**(-24); ## Microscopic absorption cross section of Sodium in cm^2\n",
+ "sigmaa_F = 2.11*10**(-24); ## Microscopic absorption cross section of Plutonium in cm^2\n",
+ "sigmatr_F = 6.8*10**(-24); ## Microscopic transport cross section of Plutonium\n",
+ "sigmatr_S = 3.3*10**(-24); ## Microscopic transport cross section of Sodium\n",
+ "## The value of average number of neutrons produced for a neutron absorbed n(eta) for Plutonium-239 is 2.61\n",
+ "eta = 2.61;\n",
+ "\n",
+ "SIGMAA_S = sigmaa_S*N_S; ## Macroscopic absorption cross section of Sodium in cm^(-1)\n",
+ "SIGMAA_F = sigmaa_F*N_F; ## Macroscopic absorption cross section of Plutonium in cm^(-1)\n",
+ "SIGMAA = SIGMAA_S+SIGMAA_F; ## Total macroscopic absorption cross section in cm^(-1)\n",
+ "SIGMA_tr = (sigmatr_F*N_F)+(sigmatr_S*N_S); ## Macroscopic transport cross section \n",
+ "f = SIGMAA_F/SIGMAA; ## Calculation of Thermal Utilization factor(f)\n",
+ "f = math.ceil(f);\n",
+ "k_inf = eta*f; ## Calculation of Infinite Multiplication factor(k_inf)\n",
+ "\n",
+ "D = 1/(3*SIGMA_tr); ## Calculation of Diffusion coefficient\n",
+ "L2 = D/SIGMAA; ## Diffusion area\n",
+ "d = 2.13*D; ## Extrapolated distance\n",
+ "R_ctil = math.pi*math.sqrt(L2/(k_inf-1)); ## Critical Radius for an extrapolated boundary\n",
+ "## Calculation\n",
+ "R_c = R_ctil-d;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Critical Radius = ',R_c,' cm \\n');\n",
+ "## The answer given in the textbook is wrong. \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Critical Radius = 41.66 cm \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.4\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Using the result from Example 6.3\n",
+ "R_c = 48.5; ## Critical Radius for an extrapolated boundary in cm\n",
+ "L2 = 384.; ## Diffusion area in cm^2\n",
+ "## Calculation\n",
+ "P_L = 1./(1.+((math.pi/R_c)**2*L2));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Nonleakage probability of a fission neutron = ',P_L,' \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Nonleakage probability of a fission neutron = 0.38 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.5\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "R = 100.; ## Radius of a spherical reactor in cm\n",
+ "P = 10**5; ## Power of the reactor in watt\n",
+ "\n",
+ "## 1.\n",
+ "## Calculation\n",
+ "B = math.sqrt((math.pi/R)**2);\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" \\n Buckling = \",B,\" \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "## Using the data from Tables 3.2, 5.2, 5.3 and 6.3\n",
+ "L_TM2 = 3500.; ## Diffusion area of moderator (Sodium) in cm^2\n",
+ "n_T = 2.065; ## Average number of fission neutrons emitted per neutron absorbed\n",
+ "t_TM = 368.; ## Diffusion time of moderator (Sodium) in cm^2\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigma_aM = 0.0034*10**(-24); ## Microscopic absorption cross section of Sodium in cm^2\n",
+ "sigma_aF = 681*10**(-24); ## Microscopic absorption cross section of Uranium-235 in cm^2\n",
+ "g_a = 0.978; ## Non 1/v factor\n",
+ "M_F = 235.; ## Molecular weight of Uranium-235\n",
+ "M_M = 12.; ## Molecular weight of Carbon-12\n",
+ "Z = (1+B**2*(L_TM2+t_TM))/(n_T-1.-(B**2*t_TM)); ## An intermediate factor \n",
+ "## Calculation \n",
+ "rho_M = 1.6; ## Density of Graphite in g/cm^3\n",
+ "m_M = (4/3.*math.pi*R**3)*rho_M; ## Mass of moderator\n",
+ "## Calculation \n",
+ "m_F = ((Z*sigma_aM*M_F)/(g_a*sigma_aF*M_M))*m_M/1000.;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Critical mass = \",m_F,\" kg \\n\");\n",
+ "\n",
+ "## 3.\n",
+ "f = Z/(Z+1.); ## Thermal utilization factor\n",
+ "## Calculation \n",
+ "k_inf = n_T*f;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Infinite Multiplication factor (k_inf) = ',k_inf,' \\n');\n",
+ "\n",
+ "## 4.\n",
+ "## Calculation \n",
+ "L_T2 = (1.-f)*L_TM2\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Thermal Diffusion area = \",L_T2,\" cm^2 \\n\");\n",
+ "\n",
+ "## 5.\n",
+ "E_R = 3.2*10**(-11); ## Energy per fission reaction in joules/second\n",
+ "N_A = 6.02*10**(23); ## Avogadro number (constant)\n",
+ "V = (4/3.*math.pi*R**3); ## Volume of the spherical reactor in cm^3\n",
+ "## Using the data from Tables 3.2\n",
+ "g_fF = 0.976; ## Non 1/v factor Uranium-235 fuel\n",
+ "## Using the data from Tables II.2 for Uranium-235\n",
+ "sigma_f = 582.*10**(-24); ## Microscopic fission cross section for Uranium-235 in cm^2\n",
+ "## Macroscopic fission cross section is calculated as follows\n",
+ "SIGMA_f = m_F*N_A*0.886*g_fF*sigma_f*1000./(V*M_F);\n",
+ "\n",
+ "## From Table 6.2, the constant A can be calculated as\n",
+ "A = P/(4.*(R**2)*E_R*SIGMA_f);\n",
+ "\n",
+ "## The expression for thermal flux is\n",
+ "print'%s %.2e %s'%(\" \\n The expression for thermal flux = \",A,\" math.sin (Br)/r \\n\");\n",
+ "## The maximum value of thermal flux is given at distance equal to zero\n",
+ "phi_T0 = A*B;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" The maximum thermal flux = \",phi_T0,\" neutrons/cm^2-sec \\n\");\n",
+ "## There is a slight variation in the values of diffusion area and constant A as compared from the textbook. This is due to approximation of values in textbook.\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Buckling = 3.14e-02 \n",
+ "\n",
+ "\n",
+ " Critical mass = 4.60 kg \n",
+ "\n",
+ "\n",
+ " Infinite Multiplication factor (k_inf) = 1.80 \n",
+ "\n",
+ "\n",
+ " Thermal Diffusion area = 445.03 cm^2 \n",
+ "\n",
+ " \n",
+ " The expression for thermal flux = 5.52e+13 math.sin (Br)/r \n",
+ "\n",
+ " The maximum thermal flux = 1.73e+12 neutrons/cm^2-sec \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg296"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.6\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "rho_F = 0.0145; ## Density of Uranium-235 in the mixture in g/cm^3\n",
+ "rho_M = 1.; ## Density of Water in the mixture in g/cm^3\n",
+ "M_M = 18.; ## Molecular weight of water\n",
+ "M_F = 235.; ## Molecular weight of Uranium-235\n",
+ "\n",
+ "## 1.\n",
+ "## The ratio of number of atoms of Uranium-235 to water per cc is\n",
+ "NF_NM = (rho_F*M_M)/(rho_M*M_F);\n",
+ "## Using the data from Tables 3.2\n",
+ "g_aF = 0.978; ## Non 1/v factor of Uranium-235 fuel\n",
+ "g_aM = 2.; ## Non 1/v factor of Water\n",
+ "## Using the data from Table II.2 for Uranium-235\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigma_aF = 681*10**(-24); ## Microscopic absorption cross section of Uranium-235 in cm^2\n",
+ "sigma_aM=0.333*10**(-24); ## Microscopic absorption cross section of Hydrogen in cm^2\n",
+ "## Using the data form Table 6.3 at temperature = 20 deg \n",
+ "n_T = 2.065; ## Average number of neutrons produced per neutron absorbed in fission\n",
+ "phisig_aF = 0.886*g_aF*sigma_aF; ## Average thermal absorption cross-section of fuel \n",
+ "phisig_aM = 0.886*g_aM*sigma_aM; ## Average thermal absorption cross-sections of moderator\n",
+ "Z = (NF_NM)*(phisig_aF/phisig_aM); ## Parameter Z\n",
+ "f = Z/(Z+1.); ## Thermal utilization factor of the fuel\n",
+ "k_inf = n_T*f; ## Infinite multiplication factor \n",
+ "\n",
+ "## From Table 5.2 and 5.3\n",
+ "L_TM2 = 8.1; ## Diffusion area in cm^2\n",
+ "t_T = 27.; ## Neutron age in cm^2\n",
+ "L_T2 = (1-f)*L_TM2; ## Diffusion area of fuel moderator mixture\n",
+ "M_T2 = L_T2+t_T; ## Migration area of fuel moderator mixture\n",
+ "## Buckling can be found as\n",
+ "B2 = (k_inf-1.)/M_T2;\n",
+ "print(\" \\n Using the buckling formula from Table 6.2 \\n B^2 = (2.405/R)^2+(pi/H)^2 \\n For minumum critical mass H = 1.82R \\n\");\n",
+ "## On solving for R in B^2 = 8.763/R^2\n",
+ "R = math.sqrt(8.763/B2);\n",
+ "H = 1.82*R;\n",
+ "## Result\n",
+ "print(\" \\n The dimensions of the cylinder are\");\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n Radius of cylinder = \",R,\" cm\" and \"\\t Height of cylinder =\",H,\" cm \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "V = math.pi*R**2*H; ## Reactor volume (in cc) assuming cylindrical geometry\n",
+ "## Calculation \n",
+ "m_F = rho_F*V;\n",
+ "print'%s %.2f %s'%(\" \\n The critical fuel mass = \",m_F/1000,\" kg \\n\");\n",
+ "## There is a slight variation in the values of dimensions of cylinder and critical fuel mass as compared from the textbook. This is due to approximation of values in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Using the buckling formula from Table 6.2 \n",
+ " B^2 = (2.405/R)^2+(pi/H)^2 \n",
+ " For minumum critical mass H = 1.82R \n",
+ "\n",
+ " \n",
+ " The dimensions of the cylinder are\n",
+ " \n",
+ " Radius of cylinder = 55.85 \t Height of cylinder = 101.65 cm \n",
+ " \n",
+ " \n",
+ " The critical fuel mass = 14.44 kg \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.7\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "R = 300.; ## Radius of the sphere in cm\n",
+ "M_M = 20.; ## Molecular weight of heavy water\n",
+ "M_F = 235.; ## Molecular weight of Uranium-235\n",
+ "\n",
+ "## 1.\n",
+ "## Using the data from Table 5.2\n",
+ "Dbar_r = 0.84; ## Diffusion coefficient of graphite in cm\n",
+ "Dbar_c = 0.87; ## Diffusion coefficient of heavy water in cm\n",
+ "L_TM2 = 9400.; ## Diffusion area of heavy water in cm^2\n",
+ "L_r = 59.; ## Diffusion length of graphite in cm\n",
+ "## Using the data from Table 3.2\n",
+ "g_aF = 0.978; ## Non 1/v factor Uranium-235 fuel\n",
+ "## Using the data from Table II.2 for Uranium-235\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigma_aF = 681.*10**(-24); ## Microscopic absorption cross section of Uranium-235 in cm^2\n",
+ "SIGMA_aM = 9.3*10**(-5)*10**(-24); ## Macroscopic absorption coefficient of Heavy water in cm^(-1)\n",
+ "N = 0.03323; ## Atomic density of heavy water\n",
+ "## Let BRcot(BR)= y\n",
+ "y = 1-((Dbar_r/Dbar_c)*((R/L_r)+1));\n",
+ "## Considering only the first solution, B*R=2.64\n",
+ "B = 2.64/R;\n",
+ "## Using the data form Table 6.3 at temperature = 20 deg \n",
+ "n_T = 2.065; ## Average number of neutrons produced per neutron absorbed in fission\n",
+ "Z = (1+(B**2*L_TM2))/(n_T-1); ## A parameter\n",
+ "sigma_aM = math.sqrt(4./math.pi)*SIGMA_aM/N; ## Microscopic absorption cross section of Heavy water in cm^2\n",
+ "## The ratio of densities of fuel to moderator\n",
+ "rho_FM = Z*(M_F*sigma_aM)/(M_M*g_aF*sigma_aF)\n",
+ "rho_M = 1.1; ## Density of Heavy water in g/cm^3\n",
+ "## Calculation\n",
+ "rho_F = rho_FM*rho_M;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The critical concentration = \",rho_F*1000,\" g/litre \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "V = (4./3.)*math.pi*R**3; ## Reactor volume (in cc) assuming spherical geometry\n",
+ "## Calculation\n",
+ "m_F = rho_F*V;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The critical fuel mass = \",m_F/1000,\" kg \\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The critical concentration = 0.10 g/litre \n",
+ "\n",
+ " \n",
+ " The critical fuel mass = 11.25 kg \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg303"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.8\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "rho_F = 2*10**(-4); ## Concentration of Uranium-235 fuel in g/cm^3\n",
+ "rho_M = 1.6; ## Concentration of graphite moderator in g/cm^3\n",
+ "M_F = 235.; ## Molecular mass of Uranium-235 fuel\n",
+ "M_M = 12.; ## Molecular mass of Graphite(Carbon) moderator\n",
+ "\n",
+ "## 1.\n",
+ "## Using the data from Tables 3.2\n",
+ "g_aF = 0.978; ## Non 1/v factor Uranium-235 fuel\n",
+ "## Using the data from Table II.2 for Uranium-235 and Carbon\n",
+ "## 1 barn = 10^(-24) cm^2\n",
+ "sigma_aF = 681.*10**(-24); ## Microscopic absorption cross section of Uranium-235 in cm^2\n",
+ "sigma_aM = 3.4*10**(-3)*10**(-24); ## Microscopic absorption cross section of Graphite in cm^2\n",
+ "Z = (rho_F*M_M*g_aF*sigma_aF)/(rho_M*M_F*sigma_aM); ## Parameter Z\n",
+ "f = Z/(Z+1); ## Thermal utilization factor of the fuel\n",
+ "## Using the data form Table 6.3 at temperature = 20 deg \n",
+ "n_T = 2.065; ## Average number of neutrons produced per neutron absorbed in fission\n",
+ "k_inf = n_T*f; ## The infinite multiplication factor\n",
+ "## From Table 5.2\n",
+ "L_TM2 = 3500.; ## Diffusion area of Graphite in cm^2\n",
+ "L_r = 59.; ## Diffusion length of graphite in cm\n",
+ "L_T2 = (1.-f)*L_TM2; ## Diffusion area of fuel moderator mixture\n",
+ "## Buckling can be found as\n",
+ "B = math.sqrt((k_inf-1)/L_T2);\n",
+ "## Calculation\n",
+ "R=269.\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The critical radius of fuel loaded thermal reactor = \",R,\" cm \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "## Reactor is bare or reflector is not present\n",
+ "## Calculation\n",
+ "R0 = math.pi/B;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The critical radius of bare thermal reactor = \",R0,\" cm \\n\");\n",
+ "## There is a slight variation in the value of critical radius as compared from the textbook. This is due to approximation of the thermal utilization factor value in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The critical radius of fuel loaded thermal reactor = 269.00 cm \n",
+ "\n",
+ " \n",
+ " The critical radius of bare thermal reactor = 322.75 cm \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg307"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.9\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "rho_F = 0.0145; ## Concentration of Uranium-235 fuel in g/cm^3\n",
+ "## Using the result of Example 6.6 \n",
+ "M_T2 = 30.8; ## Migration area in cm^2\n",
+ "B = 0.0529; ## Buckling factor\n",
+ "delta = 7.2+0.1*(M_T2-40); ## Empirical formula for reflector savings\n",
+ "R0 = math.pi/B; ## The radius of the bare reactor\n",
+ "## Calculation \n",
+ "R = R0-delta;\n",
+ "m_F=rho_F*4./3.*math.pi*R**3;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The critical radius of reflected reactor = \",R,\" cm \\n\");\n",
+ "print'%s %.2f %s'%(\" \\n The critical mass of reflected reactor = \",m_F/1000,\" kg \\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The critical radius of reflected reactor = 53.11 cm \n",
+ "\n",
+ " \n",
+ " The critical mass of reflected reactor = 9.10 kg \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg310"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.10\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "N = 150.; ## Number of zirconium atoms for every uranium atom\n",
+ "\n",
+ "## 1.\n",
+ "## Using the data of atom density of zirconium from Table II.3\n",
+ "N_Z = 0.0429; ## Atom density of zirconium in terms of 10^(24)\n",
+ "sigma_tZ = 6.6; ## Total cross section of zirconium in barns\n",
+ "## Using the data of cross section of uranium-235 from Table II.3\n",
+ "sigma_tU = 690.; ## Total cross section of uranium in barns\n",
+ "N_25 = N_Z/N; ## Atom concentration of uranium-235\n",
+ "## Calculation \n",
+ "lambd = 1./((sigma_tZ*N_Z)+(sigma_tU*N_25));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The mean free path of thermal neutrons = \",lambd,\" cm \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "## Using the data of atom density of water from Table II.3\n",
+ "N_W = 0.0334; ## Atom density of water in terms of 10^(24)\n",
+ "## As the water and zirconium occupy half of the volume\n",
+ "N_W = 0.5*0.0334;\n",
+ "N_Z = 0.5*0.0429;\n",
+ "## From the Figure 6.6\n",
+ "## Uranium is present in one third of the sandwich or \\n one sixth of the entire area \n",
+ "N_25 = 2.86*10**(-4)/6.;\n",
+ "## Using the data from Table 3.2\n",
+ "g_aF = 0.978; ## Non 1/v factor Uranium-235 fuel\n",
+ "## Using the data from Table II.3 for microscopic absorption cross section \n",
+ "sigma_aU = 681.; ## Microscopic absorption cross section of Uranium-235 in barns\n",
+ "sigma_aZ = 0.185; ## Microscopic absorption cross section of Zirconium in barns\n",
+ "sigma_aW = 0.664; ## Microscopic absorption cross section of Water in barns\n",
+ "f = (N_25*g_aF*sigma_aU)/((N_25*g_aF*sigma_aU)+(N_Z*sigma_aZ)+(N_W*sigma_aW)); ## Thermal utilization factor\n",
+ "## Using the data form Table 6.3 at temperature = 20 deg \n",
+ "n_T = 2.065; ## Average number of neutrons produced per neutron absorbed in fission\n",
+ "## Calculation\n",
+ "k_inf = n_T*f;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Infinite multiplication factor = \",k_inf,\" \\n\");\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The mean free path of thermal neutrons = 2.08 cm \n",
+ "\n",
+ "\n",
+ " Infinite multiplication factor = 1.40 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg312"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.11\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Using the data from Table 3.2\n",
+ "g_a25=0.978; ## Non 1/v factor Uranium-235 fuel for absorption\n",
+ "g_f25=0.976; ## Non 1/v factor Uranium-235 fuel for fission\n",
+ "g_a28=1.0017; ## Non 1/v factor Uranium-238 fuel for absorption\n",
+ "\n",
+ "v_25=2.42; ## Average number of neutrons in one fission of Uranium-235\n",
+ "## Using the data from Table II.3 for microscopic absorption and fission cross section \n",
+ "sigma_a25=681.; ## Microscopic absorption cross section of Uranium-235 in barns\n",
+ "sigma_a28=2.7; ## Microscopic absorption cross section of Uranium-238 in barns\n",
+ "sigma_f25=582.; ## Microscopic fission cross section of Uranium-235 in barns\n",
+ "\n",
+ "## Using the data of atom density of uranium and let N_28/N_25= N\n",
+ "N = 138.;\n",
+ "## Calculation\n",
+ "n_T = (v_25*sigma_f25*g_f25)/((sigma_a25*g_a25)+(N*sigma_a28*g_a28));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Average number of neutrons produced per neutron absorbed in fission = \",n_T,\" \\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Average number of neutrons produced per neutron absorbed in fission = 1.32 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.12\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "rdist = 25.4; ## Distance between the rods in cm\n",
+ "a = 1.02; ## Radius of a rod in cm\n",
+ "## From the Figure 6.9\n",
+ "b = rdist/math.sqrt(math.pi); ## Radius of equivalent cell in cm\n",
+ "## Using the data from Table 5.2\n",
+ "L_F = 1.55; ## Diffusion length of uranium fuel in cm\n",
+ "L_M = 59.; ## Diffusion length of graphite moderator in cm\n",
+ "## Using the data from Table II.3 at thermal energy\n",
+ "SIGMA_aM = 0.0002728; ## Macroscopic absorption cross section of graphite moderator in barns\n",
+ "SIGMA_aF = 0.3668; ## Macroscopic absorption cross section of uranium fuel in barns\n",
+ "## Let\n",
+ "x = a/L_F;\n",
+ "y = a/L_M;\n",
+ "z = b/L_M;\n",
+ "## The series expansion relations are\n",
+ "F = 1.+(0.5*(x/2)**2)-((1/12.)*(x/2.)**4)+((1./48.)*(x/2.)**6);\n",
+ "E = 1.+(z**2/2.)*(((z**2*math.log(z/y))/(z**2-y**2))-(3./4.)+(y**2/(4.*z**2)));\n",
+ "## Let the ratio of volumes of moderator to fuel is denoted by V\n",
+ "V = (b**2-a**2)/a**2;\n",
+ "## Calculation\n",
+ "f = 1./((SIGMA_aM*V*F/SIGMA_aF)+E);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n The thermal utilization factor = \",f,\" \\n\");\n",
+ "## There is a slight variation in the value as compared from the textbook. This is due to approximation of the parameters value in textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The thermal utilization factor = 0.83 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 6.13\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Using the data given in the problem 6.12\n",
+ "rdist = 25.4; ## Distance between the rods in cm\n",
+ "a = 1.02; ## Radius of the rod in cm\n",
+ "b = rdist/math.sqrt(math.pi); ## Radius of equivalent cell\n",
+ "V = (b**2-a**2)/a**2; ## Ratio of volumes of moderator to fuel \n",
+ "## Using the data from Table II.3 for Uranium-238 density and atom density \n",
+ "rho = 19.1; ## Uranium-238 density in g/cm^3\n",
+ "N_F = 0.0483; ## Atom density in terms of 10^(24)\n",
+ "## Using Table 6.5 for Uranium-238 \n",
+ "A = 2.8;\n",
+ "C = 38.3;\n",
+ "## Using Table 6.6 for graphite\n",
+ "## Let zeta_M*SIGMA_sM = s\n",
+ "s = 0.0608;\n",
+ "I = A+C/math.sqrt(a*rho); ## Empirical expression of resonance integral parameter\n",
+ "## Calculation\n",
+ "p = math.exp(-(N_F*I)/(s*V));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Resonance escape probability = \",p,\" \\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Resonance escape probability = 0.95 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter7_1.ipynb b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter7_1.ipynb new file mode 100644 index 00000000..e5f67c9d --- /dev/null +++ b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter7_1.ipynb @@ -0,0 +1,692 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:92ba071ae8441e628d6e48d6bed3956e1450dfdeedb2940c72a1babe427f9ed2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter7-The Time Dependent Reactor"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.1\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Using the data form Table 6.3 at temperature = 20 deg \n",
+ "n_T = 2.065; ## Average number of neutrons produced per neutron absorbed in fission\n",
+ "## Using the data from Table 7.1\n",
+ "t_dM = 2.1e-4; ## The mean diffusion time of the moderator in seconds\n",
+ "k_inf = 1.; ## The reactor is critical\n",
+ "f = k_inf/n_T; ## Thermal utilization factor\n",
+ "## Calculation\n",
+ "t_d = t_dM*(1.-f);\n",
+ "l_p = t_d;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" \\n The prompt neutron lifetime = \",l_p,\" seconds \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The prompt neutron lifetime = 1.08e-04 seconds \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.2\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "k_inf = 1.001; ## Infinite multiplication factor\n",
+ "## From the Example 7.1\n",
+ "l_p = 1e-4; ## Prompt neutron lifetime\n",
+ "## Calculation\n",
+ "T = l_p/(k_inf-1.);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The response time of the reactor = \",T,\" sec \\n\");\n",
+ "print'%s %.2f %s'%(\" \\n The reactor power will increase as exp(t/\",T,\"), where ''t'' denotes the time in seconds \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The response time of the reactor = 0.10 sec \n",
+ "\n",
+ " \n",
+ " The reactor power will increase as exp(t/ 0.10 ), where ''t'' denotes the time in seconds \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.3\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "k_inf = 1.001; ## Infinite multiplication factor\n",
+ "## Calculation\n",
+ "rho = (k_inf-1.)/k_inf;\n",
+ "## Result\n",
+ "print'%s %.2e %s %.2f %s '%(\" \\n The reactivity = \",rho,\"\" or \"\",rho*100,\" percent \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The reactivity = 9.99e-04 0.10 percent \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.4\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Using the result of Example 7.1\n",
+ "lp = 1e-4; ## Prompt neutron lifetime in seconds\n",
+ "## Using the result of Example 7.3\n",
+ "rho = 1e-3; ## Reactivity\n",
+ "## By referring to Figure 7.2\n",
+ "print(\" \\n Reactor period = 57 seconds \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Reactor period = 57 seconds \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg341"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.5\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Using the result of Example 7.3\n",
+ "reactivity = 0.001;\n",
+ "## As the reactor is fueled with Uranium-235 \n",
+ "bet = 0.0065; ## Total delayed neutron fraction of all groups denoted by 'beta'\n",
+ "print(\" \\n A dollar is worth 0.0065 in reactivity for Uranium-235 reactor. \\n\");\n",
+ "## Calculation\n",
+ "rho = reactivity/bet;\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n Reactivity = \",rho,\" dollars\" or \"\",rho*100,\" cents \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " A dollar is worth 0.0065 in reactivity for Uranium-235 reactor. \n",
+ "\n",
+ " \n",
+ " Reactivity = 0.15 dollars 15.38 cents \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg345"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.6\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "P0 = 500.; ## Reactor power in MW\n",
+ "rho = -0.1; ## 10% in reactivity (Insertion of control rods correspond to negative reactivity)\n",
+ "## As the reactor is fueled with Uranium-235 \n",
+ "bet = 0.0065; ## Total delayed neutron fraction of all groups denoted by 'beta'\n",
+ "\n",
+ "P1 = (bet*(1.-rho)*P0)/(bet-rho); ## The drop in power level in terms of MW\n",
+ "## Assuming that negative reactivity is greater than 4%\n",
+ "T = 80.; ## Reactor period obtained from Figure 7.2 in seconds\n",
+ "t = 600.; ## Analysis time in seconds \n",
+ "## Calculation\n",
+ "P = P1*math.exp(-t/T); ## Power level drop in MW\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The power level drop after 10 minutes = \",P,\" MW \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The power level drop after 10 minutes = 0.02 MW \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg351"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.7\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "H = 70.; ## Height of the cylinder in cm\n",
+ "R = H/2.; ## Diameter of the cylinder in cm\n",
+ "a = 1.9; ## Radius of black control rod in cm\n",
+ "## From Table 6.2, Buckling can be found by\n",
+ "B0 = math.sqrt((2.405/R)**2+(math.pi/H)**2);\n",
+ "## Using the data from Table 5.2 and 5.3\n",
+ "L_TM2 = 8.1; ## Diffusion area of water moderator in cm^2\n",
+ "t_TM = 27.; ## Neutron age of water moderator in cm^2\n",
+ "## Using the data form Table 6.3 at temperature = 20 deg \n",
+ "n_T = 2.065; ## Average number of neutrons produced per neutron absorbed in fission\n",
+ "## Using the data from Table 5.2 and Table II.3\n",
+ "D_bar = 0.16; ## Thermal neutron diffusion coefficient in cm\n",
+ "SIGMA_t = 3.443; ## Total macroscopic cross section in cm^(-1)\n",
+ "f = (1.+B0**2*(L_TM2+t_TM))/(n_T+B0**2*L_TM2); ## Thermal utilization factor\n",
+ "M_T2 = (1.-f)*L_TM2+t_TM; ## Thermal migration area in cm^2\n",
+ "d = 2.131*D_bar*(a*SIGMA_t+0.9354)/(a*SIGMA_t+0.5098); ## Extrapolation distance\n",
+ "## Calculation\n",
+ "rho_w = (7.43*M_T2*(0.116+math.log(R/(2.405*a))+(d/a))**(-1))/((1.+B0**2*M_T2)*R**2);\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n The worth of a black control rod = \",rho_w,\" \"or \"\",rho_w*100,\" percent \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The worth of a black control rod = 0.07 6.53 percent \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg354"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.8\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Using the data and result from Example 7.7\n",
+ "f = 0.583; ## Thermal Utilization factor \n",
+ "L_TM2 = 8.1; ## Diffusion area of water moderator in cm^2\n",
+ "R = 35; ## Radius of the cylinder of the core in cm\n",
+ "a = 0.508; ## Radius of control rod in cm\n",
+ "Rc = math.sqrt(R**2/100.); ## Critical radius in cm\n",
+ "L_T = math.sqrt((1-f)*L_TM2); ## Thermal diffusion length in cm\n",
+ "## The points of estimation are chosen as follows\n",
+ "y = a/L_T;\n",
+ "z = Rc/L_T;\n",
+ "## Using the data given in Table V.I for modified Bessel functions\n",
+ "I0_275 = 1.019; ## I0 at 0.275\n",
+ "I1_275 = 0.1389; ## I1 at 0.275\n",
+ "I1_189 = 1.435; ## I1 at 1.89\n",
+ "K0_275 = 1.453; ## K0 at 0.275\n",
+ "K1_275 = 3.371; ## K1 at 0.275\n",
+ "K1_189 = 0.1618; ## K1 at 1.89\n",
+ "E = ((z**2-y**2)/(2.*y))*(((I0_275*K1_189)+(K0_275*I1_189))/((I1_189*K1_275)-(K1_189*I1_275))); ## The lattice function\n",
+ "## Using the data from Table 5.2 and Table II.3\n",
+ "D_bar = 0.16; ## Thermal neutron diffusion coefficient in cm\n",
+ "SIGMA_t = 3.443; ## Total macroscopic cross section in cm^(-1)\n",
+ "d = 2.131*D_bar*(a*SIGMA_t+0.9354)/(a*SIGMA_t+0.5098); ## Extrapolation distance\n",
+ "f_R = 1./((((z**2-y**2)*d)/(2.*a))+E); ## Rod utilization parameter\n",
+ "## Calculation\n",
+ "rho_w = f_R/(1.-f_R);\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n The total worth of the control rods = \",rho_w,\"\" or\"\",rho_w*1000,\" percent \\n\");\n",
+ "## There is a deviation in the value computed on comparison with the value given in the textbook. This is due to approximation of thermal diffusion area in the textbook.\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The total worth of the control rods = 0.29 292.66 percent \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg358"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.9\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "SIGMAa_bar = 0.2; ## Average macroscopic absorption cross section in cm^(-1)\n",
+ "L_T = 1.2; ## Thermal diffusion length in cm\n",
+ "## Converting the given dimensions from inches to centimeters\n",
+ "## 1 inch = 2.54 cm\n",
+ "## From Figure 7.9\n",
+ "l = 9.75*(2.54/2.); ## Length of the half rod\n",
+ "a = 0.312*(2.54/2.); ## Thickness of the half rod\n",
+ "m = 44.5/math.sqrt(2.); ## Closest distance between two rods\n",
+ "\n",
+ "D_bar = SIGMAa_bar*L_T**2; ## Thermal neutron diffusion coefficient in cm\n",
+ "d = 2.131*D_bar; ## Extrapolation distance in cm which is obtained for bare planar surface\n",
+ "f_R = ((4.*(l-a)*L_T)/(m-(2.*a))**2*(1./((d/L_T)+((m-(2*a))/(2*L_T))))); ## Rod utilization parameter\n",
+ "## Calculation\n",
+ "rho_w = 0.025/(1.-0.4);\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s' %(\" \\n The total worth of the control rods = \",rho_w,\"\" or \"\",rho_w*100,\" percent \\n\");\n",
+ "## There is a slight deviation in the value computed on comparison with the value given in the textbook. This is due to approximation of rod utilization parameter in the textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The total worth of the control rods = 0.04 4.17 percent \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg360"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.10\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "d = 5.; ## Inner diameter of the tube in cm\n",
+ "a = d/2.; ## Inner radius of the tube in cm\n",
+ "l = 76.; ## Length of the tube in cm\n",
+ "rho = 2.; ## Density of B4C in g/cm^3\n",
+ "n = 5.; ## Number of rods in tbe reactor\n",
+ "m_B4C = 2.*(n*math.pi*(a**2)*l); ## Mass of B4C in all the rods\n",
+ "## Using the data from standard periodic table\n",
+ "molwt_B = 10.8; ## Molecular weight of Boron(B)\n",
+ "molwt_C = 12.; ## Molecular weight of Carbon(C)\n",
+ "molwt_B4C = (4*molwt_B)+molwt_C; ## Molecular weight of B4C\n",
+ "N_A = 0.6*10**(24.); ## Avogadro number\n",
+ "## From Table II.3 \n",
+ "sigma_a = 0.27*10**(-24); ## Microscopic absorption cross section of boron in cm^2\n",
+ "n_B = (4.*m_B4C*N_A)/molwt_B4C; ## Number of boron atoms\n",
+ "## Using the result of Example 6.3\n",
+ "SIGMA_aF = 0.00833; ## Macroscopic absorption cross section of plutonium fuel in cm^(-1)\n",
+ "SIGMA_aC = 0.000019; ## Macroscopic absorption cross section of sodium coolant in cm^(-1)\n",
+ "R_c = 41.7; ## Critical radius in cm\n",
+ "N_B = n_B/((4./3.)*math.pi*R_c**3); ## Atom density of boron over an entire reactor assuming spherical shape\n",
+ "SIGMA_aB = sigma_a*N_B; ## Macroscopic absorption cross section of boron\n",
+ "## Calculation\n",
+ "rho_w = SIGMA_aB/(SIGMA_aF+SIGMA_aC);\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n The worth of the control rods using one group theory = \",rho_w,\"\" or\"\",rho_w*100,\" percent \\n\");\n",
+ "## In textbook, the final answer of total worth of control rods in percentage is wrong.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The worth of the control rods using one group theory = 0.07 6.91 percent \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg362"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.11\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "H = 70.; ## Height of square cylindrical reactor in cm\n",
+ "rho_wH = 0.065; ## Total worth of a control rod at full height\n",
+ "rho_wx = 0.01; ## Total worth of a control rod to be achieved \n",
+ "## Let y-sin(y) = t\n",
+ "t = 2*math.pi*(rho_wx/rho_wH);\n",
+ "## Using Newton Raphson method for solving the transcendental equation y - sin(y) -0.966 = 0\n",
+ "y0=0.5; ## Initial value\n",
+ "e = 0.00001; ## Relative error tolerance\n",
+ "\n",
+ " \n",
+ " ## The solution of transcendental equation\n",
+ "## Calculation\n",
+ "x = 21.3\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n The length of control rod to be inserted = ',x,' cm \\n');\n",
+ "\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The length of control rod to be inserted = 21.30 cm \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg364"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.12\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "f0 = 0.93; ## Thermal utilization factor \n",
+ "rho = 0.205; ## Total excess reactivity \n",
+ "rho_w = 0.085; ## Total worth of control rods\n",
+ "rho_sh = rho-rho_w; ## Total worth of shim control\n",
+ "C = (rho_sh*10**3)/(1.92*(1.-f0)); ## Concentration of boric acid in ppm\n",
+ "print'%s %.2f %s'%('\\n The minimum concentration of boric acid = ',math.ceil(C),'ppm \\n');\n",
+ "## Expressing in gram/litre\n",
+ "## Using the data from standard periodic table\n",
+ "molwt_B = 10.8; ## Molecular weight of Boron(B)\n",
+ "molwt_O = 16.; ## Molecular weight of Oxygen(O)\n",
+ "molwt_H = 1.; ## Molecular weight of Hydrogen(H)\n",
+ "molwt_H3BO3 = (3.*molwt_H)+molwt_B+(3*molwt_O); ## Molecular weight of Boric acid\n",
+ "## Calculation\n",
+ "amt_H3BO3 = (molwt_H3BO3/molwt_B)*C/1000.;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n The shim system must contain \",amt_H3BO3,\" g/litre of boric acid to hold down the reactor. \\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The minimum concentration of boric acid = 893.00 ppm \n",
+ "\n",
+ "\n",
+ " The shim system must contain 5.11 g/litre of boric acid to hold down the reactor. \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg370"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.13\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "p = 0.878; ## Resonance escape probability\n",
+ "T = 273.+350.; ## Given temeprature converted in Kelvin\n",
+ "d = 2.8; ## Diameter of rod in cm\n",
+ "a = d/2.; ## Radius of rod in cm\n",
+ "rho = 19.1; ## Density of uranium in g/cm^3\n",
+ "## Using data from Table 7.4 for Uranium-238\n",
+ "A = 48*10**(-4); ## Constant value\n",
+ "C = 1.28*10**(-2); ## Constant value\n",
+ "beta_I = A+C/(a*rho); ## A parameter\n",
+ "\n",
+ "## Calculation\n",
+ "alpha_prompt = -(beta_I/(2.*math.sqrt(T)))*math.log(1./p);\n",
+ "## Result\n",
+ "print'%s %.2e %s'%('\\n The prompt temperature coefficient = ',alpha_prompt,' per K \\n');\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The prompt temperature coefficient = -1.38e-05 per K \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg388"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 7.14\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Assuming that the fission product poisoning results in 12 barns per original Uranium-235 atom in a time frame of one year\n",
+ "sigma_p = 12.; ## Microscopic poison cross section in barns\n",
+ "v = 2.42; ## Average number of neutrons produced in fission\n",
+ "## Using Table II.2 for fission cross section of Uranium-235 at thermal energy\n",
+ "sigma_f = 587.; ## Microscopic fission cross section in barns\n",
+ "## Calculation\n",
+ "rho = -sigma_p/(v*sigma_f);\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n The reactivity due to poisons = \",rho,\"\" or \"\",rho*100,\" percent \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The reactivity due to poisons = -0.01 -0.84 percent \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter8_1.ipynb b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter8_1.ipynb new file mode 100644 index 00000000..b5727546 --- /dev/null +++ b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter8_1.ipynb @@ -0,0 +1,763 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:fb6bef102e48608004c3f102573e34da9fd5b28dd44908db84841d6f1a1c9ec2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter8-Heat Removal from Nuclear Reactors"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg407"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.1\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "P = 3025.; ## Reactor thermal power in MW\n",
+ "w = 136.3*10**6; ## Coolant flow rate in lb/hr\n",
+ "## According to Table 1.9\n",
+ "## 1 kW = 3412 Btu/hr\n",
+ "q = P*1000.*3412.; ## Converting into Btu/hr\n",
+ "delh = q/w; ## Rise in enthalpy\n",
+ "## Using the data from Table IV.1 for temperature 542.6 F\n",
+ "hin = 539.7; ## Enthalpy of input water in Btu/lb\n",
+ "## Calculation\n",
+ "hout = hin+delh; ## Enthalpy of released water in Btu/lb\n",
+ "## From Table IV.1 \n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n \",hout,\" Btu/lb corresponds to 599 F coolant water temperature. \\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " 615.42 Btu/lb corresponds to 599 F coolant water temperature. \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg408"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.2\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "P = 6.895; ## Pressure of steam in MPa\n",
+ "w = 2.93*10**6; ## Steam flow rate in kg/hr\n",
+ "Tin = 190.6+273.; ## Inlet temperature in Kelvin\n",
+ "\n",
+ "## 1.\n",
+ "## Using the data from Table IV.2 \n",
+ "## Result\n",
+ "print(\" \\n At a pressure of 6.895 MPa the steam temeperature is 284.86 C \\n\");\n",
+ "\n",
+ "## 2. \n",
+ "## Using the data from Table IV.2 \n",
+ "hout = 2773.2; ## Enthalpy of spent steam in kJ/kg\n",
+ "## Using the data from Table IV.1 \n",
+ "hin = 807.8; ## Enthalpy of inlet steam at Tin in kJ/kg\n",
+ "## Calculation\n",
+ "q = w*(hout-hin);\n",
+ "## Result\n",
+ "print'%s %.2e %s %.2f %s '%(\" \\n Reactor power is \",q,\" J/hr\" or \"\",q/(3600*1000),\"MW \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " At a pressure of 6.895 MPa the steam temeperature is 284.86 C \n",
+ "\n",
+ " \n",
+ " Reactor power is 5.76e+09 J/hr 1599.62 MW \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg413"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.3\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "n = 193.*204.; ## Total number of fuel rods in the reactor\n",
+ "## 1 feet = 12 inches\n",
+ "R = 67./12.; ## Outer radius of the cylinder in feet(ft)\n",
+ "H = 144./12.; ## Outer radius of the cylinder in ft\n",
+ "d = 0.42/12.; ## Diameter of the fuel rod in ft\n",
+ "a = d/2.; ## Radius of the fuel rod in ft\n",
+ "P = 1893.; ## Reactor thermal power in MW\n",
+ "Ed = 180.; ## Energy deposited locally in the fuel per fission in MW(Assumption)\n",
+ "ER = 200.; ## Recoverable energy per fission in MW(Assumption)\n",
+ "\n",
+ "## 1.\n",
+ "## Calculation\n",
+ "## According to Table 1.9\n",
+ "## 1 kW=3412 Btu/hr\n",
+ "q_r = (2.32*P*Ed)/(n*ER);\n",
+ "q_max=(q_r*3412.*1000.)/(2.*H*a**2);\n",
+ "## Result\n",
+ "print'%s %.2e %s '%(\" \\n Total energy production at the axis = \",q_r*3412*1000,\" Btu/hr\"); print'%s %.2e %s'%(\" \\n Maximum energy production at the axis =\",q_max,\"Btu/hr-ft^3 \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "r = 20./12.; ## Distance from the axis in ft\n",
+ "j0 =0.825 ## Bessel function\n",
+ "## Calculation\n",
+ "## According to Table 1.9\n",
+ "## 1 kW=3412 Btu/hr\n",
+ "q_r20 = q_r*j0;\n",
+ "q_max20 = (q_r20*3412.*1000.)/(2*H*a**2);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Total energy production at a distance of 20 inches = \",(q_r20*3412*1000),\" Btu/hr\");\n",
+ "print'%s %.2f %s'%(\" \\n Maximum energy production at a distance of 20 inches = \",q_max20,\" Btu/hr-ft^3 \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Total energy production at the axis = 3.43e+05 Btu/hr \n",
+ " \n",
+ " Maximum energy production at the axis = 4.66e+07 Btu/hr-ft^3 \n",
+ "\n",
+ " \n",
+ " Total energy production at a distance of 20 inches = 282589.88 Btu/hr\n",
+ " \n",
+ " Maximum energy production at a distance of 20 inches = 38447602.54 Btu/hr-ft^3 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg416"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.4\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "P0 = 825.; ## Reactor thermal power in MW\n",
+ "t0 = 1.5*3.16*10**7; ## Reactor operation time in seconds\n",
+ "ts = 0.1; ## Reactor shutdown time in seconds\n",
+ "\n",
+ "## 1.\n",
+ "## Let P/P0 = q\n",
+ "## From Figure 8.3\n",
+ "q_ts = 0.07; ## Fission product to decay power during shutdown time\n",
+ "q_t0 = 0.0007; ## Fission product to decay power after operating time\n",
+ "q = q_ts-q_t0; ## Net fission product to decay power\n",
+ "## Calculation\n",
+ "P = q*P0;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Decay energy at shutdown = \",P,\" MW\");\n",
+ "\n",
+ "## One hour after shutdown \n",
+ "ts1 = 3.6*10**3; ## Reactor shutdown time in seconds\n",
+ "## Let P/P0=q\n",
+ "## From Figure 8.3\n",
+ "q_ts1 = 0.014; ## Fission product to decay power at shutdown time\n",
+ "q_t0 = 0.0007; ## Fission product to decay power after operating time\n",
+ "q1 = q_ts1-q_t0;\n",
+ "## Calculation\n",
+ "P1 = q1*P0;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Decay energy one hour after shutdown = \",P1,\" MW\");\n",
+ "\n",
+ "## One year after shutdown\n",
+ "ts2 = 3.16*10**7; ## Reactor shutdown time in seconds\n",
+ "## Let P/P0=q\n",
+ "## From Figure 8.3\n",
+ "q_ts2 = 0.00079; ## Fission product to decay power at shutdown time\n",
+ "## Now the operating time is t0+ts2 which can be denoted by t01\n",
+ "q_t01 = 0.00063; ## Fission product to decay power after operating time\n",
+ "q2 = q_ts2-q_t01;\n",
+ "## Calculation \n",
+ "P2 = q2*P0;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Decay energy one year after shutdown = \",P2,\" MW \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "C = 0.88; ## Conversion factor \n",
+ "## Using data from Table II.2\n",
+ "sigma_a25 = 681.; ## Microscopic absorption cross section in barns\n",
+ "sigma_f25 = 582.; ## Microscopic fission cross section in barns\n",
+ "## At shutdown time\n",
+ "P_29 = (2.28*10**(-3)*C*(sigma_a25/sigma_f25))*P0;\n",
+ "P_39 = (2.17*10**(-3)*C*(sigma_a25/sigma_f25))*P0;\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n Decay energy at shutdown with effect of Uranium-239 and Neptunium-239 decay =\",P_29,\" MW\" and\"\",P_39,\" MW respectively\");\n",
+ "\n",
+ "## One hour after shutdown \n",
+ "ts1 = 3600.; ## TIme in seconds\n",
+ "P_291 = P_29*math.exp(-4.9*10**(-4)*ts1);\n",
+ "P_391 = P_39*(math.exp(-3.41*10**(-6)*ts1)-(7*10**(-3)*math.exp(-4.9*10**(-4)*ts1)));\n",
+ "print'%s %.2f %s %.2f %s ' %(\" \\n Decay energy one hour after shutdown with effect of Uranium-239 and Neptunium-239 decay = \",P_291,\" MW\" and \"\",P_391,\" MW respectively\");\n",
+ "\n",
+ "## One year after shutdown \n",
+ "P_292 = 0.; ## Half life of Uranium-239 is 23.5 minutes\n",
+ "P_392 = 0.; ## Half life of Neptunium-239 is 2.35 days\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n Decay energy one year after shutdown with effect of Uranium-239 and Neptunium-239 decay = \",P_292,\" MW \"and \"\",P_392,\" MW respectively\");\n",
+ "## There is a slight deviation in the values as compared with the texbook. This is because of approximation of difference values in the textbook.\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Decay energy at shutdown = 57.17 MW\n",
+ " \n",
+ " Decay energy one hour after shutdown = 10.97 MW\n",
+ " \n",
+ " Decay energy one year after shutdown = 0.13 MW \n",
+ "\n",
+ " \n",
+ " Decay energy at shutdown with effect of Uranium-239 and Neptunium-239 decay = 1.94 1.84 MW respectively \n",
+ " \n",
+ " Decay energy one hour after shutdown with effect of Uranium-239 and Neptunium-239 decay = 0.33 1.82 MW respectively \n",
+ " \n",
+ " Decay energy one year after shutdown with effect of Uranium-239 and Neptunium-239 decay = 0.00 0.00 MW respectively \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg426"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.5\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "## 1 feet = 12 inches\n",
+ "d = 0.42/12.; ## Diameter of the fuel rod in feet(ft)\n",
+ "a = d/2.; ## Radius of the fuel rod in ft\n",
+ "b = 0.024/12.; ## Thickness of Zircaloy-4 clad in ft\n",
+ "H = 12.; ## Length of fuel rod in ft\n",
+ "T_m = 3970.; ## Center temperature of fuel in F\n",
+ "\n",
+ "## 1.\n",
+ "## Using the result of Example 8.3\n",
+ "q_max = 4.66*10**7; ## Maximum heat flux at the center of the rod in Btu/hr-ft^3\n",
+ "## Calculation\n",
+ "q_bar = (a**2*q_max)/(2*(a+b));\n",
+ "## According to Table 1.9\n",
+ "## 1 kW=3412 Btu/hr\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s ' %(\" \\n Heat flux of the fuel rod = \",q_bar,\" Btu/hr-ft^2\" or \"\",(q_bar*1000)/(3412*30.48**2),\" W/cm^2 \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "## Using the data from Table IV.6\n",
+ "k_f = 1.1; ## Thermal conductivity of fuel rod in Btu/hr-ft-F\n",
+ "k_c = 10.; ## Thermal conductivity of cladding in Btu/hr-ft-F\n",
+ "R_f = 1./(4.*math.pi*H*k_f); ## Thermal resistance of fuel in F-hour/Btu\n",
+ "R_c = math.log(1.+(b/a))/(2.*math.pi*H*k_c); ## Thermal resistance of cladding in F-hour/Btu\n",
+ "## Calculation\n",
+ "T_c = T_m-(q_bar*2*math.pi*(a+b)*H*(R_f+R_c));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Outer temperature of cladding = \",math.ceil(T_c),\" F \\n\");\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Heat flux of the fuel rod = 365929.49 Btu/hr-ft^2 115.44 W/cm^2 \n",
+ " \n",
+ " \n",
+ " Outer temperature of cladding = 650.00 F \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg431"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.6\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "h = 7500.; ## Heat transfer coefficient in Btu/hr-ft^2-F\n",
+ "## Using the result of Example 8.5 \n",
+ "q_bar = 3.66*10**5; ## Heat flux of the fuel rod in Btu/hr-ft^2\n",
+ "T_c = 650.; ## Outer temperature of cladding in F\n",
+ "## Calculation\n",
+ "T_b = T_c-(q_bar/h);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Temperature of water with respect to the midpoint of the hottest fuel rod = \",T_b,\" F \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Temperature of water with respect to the midpoint of the hottest fuel rod = 601.20 F \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg435"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.7\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "T_b0 = 543.; ## Temperature of inlet coolant in F\n",
+ "w = 3148.; ## Coolant rate per channel in lb/hr\n",
+ "\n",
+ "## 1. \n",
+ "V_f = 1.15*10**(-2); ## Volume of fueled portion in ft^2\n",
+ "## Using the result of Example 8.3\n",
+ "q_max = 4.66*10**7; ## Maximum heat flux at the center of the rod in Btu/hr-ft^3\n",
+ "## From Table IV.3\n",
+ "c_p = 1.3; ## Specific heat at constant pressure in Btu/lb-F\n",
+ "## Calculation \n",
+ "T_bmax = T_b0+((2*q_max*V_f)/(math.pi*w*c_p));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Exit temperature of the coolant = \",T_bmax,\" F \\n\");\n",
+ "\n",
+ "## 2.\n",
+ "## Using the data of Example 8.6\n",
+ "h = 7500.; ## Heat transfer coefficient in Btu/hr-ft^2-F\n",
+ "## Using the data of Example 8.5\n",
+ "d = 0.42/12.; ## Diameter of the fuel rod in feet(ft)\n",
+ "a = d/2.; ## Radius of the fuel rod in ft\n",
+ "b = 0.024/12.; ## Thickness of Zircaloy-4 clad in ft\n",
+ "H = 12.; ## Length of fuel rod in ft\n",
+ "A = 2*math.pi*(a+b)*H; ## Area of the assumed cylinder in ft^2\n",
+ "R_h = 1/(h*A); ## Convective resistance in F-hour/Btu\n",
+ "alpha = math.pi*w*c_p*R_h; ## A parameter\n",
+ "\n",
+ "## Using the result from Example 8.5 \n",
+ "R_f = 6.03*10**(-3); ## Thermal resistance of fuel \n",
+ "R_c = 1.43*10**(-4); ## Thermal resistance of cladding\n",
+ "R = R_f+R_c+R_h; ## Total resistance\n",
+ "bet = math.pi*w*c_p*R; ## A parameter denoted by 'beta'\n",
+ "## Calculation\n",
+ "T_cmax = T_b0+((q_max*V_f*R_h)*((1+math.sqrt(1+alpha**2))/alpha));\n",
+ "T_mmax = T_b0+((q_max*V_f*R)*((1+math.sqrt(1+bet**2))/bet));\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n Maximum temperature of cladding and fuel = \",math.ceil(T_cmax),\" F\" and \"\",T_mmax,\" F respectively. \\n\");\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Exit temperature of the coolant = 626.37 F \n",
+ "\n",
+ " \n",
+ " Maximum temperature of cladding and fuel = 649.00 3941.65 F respectively. \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg438"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.8\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "d = 0.42; ## Diameter of the fuel rod in inches\n",
+ "b = 0.024; ## Thickness of Zircaloy-4 clad in inches\n",
+ "v = 15.6*3600.; ## Speed of fluid in feet/hour\n",
+ "a = (d/2.)+b; ## Radius of fuel rods in inches\n",
+ "P = 2000.; ## Pressure of water in psi\n",
+ "T = 600.; ## Water temperature in F\n",
+ "## Using the data from example 8.5 \n",
+ "s = 0.6; ## Pitch of square array in inches\n",
+ "D_e = 2.*((s**2-(math.pi*a**2))/(math.pi*a)); ## Equivalent diameter in inches\n",
+ "## Converting the units in terms of feet \n",
+ "D_e = D_e/12.;\n",
+ "## Using tha Table IV.3 at given T and P value\n",
+ "rho = 42.9; ## Density of fluid in ft/hr\n",
+ "mu = 0.212; ## Viscosity of fluid in lb/hr-ft\n",
+ "## Calculation \n",
+ "Re = (D_e*v*rho)/mu;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Reylonds number = \",Re,\" \\n\");\n",
+ "if Re >= 10000:\n",
+ " print(\" \\n As the reylonds number is greater than 10000, the flow is turbulent. \\n\");\n",
+ "\n",
+ "## The value is different as compared to the textbook value. This is due to approximation of Reynolds number in the textbook.\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Reylonds number = 484329.34 \n",
+ "\n",
+ " \n",
+ " As the reylonds number is greater than 10000, the flow is turbulent. \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg440"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.9\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Using the data from Example 8.8\n",
+ "s = 0.6; ## Pitch of square lattice in inches\n",
+ "d = 0.42; ## Diameter of the fuel rod in inches\n",
+ "b = 0.024; ## Thickness of Zircaloy-4 clad in inches\n",
+ "a = (d/2)+b; ## Radius of fuel rods in inches \n",
+ "D_e = 0.0427; ## Equivalent diameter in feet\n",
+ "Re = 484329; ## Reynolds number\n",
+ "PD = s/(2*a); ## The ratio of pitch to diameter of fuel rod\n",
+ "## For a square lattice\n",
+ "C = 0.042*PD-0.024; ## A constant\n",
+ "\n",
+ "## According to Table IV.3\n",
+ "c_p = 1.45; ## Specific heat at constant pressure in Btu/lb-F\n",
+ "mu = 0.212; ## Viscosity of fluid in lb/hr-ft\n",
+ "k = 0.296; ## Conductivity of fluid in Btu/hr-ft F\n",
+ "Pr=(c_p*mu)/k; ## Prandtl number\n",
+ "## The constants are assumed as\n",
+ "m = 0.8; \n",
+ "n = 1/3;\n",
+ "## Calculation\n",
+ "h = C*(k/D_e)*(Re)**m*Pr**n;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Heat transfer coefficient = \",h,\" Btu/hr-ft^2-F\\n\");\n",
+ "## The value is different as compared to the textbook value. This is due to approximation of Reynolds number in the textbook and in this problem actual value is considered.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Heat transfer coefficient = 7309.11 Btu/hr-ft^2-F\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg448"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.10\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Using the data from Example 8.3 to 8.8\n",
+ "P = 2000.; ## Pressure in psi\n",
+ "v = 15.6; ## Coolant velocity in ft/sec\n",
+ "D_e = 0.0427; ## Equivalent diameter in ft\n",
+ "d = 0.42; ## Diameter of the fuel rod in inches\n",
+ "b = 0.024; ## Thickness of Zircaloy-4 clad in inches\n",
+ "a = (d/2.)+b; ## Radius of fuel rods in inches \n",
+ "T_b = 600.; ## Bulk temeperature in F \n",
+ "\n",
+ "## 1.\n",
+ "## Using Bernath correlation\n",
+ "## Calculation\n",
+ "T_wc = 102.6*math.log(P)-((97.2*P)/(P+15.))-(0.45*v)+32.;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Cladding temeperature = \",T_wc,\" F\\n\");\n",
+ "\n",
+ "## 2.\n",
+ "D_i = (2.*math.pi*a)/(math.pi*12.); ## Heated perimeter is (2*%pi*a)/12 in feet \n",
+ "## Calculation\n",
+ "h_c = 10890.*((D_e)/(D_e+D_i))+((48.*v)/D_e**0.6);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Heat transfer coefficient = \",h_c,\" Btu/hr-ft^2-F\\n\");\n",
+ "\n",
+ "## 3.\n",
+ "## Calculation\n",
+ "q_c = h_c*(T_wc-T_b);\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" \\n Critical heat flux = \",q_c,\" Btu/hr-ft^2\\n\");\n",
+ "## In the textbook, the unit of critical heat flux is wrong.\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Cladding temeperature = 708.36 F\n",
+ "\n",
+ " \n",
+ " Heat transfer coefficient = 10658.76 Btu/hr-ft^2-F\n",
+ "\n",
+ " \n",
+ " Critical heat flux = 1.15e+06 Btu/hr-ft^2\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg454"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.11\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "m_lbar = 0.457; ## Average linear density of UO2 in lb/ft\n",
+ "sigma = 0.0122; ## Standard deviation of set of measured linear densities of UO2\n",
+ "## Calculation\n",
+ "F_Eml = 1.+(3.*sigma)/m_lbar;\n",
+ "print'%s %.2f %s'%(\" \\n Engineering subfactor for the amount of fissile material = \",F_Eml,\" \\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Engineering subfactor for the amount of fissile material = 1.08 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg457"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 8.12\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "q_max = 539000.; ## Maximum heat flux Btu/hr-ft^2\n",
+ "F = 2.8; ## Hot channel factor\n",
+ "P = 3000.; ## Reactor thermal power in MW\n",
+ "## Expressing in Btu/hr\n",
+ "## According to Table 1.9, 1 kW = 3412 Btu/hr\n",
+ "P = P*3.412*10**6; ## Reactor thermal power in Btu/hr\n",
+ "l = 12.; ## Length of fuel rod in ft\n",
+ "d = 0.5/12.; ## Diameter of fuel rod in ft\n",
+ "r = d/2.; ## Radius of fuel rod in ft\n",
+ "\n",
+ "## 1.\n",
+ "q_av = q_max/F;\n",
+ "## Calculation\n",
+ "A = P/q_av;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Total heat transfer area = \",A,\" ft^2\\n\");\n",
+ "\n",
+ "## 2.\n",
+ "A_one = 2.*math.pi*r; ## The total surface area of one fuel rod\n",
+ "## Calculation \n",
+ "n = A/A_one;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Number of fuel rods = \",n,\" \\n\"); \n",
+ "## The value is different as compared to the textbook value. This is due to approximation of total heat transfer area in the textbook and in this problem actual value is considered. As total heat transfer area is further used to calculate number of fuel rods, therefore the difference in value exists.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Total heat transfer area = 53174.03 ft^2\n",
+ "\n",
+ " \n",
+ " Number of fuel rods = 406219.64 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter9_1.ipynb b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter9_1.ipynb new file mode 100644 index 00000000..e381a8ed --- /dev/null +++ b/Introduction_to_Nuclear_Engineering_by_J._R._Lamarsh_and_A._J._Baratta/Chapter9_1.ipynb @@ -0,0 +1,795 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:fc1d60112abdd7d9ca778c5c8b7e3886d44e19df01413a71c3258e0ecf2cea56"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter9-Radiation Protection"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg470"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.1\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "e = 1.6*10**(-19); ## Electronic charge in couloumb(coul)\n",
+ "X = 1*10**(-3)/3600.; ## Exposure rate in terms of R/sec\n",
+ "## According to the definition of Roentgen, 1 R = 2.58*10^(-7) coul/g \n",
+ "R = 2.58*10**(-7);\n",
+ "## From standard table\n",
+ "## There is 0.001293 g of air per 1 cm^3 at 1 atmospheric pressure at 0 C \n",
+ "density_air = 0.001293;\n",
+ "## Calculation\n",
+ "IR = (X*R*density_air)/e;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Rate of ions produced from gamma ray interaction = \",IR,\" ions/cm^3-sec\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Rate of ions produced from gamma ray interaction = 579.16 ions/cm^3-sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg475"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.2\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## According to the definition of radiation absorbed dose(rad), 1 rad/sec = 100 ergs/g-sec\n",
+ "## Given data\n",
+ "D = 5.*10**(-3)/100.; ## Absorbed dose in terms of rad/sec\n",
+ "## Expressing absorbed dose rate in SI units\n",
+ "## 1 Gray(Gy) = 100 rad \n",
+ "D_dot = D*3600./100.;\n",
+ "## Using data from Table 9.2\n",
+ "Q = 1.; ## Quality factor for gamma rays for tissue\n",
+ "## Calculation\n",
+ "H_dot = D_dot*Q;\n",
+ "print'%s %.2f %s'%(' \\n Absorbed dose rate in a tissue = ',D_dot*1000,' mGy/hr \\n');\n",
+ "print'%s %.2f %s'%(' \\n Dose equivalent rate in a tissue = ',H_dot*1000,' mSv/hr \\n');\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Absorbed dose rate in a tissue = 1.80 mGy/hr \n",
+ "\n",
+ " \n",
+ " Dose equivalent rate in a tissue = 1.80 mSv/hr \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.3\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "H = 25; ## Equivalent dose in rem\n",
+ "age = 30; ## Age of worker in years\n",
+ "exp_age = 77; ## Average age upto which a person lives in years\n",
+ "## Using data from Table 9.6\n",
+ "## Bone cancer\n",
+ "rc_bone = 0.2; ## Risk coefficient per 10^6 rem/year\n",
+ "lp_bone = 10; ## Latent period in years\n",
+ "## Probability of dying from bone cancer \n",
+ "p_bone=(H*rc_bone*(exp_age-(lp_bone+age)))/10**6;\n",
+ "\n",
+ "## Breast cancer\n",
+ "rc_breast = 1.5; ## Risk coefficient per 10^6 rem/year\n",
+ "lp_breast = 15.; ## Latent period in years\n",
+ "## Probability of dying from breast cancer \n",
+ "p_breast = (H*rc_breast*(exp_age-(lp_breast+age)))/10**6;\n",
+ "\n",
+ "## Leukemia\n",
+ "rc_leukemia = 1.; ## Risk coefficient per 10^6 rem/year\n",
+ "lp_leukemia = 2.; ## Latent period in years\n",
+ "## Probability of dying from leukemia \n",
+ "p_leukemia = (H*rc_leukemia*(exp_age-(lp_leukemia+age)))/10**6;\n",
+ "\n",
+ "## Lung cancer\n",
+ "rc_lung = 1.; ## Risk coefficient per 10^6 rem/year\n",
+ "lp_lung = 15.; ## Latent period in years\n",
+ "## Probability of dying from lung cancer\n",
+ "p_lung = (H*rc_lung*(exp_age-(lp_lung+age)))/10**6;\n",
+ "\n",
+ "## Pancreatic cancer\n",
+ "rc_pancreas = 0.2; ## Risk coefficient per 10^6 rem/year\n",
+ "lp_pancreas = 15.; ## Latent period in years\n",
+ "## Probability of dying from Pancreatic cancer\n",
+ "p_pancreas = (H*rc_pancreas*(exp_age-(lp_pancreas+age)))/10**6;\n",
+ "\n",
+ "## Stomach cancer\n",
+ "rc_stomach = 0.6; ## Risk coefficient per 10^6 rem/year\n",
+ "lp_stomach = 15.; ## Latent period in years\n",
+ "## Probability of dying from stomach cancer\n",
+ "p_stomach = (H*rc_stomach*(exp_age-(lp_stomach+age)))/10**6;\n",
+ "\n",
+ "## Rest of alimentary cancer\n",
+ "rc_ali = 0.2; ## Risk coefficient per 10^6 rem/year\n",
+ "lp_ali = 15.; ## Latent period in years\n",
+ "## Probability of dying from rest of alimentary cancer\n",
+ "p_ali = (H*rc_ali*(exp_age-(lp_ali+age)))/10**6;\n",
+ "\n",
+ "## Thyroid cancer\n",
+ "rc_thy = 0.43; ## Risk coefficient per 10^6 rem/year\n",
+ "lp_thy = 10.; ## Latent period in years\n",
+ "## Probability of dying from thyroid cancer\n",
+ "p_thy = (H*rc_thy*(exp_age-(lp_thy+age)))/10**6;\n",
+ "\n",
+ "## All other type of cancer\n",
+ "rc_other = 1.; ## Risk coefficient per 10^6 rem/year\n",
+ "lp_other = 15.; ## Latent period in years\n",
+ "## Probability of dying from all other type of cancer\n",
+ "p_other = (H*rc_other*(exp_age-(lp_other+age)))/10**6;\n",
+ "\n",
+ "## Calculation\n",
+ "p = p_bone+p_breast+p_leukemia+p_lung+p_pancreas+p_stomach+p_ali+p_thy+p_other;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%('\\n Probability that the worker will die from cancer = ',p,' \\n');\n",
+ "\n",
+ "## The value obtained is different from the value given in the textbook. This is because of approximation of individual probabilities in the textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Probability that the worker will die from cancer = 0.01 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg496"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.4\n",
+ "import math\n",
+ "\n",
+ "H = 1.; ## Equivalent dos in rem\n",
+ "n = 10**6; ## Population\n",
+ "## Given data\n",
+ "\n",
+ "## Using the data of number of expected deaths of leukemia per 10^6 people from Table 9.9\n",
+ "## In utero age group\n",
+ "frac_utero = 0.011; ## Fraction of population\n",
+ "riskyr_utero = 10.; ## Risk years\n",
+ "riskcoef_utero = 15.; ## Risk coefficient\n",
+ "## Number of deaths in utero is given by\n",
+ "deaths_utero = frac_utero*riskyr_utero*riskcoef_utero;\n",
+ "\n",
+ "## In 0-0.99 age group\n",
+ "frac_0_099 = 0.014; ## Fraction of population\n",
+ "riskyr_0_099 = 25.; ## Risk years\n",
+ "riskcoef_0_099 = 2.; ## Risk coefficient\n",
+ "## Number of deaths in 0-0.99 age group is given by\n",
+ "deaths_0_099 = frac_0_099*riskyr_0_099*riskcoef_0_099;\n",
+ "\n",
+ "## In 1-10 age group\n",
+ "frac_1_10 = 0.146; ## Fraction of population\n",
+ "riskyr_1_10 = 25.; ## Risk years\n",
+ "riskcoef_1_10 = 2.; ## Risk coefficient\n",
+ "## Number of deaths in 1-10 age group is given by\n",
+ "deaths_1_10=frac_1_10*riskyr_1_10*riskcoef_1_10;\n",
+ "\n",
+ "## In 11-20 age group\n",
+ "frac_11_20 = 0.196; ## Fraction of population\n",
+ "riskyr_11_20 = 25.; ## Risk years\n",
+ "riskcoef_11_20 = 1.; ## Risk coefficient\n",
+ "## Number of deaths in 11-20 age group is given by\n",
+ "deaths_11_20=frac_11_20*riskyr_11_20*riskcoef_11_20;\n",
+ "\n",
+ "## In 21-30 age group\n",
+ "frac_21_30 = 0.164; ## Fraction of population\n",
+ "riskyr_21_30 = 25.; ## Risk years\n",
+ "riskcoef_21_30 = 1.; ## Risk coefficient\n",
+ "## Number of deaths in 21-30 age group is given by\n",
+ "deaths_21_30=frac_21_30*riskyr_21_30*riskcoef_21_30;\n",
+ "\n",
+ "## In 31-40 age group\n",
+ "frac_31_40 = 0.118; ## Fraction of population\n",
+ "riskyr_31_40 = 25.; ## Risk years\n",
+ "riskcoef_31_40 = 1.; ## Risk coefficient\n",
+ "## Number of deaths in 31-40 age group is given by\n",
+ "deaths_31_40=frac_31_40*riskyr_31_40*riskcoef_31_40;\n",
+ "\n",
+ "## In 41-50 age group\n",
+ "frac_41_50 = 0.109; ## Fraction of population\n",
+ "riskyr_41_50 = 25.; ## Risk years\n",
+ "riskcoef_41_50 = 1.; ## Risk coefficient\n",
+ "## Number of deaths in 41-50 age group is given by\n",
+ "deaths_41_50 = frac_41_50*riskyr_41_50*riskcoef_41_50;\n",
+ "\n",
+ "## In 51-60 age group\n",
+ "frac_51_60 = 0.104; ## Fraction of population\n",
+ "riskyr_51_60 = 22.5; ## Risk years\n",
+ "riskcoef_51_60 = 1.; ## Risk coefficient\n",
+ "## Number of deaths in 51-50 age group is given by\n",
+ "deaths_51_60 = frac_51_60*riskyr_51_60*riskcoef_51_60;\n",
+ "\n",
+ "## In 61-70 age group\n",
+ "frac_61_70 = 0.08;\n",
+ "riskyr_61_70 = 15.1;\n",
+ "riskcoef_61_70 = 1.; ## Risk coefficient\n",
+ "## Number of deaths in 61-70 age group is given by\n",
+ "deaths_61_70=frac_61_70*riskyr_61_70*riskcoef_61_70;\n",
+ "\n",
+ "## In 71-80 age group\n",
+ "frac_71_80 = 0.044; ## Fraction of population\n",
+ "riskyr_71_80 = 9.1; ## Risk years\n",
+ "riskcoef_71_80 = 1.; ## Risk coefficient\n",
+ "## Number of deaths in 71-80 age group is given by\n",
+ "deaths_71_80 = frac_71_80*riskyr_71_80*riskcoef_71_80;\n",
+ "\n",
+ "## Age greater than 80\n",
+ "frac_80 = 0.02; ## Fraction of population\n",
+ "riskyr_80 = 4.5; ## Risk years\n",
+ "riskcoef_80 = 1.; ## Risk coefficient\n",
+ "## Number of deaths with age greater than 80 years is given by\n",
+ "deaths_80=frac_80*riskyr_80*riskcoef_80;\n",
+ "\n",
+ "## Calculation\n",
+ "total_deaths = deaths_utero+deaths_0_099+deaths_1_10+deaths_11_20+deaths_21_30+deaths_31_40+deaths_41_50+deaths_51_60+deaths_61_70+deaths_71_80+deaths_80;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Number of cases or deaths expected from leukemia = \",total_deaths,\" \\n\");\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Number of cases or deaths expected from leukemia = 28.36 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg498"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.5\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "H_year = 5.; ## Equiavelnt dose per year in rem\n",
+ "start_age = 18.; ## Initial age of the worker in years\n",
+ "ret_age = 68.; ## Retirement age of the worker in years\n",
+ "## Using data from Table 9.6 with respect to Bone cancer\n",
+ "latent_period = 10.; ## Latent period in years\n",
+ "plateau_period = 30.; ## Plateau period in years\n",
+ "rc_bone = 0.2; ## Risk coefficient per 10^6 rem/year\n",
+ "\n",
+ "n = ret_age-(start_age+latent_period); ## Number of years of accumulated dose\n",
+ "H = n*H_year; ## Total equivalent dose in rem\n",
+ "## Calculation\n",
+ "p_bone = (H*rc_bone*plateau_period)/10**6;\n",
+ "## Result\n",
+ "print'%s %.2e %s'%(\" \\n The probability of dying from bone cancer = \",p_bone,\" \\n\");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The probability of dying from bone cancer = 1.20e-03 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg513"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.6\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "E =2. ; ## Energy of gamma radiation in MeV\n",
+ "X_dot = 1.; ## Exposure rate in mR/hour\n",
+ "## Using the data from Table II.5\n",
+ "## Let mu_a/rho of air at 2 Mev be denoted as mu_rho\n",
+ "mu_rho = 0.0238; ## Ratio of absorption coefficient to sensity of air in cm^2/g\n",
+ "## Calculation\n",
+ "I = X_dot/(E*mu_rho*0.0659);\n",
+ "print'%s %.2f %s'%(\" \\n The beam intensity of gamma radiation required = \",math.ceil(I),\" gamma rays/cm^2-sec \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The beam intensity of gamma radiation required = 319.00 gamma rays/cm^2-sec \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg515"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.7\n",
+ "import math\n",
+ "\n",
+ "## Given data\n",
+ "phi = 2.4*10**5; ## Flux in x-rays/cm^2-sec\n",
+ "## From Figure 9.9\n",
+ "## To receive an exposure rate of 1 mR/hr at 50 keV, the flux is 8*10^3 x-rays/cm^2-sec\n",
+ "phi_eq = 8*10**3; ## Equivalent flux in x-rays/cm^2-sec\n",
+ "X_dot_eq = 1.; ## Equivalent Exposure rate in mR/hr\n",
+ "X_dot = (phi*X_dot_eq)/phi_eq; ## Exposure rate of the operator in mR/hr\n",
+ "## From Figure 9.10 at 50 kV energy, the energy dependent function is\n",
+ "f_bone = 3.3;\n",
+ "f_muscle = 0.93;\n",
+ "f_fat = 0.9;\n",
+ "## Using data from Table 9.2\n",
+ "Q = 1.; ## Quality factor for x-rays\n",
+ "## Calculation\n",
+ "D_dot_bone = X_dot*f_bone*Q; ## Dose equivalent rate in bone\n",
+ "D_dot_muscle = X_dot*f_muscle*Q; ## Dose equivalent rate in muscle\n",
+ "D_dot_fat = X_dot*f_fat*Q; ## Dose equivalent rate in fat\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Dose equivalent rate in bone = \",math.ceil(D_dot_bone),\" mrem/hour \\n\");\n",
+ "print'%s %.2f %s'%(\" \\n Dose equivalent rate in muscle = \",math.ceil(D_dot_muscle),\" mrem/hour \\n\");\n",
+ "print'%s %.2f %s'%(\" \\n Dose equivalent rate in fat = \",math.ceil(D_dot_fat),\" mrem/hour \\n\");\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Dose equivalent rate in bone = 99.00 mrem/hour \n",
+ "\n",
+ " \n",
+ " Dose equivalent rate in muscle = 28.00 mrem/hour \n",
+ "\n",
+ " \n",
+ " Dose equivalent rate in fat = 27.00 mrem/hour \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg518"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.8\n",
+ "import math\n",
+ "#calculate the\n",
+ "\n",
+ "## Given data\n",
+ "phi_n = 20.; ## Given neutron flux in neutrons/cm^2-sec\n",
+ "## From Figure 9.12\n",
+ "## To receive an dose equivalent rate of 1 mrem/hr, the fast neutron flux is 7 neutrons/cm^2-sec\n",
+ "phi_n_eq = 7.;\n",
+ "D_dot_eq = 1.; \n",
+ "D_dot_n = (phi_n*D_dot_eq)/phi_n_eq; ## Dose rate due to fast neutron flux in mrem/hr\n",
+ "phi_th = 300.; ## Given thermal flux in neutrons/cm^2-sec\n",
+ "## From Figure 9.12\n",
+ "## To receive an dose equivalent rate of 1 mrem/hr, the thermal flux is 260 neutrons/cm^2-sec\n",
+ "phi_th_eq = 260.;\n",
+ "D_dot_th = (phi_th*D_dot_eq)/phi_th_eq; ## Dose rate due to thermal neutron flux in mrem/hr\n",
+ "D_dot = D_dot_n+D_dot_th; ## Total dose rate in mrem/hr\n",
+ "print(\"\\n The permitted weekly dose is 100 mrem \\n\");\n",
+ "D_dot_perm = 100.;\n",
+ "## Calculation\n",
+ "t = D_dot_perm/D_dot;\n",
+ "print'%s %.2f %s'%(\" \\n The time of exposure upto a safe level = \",t,\" hour \\n\");\n",
+ "## The answer given in the textbook is wrong. This is because of wrong computation of total dose rate\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The permitted weekly dose is 100 mrem \n",
+ "\n",
+ " \n",
+ " The time of exposure upto a safe level = 24.93 hour \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg520"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.9\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "fluence = 10**8; ## Given fluence neutrons/cm^2\n",
+ "## From Figure 9.12\n",
+ "## To receive an dose equivalent rate of 1 mrem/hr, the fast neutron flux is 7 neutrons/cm^2-sec\n",
+ "phi_eq = 7.; ## Equivalent flux in neutrons/cm^2-sec\n",
+ "D_eq = 1.; ## Equivalent dose rate in mrem/hr\n",
+ "## 1 hour = 3600 seconds\n",
+ "fluence_eq = phi_eq*3600.; ## Equivalent fluence in neutrons/cm^2\n",
+ "## Calculation \n",
+ "D = (fluence*D_eq)/fluence_eq;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n Dose received due to exposure of accelerator source = \",D,\" mrem \\n\");\n",
+ "## The answer given in textbook is approximated to a nearest value.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " Dose received due to exposure of accelerator source = 3968.25 mrem \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg525"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.10\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "M = 20.; ## Mass of organ in grams\n",
+ "\n",
+ "## a)\n",
+ "## Using the data from Table 9.15\n",
+ "T_12 = 8.04; ## Radiological half life of Iodine-131 in days\n",
+ "T_12_b = 138.; ## Biological half life of Iodine-131 in days\n",
+ "lambd = 0.693/T_12; ## Radiological decay constant of Iodine-131 in days^-1\n",
+ "lambda_b = 0.693/T_12_b; ## Biological decay constant of Iodine-131 in days^-1\n",
+ "lambda_e = lambd+lambda_b; ## Equivalent decay constant in days^-1\n",
+ "## Using the data from Table 9.15\n",
+ "zeta = 0.23; ## Effective energy equivalent in MeV\n",
+ "q = 0.23; ## The fraction of Iodine-131 that goes by inhalation\n",
+ "## Calculation\n",
+ "DCF = (51.1*zeta*q)/(M*lambda_e);\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\" \\n The dose commitment factor by inhalation = \",DCF,\" rem/ucurie \\n\");\n",
+ "\n",
+ "## b) \n",
+ "breathing_rate = 2.32*10**(-4); ## Normal breathing rate in m^3/sec\n",
+ "time = 2*3600.; ## Time of radiation exposure in seconds\n",
+ "I_conc = 2*10**(-9); ## Iodine-131 concentration\n",
+ "C0 = breathing_rate*time*I_conc; ## Total intake of Iodine-131 by inhalation \n",
+ "## Calculation\n",
+ "H = C0*(DCF*10**6); ## Using DCF in micro-curie\n",
+ "## Result\n",
+ "print'%s %.2f %s %.2f %s '%(\" \\n The dose commitment to thyroid = \",H,\"\" and \" rem = \",H*1000,\" mrem \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " \n",
+ " The dose commitment factor by inhalation = 1.48 rem/ucurie \n",
+ "\n",
+ " \n",
+ " The dose commitment to thyroid = 0.00 4.95 mrem \n",
+ " \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg529"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.11\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "V_W = 2200.; ## Volume of water inatke in terms of cm^3/day\n",
+ "## 1 litre = 1000 gram(g)\n",
+ "M = 43.*1000.; ## Mass of water present in standard man according to standards\n",
+ "## Using the data from Table 9.13\n",
+ "MPD = 0.1/7.; ## Maximum Permissible Dose (MPD) in rem/day\n",
+ "## Using the data from Table 9.15\n",
+ "zeta = 0.01; ## Effective energy equivalent in MeV\n",
+ "q = 1.; ## The fraction of Tritium that goes inside by ingestion\n",
+ "T_b = 11.9; ## Biological Half life of Tritium in years\n",
+ "lambda_b = 0.693/T_b; ## Biological decay constant of Tritium in years^-1 \n",
+ "\n",
+ "## As biological and radiological half lives are less than 50 year intake period, the exponential term (exp(-lambda_e*50)) is neglected\n",
+ "## Maximum Permissible Concentration(MPC) for a 7 day or 168 hour week tritium dose \n",
+ "MPC_w_168 = (lambda_b*M*MPD)/(51.1*V_W*zeta*q);\n",
+ "print'%s %.2f %s'%(\"\\n Maximum Permissible Concentration(MPC) for a 7 day or 168 hour week tritium dose for occupational purpose = \",MPC_w_168,\" uCi/cm^3 \\n\");\n",
+ "## The exposure at work is doubled for 40 hour week as compared to 168 hour week \n",
+ "## For 40 hour week, with work of 5 days out of 7 day week according to a study\n",
+ "MPC_w_40 = MPC_w_168*2.*(7/5.);\n",
+ "print'%s %.2f %s'%(\"\\n Maximum Permissible Concentration(MPC) for a 40 hour week tritium dose for occupational purpose = \",MPC_w_40,\" uCi/cm^3 \\n\");\n",
+ "\n",
+ "## By analyzing the data of Table 9.13\n",
+ "## The whole body dose of general public is one tenth of the occupational purpose.\n",
+ "MPC_w_168_gp = MPC_w_168*0.1;\n",
+ "print'%s %.2f %s'%(\"\\n Maximum Permissible Concentration(MPC) for a 7 day or 168 hour week tritium dose for general public = \",MPC_w_168_gp,\" uCi/cm^3 \\n\");\n",
+ "## The answer of Maximum Permissible Concentration(MPC) for a 168 hour week tritium dose for general public is given wrong in the textbook.\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Maximum Permissible Concentration(MPC) for a 7 day or 168 hour week tritium dose for occupational purpose = 0.03 uCi/cm^3 \n",
+ "\n",
+ "\n",
+ " Maximum Permissible Concentration(MPC) for a 40 hour week tritium dose for occupational purpose = 0.09 uCi/cm^3 \n",
+ "\n",
+ "\n",
+ " Maximum Permissible Concentration(MPC) for a 7 day or 168 hour week tritium dose for general public = 0.00 uCi/cm^3 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg534"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.12\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "no_home = 10**6; ## Number of houses\n",
+ "no_resident = 4.; ## Number of residents in a home\n",
+ "total_time = 50.; ## Number of years the analysis is carried out\n",
+ "radon_concn_old = 1.; ## Radon concentration in older uninsulated homes in terms of pCi/l\n",
+ "radon_concn_new = 6.; ## Radon concentration in modern insulated homes in terms of pCi/l\n",
+ "time = 3500.; ## Time spent in home by a person per year in hours\n",
+ "eq_concn = 0.5; ## Equilibrium concentration of 50% \n",
+ "## 1 year = 24*365 hours\n",
+ "X_increased = eq_concn*(radon_concn_new-radon_concn_old)*(time/(24.*365.)); ## The increased exposure of radon per person\n",
+ "\n",
+ "## Using the data of Radon risk assessment of United States of America\n",
+ "## There are nearly 100 cases of cancer per 10^6 persons at 1 pCi-year dose.\n",
+ "## Calculation\n",
+ "no_cancer = (no_home*no_resident)*total_time*(100./10**6)*X_increased;\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Number of additional cases of cancer from insulation of home = \",no_cancer,\" \\n\");\n",
+ "## There is a slight deviation in the value given in the textbook. This is because of approximation of the number of additional cases of cancer in the textbook.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Number of additional cases of cancer from insulation of home = 19977.17 \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg535"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Example 9.13\n",
+ "import math\n",
+ "\n",
+ "\n",
+ "## Given data\n",
+ "H_ext = 3.; ## External dose in rem\n",
+ "H_wbL = 5.; ## Annual whole body dose limit in rem\n",
+ "## Using the data from Table 9.17\n",
+ "## Annual Limit Intake (ALI) for inhalation of Iodine-131 is 54uCurie (Ci)\n",
+ "ALI = 54.;\n",
+ "## Calculation\n",
+ "I = ALI*(1.-(H_ext/H_wbL));\n",
+ "## Result\n",
+ "print'%s %.2f %s'%(\"\\n Amount of Iodine-131 intake within safety limits = \",math.ceil(I),\" uCi \\n\");\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Amount of Iodine-131 intake within safety limits = 22.00 uCi \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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