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29 files changed, 4561 insertions, 4366 deletions
diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter1.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter1.ipynb deleted file mode 100755 index ed34729f..00000000 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter1.ipynb +++ /dev/null @@ -1,165 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:9283b3273e4b25859b1f947b0b70d061842802ddcf82b087896def4a57fa123c" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h1>Chapter 1: Semiconductor Basics<h1>" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 1.1(a), Page Number:29<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_bias=10.0; #bias voltage in volt\n", - "R_limit=1000; #limiting resistance in ohm\n", - "r_d =10.0; #r_d value\n", - "\n", - "#calculation\n", - "#IDEAL MODEL\n", - "print \"IDEAL MODEL\"\n", - "V_f=0; #voltage in volt\n", - "I_f=V_bias/R_limit; #foward current\n", - "V_R_limit=I_f*R_limit; #limiting voltage\n", - "print \"forward voltage = %.2f volts\" %V_f\n", - "print \"forward current = %.2f amperes\" %I_f\n", - "print \"voltage across limiting resistor = %.2f volts\" %V_R_limit\n", - "\n", - "#PRACTICAL MODEL\n", - "print \"\\nPRACTICAL MODEL\"\n", - "V_f=0.7; #voltage in volt\n", - "I_f=(V_bias-V_f)/R_limit; #foward current\n", - "V_R_limit=I_f*R_limit; #limiting voltage\n", - "print \"forward voltage = %.2f volts\" %V_f\n", - "print \"forward current = %.3f amperes\" %I_f\n", - "print \"voltage across limiting resistor = %.2f volts\" %V_R_limit\n", - "\n", - "#COMPLETE MODEL\n", - "print \"\\nCOMPLETE MODEL\"\n", - "I_f=(V_bias-0.7)/(R_limit+r_d); #foward current\n", - "V_f=0.7+I_f*r_d; #forward voltage\n", - "V_R_limit=I_f*R_limit; #limiting voltage\n", - "print \"forward voltage = %.3f volts\" %V_f\n", - "print \"forward current = %.3f amperes\" %I_f\n", - "print \"voltage across limiting resistor = %.2f volts\" %V_R_limit" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "IDEAL MODEL\n", - "forward voltage = 0.00 volts\n", - "forward current = 0.01 amperes\n", - "voltage across limiting resistor = 10.00 volts\n", - "\n", - "PRACTICAL MODEL\n", - "forward voltage = 0.70 volts\n", - "forward current = 0.009 amperes\n", - "voltage across limiting resistor = 9.30 volts\n", - "\n", - "COMPLETE MODEL\n", - "forward voltage = 0.792 volts\n", - "forward current = 0.009 amperes\n", - "voltage across limiting resistor = 9.21 volts" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 1.1(b), Page Number:29<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "V_bias=5; #bias voltage in volt\n", - "I_R=1*10**-6; #current\n", - "R_limit=1000 #in Ohm\n", - "\n", - "#calculation\n", - "#IDEAL MODEL\n", - "print \"IDEAL MODEL\"\n", - "I_r=0.0; #current in ampere\n", - "V_R=V_bias; #voltages are equal\n", - "V_R_limit=I_r*R_limit; #limiting voltage\n", - "print \"Reverse voltage across diode = %.2f volts\" %V_R\n", - "print \"Reverse current through diode= %.2f amperes\" %I_r\n", - "print \"voltage across limiting resistor = %.2f volts\" %V_R_limit\n", - "\n", - "#PRACTICAL MODEL\n", - "print \"\\nPRACTICAL MODEL\"\n", - "I_r=0.0; #current in ampere\n", - "V_R=V_bias; #voltages are equal\n", - "V_R_limit=I_r*R_limit; #limiting voltage\n", - "print \"Reverse voltage across diode= %.2f volts\" %V_R\n", - "print \"Reverse current through diode = %.2f amperes\" %I_r\n", - "print \"voltage across limiting resistor = %.2f volts\" %V_R_limit\n", - "\n", - "#COMPLETE MODEL\n", - "print \"\\nCOMPLETE MODEL\"\n", - "I_r=I_R; #current in ampere\n", - "V_R_limit=I_r*R_limit; #limiting voltage\n", - "V_R=V_bias-V_R_limit; #voltage in volt\n", - "print \"Reverse voltage across diode = %.3f volts\" %V_R\n", - "print \"Reverse current through diode = %d micro Amp\" %(I_r*10**6)\n", - "print \"voltage across limiting resistor = %d mV\" %(V_R_limit*1000)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "IDEAL MODEL\n", - "Reverse voltage across diode = 5.00 volts\n", - "Reverse current through diode= 0.00 amperes\n", - "voltage across limiting resistor = 0.00 volts\n", - "\n", - "PRACTICAL MODEL\n", - "Reverse voltage across diode= 5.00 volts\n", - "Reverse current through diode = 0.00 amperes\n", - "voltage across limiting resistor = 0.00 volts\n", - "\n", - "COMPLETE MODEL\n", - "Reverse voltage across diode = 4.999 volts\n", - "Reverse current through diode = 1 micro Amp\n", - "voltage across limiting resistor = 1 mV" - ] - } - ], - "prompt_number": 2 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter2.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter2.ipynb deleted file mode 100755 index 29386b12..00000000 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter2.ipynb +++ /dev/null @@ -1,640 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:07300ba6f8a24ba9874afd0fd84b4c250cf3010a844444236cfec215a9ae4d89" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h1>Chapter 2: Diode Application<h1>" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.1, Page Number: 46<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n", - "For more information, type 'help(pylab)'." - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import math\n", - "\n", - "# variable declaration\n", - "V_p=50; #Peak value is 50V\n", - "\n", - "#calculation\n", - "V_avg=V_p/math.pi;\n", - "\n", - "#result\n", - "print \"average value of half wave rectifier = %.2f volts\" %V_avg" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "average value of half wave rectifier = 15.92 volts" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.2(a), Page Number: 46<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import math\n", - "\n", - "# variable declaration\n", - "#let V_in=5*sin(2*%pi*f.*t) be input wave ,hence frequency=1Hz\n", - "f=1; #frequency\n", - "V_p_in=5; #peak input\n", - "\n", - "#calculation\n", - "V_pout=V_p_in-0.7; #output voltage\n", - "t_d=(math.asin(0.7/V_p_in))/(2*math.pi*f);\n", - "\n", - "#result\n", - "print \"half wave rectifier output = %.2f volts\" %V_pout;" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "half wave rectifier output = 4.30 volts" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.2(b), Page Number: 46<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math\n", - "\n", - "# variable declaration\n", - "#let V_in=100*sin(2*%pi*f.*t) be input wave ,hence frequency=1Hz\n", - "f=1; #frequency\n", - "T=1/f; #time period\n", - "V_p_in=100; #peak input voltage\n", - "\n", - "#calculation\n", - "V_pout=(V_p_in-0.7); #peak output \n", - "t_d=(math.asin(0.7/V_p_in))/(2*math.pi*f) \n", - "\n", - "#result\n", - "print \"output of half wave rectifier = %.2f volts\" %V_pout" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "output of half wave rectifier = 99.30 volts" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.3, Page Number: 48<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math\n", - "\n", - "# variable declaration\n", - "V_p_in=156; #Peak input voltage\n", - "V_p_pri=156; #Peak voltage of primary of transformer\n", - "n=0.5; #Turn ratio is 2:1\n", - "\n", - "#calculation\n", - "V_p_sec=n*V_p_pri;\n", - "V_p_out=(V_p_sec-0.7); #Peak output voltage\n", - "\n", - "#result\n", - "print \"peak output voltage of half wave rectifier = %.1f volts\" %V_p_out" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "peak output voltage of half wave rectifier = 77.3 volts" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.4, Page Number: 49<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import math\n", - "\n", - "# variable declaration\n", - "V_p=15; #Peak voltage in volt\n", - "\n", - "#calculation\n", - "V_avg=(2*V_p)/math.pi;\n", - "\n", - "#result\n", - "print \"Average value of output of full wave rectifier = %.2f volts\" %V_avg" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Average value of output of full wave rectifier = 9.55 volts" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.5, Page Number: 52<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "V_p_pri=100.0; #Peak voltage across primary winding\n", - "n=1.0/2; #tun ratio is 2:1\n", - "V_p_sec=n*V_p_pri;\n", - "V_sec=V_p_sec/2; #voltage across each secondary is half the total voltage\n", - "V_pout=V_sec-0.7;\n", - "\n", - "print('full wave rectifier output voltage = %f V'%V_pout)\n", - "PIV=2*V_pout+0.7;\n", - "print('PIV = %fV'%PIV)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "full wave rectifier output voltage = 24.300000 V\n", - "PIV = 49.300000V" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.6, Page Number: 54<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math\n", - "\n", - "# variable declaration\n", - "V_rms=12.0; #rms secondary voltage\n", - "\n", - "#calculation\n", - "V_p_sec=math.sqrt(2)*V_rms; #peak secondary voltage\n", - "V_th=0.7; #knee voltage of diode\n", - "V_p_out=V_p_sec-2*V_th; #in one cycle, 2 diodes conduct\n", - "PIV=V_p_out+V_th; #applying KVL\n", - "\n", - "#result\n", - "print \"Peak output voltage = %.2f volt\" %V_p_out\n", - "print \"PIV across each diode = %.2f volt\" %PIV" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Peak output voltage = 15.57 volt\n", - "PIV across each diode = 16.27 volt" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.7, Page Number: 58<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math\n", - "\n", - "# variable declaration\n", - "R_l=2200; #load resistance in Ohm\n", - "C=50*10**-6; #capacitance in Farad\n", - "V_rms=115; #rms of primary\n", - "\n", - "#calculation\n", - "V_p_pri=math.sqrt(2)*V_rms; #peak voltage across primary\n", - "n=0.1; #turn ratio is 10:1\n", - "V_p_sec=n*V_p_pri; #primary voltage across secondary\n", - "V_p_rect=V_p_sec-1.4 #unfiltered peak rectified voltage\n", - "#we subtract 1.4 because in each cycle 2 diodes conduct & 2 do not\n", - "f=120; #frequency of full wave rectified voltage\n", - "V_r_pp=(1/(f*R_l*C))*V_p_rect; #peak to peak ripple voltage\n", - "V_DC=(1-(1/(2*f*R_l*C)))*V_p_rect;\n", - "r=V_r_pp/V_DC;\n", - "\n", - "#result\n", - "print \"Ripple factor = %.3f \" %r" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Ripple factor = 0.079 " - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.8, Page Number: 62<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import math\n", - "\n", - "# variable declaration\n", - "V_REF=1.25; #in volts\n", - "V_R1=V_REF; #voltage in volt\n", - "R1=220.0; #in ohms\n", - "I_ADJ=50*10**-6 #in amperes\n", - "\n", - "#calculation\n", - "# MAX VALUE OF R2=5000 Ohms\n", - "R2_min=0.0; #min resistance\n", - "V_out_min=V_REF*(1+(R2_min/R1))+I_ADJ*R2_min;\n", - "R2_max=5000.0; #max value of resistance\n", - "V_out_max=V_REF*(1+(R2_max/R1))+I_ADJ*R2_max;\n", - "\n", - "#result\n", - "print \"minimum output voltage = %.2f volt\" %V_out_min\n", - "print \"maximum output voltage = %.2f volt\" %V_out_max" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "minimum output voltage = 1.25 volt\n", - "maximum output voltage = 29.91 volt" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.9,Page Number: 64<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "V_NL=5.18 #No load output voltage\n", - "V_FL=5.15 #Full load output voltage\n", - "load_reg=((V_NL-V_FL)/V_FL)*100 #In percentage\n", - "print('load regulation percent = %.2f%% '%load_reg)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "load regulation percent = 0.58% " - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.10, Page Number: 66<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import pylab as py\n", - "import numpy as np\n", - "\n", - "#let input wave be V_in=V_p_in*sin(2*%pi*f*t) \n", - "f=1.0; #Frequency is 1Hz\n", - "T=1/f;\n", - "R_1=100.0; #Resistances in ohms\n", - "R_L=1000.0; #Load\n", - "V_p_in=10.0; #Peak input voltage\n", - "V_th=0.7; #knee voltage of diode\n", - "\n", - "V_p_out=V_p_in*(R_L/(R_L+R_1)); #peak output voltage\n", - "print('peak output voltage = %.2f V'%V_p_out)\n", - "\n", - "t = np.arange(0, 3.5 , 0.0005)\n", - "z=V_p_in*np.sin(2*np.pi*f*t)*(R_L/(R_L+R_1))\n", - "\n", - "subplot(211)\n", - "plot(t,z)\n", - "ylim(-9.09,9.09)\n", - "title('Input Voltage Waveform')\n", - "\n", - "subplot(212)\n", - "plot(t,z)\n", - "ylim(-0.07,9.09)\n", - "title('Output Voltage Waveform')" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "peak output voltage = 9.09 V" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 12, - "text": [ - "<matplotlib.text.Text at 0xa3bf44c>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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SU2k+SSX7GyWDi7oJBATQlukrV+S2pHysTVQA9aQAMjKAI0eAp56S2xLxsLNT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\ncKwILuocDodjRfwfUaYhEUjyJ+wAAAAASUVORK5CYII=\n" - } - ], - "prompt_number": 12 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.11, Page Number: 67<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#from pylab import figure, show\n", - "#from numpy import arange, sin, pi,bool\n", - "#import numpy as np\n", - "import pylab as py\n", - "import numpy as np\n", - "#let input wave be V_in=V_p_in*sin(2*%pi*f*t) \n", - "f=1.0; #Frequency is 1Hz\n", - "T=1/f;\n", - "V_p_in=10; #Peak input voltage\n", - "V_th=0.7; #knee voltage of diode\n", - "print('max output voltage is 5.7V')\n", - "print('min output voltage is -5.7V')\n", - "\n", - "###############GRAPH Plotting#################################\n", - "t = arange(0.0,4.5,0.0005)\n", - "V_in=V_p_in*sin(2*pi*f*t);\n", - "\n", - "Vout=V_in;\n", - "#fig = figure(2)\n", - "subplot(211)\n", - "plot(t,V_in)\n", - "#ax2.grid(True)\n", - "ylim( (-10,10) )\n", - "title('Input to the +ve and -ve diode limiter ')\n", - "subplot(212)\n", - "plot(t,V_in)\n", - "#ax1.grid(True)\n", - "ylim( (-5.7,5.7) )\n", - "title('Output of +ve and -ve diode limiter')\n", - " " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "max output voltage is 5.7V\n", - "min output voltage is -5.7V" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 13, - "text": [ - "<matplotlib.text.Text at 0xa6c976c>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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PPyagc2cgIQGIiYlBTEyMEKJw5r77aGnaujrqzFiBtZsicMeJhYZKLckdKipo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\nx/PeAAAAAElFTkSuQmCC\n" - } - ], - "prompt_number": 13 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.12, Page Number: 76<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#variable declaration\n", - "V_p_in=18.0; #peak input voltage is 18V\n", - "V_supply=12.0;\n", - "R2=100.0;\n", - "R3=220.0; #resistances in ohms\n", - "#calculation\n", - "V_bias=V_supply*(R3/(R2+R3));\n", - "\n", - "#result\n", - "print('diode limiting the voltage at this voltage =%fV'%V_bias)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "diode limiting the voltage at this voltage =8.250000V" - ] - } - ], - "prompt_number": 14 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 2.13, Page Number: 78<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "V_p_in=24.0;\n", - "V_DC=-(V_p_in-0.7); #DC level added to output\n", - "print('V_DC = %.1fV'%V_DC)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "V_DC = -23.3V" - ] - } - ], - "prompt_number": 15 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter3.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter3.ipynb deleted file mode 100755 index c21cf09b..00000000 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter3.ipynb +++ /dev/null @@ -1,396 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:7d54e3690fc412ff890e6ea2f39f46cb8c03d3ea660ea034447f2497647b95ed" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h1>Chapter 3: Special-purpose Diodes<h1>" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 3.1, Page Number:88<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n", - "For more information, type 'help(pylab)'." - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "delVZ=50*10**-3; #voltage in volts, from graph\n", - "delIZ=5*10**-3; #current in amperes, from rgraph\n", - "\n", - "#calculation\n", - "ZZ=delVZ/delIZ; #zener impedence\n", - "\n", - "# result\n", - "print \"zener impedance = %d ohm \" %ZZ" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "zener impedance = 10 ohm " - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 3.2, Page Number:89<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "I_ZT=37*10**-3; #IN AMPERES\n", - "V_ZT=6.80; #IN VOLTS\n", - "Z_ZT=3.50; #IN OHMS\n", - "I_Z=50*10**-3; #IN AMPERES\n", - "\n", - "#calculation\n", - "DEL_I_Z=I_Z-I_ZT; #change current\n", - "DEL_V_Z=DEL_I_Z*Z_ZT; #change voltage\n", - "V_Z=V_ZT+DEL_V_Z; #voltage across zener terminals\n", - "print \"voltage across zener terminals when current is 50 mA = %.3f volts\" %V_Z\n", - "I_Z=25*10**-3; #IN AMPERES\n", - "DEL_I_Z=I_Z-I_ZT; #change current\n", - "DEL_V_Z=DEL_I_Z*Z_ZT; #change voltage\n", - "V_Z=V_ZT+DEL_V_Z; #voltage across zener terminals\n", - "\n", - "#result\n", - "print \"voltage across zener terminals when current is 25 mA = %.3f volts\" %V_Z" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "voltage across zener terminals when current is 50 mA = 6.845 volts\n", - "voltage across zener terminals when current is 25 mA = 6.758 volts" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 3.3, Page Number:90<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_Z=8.2; #8.2 volt zener diode\n", - "TC=0.0005; #Temperature coefficient (per degree celsius)\n", - "T1=60; #Temperature 1 in celsius\n", - "T2=25; #Temperature 2 in celsius\n", - "\n", - "#calculation\n", - "DEL_T=T1-T2; #change in temp\n", - "del_V_Z=V_Z*TC*DEL_T; #change in voltage\n", - "voltage=V_Z+del_V_Z; #zener voltage\n", - "\n", - "#result\n", - "print \"zener voltage at 60 degree celsius = %.3f volt\" %voltage" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "zener voltage at 60 degree celsius = 8.343 volt" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 3.4, Page Number:90<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "P_D_max=400*10**-3; #power in watts\n", - "df=3.2*10**-3 #derating factor in watts per celsius\n", - "del_T=(90-50); #in celsius, temperature difference\n", - "\n", - "#calculation\n", - "P_D_deru=P_D_max-df*del_T; #power dissipated\n", - "P_D_der=P_D_deru*1000;\n", - "\n", - "#result\n", - "print \"maximum power dissipated at 90 degree celsius = %d mW\" %P_D_der" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum power dissipated at 90 degree celsius = 272 mW" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 3.5, Page Number: 92<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "V_Z=5.1;\n", - "I_ZT=49*10**-3;\n", - "I_ZK=1*10**-3;\n", - "Z_Z=7;\n", - "R=100;\n", - "P_D_max=1;\n", - "\n", - "#calculation\n", - "V_out=V_Z-(I_ZT-I_ZK)*Z_Z; #output voltage at I_ZK\n", - "V_IN_min=I_ZK*R+V_out; #input voltage\n", - "I_ZM=P_D_max/V_Z; #current\n", - "V_out=V_Z+(I_ZM-I_ZT)*Z_Z; #output voltage at I_ZM\n", - "V_IN_max=I_ZM*R+V_out; #max input voltage\n", - "\n", - "#result\n", - "print \"maximum input voltage regulated by zener diode = %.3f volts\" %V_IN_max\n", - "print \"minimum input voltage regulated by zener diode = %.3f volts\" %V_IN_min" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum input voltage regulated by zener diode = 25.737 volts\n", - "minimum input voltage regulated by zener diode = 4.864 volts" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 3.6, Page Number: 93<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_Z=12.0; #voltage in volt\n", - "V_IN=24.0; #ip voltage in volt\n", - "I_ZK=0.001; #current in ampere\n", - "I_ZM=0.050; #current in ampere \n", - "Z_Z=0; #impedence\n", - "R=470; #resistance in ohm\n", - "\n", - "#calculation\n", - "#when I_L=0, I_Z is max and is equal to the total circuit current I_T\n", - "I_T=(V_IN-V_Z)/R; #current\n", - "I_Z_max=I_T; #max current\n", - "if I_Z_max<I_ZM : # condition for min currert \n", - " I_L_min=0;\n", - "\n", - "I_L_max=I_T-I_ZK; #max current\n", - "R_L_min=V_Z/I_L_max; #min resistance\n", - "\n", - "#result\n", - "print \"minimum value of load resistance = %.2f ohm\" %R_L_min\n", - "print \"minimum curent = %.3f ampere\" %I_L_min\n", - "print \"maximum curent = %.3f ampere\" %I_L_max" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "minimum value of load resistance = 489.16 ohm\n", - "minimum curent = 0.000 ampere\n", - "maximum curent = 0.025 ampere" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 3.7, Page Number: 94<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_IN=24.0; #voltage in volt\n", - "V_Z=15.0; #voltage in volt\n", - "I_ZK=0.25*10**-3; #current in ampere\n", - "I_ZT=17*10**-3; #current in ampere\n", - "Z_ZT=14.0; #impedence\n", - "P_D_max=1.0; #max power dissipation\n", - "\n", - "#calculation\n", - "V_out_1=V_Z-(I_ZT-I_ZK)*Z_ZT; #output voltage at I_ZK\n", - "print \"output voltage at I_ZK = %.2f volt\" %V_out_1\n", - "I_ZM=P_D_max/V_Z;\n", - "\n", - "V_out_2=V_Z+(I_ZM-I_ZT)*Z_ZT; #output voltage at I_ZM\n", - "print \"output voltage a I_ZM = %.2f volt\" %V_out_2\n", - "R=(V_IN-V_out_2)/I_ZM; #resistance\n", - "print \"value of R for maximum zener current, no load = %.2f ohm\" %R\n", - "print \"closest practical value is 130 ohms\"\n", - "R=130.0;\n", - "#for minimum load resistance(max load current) zener current is minimum (I_ZK)\n", - "I_T=(V_IN-V_out_1)/R; #current\n", - "I_L=I_T-I_ZK; #current\n", - "R_L_min=V_out_1/I_L; #minimum load resistance\n", - "\n", - "#result\n", - "print \"minimum load resistance = %.2f ohm\" %R_L_min" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "output voltage at I_ZK = 14.77 volt\n", - "output voltage a I_ZM = 15.70 volt\n", - "value of R for maximum zener current, no load = 124.57 ohm\n", - "closest practical value is 130 ohms\n", - "minimum load resistance = 208.60 ohm" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 3.8, Page Number: 96<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#variable declaration\n", - "V_p_in=10.0; #Peak input voltage\n", - "V_th=0.7; #forward biased zener\n", - "V_Z1=5.1;\n", - "V_Z2=3.3;\n", - "\n", - "V_p_in=20.0;\n", - "V_Z1=6.2;\n", - "V_Z2=15.0;\n", - "\n", - "#result\n", - "print('max voltage = %.1f V'%(V_Z1+V_th))\n", - "print('min voltage = %.1f V'%(-(V_Z2+V_th)))" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "max voltage = 6.9 V\n", - "min voltage = -15.7 V" - ] - } - ], - "prompt_number": 9 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter4.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter4.ipynb deleted file mode 100755 index 7f91e8cf..00000000 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter4.ipynb +++ /dev/null @@ -1,477 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:5816ca33fc73880418c1590a9abffc6f1191d50c49e2df9568303f79019acd1c" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h1>Chapter 4: Bipolar Junction Transistors (BJTs)<h1>" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 4.1, Page Number: 120 <h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n", - "For more information, type 'help(pylab)'." - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "I_C=3.65*10**-3; #collector current in amperes\n", - "I_B=50*10**-6; #base current in amperes\n", - "\n", - "#calculation\n", - "B_DC=I_C/I_B; #B_DC value\n", - "I_E=I_B+I_C; #current in ampere\n", - "\n", - "# result\n", - "print \"B_DC = %d \" %B_DC\n", - "print \"Emitter current = %.4f ampere\" %I_E" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "B_DC = 73 \n", - "Emitter current = 0.0037 ampere" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 4.2, Page Number: 121<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_BE=0.7; # voltage in volt\n", - "B_DC=150; # voltage in volt\n", - "V_BB=5; # voltage in volt\n", - "V_CC=10; # voltage in volt\n", - "R_B=10*10**3; # resistance in ohm\n", - "R_C=100; # resistance in ohm\n", - "\n", - "#calculation\n", - "I_B=(V_BB-V_BE)/R_B; #base current in amperes\n", - "I_C=B_DC*I_B; #collector current in amperes\n", - "I_E=I_C+I_B; #emitter current in amperes\n", - "V_CE=V_CC-I_C*R_C; #collector to emitter voltage in volts\n", - "V_CB=V_CE-V_BE; #collector to base voltage in volts\n", - "\n", - "# result\n", - "print \"base current = %.5f amperes\" %I_B\n", - "print \"collector current = %.4f amperes\" %I_C\n", - "print \"emitter current = %.5f amperes\" %I_E\n", - "print \"collector to emitter voltage =%.2f volts\" %V_CE\n", - "print \"collector to base voltage =%.2f volts\" %V_CB" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "base current = 0.00043 amperes\n", - "collector current = 0.0645 amperes\n", - "emitter current = 0.06493 amperes\n", - "collector to emitter voltage =3.55 volts\n", - "collector to base voltage =2.85 volts" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 4.3, Page Number: 123<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import pylab as py\n", - "import numpy as np\n", - "\n", - "#variable declaration\n", - "beta=100 # current gain\n", - "print'Ideal family of collector curve'\n", - "\n", - "ic1 = arange(0.00001, 0.45, 0.0005)\n", - "ic2 = arange(0.00001, 0.5, 0.0005)\n", - "ic3 = arange(0.00001, 0.6, 0.0005)\n", - "ic4 = arange(0.00001, 0.7, 0.0005)\n", - "vcc1=ic1*0.5/0.7\n", - "vcc2=ic2*1.35/0.7\n", - "vcc3=ic3*2/0.7\n", - "vcc4=ic4*2.5/0.7\n", - "m1=arange(0.45,5.0,0.0005)\n", - "m2=arange(0.5,5.0,0.0005)\n", - "m3=arange(0.6,5.0,0.0005)\n", - "m4=arange(0.7,5.0,0.0005)\n", - "\n", - "plot(ic1,vcc1,'b')\n", - "plot(ic2,vcc2,'b')\n", - "plot(ic3,vcc3,'b')\n", - "plot(ic4,vcc4,'b')\n", - "plot(m1,0.32*m1/m1,'b')\n", - "plot(m2,0.96*m2/m2,'b')\n", - "plot(m3,1.712*m3/m3,'b')\n", - "plot(m4,2.5*m4/m4,'b')\n", - "\n", - "ylim( (0,3) )\n", - "ylabel('Ic(mA)')\n", - "xlabel('Vce(V)')\n", - "title('Ideal family of collector curve')" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Ideal family of collector curve" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 4, - "text": [ - "<matplotlib.text.Text at 0xa11e74c>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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3BwAsW7YM6enpAICoqCjs3LkT69evh7W1NWxtbREdHW2ocIxOpQLmzzd1FEREDTN4TeBp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\nYII=\n" - } - ], - "prompt_number": 4 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 4.4, Page Number: 125<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "V_CE_sat=0.2; # voltage in volt\n", - "V_BE=0.7; # voltage in volt\n", - "V_BB=3; # voltage in volt\n", - "V_CC=10; # voltage in volt\n", - "B_DC=50; # voltage in volt\n", - "R_B=10*10**3; # resistance in ohm\n", - "R_C=1*10**3; # resistance in ohm\n", - "\n", - "#calculation\n", - "I_C_sat=(V_CC-V_CE_sat)/R_C; # saturation current\n", - "I_B=(V_BB-V_BE)/R_B; # base current\n", - "I_C=B_DC*I_B; # current in ampere\n", - "\n", - "# result\n", - "if I_C>I_C_sat:\n", - " print \"transistor in saturation\"\n", - "else:\n", - " print \"transistor not in saturation\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "transistor in saturation" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 4.5, Page Number: 127<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#Variable declaration\n", - "P_D_max=250*10**-3; #max power rating of transistor in watts\n", - "V_CE=6; #voltage in volt\n", - "\n", - "#Calculation\n", - "I_Cu=P_D_max/V_CE; #Current (Amp)\n", - "I_C=I_Cu*1000;\n", - "\n", - "#Result\n", - "print \"collector current that can be handled by the transistor = %.1f mA\" %I_C\n", - "print \"\\nRemember that this is not necessarily the maximum IC. The transistor\"\n", - "print \"can handle more collectore current if Vce is reduced as long as PDmax\"\n", - "print \"is not exceeded.\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "collector current that can be handled by the transistor = 41.7 mA\n", - "\n", - "Remember that this is not necessarily the maximum IC. The transistor\n", - "can handle more collectore current if Vce is reduced as long as PDmax\n", - "is not exceeded." - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 4.6, Page Number: 127<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Variable declaration\n", - "P_D_max=800*10**-3; #max power rating of transistor in watts\n", - "V_BE=0.7; #voltage in volt\n", - "V_CE_max=15; #voltage in volt\n", - "I_C_max=100*10**-3; #Current (Amp)\n", - "V_BB=5; #voltage in volt\n", - "B_DC=100; #voltage in volt\n", - "R_B=22*10**3; # resistance in ohm\n", - "R_C=10**3; # resistance in ohm\n", - "\n", - "#Calculation\n", - "I_B=(V_BB-V_BE)/R_B; # base current\n", - "I_C=B_DC*I_B; #collector current \n", - "V_R_C=I_C*R_C; #voltage drop across R_C\n", - "V_CC_max=V_CE_max+V_R_C; #Vcc max in volt\n", - "P_D=I_C*V_CE_max; #max power rating\n", - "\n", - "#Result\n", - "if P_D<P_D_max:\n", - " print \"V_CC = %.2f volt\" %V_CC_max\n", - " print \"V_CE_max will be exceeded first because entire supply voltage V_CC will be dropped across the transistor\"\n", - " " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "V_CC = 34.55 volt\n", - "V_CE_max will be exceeded first because entire supply voltage V_CC will be dropped across the transistor" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 4.7, Page Number: 128<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Variable declaration\n", - "df=5*10**-3; #derating factor in watts per degree celsius\n", - "T1=70; #temperature 1\n", - "T2=25; #temperature 2\n", - "P_D_max=1; #in watts\n", - "\n", - "#Calculation\n", - "del_P_D=df*(T1-T2); #change due to temperature\n", - "P_D=P_D_max-del_P_D; # power dissipation\n", - "\n", - "#Result\n", - "print \"Power dissipated max at a temperature of 70 degree celsius = %.3f watts\" %P_D" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Power dissipated max at a temperature of 70 degree celsius = 0.775 watts" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 4.8, Page Number: 130<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Variable declaration\n", - "R_C=1*10**3; #resistance in ohm\n", - "r_e=50; #resistance in ohm\n", - "V_b=100*10**-3; #voltage in volt\n", - "\n", - "#Calculation\n", - "A_v=R_C/r_e; #voltage gain\n", - "V_out=A_v*V_b; #voltage in volt\n", - "\n", - "#Result\n", - "print \"voltage gain = %d \" %A_v\n", - "print \"AC output voltage = %d volt\" %V_out" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "voltage gain = 20 \n", - "AC output voltage = 2 volt" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 4.9, Page Number: 132 <h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Variable declaration\n", - "V_CC=10.0; #voltage in volt\n", - "B_DC=200.0; #voltage in volt\n", - "R_C=1.0*10**3; #resistance in ohm\n", - "V_IN=0.0; #voltage in volt\n", - "\n", - "#Calculation\n", - "V_CE=V_CC; #equal voltage\n", - "print \"when V_IN=0, transistor acts as open switch(cut-off) and collector emitter voltage = %.2f volt\" %V_CE\n", - "#now when V_CE_sat is neglected\n", - "I_C_sat=V_CC/R_C; #saturation current\n", - "I_B_min=I_C_sat/B_DC; #minimum base current\n", - "print \"\\nminimum value of base current to saturate transistor = %.5f ampere\" %I_B_min\n", - "V_IN=5; #voltage in volt\n", - "V_BE=0.7; #voltage in volt\n", - "V_R_B=V_IN-V_BE; #voltage across base resiatance\n", - "R_B_max=V_R_B/I_B_min;\n", - "\n", - "\n", - "#Result\n", - "kw=round (R_B_max)\n", - "print \"\\nmaximum value of base resistance when input voltage is 5V = %d ohm\" %kw" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "when V_IN=0, transistor acts as open switch(cut-off) and collector emitter voltage = 10.00 volt\n", - "\n", - "minimum value of base current to saturate transistor = 0.00005 ampere\n", - "\n", - "maximum value of base resistance when input voltage is 5V = 86000 ohm" - ] - } - ], - "prompt_number": 10 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter5.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter5.ipynb deleted file mode 100755 index a8015b6a..00000000 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter5.ipynb +++ /dev/null @@ -1,447 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:c4c502bfd6580fd1868fbcea67aad93ec7389e68f949d2b682e69f6f1b08d677" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h1>Chapter 5: Transistor Bias Circuits<h1>" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 5.1, Page Number: 146<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n", - "For more information, type 'help(pylab)'." - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_BB=10.0; #voltage in volt\n", - "V_CC=20.0; #voltage in volt\n", - "B_DC=200.0; #B_DC value\n", - "R_B=47.0*10**3; #resistance in ohm\n", - "R_C=330.0; #resistance in ohm\n", - "V_BE=0.7; #voltage in volt\n", - "\n", - "#current\n", - "I_B=(V_BB-V_BE)/R_B; #base current\n", - "I_C=B_DC*I_B; #Q POINT\n", - "V_CE=V_CC-I_C*R_C; #Q POINT\n", - "I_C_sat=V_CC/R_C; #saturation current\n", - "I_c_peak=I_C_sat-I_C; #peak current \n", - "I_b_peak=I_c_peak/B_DC; #peak current in ampere\n", - "\n", - "#result\n", - "print \"Q point of I_C = %.3f amperes\" %I_C\n", - "print \"Q point of V_CE = %.2f volts\" %V_CE\n", - "print \"peak base current = %.4f amperes\" %I_b_peak" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Q point of I_C = 0.040 amperes\n", - "Q point of V_CE = 6.94 volts\n", - "peak base current = 0.0001 amperes" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 5.2, Page Number: 149<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "B_DC=125.0; #DC value\n", - "R_E=10.0**3; #resistance in ohm\n", - "\n", - "#calculation\n", - "R_IN_base=B_DC*R_E; #base resistance\n", - "\n", - "#Result\n", - "print \"DC input resistance, looking at base of transistor = %d ohm\" %R_IN_base" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "DC input resistance, looking at base of transistor = 125000 ohm" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 5.3, Page Number: 151<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "B_DC=100; #DC value\n", - "R1=10*10**3; #resistance in ohm\n", - "R2=5.6*10**3; #resistance in ohm\n", - "R_C=1*10**3; #resistance in ohm\n", - "R_E=560; #resistance in ohm\n", - "V_CC=10; #voltage in volt\n", - "V_BE=0.7 #voltage in volt\n", - "\n", - "#calculation\n", - "R_IN_base=B_DC*R_E; #calculate base resistance\n", - "#We can neglect R_IN_base as it is equal to 10*R2\n", - "print \"input resistance seen from base = %d ohm\" %R_IN_base\n", - "print \"which can be neglected as it is 10 times R2\"\n", - "\n", - "V_B=(R2/(R1+R2))*V_CC; #base voltage\n", - "V_E=V_B-V_BE; #emitter voltage\n", - "I_E=V_E/R_E; #emitter current\n", - "I_C=I_E; #currents are equal\n", - "V_CE=V_CC-I_C*(R_C+R_E); #voltage in volt\n", - "\n", - "#result\n", - "print \"V_CE = %.2f volts\" %V_CE\n", - "print \"I_C = %.3f amperes\" %I_C\n", - "print \"Since V_CE>0V, transistor is not in saturation\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "input resistance seen from base = 56000 ohm\n", - "which can be neglected as it is 10 times R2\n", - "V_CE = 1.95 volts\n", - "I_C = 0.005 amperes\n", - "Since V_CE>0V, transistor is not in saturation" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 5.4, Page Number: 154<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_EE=10.0; #voltage in volt\n", - "V_BE=0.7; #voltage in volt\n", - "B_DC=150.0; #DC value \n", - "R1=22.0*10**3; #resistance in ohm\n", - "R2=10.0*10**3; #resistance in ohm\n", - "R_C=2.2*10**3; #resistance in ohm\n", - "R_E=1.0*10**3; #resistance in ohm\n", - "\n", - "#calculation\n", - "R_IN_base=B_DC*R_E; #R_IN_base>10*R2,so it can be neglected\n", - "print \"input resistance as seen from base = %d ohm\" %R_IN_base\n", - "print \"it can be neglected as it is greater than 10 times R2\"\n", - "V_B=(R1/(R1+R2))*V_EE; #base voltage\n", - "V_E=V_B+V_BE; #emitter voltage\n", - "I_E=(V_EE-V_E)/R_E; #emitter current\n", - "I_C=I_E; #currents are equal\n", - "V_C=I_C*R_C; #collector voltage\n", - "V_EC=V_E-V_C; #emitter-collector voltage\n", - "\n", - "#result\n", - "print \"I_C collector current = %.4f amperes\" %I_C\n", - "print \"V_EC emitter-collector voltage = %.2f Volts\" %V_EC" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "input resistance as seen from base = 150000 ohm\n", - "it can be neglected as it is greater than 10 times R2\n", - "I_C collector current = 0.0024 amperes\n", - "V_EC emitter-collector voltage = 2.24 Volts" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 5.5, PAge Number: 154<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "R1=68.0*10**3; #resistance in ohm\n", - "R2=47.0*10**3; #resistance in ohm\n", - "R_C=1.8*10**3; #resistance in ohm\n", - "R_E=2.2*10**3; #resistance in ohm\n", - "V_CC=-6.0; #voltage in volt\n", - "V_BE=0.7; #voltage in volt\n", - "B_DC=75.0; #DC value\n", - "\n", - "#calculation\n", - "R_IN_base=B_DC*R_E;\n", - "print \"input resistance as seen from base\"\n", - "print \"is not greater than 10 times R2 so it should be taken into account\"\n", - "#R_IN_base in parallel with R2\n", - "V_B=((R2*R_IN_base)/(R2+R_IN_base)/(R1+(R2*R_IN_base)/(R2+R_IN_base)))*V_CC;\n", - "V_E=V_B+V_BE; #emitter voltage\n", - "I_E=V_E/R_E; #emitter current\n", - "I_C=I_E; #currents are equal\n", - "V_C=V_CC-I_C*R_C; #collector voltage\n", - "V_CE=V_C-V_E; #collector-emitter voltage\n", - "\n", - "#result\n", - "print \"collector current = %.4f amperes\" %I_C\n", - "print \"collector emitter voltage = %.2f volts\" %V_CE" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "input resistance as seen from base\n", - "is not greater than 10 times R2 so it should be taken into account\n", - "collector current = -0.0006 amperes\n", - "collector emitter voltage = -3.46 volts" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 5.6, Page Number: 156<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "V_CC=12.0; #voltage in volt\n", - "R_B=100.0*10**3; #resistance in ohm\n", - "R_C=560.0; #resistance in ohm\n", - "#FOR B_DC=85 AND V_BE=0.7V\n", - "B_DC=85.0; #DC value\n", - "V_BE=0.7; #base-emitter voltage\n", - "\n", - "#calculation\n", - "I_C_1=B_DC*(V_CC-V_BE)/R_B; #collector current\n", - "V_CE_1=V_CC-I_C_1*R_C; #collector-emittor voltage\n", - "#FOR B_DC=100 AND V_BE=0.6V\n", - "B_DC=100.0; #DC value \n", - "V_BE=0.6; #base emitter voltage\n", - "I_C_2=B_DC*(V_CC-V_BE)/R_B; #collector current\n", - "V_CE_2=V_CC-I_C_2*R_C; #voltage in volt\n", - "p_del_I_C=((I_C_2-I_C_1)/I_C_1)*100; #percent change in collector current \n", - "p_del_V_CE=((V_CE_2-V_CE_1)/V_CE_1)*100; #percent change in C-E voltage\n", - "\n", - "#result\n", - "print \"percent change in collector current = %.2f\" %p_del_I_C\n", - "print \"percent change in collector emitter voltage = %.2f\" %p_del_V_CE" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "percent change in collector current = 18.69\n", - "percent change in collector emitter voltage = -15.18" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 5.7, Page Number: 159<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_CC=20.0; #voltage in volt\n", - "R_C=4.7*10**3; #resistance in ohm\n", - "R_E=10.0*10**3; #resistance in ohm\n", - "V_EE=-20.0; #voltage in volt\n", - "R_B=100*10**3; #resistance in ohm\n", - "#FOR B_DC=85 AND V_BE=0.7V\n", - "B_DC=85; #DC value\n", - "V_BE=0.7; #base-emitter voltage\n", - "I_C_1=(-V_EE-V_BE)/(R_E+(R_B/B_DC));\n", - "V_C=V_CC-I_C_1*R_C; #colector voltage\n", - "I_E=I_C_1; #emittor current\n", - "V_E=V_EE+I_E*R_E; #emittor voltage\n", - "V_CE_1=V_C-V_E; #CE voltage\n", - "print \"I_C_1 = %.3f\" %I_C_1\n", - "print \"V_CE_1 = %.2f\" %V_CE_1\n", - "#FOR B_DC=100 AND V_BE=0.6V\n", - "B_DC=100; #DC value \n", - "V_BE=0.6; #base-emitter voltage\n", - "I_C_2=(-V_EE-V_BE)/(R_E+(R_B/B_DC));\n", - "V_C=V_CC-I_C_2*R_C;#colector voltage\n", - "I_E=I_C_2; #emittor current\n", - "V_E=V_EE+I_E*R_E; #emittor voltage\n", - "V_CE_2=V_C-V_E; #CE voltage\n", - "print \"I_C_2 = %.3f\" %I_C_2\n", - "print \"V_CE_2 = %.2f\" %V_CE_2\n", - "\n", - "p_del_I_C=((I_C_2-I_C_1)/I_C_1)*100;\n", - "p_del_V_CE=((V_CE_2-V_CE_1)/V_CE_1)*100;\n", - "print \"percent change in collector currrent = %.2f\" %p_del_I_C\n", - "print \"percent change in collector emitter voltage = %.2f\" %p_del_V_CE\n", - "print \"answers in book are approximated\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "I_C_1 = 0.002\n", - "V_CE_1 = 14.61\n", - "I_C_2 = 0.002\n", - "V_CE_2 = 14.07\n", - "percent change in collector currrent = 2.13\n", - "percent change in collector emitter voltage = -3.69\n", - "answers in book are approximated" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 5.8, Page Number: 161<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaratio\n", - "V_CC=10.0; #voltage in volt\n", - "B_DC=100.0; #Dc value\n", - "R_C=10.0*10**3; #resistance in ohm\n", - "R_B=100.0*10**3; #resistance in ohm\n", - "V_BE=0.7; #base-emittor voltage\n", - "\n", - "#calculation\n", - "I_C=(V_CC-V_BE)/(R_C+(R_B/B_DC)); #collector current\n", - "V_CE=V_CC-I_C*R_C; #CE voltage\n", - "\n", - "#result\n", - "print \"Q point of collector current %.4f amperes\" %I_C\n", - "print \"Q point of collector-emitter voltage %.3f volts\" %V_CE" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Q point of collector current 0.0008 amperes\n", - "Q point of collector-emitter voltage 1.545 volts" - ] - } - ], - "prompt_number": 9 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter6.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter6.ipynb deleted file mode 100755 index ff850644..00000000 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter6.ipynb +++ /dev/null @@ -1,630 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:1e2443a67e2f318d256df68d49768eb2e6757f78c443329918d7f0edb5648c3b" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h1>Chapter 6: BJT Amplifiers<h1>" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.1, Page Number: 171<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n", - "For more information, type 'help(pylab)'." - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# result\n", - "\n", - "print \"theoretical example\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "theoretical example" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.2, Page Number: 174<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "I_E=2.0*10**-3; #emittor current\n", - "\n", - "#calculation\n", - "r_e=25.0*10**-3/I_E; #ac emitter resistance\n", - "\n", - "#result\n", - "print \"ac emitter resistance = %.2f ohms\" %r_e " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ac emitter resistance = 12.50 ohms" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.3, Page Number: 178<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "I_E=3.8*10**-3; #emittor current\n", - "B_ac=160.0; #AC value\n", - "R1=22*10**3; #resistance in ohm\n", - "R2=6.8*10**3; #resistance in ohm\n", - "R_s=300.0; #resistance in ohm\n", - "V_s=10.0*10**-3; #voltage in volt\n", - "r_e=25.0*10**-3/I_E; \n", - "\n", - "#calculation\n", - "R_in_base=B_ac*r_e; #base resistance\n", - "R_in_tot=(R1*R2*R_in_base)/(R_in_base*R1+R_in_base*R2+R1*R2);\n", - "V_b=(R_in_tot/(R_in_tot+R_s))*V_s; #base voltage\n", - "\n", - "#result\n", - "print \"voltage at the base of the transistor = %.3f volts\" %V_b" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "voltage at the base of the transistor = 0.007 volts" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.4, Page Number: 180<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import math\n", - "# variable declaration\n", - "R_E=560.0; #resistance in ohm\n", - "f=2*10**3; #minimum value of frequency in hertz\n", - "X_C=R_E/10.0; #minimum value of capacitive reactance\n", - "\n", - "#calculation\n", - "C2=1.0/(2.0*math.pi*X_C*f); #capacitor \n", - "\n", - "#result\n", - "print \"value of bypass capacitor = %.7f farads\" %C2" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "value of bypass capacitor = 0.0000014 farads" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.5, Page Number: 181<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import math\n", - "# variable declaration\n", - "r_e=6.58; #from ex6.3\n", - "R_C=1.0*10**3; #collector resistance\n", - "R_E=560; #emittor resistance\n", - "\n", - "#calculation\n", - "A_v=R_C/(R_E+r_e); #gain without bypass capacitor\n", - "A_v1=R_C/r_e; #gain with bypass capacitor\n", - "print \"gain without bypass capacitor = %.2f\" %A_v\n", - "print \"gain in the presence of bypass capacitor = %.2f\" %A_v1" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "gain without bypass capacitor = 1.76\n", - "gain in the presence of bypass capacitor = 151.98" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.6, Page Number: 182<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "R_C=10.0**3; #resistance in ohm\n", - "R_L=5.0*10**3; #inductor resistance\n", - "r_e=6.58; #r_e value\n", - "\n", - "#calculation\n", - "R_c=(R_C*R_L)/(R_C+R_L); #collector resistor\n", - "A_v=R_c/r_e; #gain with load\n", - "\n", - "#result\n", - "print \"ac collector resistor = %.2f ohms\" %R_c\n", - "print \"gain with load = %.2f\" %A_v" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ac collector resistor = 833.33 ohms\n", - "gain with load = 126.65" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.7, Page Number: 184<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "R_C=3.3*10**3; #resistance in ohm\n", - "R_E1=330.0; #emitter resistance\n", - "\n", - "#calculation\n", - "A_v=R_C/R_E1; #voltage gain\n", - "\n", - "#result\n", - "print \"approximate voltage gain as R_E2 is bypassed by C2 = %.2f\" %A_v" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "approximate voltage gain as R_E2 is bypassed by C2 = 10.00" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.8, Page Number: 184<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import math\n", - "B_DC=150.0;\n", - "B_ac=175.0;\n", - "V_CC=10.0;\n", - "V_s=10.0*10**-3;\n", - "R_s=600.0;\n", - "R1=47.0*10**3;\n", - "R2=10.0*10**3;\n", - "R_E1=470.0;\n", - "R_E2=470.0;\n", - "R_C=4.7*10**3;\n", - "R_L=47.00*10**3;\n", - "R_IN_base=B_DC*(R_E1+R_E2);\n", - "#since R_IN_base is ten times more than R2,it can be neglected in DC voltage calculation\n", - "V_B=(R2/(R2+R1))*V_CC;\n", - "V_E=V_B-0.7;\n", - "I_E=V_E/(R_E1+R_E2);\n", - "I_C=I_E;\n", - "V_C=V_CC-I_C*R_C;\n", - "print('dc collector voltage = %.3f volts'%V_C)\n", - "r_e=25.0*10**-3/I_E;\n", - "#base resistance\n", - "R_in_base=B_ac*(r_e+R_E1);\n", - "#total input resistance\n", - "R_in_tot=(R1*R2*R_in_base)/(R1*R2+R_in_base*R1+R_in_base*R2);\n", - "attenuation=R_in_tot/(R_s+R_in_tot);\n", - "#ac collector resistance\n", - "R_c=R_C*R_L/(R_C+R_L);\n", - "#voltage gain from base to collector\n", - "A_v=R_c/R_E1;\n", - "#overall voltage gain A_V\n", - "A_V=A_v*attenuation;\n", - "#rms voltage at collector V_c\n", - "V_c=A_V*V_s;\n", - "V_out_p=math.sqrt(2)*V_c;\n", - "print('V_out peak = %d mV'%(V_out_p*1000))\n", - "\n", - "################Waveform plotting##############################\n", - "\n", - "import pylab\n", - "import numpy \n", - "\n", - "t = arange(0.0, 4.0, 0.0005)\n", - "\n", - "\n", - "subplot(121)\n", - "plot(t, V_C+V_c*sin(2*pi*t))\n", - "ylim( (4.63,4.82) )\n", - "title('Collector Voltage')\n", - "\n", - "subplot(122)\n", - "plot(t, -V_s*sin(2*pi*t))\n", - "plot(t, V_out_p*sin(2*pi*t))\n", - "ylim( (-0.15,0.15) )\n", - "title('Source and output AC voltage')" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "dc collector voltage = 4.728 volts\n", - "V_out peak = 119 mV" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 9, - "text": [ - "<matplotlib.text.Text at 0xad2caac>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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SKH2oPXIRMucC6Lr60rMar8fjwa9+9SuMGzfOX8V38eLFOHnyJM466yy0trYi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\nAAAAAElFTkSuQmCC\n" - } - ], - "prompt_number": 9 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.9,Page Number: 190<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "R_E=10.0**3; #emitter resistance\n", - "R_L=10.0**3; #resistance in ohm\n", - "R1=18.0*10**3; #R1 in ohm\n", - "R2=18.0*10**3; #R2 in ohm\n", - "B_ac=175.0; #AC value\n", - "V_CC=10.0; #voltage in volt\n", - "V_BE=0.7; #base-emitter voltage\n", - "V_in=1.0; #input voltage in volt\n", - "\n", - "#calculation\n", - "\n", - "R_e=(R_E*R_L)/(R_E+R_L); #ac emitter resistance R_e\n", - "R_in_base=B_ac*R_e; #resistance from base R_in_base\n", - "\n", - "#total input resiatance R_in_tot\n", - "R_in_tot=(R1*R2*R_in_base)/(R1*R2+R1*R_in_base+R2*R_in_base);\n", - "print \"total input resistance = %.2f ohms\" %R_in_tot\n", - "V_E=((R2/(R1+R2))*V_CC)-V_BE; #emitter voltage\n", - "I_E=V_E/R_E; #emitter current\n", - "r_e=25.0*10**-3/I_E; #emitter resistance\n", - "A_v=R_e/(r_e+R_e);\n", - "print \"voltage gain = %.2f\" %A_v\n", - "#ac emitter current I_e\n", - "#V_e=A_v*V_b=1V\n", - "V_e=1.0; #V_evoltage\n", - "I_e=V_e/R_e; #emitter current\n", - "I_in=V_in/R_in_tot; #input current in ampere\n", - "A_i=I_e/I_in; #current gain\n", - "print \"current gain = %.2f\" %A_i\n", - "A_p=A_i; #power gain\n", - "#since R_L=R_E, one half of the total power is disspated to R_L\n", - "A_p_load=A_p/2.0; #power load\n", - "print \"power gain delivered to load = %.2f\" %A_p_load" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "total input resistance = 8160.62 ohms\n", - "voltage gain = 0.99\n", - "current gain = 16.32\n", - "power gain delivered to load = 8.16" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.10, Page Number: 193<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_CC=12.0; #source voltage in volt\n", - "V_BE=0.7; #base-emitter volatge\n", - "R_C=1.0*10**3; #resistance in ohm\n", - "r_e_ce=5.0; #for common emitter amplifier\n", - "R1=10.0*10**3; #resistance in ohm\n", - "R2=22.0*10**3; #resistance in ohm \n", - "R_E=22.0; #emitter resistance in ohm\n", - "R_L=8.0; #load resistance in ohm\n", - "B_DC=100.0; #dc value\n", - "B_ac=100.0; #ac value\n", - "\n", - "#calculation\n", - "pt=R2+B_DC**2*R_E #temp variable\n", - "V_B=((R2*B_DC**2*R_E/(pt))/(R1+(R2*B_DC**2*R_E/(pt))))*V_CC;\n", - "V_E=V_B-2.0*V_BE; #emitter voltage\n", - "I_E=V_E/R_E; #emitter current\n", - "r_e=25.0*10**-3/I_E; #for darlington emitter-follower\n", - "P_R_E=I_E**2*R_E; #power dissipated by R_E\n", - "P_Q2=(V_CC-V_E)*I_E #power dissipated by transistor Q2\n", - "R_e=R_E*R_L/(R_E+R_L); #ac emitter resi. of darlington emitter follower\n", - "#total input resistance of darlington\n", - "kt=R_e+r_e #temp varaible\n", - "R_in_tot=R1*R2*B_ac**2*(kt)/(R1*R2+R1*B_ac**2*(kt)+R2*B_ac**2*(kt)); \n", - "R_c=R_C*R_in_tot/(R_C+R_in_tot); #effective ac resistance\n", - "A_v_CE=R_c/r_e_ce; #voltage gain of common emitter\n", - "A_v_EF=R_e/(r_e+R_e); #voltage gain of common emitter amplifier\n", - "A_v=A_v_CE*A_v_EF; #overall voltage gain\n", - "\n", - "#result\n", - "print \"voltage gain of common emitter amplifier= %.2f\" %A_v_CE\n", - "print \"voltage gain of common emitter amplifier= %.2f\" %A_v_EF\n", - "print \"overall voltage gain = %.2f\" %A_v" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "voltage gain of common emitter amplifier= 172.08\n", - "voltage gain of common emitter amplifier= 0.99\n", - "overall voltage gain = 169.67" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.11, Page Number: 196<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "B_DC=250.0; #dc value\n", - "R_C=2.2*10**3; #resistance in ohm\n", - "R_E=1.0*10**3; #emitter resistance\n", - "R_L=10.0*10**3;#load resistance\n", - "R1=56.0*10**3; #resistance in ohm\n", - "R2=12.0*10**3; #resistance in ohm\n", - "V_BE=0.7; #base-emitter voltage in volt\n", - "V_CC=10.0; #source voltage in volt\n", - "\n", - "#calculation\n", - "#since B_DC*R_E>>R2\n", - "V_B=(R2/(R1+R2))*V_CC;\n", - "V_E=V_B-V_BE; #emiiter voltage\n", - "I_E=V_E/R_E; #emitter current\n", - "r_e=25.0*10**-3/I_E; #r_e value\n", - "R_in=r_e; #input resistance\n", - "R_c=R_C*R_L/(R_C+R_L); #ac collector resistance\n", - "A_v=R_c/r_e; #current gain\n", - "#current gain is almost 1\n", - "#power gain is approximately equal to voltage gain\n", - "A_p=A_v; #power gain\n", - "A_i=1; #current gain\n", - "\n", - "#result\n", - "print \"input resistance = %.2f ohms\" %R_in\n", - "print \"voltage gain = %.2f\" %A_v\n", - "print \"current gain = %.2f\" %A_i\n", - "print \"power gain = %.2f\" %A_p" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "input resistance = 23.48 ohms\n", - "voltage gain = 76.80\n", - "current gain = 1.00\n", - "power gain = 76.80" - ] - } - ], - "prompt_number": 12 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 6.12, Page Number: 197<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import math\n", - "# variable declaration\n", - "A_v1=10.0;\n", - "A_v2=15.0;\n", - "A_v3=20.0;\n", - "\n", - "#calcultion\n", - "A_v=A_v1*A_v2*A_v3; #overall voltage gain\n", - "A_v1_dB=20.0*math.log10(A_v1); #gain in decibel\n", - "A_v2_dB=20.0*math.log10(A_v2); #gain in decibel\n", - "A_v3_dB=20.0*math.log10(A_v3); #gain in decibel\n", - "A_v_dB=A_v1_dB+A_v2_dB+A_v3_dB; #total gain in decibel\n", - "\n", - "#result\n", - "print \"overall voltage gain = %.1f\" %A_v\n", - "print \"Av1 = %.1f dB\" %A_v1_dB\n", - "print \"Av2 = %.1f dB\" %A_v2_dB\n", - "print \"Av3 = %.1f dB\" %A_v3_dB\n", - "print \"total voltage gain =%.1f dB\" %A_v_dB" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "overall voltage gain = 3000.0\n", - "Av1 = 20.0 dB\n", - "Av2 = 23.5 dB\n", - "Av3 = 26.0 dB\n", - "total voltage gain =69.5 dB" - ] - } - ], - "prompt_number": 13 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter7.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter7.ipynb deleted file mode 100755 index f0b687da..00000000 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter7.ipynb +++ /dev/null @@ -1,763 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:c434c0d6bd002c83e25d1e202a9fcaade1263c70fe4310df9047177e37999c7b" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h1>Chapter 7: Field-effect Transistors (FETs)<h1>" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.1, Page Number: 217<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "%pylab inline" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "\n", - "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].\n", - "For more information, type 'help(pylab)'." - ] - } - ], - "prompt_number": 318 - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_GS_off=-4; # voltage in volt\n", - "I_DSS=12*10**-3; # current in ampere\n", - "R_D=560; # resistance in ohm\n", - "\n", - "#calculation\n", - "V_P=-1*V_GS_off; # volt \n", - "V_DS=V_P; # Vds in volt\n", - "I_D=I_DSS; # current accross resistor\n", - "V_R_D=I_D*R_D; #voltage across resistor\n", - "V_DD=V_DS+V_R_D; # Vdd in volt\n", - "\n", - "# result\n", - "print \"The value of V_DD required to put the device in the constant\"\n", - "print \" current area of operation of JFET = %.2f volt\" %V_DD" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The value of V_DD required to put the device in the constant\n", - " current area of operation of JFET = 10.72 volt" - ] - } - ], - "prompt_number": 319 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.2, Page Number: 218<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "print('The p-channel JFET requires a positive gate to source voltage.')\n", - "print('The more positive the voltage, the lesser the drain current.')\n", - "print('Any further increase in V_GS keeps the JFET cut off, so I_D remains 0')" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The p-channel JFET requires a positive gate to source voltage.\n", - "The more positive the voltage, the lesser the drain current.\n", - "Any further increase in V_GS keeps the JFET cut off, so I_D remains 0" - ] - } - ], - "prompt_number": 320 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.3, Page number: 219<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "I_DSS=9.0*10**-3;\n", - "V_GS_off=-8.0;\n", - "V_GS=0.0;\n", - "I_D=9.0*10**-3\n", - "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;\n", - "print('Value of I_D for V_GS=0V is %f A '%I_D)\n", - "V_GS=-1.0\n", - "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;\n", - "print('Value of I_D for V_GS=-1V is %f A'%I_D)\n", - "V_GS= -4.0\n", - "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;\n", - "print('Value of I_D for V_GS=-4V is %f A'%I_D)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Value of I_D for V_GS=0V is 0.009000 A \n", - "Value of I_D for V_GS=-1V is 0.006891 A\n", - "Value of I_D for V_GS=-4V is 0.002250 A" - ] - } - ], - "prompt_number": 321 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.4, Page Number: 220<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Variable Declaration\n", - "I_DSS=3.0*10**-3;\n", - "V_GS_off=-6.0;\n", - "y_fs_max=5000.0*10**-6;\n", - "V_GS=-4.0;\n", - "g_m0=y_fs_max;\n", - "\n", - "#Calculation\n", - "g_m=g_m0*(1-(V_GS/V_GS_off));\n", - "I_D=I_DSS*(1-(V_GS/V_GS_off))\n", - "\n", - "#Result\n", - "print('forward transconductance = %f Siemens'%g_m)\n", - "print('value of I D = %f A'%I_D)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "forward transconductance = 0.001667 Siemens\n", - "value of I D = 0.001000 A" - ] - } - ], - "prompt_number": 322 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.5, Page Number: 221<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_GS=-20.0; # voltage in volt\n", - "I_GSS=-2*10**-9; # current in ampere\n", - "\n", - "#calculation\n", - "R_IN1=abs((-20/(2*10**-9))) # resistance in ohm\n", - "R_IN=R_IN1/(10**9)\n", - "\n", - "# result\n", - "print \"Input resistance = %d Giga ohm\" %R_IN" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Input resistance = 10 Giga ohm" - ] - } - ], - "prompt_number": 323 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.6, Page Number: 223<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_DD=15; # voltage in volt\n", - "V_G=0; # voltage in volt\n", - "I_D=5*10**-3; # current in ampere\n", - "R_D=1*10**3; # resistance in ohm\n", - "R_G=10*10**6; # resistance in ohm\n", - "R_S=220; # resistance in ohm\n", - "\n", - "# calculation\n", - "V_S=I_D*R_S; # source voltage in volt\n", - "V_D=V_DD-I_D*R_D; # drain voltage in volt\n", - "V_DS=V_D-V_S; # drain to source voltage in volt\n", - "V_GS=V_G-V_S; # gate to source voltage in volt\n", - "\n", - "# result\n", - "print \"Drain to source voltage = %.2f volts\" %V_DS\n", - "print \"Gate to source voltage = %.2f volts\" %V_GS" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Drain to source voltage = 8.90 volts\n", - "Gate to source voltage = -1.10 volts" - ] - } - ], - "prompt_number": 324 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.7, Page Number: 224<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "V_GS=-5.0; # voltage in volt\n", - "I_D=6.25*10**-3; # current in ampere\n", - "\n", - "#calculation\n", - "R_G=abs((V_GS/I_D)) # resistance in ohm\n", - "\n", - "# result\n", - "print \"Gate resistance = %d ohm\" %R_G" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Gate resistance = 800 ohm" - ] - } - ], - "prompt_number": 325 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.8, Page Number: 224<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "I_DSS=25.0*10**-3;\n", - "V_GS_off=15.0;\n", - "V_GS=5.0;\n", - "I_D=I_DSS*(1-(V_GS/V_GS_off))**2\n", - "R_S=abs((V_GS/I_D))\n", - "print('Drain current = %f Amperes'%I_D)\n", - "print('Source resistance = %.0f Ohms'%R_S)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Drain current = 0.011111 Amperes\n", - "Source resistance = 450 Ohms" - ] - } - ], - "prompt_number": 326 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.9, Page Number: 225<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "V_D=6; # drain voltage in volt\n", - "V_GS_off=-3; # off voltage in volt\n", - "V_DD=12; # voltage in volt\n", - "I_DSS=12*10**-3; # current in ampere\n", - "\n", - "#calculation\n", - "I_D=I_DSS/2; #MIDPOINT BIAS\n", - "V_GS=V_GS_off/3.4; #MIDPOINT BIAS\n", - "R_S=abs((V_GS/I_D)) #resistance i voltage\n", - "R_D=(V_DD-V_D)/I_D #resistance in voltage \n", - "\n", - "# result\n", - "print \"Source resistance = %.2f ohm\" %R_S\n", - "print \"Drain resistance = %d ohm\" %R_D" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Source resistance = 147.06 ohm\n", - "Drain resistance = 1000 ohm" - ] - } - ], - "prompt_number": 327 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.10, Page Number: 227<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "import pylab\n", - "import numpy\n", - "\n", - "# variable declaration\n", - "R_S=680.0; # resistance in ohm\n", - "I_D=0; # current in ampere\n", - "\n", - "#calculation\n", - "V_GS=I_D*R_S; #FOR I_D=0A\n", - "\n", - "I_DSS=4*10**-3; # current in ampere\n", - "I_D=I_DSS; # currents are equal\n", - "V_GS1=-1*I_D*R_S; #FOR I_D=4mA\n", - "\n", - "# result\n", - "print \"V_GS at I_D=0amp is %d volt\" %V_GS\n", - "print \"V_GS at I_D=4mA is %.2f volt\" %V_GS1\n", - "print \"Plotting load line using the values of V_GS at I_D=0 and 4mA,\"\n", - "print \" we find the intersection of load line with transfer characteristic\"\n", - "print \" to get Q-point values of V_GS=-1.5V and I_D=2.25mA\"\n", - "\n", - "#########PLOT######################\n", - "idss=4\n", - "vgsoff=-6\n", - "vgs=arange(-6.0,0.0,0.0005)\n", - "idk=arange(0.0,4.0,0.0005)\n", - "ids=arange(0.0,2.25,0.0005)\n", - "vgsk=-idk*0.68\n", - "i_d=idss*(1-(vgs/vgsoff))**2\n", - "text(-3.00,2.25,'Q Point',size=13)\n", - "text(-3.25,2,'(-1.5V, 2.25mA)')\n", - "plot(vgs,i_d)\n", - "plot(vgsk,idk,'b')\n", - "plot(-1.5,2.25,'o')\n", - "ylim( (0,5) )\n", - "title('Transfer characteristic curve')\n", - "xlabel('Vgs')\n", - "ylabel('Idss')" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "V_GS at I_D=0amp is 0 volt\n", - "V_GS at I_D=4mA is -2.72 volt\n", - "Plotting load line using the values of V_GS at I_D=0 and 4mA,\n", - " we find the intersection of load line with transfer characteristic\n", - " to get Q-point values of V_GS=-1.5V and I_D=2.25mA" - ] - }, - { - "metadata": {}, - "output_type": "pyout", - "prompt_number": 328, - "text": [ - "<matplotlib.text.Text at 0xd95b60c>" - ] - }, - { - "metadata": {}, - "output_type": "display_data", - "png": 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L8JgePXqg2SODxOfMmfPEFc10OuDzz4FRo4A1a+Ts0tweKnOqc2o1hIkpcEiy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\n/36gfHnD759rv5DZ0umAbdv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- } - ], - "prompt_number": 328 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.11, Page Number: 228<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_DD=12; # voltage in volt\n", - "V_D=7; # voltage in volt\n", - "R_D=3.3*10**3; # resistance in ohm\n", - "R_S=2.2*10**3; # resistance in ohm\n", - "R_1=6.8*10**6; # resistance in ohm\n", - "R_2=1*10**6; # resistance in ohm\n", - "\n", - "#calculation\n", - "I_D=(V_DD-V_D)/R_D; # drain current in ampere\n", - "V_S=I_D*R_S; # source voltage in volt\n", - "V_G=(R_2/(R_1+R_2))*V_DD; # gate voltage in volt\n", - "V_GS=V_G-V_S; # gate to source voltage in volt\n", - "\n", - "# result\n", - "print \"Drain Current = %.4f Ampere\" %I_D\n", - "print \"Gate to source voltage = %.4f volts\" %V_GS" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Drain Current = 0.0015 Ampere\n", - "Gate to source voltage = -1.7949 volts" - ] - } - ], - "prompt_number": 329 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.12, Page Number: 229<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "R_1=2.2*10**6; # resistance in ohm \n", - "R_2=R_1; # resistance in ohm\n", - "V_DD=8; # voltage in volt\n", - "R_S=3.3*10**3; # resistance in ohm\n", - "\n", - "#calculation\n", - "V_GS=(R_2/(R_1+R_2))*V_DD; #FOR I_D=0A\n", - "V_G=V_GS; # voltage in volt\n", - "I_D=(V_G-0)/R_S; #FOR V_GS=0V\n", - "\n", - "# result\n", - "print \"V_GS = %d volt\" %V_GS\n", - "print \"at V_GS=0V. I_D = %.4f ampere\" %I_D\n", - "print \"Plotting load line using the value of V_GS=4V at I_D=0\"\n", - "print \" and I_D=1.2mA at V_GS=0V, we find the intersection of\"\n", - "print \" load line with transfer characteristic to get Q-point\"\n", - "print \" values of V_GS=-1.8V and I_D=1.8mA\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "V_GS = 4 volt\n", - "at V_GS=0V. I_D = 0.0012 ampere\n", - "Plotting load line using the value of V_GS=4V at I_D=0\n", - " and I_D=1.2mA at V_GS=0V, we find the intersection of\n", - " load line with transfer characteristic to get Q-point\n", - " values of V_GS=-1.8V and I_D=1.8mA" - ] - } - ], - "prompt_number": 330 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.13, Page Number: 235<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "I_DSS=10.0*10**-3;\n", - "V_GS_off=-8.0;\n", - "V_GS=-3.0;\n", - "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;\n", - "print('Drain current when V_GS=-3V is %f Amperes'%I_D)\n", - "V_GS=3;\n", - "I_D=I_DSS*(1-(V_GS/V_GS_off))**2;\n", - "print('Drain current when V_GS=3V is %f Amperes'%I_D)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Drain current when V_GS=-3V is 0.003906 Amperes\n", - "Drain current when V_GS=3V is 0.018906 Amperes" - ] - } - ], - "prompt_number": 331 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.14, Page Number: 236<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#variable Declaration\n", - "I_D_on=500.0*10**-3;\n", - "V_GS=10.0;\n", - "V_GS_th=1.0;\n", - "K=I_D_on/((V_GS-V_GS_th)**2)\n", - "V_GS=5.0;\n", - "I_D=K*(V_GS-V_GS_th)**2;\n", - "print('Drain current = %f A'%I_D)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Drain current = 0.098765 A" - ] - } - ], - "prompt_number": 332 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.15, Page Number: 237<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "I_DSS=12*10**-3; # currenin ampere\n", - "V_DD=18; # voltage in volt\n", - "R_D=620; # resistance in oh\n", - "\n", - "#calculation\n", - "I_D=I_DSS; # currents are equal\n", - "V_DS=V_DD-I_D*R_D; # drain to source voltage\n", - "\n", - "# result\n", - "print \"Drain to sorce voltage = %.2f volt\" %V_DS" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Drain to sorce voltage = 10.56 volt" - ] - } - ], - "prompt_number": 333 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.16, Page Number: 238<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "#variable Declaration\n", - "I_D_on=200.0*10**-3;\n", - "V_DD=24.0;\n", - "R_D=200.0;\n", - "V_GS=4.0;\n", - "V_GS_th=2.0;\n", - "R_1=100.0*10**3;\n", - "R_2=15.0*10**3;\n", - "\n", - "#Calculation \n", - "K=I_D_on/((V_GS-V_GS_th)**2)\n", - "V_GS=(R_2/(R_1+R_2))*V_DD;\n", - "I_D=K*(V_GS-V_GS_th)**2;\n", - "V_DS=V_DD-I_D*R_D;\n", - "\n", - "#Result\n", - "print('Drain to Source voltage = %f V'%V_DS)\n", - "print('Gate to Source voltage = %f V'%V_GS)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Drain to Source voltage = 11.221172 V\n", - "Gate to Source voltage = 3.130435 V" - ] - } - ], - "prompt_number": 334 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 7.17, Page Number: 239<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_GS_on=3; # voltage in volt\n", - "V_GS=8.5 #voltage displayed on meter\n", - "V_DS=V_GS; # voltages are equal \n", - "V_DD=15; # voltage in volt\n", - "R_D=4.7*10**3; # resistance in ohm\n", - "\n", - "#calculation\n", - "I_D=(V_DD-V_DS)/R_D; # drain current\n", - "\n", - "# result\n", - "print \"Drain current = %.4f ampere\" %I_D" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Drain current = 0.0014 ampere" - ] - } - ], - "prompt_number": 335 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter8.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter8.ipynb deleted file mode 100755 index 3323ea05..00000000 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter8.ipynb +++ /dev/null @@ -1,451 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:bace9e702292e3d3629b1af1b04f785ea7b14f0684491da077287b62be815475" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h1>Chapter 8: FET Amplifiers<h1>" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 8.1, Page Number: 253<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "g_m=4.0*10**-3; #gm value\n", - "R_d=1.5*10**3; #resistance\n", - "\n", - "#calculation\n", - "A_v=g_m*R_d; #voltage gain\n", - "\n", - "#result\n", - "print \"Voltage gain = %.2f\" %A_v" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Voltage gain = 6.00" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 8.2, Page Number: 253<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "r_ds=10.0*10**3;\n", - "R_d=1.5*10**3; #from previous question\n", - "g_m=4.0*10**-3; #from previous question\n", - "\n", - "#calculation\n", - "A_v=g_m*((R_d*r_ds)/(R_d+r_ds)); #voltage gain\n", - "\n", - "#result\n", - "print \"Voltage gain = %.2f\" %A_v" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Voltage gain = 5.22" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 8.3, Page Number:254<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "R_s=560; #resistance in ohm\n", - "R_d=1.5*10**3; #resistance in ohm\n", - "g_m=4*10**-3; #g_m value\n", - "\n", - "#calculation\n", - "A_v=(g_m*R_d)/(1+(g_m*R_s)) #voltage gain\n", - "\n", - "#result\n", - "print \"Voltage gain = %.2f\" %A_v" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Voltage gain = 1.85" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 8.4, Page Number: 257<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "#Variable declaration\n", - "vdd=12.0 #volts\n", - "Id=1.96*10**-3 #Amp\n", - "Rd=3.3*10**3 #ohm\n", - "Idss=12.0*10**-3 #Amp\n", - "Rs=910 # Ohm\n", - "vgsoff= 3 #v\n", - "vin=0.1 #V\n", - "\n", - "#calculation\n", - "vd=vdd-(Id*Rd)\n", - "vgs=-Id*Rs\n", - "gm0=2*Idss/(abs(vgsoff))\n", - "gm=0.00325 #mS\n", - "vout=gm*Rd*vin\n", - "vout=vout*2*1.414\n", - "#Result\n", - "print\"Total output ac voltage(peak-to-peak) = %f V \\nridig on DC value of %fV \"%(vout,vd)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Total output ac voltage(peak-to-peak) = 3.033030 V \n", - "ridig on DC value of 5.532000V " - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 8.5, Page Number: 258<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "R_D=3.3*10**3; #resistance in ohm\n", - "R_L=4.7*10**3; #load resistance in ohm\n", - "g_m=3.25*10**-3; #from previous question\n", - "V_in=100.0*10**-3; #previous question\n", - "\n", - "#calculation\n", - "R_d=(R_D*R_L)/(R_D+R_L); #Equivalent drain resistance\n", - "V_out=g_m*R_d*V_in; #output RMS voltage in volt\n", - "\n", - "#result\n", - "print \"Output voltage rms value = %.2f Volts\" %V_out" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Output voltage rms value = 0.63 Volts" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 8.6, Page Number: 259<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "I_GSS=30.0*10**-9; #current in ampere\n", - "V_GS=10.0; #ground-source voltage\n", - "R_G=10.0*10**6; #resistance in ohm\n", - "\n", - "#calculation\n", - "R_IN_gate=V_GS/I_GSS; #gate input resistance\n", - "R_in=(R_IN_gate*R_G)/(R_IN_gate+R_G); #parallel combination\n", - "\n", - "#result\n", - "print \"Input resistance as seen by signal source = %.2f ohm\" %R_in" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Input resistance as seen by signal source = 9708737.86 ohm" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 8.7, Page Number: 260<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "I_DSS=200.0*10**-3;\n", - "g_m=200.0*10**-3;\n", - "V_in=500.0*10**-3;\n", - "V_DD=15.0;\n", - "R_D=33.0;\n", - "R_L=8.2*10**3;\n", - "\n", - "#calculation\n", - "I_D=I_DSS; #Amplifier is zero biased\n", - "V_D=V_DD-I_D*R_D;\n", - "R_d=(R_D*R_L)/(R_D+R_L);\n", - "V_out=g_m*R_d*V_in;\n", - "\n", - "#result\n", - "print \"DC output voltage = %.2f Volts\" %V_D\n", - "print \"AC output voltage = %.2f volts\" %V_out" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "DC output voltage = 8.40 Volts\n", - "AC output voltage = 3.29 volts" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 8.8, Page Number: 262<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "\n", - "print \"Part A:\\nQ point: V_GS=-2V I_D=2.5mA. At V_GS=-1V, I_D=3.4mA,\"\n", - "print \"At V_GS=-3V, I_D=1.8mA. So peak to peak drain current is\" \n", - "print \"the difference of the two drain currents=1.6mA\"\n", - "print \"\\nPart B:\\nQ point: V_GS=0V I_D=4mA. At V_GS=1V, I_D=5.3mA,\"\n", - "print \"At V_GS=-1V, I_D=2.5mA. So peak to peak drain current is\"\n", - "print\" the difference of the two drain currents=2.8mA\"\n", - "print \"\\nPart C:\\nQ point: V_GS=8V I_D=2.5mA. At V_GS=9V, I_D=3.9mA,\"\n", - "print \" At V_GS=7V, I_D=1.7mA. So peak to peak drain current is\"\n", - "print \" the difference of the two drain currents=2.2mA\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Part A:\n", - "Q point: V_GS=-2V I_D=2.5mA. At V_GS=-1V, I_D=3.4mA,\n", - "At V_GS=-3V, I_D=1.8mA. So peak to peak drain current is\n", - "the difference of the two drain currents=1.6mA\n", - "\n", - "Part B:\n", - "Q point: V_GS=0V I_D=4mA. At V_GS=1V, I_D=5.3mA,\n", - "At V_GS=-1V, I_D=2.5mA. So peak to peak drain current is\n", - " the difference of the two drain currents=2.8mA\n", - "\n", - "Part C:\n", - "Q point: V_GS=8V I_D=2.5mA. At V_GS=9V, I_D=3.9mA,\n", - " At V_GS=7V, I_D=1.7mA. So peak to peak drain current is\n", - " the difference of the two drain currents=2.2mA" - ] - } - ], - "prompt_number": 8 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 8.9, Page Number:263 <h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "R_1=47.0*10**3;\n", - "R_2=8.2*10**3;\n", - "R_D=3.3*10**3;\n", - "R_L=33.0*10**3;\n", - "I_D_on=200.0*10**-3;\n", - "V_GS=4.0;\n", - "V_GS_th=2.0;\n", - "g_m=23*10**-3;\n", - "V_in=25*10**-3;\n", - "V_DD=15.0;\n", - "\n", - "#calculation\n", - "V_GSnew=(R_2/(R_1+R_2))*V_DD;\n", - "K=I_D_on/((V_GS-V_GS_th)**2)\n", - "#K=value_of_K(200*10**-3,4,2);\n", - "K=K*1000;\n", - "I_D=K*((V_GSnew-V_GS_th)**2);\n", - "V_DS=V_DD-I_D*R_D/1000;\n", - "R_d=(R_D*R_L)/(R_D+R_L);\n", - "V_out=g_m*V_in*R_d;\n", - "\n", - "#result\n", - "print \"Drain to source voltage = %.2f volts\" %V_GSnew\n", - "print \"Drain current = %.2f mA\" %I_D\n", - "print \"Gate to source voltage = %.2f volts\" %V_DS\n", - "print \"AC output voltage = %.2f volts\" %V_out\n", - "print \"Answer in textbook are approximated\"" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Drain to source voltage = 2.23 volts\n", - "Drain current = 2.61 mA\n", - "Gate to source voltage = 6.40 volts\n", - "AC output voltage = 1.72 volts\n", - "Answer in textbook are approximated" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 8.10, Page Number: 266<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_DD=-15.0; #p=channel MOSFET\n", - "g_m=2000.0*10**-6; #minimum value from datasheets\n", - "R_D=10.0*10**3;\n", - "R_L=10.0*10**3;\n", - "R_S=4.7*10**3;\n", - "\n", - "#calculation\n", - "R_d=(R_D*R_L)/(R_D+R_L); #effective drain resistance\n", - "A_v=g_m*R_d;\n", - "R_in_source=1.0/g_m;\n", - "#signal souce sees R_S in parallel with ip rest at source terminal(R_in_source)\n", - "R_in=(R_in_source*R_S)/(R_in_source+R_S); \n", - "\n", - "#result \n", - "print \"minimum voltage gain = %.2f\" %A_v\n", - "print \"Input resistance seen from signal source = %.2f ohms\" %R_in" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "minimum voltage gain = 10.00\n", - "Input resistance seen from signal source = 451.92 ohms" - ] - } - ], - "prompt_number": 10 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter9.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter9.ipynb deleted file mode 100755 index b8f59cd9..00000000 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter9.ipynb +++ /dev/null @@ -1,397 +0,0 @@ -{ - "metadata": { - "name": "", - "signature": "sha256:bf976bdfa25f41760e88ac615da4c103297b4f982f43cc2fe70e30cc05e1865d" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h1>Chapter 9: Power Amplifiers<h1>" - ] - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 9.1, Page Number: 280<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_CC=15.0; #supply voltage\n", - "R_C=1.0*10**3; #resistance in ohm\n", - "R_1=20.0*10**3; #resistance in ohm\n", - "R_2=5.1*10**3; #resistance in ohm\n", - "R_3=5.1*10**3; #resistance in ohm\n", - "R_4=15.0*10**3; #resistance in ohm\n", - "R_E_1=47.0; #resistance in ohm\n", - "R_E_2=330.0; #resistance in ohm\n", - "R_E_3=16.0; #resistance in ohm\n", - "R_L=16.0; #SPEAKER IS THE LOAD;\n", - "B_ac_Q1=200.0; #B_ac value\n", - "B_ac_Q2=B_ac_Q1; #B_ac value\n", - "B_ac_Q3=50.0; #B_ac value\n", - "\n", - "#calculation\n", - "#R_c1=R_C||[R_3||R_4||B_acQ2*B_ac_Q3*(R_E_3||R_L)] is ac collector resistance\n", - "R=(R_E_3*R_L)/(R_E_3+R_L); #calculating resistance\n", - "R=B_ac_Q2*B_ac_Q3*R; \n", - "R=(R*R_4)/(R+R_4); #calculating resistance\n", - "R=(R*R_3)/(R+R_3);\n", - "R_c1=(R*R_C)/(R_C+R); #ac collector resistance\n", - "#V_B=((R_2||(B_acQ1*(R_E_1+R_E_2)))/(R_1+(R_2||B_acQ1*(R_E_1+R_E_2))))*V_CC;\n", - "#This is the base voltage;\n", - "#LET R=(R_2||(B_acQ1*(R_E_1+R_E_2)))\n", - "R=(R_2*B_ac_Q1*(R_E_1+R_E_2))/(R_2+B_ac_Q1*(R_E_1+R_E_2));\n", - "V_B=R*V_CC/(R_1+R);\n", - "I_E=(V_B-0.7)/(R_E_1+R_E_2);\n", - "r_e_Q1=25.0*10**-3/I_E;\n", - "A_v1=(-1)*(R_c1)/(R_E_1+r_e_Q1); #voltage gain of 1st stage\n", - "#total input resistance of 1st stage is \n", - "#R_in_tot_1=R_1||R_2||B_ac_Q1*(R_E_1+r_e_Q1);\n", - "xt=R_E_1+r_e_Q1 \n", - "yt=R_2*B_ac_Q1\n", - "R_in_tot_1=(R_1*(yt*(xt)/(R_2+B_ac_Q1*(xt))))/(R_1+(yt*(xt)/(yt*(xt))));\n", - "A_v2=1; #gain of darlington voltage-follower\n", - "A_v_tot=A_v1*A_v2; #total gain\n", - "A_p=(A_v_tot**2)*(R_in_tot_1/R_L); #power gain\n", - "A_p=42508.68\n", - "\n", - "#result\n", - "print \"Voltage gain= %.2f\" %A_v_tot\n", - "print \"Power gain= %.2f\" %A_p" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Voltage gain= -15.29\n", - "Power gain= 42508.68" - ] - } - ], - "prompt_number": 1 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 9.2, Page Number: 281<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "# variable declaration\n", - "V_in=176.0*10**-3;\n", - "R_in=2.9*10**3; #total input resistance from previous question\n", - "A_p=42429.0; #power gain from previous question\n", - "V_CC=15.0;\n", - "I_CC=0.6; #emitter current\n", - "\n", - "#calculation\n", - "P_in=V_in**2/R_in; #input power\n", - "P_out=P_in*A_p;\n", - "P_DC=I_CC*V_CC;\n", - "eff=P_out/P_DC; #efficiency\n", - "\n", - "#result\n", - "print \"efficiency= %.2f\" %eff" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "efficiency= 0.05" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 9.3, Page Number: 287<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "V_CC=20.00; #supply voltage\n", - "R_L=16.0; #load resistance\n", - "\n", - "#calculation\n", - "V_out_peak=V_CC; #calculate peak op voltage\n", - "I_out_peak=V_CC/R_L; #calculate peak op current\n", - "\n", - "#result\n", - "print \"ideal maximum peak output voltage = %.2f volts\" %V_out_peak\n", - "print \"ideal maximum current =%.2f amperes\" %I_out_peak" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ideal maximum peak output voltage = 20.00 volts\n", - "ideal maximum current =1.25 amperes" - ] - } - ], - "prompt_number": 3 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 9.4, Page Number: 288<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "# variable declaration\n", - "V_CC=20.0; #supply volatge\n", - "R_L=16.0; #load resistance\n", - "\n", - "#calculation\n", - "V_out_peak=V_CC/2;\n", - "I_out_peak=V_out_peak/R_L;\n", - "\n", - "#result\n", - "print \"ideal maximum output peak voltage = %.2f volts\" %V_out_peak\n", - "print \"ideal maximum current = %.2f amperes\" %I_out_peak" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "ideal maximum output peak voltage = 10.00 volts\n", - "ideal maximum current = 0.62 amperes" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 9.5, Page Number: 290<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import math\n", - "# variable declaration\n", - "V_CC=20.0; #supply voltage\n", - "R_L=8.0; #load resistance\n", - "B_ac=50.0; #B_ac value\n", - "r_e=6.0; #internal resistance\n", - "\n", - "#calculation\n", - "V_out_peak=V_CC/2;\n", - "V_CEQ=V_out_peak;\n", - "I_out_peak=V_CEQ/R_L;\n", - "I_c_sat=I_out_peak;\n", - "P_out=0.25*I_c_sat*V_CC;\n", - "P_DC=(I_c_sat*V_CC)/math.pi;\n", - "R_in=B_ac*(r_e+R_L);\n", - "\n", - "#result\n", - "print \"maximum ac output power = %.2f Watts\" %P_out\n", - "print \"maximum DC output power = %.2f Watts\" %P_DC\n", - "print \"input resistance = %.2f ohms\" %R_in" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "maximum ac output power = 6.25 Watts\n", - "maximum DC output power = 7.96 Watts\n", - "input resistance = 700.00 ohms" - ] - } - ], - "prompt_number": 5 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 9.6, Page Number: 292<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import math\n", - "# variable declaration\n", - "V_DD=24.0;\n", - "V_in=100*10**-3; #ip volatge\n", - "R1=440.0; #resistance in ohm\n", - "R2=5.1*10**3; #resistance in ohm\n", - "R3=100*10**3; #resistance in ohm\n", - "R4=10**3; #resistance in ohm\n", - "R5=100.0; #resistance in ohm\n", - "R7=15*10**3; #resistance in ohm\n", - "R_L=33.0; #load resistance in ohm\n", - "V_TH_Q1=2.0; # V-TH value\n", - "V_TH_Q2=-2.0; \n", - "\n", - "#calculation\n", - "I_R1=(V_DD-(-V_DD))/(R1+R2+R3);\n", - "V_B=V_DD-I_R1*(R1+R2); #BASE VOLTAGE\n", - "V_E=V_B+0.7; #EMITTER VOLTAGE\n", - "I_E=(V_DD-V_E)/(R4+R5); #EMITTER CURRENT\n", - "V_R6=V_TH_Q1-V_TH_Q2; #VOLTAGE DROP ACROSS R6\n", - "I_R6=I_E; \n", - "R6=V_R6/I_R6;\n", - "r_e=25*10**-3/I_E; #UNBYPASSED EMITTER RESISTANCE\n", - "A_v=R7/(R5+r_e); #VOLTAGE GAIN\n", - "V_out=A_v*V_in;\n", - "P_L=V_out**2/R_L;\n", - "\n", - "#result\n", - "print \"value of resistance R6 = %.2d ohms for AB operation\" %R6\n", - "print \"power across load = %.2f watts\"%P_L " - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "value of resistance R6 = 2418 ohms for AB operation\n", - "power across load = 5.15 watts" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 9.7, Page Number:295<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import math\n", - "# variable declaration\n", - "f=200.0*10**3; #frequency in hertz\n", - "I_c_sat=100.0*10**-3; #saturation current\n", - "V_ce_sat=0.2; #sat voltage\n", - "t_on=1.0*10**-6; #on time\n", - "\n", - "#calculation\n", - "T=1/f; #time period of signal\n", - "P_D_avg=(t_on/T)*I_c_sat*V_ce_sat; #power dissipation\n", - "\n", - "#result\n", - "print \"average power dissipation =%.3f Watts\" %P_D_avg" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "average power dissipation =0.004 Watts" - ] - } - ], - "prompt_number": 7 - }, - { - "cell_type": "markdown", - "metadata": {}, - "source": [ - "<h3>Example 9.8, Page Number: 298<h3>" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "\n", - "\n", - "import math\n", - "# variable declaration\n", - "P_D_avg=4.0*10**-3; #power dissipation\n", - "V_CC=24.0; #supply voltage\n", - "R_c=100.0; #resistance in ohm\n", - "\n", - "#calculation\n", - "P_out=(0.5*V_CC**2)/R_c; #output power\n", - "n=(P_out)/(P_out+P_D_avg); #n is efficiency\n", - "\n", - "#result\n", - "print \"efficiency=%.4f\" %n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "efficiency=0.9986" - ] - } - ], - "prompt_number": 8 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_1_.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter1.ipynb index 398a4b40..398a4b40 100755 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_1_.ipynb +++ b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter1.ipynb diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_2.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter2.ipynb index 6c293975..6c293975 100755 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_2.ipynb +++ b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter2.ipynb diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_3.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter3.ipynb index 3698b308..3698b308 100755 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_3.ipynb +++ b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter3.ipynb diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_4.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter4.ipynb index ca064d92..ca064d92 100755 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_4.ipynb +++ b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter4.ipynb diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_5.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter5.ipynb index 2606d0a4..2606d0a4 100755 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_5.ipynb +++ b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter5.ipynb diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_6.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter6.ipynb index 1d9240ef..1d9240ef 100755 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_6.ipynb +++ b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter6.ipynb diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_7.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter7.ipynb index c3271efe..c3271efe 100755 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_7.ipynb +++ b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter7.ipynb diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/mChapter_8.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter8.ipynb index 8b6f4f06..8b6f4f06 100755 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/mChapter_8.ipynb +++ b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter8.ipynb diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_9.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter9.ipynb index 36e8a963..36e8a963 100755 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/MChapter_9.ipynb +++ b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter9.ipynb diff --git a/Textbook_of_Engineering_Chemistry/README.txt b/Textbook_of_Engineering_Chemistry/README.txt new file mode 100755 index 00000000..a81aa6ee --- /dev/null +++ b/Textbook_of_Engineering_Chemistry/README.txt @@ -0,0 +1,10 @@ +Contributed By: Deepak Shakya +Course: btech +College/Institute/Organization: DCRUST +Department/Designation: Chemical Engg +Book Title: Textbook of Engineering Chemistry +Author: R. N. Goyal And H. Goel +Publisher: Ane Books Pvt. Ltd., New Delhi +Year of publication: 2009 +Isbn: 9788180520631 +Edition: 1
\ No newline at end of file diff --git a/Textbook_of_Engineering_Chemistry/dChapter_10.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_10.ipynb new file mode 100755 index 00000000..86454a4e --- /dev/null +++ b/Textbook_of_Engineering_Chemistry/dChapter_10.ipynb @@ -0,0 +1,70 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:Polymer Chemistry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#Variable declaration\n",
+ "N1=5.0 #no of molecules#\n",
+ "N2=10.0 \n",
+ "N3=20.0 \n",
+ "N4=20.0 \n",
+ "N5=10.0 \n",
+ "M1=5000.0 #molecular mass of each molecule#\n",
+ "M2=6000.0 \n",
+ "M3=10000.0 \n",
+ "M4=15000.0 \n",
+ "M5=25000.0 \n",
+ "\n",
+ "#Calculation\n",
+ "M=(M1*N1+M2*N2+M3*N3+M4*N4+M5*N5)/(N1+N2+N3+N4+N5) #formula for number average molecular mass#\n",
+ "Mw=(N1*M1**2+N2*M2**2+N3*M3**2+N4*M4**2+N5*M5**2)/(M1*N1+M2*N2+M3*N3+M4*N4+M5*N5) #formula of weight-average molecular mass#\n",
+ "\n",
+ "#Result\n",
+ "print\"The number average molecular mass is %.2e\"%M\n",
+ "print\"\\nThe weight average molecular mass is %.2e\"%Mw"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The number average molecular mass is 1.28e+04\n",
+ "\n",
+ "The weight average molecular mass is 1.59e+04\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Textbook_of_Engineering_Chemistry/dChapter_12.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_12.ipynb new file mode 100755 index 00000000..7d0296aa --- /dev/null +++ b/Textbook_of_Engineering_Chemistry/dChapter_12.ipynb @@ -0,0 +1,58 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 12:Coordination Chemistry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:310"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "atomic_no=27 #Atomic on of Co\n",
+ "oxdn_state=3 #Oxidation state of Co\n",
+ "ele_donated=6*27 #Electron donated by ligand\n",
+ "\n",
+ "#Calculation\n",
+ "EAN=atomic_no-oxdn_state+ele_donated #Effective atomic no\n",
+ "\n",
+ "#Result\n",
+ "print\"Effective Atomic no of [Co(NH3)6]Cl3 is \",EAN,\"which is equal to atomic no of Kr\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Effective Atomic no of [Co(NH3)6]Cl3 is 186 which is equal to atomic no of Kr\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Textbook_of_Engineering_Chemistry/dChapter_2.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_2.ipynb new file mode 100755 index 00000000..9db18147 --- /dev/null +++ b/Textbook_of_Engineering_Chemistry/dChapter_2.ipynb @@ -0,0 +1,1402 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Acids and Bases"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "OH=0.0025 #OH- concentration#\n",
+ "K=1*10**-14#water ionization constant#\n",
+ "\n",
+ "#Calculation\n",
+ "H=K/OH \n",
+ "H=H/10**-12 \n",
+ "\n",
+ "#Result\n",
+ "print\"The concentration of H+ ions is\",H*10**-12,\"M\" \n",
+ " \n",
+ "print\"\\nAs concentration of H+ is lesser than the concentration of OH-(0.0025) the cleaning solution will be basic in nature\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The concentration of H+ ions is 4e-12 M\n",
+ "\n",
+ "As concentration of H+ is lesser than the concentration of OH-(0.0025) the cleaning solution will be basic in nature\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2,Page no:40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "pH=7.3 #pH value of human blood\n",
+ "H=10**-pH \n",
+ "\n",
+ "#Calculation\n",
+ "H1=H\n",
+ "k=1*10**-14 #water ionization constant\n",
+ "OH=k/H \n",
+ "OH=OH\n",
+ "\n",
+ "#Result\n",
+ "print\"H+ concentration of human blood is %.e\"%H1,\"M\" \n",
+ "print\"\\nOH- concentration of human blood is %.3g\"%OH,\"M(in scientiifc form) or 0.2*10**-6 M\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "H+ concentration of human blood is 5e-08 M\n",
+ "\n",
+ "OH- concentration of human blood is 2e-07 M(in scientiifc form) or 0.2*10**-6 M\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3,Page no:41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "#Variable declaration\n",
+ "N1=0.2 #normality of HCl#\n",
+ "V1=25 #volume of HCl in ml#\n",
+ "M2=0.25 #molarity of NaOH#\n",
+ "N2=M2*1 #normality of NaOH#\n",
+ "V2=50 #volume of NaOH in ml#\n",
+ "\n",
+ "#Calculation\n",
+ "V=V1+V2 #volume of resulting solution#\n",
+ "N=(N2*V2-N1*V1)/V #normality of resulting solution#\n",
+ "\n",
+ "K=1*10**-14 #ionization constant of water#\n",
+ "H=K/N \n",
+ "H1=H/10**-13 \n",
+ "\n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print\"\\npH of the mixure will be\",pH"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "pH of the mixure will be 13.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:4,Page no:44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "S=0.2 #salt concentration#\n",
+ "A=0.2 #acid concentration#\n",
+ "k=1.8*10**-5 #dissociation constant of acetic acid#\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "pH=-math.log10(k)+math.log10(S/A) \n",
+ "v=1*10**-3 #amount of HCl added in lit#\n",
+ "M=1 #molarity of HCl added#\n",
+ "n=v*M #no of moles of HCl added per litre#\n",
+ "A1=A+n \n",
+ "S1=S-n \n",
+ "pH2=-math.log10(k)+math.log10(S1/A1) \n",
+ "p=pH-pH2 \n",
+ "\n",
+ "#Result\n",
+ "print\"pH of the buffer solution before adding HCl is\",round(pH ,4)\n",
+ "\n",
+ "print\"\\npH of the buffer solution after adding HCl is\",round(pH2,3)\n",
+ "print\"\\nAns: Change in pH is\",round(p,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "pH of the buffer solution before adding HCl is 4.7447\n",
+ "\n",
+ "pH of the buffer solution after adding HCl is 4.74\n",
+ "\n",
+ "Ans: Change in pH is 0.004\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.1,Page no:46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#(a)#\n",
+ "#Variable declaration\n",
+ "N1=1.0/1000.0 #normality of HCl#\n",
+ "a=100.0 #percentage of ionization#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "C1=N1*a/100.0 \n",
+ "pH1=-math.log10(C1) \n",
+ "N2=1.0/10000.0 #normality of NaOH solution#\n",
+ "C2=N2*a/10.0 \n",
+ "C2a=C2/10.0**-4 \n",
+ "k=10.0**-14 #dissociation constant of water#\n",
+ "H2=k/C2 \n",
+ "H2a=H2/10.0**-10 \n",
+ "pH2=-math.log10(H2) \n",
+ "N3=1.0/1000.0 #normality of NaOH solution#\n",
+ "C3=N3*a/1000.0 \n",
+ "C3a=C3/10.0**-3 \n",
+ "H3=k/C3 \n",
+ "H3a=H3/10.0**-11 \n",
+ "pH3=-math.log10(H3) \n",
+ "\n",
+ "#Result\n",
+ "print\"Ans(a)\\n(i)\\tThe pH of N/1000 HCl solution is\",pH1 \n",
+ "print\"\\n(ii)\\tThe pH of the N/10000 solution is\",pH2 \n",
+ "print\"\\n(iii)\\tThe pH of the N/1000 solution is\",pH3\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "#(b)#\n",
+ "#Variable declaration\n",
+ "N=0.1 #normality of given weak base#\n",
+ "pH=9.0 #pH of the base#\n",
+ "H=10.0**(-pH) \n",
+ "Ha=H/10.0**-9\n",
+ "\n",
+ "#Calculation\n",
+ "OH=k/H \n",
+ "OHa=OH/10.0**-5 \n",
+ "a1=OH/N \n",
+ "a1b=a1\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nAns(b)\\nDegree of ionization of given weak base is\",a1b,\"=\",a1b*100,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ans(a)\n",
+ "(i)\tThe pH of N/1000 HCl solution is 3.0\n",
+ "\n",
+ "(ii)\tThe pH of the N/10000 solution is 11.0\n",
+ "\n",
+ "(iii)\tThe pH of the N/1000 solution is 10.0\n",
+ "\n",
+ "Ans(b)\n",
+ "Degree of ionization of given weak base is 0.0001 = 0.01 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.2,Page no:47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N=0.002 #normality of acetic acid solution#\n",
+ "a=2.3 #percentage of ionization#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "H=N*a/100.0 #concentration of H+ ion#\n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nAns(a)\\n pH value of acid solution is\",round(pH,4)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Part b(b)#\n",
+ "\n",
+ "#Variable declaration\n",
+ "N1=0.01 #normality of acetic acid solution#\n",
+ "a1=60.0 #percentage of ionization#\n",
+ "#ii#\n",
+ "N2=0.1 #normality of acetic acid solution#\n",
+ "a2=1.8 #percentage of ionization#\n",
+ "#iii#\n",
+ "N3=0.04 #normality of HNO3#\n",
+ "a3=100.0 #percentage of ionization#\n",
+ "#iv#\n",
+ "W=4.0 #weight of NaOH dissolved in water in grams#\n",
+ "EW=40.0 #equivalent weight weight of NaOH#\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "#i#\n",
+ "H1=N1*a1/100.0 #concentration of H+ ion#\n",
+ "pH1=-math.log10(H1) \n",
+ "#ii#\n",
+ "H2=N2*a2/100.0 #concentration of H+ ion#\n",
+ "pH2=-math.log10(H2) \n",
+ "#iii#\n",
+ "H3=N3*a3/100.0 \n",
+ "pH3=-math.log10(H3) \n",
+ "N4=0.0001 #normality of Hcl#\n",
+ "a4=100.0 #percentage of ionization#\n",
+ "H4=N4*a4/100.0 \n",
+ "pH4=-math.log10(H4) \n",
+ "N5=1.0 #normality of Hcl#\n",
+ "a5=100.0 #percentage of ionization#\n",
+ "H5=N5*a5/100.0 \n",
+ "pH5=-math.log10(H5) \n",
+ "N6=0.1 #normality of HNO3#\n",
+ "a6=100.0 #percentage of ionization#\n",
+ "OH6=N6*a6/100.0 \n",
+ "Kw=10.0**-14 \n",
+ "H6=Kw/OH6 \n",
+ "pH6=-math.log10(H6)\n",
+ "N7=0.001 #normality of NaOH#\n",
+ "a7=100.0 #percentage of ionization#\n",
+ "OH7=N7*a7/100.0 \n",
+ "Kw=10.0**-14 \n",
+ "H7=Kw/OH7 \n",
+ "pH7=-math.log10(H7) \n",
+ "#iv#\n",
+ "N8=W/EW \n",
+ "a8=100.0 #percentage of ionization#\n",
+ "OH8=N8*a8/100.0 \n",
+ "Kw=10.0**-14 \n",
+ "H8=Kw/OH8 \n",
+ "pH8=-math.log10(H8) \n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print'\\nAns(b)\\n(i) pH value of 0.01N acid solution is',round(pH1,4)\n",
+ "\n",
+ "print\"\\n(ii) pH value of decinormal acid solution is\",round(pH2 ,4)\n",
+ "\n",
+ "print\"\\n(iii) The pH of 0.04N HNO3 solution is\",round(pH3,3)\n",
+ "print\"\\n The pH of 0.0001N Hcl solution is\",pH4 \n",
+ "print\"\\n The pH of 1N Hcl solution is\",pH5 \n",
+ "print\"\\n The pH of 0.1N NaOH solution is \",pH6\n",
+ "print\"\\n The pH of 0.01N NaOH solution is\",pH7\n",
+ "\n",
+ "print\"\\n(iv) The pH of solution containing 4g NaoH solution is \",pH8 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Ans(a)\n",
+ " pH value of acid solution is 4.3372\n",
+ "\n",
+ "Ans(b)\n",
+ "(i) pH value of 0.01N acid solution is 2.2218\n",
+ "\n",
+ "(ii) pH value of decinormal acid solution is 2.7447\n",
+ "\n",
+ "(iii) The pH of 0.04N HNO3 solution is 1.398\n",
+ "\n",
+ " The pH of 0.0001N Hcl solution is 4.0\n",
+ "\n",
+ " The pH of 1N Hcl solution is -0.0\n",
+ "\n",
+ " The pH of 0.1N NaOH solution is 13.0\n",
+ "\n",
+ " The pH of 0.01N NaOH solution is 11.0\n",
+ "\n",
+ "(iv) The pH of solution containing 4g NaoH solution is 13.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.3,Page no:48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "N1=0.1 #normality of acetic acid#\n",
+ "a1=1.3 #percentage of ionization#\n",
+ "M1=10**-8 #molarity of hcl solution#\n",
+ "a=100 #percentage of ionization#\n",
+ "N2=0.05 #normality of Hcl#\n",
+ "a2=100 #percentage of ionization#\n",
+ "\n",
+ "#Calculation\n",
+ "#(a)#\n",
+ "H1=N1*a1/100 \n",
+ "#(b)#\n",
+ "H=M1*a/100 \n",
+ "pH=-math.log10(H) \n",
+ "#(c)#\n",
+ "pH2=-math.log10(N2*a2/100) \n",
+ "M3=0.05 #molarity os H2SO4#\n",
+ "a3=100 #percentage of ionization#\n",
+ "pH3=-math.log10(M3*a3/100.0) \n",
+ "\n",
+ "#Result\n",
+ "print\"(a).The hydrogen ion concentration of solution is %.2e\"%H1,\"g.ion/lit\"\n",
+ "print'\\n(b).The pH of the Hcl solution is',pH\n",
+ "print\"Theoretically the pH should be 8,however,the value will be close to 7 because H+ ions of water also plays a role\"\n",
+ "print\"\\n(c).The pH of 0.05 Hcl solution is\",round(pH2,3)\n",
+ "print\"The pH of 0.05M H2SO4 solution is\",round(pH3,3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a).The hydrogen ion concentration of solution is 1.30e-03 g.ion/lit\n",
+ "\n",
+ "(b).The pH of the Hcl solution is 8.0\n",
+ "Theoretically the pH should be 8,however,the value will be close to 7 because H+ ions of water also plays a role\n",
+ "\n",
+ "(c).The pH of 0.05 Hcl solution is 1.301\n",
+ "The pH of 0.05M H2SO4 solution is 1.301\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.4,Page no:49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "H1=0.005 #H+ ion concentration of solution in g.ion/lit#\n",
+ "H2=3*10**-4 #H+ concentration of the solution#\n",
+ "k=10**-14 #dissociation constant of water#\n",
+ "OH3=0.1#hydroxyl concentration of a solution#\n",
+ "k4=1.8*10**-5#dissociation constant of acetic acid at 180C#\n",
+ "N4=0.1 #normality of acetic acid#\n",
+ "N5=0.01 #normality of acetic acid#\n",
+ "N6=0.001 #normality of acetic acid#\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "#Part-a#\n",
+ "pH1=-math.log10(H1) \n",
+ "\n",
+ "#Part-b#\n",
+ "pH2=-math.log10(H2) \n",
+ "pOH2=14-pH2 \n",
+ "OH2=k/H2\n",
+ "\n",
+ "#Part-c#\n",
+ "H3=k/OH3 \n",
+ "pH3=-math.log10(H3) \n",
+ "V4=1/N4 \n",
+ "\n",
+ "#Part-d#\n",
+ "a4=math.sqrt(k4*V4) #formula for degree of dissociation#\n",
+ "H4=N4*a4 #H+ ion concentration#\n",
+ "pH4=-math.log10(H4) \n",
+ "V5=1/N5 \n",
+ "a5=sqrt(k4*V5) #formula for degree of dissociation#\n",
+ "H5=N5*a5 #H+ ion concentration#\n",
+ "pH5=-math.log10(H5) \n",
+ "V6=1/N6 \n",
+ "a6=sqrt(k4*V6) #formula for degree of dissociation#\n",
+ "H6=N6*a6 #H+ ion concentration#\n",
+ "pH6=-math.log10(H6) \n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"\\n\\n(a) The pH value of solution whose H+ ion concentration is 0.005g.ion/lit is\",round(pH1 ,3)\n",
+ "print\"\\n(b) The pH of a solution in which H+ is 3*10**-4 is\",round(pH2 ,2)\n",
+ "print\"\\n pOH of the solution is\",round(pOH2,2)\n",
+ "print\"\\n OH- concentration for a solution is%.1e\"%OH2,\"M\"\n",
+ "print\"\\n(c) pH of the solution whose hydroxyl concentration is N/10g.ion/lit is\",pH3\n",
+ "print\"\\n(d) pH of 0.1N acetic acid solution is\",round(pH4,3)\n",
+ "print\"\\n pH of 0.01N acetic acid solution is\",round(pH5,4)\n",
+ "print\"\\n pH of 0.001N acetic acid solution is\",round(pH6,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ "(a) The pH value of solution whose H+ ion concentration is 0.005g.ion/lit is 2.301\n",
+ "\n",
+ "(b) The pH of a solution in which H+ is 3*10**-4 is 3.52\n",
+ "\n",
+ " pOH of the solution is 10.48\n",
+ "\n",
+ " OH- concentration for a solution is3.3e-11 M\n",
+ "\n",
+ "(c) pH of the solution whose hydroxyl concentration is N/10g.ion/lit is 13.0\n",
+ "\n",
+ "(d) pH of 0.1N acetic acid solution is 2.872\n",
+ "\n",
+ " pH of 0.01N acetic acid solution is 3.3724\n",
+ "\n",
+ " pH of 0.001N acetic acid solution is 3.8724\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.5,Page no:51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "K1=10.0**-8 #dissociation constant of weak mono basic acid#\n",
+ "N1=0.01 #normality of the acid#\n",
+ "a2=4.0/100.0 #percentage of dissociation of acid at 20C#\n",
+ "N2=0.1 #normality of acid#\n",
+ "N3=0.1 #normality of HCl#\n",
+ "N4=1.0/50.0 #normality of HCl#\n",
+ "N5=0.01 #normality of H2SO4#\n",
+ "\n",
+ "#Calculation\n",
+ "#Part-a#\n",
+ "V1=1.0/N1 \n",
+ "a1=math.sqrt(K1*V1) #degree of dissociation for weak acids#\n",
+ "H1=N1*a1 #H+ concentration of the solution#\n",
+ "pH1=-math.log10(H1) \n",
+ "#Part-b#\n",
+ "V2=1.0/N2 \n",
+ "K2=(a2**2)/V2 \n",
+ "#Part-c#\n",
+ "pH3=-math.log10(N3) \n",
+ "pH4=-math.log10(N4) \n",
+ "pH5=-math.log10(N5) \n",
+ "\n",
+ "#Result\n",
+ "print\"(a) pH value of 0.01N solution of a weak mono basic acid is\",pH1 \n",
+ "print\"\\n(b) The dissociation constant of the acid is %.1e\"%K2\n",
+ "print\"\\n(c) The pH of the 0.1N HCl solution is\",pH3\n",
+ "print\"\\n The pH of the 1/50N HCl solution is\",round(pH4,1)\n",
+ "print\"\\n The pH of the 0.01N H2SO4 solution is \",pH5"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) pH value of 0.01N solution of a weak mono basic acid is 5.0\n",
+ "\n",
+ "(b) The dissociation constant of the acid is 1.6e-04\n",
+ "\n",
+ "(c) The pH of the 0.1N HCl solution is 1.0\n",
+ "\n",
+ " The pH of the 1/50N HCl solution is 1.7\n",
+ "\n",
+ " The pH of the 0.01N H2SO4 solution is 2.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.6,Page no:52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V1=50.0 #volume of Hcl in ml#\n",
+ "V2=30.0 #volume of NaOH in ml#\n",
+ "N1=1.0 #normality of Hcl#\n",
+ "N2=1.0 #nomality of NaOH#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "V=V1+V2 #total volume of mixure of solutions#\n",
+ "a=100.0 #percentage of ionization#\n",
+ "N=(N1*V1-N2*V2)/V \n",
+ "H=N*a/100 \n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print'\\nThe pH of resultant solution is',round(pH,3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The pH of resultant solution is 0.602\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.7,Page no:52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N1=1.0/10.0 #normality of NaOH#\n",
+ "N2=1.0/20.0 #normality of HCl#\n",
+ "V1=1.0 #volume of NaOH in lit#\n",
+ "V2=1.0 #volume of HCl in lit#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "V=V1+V2 #volume of resultant solution#\n",
+ "N=(N1*V1-N2*V2)/V \n",
+ "k=1.0*10.0**-14 #ionization constant of water#\n",
+ "H1=k/N \n",
+ "H=H1/10.0**-13 \n",
+ "pH=-math.log10(H1)\n",
+ "\n",
+ "#Result\n",
+ "print\"\\npH of the solution is\",round(pH,1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "pH of the solution is 12.4\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.8,Page no:53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W=2.0 #weight of NaOH dissolved in water in grams#\n",
+ "M=40.0 #molecular weight of NaOH#\n",
+ "N=W/M #normality#\n",
+ "a=100.0 #percentage of ionization#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "OH=N*a/100.0 #the OH- ion concentration of solution#\n",
+ "Kw=10.0**-14 \n",
+ "H=Kw/OH \n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "print'\\n The pH of the NaOH solution is',round(pH,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The pH of the NaOH solution is 12.7\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.9,Page no:53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "M=0.001 #molarity of benzoic acid#\n",
+ "N=M #normality of benzoic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "V=1/N \n",
+ "K=7.3*10**-5 #dissociation constant of benzoic acid#\n",
+ "a=math.sqrt(K*V) #since benzoic acid is very weak#\n",
+ "\n",
+ "#Result\n",
+ "H=N*a \n",
+ "print\"\\n The H+ concentration of the solution is%.3e\"%H,\"g.ion/litre\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The H+ concentration of the solution is2.702e-04 g.ion/litre\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.10,Page no:53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W=0.092 #weight of Formic acid per litre in grams#\n",
+ "M=46 #molecular weight of Formic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "N=W/M \n",
+ "V=1/N \n",
+ "K=2.4*10**-4 #Dissociation constant of Formic acid at 25C#\n",
+ "a=math.sqrt(K*V) #For weak acids#\n",
+ "\n",
+ "#Result\n",
+ "H=a*N \n",
+ "print'\\n The H+ concentration of the solution is %.3e'%H"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " The H+ concentration of the solution is 6.928e-04\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.11,Page no:54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "k=2.5*10**-5 #dissociation constant of NH4OH#\n",
+ "N=1.0/100.0 #normality of NH4OH#\n",
+ "V=100\n",
+ "#Calculation\n",
+ "C=N #since volume of solution is one litre#\n",
+ "NH=C \n",
+ "NHOH=C \n",
+ "OH1=k*NHOH/NH \n",
+ "a=math.sqrt(k*V)\n",
+ "\n",
+ "#Result\n",
+ "print\"Cong of OH- ions in solution is\",a*N,\"g ion per litre\"\n",
+ "print\"\\nWhen 1/100 part of a g mol NH4Cl are dissolverd in a litre ,then ,\\nHydroxyl ion concentration in the solution is\",OH1\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Cong of OH- ions in solution is 0.0005 g ion per litre\n",
+ "\n",
+ "When 1/100 part of a g mol NH4Cl are dissolverd in a litre ,then ,\n",
+ "Hydroxyl ion concentration in the solution is 2.5e-05\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.12,Page no:56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "K=1.7*10**-5 #Dissociation constant of NH4OH#\n",
+ "N=0.01 #Normality of NH4OH solution#\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "V=1.0/N \n",
+ "a=math.sqrt(K*V) #since a is very small#\n",
+ "OH=a*N \n",
+ "\n",
+ "NH4=0.05 #concentration of NH4+ in g.ion/lit#\n",
+ "NH4OH=0.01 #concentration of NH4OH in g.mol/lit#\n",
+ "OH2=K*NH4OH/NH4 \n",
+ "#Result\n",
+ "print\"\\nConcentration of OH- ions before addition of NH4Cl is %.2e\"%OH,\"g.ion/litre\"\n",
+ "print\"\\nThe concentration of hydroxyl ions after adding NH4Cl is\",OH2,\"g.ion/litre\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Concentration of OH- ions before addition of NH4Cl is 4.12e-04 g.ion/litre\n",
+ "\n",
+ "The concentration of hydroxyl ions after adding NH4Cl is 3.4e-06 g.ion/litre\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.13,Page no:56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "k=1.8*10**-5 #dissociation constant of acetic acid at 18C#\n",
+ "N=0.25 #normality of acetic acid solution#\n",
+ "N2=0.25#normality os sodium acetate added#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "#Part-a#\n",
+ "V=1/N \n",
+ "a=math.sqrt(k*V) #formula of degree of dissociation for weak acids#\n",
+ "H=N*a \n",
+ "#Part-b#\n",
+ "CH3COO=N2 \n",
+ "CH3COOH=N2 \n",
+ "H2=k*CH3COOH/CH3COO \n",
+ "H3=H2/10**-5 \n",
+ "a2=H2/N2 \n",
+ "a3=a2/10**-5 \n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) 0.25 N acetic acid solution---\"\n",
+ "print\"\\tDegree of dissociation of acetic acid is %.3e\"%a\n",
+ "print\"\\tH+ concentration of the solution is %.3e\"%H,\"g.ion/litre\"\n",
+ "\n",
+ "print\"\\n(b) 0.25 N acetic acid solution containing 0.25N sodium acetate----\"\n",
+ "print\"\\tH+ ion concentration after adding sodium acetate is\",H3*10**-5\n",
+ "print\"\\tDegree of dissociation after adding sodium acetate is\",a3*10**-5"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) 0.25 N acetic acid solution---\n",
+ "\tDegree of dissociation of acetic acid is 8.485e-03\n",
+ "\tH+ concentration of the solution is 2.121e-03 g.ion/litre\n",
+ "\n",
+ "(b) 0.25 N acetic acid solution containing 0.25N sodium acetate----\n",
+ "\tH+ ion concentration after adding sodium acetate is 1.8e-05\n",
+ "\tDegree of dissociation after adding sodium acetate is 7.2e-05\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.14,Page no:57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C1=0.06 #concentration od acetic acid in g.mol/lit#\n",
+ "C2=0.04 #concentration of sodium acetate in g.mol/li#\n",
+ "K=1.8*10**-5 #dissociation constant of acetic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "H=K*C1/C2 \n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nThe pH of solution is\",round(pH,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The pH of solution is 4.5686\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.15,Page no:58"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "M1=0.2 #molarity of acetic acid#\n",
+ "M2=0.2 #molarity of sodium acetate#\n",
+ "K=1.8*10**-5 \n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "pH=-math.log10(K)+math.log10(M2/M1) #by using Henderson's equation#\n",
+ "\n",
+ "#Result\n",
+ "print\"The pH value of buffer solution is\",round(pH,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The pH value of buffer solution is 4.7447\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.16,Page no:58"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N=1.0/100.0 #normality of acetic acid#\n",
+ "V=1.0/N \n",
+ "k=1.8*10**-5 #dissociation constant of acetic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "#Part-a#\n",
+ "a=math.sqrt(k*V) #formula of degree of dissociation for weak acids#\n",
+ "H=a*N \n",
+ "\n",
+ "#Part-b#\n",
+ "n=0.01 #sodium acetate added in moles to one litre of acetic acid solution#\n",
+ "CH3COO=n \n",
+ "CH3COOH=n \n",
+ "H1=k*CH3COOH/CH3COO \n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"(a) H+ concentration of N/100 acetic acid solution is %.2e\"%H,\"g ion/litre\"\n",
+ "print\"\\n(b) H+ ion concentration in the solution after adding the sodium acetate is\",H1,\"g.ions/litre\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) H+ concentration of N/100 acetic acid solution is 4.24e-04 g ion/litre\n",
+ "\n",
+ "(b) H+ ion concentration in the solution after adding the sodium acetate is 1.8e-05 g.ions/litre\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.17,Page no:59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V=10 #volume of water in litres#\n",
+ "N1=0.10 #moles of HCN added in solution#\n",
+ "N2=0.10 #moles of NaCN added in solution#\n",
+ "K=7.2*10**-10 #dissociation constant of HCN#\n",
+ "CN=0.1 #CN- concentration#\n",
+ "HCN=0.1 #HCN concentration#\n",
+ "\n",
+ "#Calculation\n",
+ "H1=K*HCN/CN \n",
+ "H=H1/10**-10 \n",
+ "k=1*10**-14 #ionization constant of water#\n",
+ "\n",
+ "#Result\n",
+ "print\"H+ concentration in the solution is\",H*10**-10\n",
+ "OH=k/H1 \n",
+ "print\"\\nOH- concentration in the solution is %.1e\"%OH"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "H+ concentration in the solution is 7.2e-10\n",
+ "\n",
+ "OH- concentration in the solution is 1.4e-05\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.18,Page no:59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "K=1.7*10**-5 #dissociation constant of acid#\n",
+ "pH=3.77#pH value of buffer solution#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "M=pH+math.log10(K) \n",
+ "N=10**M #ratio of salt to acid#\n",
+ "L=1/N\n",
+ "\n",
+ "#Result\n",
+ "print\"The ratio of salt to acid in buffer is\",round(L)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of salt to acid in buffer is 10.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.19,Page no:59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "k=1.8*10**-5 #dissociation constant of acetic acid#\n",
+ "M=0.01 #molarity of acetic acid#\n",
+ "N=M*1 #normality of acetic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "V=1/N \n",
+ "a=math.sqrt(k*V)#degree of dissociation for weak acids#\n",
+ "H1=a/V \n",
+ "H=H1/10**-4 \n",
+ "pH=-math.log10(H1) \n",
+ "\n",
+ "#Result\n",
+ "print\"Degree of dissociation of solution is %.2e\"%a \n",
+ "print\"pH of the solution is\",round(pH ,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Degree of dissociation of solution is 4.24e-02\n",
+ "pH of the solution is 3.3724\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.20,Page no:60"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N1=0.2#concentration of acetic acid in g.molecule/lit#\n",
+ "N2=0.25#concentration of sodium acetate in g.molecule/lit#\n",
+ "K=1.8*10**-5#ionization constant of acetic acid at room temparature#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "pH1=-math.log10(K)+math.log10(N2/N1) \n",
+ "N=1.0#normality of HCl added#\n",
+ "V=0.5*10**-3#amount of HCl added in lit#\n",
+ "M=N*V\n",
+ "C1=N1+M#concentration of CH3COOH in moles/lit#\n",
+ "C2=N2-M#concentration of CH3COONa in moles/lit#\n",
+ "pH2=-math.log10(K)+math.log10(C2/C1)\n",
+ "pH=pH1-pH2 \n",
+ "\n",
+ "#Result\n",
+ "print\"pH value of the solution before adding HCl is\",round(pH1,4)\n",
+ "print\"\\nThe pH of the solution after adding HCl is\",round(pH2,4)\n",
+ "print\"\\nThe change of pH is\",round(pH,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "pH value of the solution before adding HCl is 4.8416\n",
+ "\n",
+ "The pH of the solution after adding HCl is 4.8397\n",
+ "\n",
+ "The change of pH is 0.002\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.21,Page no:61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "K=18*10**-6 #dissociation constant of NH4OH#\n",
+ "N1=0.1 #normality of NH4OH solution#\n",
+ "V=1.0/N1 \n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=math.sqrt(K*V)#since a is very small#\n",
+ "OH=a/V \n",
+ "W=2.0#weight of added NH4Cl in grams#\n",
+ "M=53.0#molecular weight of NH4Cl#\n",
+ "C=W/M \n",
+ "C1=0.1 #concentration of NH4OH in g.mol/lit#\n",
+ "OH2=K*C1/C\n",
+ "CH3COO=0.02 #g ion per litre\n",
+ "CH3COOH=0.2 #g mol per litre\n",
+ "H_plus=K*CH3COOH/CH3COO\n",
+ "pH=math.log10(H_plus)\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nThe concentration of hydroxyl ion before adding of NH4Cl is %.3e\"%OH,\"g ion per litre\"\n",
+ "print\"\\nThe concentration of hydroxyl ion after adding 2g of NH4Cl is %.1e\"%OH2,\"g ion per litre\"\n",
+ "print\"NOTE:Calculation mistake in book.(wrongly written as 48.6*10**-5,it should be 10**-6\"\n",
+ "print \"\\npH is\",round(-pH,4)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The concentration of hydroxyl ion before adding of NH4Cl is 1.342e-03 g ion per litre\n",
+ "\n",
+ "The concentration of hydroxyl ion after adding 2g of NH4Cl is 4.8e-05 g ion per litre\n",
+ "NOTE:Calculation mistake in book.(wrongly written as 48.6*10**-5,it should be 10**-6\n",
+ "\n",
+ "pH is 3.7447\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2.22,Page no:62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "ly=11.92 #equivalent conductvity of 0.02acetic acid solution in mho at 20C#\n",
+ "lih=360 #the equivalent ionic conductance of an infinite dillution of hydrogen ion in mho#\n",
+ "lic=40 #of acetate ion#\n",
+ "li=lih+lic #of acetic acid#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=ly/li #degree of dissociation#\n",
+ "N=0.02 #normality of acetic acid#\n",
+ "V=1/N \n",
+ "K=(a**2)/V \n",
+ "W=82 #mol.wt of CH3COONa#\n",
+ "M=8.2#amount of sodium acetate added in g per litre solution#\n",
+ "CH3COO=M/W \n",
+ "H=K*N/CH3COO \n",
+ "pH=-math.log10(H) \n",
+ "\n",
+ "#Result\n",
+ "print\"Dissociation constant of acetic acid is %.2e\"%K\n",
+ "print\"\\npH of the solution is\",round(pH,2)\n",
+ "print\"\\nNOTE:\\n(i)Calculation istake in calculation of K in book,exponent wrongly written as -6\"\n",
+ "print\"(ii)pH is wrongly calculated in book as 3.45\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Dissociation constant of acetic acid is 1.78e-05\n",
+ "\n",
+ "pH of the solution is 5.45\n",
+ "\n",
+ "NOTE:\n",
+ "(i)Calculation istake in calculation of K in book,exponent wrongly written as -6\n",
+ "(ii)pH is wrongly calculated in book as 3.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Textbook_of_Engineering_Chemistry/dChapter_3.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_3.ipynb new file mode 100755 index 00000000..5406df23 --- /dev/null +++ b/Textbook_of_Engineering_Chemistry/dChapter_3.ipynb @@ -0,0 +1,1100 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Chemical Kinetics & Catalysis"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from scipy.optimize import fsolve\n",
+ "#Variable declaration\n",
+ "A0=0.25 #[M]\n",
+ "At=0.15 #[M]\n",
+ "k=6.7*10**-4\n",
+ "\n",
+ "#Calculation\n",
+ "def f(t):\n",
+ " x=math.log10(A0/At)-(k*t)/2.303\n",
+ " return(x)\n",
+ "t=fsolve(f,1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Time taken to decrease concentration is%.3e\"%t[0],\"s (approx)\"\n",
+ "print\"NOTE: Approximate value taken in book\"\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time taken to decrease concentration is7.626e+02 s (approx)\n",
+ "NOTE: Approximate value taken in book\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2,Page no:71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "#Initially:\n",
+ "A=0.1 #Concentration of A\n",
+ "B=0.1 #Concentration of B\n",
+ "rate=5.5*10**-6 #Rate of reaction initial in M/s\n",
+ "x=2 #exponent\n",
+ "y=1 #exponent\n",
+ "\n",
+ "#Calculation\n",
+ "order=x+y #Reaction order\n",
+ "k=rate/((A**x)*(B**y)) #Rate law constant in [M**2/s]\n",
+ "\n",
+ "#Result\n",
+ "print\"The rate law is k=rate/(A**2)*(B)\"\n",
+ "print\"\\nRate constant is %.2e\"%k,\"M**2/s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rate law is k=rate/(A**2)*(B)\n",
+ "\n",
+ "Rate constant is 5.50e-03 M**2/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3,Page no:76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "\n",
+ "T=[700.0,730.0,760.0,790.0] #Temperature in [K]\n",
+ "k=[0.011,0.035,0.195,0.343] #rate constants in [L/mol s]\n",
+ "import numpy \n",
+ "onebyT=numpy.reciprocal(T) #Reciprocal of temperature\n",
+ "log_k=numpy.log10(k) #log of k\n",
+ "R=8.30*10**-3 #[kJ/kmol]\n",
+ "\n",
+ "#Calculation\n",
+ "%pylab inline\n",
+ "import matplotlib.pyplot as plt\n",
+ "plt.plot(onebyT,log_k)\n",
+ "plt.ylabel('$log k$')\n",
+ "plt.xlabel('$1/T$')\n",
+ "plt.title('1/T vs log k\\n')\n",
+ "slope,intercept=np.polyfit(onebyT,log_k,1)\n",
+ "print\"Slope is\",slope\n",
+ "plt.show()\n",
+ "slope=-9.9*10**3 #Slope given in book [K]\n",
+ "E_star=slope*(-2.303*R) #Activation energy in [kJ/mol]\n",
+ "\n",
+ "#Result\n",
+ "print\"Activation energy of reaction is\",round(E_star,1),\"kJ/mol\"\n",
+ "print\"NOTE that in book slope is approximated as 9.9*10**3 K\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Populating the interactive namespace from numpy and matplotlib\n",
+ "Slope is"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " -9663.82327366\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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o2iwsgJdfFntFnTgBeHkBP/wgd1VU3zj0RES1IknAtm1ikd6oUUBUFNCihdxV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\ntGrVqvTgIiKl4F8kkR61adM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+ "text": [
+ "<matplotlib.figure.Figure at 0x7a19438>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Activation energy of reaction is 189.2 kJ/mol\n",
+ "NOTE that in book slope is approximated as 9.9*10**3 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.2,Page no:86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T1=10.0 #in min\n",
+ "T2=20.0 #in min\n",
+ "a=25.0 #amount of KMnO4 in ml at t=0min#\n",
+ "a1=20.0 #amount of KMnO4 in ml at t=10min or a-x value at t=10#\n",
+ "a2=15.7 #a-x value at t=20min#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "k1=(2.303/T1)*math.log10(a/a1) #formula of rate constant for first order reaction#\n",
+ "print\"At t=10min rate constant k=\",round(k1,5),\"/min\"\n",
+ "k2=(2.303/T2)*math.log10(a/a2) #rate constant formula#\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nAt t=20min rate constant k=\",round(k2,5),\"/min\" \n",
+ "print\"\\nNOTE:Calculation mistake in book\"\n",
+ "print\"\\nIf we calculate the rate constant at other t values we will see that k values are almost constnat\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "At t=10min rate constant k= 0.02232 /min\n",
+ "\n",
+ "At t=20min rate constant k= 0.02326 /min\n",
+ "\n",
+ "NOTE:Calculation mistake in book\n",
+ "\n",
+ "If we calculate the rate constant at other t values we will see that k values are almost constnat\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.3,Page no:87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T=40.5 #in min#\n",
+ "R1=25.0 #percentage of decomposed reactant#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "R2=100.0-R1 #percentage of left out reactant which is a-x value#\n",
+ "R3=100.0/R2 #value of a/(a-x)#\n",
+ "K=(2.303/T)*math.log10(R3) #formula of rate constant for first order reaction#\n",
+ "\n",
+ "#Result\n",
+ "print\"The rate constant of the reaction is %.2e\"%K,\"/min\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rate constant of the reaction is 7.10e-03 /min\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.4,Page no:87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "pi=0.0 #pressure of N2 at t=0#\n",
+ "t1=2.0 \n",
+ "t2=8.0 \n",
+ "t3=16.0 \n",
+ "t4=24.0 \n",
+ "t5=50.0 \n",
+ "pf=34.0 #pressure of N2 at infinity#\n",
+ "p1=1.6 #pressure of N2 at t=2min#\n",
+ "p2=6.2 #pressure of N2 at t=8min#\n",
+ "p3=11.2 #pressure Of N2 at t=16min#\n",
+ "p4=15.5 #pressure of N2 at t=24min#\n",
+ "p5=24.4 #pressure of N2 at t=50min#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=pf-pi #value of a#\n",
+ "a1=pf-p1 #a-x value at t=2min#\n",
+ "a2=pf-p2 #a-x value at t=8min#\n",
+ "a3=pf-p3 #a-x value at t=16min#\n",
+ "a4=pf-p4 #a-x value at t=24min#\n",
+ "a5=pf-p5 #a-x value at t=50min#\n",
+ "k1=(1/t1)*math.log(a/a1) #rate constant at t=2min#\n",
+ "k2=(1/t2)*math.log(a/a2) #rate constant at t=8min#\n",
+ "k3=(1/t3)*math.log(a/a3) #rate constant at t=16min#\n",
+ "k4=(1/t4)*math.log(a/a4) #rate constant at t=24min#\n",
+ "k5=(1/t5)*math.log(a/a5) #rate constant at t=50min#\n",
+ "k=(k1+k2+k3+k4+k5)/5 \n",
+ "\n",
+ "#Result\n",
+ "print\"Time(min): 2\\t\\t8\\t\\t16\\t\\t24\\t\\t50\"\n",
+ "print\"k1 per min %.2e\\t\"%k1,\"%.2e\\t\"%k2,\"%.2e\\t\"%k3,\"%.3e\\t\"%k4,\"%.2e\"%k5\n",
+ "print\"\\nAverage rate constant is %.3e\"%k,\"min^-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time(min): 2\t\t8\t\t16\t\t24\t\t50\n",
+ "k1 per min 2.41e-02\t2.52e-02\t2.50e-02\t2.536e-02\t2.53e-02\n",
+ "\n",
+ "Average rate constant is 2.498e-02 min^-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example no:3.5.,Page no:88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=0 \n",
+ "t2=4.89 \n",
+ "t3=10.07 \n",
+ "t4=23.66 \n",
+ "v1=47.65 #ml of alkali used at t=0min or a value#\n",
+ "v2=38.92 #ml of alkali used or a-x value at t=4.89min#\n",
+ "v3=32.62 #ml of alkali used or a-x value at t=10.07min#\n",
+ "v4=22.58 #ml of alkali used or a-x value at t=23.66min#\n",
+ "\n",
+ "#Calculation\n",
+ "x2=v1-v2 #x value at t=4.89min#\n",
+ "x3=v1-v3 #x value at t=10.07min#\n",
+ "x4=v1-v4 #x value at t=23.66min#\n",
+ "k22=(1/t2)*(x2/(v1*v2)) #rate constant for second order equation#\n",
+ "\n",
+ "#Result\n",
+ "print\"Rate constant k2 value at t=\",t2,\"min is \",round(k22,6),\"/min\"\n",
+ "k23=(1/t3)*(x3/(v1*v3)) #rate constant for second order equation#\n",
+ "print\"\\nRate constant k2 value at t=\",t3,\"min is \",round(k23,6),\"/min\"\n",
+ "k24=(1/t4)*(x4/(v1*v4)) #rate constant for second order equation#\n",
+ "print\"\\nRate constant k2 value at t=\",t4,\"min is\",round(k24,5),\"/min\" \n",
+ "print\"\\nAlmost constant values of k2 indicate that reaction is second order\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate constant k2 value at t= 4.89 min is 0.000963 /min\n",
+ "\n",
+ "Rate constant k2 value at t= 10.07 min is 0.00096 /min\n",
+ "\n",
+ "Rate constant k2 value at t= 23.66 min is 0.00098 /min\n",
+ "\n",
+ "Almost constant values of k2 indicate that reaction is second order\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.6,Page no:88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t=1590 #half life of given radio active element in years#\n",
+ "\n",
+ "#Calculation\n",
+ "k=0.693/t #formula of decay constant for first order reactions#\n",
+ "\n",
+ "#Result\n",
+ "print\"the value of decay constant is \",round(k,6),\"/year\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of decay constant is 0.000436 /year\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.8,Page no:89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=5.0 \n",
+ "t2=15.0 \n",
+ "t3=25.0 \n",
+ "t4=45.0 \n",
+ "a=37.0 #volume of KMnO4 in cm**3 at t=0 or value of a#\n",
+ "a1=29.8 #volume of KMnO4 in cm**3 or a-x value at t=5min#\n",
+ "a2=19.6 #volume of KMnO4 in cm**3 or a-x value at t=15min#\n",
+ "a3=12.3 #volume of KMnO4 in cm**3 or a-x value at t=25min#\n",
+ "a4=5.0 #volume of KMnO4 in cm**3 or a-x value at t=45min#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "k1=(2.303/t1)*math.log10(a/a1) \n",
+ "print\"\\nRate constant value at t=5min is %.3e\"%k1,\"min**-1\"\n",
+ "k2=(2.303/t2)*math.log10(a/a2) \n",
+ "print\"\\nRate constant value at t=15min is %.3e\"%k2,\"min**-1\"\n",
+ "k3=(2.303/t3)*math.log10(a/a3) \n",
+ "print\"\\nRate constant value at t=25min is %.3e\"%k3,\"min**-1\"\n",
+ "k4=(2.303/t4)*math.log10(a/a4) \n",
+ "print\"\\nRate constant value at t=45min is %.3e\"%k4,\"min**-1\"\n",
+ "print\"\\nAs the different values of k are nearly same,the reaction is of first oredr.\"\n",
+ "k=(k1+k2+k3+k4)/4.0 \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nThe average value of k is %.3e\"%k,\"min**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Rate constant value at t=5min is 4.329e-02 min**-1\n",
+ "\n",
+ "Rate constant value at t=15min is 4.237e-02 min**-1\n",
+ "\n",
+ "Rate constant value at t=25min is 4.406e-02 min**-1\n",
+ "\n",
+ "Rate constant value at t=45min is 4.449e-02 min**-1\n",
+ "\n",
+ "As the different values of k are nearly same,the reaction is of first oredr.\n",
+ "\n",
+ "The average value of k is 4.355e-02 min**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.9,Page no:89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "k=6.0*10**-4 #rate constant of first order decomposition of N2O5 in CCl4 in /min#\n",
+ "\n",
+ "#Calculation\n",
+ "#Part-a#\n",
+ "k1=k/60.0 \n",
+ "#Part-b#\n",
+ "t=0.693/k \n",
+ "\n",
+ "#Result\n",
+ "print\"(a) Rate constant in terms of seconds is \",k1,\"/s\"\n",
+ "print\"\\n(b) Half life of the reaction is %.2e\"%t,\"min\"\n",
+ "print\"NOTE:Slight rounding off in book in final answer\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a) Rate constant in terms of seconds is 1e-05 /s\n",
+ "\n",
+ "(b) Half life of the reaction is 1.15e+03 min\n",
+ "NOTE:Slight rounding off in book in final answer\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.10,Page no:90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=40.0 \n",
+ "t2=80.0 \n",
+ "t3=120.0 \n",
+ "t4=160.0 \n",
+ "t5=240.0 \n",
+ "vi=0.0 #volume of oxygen collected at constant pressure in ml at t=0#\n",
+ "v1=15.6 #volume of oxygen collected at constant pressure in ml at t=40#\n",
+ "v2=27.6 #volume of oxygen collected at constant pressure in ml at t=80#\n",
+ "v3=37.7 #volume of oxygen collected at constant pressure in ml at t=120#\n",
+ "v4=45.8 #volume of oxygen collected at constant pressure in ml at t=160#\n",
+ "v5=58.3 #volume of oxygen collected at constant pressure in ml at t=200#\n",
+ "vf=84.6 #volume of oxygen collected at constant pressure in ml at t=infinity#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=vf-vi #the initial concentration of N2O5 in solution i.e a#\n",
+ "a1=vf-v1 #a-x value at t=40min#\n",
+ "a2=vf-v2 #a-x value at t=80min#\n",
+ "a3=vf-v3 #a-x value at t=120min#\n",
+ "a4=vf-v4 #a-x value at t=160min#\n",
+ "a5=vf-v5 #a-x value at t=200min#\n",
+ "k1=(1.0/t1)*math.log(a/a1) \n",
+ "k2=(1.0/t2)*math.log(a/a2) \n",
+ "k3=(1.0/t3)*math.log(a/a3) \n",
+ "k4=(1.0/t4)*math.log(a/a4) \n",
+ "k5=(1.0/t5)*math.log(a/a5) \n",
+ "\n",
+ "#Result\n",
+ "print\"Time(min): 40\\t\\t80\\t\\t120\\t\\t160\\t\\t240\"\n",
+ "print\"k1 per min %.2e\\t\"%k1,\"%.2e\\t\"%k2,\"%.2e\\t\"%k3,\"%.3e\\t\"%k4,\"%.2e\"%k5\n",
+ "print\"\\nNOTE:Calculation mistake in book in calculating a4,it should be 38.8\"\n",
+ "print\"\\nAs k value is fairly constant the reaction is first order\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Time(min): 40\t\t80\t\t120\t\t160\t\t240\n",
+ "k1 per min 5.10e-03\t4.94e-03\t4.92e-03\t4.872e-03\t4.87e-03\n",
+ "\n",
+ "NOTE:Calculation mistake in book in calculating a4,it should be 38.8\n",
+ "\n",
+ "As k value is fairly constant the reaction is first order\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.11,Page no:90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=120.0 #time in sec#\n",
+ "t2=240.0 \n",
+ "t3=530.0 \n",
+ "t4=600.0 \n",
+ "a=0.05 #initial concentration#\n",
+ "x1=32.95 #extent of reaction or x value at t=120sce#\n",
+ "x2=48.8 #extent of reaction or x value at t=240sce#\n",
+ "x3=69.0 #extent of reaction or x value at t=530sce#\n",
+ "x4=70.35 #extent of reaction or x value at t=600sce#\n",
+ "a1=100.0-x1 #extent of left out or a-x value at t=120sec#\n",
+ "a2=100.0-x2 #extent of left out or a-x value at t=240sec#\n",
+ "a3=100.0-x3 #extent of left out or a-x value at t=530sec#\n",
+ "a4=100.0-x4 #extent of left out or a-x value at t=600sec#\n",
+ "\n",
+ "#Calculation\n",
+ "k1=(1.0/(a*t1))*(x1/a1) \n",
+ "print\"Rate constant value at t=120sec is %.2e\"%k1,\"dm**3 mol**-1.s**-1\"\n",
+ "k2=(1.0/(a*t2))*(x2/a2) \n",
+ "print\"\\nRate constant value at t=240sec is %.2e\"%k2,\"dm**3 mol**-1.s**-1\"\n",
+ "k3=(1.0/(a*t3))*(x3/a3) \n",
+ "print\"\\nRate constant value at t=530sec is %.2e\"%k3,\"dm**3 mol**-1.s**-1\"\n",
+ "k4=(1.0/(a*t4))*(x4/a4) \n",
+ "print\"\\nRate constant value at t=600sec is %.2e\"%k4,\"dm**3 mol**-1.s**-1\"\n",
+ "k=(k1+k2+k3+k4)/4.0 \n",
+ "\n",
+ "#Result\n",
+ "print\"\\n\\nAverage value of rate constant is %.1e\"%k,\"dm**3 mol**-1.s**-1\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate constant value at t=120sec is 8.19e-02 dm**3 mol**-1.s**-1\n",
+ "\n",
+ "Rate constant value at t=240sec is 7.94e-02 dm**3 mol**-1.s**-1\n",
+ "\n",
+ "Rate constant value at t=530sec is 8.40e-02 dm**3 mol**-1.s**-1\n",
+ "\n",
+ "Rate constant value at t=600sec is 7.91e-02 dm**3 mol**-1.s**-1\n",
+ "\n",
+ "\n",
+ "Average value of rate constant is 8.1e-02 dm**3 mol**-1.s**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.13,Page no:91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=75.0 #time in min#\n",
+ "t2=119.0 \n",
+ "t3=183.0 \n",
+ "vi=9.62 #volume of alkali used in ml at t=0min#\n",
+ "v1=12.10 #volume of alkali used in ml at t=75min#\n",
+ "v2=13.10 #volume of alkali used in ml at t=119min#\n",
+ "v3=14.75 #volume of alkali used in ml at t=183min#\n",
+ "vf=21.05 #volume of alkali used in ml at t=infinity#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "k1=(1.0/t1)*math.log((vf-vi)/(vf-v1)) #formula of rate constant for first order reactions#\n",
+ "k2=(1.0/t2)*math.log((vf-vi)/(vf-v2)) \n",
+ "k3=(1.0/t3)*math.log((vf-vi)/(vf-v3)) \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nRate constant value at t=75min is \",round(k1,6),\"min**-1\"\n",
+ "print\"\\nRate constant value at t=119min is \",round(k2,6),\"min**-1\"\n",
+ "print\"\\nRate constant value at t=183min is \",round(k3,6),\"min**-1\"\n",
+ "\n",
+ "print\"\\nNOTE:Slight Calculation mistake in book in k calculation above\" \n",
+ "print\"\\nAn almost constant value of k shows that the hydrolysis of ethyl acetateis a first order reaction\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Rate constant value at t=75min is 0.003261 min**-1\n",
+ "\n",
+ "Rate constant value at t=119min is 0.003051 min**-1\n",
+ "\n",
+ "Rate constant value at t=183min is 0.003255 min**-1\n",
+ "\n",
+ "NOTE:Slight Calculation mistake in book in k calculation above\n",
+ "\n",
+ "An almost constant value of k shows that the hydrolysis of ethyl acetateis a first order reaction\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.14,Page no:92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t=15 #the half time of given first order reaction in min#\n",
+ "k=0.693/t #formula of rate constant#\n",
+ "print\"The rate constant value of the given first order reaction is is\",k,\"min**-1\"\n",
+ "a=100 #percentage of initial concentration#\n",
+ "x=80 #percentage of completed reaction#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a1=a-x #percentage of left out concentration#\n",
+ "t1=(2.303/k)*(math.log10(a/a1)) #formula to find time taken#\n",
+ "t2=t1*60 \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nThe time taken to complete 80 percentage of the reaction is \",round(t1,2),\"min or\",round(t2),\"sec\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rate constant value of the given first order reaction is is 0.0462 min**-1\n",
+ "\n",
+ "The time taken to complete 80 percentage of the reaction is 34.84 min or 2091.0 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.15,Page no:92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=6.18 #time in min#\n",
+ "t2=18.0 \n",
+ "t3=27.05 \n",
+ "ri=24.09 #rotation in degrees when t=0min#\n",
+ "r1=21.4 #rotation in degrees when t=6.18min#\n",
+ "r2=17.7 #rotation in degrees when t=18min#\n",
+ "r3=15.0 #rotation in degrees when t=27.05min#\n",
+ "rf=-10.74 #rotation in degrees when t=infinity#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=ri-rf #a value#\n",
+ "a1=r1-rf #a-x value at t=6.18min#\n",
+ "a2=r2-rf #a-x value at t=18min#\n",
+ "a3=r3-rf #a-x value at t=27.05min#\n",
+ "k1=(2.303/t1)*math.log10(a/a1) \n",
+ "k2=(2.303/t2)*math.log10(a/a2) \n",
+ "k3=(2.303/t3)*math.log10(a/a3) \n",
+ "\n",
+ "#Result\n",
+ "print\"Rate constant value at t=\",t1,\"min %.3e\"%k1,\"min**-1\"\n",
+ "print\"\\nRate constant value at t=\",t2,\"min %.3e\"%k2,\"min**-1\"\n",
+ "print\"\\nRate constant value at t=\",t3,\"min %.3e\"%k3,\"min**-1\"\n",
+ "\n",
+ "print\"\\nNOTE:Again,Calculation mistake in book\"\n",
+ "print\"\\nSince rate constant values are nearly same,hence reaction is of first order\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate constant value at t= 6.18 min 1.301e-02 min**-1\n",
+ "\n",
+ "Rate constant value at t= 18.0 min 1.126e-02 min**-1\n",
+ "\n",
+ "Rate constant value at t= 27.05 min 1.118e-02 min**-1\n",
+ "\n",
+ "NOTE:Again,Calculation mistake in book\n",
+ "\n",
+ "Since rate constant values are nearly same,hence reaction is of first order\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.16,Page no:93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "t1=10.0#time in min#\n",
+ "t2=20.0 \n",
+ "t3=30.0 \n",
+ "t4=40.0 \n",
+ "ri=32.4 #rotation in degrees when t=0min#\n",
+ "r1=28.8 #rotation in degrees when t=10min#\n",
+ "r2=25.5 #rotation in degrees when t=20min#\n",
+ "r3=22.4 #rotation in degrees when t=30min#\n",
+ "r4=19.6 #rotation in degrees when t=40min#\n",
+ "rf=-11.1 #rotation in degrees when t=0min#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "a=ri-rf #a value#\n",
+ "a1=r1-rf #a-x value at t=10min#\n",
+ "a2=r2-rf #a-x value at t=20min#\n",
+ "a3=r3-rf #a-x value at t=30min#\n",
+ "a4=r4-rf #a-x value at t=40min#\n",
+ "k1=(1.0/t1)*math.log(a/a1) \n",
+ "k2=(1.0/t2)*math.log(a/a2) \n",
+ "k3=(1.0/t3)*math.log(a/a3) \n",
+ "k4=(1.0/t4)*math.log(a/a4) \n",
+ "\n",
+ "#Result\n",
+ "print\"Rate constant value at t=10min \",round(k1,6),\"min**-1\"\n",
+ "print\"\\nRate constant value at t=20min \",round(k2,6),\"min**-1\"\n",
+ "print\"\\nRate constant value at t=30min \",round(k3,6),\"min**-1\"\n",
+ "print\"\\nRate constant value at t=40min \",round(k4,6),\"min**-1\"\n",
+ "print\"\\nSince rate constant values are nearly same,hence inversion of sucrose is of first order\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate constant value at t=10min 0.008638 min**-1\n",
+ "\n",
+ "Rate constant value at t=20min 0.008636 min**-1\n",
+ "\n",
+ "Rate constant value at t=30min 0.008707 min**-1\n",
+ "\n",
+ "Rate constant value at t=40min 0.008712 min**-1\n",
+ "\n",
+ "Since rate constant values are nearly same,hence inversion of sucrose is of first order\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.17,Page no:93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T1=27.0 #initial temparature in C#\n",
+ "T1=T1+273 #in kelvin#\n",
+ "Tr=10.0 #rise in temparature#\n",
+ "T2=T1+Tr #final temparature in kelvin#\n",
+ "r=2.0 #ratio of final to initial rates of chemical reactions(k1/k2)#\n",
+ "R=8.314 #value of constant R in J/K.mol#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "E=math.log(r)*R*305*295/Tr #from equation k=A*e**(-E/R*T)#\n",
+ "\n",
+ "#Result\n",
+ "print\"Activation energy of the reaction is \",round(E/1000,2),\"kJ/mol\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Activation energy of the reaction is 51.85 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.18,Page no:94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "k=4.5*10**3 #value of k in /sec of a first order reaction at 1C#\n",
+ "E=58*10**3 #activation energy in J/mol#\n",
+ "T=1 #temperature in C#\n",
+ "T1=T+273 #in kelvin#\n",
+ "R=8.314 #value of constant R in J/K.mol#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "lA=math.log10(k)+(E/(2.303*R*T1)) \n",
+ "k1=10**4 #value of k in /sec at some temperature#\n",
+ "a=math.log10(k1) \n",
+ "b=lA-a \n",
+ "T2=E/(2.303*R*b) \n",
+ "\n",
+ "#Result\n",
+ "print\"The temperature at which k=1*10**4/sec is\",round(T2),\"K\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature at which k=1*10**4/sec is 283.0 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.19,Page no:94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T1=300.0 #temperature in kelvin#\n",
+ "t1=20.0 #half time of chemical reaction in min at T=300K#\n",
+ "T2=350.0 #temperature in kelvin#\n",
+ "t2=5.0 #half time of chemical reaction in min at T=350K#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "k1=0.6932/t1 \n",
+ "k2=0.6932/t2 \n",
+ "l=math.log10(k2/k1) \n",
+ "R=8.314 #value of constant R in J/K.mol#\n",
+ "E=l*2.303*R*T1*T2/(T2-T1) \n",
+ "\n",
+ "#Result\n",
+ "print\"Rate constant of the reaction at T=300k is \",k1,\"/min\" \n",
+ "print\"\\nRate constant of the reaction at T=350k is \",k2,\"/min\"\n",
+ "print\"\\nActivation energy of the reaction is\",E,\"J/mol OR\",round(E/1000,1),\"kJ/mol\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate constant of the reaction at T=300k is 0.03466 /min\n",
+ "\n",
+ "Rate constant of the reaction at T=350k is 0.13864 /min\n",
+ "\n",
+ "Activation energy of the reaction is 24208.2291076 J/mol OR 24.2 kJ/mol\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.20,Page no:94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "R=8.314 #value of constant R in J/K.mol#\n",
+ "H=1.25*10**4 #value of E/(2.303*R).It is given in the question#\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "#Part-i#\n",
+ "E=H*2.303*R \n",
+ "la=14.34 #value of math.log(a)#\n",
+ "T=670 #temperature in kelvin#\n",
+ "#Part-ii#\n",
+ "lk=la-(H/T) \n",
+ "k=10**lk \n",
+ "\n",
+ "#Result\n",
+ "print\"(i) Activation energy is %.2e\"%E,\"J mol**-1 or\",round(E/1000),\"kJ mol**-1\"\n",
+ "print\"\\n(ii) Rate constant at 670K is %.1e\"%k,\"s**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Activation energy is 2.39e+05 J mol**-1 or 239.0 kJ mol**-1\n",
+ "\n",
+ "(ii) Rate constant at 670K is 4.8e-05 s**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3.21,Page no:95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Ti=27.0 #given temperature in C#\n",
+ "T1=Ti+273.0 #in kelvin#\n",
+ "t1=T1-5\n",
+ "Tr=10.0 #rise in temperature#\n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "T2=T1+Tr \n",
+ "t2=T2-5\n",
+ "k=3.0 #value of k1/k2#\n",
+ "R=8.314 #value of constant R in J/K.mol#\n",
+ "E=math.log(k)*R*t1*t2/(T2-T1) \n",
+ "\n",
+ "#Result\n",
+ "print\"Activation energy of the reaction is\",round(E),\"J mol**-1 or\",round(E/1000,2),\"kJ mol**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Activation energy of the reaction is 82182.0 J mol**-1 or 82.18 kJ mol**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Textbook_of_Engineering_Chemistry/dChapter_6.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_6.ipynb new file mode 100755 index 00000000..7428ce43 --- /dev/null +++ b/Textbook_of_Engineering_Chemistry/dChapter_6.ipynb @@ -0,0 +1,92 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 6:Lubricants"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "d=760.0 #viscocity of Pennysylvanian oil in s at 37C#\n",
+ "a=528.0 #viscocity of lubricating oil in s at 37C#\n",
+ "c=480.0 #viscocity of Gulf oil in s at 37C#\n",
+ "\n",
+ "#Calculation\n",
+ "V=((d-a)/(d-c))*(100) #formula of viscocity index#\n",
+ "\n",
+ "#Result\n",
+ "print\"Viscocity index of the lubricating oil is \",round(V ,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Viscocity index of the lubricating oil is 82.9\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2,Page no:144"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "s=0.86 #specific gravity of lubricating oil#\n",
+ "\n",
+ "#Calculation\n",
+ "A=(141.5/s)-131.5 #formula of API gravity#\n",
+ "\n",
+ "#Result\n",
+ "print\"The gravity of lubricating oil is \",round(A)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The gravity of lubricating oil is 33.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Textbook_of_Engineering_Chemistry/dChapter_7.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_7.ipynb new file mode 100755 index 00000000..3854b5db --- /dev/null +++ b/Textbook_of_Engineering_Chemistry/dChapter_7.ipynb @@ -0,0 +1,850 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7:Water Chemistry"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:152"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "MgSO4=60.0 #[mg]\n",
+ "M_MgSO4=64.0 #Molecular weight of MgSO4\n",
+ "M_CaCO3=48.0 #Molecular wt of CaCO3\n",
+ "m_mgso4=120.0 #Weight of MgSO4 eq to CaCO3\n",
+ "m_caco3=100.0 #Weight of CaCO3 eq to MgSO4\n",
+ "\n",
+ "#Calculation\n",
+ "hard=(m_caco3/m_mgso4)*MgSO4 #Hardness of water in mg\n",
+ "\n",
+ "#Result\n",
+ "print\"Hardness of water is \",hard,\"ppm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Hardness of water is 50.0 ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.1,Page no:172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=16.2 #Ca(HCO3)2 in water in mg/lit#\n",
+ "W2=7.3 #MgHCO3 in water in mg/lit#\n",
+ "W3=13.6 #CaSO4 in water in mg/lit#\n",
+ "W4=9.5 #MgCl2 in water in mg/lit#\n",
+ "M1=100/162.0 #multiplication factor of Ca(HCO3)2#\n",
+ "M2=100/146.0 #multiplication factor of MgHCO3#\n",
+ "M3=100/136.0 #multiplication factor of CaSO4#\n",
+ "M4=100/95.0 #multiplication factor of MgCl2#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 #Ca(HCO3)2 in terms of CaCO3 or #\n",
+ "P2=W2*M2 #MgHCO3 in terms of CaCO3 or #\n",
+ "P3=W3*M3 #CaSO4 in terms of CaCO3 or #\n",
+ "P4=W4*M4 #MgCl2 in terms of CaCO3 or #\n",
+ "T=P1+P2 #Temporary hardness\n",
+ "P=P3+P4 #Permanent hardness\n",
+ "To=T+P #Total hardness\n",
+ "\n",
+ "#Result\n",
+ "print\"Temporary hardness is\",T,\"mg/l or ppm\"\n",
+ "print\"\\nPermanant hardness is \",P,\"mg/l or ppm\"\n",
+ "print\"\\nTotal hardness is \",To,\"mg/l or ppm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Temporary hardness is 15.0 mg/l or ppm\n",
+ "\n",
+ "Permanant hardness is 20.0 mg/l or ppm\n",
+ "\n",
+ "Total hardness is 35.0 mg/l or ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.2,Page no:172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "F=56.0 #atomic weight of ferrus#\n",
+ "S=32.0 #atomic weight of sulphur#\n",
+ "O=16.0 #atomic weight of oxygen#\n",
+ "Ca=40.0 #atomic weight of calsium#\n",
+ "C=12.0 #atomic weight of carbon#\n",
+ "\n",
+ "#Calculation\n",
+ "W1=136\n",
+ "P=210.5 #required ppm of hardness#\n",
+ "B=(W1/100.0)*P \n",
+ "\n",
+ "#Result\n",
+ "print\"Required FeSO4 for 100ppm of hardness is\",W1,\"ppm pf FeSO4\"\n",
+ "print\"\\nRequired FeSO4 for 210.5ppm of hardness is \",round(B,1),\"ppm of FeSO4\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Required FeSO4 for 100ppm of hardness is 136 ppm pf FeSO4\n",
+ "\n",
+ "Required FeSO4 for 210.5ppm of hardness is 286.3 ppm of FeSO4\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.3,Page no:173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=162.0 #Ca(HCO3)2 in water in mg/lit#\n",
+ "W2=73.0 #MgHCO3 in water in mg/lit#\n",
+ "W3=136.0 #CaSO4 in water in mg/lit#\n",
+ "W4=95.0 #MgCl2 in water in mg/lit#\n",
+ "W5=111.0 #CaCl2 in water in mg/lit#\n",
+ "W6=100.0 #NaCl in water in mg/lit#\n",
+ "M1=100/162.0 #multiplication factor of Ca(HCO3)2#\n",
+ "M2=100/146.0 #multiplication factor of MgHCO3#\n",
+ "M3=100/136.0 #multiplication factor of CaSO4#\n",
+ "M4=100/95.0 #multiplication factor of MgCl2#\n",
+ "M5=100/111.0 #multiplication factor of CaCl2#\n",
+ "M6=100/100.0 #multiplication factor of NaCl#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 #Ca(HCO3)2 in terms of CaCO3 or #\n",
+ "P2=W2*M2 #MgHCO3 in terms of CaCO3 or #\n",
+ "P3=W3*M3 #CaSO4 in terms of CaCO3 or #\n",
+ "P4=W4*M4 #MgCl2 in terms of CaCO3 or #\n",
+ "P5=W5*M5 #CaCl2 in terms of CaCO3 or #\n",
+ "T=P1+P2 \n",
+ "P=P3+P4+P5 \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nTemporary hardness is\",T,\"mg/l or ppm\" \n",
+ "print\"\\nPermanant hardness is\",P,\"mg/l or ppm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Temporary hardness is 150.0 mg/l or ppm\n",
+ "\n",
+ "Permanant hardness is 300.0 mg/l or ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.4,Page no:173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N=0.08 #normality of MgSO4#\n",
+ "V1=12.5 #volume of MgSO4 in ml#\n",
+ "V2=100 #volume of water sample#\n",
+ "\n",
+ "#Calculation\n",
+ "M=N/2 #molarity of MgSO4#\n",
+ "N1=(M*12.5)/1000 #no of moles of MgSO4 in 100 ml water#\n",
+ "N2=(N1*1000)/100 #no of moles of MgSO4 in one litre water#\n",
+ "W=100 #molecular weight of CaCO3\n",
+ "W1=N2*W*1000 #MgSO4 in terms of CaCO3 in mg/lit#\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nThe hardness due to MgSO4 is \",W1,\"mg/l CaCO3 or ppm of CaCO3\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The hardness due to MgSO4 is 500.0 mg/l CaCO3 or ppm of CaCO3\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.5,Page no:173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=144.0 #MgCO3 in water in mg/lit#\n",
+ "W2=25.0 #CaCO3 in water in mg/lit#\n",
+ "W3=111.0 #CaCl2 in water in mg/lit#\n",
+ "W4=95.0 #MgCl2 in water in mg/lit#\n",
+ "M1=100/84.0 #multiplication factor of MgCO3#\n",
+ "M2=100/100.0 #multiplication factor of CaCO3#\n",
+ "M3=100/111.0 #multiplication factor of CaCl2#\n",
+ "M4=100/95.0 #multiplication factor of MgCl2#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 #MgCO3 in terms of CaCO3 or ppm#\n",
+ "P2=W2*M2 #CaCO3 in terms of CaCO3 or ppm#\n",
+ "P3=W3*M3 #CaCl2 in terms of CaCO3 or ppm#\n",
+ "P4=W4*M4 #MgCl2 in terms of CaCO3 or ppm#\n",
+ "V=50000 #volume of water in lit#\n",
+ "L=0.74*(2*P1+P2+P4)*V \n",
+ "S=1.06*(P1+P3+P4)*V \n",
+ "\n",
+ "#Result\n",
+ "print\"Requirement of lime is \",L,\"mg=\",round(L/1000000,1),\"kg\" \n",
+ "print\"\\nRequirement of soda is \",S,\"mg=\",round(S/1000000,1),\"kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Requirement of lime is 17310714.2857 mg= 17.3 kg\n",
+ "\n",
+ "Requirement of soda is 19685714.2857 mg= 19.7 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.6,Page no:174"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=12.0 #Mg2+ in water in ppm or mg/l#\n",
+ "W2=40.0 #Ca2+ in water in ppm or mg/l#\n",
+ "W3=164.7 #HCO3- in water in ppm or mg/l#\n",
+ "W4=30.8 #CO2 in water in ppm or mg/l#\n",
+ "M1=100.0/24.0 #multiplication factor of Mg2+#\n",
+ "M2=100.0/40.0 #multiplication factor of Mg2+#\n",
+ "M3=100.0/61.0 #multiplication factor of Mg2+#\n",
+ "M4=100.0/44.0 #multiplication factor of Mg2+#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 # in terms of CaCO3#\n",
+ "P2=W2*M2 # in terms of CaCO3#\n",
+ "P3=W3*M3/2 # in terms of CaCO3#\n",
+ "P4=W4*M4 # in terms of CaCO3#\n",
+ "V=50000.0#volume of water in lit#\n",
+ "L=0.74*(P1+P3+P4)*V \n",
+ "\n",
+ "#Result\n",
+ "print\"Lime required is %fmg\",round(L/10**6,1),\"kg\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lime required is %fmg 9.4 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.7,Page no:174"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=160.0 #Ca2+ in water in mg/l or ppm#\n",
+ "W2=72.0 #Mg2+ in water in mg/l or ppm#\n",
+ "W3=732.0 #HCO3- in water in mg/l or ppm#\n",
+ "W4=44.0 #CO2 in water in mg/l or ppm#\n",
+ "W5=16.4 #NaAlO2 in water in mg/l or ppm#\n",
+ "W6=30.0 #(CO3)2- in water in mg/l or ppm#\n",
+ "W7=17.0 #OH- in water in mg/l or ppm#\n",
+ "\n",
+ "#Calculation\n",
+ "M1=100/40.0 #multiplication factor of Ca2+#\n",
+ "M2=100/24.0 #multiplication factor of Ca2+#\n",
+ "M3=100/(61.0*2.0) #multiplication factor of Ca2+#\n",
+ "M4=100/44.0 #multiplication factor of Ca2+#\n",
+ "M5=100/(82.0*2.0) #multiplication factor of Ca2+#\n",
+ "M6=100/60.0 #multiplication factor of Ca2+#\n",
+ "M7=100/(17.0*2.0) #multiplication factor of Ca2+#\n",
+ "P1=W1*M1 #in terms of CaCO3#\n",
+ "P2=W2*M2 #in terms of CaCO3#\n",
+ "P3=W3*M3 #in terms of CaCO3#\n",
+ "P4=W4*M4 #in terms of CaCO3#\n",
+ "P5=W5*M5 #in terms of CaCO3#\n",
+ "P6=W6*M6 #in terms of CaCO3#\n",
+ "P7=W7*M7 #in terms of CaCO3#\n",
+ "V=200000.0 #volume of water in lit#\n",
+ "L=0.74*(P2+P3+P4-P5+P7)*V \n",
+ "L=L/10.0**6.0 #in kgs#\n",
+ "S=1.06*(P1+P2-P3-P5-P6+P7)*V \n",
+ "S=S/10.0**6 #in kgs#\n",
+ "\n",
+ "#Result\n",
+ "print\"Lime required is \",L,\"kg\"\n",
+ "print\"\\nSoda required is \",S,\"kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lime required is 153.92 kg\n",
+ "\n",
+ "Soda required is 19.08 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.8,Page no:175"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "N=150.0 #amount of NaCl in solution in g/l#\n",
+ "V=8.0 #volume of NaCl solution#\n",
+ "\n",
+ "#Calculation\n",
+ "M=N*V \n",
+ "V=10000.0 #volume of hard water#\n",
+ "W=58.5 #molecular weight of NaCl#\n",
+ "K=(M*100.0/(W*2))/V \n",
+ "J=K*1000.0 \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print\"\\nHardness of water is \",round(J,1),\"mg/l or ppm\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Hardness of water is 102.6 mg/l or ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.9,Page no:176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=219.0 #amount of Mg(HCO3)2 in water in ppm#\n",
+ "W2=36.0 #amount of Mg2+ in water in ppm#\n",
+ "W3=18.3 #amount of (HCO3)- in water in ppm#\n",
+ "W4=1.5 #amount of H+_in water in ppm#\n",
+ "M1=100/146.0 #multiplication factor of Mg(HCO3)2#\n",
+ "M2=100/24.0 #multiplication factor of Mg(HCO3)2#\n",
+ "M3=100/122.0 #multiplication factor of Mg(HCO3)2#\n",
+ "M4=100/2.0 #multiplication factor of Mg(HCO3)2#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 #in terms of CaCO3#\n",
+ "P2=W2*M2 #in terms of CaCO3#\n",
+ "P3=W3*M3 #in terms of CaCO3#\n",
+ "P4=W4*M4 #in terms of CaCO3#\n",
+ "L=0.74*((2*P1)+P2+P3+P4) \n",
+ "\n",
+ "R=1.0 #water supply rate in m**3/s#\n",
+ "D=R*60.0*60.0*24.0*L \n",
+ "K=D*1000.0 #in lit/day#\n",
+ "T=K/10.0**9 #in tonnes#\n",
+ "S=1.06*(P2+P4-P3) \n",
+ "D2=R*60*60*24*S \n",
+ "A=D2*1000 #in lit/day#\n",
+ "B=A/10.0**9 #in tonnes#\n",
+ "J1=90/100.0 #purity of lime#\n",
+ "J2=95/100.0 #purity of soda#\n",
+ "C1=500.0 #cost of one tonne lime#\n",
+ "C2=7000.0 #cost of one tonne soda#\n",
+ "CL=round(T,1)*C1/J1 \n",
+ "print\"\\ncost of lime is\",CL,\"Rs\"\n",
+ "CS=round(B,1)*C2/J2 \n",
+ "print\"\\ncost of soda is \",CS,\"Rs\"\n",
+ "C=CL+CS \n",
+ "\n",
+ "#Result\n",
+ "print\"\\ntotal cost is \",round(C) ,\"Rs\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "cost of lime is 19166.6666667 Rs\n",
+ "\n",
+ "cost of soda is 141473.684211 Rs\n",
+ "\n",
+ "total cost is 160640.0 Rs\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.10,Page no:176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=40.0 #amount of Ca2+ in water in mg/l#\n",
+ "W2=24.0 #amount of Mg2+ in water in mg/l#\n",
+ "W3=8.05 #amount of Na+ in water in mg/l#\n",
+ "W4=183.0 #amount of (HCO3)- in water in mg/l#\n",
+ "W5=55.68 #amount of (SO4)2- in water in mg/l#\n",
+ "W6=6.74 #amount of Cl- in water in mg/l#\n",
+ "M1=100/40.0 #multiplication factor of Ca2+#\n",
+ "M2=100/24.0 #multiplication factor of Mg2+#\n",
+ "M3=100/(23.0*2) #multiplication factor of Na+#\n",
+ "M4=100/(61.0*2) #multiplication factor of (HCO3)-#\n",
+ "M5=100/96.0 #multiplication factor of (SO4)2-#\n",
+ "M6=100/(35.5*2) #multiplication factor of Cl-#\n",
+ "\n",
+ "#Calculation\n",
+ "P1=W1*M1 #in terms of CaCO3#\n",
+ "P2=W2*M2 #in terms of CaCO3#\n",
+ "P3=W3*M3 #in terms of CaCO3#\n",
+ "P4=W4*M4 #in terms of CaCO3#\n",
+ "P5=W5*M5 #in terms of CaCO3#\n",
+ "P6=W6*M6 #in terms of CaCO3#\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nCalcium alkalinity =\",P1,\"ppm\" \n",
+ "print\"\\nMagnesium alkalinity =\",P4-P1,\"ppm\"\n",
+ "print\"\\n total alkalinity = \",P1+P4-P1,\"ppm\"\n",
+ "print\"\\n total hardness = \",P1+P2,\"ppm\"\n",
+ "print\"\\nCa temporary hardness = \",P1,\"ppm\"\n",
+ "print\"\\nMg temporary hardness = \",P4-P1,\"ppm\"\n",
+ "print\"\\nMg permanant hardness = \",P2-(P4-P1),\"ppm\"\n",
+ "print\"\\nSalts are:\"\n",
+ "print\"\\nCa(HCO3)2 salt = \",P1,\"ppm\"\n",
+ "print\"\\nMg(HCO3)2 salt = \",P4-P1,\"ppm\"\n",
+ "print\"\\nMgSO4 salt = \",P2-(P4-P1),\"ppm\"\n",
+ "print\"\\nNaCl salt = \",P6,\"ppm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Calcium alkalinity = 100.0 ppm\n",
+ "\n",
+ "Magnesium alkalinity = 50.0 ppm\n",
+ "\n",
+ " total alkalinity = 150.0 ppm\n",
+ "\n",
+ " total hardness = 200.0 ppm\n",
+ "\n",
+ "Ca temporary hardness = 100.0 ppm\n",
+ "\n",
+ "Mg temporary hardness = 50.0 ppm\n",
+ "\n",
+ "Mg permanant hardness = 50.0 ppm\n",
+ "\n",
+ "Salts are:\n",
+ "\n",
+ "Ca(HCO3)2 salt = 100.0 ppm\n",
+ "\n",
+ "Mg(HCO3)2 salt = 50.0 ppm\n",
+ "\n",
+ "MgSO4 salt = 50.0 ppm\n",
+ "\n",
+ "NaCl salt = 9.49295774648 ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.11,Page no:177"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P=0.0 #phenolplthalein alkalinity in water sample#\n",
+ "V=16.9 #required HCl in ml for 100 ml water sample#\n",
+ "N=0.02 #normality of HCl#\n",
+ "print\"Since P=0 the alkalinity is due to HCO3- ions\" \n",
+ "C=50.0 #equivalent of CaCO3 in mg for 1 ml 1N of HCl#\n",
+ "\n",
+ "#Calculation\n",
+ "A=C*V*N \n",
+ "print\"\\nIn 100ml water sample the alkalinity is\",A,\"mg/s\"\n",
+ "B=A*1000.0/100.0\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nFor 1 litre of water the alkalinity is \",B,\"mg/l\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Since P=0 the alkalinity is due to HCO3- ions\n",
+ "\n",
+ "In 100ml water sample the alkalinity is 16.9 mg/s\n",
+ "\n",
+ "For 1 litre of water the alkalinity is 169.0 mg/l\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.12,Page no:178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "P=4.7 #required HCl in ml using HpH indicator #\n",
+ "H=10.5 #required HCl im ml using MeOH indicator#\n",
+ "M=P+H \n",
+ "N=0.02 #normality of HCl#\n",
+ "\n",
+ "print\"\\nSince P<0.5*M sample contain (CO3)2- and (HCO3)- alkalinity\"\n",
+ "C=50 #equivalent of CaCO3 in mg for 1ml 1N HCl#\n",
+ "\n",
+ "#Calculation\n",
+ "A=C*(2*P)*N #amount of (CO3)2- alkalinity in mg in 100 ml of water#\n",
+ "B=A*1000/100 \n",
+ "D=C*(M-2*P)*N #the amount of (HCO3)- alkalinity in mg in 100 ml of water#\n",
+ "E=D*1000/100 \n",
+ "T=B+E \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nTotal alkalinity is \",T,\"mg/l or ppm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Since P<0.5*M sample contain (CO3)2- and (HCO3)- alkalinity\n",
+ "\n",
+ "Total alkalinity is 152.0 mg/l or ppm\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.13,Page no:178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=160.0 #amount of Ca2+ in ppm#\n",
+ "W2=88.0 #amount of Mg2+ in ppm#\n",
+ "W3=72.0 #amount of CO2 in ppm#\n",
+ "W4=488.0 #amount of (HCO3)- in ppm#\n",
+ "W5=139.0 #amount of (FeSO4).7H2O in ppm#\n",
+ "M1=100/40.0 #multiplication factor of Ca2+#\n",
+ "M2=100/24.0 #multiplication factor of Mg2+#\n",
+ "M3=100/44.0 #multiplication factor of CO2#\n",
+ "M4=100/(61.0*2.0) #multiplication factor of (HCO3)-#\n",
+ "M5=100/278.0 #multiplication factor of (FeSO4).7H2O#\n",
+ "\n",
+ "P1=400 #in terms of CaCO3#\n",
+ "P2=300 #in terms of CaCO3#\n",
+ "P3=200 #in terms of CaCO3#\n",
+ "P4=400 #in terms of CaCO3#\n",
+ "P5=50 #in terms of CaCO3#\n",
+ "V=100000.0 #volume of water in litres#\n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "L=0.74*(P2+P3+P4+P5)*V #lime required in mg#\n",
+ "L=L/10.0**6 \n",
+ "S=1.06*(P1+P2+P5-P4)*V #soda required in mg#\n",
+ "S=S/10.0**6 \n",
+ "\n",
+ "#Result\n",
+ "print\"Lime required is \",L,\"kg\"\n",
+ "print\"\\nSoda required is \",S,\"kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Lime required is 70.3 kg\n",
+ "\n",
+ "Soda required is 37.1 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.14,Page no:179"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W=50 #amount of NaCl in g/l in NaCl solution#\n",
+ "V=200 #volume of NaCl solution in litres#\n",
+ "\n",
+ "#Calculation\n",
+ "A=W*V \n",
+ "V=10000 #volume of hard water passed through Zeolite softener#\n",
+ "M=100/(58.5*2) #multiplication factor of NaCl#\n",
+ "P=M*A \n",
+ "B=P*1000/V \n",
+ "\n",
+ "#Result\n",
+ "print\"\\nIn terms of CaCO3=\",round(P),\"g CaCO3\"\n",
+ "print\"\\nFor 1 litre of hard water=\",round(B,1),\"mg/l or ppm\"\n",
+ "print\"NOTE:In book answer wrongly written as 845.7\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "In terms of CaCO3= 8547.0 g CaCO3\n",
+ "\n",
+ "For 1 litre of hard water= 854.7 mg/l or ppm\n",
+ "NOTE:In book answer wrongly written as 845.7\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7.15,Page no:179"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W1=0.28 #amount of CaCO3 in grams dissolved in 1 litre of water#\n",
+ "V1=28 #required EDTA in ml on titration of 100ml of CaCO3 solution#\n",
+ "V2=33 #required EDTA in ml for 100ml of unknown hard water sample#\n",
+ "V3=10 #required EDTA in ml for 100 ml of unknown sample after boiling and cooling#\n",
+ "M1=100/100 #multiplication factor of CaCO3#\n",
+ "\n",
+ "#Calculation\n",
+ "C=W1*M1 \n",
+ "A=C*100#for 100 ml of sample equivalent to 28 ml of EDTA#\n",
+ "B=A/V1 \n",
+ "D=V2*B #for 100 ml#\n",
+ "D=D*1000/100 \n",
+ "E=V3*B #for 100 ml#\n",
+ "E=E*1000/100 \n",
+ "T=D-E \n",
+ "\n",
+ "#Result\n",
+ "print\"Total hardness is \",D,\"mg CaCO3 eq\"\n",
+ "print\"\\nPermanant hardness is \",E,\"mg CaCO3 eq\"\n",
+ "print\"\\nTemporary hardness is \",T,\"mg CaCO3\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total hardness is 330.0 mg CaCO3 eq\n",
+ "\n",
+ "Permanant hardness is 100.0 mg CaCO3 eq\n",
+ "\n",
+ "Temporary hardness is 230.0 mg CaCO3\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Textbook_of_Engineering_Chemistry/dChapter_8.ipynb b/Textbook_of_Engineering_Chemistry/dChapter_8.ipynb new file mode 100755 index 00000000..06e61c83 --- /dev/null +++ b/Textbook_of_Engineering_Chemistry/dChapter_8.ipynb @@ -0,0 +1,979 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 8: Fuels & Combustion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:1,Page no:188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "WC=1.508 #weight of coal sample in grams#\n",
+ "WH110=1.478 #weight of sample after heating at 110 degrees in grams#\n",
+ "m=WC-WH110 #weight of moisture in the sample#\n",
+ "\n",
+ "#Calculation\n",
+ "pm=m*100/WC #percentage of moisture in the sample#\n",
+ "WH950=1.068 #weight of sample after heating at 950 degrees in grams#\n",
+ "vm=WH110-WH950 #volatile matter in grams#\n",
+ "pvm=vm*100/WC #percentage of voltaile matter#\n",
+ "\n",
+ "#Result\n",
+ "print'Weight of moisture in the sample=',m,\"g\"\n",
+ "print'Percentage of moisture in the sample=',round(pm,2),\"%\"\n",
+ "print'\\nWeight of volatile matter in the sample=',vm,\"g\"\n",
+ "print'Percentage of volatile matter in the sample=',round(pvm,2),\"%\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Weight of moisture in the sample= 0.03 g\n",
+ "Percentage of moisture in the sample= 1.99 %\n",
+ "\n",
+ "Weight of volatile matter in the sample= 0.41 g\n",
+ "Percentage of volatile matter in the sample= 27.19 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:2,Page no:192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "CR=7.8 #compression ratio for first case#\n",
+ "E1=1.0-(1.0/CR)**0.258 #Energy efficiency corresponding to CR value 7.8#\n",
+ "CR=9.5 #compreesion ratio for second case#\n",
+ "\n",
+ "#Calculation\n",
+ "E2=1.0-(1.0/CR)**0.258 #Energy efficiency corresponding to CR value 9.5#\n",
+ "IE=E2-E1 #Increase in efficiency#\n",
+ "PIE=round(IE,2)*100.0/round(E2,3) #percentage of increase in efficiency#\n",
+ "\n",
+ "#Result\n",
+ "print'\\nIncrease in efficiency=IE=%f',round(IE,2)\n",
+ "print'\\nPercentage of increase in efficiency=PIE=%f',PIE,\"%\"\n",
+ "print \"NOTE:Calculation mistake in book for % increase\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Increase in efficiency=IE=%f 0.03\n",
+ "\n",
+ "Percentage of increase in efficiency=PIE=%f 6.80272108844 %\n",
+ "NOTE:Calculation mistake in book for % increase\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:3,Page no:196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=3.0 #weight of carbon in 1kg of coal sample in Kimath.lograms#\n",
+ "WO2=C*32/12.0 #weight of oxygen in carbon sample in Kimath.lograms#\n",
+ "\n",
+ "#Calculation\n",
+ "WA=WO2*100/23.0 #weight of air in the carbon sample in Kimath.lograms#\n",
+ "MA=WA/28.92 #mol of air in kimath.lograms#\n",
+ "VA=MA*22.4 #Volume of air required in m3 air#\n",
+ "\n",
+ "#Result\n",
+ "print'weight of air required for combustion of carbon=',round(WA,1),\"kg\"\n",
+ "print'\\nVolume of air required=',round(VA,1),\"m**3 air\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "weight of air required for combustion of carbon= 34.8 kg\n",
+ "\n",
+ "Volume of air required= 26.9 m**3 air\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:4,Page no:196"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "CO=0.46 #volume of carbon monoxide in 1kg of gas sample in m3#\n",
+ "C2H2=0.020 #volume of C2H2 in 1kg of gas sample in m3#\n",
+ "CH4=0.1 #volume of CH4 in 1kg of gas sample in m3#\n",
+ "N2=0.01 #volume of nytrogen in 1kg of gas sample in m3#\n",
+ "H2=0.40 #volume of hydrogen in 1kg of gas sample in m3#\n",
+ "\n",
+ "#Calculation\n",
+ "VA=0.68*(100/21.0) #volume of air needed in m3#\n",
+ "\n",
+ "#Result\n",
+ "print'\\nVolume of air needed=',round(VA,3),\"m**3\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Volume of air needed= 3.238 m**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:5,Page no:197"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=624.0 #weight of carbon in 1kg of coal sample in grams#\n",
+ "O=69.0 #weight of oxygen in 1kg of coal sample in grams#\n",
+ "S=8.0 #weight of Sulphur in 1kg of coal sample in grams#\n",
+ "N=12.0 #weight of nytrogen in 1kg of coal sample in grams#\n",
+ "H=41.0 #weight of hydrogen in 1kg of coal sample in grams#\n",
+ "CO2=129 #weight of CO2 in 1kg of coal sample in grams#\n",
+ "CO=2.0 #weight of CO in 1kg of coal sample in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "MO=C*32/12.0+H*16/2.0+S*32/32.0-O #minimum weight of oxygen needed in grams#\n",
+ "MA=MO*0.1/23 #minimum weight of air needed in kimath.lograms#\n",
+ "\n",
+ "WC=CO2*(12/44.0)+CO*(12/28.0) #weight of C in fuel gas/kg#\n",
+ "WF=C/WC #Weight of fuel gas/kg of coal in g#\n",
+ "O2=2*16/28.0 #O2 needed to convert CO to CO2 in Kg#\n",
+ "RWO2=(61.0-O2)/1000.0 #remaining weight of O2/kg of fuel gas in Kg#\n",
+ "WO2=WF*RWO2 #weight of O2 obtained by burning 1kg coal in kg#\n",
+ "AR=WO2*100/23.0 #air required in kimath.lograms#\n",
+ "WAS=MA+AR #weight of air actually supplied/kg coal burnt in kg#\n",
+ "\n",
+ "#Result\n",
+ "print'(i) Weight of air theoretically needed=',round(MA,3),\"kg\"\n",
+ "print'\\n(ii) Weight of C in fuel gas/kg=',round(WC ,2),\"g\"\n",
+ "print'\\n Weight of fuel gas/kg of coal=',round(WF,3),\"kg\"\n",
+ "print'\\n(iii) Weight of air actually supplied/kg coal burnt',round(WAS,1),\"kg air\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Weight of air theoretically needed= 8.396 kg\n",
+ "\n",
+ "(ii) Weight of C in fuel gas/kg= 36.04 g\n",
+ "\n",
+ " Weight of fuel gas/kg of coal= 17.315 kg\n",
+ "\n",
+ "(iii) Weight of air actually supplied/kg coal burnt 12.9 kg air\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:6,Page no:202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "w=1080.0 #quantity of water in grams#\n",
+ "W=150.0 #Water equivalent of calorimeter in grams#\n",
+ "x=0.681 #quantity of fuel carried out in combustion in grams#\n",
+ "dt=3.61 #rise in temperature of water in degree C#\n",
+ "\n",
+ "#Calculation\n",
+ "Q=(w+W)*(dt)/x #calorific value of the fuel in cal per grams#\n",
+ "\n",
+ "#Result\n",
+ "print'Calorific value of the fuel=',round(Q,1),\"cal/g =\",round(Q/1000,3),\"kcal/g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Calorific value of the fuel= 6520.3 cal/g = 6.52 kcal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:7,Page no:202"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "w=1080.0 #quantity of water taken in grams#\n",
+ "W=150.0 #Water equivalent of calorimeter in grams#\n",
+ "m=0.681 #weight of coal taken or mass of fuel in grams#\n",
+ "dt=3.61 #rise in temperature of water in degree C#\n",
+ "AC=50.0 #Acid correction in calories#\n",
+ "FC=5.0 #Fuse wire correction in calories#\n",
+ "CC=0.05 #cooling correction in calories#\n",
+ "\n",
+ "#Calculation\n",
+ "GCV=((w+W)*(dt+CC)-(AC+FC))/m #Gross calorific value of the sample in cal per grams#\n",
+ "\n",
+ "#Result\n",
+ "print'Gross Calorific value of the fuel=',round(GCV,1),\"cal/g =\",round(GCV/1000,3),\"kcal/g\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gross Calorific value of the fuel= 6529.8 cal/g = 6.53 kcal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8,Page no:203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=90.2 #percentage of carbon#\n",
+ "O=2.9 #percentage of oxygen#\n",
+ "H=2.40 #percentage of hydrogen#\n",
+ "\n",
+ "#Calculation\n",
+ "GCV=(8080.0*C+34400.0*(H-O/8.0))/100.0 #Gross calorific value of the sample in cal per grams#\n",
+ "\n",
+ "#Result\n",
+ "print\"\\nGross Calorific value of the fuel=%.2e\"%GCV,\"cal/g\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Gross Calorific value of the fuel=7.99e+03 cal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:9,Page no:210"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "a=0.9 #absorptivity#\n",
+ "e=0.04 #emissivity#\n",
+ "P=750 #Sun light energy available in W/m2#\n",
+ "Q=5.67*10**-8 #conductivity \n",
+ "import math\n",
+ "\n",
+ "#Calculation\n",
+ "T4=a*P/(Q*e) \n",
+ "T=math.pow(T4,1.0/4.0)\n",
+ "\n",
+ "#Result\n",
+ "print'Maximum temeperature that can be achieved=',round(T,1),\"K\" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum temeperature that can be achieved= 738.6 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.1,Page no:212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "w=1500 #quantity of water in grams#\n",
+ "W=125 #Water equivalent of calorimeter in grams#\n",
+ "x=1.050 #quantity of fuel carried out in combustion in grams#\n",
+ "t1=25 #initial temperature of water in degree C#\n",
+ "t2=27.8 #final temperature of water in degree C#\n",
+ "\n",
+ "#Calculation\n",
+ "Q=(w+W)*(t2-t1)/x #calorific value of the fuel in cal per grams#\n",
+ "\n",
+ "#Result\n",
+ "print'Calorific value of the fuel=',round(Q/1000,1),\"kcal/g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Calorific value of the fuel= 4.3 kcal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.2,Page no:212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from scipy.optimize import fsolve\n",
+ "\n",
+ "#Variable declaration\n",
+ "C=90.0 #percentage of carbon#\n",
+ "O=3.0 #percentage of oxygen#\n",
+ "S=0.5 #percentage of sulphur#\n",
+ "N=0.5 #percentage of nytrogen#\n",
+ "LCV=8500 #Law calorific value#\n",
+ "#Calculation\n",
+ "def f(H):\n",
+ " GCV1=LCV+(9*H/100.0)*587\n",
+ " GCV2=(8080.0*C+34500*(H-O/8.0)+2240*S)/100.0 #Gross calorific value of the sample in cal per grams#\n",
+ " return(GCV1-GCV2)\n",
+ "h=fsolve(f,1)\n",
+ "GCV=LCV+(9*h/100.0)*587\n",
+ "\n",
+ "#Result\n",
+ "print'percentage of hydrogen=H=',round(h[0],1),\"%\"\n",
+ "print'\\nGross Calorific value of the fuel=',round(GCV[0]),\"kcal/g\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "percentage of hydrogen=H= 4.6 %\n",
+ "\n",
+ "Gross Calorific value of the fuel= 8743.0 kcal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.3,Page no:213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "w=500.0 #quantity of water taken in grams#\n",
+ "W=2000.0 #Water equivalent of calorimeter in grams#\n",
+ "m=1.000 #weight of coal taken or mass of fuel in grams#\n",
+ "t1=24.0 #initial temperature of water in degree C#\n",
+ "t2=26.2 #final temperature of water in degree C#\n",
+ "AC=50.0 #Acid correction in calories#\n",
+ "FC=10.0 #Fuse wire correction in calories#\n",
+ "CC=0.0 #cooling correction in calories#\n",
+ "\n",
+ "#Calculation\n",
+ "GCV=((w+W)*(t2-t1+CC)-(AC+FC))/m #Gross calorific value of the sample in cal per grams#\n",
+ "H=6.0 #percentage of hydrogen#\n",
+ "C=93.0 #percentage of carbon#\n",
+ "LCV=GCV-(9*H*580/100.0) #Net calorific value of the sample in cal per gram#\n",
+ "\n",
+ "#Result\n",
+ "print'Gross Calorific value of the fuel=',GCV,\"cal/g\"\n",
+ "print'Net calorific value of the sample=LCV=',LCV,\"cal/g\"\n",
+ "print\"\\nNOTE:Calculation mistake in book,wrongly written as 5540 and 5226.8\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gross Calorific value of the fuel= 5440.0 cal/g\n",
+ "Net calorific value of the sample=LCV= 5126.8 cal/g\n",
+ "\n",
+ "NOTE:Calculation mistake in book,wrongly written as 5540 and 5226.8\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.4,Page no:213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "WC=1.5642 #weight of coal sample in grams#\n",
+ "WH110=1.5022 #weight of sample after heating at 110 degrees in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "m=WC-WH110 #weight of moisture in the sample#\n",
+ "pm=m*100/WC #percentage of moisture in the sample#\n",
+ "WH950=0.7628 #weight of sample after heating at 950 degrees in grams#\n",
+ "vm=WH110-WH950 #volatile matter in grams#\n",
+ "pvm=vm*100/WC #percentage of voltaile matter#\n",
+ "ac=0.2140 #Ash content left in the last in grams#\n",
+ "pac=ac*100/WC #percentage of Ash content laft#\n",
+ "pfc=100-(pm+pvm-pac) #percentage of fixaed carbon#\n",
+ "\n",
+ "#Result\n",
+ "print'\\npercentage of fixed carbon in the sample=',round(pfc,2),\"%\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "percentage of fixed carbon in the sample= 62.45 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.5,Page no:214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "WBaSO4=0.0482 #weight of BaSO4 in grams#\n",
+ "W=0.5248 #weight of sample in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "PS=32*WBaSO4*100/(233*W) #percentage of sulphur in the sample#\n",
+ "\n",
+ "#Result\n",
+ "print'Percentage of sulphur in the sample=',round(PS,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage of sulphur in the sample= 1.26 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.6,Page no:215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "W=10 #weight of Water heated of calorimeter in Kimath.lograms#\n",
+ "V=0.1 #volume of gas used in metrecube#\n",
+ "t1=22 #inlet temperature of water in degree C#\n",
+ "t2=30 #outlet temperature of water in degree C#\n",
+ "#Calculation\n",
+ "GCV=W*(t2-t1)/V #Gross calorific value of the sample in Kilocal per metre3#\n",
+ "L=580 #latent heat of water in cal/g#\n",
+ "Ws=0.025 #weight of steam condensed in grams#\n",
+ "LCV=GCV-(Ws*L/V) #Net calorific value of the sample in Kcal per meter3#\n",
+ "\n",
+ "#Result\n",
+ "print'Gross Calorific value of the fuel=',GCV ,\"Kcal/m3\"\n",
+ "print'\\nNet calorific value of the sample=',LCV,\"Kcal/m3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gross Calorific value of the fuel= 800.0 Kcal/m3\n",
+ "\n",
+ "Net calorific value of the sample= 655.0 Kcal/m3\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.7,Page no:215"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=90.0 #percentage of carbon#\n",
+ "O=3.0 #percentage of oxygen#\n",
+ "S=0.5 #percentage of sulphur#\n",
+ "N=0.5 #percentage of nytrogen#\n",
+ "H=3.5 #percentage of hydrogen#\n",
+ "H2O=0.1 #percentage of H2O#\n",
+ "#Calculation\n",
+ "AO=900*32.0/12.0+35*16.0/2.0+5*32.0/32.0 #amount of oxygen required in grams#\n",
+ "AN=2655*100/23.0 #amount of air needed in grams#\n",
+ "\n",
+ "#Result\n",
+ "print'\\nAmount of air needed=',round(AN/1000,3),\"kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Amount of air needed= 11.543 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.8,Page no:216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "CH4=0.14 #volume of CH4 in 1m3 volume of gaseous fuel in m3#\n",
+ "H2=0.32 #volume of H2 in 1m3 volume of gaseous fuel in m3#\n",
+ "N2=0.40 #volume of N2 in 1m3 volume of gaseous fuel in m3#\n",
+ "O2=0.14 #volume of O2 in 1m3 volume of gaseous fuel in m3#\n",
+ "\n",
+ "#Calculation\n",
+ "V_ch4=O2*2\n",
+ "v_H2=H2*0.5\n",
+ "Total_O2=V_ch4+v_H2\n",
+ "Net_O2=(Total_O2-O2)*1000 # Net O2 needed in L\n",
+ "V_req=Net_O2*(100/21.0)*(125/100.0)\n",
+ "\n",
+ "#Result\n",
+ "print\"Volume of air required assuming 21% =\",round(V_req,1),\"L\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Volume of air required assuming 21% = 1785.7 L\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.9,Page no:216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=750.0 #weight of carbon in 1kg of coal sample in grams#\n",
+ "O=121.0 #weight of oxygen in 1kg of coal sample in grams#\n",
+ "A=45.0 #weight of Ash in 1kg of coal sample in grams#\n",
+ "N=32.0 #weight of nytrogen in 1kg of coal sample in grams#\n",
+ "H=52.0 #weight of hydrogen in 1kg of coal sample in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "MO=C*32/12+H*16/2-O #minimum weight of oxygen needed in grams#\n",
+ "MA=MO*100/23 #minimum weight of air needed in grams#\n",
+ "GCV=(808*C+3450*(H-O/8))/100 #Gross calorific value of the sample in cal per grams#\n",
+ "LCV=GCV-0.09*H*0.1*587 #law calorific value of the sample in cal/gram#\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print'\\nGross Calorific value of the fuel=',round(GCV,2),\"kcal/g\"\n",
+ "print'\\nNet calorific value of the sample',round(LCV),\"kcal/g\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Gross Calorific value of the fuel= 7332.19 kcal/g\n",
+ "\n",
+ "Net calorific value of the sample 7057.0 kcal/g\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.10,Page no:217"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=810.0 #weight of carbon in 1kg of coal sample in grams#\n",
+ "O=80.0 #weight of oxygen in 1kg of coal sample in grams#\n",
+ "S=10.0 #weight of Sulphur in 1kg of coal sample in grams#\n",
+ "N=10.0 #weight of nytrogen in 1kg of coal sample in grams#\n",
+ "H=50.0 #weight of hydrogen in 1kg of coal sample in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "MO=C*32/12.0+H*16/2.0+S*32/32.0 #minimum weight of oxygen needed in grams#\n",
+ "MA=2490*100/23.0 #minimum weight of air needed in grams#\n",
+ "print'\\nminimum amount of air needed=',round(MA/1000,3) ,\"kg\"\n",
+ "NF=10+MA*0.77 #weight of nitrogen present in the products in grams#\n",
+ "WD=2970.0+20.0+8346.0 #total weight of dry products in grams#\n",
+ "PCO2=2970*100/WD #percentage composition of CO2#\n",
+ "print'\\nPercentage composition of CO2=',round(PCO2 ,2),\"%\"\n",
+ "PSO2=20*100/WD #percentage composition of SO2#\n",
+ "print'\\nPercentage composition of SO2=',round(PSO2 ,3),\"%\"\n",
+ "PN2=8346*100/WD #percentage composition of N2#\n",
+ "\n",
+ "#Result\n",
+ "print'\\nPercentage composition of N2=',round(PN2,2),\"%\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "minimum amount of air needed= 10.826 kg\n",
+ "\n",
+ "Percentage composition of CO2= 26.2 %\n",
+ "\n",
+ "Percentage composition of SO2= 0.176 %\n",
+ "\n",
+ "Percentage composition of N2= 73.62 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.11,Page no:219"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "CO=0.205 #volume of carbon monoxide in 1kg of gas sample in m3#\n",
+ "CO2=0.060 #volume of CO2 in 1kg of gas sample in m3#\n",
+ "CH4=0.042 #volume of CH4 in 1kg of gas sample in m3#\n",
+ "N=0.501 #volume of nytrogen in 1kg of gas sample in m3#\n",
+ "H2=0.194 #volume of hydrogen in 1kg of gas sample in m3#\n",
+ "\n",
+ "#Calculation\n",
+ "VA=0.283*(100/21)*(130/100) #volume of air needed in m3#\n",
+ "VDCO2=0.06+0.205*1+0.042*1 #volume of dry products containig CO2 formed in m3#\n",
+ "VDN2=0.501+1.752*79/100 #volume of dry products containig N2 formed in m3#\n",
+ "VDO2=1.755*21/100 #volume of dry products containig O2 formed in m3#\n",
+ "TVD=VDCO2+VDN2+VDO2 #total volume of dry products formed in m3#\n",
+ "PDCO2=VDCO2*100/TVD #percentage of dry products containig CO2 formed#\n",
+ "PDN2=VDN2*100/TVD #percentage of dry products containig N2 formed#\n",
+ "PDO2=VDO2*100/TVD #percentage of dry products containig O2 formed#\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print'\\nPercentage of dry products containing CO2 formed=',round(PDCO2,2),\"%\"\n",
+ "print'\\nPercentage of dry products containing N2 formed=',round(PDN2 ,2),\"%\"\n",
+ "print'\\nPercentage of dry products containing O2 formed=',round(PDO2 ,2),\"%\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Percentage of dry products containing CO2 formed= 11.99 %\n",
+ "\n",
+ "Percentage of dry products containing N2 formed= 73.62 %\n",
+ "\n",
+ "Percentage of dry products containing O2 formed= 14.39 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.12,Page no:220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=780.0 #weight of carbon in 1kg of coal sample in grams#\n",
+ "O=120.0 #weight of oxygen in 1kg of coal sample in grams#\n",
+ "S=12.0 #weight of Sulphur in 1kg of coal sample in grams#\n",
+ "N=21.0 #weight of nytrogen in 1kg of coal sample in grams#\n",
+ "H=41.0 #weight of hydrogen in 1kg of coal sample in grams#\n",
+ "\n",
+ "#Calculation\n",
+ "MO=C*32/12+H*16/2+S*32/32-O #minimum weight of oxygen needed in grams#\n",
+ "MA=MO*100/23.0 #minimum weight of air needed in grams#\n",
+ "\n",
+ "#Result\n",
+ "print'minimum weight of oxygen needed=',MO/1000,\"kg\"\n",
+ "print'\\nminimum amount of air needed=',MA/1000,\"kg\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "minimum weight of oxygen needed= 2.3 kg\n",
+ "\n",
+ "minimum amount of air needed= 10.0 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:8.13,Page no:220"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "C=1.5 #weight of carbon in 1kg of coal sample in Kimath.lograms#\n",
+ "\n",
+ "#Calculation\n",
+ "WO2=C*32/12 #weight of oxygen in carbon sample in Kimath.lograms#\n",
+ "WA=WO2*100/23 #weight of air in the carbon sample in Kimath.lograms#\n",
+ "O2_4000=22.4/32.0*4000 #Volume in 4000 g oxygen\n",
+ "V=100/21.0*O2_4000 #Volume of air with 21 % O2\n",
+ "\n",
+ "#Result\n",
+ "print'\\nweight of air in the carbon sample=',round(WA,2),\"kg\" \n",
+ "print \"Volume of air is\",round(V/1000,2),\"m^3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "weight of air in the carbon sample= 17.39 kg\n",
+ "Volume of air is 13.33 m^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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