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| author | hardythe1 | 2014-08-06 17:26:32 +0530 |
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| committer | hardythe1 | 2014-08-06 17:26:32 +0530 |
| commit | f3e94078a83634b4353ab0cd2de3b0e204a48ac7 (patch) | |
| tree | f4d42d3a77bed54d4b4e18a347bb860a39dc77d3 /Elementary_Principles_of_Chemical_Processes | |
| parent | 34186106dd668bb95f7c60ba60942001a490a20b (diff) | |
| download | Python-Textbook-Companions-f3e94078a83634b4353ab0cd2de3b0e204a48ac7.tar.gz Python-Textbook-Companions-f3e94078a83634b4353ab0cd2de3b0e204a48ac7.tar.bz2 Python-Textbook-Companions-f3e94078a83634b4353ab0cd2de3b0e204a48ac7.zip | |
adding books
Diffstat (limited to 'Elementary_Principles_of_Chemical_Processes')
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/Chapter11.ipynb | 155 | ||||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/Chapter2.ipynb | 501 | ||||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/Chapter3.ipynb | 835 | ||||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/Chapter4.ipynb | 1771 | ||||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/Chapter5.ipynb | 807 | ||||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/Chapter6.ipynb | 1137 | ||||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/Chapter7.ipynb | 893 | ||||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/Chapter8.ipynb | 1053 | ||||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/Chapter9.ipynb | 822 | ||||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/README.txt | 10 | ||||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/screenshots/dev.png | bin | 0 -> 80892 bytes | |||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/screenshots/mbtT.png | bin | 0 -> 19516 bytes | |||
| -rwxr-xr-x | Elementary_Principles_of_Chemical_Processes/screenshots/vbyl.png | bin | 0 -> 15350 bytes |
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diff --git a/Elementary_Principles_of_Chemical_Processes/Chapter11.ipynb b/Elementary_Principles_of_Chemical_Processes/Chapter11.ipynb new file mode 100755 index 00000000..bec15476 --- /dev/null +++ b/Elementary_Principles_of_Chemical_Processes/Chapter11.ipynb @@ -0,0 +1,155 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9a8a7685dfe3ac94231d04568a5fa2ca80f262b83285bc314eef1bc912a7ecc6" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 11: Balances on Transient Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.1-2, page no. 547" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "v0=math.pow(10.,9)\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "def dv(t):\n", + " dv=math.pow(10,6)*math.exp(-t/100) -math.pow(10,7)\n", + " return dv\n", + "\n", + "vol, err=scipy.integrate.quad(dv,0,60) #integrate.quad is an inbult function used for definite integration\n", + "print '%s %.3E' %(\" \\n Volume at the end of 60 days=\",vol+v0)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Volume at the end of 60 days= 4.451E+08\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 1, + "text": [ + "''" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.3-1, page no. 556" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "m1=1.50 #kg\n", + "m2=3.0 #kg\n", + "Cv1=0.9 #cal/g C\n", + "Cv2=0.12 #cal/g C\n", + "Qdot=500.0 #W\n", + "T1=250.0 #C\n", + "T2=25.0 #C\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "MCv=(m1*Cv1 + m2*Cv2)*1000*4.184\n", + "print(MCv)\n", + "tf=(T1-T2)*MCv/Qdot\n", + "print '%s %d %s %.3f' %(\" \\n Time required\",tf+1,\" in sec and in min =\",tf/60)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "7154.64\n", + " \n", + " Time required 3220 in sec and in min = 53.660\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 2, + "text": [ + "''" + ] + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Principles_of_Chemical_Processes/Chapter2.ipynb b/Elementary_Principles_of_Chemical_Processes/Chapter2.ipynb new file mode 100755 index 00000000..6f35a4a4 --- /dev/null +++ b/Elementary_Principles_of_Chemical_Processes/Chapter2.ipynb @@ -0,0 +1,501 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1e57dcb73985a42b8c9ee1409e50d4315b6a10f8a73cdae8a91714580c2fe346" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2: Introduction to Engineering Calculations" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2-1, page no. 10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "AcclInitial=1.0 #cm/s^2\n", + "\n", + "#Calculations and printing :\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "AcclFinal=AcclInitial*math.pow(3600*24*365,2)/math.pow(10,5)\n", + "#the calculations involved are the conversion factors\n", + "print '%s %.3E' %(\" \\n final acceleration (Km/Yr^2) = \",AcclFinal)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " final acceleration (Km/Yr^2) = 9.945E+09\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 2, + "text": [ + "''" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3-1, page no. 11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "Initial=23.0 #lb.ft/min^2\n", + "\n", + "#Calculations and printing :\n", + "\n", + "Final=Initial*0.453593*100/(3.281*60*60)\n", + "#the calculations involved are conversion factors\n", + "print '%s %.4f' %(\"final (kg.cm/s^2) = \",Final)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "final (kg.cm/s^2) = 0.0883\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 4, + "text": [ + "''" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4-1, page no. 13" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "density=62.4 #lbm/ft^3\n", + "volume=2.0 #ft^3\n", + "\n", + "#Calculations and printing :\n", + "mass=volume*density\n", + "print '%s %.3f' %(\"mass of the water = volume x density(lbm) = \",mass)\n", + "print(\" \\n At sealevel, g=32.174 ft/s^2\")\n", + "g=32.174\n", + "weight=mass*g/32.174\n", + "print '%s %.3f' %(\" \\n weight at sealevel (lbf) = \",weight)\n", + "print(\" \\n At denver, g=32.139 ft/s^2\")\n", + "g=32.139\n", + "weight=mass*g/32.174\n", + "print '%s %.3f' %(\"\\n weight at denver (lbf) = \",weight)\n", + "#the division with 32.174 is to convert lbm.math.pow(ft/s,2) to lbf\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "mass of the water = volume x density(lbm) = 124.800\n", + " \n", + " At sealevel, g=32.174 ft/s^2\n", + " \n", + " weight at sealevel (lbf) = 124.800\n", + " \n", + " At denver, g=32.139 ft/s^2\n", + "\n", + " weight at denver (lbf) = 124.664\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 7, + "text": [ + "''" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5-2, page no. 19" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + " \n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "#the badbatches per week are taken as elements of a vector y\n", + "\n", + "y=[17, 27, 18, 18, 23, 19, 18, 21, 20, 19, 21, 18]\n", + "\n", + "#Calculations and printing :\n", + "#Here We used standard library functions mean and st_deviation\n", + "ybar=numpy.mean(y)\n", + "sy=numpy.std(y)\n", + "defaultvalue=ybar+3*sy+1\n", + "print '%s %d' %(\"the maximum allowed value of y i.e. bad batches in a week is \", defaultvalue+1)\n", + "print(\" \\n in case of 2 standard deviations\")\n", + "defaultvalue=ybar+2*sy+1\n", + "print '%s %d' %(\"the limiting value of y i.e. bad batches in a week is \",defaultvalue)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the maximum allowed value of y i.e. bad batches in a week is 29\n", + " \n", + " in case of 2 standard deviations\n", + "the limiting value of y i.e. bad batches in a week is 26\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 3, + "text": [ + "''" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7-1, page no. 24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "from matplotlib import pyplot\n", + "%pylab inline\n", + "x=numpy.array([10., 30., 50., 70., 90.])\n", + "y=numpy.array([20., 52.1, 84.6, 118.3, 151.0])\n", + "\n", + "#Calculations and printing :\n", + "#this program uses least squares fit to solve for slope and intercept.\n", + "#hence the value differs from textbook a bit.\n", + "sx=sum(x)\n", + "sx2=sum(x*x)\n", + "sy=sum(y)\n", + "D=numpy.transpose(y)\n", + "sxy=sum(x*D)\n", + "n=len(x)\n", + "A=([[sx,n],[sx2,sx]])\n", + "B=([[sy],[sxy]])\n", + "p=numpy.dot(linalg.inv(A),B)\n", + "m=p[0,0]\n", + "b=p[1,0]\n", + "yf=m*x+b\n", + "pyplot.plot(x,yf)\n", + "pyplot.xlim(min(x)-1,max(x)+1)\n", + "pyplot.ylim(min(yf)-1,max(yf)+1)\n", + "pyplot.xlabel('R')\n", + "pyplot.ylabel('V')\n", + "pyplot.show()\n", + "print(\"in case 2, R=36\")\n", + "R=36\n", + "V=m*R+b\n", + "print '%s %.3f' %(\"then V (L/min) =\",V),\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Populating the interactive namespace from numpy and matplotlib\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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YMwcGD4Zp0yArS4NAROR62PqcQXFxMfn5+dxzzz2Ul5fTqlUrAMLCwjh27Fi9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\n", + "text": [ + "<matplotlib.figure.Figure at 0x2da0750>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "in case 2, R=36\n", + "then V (L/min) = 62.226\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7-2, page no. 26" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "from matplotlib import pyplot\n", + "%pylab inline\n", + "\n", + "T=[10., 20., 40., 80.]\n", + "M=[14.76, 20.14, 27.73, 38.47]\n", + "sqrtT=numpy.sqrt(T)\n", + "\n", + "\n", + "#Calculations and printing :\n", + "sx=sum(sqrtT)\n", + "sx2=sum(T)\n", + "sy=sum(M)\n", + "D=numpy.transpose(M)\n", + "sxy=sum(sqrtT*D)\n", + "n=len(T)\n", + "A=([[sx,n],[sx2,sx]]) \n", + "B=([[sy],[sxy]])\n", + "p=numpy.dot(linalg.inv(A),B)\n", + "a=p[0,0]\n", + "b=p[1,0]\n", + "yf=a*sqrtT+b\n", + "x=sqrtT\n", + "pyplot.plot(x,yf)\n", + "pyplot.xlim(min(x)-1,max(x)+1)\n", + "pyplot.ylim(min(yf)-1,max(yf)+1)\n", + "pyplot.xlabel('sqrtT')\n", + "pyplot.ylabel('M')\n", + "pyplot.show()\n", + "print '%s %.3f' %(\"slope (g/s*K^.5) =\",a)\n", + "print '%s %.3f' %(\"\\n intercept (g/s) =\",b)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Populating the interactive namespace from numpy and matplotlib\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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7O7to58Xx48dJTEz02espGzduTL169di+fTtgZuG+/i/6rFmzip4x+KrLLruM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\nEFDZi4gEAZW9iEgQUNmLiASB/wN8FSViNpJblAAAAABJRU5ErkJggg==\n", + "text": [ + "<matplotlib.figure.Figure at 0x2d9a8d0>" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "slope (g/s*K^.5) = 4.100\n", + "\n", + " intercept (g/s) = 1.798\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7-3, page no. 29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "#Units are not mentioned in this example\n", + "import math\n", + "t1=15.\n", + "t2=30.\n", + "F1=0.298\n", + "F2=0.0527\n", + "#Calculations and printing:\n", + "print(\"Semi-log plot\")\n", + "b1=(math.log(F2/F1))/(t2-t1)\n", + "a1=math.exp(math.log(F1) -b1*t1)\n", + "print '%s %.3f %s %.4f %s' %(\" \\n F=\",a1,\"exp(\",b1,\"t)\")\n", + "print(\"\\n Log plot\")\n", + "b2=(math.log(F2/F1))/(math.log(t2/t1))\n", + "a2=math.exp(math.log(F1) -b2*math.log(t1))\n", + "print '%s %.1f %s %.1f %s' %(\" \\n F=\",a2,\"exp(\",b2,\"t)\")\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Semi-log plot\n", + " \n", + " F= 1.685 exp( -0.1155 t)\n", + "\n", + " Log plot\n", + " \n", + " F= 259.3 exp( -2.5 t)\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 15, + "text": [ + "''" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Principles_of_Chemical_Processes/Chapter3.ipynb b/Elementary_Principles_of_Chemical_Processes/Chapter3.ipynb new file mode 100755 index 00000000..c4f34658 --- /dev/null +++ b/Elementary_Principles_of_Chemical_Processes/Chapter3.ipynb @@ -0,0 +1,835 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9d66b3fe8b5c5765d2f891791ade4779b402bd1eb1946b5f370c5174c61892cb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3: Processes and Process Variables" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1-1, page no. 44" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "mass=215.0 #kg\n", + "\n", + "#Calculations and printing :\n", + "\n", + "density=13.546*62.43\n", + "print '%s %.2f' %(\"density of mercury (lbm/ft^3) = \",density)\n", + "#the multiplication factor is to convert density from gm/cc to lbm/ft^3.\n", + "volume=mass/(.454*density) #ft^3\n", + "#the division by 0.454 is to convert mass in kg to lbm.\n", + "print '%s %.2f %s %.2f' %(\" \\n The volume of \",mass,\"kg of mercury is(ft^3) = \",volume)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "density of mercury (lbm/ft^3) = 845.68\n", + " \n", + " The volume of 215.00 kg of mercury is(ft^3) = 0.56\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 1, + "text": [ + "''" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1-2, page no. 45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "T1=20.0 # \u02daC\n", + "T2=100.0 # \u02daC\n", + "Vat20=0.560 #ft^3\n", + "D=0.0208333 #ft\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\"we know that V(T)=Vo[1+0.math.pow(18182x10,()-3)xT +0.math.pow(0078x10,()-6)xTxT]\")\n", + "Vat0=Vat20/(1+0.18182*math.pow(10,-3)*T1 +0.0078*math.pow(10,-6)*T1*T1)\n", + "#the function is defined with the variable as temperature\n", + "def volume(T):\n", + " volume=Vat0*(1+0.18182* math.pow(10,-3) *T +0.0078* math.pow(10,-6) *T*T)\n", + " return volume\n", + "\n", + "print '%s %.3f' %(\" vat20 (ft^3) = \",volume(T1))\n", + "print '%s %.3f' %(\" \\n vat100 (ft^3) =\",volume(T2))\n", + "change=((volume(T2))-(volume(T1)))*4/(math.pi*D*D)\n", + "print '%s %.3f' %(\" \\n change in the height of mercury level (ft) = \",change)\n", + "#the answer is a bit different due to rounding off of volume(T2) in textbook\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "we know that V(T)=Vo[1+0.math.pow(18182x10,()-3)xT +0.math.pow(0078x10,()-6)xTxT]\n", + " vat20 (ft^3) = 0.560\n", + " \n", + " vat100 (ft^3) = 0.568\n", + " \n", + " change in the height of mercury level (ft) = 23.931\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 2, + "text": [ + "''" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-1, page no. 48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "mass=100.0 #g of CO2\n", + "M=44.01 #molecular weight\n", + "\n", + "#Calculations and print '%s %.3f' %ing :\n", + "\n", + "moles=mass/M\n", + "print '%s %.3f' %(\"\\n no.of moles=\",moles)\n", + "lbmole=moles/453.6\n", + "print '%s %.3f' %(\"\\n no.of lb moles=\",lbmole)\n", + "Cmoles=moles\n", + "print '%s %.3f' %(\"\\n no.of moles of carbon=\",Cmoles)\n", + "Omoles=2*moles\n", + "print '%s %.3f' %(\"\\n no.of moles of oxygen=\",Omoles)\n", + "O2moles=moles\n", + "print '%s %.3f' %(\"\\n no.of moles of dioxygen=\",O2moles)\n", + "gramsO=Omoles*16\n", + "print '%s %.3f' %(\"\\n no.of grams of oxygen=\",gramsO)\n", + "gramsO2=O2moles*32\n", + "print '%s %.3f' %(\"\\n no.of grams of oxygen=\",gramsO2)\n", + "moleculesCO2=moles*6.02*math.pow(10,(23))\n", + "print '%s %.3E' %(\"\\n no.of molecules of CO2 =\",moleculesCO2)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + " no.of moles= 2.272\n", + "\n", + " no.of lb moles= 0.005\n", + "\n", + " no.of moles of carbon= 2.272\n", + "\n", + " no.of moles of oxygen= 4.544\n", + "\n", + " no.of moles of dioxygen= 2.272\n", + "\n", + " no.of grams of oxygen= 72.711\n", + "\n", + " no.of grams of oxygen= 72.711\n", + "\n", + " no.of molecules of CO2 = 1.368E+24\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 3, + "text": [ + "''" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-2, page no. 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "xA=0.15 #mass fraction\n", + "yB=0.20 #mole fraction\n", + "mass=175.0 #kg of solution\n", + "flowrate1=53.0 #lbm/h\n", + "flowrate2=1000.0 #mol/min\n", + "massofA=300.0 #lbm\n", + "molarB=28.0 #kmolB/s\n", + "\n", + "#Calculations and printing :\n", + "\n", + "massA=mass*xA\n", + "print '%s %d %s %.3f'%(\" \\n Mass of A in\",mass,\" kg of solution (kg A) =\",massA)\n", + "flowrateA=flowrate1*xA\n", + "print '%s %d %s %.3f'%(\" \\n Mass flow rate of A in a stream flowing at\",flowrate1,\" lbm/h (lbm A/h)=\",flowrateA )\n", + "flowrateB=flowrate2*yB\n", + "print '%s %d %s %.3f'%(\" \\n Molar flowrate of B in a stream flowing at\",flowrate2,\" mol/min (molB/min)= \",flowrateB)\n", + "Totalflowrate=molarB/yB\n", + "print '%s %d %s %.3f'%(\" \\n Total flow rate of a solution with\",molarB,\"kmolB/s=\",Totalflowrate)\n", + "MassSolution=massofA/xA\n", + "print '%s %d %s %.3f'%(\" \\n Mass of solution that contains\",massofA,\" lbm of A =\",MassSolution)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " \n", + " Mass of A in 175 kg of solution (kg A) = 26.250\n", + " \n", + " Mass flow rate of A in a stream flowing at 53 lbm/h (lbm A/h)= 7.950\n", + " \n", + " Molar flowrate of B in a stream flowing at 1000 mol/min (molB/min)= 200.000\n", + " \n", + " Total flow rate of a solution with 28 kmolB/s= 140.000\n", + " \n", + " Mass of solution that contains 300 lbm of A = 2000.000\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 4, + "text": [ + "''" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-3, page no. 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "#let the total mass be 100\n", + "massO2=16.0\n", + "massCO=4.0\n", + "massCO2=17.0\n", + "massN2=63.0\n", + "MO2=32.0\n", + "MCO=28.0\n", + "MCO2=44.0\n", + "MN2=28.0\n", + "\n", + "#Calculations and print '%s %.3f' %ing :\n", + "\n", + "molO2=massO2/MO2\n", + "molCO=massCO/MCO\n", + "molCO2=massCO2/MCO2\n", + "molN2=massN2/MN2\n", + "TotalMol=molO2+molCO+molCO2+molN2\n", + "print '%s %.3f' %(\" \\n molefraction of O2=\",molO2/TotalMol)\n", + "print '%s %.3f' %(\" \\n molefraction of CO=\",molCO/TotalMol)\n", + "print '%s %.3f' %(\" \\n molefraction of CO2=\",molCO2/TotalMol)\n", + "print '%s %.3f' %(\" \\n molefraction of N2=\",molN2/TotalMol)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " \n", + " molefraction of O2= 0.152\n", + " \n", + " molefraction of CO= 0.044\n", + " \n", + " molefraction of CO2= 0.118\n", + " \n", + " molefraction of N2= 0.686\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 5, + "text": [ + "''" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-4, page no. 51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "yN2=0.79\n", + "MN2=28.0 #g/mol\n", + "MO2=32.0 g/mol\n", + "xN2=0.767\n", + "\n", + "#Calculations and printing :\n", + "\n", + "Mbar=yN2*MN2+(1-yN2)*MO2\n", + "print '%s %.3f' %(\" \\n average molecular weight of air from molar composition (g/mol) = \",Mbar)\n", + "InvMbar=xN2/28 + (1-xN2)/32\n", + "print '%s %.3f' %(\" \\n average molecular weight of air from mass composition (g/mol) = \",1./InvMbar)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " \n", + " average molecular weight of air from molar composition (g/mol) = 28.840\n", + " \n", + " average molecular weight of air from mass composition (g/mol) = 28.840\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 6, + "text": [ + "''" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3-5, page no. 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "conc=0.5 #molar\n", + "rate=1.25 #m^3/min\n", + "D=1.03 #no units\n", + "M=98.0\n", + "\n", + "#Calculations and print '%s %.3f' %ing :\n", + "\n", + "mass_conc=conc*98.\n", + "print '%s %.3f' %(\"mass concentration of sulfuric acid (kg/m^3) = \",mass_conc)\n", + "mass_flowrate=rate*mass_conc/60.\n", + "print '%s %.3f' %(\" \\n Mass flow rate of sulfuric acid (kg/s) = \",mass_flowrate)\n", + "massfraction=1/(rate*D*1000/60)\n", + "print '%s %.3f' %(\" \\n Mass fraction of sulfuric acid=\",massfraction)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "mass concentration of sulfuric acid (kg/m^3) = 49.000\n", + " \n", + " Mass flow rate of sulfuric acid (kg/s) = 1.021\n", + " \n", + " Mass fraction of sulfuric acid= 0.047\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 8, + "text": [ + "''" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4-1, page no. 55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "Pressure=2.00*100000. #Pa\n", + "\n", + "#Calculations and printing :\n", + "\n", + "Pressure=Pressure*1000./(13600.*9.807)\n", + "print '%s %.3E' %(\"Pressure (mm of Hg) = \",Pressure)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pressure (mm of Hg) = 1.500E+03\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 9, + "text": [ + "''" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4-2, page no. 55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "P0=10.4*1.013*100000. /10.33 #m H2O\n", + "D=1000.0 #kg/m^3\n", + "g=9.807 #m/s^2\n", + "h=30.0 #m\n", + "\n", + "#Calculations and printing :\n", + "\n", + "Ph=P0+D*g*h\n", + "print '%s %.3E' %(\"Pressure at the bottom of the lake (N/m^2) = \",Ph)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pressure at the bottom of the lake (N/m^2) = 3.962E+05\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 12, + "text": [ + "''" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4-3, page no. 59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "import math\n", + "spg=1.05\n", + "h1=382. #mm of fluid\n", + "h2=374. #mm of fluid\n", + "reading= -2 #in of Hg\n", + "Patm=30. #in of Hg\n", + "#Calculations and printing:\n", + "deltaP=(spg-1)*980.7*(h1-h2)/10.\n", + "print '%s %.1f' %(\"Pressure drop between points 1 and 2 (dynes/cm^2) = \",deltaP)\n", + "P1=Patm+reading\n", + "print '%s %.1f' %(\"\\n Absolute Pressure (in Hg) = \",P1)\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Pressure drop between points 1 and 2 (dynes/cm^2) = 39.2\n", + "\n", + " Absolute Pressure (in Hg) = 28.0\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 13, + "text": [ + "''" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5-2, page no. 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "T1=20.0 #F\n", + "T2=80.0 #F\n", + "\n", + "#Calculations and printing :\n", + "\n", + "#In this code I used a function to achieve the conversion\n", + "def conversion(fahrenheit):\n", + " conversion=(fahrenheit-32)/1.8\n", + " return conversion;\n", + "\n", + "difference=conversion(T2)-conversion(T1)\n", + "print '%s %.2f %s %.2f' %(\"Equivalent temperature of\",T2-T1,\" temperature in C =\",difference)\n", + "deltaTF=T2-T1\n", + "deltaTC=deltaTF/1.8\n", + "print '%s %.2f' %(\" \\n By second method, result=\",deltaTC)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent temperature of 60.00 temperature in C = 33.33\n", + " \n", + " By second method, result= 33.33\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 14, + "text": [ + "''" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5-3, page no. 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "import math\n", + "a=0.487\n", + "b=2.29*math.pow(10, -4)\n", + "print '%s %.2f %s %.1E %s' %(\"Given Cp(Btu/lbm.F) = \",a,\"+\",b,\"T (F)\")\n", + "print(\"\\n Step 1\")\n", + "af1=a+b*32\n", + "bf1=b*1.8\n", + "print(\"\\n Step 2\")\n", + "af2=af1*1.8/(454*9.486*math.pow(10, -4))\n", + "bf2=bf1*1.8/(454*9.486*math.pow(10, -4))\n", + "print '%s %.2f %s %.1E %s' %(\"Final Cp(J/g.C) = \",af2,\"+\",bf2,\"T (C)\")\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Given Cp(Btu/lbm.F) = 0.49 + 2.3E-04 T (F)\n", + "\n", + " Step 1\n", + "\n", + " Step 2\n", + "Final Cp(J/g.C) = 2.07 + 1.7E-03 T (C)\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 15, + "text": [ + "''" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Principles_of_Chemical_Processes/Chapter4.ipynb b/Elementary_Principles_of_Chemical_Processes/Chapter4.ipynb new file mode 100755 index 00000000..0ee3fbef --- /dev/null +++ b/Elementary_Principles_of_Chemical_Processes/Chapter4.ipynb @@ -0,0 +1,1771 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:cf9c0d91c794ab27af5e7da425a28b836daf2f988940f28c4a10f4b42bda95d5" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4: Fundamentals of Material Balances" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2-1, page no. 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "input=50000 #ppl/yr\n", + "generation=22000 #ppl/yr\n", + "consumption=19000 #ppl/yr\n", + "output=75000 #ppl/yr\n", + "print (\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "#Calculations and printing:\n", + "accumulation = input+generation-output-consumption\n", + "print (\" We know that accumulation=input+generation-output-consumption\")\n", + "print '%s %d %s' %('Hence, Each year population decreases by', -accumulation, 'people')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " We know that accumulation=input+generation-output-consumption\n", + "Hence, Each year population decreases by 22000 people\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 1, + "text": [ + "''" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2-2, page no. 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "inputBenzene=500 #kg/h\n", + "inputToluene=500 #kg/h\n", + "UpStreamBenzene=450 #kg/h\n", + "DownStreamToluene=475 #kg/h\n", + "#Calculations and printing:\n", + "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Using benzene balance,\")\n", + "m2dot=inputBenzene-UpStreamBenzene\n", + "print ' %s %d %s' %(' m2dot= ', m2dot, ' kg B/h')\n", + "print(\"Using Toluene balance\")\n", + "m1dot=inputToluene-DownStreamToluene\n", + "print ' %s %d %s' %(' m1dot= ', m1dot, ' kg T/h')\n", + "print(\"To check we can perform Overall mass balance\")\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Using benzene balance,\n", + " m2dot= 50 kg B/h\n", + "Using Toluene balance\n", + " m1dot= 25 kg T/h\n", + "To check we can perform Overall mass balance\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 2, + "text": [ + "''" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2-3, page no. 87" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "m1=200 #g\n", + "m2=150 #g\n", + "x1=0.4 #methanol/g\n", + "x2=0.7 #methanol/g\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "m=m1+m2\n", + "print '%s %d %s' %(' Total mass after mixing m=',m, 'g',)\n", + "x=(m1*x1 + m2*x2)/m\n", + "print'%s %.2f %s %.2f' %(' The composition of the methanol in the product is', x ,' and water is', 1-x)\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " Total mass after mixing m= 350 g\n", + " The composition of the methanol in the product is 0.53 and water is 0.47\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 3, + "text": [ + "''" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2-4, page no. 88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "rate=0.1 #kmol/min\n", + "x1=0.1 #mole fraction hexane vapour\n", + "vol=10 #m^3\n", + "d=0.659 #kg/L\n", + "M=86.2 #Kg/kmol\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "ndot=rate/(1-x1)\n", + "deltaN= -vol*d*1000 /M\n", + "tf=deltaN/(-0.1 * ndot)\n", + "print '%s %.2f %s' %(' The time Required for the Total process= ',tf, 'min')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " The time Required for the Total process= 6880.51 min\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 5, + "text": [ + "''" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3-1, page no. 92" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "Vdot=20 #CC/min\n", + "x=0.015\n", + "MH2O=18.02 #g\n", + "DH2O=1 #g/CC\n", + "x1=0.2\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "n2=Vdot*DH2O/MH2O\n", + "print'%s %.2f %s' %('n2=', n2, ' mol/min')\n", + "print(\"Using Water Balance,\")\n", + "n3=n2/x\n", + "print'%s %.2f %s' %('n3=', n3, ' mol/min')\n", + "print(\"Using total mole balance,\")\n", + "n1=(n3-n2)/(1+x1)\n", + "print'%s %.2f %s' %('n1=', n1, ' mol/min')\n", + "print(\"Using N2 balance,\")\n", + "y=1-x-0.79*n1/n3\n", + "print '%s %.2f %s' %('y=',y,' mol O2/mol')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "n2= 1.11 mol/min\n", + "Using Water Balance,\n", + "n3= 73.99 mol/min\n", + "Using total mole balance,\n", + "n1= 60.73 mol/min\n", + "Using N2 balance,\n", + "y= 0.34 mol O2/mol\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 6, + "text": [ + "''" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3-2, page no. 94" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "basis=100 #mol\n", + "FinalBasis=1250.0 #lb-moles/h\n", + "#Calulations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "multiply=FinalBasis/basis\n", + "Feed=100*multiply\n", + "TopStream=50*multiply\n", + "BottomStream1=12.5*multiply\n", + "BottomStream2=37.5*multiply\n", + "print '%s %.2f %s' %(' Final Basis=',Feed,' lb-moles /h')\n", + "print '%s %.2f %s' %(' Final Top Stream Feed=',TopStream, 'lb-moles /h')\n", + "print '%s %.2f %s' %(' Final Bottom Stream Feed 1 =', BottomStream1,'lb-moles A/h')\n", + "print '%s %.2f %s' %(' Final Bottom Stream Feed 2 =' ,BottomStream2,' lb-moles B/h')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " Final Basis= 1250.00 lb-moles /h\n", + " Final Top Stream Feed= 625.00 lb-moles /h\n", + " Final Bottom Stream Feed 1 = 156.25 lb-moles A/h\n", + " Final Bottom Stream Feed 2 = 468.75 lb-moles B/h\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 7, + "text": [ + "''" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3-3, page no. 97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "basis=100 #kg\n", + "inputx=0.2\n", + "outputx=0.08\n", + "D=1 #kg/L\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Using NaOH balance\")\n", + "m2=inputx*basis/outputx\n", + "print '%s %.2f %s' %('m2=',m2 ,'Kg NaOH')\n", + "print(\"Using Total mass balance\")\n", + "m1=m2-basis\n", + "print '%s %.2f %s' %('m1=',m1, 'Kg Water')\n", + "V1=m1/D\n", + "print '%s %.2f %s' %(' V1=',V1 ,'Litres')\n", + "Ratio1=V1/basis\n", + "Ratio2=m2/basis\n", + "print '%s %.2f %s' %(' Ratio of lt water/Kg Feed =', Ratio1,' lt water/Kg Feed')\n", + "print '%s %.2f %s' %(' Ratio of Kg product/Kg Feed =', Ratio2 ,'Kg product/Kg Feed')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Using NaOH balance\n", + "m2= 250.00 Kg NaOH\n", + "Using Total mass balance\n", + "m1= 150.00 Kg Water\n", + " V1= 150.00 Litres\n", + " Ratio of lt water/Kg Feed = 1.50 lt water/Kg Feed\n", + " Ratio of Kg product/Kg Feed = 2.50 Kg product/Kg Feed\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 8, + "text": [ + "''" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3-5, page no. 102" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "inputx=0.45\n", + "outputx=0.95\n", + "basis=2000 #L/h\n", + "outputBasis=100 #Kmol\n", + "M1=78.11\n", + "M2=92.13\n", + "D=0.872\n", + "z=0.08\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "mass1=outputBasis*M1*outputx\n", + "mass2=outputBasis*M2*(1-outputx)\n", + "mass=mass1+mass2\n", + "yB2=mass1/mass\n", + "m1=basis*D\n", + "print '%s %.2f %s' %(' m1=',m1,' Kg/h')\n", + "mB3=z*inputx*m1\n", + "print '%s %.2f %s' %(' mB3=',mB3,' Kg/h')\n", + "print(\"Using Benzene balance,\")\n", + "m2=(inputx*m1-mB3)/yB2\n", + "print '%s %.2f %s' %(' m2=',m2, 'Kg/h')\n", + "print(\"Using Toluene balance,\")\n", + "mT3=(1-inputx)*m1-(1-yB2)*m2\n", + "print '%s %.2f %s' %(' mT3=',mT3,' Kg/h')\n", + "m3=mB3+mT3\n", + "print '%s %.2f %s' %(' m3=',m3,' Kg/h')\n", + "yB3=mB3/m3\n", + "print '%s %.2f %s' %(' yB3=',yB3,' kg B/kg')\n", + "yT3=1-yB3\n", + "print '%s %.2f %s' %(' yT3=',yT3, 'kg T/kg')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " m1= 1744.00 Kg/h\n", + " mB3= 62.78 Kg/h\n", + "Using Benzene balance,\n", + " m2= 766.84 Kg/h\n", + "Using Toluene balance,\n", + " mT3= 914.38 Kg/h\n", + " m3= 977.16 Kg/h\n", + " yB3= 0.06 kg B/kg\n", + " yT3= 0.94 kg T/kg\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 10, + "text": [ + "''" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4-1, page no. 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "inputMass1=100 #Kg/h\n", + "inputMass2=30 #Kg/h\n", + "outputMass1=40 #Kg/h\n", + "outputMass2=30 #Kg/h\n", + "inputx1=0.5\n", + "inputx2=0.3\n", + "outputx1=0.9\n", + "outputx2=0.6\n", + "#Calculations and printing\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"using Overall Mass balance, \")\n", + "m3=inputMass1+inputMass2-outputMass1-outputMass2\n", + "print '%s %d %s' %('m3=',m3,' Kg/h')\n", + "print(\"using Overall balance on A, \")\n", + "x3=(inputMass1*inputx1+inputMass2*inputx2-outputMass1*outputx1-outputMass2*outputx2)/m3\n", + "print '%s %.2f %s' %('x3=',x3,' Kg A/kg')\n", + "print(\"using Mass balance on Unit 1, \")\n", + "m1=inputMass1-outputMass1\n", + "print '%s %d %s' %('m1=',m1,' Kg/h')\n", + "print(\"using A balance on Unit 1, \")\n", + "x1=(inputMass1*inputx1-outputMass1*outputx1)/m1\n", + "print '%s %.2f %s' %('x1=',x1,' Kg A/kg')\n", + "print(\"using Mass balance on mixing point, \")\n", + "m2=inputMass2+m1\n", + "print '%s %d %s' %('m2=',m2,'Kg/h')\n", + "print(\"using A balance on mixing point, \")\n", + "x2=(inputMass2*inputx2+m1*x1)/m2\n", + "print '%s %.2f %s' %('x2=',x2 ,'Kg A/kg')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "using Overall Mass balance, \n", + "m3= 60 Kg/h\n", + "using Overall balance on A, \n", + "x3= 0.08 Kg A/kg\n", + "using Mass balance on Unit 1, \n", + "m1= 60 Kg/h\n", + "using A balance on Unit 1, \n", + "x1= 0.23 Kg A/kg\n", + "using Mass balance on mixing point, \n", + "m2= 90 Kg/h\n", + "using A balance on mixing point, \n", + "x2= 0.26 Kg A/kg\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 12, + "text": [ + "''" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4-2, page no. 107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "from numpy import matrix, array, dot\n", + "from numpy.linalg import inv\n", + "import math\n", + "massin=100 #kg\n", + "M1=100 #kg\n", + "M2=75 #Kg\n", + "massout=43.1 #kg\n", + "inputx=0.5\n", + "outputxA=0.053\n", + "outputxM=0.016\n", + "m1xA=0.275\n", + "m1xM=0.725\n", + "m3xW=0.03\n", + "m3xA=0.09\n", + "m3xM=0.88\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Using Balances around two-extraxtor system,\")\n", + "print(\"Balance on total mass,\")\n", + "\n", + "A=array([[1,1],[m1xA,m3xA]])\n", + "b=array([[massin+M1+M2-massout],[massin*inputx - massout*outputxA]])\n", + "C=dot(inv(A),b)\n", + "m1=C[0,0]\n", + "m3=C[1,0]\n", + "print '%s %.2f %s' %(' m1=',m1,' Kg')\n", + "print '%s %.2f %s' %(' m3=',m3,' Kg')\n", + "print(\"Balance on M\")\n", + "xM1=(massin+M2-massout*outputxM-m3*m3xM)/m1\n", + "print '%s %.2f %s' %(' xM1=',xM1,' kg MIBK/kg')\n", + "print(\"Balances around Extract mixing point, \")\n", + "print(\"Balance on A\")\n", + "mA4=m1*m1xA+m3*m3xA\n", + "print '%s %.2f %s' %(' mA4=',mA4 ,'Kg Acetone')\n", + "print(\"Balance on M\")\n", + "mM4=m1*xM1+m3*m3xM\n", + "print '%s %.2f %s' %(' mM4=',mM4,' Kg MIBK')\n", + "print(\"Balance on W\")\n", + "mW4=m1*(m1xM-xM1) + m3*m3xW\n", + "print '%s %.2f %s' %(' mW4=',mW4 ,'Kg Water')\n", + "print(\"Balances around the First extractor\")\n", + "print(\"Balance on A\")\n", + "mA2=massin*inputx-m1*m1xA\n", + "print '%s %.2f %s' %(' mA2=',mA2,' Kg Acetone')\n", + "print(\"Balance on M\")\n", + "mM2=massin-m1*xM1\n", + "print '%s %.2f %s' %(' xM1=',xM1, 'Kg MIBK')\n", + "print(\"Balance on W\")\n", + "mW2=massin*inputx- m1*(m1xM-xM1)\n", + "print '%s %.2f %s' %(' mW2=',mW2,'Kg Water')\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Using Balances around two-extraxtor system,\n", + "Balance on total mass,\n", + " m1= 145.11 Kg\n", + " m3= 86.79 Kg\n", + "Balance on M\n", + " xM1= 0.67 kg MIBK/kg\n", + "Balances around Extract mixing point, \n", + "Balance on A\n", + " mA4= 47.72 Kg Acetone\n", + "Balance on M\n", + " mM4= 174.31 Kg MIBK\n", + "Balance on W\n", + " mW4= 9.87 Kg Water\n", + "Balances around the First extractor\n", + "Balance on A\n", + " mA2= 10.10 Kg Acetone\n", + "Balance on M\n", + " xM1= 0.67 Kg MIBK\n", + "Balance on W\n", + " mW2= 42.73 Kg Water\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5-1, page no. 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "x1=0.960\n", + "x2=0.977\n", + "x3=0.983\n", + "basis=100 #mol\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Overall dry Air balance,\")\n", + "n1=x3*basis/x1\n", + "print '%s %.2f %s' %('n1=',n1,' mol Fresh feed')\n", + "print(\"Overall mole balance,\")\n", + "n3=n1-basis\n", + "print '%s %.2f %s' %('n3=',n3,' mol Water condensed')\n", + "print(\"Mole balance on mixing point,\")\n", + "print(\"n1+n5=n2\")\n", + "print(\"Water balance on mixing point\")\n", + "print ' %.2f %s %.2f %s %.2f %s' %(1-x1,' n1+ ', 1-x3,' n5 =', 1-x2,'n2')\n", + "A=([[1 ,-1],[1-x2,-(1-x3)]])\n", + "b=([[n1],[(1-x1)*n1]])\n", + "C=numpy.dot(linalg.inv(A),b)\n", + "n2=C[0,0]\n", + "n5=C[1,0]\n", + "print '%s %.2f %s' %(' \\n n2=',n2,' mol')\n", + "print '%s %.2f %s' %(' \\n n5=',n5,' mol Recycled')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Overall dry Air balance,\n", + "n1= 102.40 mol Fresh feed\n", + "Overall mole balance,\n", + "n3= 2.40 mol Water condensed\n", + "Mole balance on mixing point,\n", + "n1+n5=n2\n", + "Water balance on mixing point\n", + " 0.04 n1+ 0.02 n5 = 0.02 n2\n", + " \n", + " n2= 392.52 mol\n", + " \n", + " n5= 290.12 mol Recycled\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 15, + "text": [ + "''" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5-2, page no. 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "feed=4500 #kg/h\n", + "feedx=0.333\n", + "m3x=0.494\n", + "m5x=0.364\n", + "x=0.95\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "m6x=m5x\n", + "print '%s %.2f %s' %('Given, m4=',x,' (m4+m5)')\n", + "print(\"using Overall K2CrO4 balance, \")\n", + "print ' %.2f %s %.2f %s %.2f %s' %(feedx,' *', feed ,'= m4+', m6x,'m5')\n", + "A=([[1-x,-x],[1,m6x]])\n", + "b=([[0],[feedx*feed]])\n", + "C=numpy.dot(linalg.inv(A),b)\n", + "m4=C[0,0]\n", + "m5=C[1,0]\n", + "print '%s %.2f %s' %(' \\n m4=',m4,' K2CrO4 crystals/h')\n", + "print '%s %.2f %s' %(' \\n m5=',m5,' entrained solution/h')\n", + "print(\"Overall Total mass balance , \")\n", + "m2=feed-m4-m5\n", + "print '%s %.2f %s' %('m2=',m2,' Kg H2O evaporated/h')\n", + "print(\"Mass balance around the crystallizer,\")\n", + "print(\"m3=m4+m5+m6\")\n", + "print(\"Water balance around the crystallizer, \")\n", + "print ' %.2f %s %.2f %s %.2f %s' %(1-m3x, 'm3=', 1-m5x,' m5 + ',1-m6x,' m6')\n", + "D=([[1,-1],[1,(-1+m6x)/(1-m3x)]])\n", + "e=([[m4+m5],[(1-m5x)*m5/(1-m3x)]])\n", + "F=numpy.dot(linalg.inv(D),e)\n", + "m3=F[0,0]\n", + "m6=F[1,0]\n", + "print '%s %.2f %s' %(' \\n m3=',m3,' Kg/h fed to the crystallizer')\n", + "print '%s %.2f %s' %(' \\n m6=',m6,' Kg/h')\n", + "ratio=m6/feed\n", + "print '%s %.2f %s' %(' \\n ratio=',ratio,' Kg recycle/ Kg fresh feed')\n", + "print(\"mass balance around Recycle-fresh feed mixing point, \")\n", + "m1=feed+m6\n", + "print '%s %.2f %s' %('m1=',m1,' kg/h feed to the evaporator')\n", + "print(\"With out recycle, \")\n", + "print(\"m3=622 Kg/h\")\n", + "print(\"m5=2380 Kg/h\")\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Given, m4= 0.95 (m4+m5)\n", + "using Overall K2CrO4 balance, \n", + " 0.33 * 4500.00 = m4+ 0.36 m5\n", + " \n", + " m4= 1470.33 K2CrO4 crystals/h\n", + " \n", + " m5= 77.39 entrained solution/h\n", + "Overall Total mass balance , \n", + "m2= 2952.28 Kg H2O evaporated/h\n", + "Mass balance around the crystallizer,\n", + "m3=m4+m5+m6\n", + "Water balance around the crystallizer, \n", + " 0.51 m3= 0.64 m5 + 0.64 m6\n", + " \n", + " m3= 7193.31 Kg/h fed to the crystallizer\n", + " \n", + " m6= 5645.60 Kg/h\n", + " \n", + " ratio= 1.25 Kg recycle/ Kg fresh feed\n", + "mass balance around Recycle-fresh feed mixing point, \n", + "m1= 10145.60 kg/h feed to the evaporator\n", + "With out recycle, \n", + "m3=622 Kg/h\n", + "m5=2380 Kg/h\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 16, + "text": [ + "''" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6-1, page no. 120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "basis=100 #mol\n", + "xP=0.1\n", + "xN=0.12\n", + "xA=0.78\n", + "x=0.3\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "nP=basis*xP\n", + "nN=basis*xN\n", + "nO2=basis*xA*0.21\n", + "nN2=basis-nP-nN-nO2\n", + "if nN/nP>1:\n", + " print(\"NH3 is in excess\")\n", + "else:\n", + " print(\"Propene is in excess\")\n", + "\n", + "if nO2/nP>1:\n", + " print(\"O2 is in excess\")\n", + "else:\n", + " print(\"propene is in excess\")\n", + "\n", + "nO2reacted=nP*1.5\n", + "nNreacted=nP*1\n", + "ExcessAmmonia=(nN-nNreacted)*100/nNreacted\n", + "ExcessO2=(nO2-nO2reacted)*100/nO2reacted\n", + "print '%s %.2f' %(' \\n percentage excess Ammonia=',ExcessAmmonia)\n", + "print '%s %.2f' %(' \\n percentage excess Oxygen=',ExcessO2)\n", + "nPout=(1-x)*nP\n", + "print '%s %.2f %s' %(' \\n no.of moles of Propylene left=', nPout,' mol')\n", + "E=nP-nPout\n", + "nNout=nN-E\n", + "nO2out=nO2-1.5*E\n", + "nAc=E\n", + "nW=3*E\n", + "print'%s %.2f %s' %(' \\n no.of moles of Ammonia left=', nNout,' mol')\n", + "print'%s %.2f %s' %(' \\n no.of moles of oxygen left=' ,nO2out,' mol')\n", + "print'%s %.2f %s' %(' \\n no.of moles of ACN formed= ',nAc ,'mol')\n", + "print'%s %.2f %s' %(' \\n no.of moles of water formed=', nW ,'mol')\n", + "print'%s %.2f %s' %(' \\n no.of moles of Nirtogen present= ',nN2 ,'mol')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "NH3 is in excess\n", + "O2 is in excess\n", + " \n", + " percentage excess Ammonia= 20.00\n", + " \n", + " percentage excess Oxygen= 9.20\n", + " \n", + " no.of moles of Propylene left= 7.00 mol\n", + " \n", + " no.of moles of Ammonia left= 9.00 mol\n", + " \n", + " no.of moles of oxygen left= 11.88 mol\n", + " \n", + " no.of moles of ACN formed= 3.00 mol\n", + " \n", + " no.of moles of water formed= 9.00 mol\n", + " \n", + " no.of moles of Nirtogen present= 61.62 mol\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 17, + "text": [ + "''" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6-2, page no. 122" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "print(\"Given, Reaction is CO(g)+ H2O(g) --> CO2(g)+ H2(g)\")\n", + "nCO=1.\n", + "nH2O=2.\n", + "#Calculations and printing:\n", + "nt=nCO+nH2O\n", + "print(\"\\n It is found that Ee=0.667\")\n", + "Ee=0.667\n", + "yCO=(nCO-Ee)/nt\n", + "yH2O=(nH2O-Ee)/nt\n", + "yCO2=Ee/nt\n", + "yH2=Ee/nt\n", + "print(\"\\n Ee is obtained by solving the quadratic yCO2*yH2/(yCO*yH2O)=1\")\n", + "print(\"\\n Limiting reactant is CO\")\n", + "print '%s %.3f' %(\"\\n yCO = \",yCO)\n", + "print '%s %.3f' %(\"\\n yCO2 = \",yCO2)\n", + "print '%s %.3f' %(\"\\n yH2O = \",yH2O)\n", + "print '%s %.3f' %(\"\\n yH2 = \",yH2)\n", + "nCOF=yCO*nt\n", + "fCO=(nCO-nCOF)/nCO\n", + "print '%s %.3f' %(\"\\n Fractional conversion of CO at equilibrium is \", fCO)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Given, Reaction is CO(g)+ H2O(g) --> CO2(g)+ H2(g)\n", + "\n", + " It is found that Ee=0.667\n", + "\n", + " Ee is obtained by solving the quadratic yCO2*yH2/(yCO*yH2O)=1\n", + "\n", + " Limiting reactant is CO\n", + "\n", + " yCO = 0.111\n", + "\n", + " yCO2 = 0.222\n", + "\n", + " yH2O = 0.444\n", + "\n", + " yH2 = 0.222\n", + "\n", + " Fractional conversion of CO at equilibrium is 0.667\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 18, + "text": [ + "''" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6-3, page no. 124" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "basis=100 #mol\n", + "x=0.850\n", + "conv1=0.501\n", + "conv2=0.471\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "n1=(1-conv1)*basis*x\n", + "n2=conv2*basis*x\n", + "E1=n2\n", + "E2=basis*x - E1-n1\n", + "n3=E1-E2\n", + "n4=2*E2\n", + "n5=basis*(1-x)\n", + "nt=n1+n2+n3+n4+n5\n", + "selectivity=n2/n4\n", + "print '%s %.2f %s ' %('selectivity=',selectivity,' mol Ethene/mol methane')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "selectivity= 7.85 mol Ethene/mol methane \n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 19, + "text": [ + "''" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7-2, page no. 135" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "basis=100 #mol\n", + "x=0.850\n", + "conv1=0.501\n", + "conv2=0.471\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "n1=(1-conv1)*basis*x\n", + "n2=conv2*basis*x\n", + "E1=n2\n", + "E2=basis*x - E1-n1\n", + "n3=E1-E2\n", + "n4=2*E2\n", + "n5=basis*(1-x)\n", + "nt=n1+n2+n3+n4+n5\n", + "selectivity=n2/n4\n", + "print '%s %.2f %s ' %('selectivity=',selectivity,' mol Ethene/mol methane')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "selectivity= 7.85 mol Ethene/mol methane \n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 20, + "text": [ + "''" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7-3, page no. 139" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "x0=0.996\n", + "basis=100 #mol combined feed to the reactor\n", + "inputxH2=0.7\n", + "single_pass=0.6\n", + "inputxCO2=0.28\n", + "molI=2\n", + "Ix=0.004\n", + "final=155 #kmol/h\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Reactor analysis, \")\n", + "n2=(1-single_pass)*basis*inputxH2\n", + "print '%s %.2f %s' %('n2=',n2,' mol H2')\n", + "print(\"H2 balance\")\n", + "consH2=basis*inputxH2-n2\n", + "print '%s %.2f %s' %('H2 moles consumed=',consH2,' mol H2')\n", + "print(\"CO2 balance\")\n", + "n1=basis*inputxCO2-consH2/3\n", + "print '%s %.2f %s' %('n1=',n1,'mol CO2')\n", + "print(\"Methanol balance\")\n", + "n3=consH2/3\n", + "print '%s %.2f %s' %('n3=',n3,'mol Methanol')\n", + "print(\"H2O balance\")\n", + "n4=consH2/3\n", + "print '%s %.2f %s' %('n4=',n4,' mol H2O')\n", + "print(\"condenser analysis\")\n", + "print(\"Total mole balance\")\n", + "n5=n1+n2+molI\n", + "print '%s %.2f %s' %('n5=',n5 ,'mol')\n", + "print(\"CO2 balance\")\n", + "x5C=n1/n5\n", + "print '%s %.2f %s' %('x5C=',x5C,' mol CO2/mol')\n", + "print(\"H2 balance\")\n", + "x5H=n2/n5\n", + "print '%s %.2f %s' %('x5H=', x5H,'mol CO2/mol')\n", + "x1=1-x5C-x5H\n", + "print '%s %.2f %s' %('x1=',x1,' mol I/mol')\n", + "print(\"Fresh Feed-Recycle mixing point analysis\")\n", + "print(\"Total mole balance\")\n", + "print '%s %.2f' %('n0+nr=',basis)\n", + "print(\"I balance\")\n", + "print '%s %.2f %s %.2f %s %.2f' %('n0', Ix,' + nr ',x1,' = ',molI)\n", + "A=([[1, 1],[Ix, x1]])\n", + "b=([[basis],[molI]])\n", + "C=numpy.dot(linalg.inv(A),b) #Here We solve two linear equations simultaneously\n", + "n0=C[0,0]\n", + "nr=C[1,0]\n", + "print '%s %.2f %s' %(' \\n n0=',n0,' mol fresh feed')\n", + "print '%s %.2f %s' %(' \\n nr=', nr ,'mol recycle')\n", + "x0C=(basis*inputxCO2-nr*x5C)/n0\n", + "print '%s %.2f %s' %(' \\n x0C=',x0C,' mol CO2/mol')\n", + "x0H=1-x0C-Ix\n", + "print '%s %.2f %s' %(' \\n x0H=',x0H ,'mol H2/mol')\n", + "print(\"Recycle-Purge splitting Analysis\")\n", + "print(\"Total mole balance\")\n", + "np=n5-nr\n", + "print '%s %.2f %s' %('np=',np,' mol purge')\n", + "print(\"Flow chart scaling\")\n", + "Factor=final/n3\n", + "print '%s %.2f %s' %('Factor for scaling=',Factor,' Kmol/h/mol')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Reactor analysis, \n", + "n2= 28.00 mol H2\n", + "H2 balance\n", + "H2 moles consumed= 42.00 mol H2\n", + "CO2 balance\n", + "n1= 14.00 mol CO2\n", + "Methanol balance\n", + "n3= 14.00 mol Methanol\n", + "H2O balance\n", + "n4= 14.00 mol H2O\n", + "condenser analysis\n", + "Total mole balance\n", + "n5= 44.00 mol\n", + "CO2 balance\n", + "x5C= 0.32 mol CO2/mol\n", + "H2 balance\n", + "x5H= 0.64 mol CO2/mol\n", + "x1= 0.05 mol I/mol\n", + "Fresh Feed-Recycle mixing point analysis\n", + "Total mole balance\n", + "n0+nr= 100.00\n", + "I balance\n", + "n0 0.00 + nr 0.05 = 2.00\n", + " \n", + " n0= 61.40 mol fresh feed\n", + " \n", + " nr= 38.60 mol recycle\n", + " \n", + " x0C= 0.26 mol CO2/mol\n", + " \n", + " x0H= 0.74 mol H2/mol\n", + "Recycle-Purge splitting Analysis\n", + "Total mole balance\n", + "np= 5.40 mol purge\n", + "Flow chart scaling\n", + "Factor for scaling= 11.07 Kmol/h/mol\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 21, + "text": [ + "''" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8-1, page no. 143" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "xN2wet=0.6\n", + "xCO2wet=0.15\n", + "xO2wet=0.1\n", + "xH2O=0.15\n", + "basis=100 #mol Wet gas \n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"part 1\")\n", + "wet=xN2wet+xCO2wet+xO2wet\n", + "xN2dry=xN2wet/wet\n", + "xCO2dry=xCO2wet/wet\n", + "xO2dry=xO2wet/wet\n", + "print '%s %f %s' %(' \\n xN2 dry =', xN2dry,' mol N2/mol dry gas')\n", + "print '%s %f %s' %(' \\n xO2 dry =',xO2dry,' mol O2/mol dry gas')\n", + "print '%s %f %s' %(' \\n xCO2 dry =', xCO2dry ,'mol CO2/mol dry gas')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "part 1\n", + " \n", + " xN2 dry = 0.705882 mol N2/mol dry gas\n", + " \n", + " xO2 dry = 0.117647 mol O2/mol dry gas\n", + " \n", + " xCO2 dry = 0.176471 mol CO2/mol dry gas\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 22, + "text": [ + "''" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8-2, page no. 145" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "basisButane=100. #mol/h butane\n", + "basisAir=5000. #mol/h\n", + "#calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "nO2Theoretical=basisButane*6.5\n", + "nAirTheoretical=nO2Theoretical*4.76\n", + "percent=(basisAir-nAirTheoretical)*100/nAirTheoretical\n", + "print '%s %.2f' %(' \\n percent excess air=',percent)\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " percent excess air= 61.60\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 23, + "text": [ + "''" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8-3, page no. 147" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "basis=100. #mol ethane feed\n", + "E1=0.9\n", + "E2=0.25\n", + "excess=0.5\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\" 50percent excess air\")\n", + "nO2theoretical=basis*3.5\n", + "n0=nO2theoretical*(1+excess)/0.21\n", + "print '%s %.4f %s' %(' n0=',n0,' mol air fed')\n", + "print(\"90% ethane conversion\")\n", + "n1=(1-E1)*basis\n", + "print '%s %.4f' %('No. of moles of ethane unreacted=',n1)\n", + "print(\"25% conversion to CO\")\n", + "n4=E2*(basis-n1)*2\n", + "print '%s %.4f %s' %('n4=', n4 ,'mol CO')\n", + "print(\"nitrogen balance\")\n", + "n3=0.79*n0\n", + "print '%s %.4f %s' %('n3=', n3,' mol N2')\n", + "print(\"Atomic carbon balance\")\n", + "n5=2*basis-2*n1-n4\n", + "print '%s %.4f %s' %('n5=', n5 ,'mol CO2')\n", + "print(\"Atomic hydrogen balance\")\n", + "n6=(basis*6-n1*6)/2\n", + "print '%s %.4f %s' %('n6=', n6,' mol H2O')\n", + "print(\"Atomic oxygen balance\")\n", + "n2=(nO2theoretical*1.5*2-n4-n5*2-n6)/2\n", + "print '%s %.4f %s' %('n2=', n2 ,'mol O2')\n", + "dry=n1+n2+n3+n4+n5\n", + "wet=dry+n6\n", + "y1=n1/dry\n", + "print '%s %.4f %s' %('\\n y1=', y1,' mol C2H6/mol')\n", + "y2=n2/dry\n", + "print '%s %.4f %s' %('\\n y2=', y2 ,'mol O2/mol')\n", + "y3=n3/dry\n", + "print '%s %.4f %s' %('\\n y3=', y3,' mol N2/mol')\n", + "y4=n4/dry\n", + "print '%s %.4f %s' %('\\n y4=', y4 ,'mol CO/mol')\n", + "y5=n5/dry\n", + "print '%s %.4f %s' %('\\n y5=',y5,' mol CO2/mol')\n", + "ratio=n6/dry\n", + "print '%s %.4f %s' %(' \\n ratio=',ratio,' mol H2O/mol dry stack gas')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " 50percent excess air\n", + " n0= 2500.0000 mol air fed\n", + "90% ethane conversion\n", + "No. of moles of ethane unreacted= 10.0000\n", + "25% conversion to CO\n", + "n4= 45.0000 mol CO\n", + "nitrogen balance\n", + "n3= 1975.0000 mol N2\n", + "Atomic carbon balance\n", + "n5= 135.0000 mol CO2\n", + "Atomic hydrogen balance\n", + "n6= 270.0000 mol H2O\n", + "Atomic oxygen balance\n", + "n2= 232.5000 mol O2\n", + "\n", + " y1= 0.0042 mol C2H6/mol\n", + "\n", + " y2= 0.0970 mol O2/mol\n", + "\n", + " y3= 0.8238 mol N2/mol\n", + "\n", + " y4= 0.0188 mol CO/mol\n", + "\n", + " y5= 0.0563 mol CO2/mol\n", + " \n", + " ratio= 0.1126 mol H2O/mol dry stack gas\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 24, + "text": [ + "''" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8-4, page no. 149" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "basis=100 #mol pf product gas\n", + "xCO=0.015\n", + "xCO2=0.060\n", + "xO2=0.082\n", + "xN2=0.843\n", + "#Calculations and printing:\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"N2 balance\")\n", + "na=basis*xN2/0.79\n", + "print '%s %.2f %s' %('na=',na,' mol air')\n", + "print(\"Atomic C balance\")\n", + "nc=basis*xCO + basis*xCO2\n", + "print '%s %.2f %s' %('nc=',nc ,'mol C')\n", + "print(\"Atomic O balance\")\n", + "nw=0.21*na*2-basis*(xCO + xCO2*2 + xO2*2)\n", + "print '%s %.2f %s' %('nw=',nw,' mol oxygen')\n", + "print(\"Atomic H2 balance\")\n", + "nh=nw*2\n", + "print '%s %.2f %s' %('nh=',nh ,'mol H2')\n", + "ratio=nh/nc\n", + "print '%s %.2f %s' %('\\n C/H ratio in fuel=',ratio,' mol H/mol C')\n", + "print(\"percent excess air\")\n", + "nO2theoretical=nc + nh/4\n", + "print '%s %.2f %s' %('nO2 theoretical=',nO2theoretical ,'mol O2')\n", + "nO2fed=0.21*na\n", + "print '%s %.2f %s' %(' \\n nO2fed=',nO2fed,' mol O2')\n", + "percent=(nO2fed-nO2theoretical)*100/nO2theoretical\n", + "print '%s %.2f %s' %('\\n percentage excess air=',percent ,'excess air')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "N2 balance\n", + "na= 106.71 mol air\n", + "Atomic C balance\n", + "nc= 7.50 mol C\n", + "Atomic O balance\n", + "nw= 14.92 mol oxygen\n", + "Atomic H2 balance\n", + "nh= 29.84 mol H2\n", + "\n", + " C/H ratio in fuel= 3.98 mol H/mol C\n", + "percent excess air\n", + "nO2 theoretical= 14.96 mol O2\n", + " \n", + " nO2fed= 22.41 mol O2\n", + "\n", + " percentage excess air= 49.80 excess air\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 25, + "text": [ + "''" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9-1, page no. 151" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "feed=59.6 #mol/s\n", + "x=0.2\n", + "TopFlow1=48.7 #mol/s\n", + "outputx1=0.021\n", + "BottomFlow1=10.9\n", + "TopFlow2=48.3\n", + "outputx2=0.063\n", + "BottomFlow2=6.4\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Design\")\n", + "MEKin1=feed*x\n", + "MEKout1=TopFlow1*outputx1 + BottomFlow1\n", + "closure1=MEKout1*100/MEKin1\n", + "print '%s %d %s' %('closure1=',closure1,' percent')\n", + "print(\"Experiment\")\n", + "MEKin2=feed*x\n", + "MEKout2=TopFlow2*outputx2 + BottomFlow2\n", + "closure2=MEKout2*100/MEKin2\n", + "print '%s %d %s' %('closure2=',closure2, 'percent')\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Design\n", + "closure1= 100 percent\n", + "Experiment\n", + "closure2= 79 percent\n" + ] + } + ] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Principles_of_Chemical_Processes/Chapter5.ipynb b/Elementary_Principles_of_Chemical_Processes/Chapter5.ipynb new file mode 100755 index 00000000..344fae37 --- /dev/null +++ b/Elementary_Principles_of_Chemical_Processes/Chapter5.ipynb @@ -0,0 +1,807 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:dfc2bc36e1576488bcd07ee5435a22145b28c649187a0989b470c7cdd7161f42" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 5: Single-Phase Systems" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1-1, page no. 190" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "wtperct=0.5\n", + "Dwater=0.998 #g/cm^3\n", + "Dsulfuric=1.834 #g/cm^3\n", + "\n", + "\n", + "#Calculations and printing :\n", + "\n", + "invPbar=wtperct/Dwater + (1-wtperct)/Dsulfuric\n", + "print '%s %.3f' %(\"Density calculated using volume additvity (g/cm^3) =\",1/invPbar)\n", + "Pbar=wtperct*Dwater + (1-wtperct)*Dsulfuric\n", + "print '%s %.3f' %(\" \\n Density calculated using mass additivity (g/cm^3) =\",Pbar)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Density calculated using volume additvity (g/cm^3) = 1.293\n", + " \n", + " Density calculated using mass additivity (g/cm^3) = 1.416\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 1, + "text": [ + "''" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2-1, page no. 192" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "T=23.0+273 #kelvin\n", + "P=3.0+14.7 #psi \n", + "#conversion of pressure from psig to psi requires addition of 14.67 which is 1 atm\n", + "R=0.08206 #lt-atm\n", + "MN2=28.0 #molecular wt.\n", + "weight=100.0 #grams\n", + "\n", + "#Calculations and printing :\n", + "\n", + "n=weight/MN2 #mol\n", + "V=n*R*T*14.7/P #lt\n", + "print '%s %.3f' %(\"assuming ideal gas behaviour, volume (litres) = \",V)\n", + "Vcap=V/n\n", + "if(Vcap>5):\n", + " print(\"ideal gas equation yields error less than 1 percent for diatomic gas\")\n", + "else:\n", + " print(\"ideal gas equation yields error greater than 1 percent for diatomic gases\")\n", + "\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "assuming ideal gas behaviour, volume (litres) = 72.046\n", + "ideal gas equation yields error less than 1 percent for diatomic gas\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 2, + "text": [ + "''" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2-2, page no. 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "T=360.0+273 #Kelvin\n", + "P=3.0 #atm\n", + "Vdot=1100.0 #kg/h\n", + "M=58.1 \n", + "\n", + "#Calculations and printing :\n", + "\n", + "ndot=Vdot/M #kmol/h\n", + "vdot=ndot*22.4*T/(273*P)\n", + "print '%s %.3f' %(\"The volumetric flow rate of the stream (m^3/h) = \",vdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The volumetric flow rate of the stream (m^3/h) = 327.781\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 3, + "text": [ + "''" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2-3, page no. 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "V1=10.0 #ft^3\n", + "T1=70.0+460 #R\n", + "P1=1.0 #atm\n", + "P2=2.5 #atm\n", + "T2=610.0+460 #R\n", + "\n", + "#Calculations and printing :\n", + "\n", + "V2=V1*P1*T2/(P2*T1)\n", + "print '%s %.3f' %(\"Volume in final state (ft^3) = \",V2)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volume in final state (ft^3) = 8.075\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 4, + "text": [ + "''" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2-4, page no. 196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "#all the calculations are done in R scale\n", + "T2=285.0+460 #R\n", + "T1=32.0+460 #R\n", + "P2=1.30 #atm\n", + "P1=1.0 #atm\n", + "V1dot=3.95*100000. #ft^3/h\n", + "\n", + "\n", + "#Calculations and printing :\n", + "\n", + "#SCFH means ft^3(STP/h)\n", + "ndot=3.95*math.pow(10,5)/359.\n", + "print '%s %.3f' %(\" Molar flowrate (lb-moles/hr) = \",ndot)\n", + "V2dot=V1dot*T2*P1/(T1*P2)\n", + "print '%s %.3E' %(\" \\n True volumetric flowrate (ft^3/h) = \",V2dot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Molar flowrate (lb-moles/hr) = 1100.279\n", + " \n", + " True volumetric flowrate (ft^3/h) = 4.601E+05\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 5, + "text": [ + "''" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2-5, page no. 197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "flowinA=400.0 #L/min\n", + "flowinN=419.0 # m^3 STP /min\n", + "Pfinal=6.3 #gauge\n", + "Tfinal=325.0 # C\n", + "Pacetone=501.0 #mm of Hg\n", + "Dacetone=791.0 #g/L\n", + "Macetone=58.08 #g\n", + "T1=300.0 #k\n", + "P1=1238.0 #mm Hg original\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "n2cap=flowinA*Dacetone/Macetone\n", + "print '%s %.3f' %(\" \\n Molar flowrate of Acetone (mol Acetone/min) = \",n2cap)\n", + "P=Pfinal*760 + 763\n", + "y4=Pacetone/P\n", + "print '%s %.3f' %(\" \\n Mole fraction of Acetone in the final flow (mol Acetone/mol) = \",y4)\n", + "print '%s %.3f' %(\" \\n Mole fraction of Nitrogen in the final flow (mol Nitrogen/mol) = \",1-y4)\n", + "n3cap=flowinN/0.0224\n", + "n4cap=n2cap/y4\n", + "print(\"By using Overall Molar balance,\")\n", + "n1cap=n4cap-n2cap-n3cap\n", + "V1cap=n1cap*0.0224*T1*760/(1*273*P1)\n", + "print '%s %.3f' %(\"Volumetric Flowrate of Nitrogen (Nitrogen/min) = \",V1cap)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Molar flowrate of Acetone (mol Acetone/min) = 5447.658\n", + " \n", + " Mole fraction of Acetone in the final flow (mol Acetone/mol) = 0.090\n", + " \n", + " Mole fraction of Nitrogen in the final flow (mol Nitrogen/mol) = 0.910\n", + "By using Overall Molar balance,\n", + "Volumetric Flowrate of Nitrogen (Nitrogen/min) = 547.119\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 6, + "text": [ + "''" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3-1, page no. 202" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "T= -150.8 + 273.2 #k\n", + "Vcap= 3./2. #L/mol\n", + "Tc=126.2 #k\n", + "Pc=33.5 #atm\n", + "w=0.040\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "Pideal=0.08206*T/Vcap\n", + "print '%s %.3f' %(\" \\n The value of pressure as per Ideal gas equation (atm) = \",Pideal)\n", + "Tr=T/Tc\n", + "B0=0.083 - (0.422/math.pow(Tr,1.6))\n", + "B1=0.139 - (0.172/math.pow(Tr,4.2))\n", + "B=0.08206*Tc*(B0+w*B1)/Pc\n", + "Pvirial=0.08206*T*(1+ B/Vcap)/Vcap\n", + "print '%s %.3f' %(\"\\n The value of pressure as per Virial gas equation (atm) = \",Pvirial)\n", + "e=(Pideal-Pvirial)*100/Pvirial\n", + "print '%s %.3f' %(\" \\n Percentage error due to Ideal gas Equation =\",e)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " The value of pressure as per Ideal gas equation (atm) = 6.696\n", + "\n", + " The value of pressure as per Virial gas equation (atm) = 6.196\n", + " \n", + " Percentage error due to Ideal gas Equation = 8.072\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 7, + "text": [ + "''" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3-2, page no. 203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "V=2.5 #m^3\n", + "n=1.00 #Kmol\n", + "T= 300 #K\n", + "Tc=304.2 #K\n", + "Pc=72.9 #atm\n", + "w=0.225\n", + "R=0.08206\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "Vcap=V/n\n", + "a=0.42747*math.pow(R*Tc,2) /Pc\n", + "b=0.08664*R*Tc/Pc\n", + "m=0.48508+ 1.5171*w -0.1561*w*w\n", + "Tr=T/Tc\n", + "alpha=math.pow(1+ m*(1-math.sqrt(Tr)),2)\n", + "P=(R*T/(Vcap-b))-(alpha*a/(Vcap*(Vcap+b)))\n", + "print '%s %.3f' %(\" \\n Pressure of gas calculated using SRK equation (atm) = \",P)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Pressure of gas calculated using SRK equation (atm) = 9.381\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 9, + "text": [ + "''" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4-1, page no. 206" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "Vcap=50.0 #M^3/hr\n", + "P=40.0 #bar\n", + "T=300.0 #K\n", + "R=8.314\n", + "M=16.04 #kg/kmol\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "z=0.934\n", + "print '%s %.3f' %(\" \\n From the Table, z= \",z)\n", + "ncap=P*Vcap*101.325/(z*R*T*1.01325)\n", + "mcap=ncap*M\n", + "print '%s %.3f' %(\" \\n Mass flow rate of Methane (Kg/hr) = \",mcap)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " From the Table, z= 0.934\n", + " \n", + " Mass flow rate of Methane (Kg/hr) = 1377.071\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 10, + "text": [ + "''" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4-2, page no. 209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "n=100.0 #gm-moles\n", + "V=5.0 #ltr\n", + "T= -20.6 + 273.2 #K\n", + "Tc=126.2 #K\n", + "Pc=33.5 #atm\n", + "R=0.08206\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "Tr=T/Tc\n", + "Vrideal=V*Pc/(n*R*Tc)\n", + "print '%s %.3f' %(\" \\n Tr= \",Tr)\n", + "print '%s %.3f' %(\"\\n Vrideal=\",Vrideal)\n", + "z=1.77\n", + "print '%s %.3f' %(\"\\n From the graphs, z=\",z)\n", + "P=z*R*T*n/V\n", + "Pr=P/Pc\n", + "print '%s %.3f' %(\" \\n Pr= \",Pr)\n", + "print '%s %.3f' %(\" \\n Pressure in the cylinder (atm) = \",P)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Tr= 2.002\n", + "\n", + " Vrideal= 0.162\n", + "\n", + " From the graphs, z= 1.770\n", + " \n", + " Pr= 21.904\n", + " \n", + " Pressure in the cylinder (atm) = 733.784\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 11, + "text": [ + "''" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4-3, page no. 212" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "yN2=0.25\n", + "yH2=1.0-yN2\n", + "P=800.0 #atm\n", + "T= -70+273.2 #K\n", + "TcH2=33.0 #K\n", + "TcN2=126.2 #K\n", + "PcH2=12.8 #atm\n", + "PcN2=33.5 #atm\n", + "R=0.0083\n", + "#Calculations and printing :\n", + "\n", + "print(\"All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Applying newton corrections for Hydrogen,\")\n", + "TcaH2=TcH2+8.\n", + "PcaH2=PcH2+8.\n", + "Tcbar=yH2*TcaH2 + yN2*TcN2\n", + "Pcbar=yH2*PcaH2 + yN2*PcN2\n", + "Trbar=T/Tcbar\n", + "Prbar=P/Pcbar\n", + "print '%s %.3f' %(\" \\n Trbar = \",Trbar)\n", + "print '%s %.3f' %(\" \\n Prbar = \",Prbar)\n", + "Zm=1.86\n", + "print '%s %.4f' %(\" \\n From the graph, Zm = \",Zm)\n", + "Vcap=Zm*R*T/P\n", + "print '%s %.5f' %(\" \\n Specific Volume of Mixture (L/mol) = \",Vcap)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Applying newton corrections for Hydrogen,\n", + " \n", + " Trbar = 3.262\n", + " \n", + " Prbar = 33.368\n", + " \n", + " From the graph, Zm = 1.8600\n", + " \n", + " Specific Volume of Mixture (L/mol) = 0.00392\n" + ] + } + ] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Principles_of_Chemical_Processes/Chapter6.ipynb b/Elementary_Principles_of_Chemical_Processes/Chapter6.ipynb new file mode 100755 index 00000000..e1ddd7be --- /dev/null +++ b/Elementary_Principles_of_Chemical_Processes/Chapter6.ipynb @@ -0,0 +1,1137 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:27df12dd9928a7234ef95007770cbb19da31332c34bb007953024d3646517da2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6: Multiphase Systems" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1-1, page no. 244" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "T2=15.4+273.2 #K\n", + "T1=7.6+273.2 #K\n", + "P1=40.0 #mm of Hg\n", + "P2=60.0 #mm of Hg\n", + "T=42.2+273.2 #K\n", + "R=8.314 #J/mol.k\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"let deltaHv/R = S\")\n", + "S= - (T1*T2* math.log(P2/P1))/(T1-T2)\n", + "deltahv=S*R\n", + "print '%s %d' %(\" \\n Latent Heat of Vaporization (J/mol) =\",deltahv)\n", + "B=math.log(P1) + S/T1\n", + "print '%s %.3f' %(\"\\n B=\",B)\n", + "P=math.exp(-S/T + B)\n", + "print '%s %.3f %s %.3f' %(\"\\n P* at\",T,\" K is (mm Hg)= \",P)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "let deltaHv/R = S\n", + " \n", + " Latent Heat of Vaporization (J/mol) = 35023\n", + "\n", + " B= 18.691\n", + "\n", + " P* at 315.400 K is (mm Hg)= 207.400\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 1, + "text": [ + "''" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3-1, page no. 250" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "P=760.0 #mm of Hg\n", + "Pstar=289.0 #mm of Hg\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "y=Pstar/P\n", + "print '%s %.3f %s %.3f' %(\" \\n Molar composition of Water is\",y,\" and Air is\",1-y)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Molar composition of Water is 0.380 and Air is 0.620\n" + ] + } + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3-2, page no. 251" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "T=100.0+273.2 #K\n", + "PT=5260.0 #mm of Hg\n", + "y=0.1 #by volume\n", + "basis= 100 #mol of feed gas\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "P=y*PT\n", + "if(P<760):\n", + " print(\"The Vapour is Super heated\")\n", + "elif(P==760): \n", + " print(\"The vapour is At Dew point\")\n", + "else:\n", + " print(\"The vapour is not Super heated\")\n", + "\n", + "print(\"From tables Tdp=90 C\")\n", + "print(\"Superheat = 100-90=10 C \")\n", + "print(\"Using Raoult law at the outlet\")\n", + "y1=355/PT\n", + "print '%s %.3f' %(\"y1=\",y1)\n", + "print(\"Balance on Dry Air\")\n", + "n2=basis*(1-y)/(1-y1)\n", + "print '%s %.3f' %(\"n2 (mol) = \",n2)\n", + "print(\"Total mole balance\")\n", + "n1=basis-n2\n", + "print '%s %.3f' %(\"n1 (mol) = \",n1)\n", + "print '%s %.3f' %(\" \\n Percentage condesation=\",n1*100/(y*basis))\n", + "Psaturation=760/y\n", + "print '%s %.3f %s' %(\"\\n Any increase in pressure above\", Psaturation,\" mm of Hg must cause condensation \")\n", + "print(\" \\n For the next part of the problem use the same code by modifying PT to be 8500 mm of Hg\")\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "The Vapour is Super heated\n", + "From tables Tdp=90 C\n", + "Superheat = 100-90=10 C \n", + "Using Raoult law at the outlet\n", + "y1= 0.067\n", + "Balance on Dry Air\n", + "n2 (mol) = 96.514\n", + "Total mole balance\n", + "n1 (mol) = 3.486\n", + " \n", + " Percentage condesation= 34.862\n", + "\n", + " Any increase in pressure above 7600.000 mm of Hg must cause condensation \n", + " \n", + " For the next part of the problem use the same code by modifying PT to be 8500 mm of Hg\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 1, + "text": [ + "''" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3-3, page no. 254" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "T=75.0 +273 #K\n", + "P75=289.0 #mm of Hg\n", + "hr=0.3\n", + "Porig=825.0 #mm of Hg\n", + "PorigBar=1.1 #bar\n", + "Vdot=1000.0 #M^3/h\n", + "R=0.0831\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print (\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "P=hr*P75\n", + "y=P/Porig\n", + "ndot=PorigBar*Vdot/(R*T)\n", + "ndotWater=ndot*y\n", + "print '%s %.3f' %(\" \\n Molar flowrate of Water (Kmol/h) = \",ndotWater)\n", + "ndotBDA=ndot*(1-y)\n", + "print '%s %.3f' %(\" \\n Molar flowrate of Dry Air (Kmol/h) = \",ndotBDA)\n", + "ndotO2=ndotBDA*0.21\n", + "print '%s %.3f' %(\" \\n Molar flowrate of Oxygen (Kmol/h) = \",ndotO2)\n", + "hm=P/(Porig-P)\n", + "ha=hm*18/29\n", + "hmdot=P75/(Porig-P75)\n", + "hp=100*hm/hmdot\n", + "print '%s %.3f' %(\" \\n Molal Humidity (mol water/mol BDA) = \",hm)\n", + "print '%s %.3f' %(\" \\n Absolute Humidity (kg water/kg BDA) = \",ha)\n", + "print '%s %.3f' %(\" \\n Percentage Humidity=\",hp)\n", + "print(\"\\n From table B.3, Tdp=48.7 C\")\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Molar flowrate of Water (Kmol/h) = 3.997\n", + " \n", + " Molar flowrate of Dry Air (Kmol/h) = 34.040\n", + " \n", + " Molar flowrate of Oxygen (Kmol/h) = 7.148\n", + " \n", + " Molal Humidity (mol water/mol BDA) = 0.117\n", + " \n", + " Absolute Humidity (kg water/kg BDA) = 0.073\n", + " \n", + " Percentage Humidity= 21.780\n", + "\n", + " From table B.3, Tdp=48.7 C\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 2, + "text": [ + "''" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4-1, page no. 255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "basis=100.0 #lb-mole/h\n", + "x=0.45\n", + "PH2O=31.6 #mm of Hg\n", + "PSO2=176.0 #mm of Hg\n", + "P=760.0 #mm of Hg\n", + "y=2.0\n", + "M1=64.0 \n", + "M2=18.0\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "yH2O=PH2O/P\n", + "ySO2=PSO2/P\n", + "yAir=1-yH2O-ySO2\n", + "print(\"Using Air balance, \")\n", + "nG2=(1-x)*basis/yAir\n", + "print '%s %.3f' %(\"nG2 (lbm/h) = \",nG2)\n", + "xSO2=y/102\n", + "xH2O=1-xSO2\n", + "print(\"Using SO2 balance, \")\n", + "nL2=(basis*x-nG2*ySO2)*M1/(xSO2)\n", + "print '%s %.3f' %(\"nL2 (lbm/h) = \",nL2)\n", + "print(\"Using H2O balance, \")\n", + "nL1=nG2*yH2O*M2 + nL2*xH2O\n", + "print '%s %.3f' %(\"nL1 (lbm H2O/h) = \",nL1)\n", + "SO2Absorbed=nL2*xSO2\n", + "SO2Fed=basis*x*M1\n", + "Fraction=SO2Absorbed/SO2Fed\n", + "print '%s %.3f' %(\" \\n Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) = \",Fraction)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Using Air balance, \n", + "nG2 (lbm/h) = 75.670\n", + "Using SO2 balance, \n", + "nL2 (lbm/h) = 89683.186\n", + "Using H2O balance, \n", + "nL1 (lbm H2O/h) = 87981.325\n", + " \n", + " Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) = 0.611\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 3, + "text": [ + "''" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4-2, page no. 258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "basis=100.0 #lb-mole/h\n", + "x=0.45\n", + "PH2O=31.6 #mm of Hg\n", + "PSO2=176.0 #mm of Hg\n", + "P=760.0 #mm of Hg\n", + "y=2.0\n", + "M1=64.0 \n", + "M2=18.0\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "yH2O=PH2O/P\n", + "ySO2=PSO2/P\n", + "yAir=1-yH2O-ySO2\n", + "print(\"Using Air balance, \")\n", + "nG2=(1-x)*basis/yAir\n", + "print '%s %.3f' %(\"nG2 (lbm/h) = \",nG2)\n", + "xSO2=y/102\n", + "xH2O=1-xSO2\n", + "print(\"Using SO2 balance, \")\n", + "nL2=(basis*x-nG2*ySO2)*M1/(xSO2)\n", + "print '%s %.3f' %(\"nL2 (lbm/h) = \",nL2)\n", + "print(\"Using H2O balance, \")\n", + "nL1=nG2*yH2O*M2 + nL2*xH2O\n", + "print '%s %.3f' %(\"nL1 (lbm H2O/h) = \",nL1)\n", + "SO2Absorbed=nL2*xSO2\n", + "SO2Fed=basis*x*M1\n", + "Fraction=SO2Absorbed/SO2Fed\n", + "print '%s %.3f' %(\" \\n Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) = \",Fraction)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Using Air balance, \n", + "nG2 (lbm/h) = 75.670\n", + "Using SO2 balance, \n", + "nL2 (lbm/h) = 89683.186\n", + "Using H2O balance, \n", + "nL1 (lbm H2O/h) = 87981.325\n", + " \n", + " Fraction SO2 absorbed (lbm SO2 absorbed/lbm SO2 fed) = 0.611\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 4, + "text": [ + "''" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5-1, page no. 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "basis=150.0 #kg feed\n", + "S100=0.905 #g AgNO3/g\n", + "S20=0.689 #g AgNO3/g\n", + "inputx=0.095 #kg water/kg\n", + "outputx=0.311 #kg water/kg\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"\\n Composition of Filter cake,\")\n", + "print(\"\\n m2=4m3\")\n", + "print(\"\\n Water balance around the crystallizer,\")\n", + "print '%s' %(\"\\n basis*inputx* kg H2O = outputx m1 + outputx m1 \\n\")\n", + "print(\"Mass balance around crystallizer, \\n\")\n", + "print '%d %s' %(basis,\"=m1+m2+m3\")\n", + "A=([[0, 1, -4],[outputx, 0, outputx],[1, 1, 1]])\n", + "b=([[0],[basis*inputx],[basis]])\n", + "C=numpy.dot(linalg.inv(A),b)\n", + "#Here we solved two linear equations simultaneously\n", + "m1=C[0,0]\n", + "print '%s %.2f' %(\" \\n m1 (Kg) = \",m1)\n", + "m2=C[1,0]\n", + "print '%s %d' %(\" \\n m2 (Kg) = \",m2)\n", + "m3=C[2,0]\n", + "print '%s %d' %(\" \\n m3 (Kg) = \",m3)\n", + "print(\"\\n Overall AgNO3 balance,\")\n", + "m5=(1-inputx)*basis - (1-outputx)*m1\n", + "print '%s %d' %(\"\\n m5 (kg AgNO3 crystals recovered) = \",m5)\n", + "percentage=m5*100/(basis*(1-inputx))\n", + "print '%s %.3f' %(\" \\n Percentage recovery=\",percentage)\n", + "print(\"\\n Overall mass balance\")\n", + "m4=basis-m1-m5\n", + "print '%s %d' %(\"\\n m4 (Kg water removed in the Dryer) = \",m4)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "\n", + " Composition of Filter cake,\n", + "\n", + " m2=4m3\n", + "\n", + " Water balance around the crystallizer,\n", + "\n", + " basis*inputx* kg H2O = outputx m1 + outputx m1 \n", + "\n", + "Mass balance around crystallizer, \n", + "\n", + "150 =m1+m2+m3\n", + " \n", + " m1 (Kg) = 19.77\n", + " \n", + " m2 (Kg) = 104\n", + " \n", + " m3 (Kg) = 26\n", + "\n", + " Overall AgNO3 balance,\n", + "\n", + " m5 (kg AgNO3 crystals recovered) = 122\n", + " \n", + " Percentage recovery= 89.963\n", + "\n", + " Overall mass balance\n", + "\n", + " m4 (Kg water removed in the Dryer) = 8\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 5, + "text": [ + "''" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5-2, page no. 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "inputx=0.6\n", + "basis=100.0 #kg Feed\n", + "S=63.0 #Kg KNO3/100 Kg H2O\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "outputx=S/(S+100)\n", + "print '%s %.3f' %(\"x (Kg KNO3/Kg) = \",outputx)\n", + "print(\"Water balance\")\n", + "m1=basis*(1-inputx)/(1-outputx)\n", + "print '%s %.3f' %(\" \\n m1 (Kg) = \",m1)\n", + "print(\"Mass balance\")\n", + "m2=basis-m1\n", + "print '%s %.3f' %(\" \\n m2 (kg) = \",m2)\n", + "percentage=m2*100/(basis*inputx)\n", + "print '%s %.3f' %(\" \\n Percentage of KNO3 in the feed that crystallizes is \",percentage)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "x (Kg KNO3/Kg) = 0.387\n", + "Water balance\n", + " \n", + " m1 (Kg) = 65.200\n", + "Mass balance\n", + " \n", + " m2 (kg) = 34.800\n", + " \n", + " Percentage of KNO3 in the feed that crystallizes is 58.000\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 6, + "text": [ + "''" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5-3, page no. 267" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "basis=1.0 #Tonne Epsom salt produced/h\n", + "inputx=0.301 #Tonne MgSO4/tonne\n", + "outputx=0.232 #Tonne MgSO4/tonne\n", + "M=120.4\n", + "M1=246.4\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Total mass balance\")\n", + "print(\"m1=1+m2\")\n", + "print(\"MgSO4 balance\")\n", + "print(\" \\n inputx m1 = bassis* M/ M1 + m2 outputx\")\n", + "A=([[1, -1],[inputx, -outputx]])\n", + "b=([[1],[basis*M/M1]])\n", + "C=numpy.dot(linalg.inv(A),b)\n", + "#Here we solved two linear equations simultaneously\n", + "m1=C[0,0]\n", + "m2=C[1,0]\n", + "print '%s %.3f' %(\" \\n m1 (Tonne/h) = \",m1)\n", + "print '%s %.3f' %(\" \\n m2 (Tonne/h) = \",m2)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Total mass balance\n", + "m1=1+m2\n", + "MgSO4 balance\n", + " \n", + " inputx m1 = bassis* M/ M1 + m2 outputx\n", + " \n", + " m1 (Tonne/h) = 3.719\n", + " \n", + " m2 (Tonne/h) = 2.719\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 7, + "text": [ + "''" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5-4, page no. 270" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "m1=5.0 #g of solute\n", + "m2=100.0 #g of Water\n", + "P=1.0 #atm\n", + "Tf=100.421 #C\n", + "Ti=25.0 #C\n", + "R=8.314 #J/mol.K\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "x=(Tf-100.)*40656./(R*373.16*373.16)\n", + "y=m2/18.016\n", + "Ms=m1*(1-x)/(y*x)\n", + "print '%s %.3f' %(\" \\n Ms (g/mol) =\",Ms)\n", + "deltaTm=R*273.16*273.16*x/6009.5\n", + "Tms=0-deltaTm\n", + "print '%s %.3f' %(\" \\n Tms (C)=\",Tms)\n", + "Pstar=(1-x)*23.756\n", + "print '%s %.3f' %(\" \\n Solvent Vapour pressure (mm Hg) = \",Pstar)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Ms (g/mol) = 60.028\n", + " \n", + " Tms (C)= -1.526\n", + " \n", + " Solvent Vapour pressure (mm Hg) = 23.405\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 8, + "text": [ + "''" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6-1, page no. 272" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "V1=200.0 #CC Acetone\n", + "x=0.1 #Wt acetone \n", + "V2=400.0 #CC chloroform\n", + "DA=0.792 #g/cc\n", + "DC=1.489 #g/cc\n", + "DW=1.0 #g/cc\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "Dbar=DA*DW/(x*DW + (1-x)*DA)\n", + "mass1=V1*Dbar\n", + "mass2=V2*DC\n", + "print(\"C balance\")\n", + "m4=mass2\n", + "print '%s %.3f' %(\" \\n m4=\",m4)\n", + "print(\"W balance\")\n", + "m2=(1-x)*mass1\n", + "print '%s %.3f' %(\" \\n m2=\",m2)\n", + "print(\"A balance\")\n", + "print(\"m1+m3=x * mass1\")\n", + "print(\"Distribution Cooefficient ,K=m3*(m1+m2)/m1*(m3+m4)\")\n", + "print(\"On solving, \")\n", + "m1=2.7\n", + "m3=16.8\n", + "percentage=m3*100/(x*mass1)\n", + "print '%s %.3f' %(\" \\n m1=\",m1)\n", + "print '%s %.3f' %(\" \\n m3=\",m3)\n", + "print '%s %.3f' %(\" \\n percentage of acetone transferred to chloroform=\",percentage)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "C balance\n", + " \n", + " m4= 595.600\n", + "W balance\n", + " \n", + " m2= 175.394\n", + "A balance\n", + "m1+m3=x * mass1\n", + "Distribution Cooefficient ,K=m3*(m1+m2)/m1*(m3+m4)\n", + "On solving, \n", + " \n", + " m1= 2.700\n", + " \n", + " m3= 16.800\n", + " \n", + " percentage of acetone transferred to chloroform= 86.206\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 9, + "text": [ + "''" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6-2, page no. 274" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "basis=1000.0 #Kg of solution\n", + "inputxA=0.3 #Wt. fraction of acetone\n", + "outputxA1=0.05\n", + "outputxM1=0.02\n", + "outputxA2=0.1\n", + "outputxM2=0.87\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Mass balance\")\n", + "print(\"ms + basis = mE + mR\")\n", + "print(\"acetone balance\")\n", + "print(\"inputxA *basis = outputxA2 *mE + outputxA1* mR\")\n", + "print(\"Water balance\")\n", + "print(\" \\n 1-inputxA *basis = 1-outputxA2-outputxM2 *mE + 1-outputxA1-outputxM1* mR\")\n", + "A=([[1, 1, -1],[outputxA2, outputxA1, 0],[1-outputxA2-outputxM2, 1-outputxA1-outputxM1, 0]])\n", + "b=([[basis],[inputxA*basis],[(1-inputxA)*basis]])\n", + "C=numpy.dot(linalg.inv(A),b)\n", + "#Here We solved three linear equations simultaneously\n", + "mE=C[0,0]\n", + "mR=C[1,0]\n", + "mS=C[2,0]\n", + "print '%s %.3f' %(\" \\n mE (Kg) = \",mE)\n", + "print '%s %.3f' %(\" \\n mR (Kg) = \",mR)\n", + "print '%s %.3f' %(\" \\n mS (Kg MIBK) = \",mS)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Mass balance\n", + "ms + basis = mE + mR\n", + "acetone balance\n", + "inputxA *basis = outputxA2 *mE + outputxA1* mR\n", + "Water balance\n", + " \n", + " 1-inputxA *basis = 1-outputxA2-outputxM2 *mE + 1-outputxA1-outputxM1* mR\n", + " \n", + " mE (Kg) = 2666.667\n", + " \n", + " mR (Kg) = 666.667\n", + " \n", + " mS (Kg MIBK) = 2333.333\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 10, + "text": [ + "''" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7-1, page no. 276" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "V=50.0 #L\n", + "P=1.0 #atm\n", + "T=34.0+273.2 #K\n", + "y=0.3\n", + "xF=0.001\n", + "R=0.08206\n", + "Pstar=169.0 #mm of Hg\n", + "Pmm=760.0 #mm of Hg\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "n=P*V/(R*T)\n", + "print '%s %.3f' %(\" \\n No.of moles (mol) = \",n)\n", + "y0=y*Pstar/Pmm\n", + "print '%s %.3f' %(\" \\n Y0= (mol CCl4/mol) = \",y0)\n", + "Pfinal=xF*Pmm\n", + "b=0.096*Pfinal\n", + "Xstar=0.794*b/(1+b)\n", + "print '%s %.3f' %(\" \\n Mass of CCl4 adsorbed to Carbon at equilibrium (g CCl4 ads/g C) = \",Xstar)\n", + "Mads=(y0*n- xF*n)*154\n", + "print '%s %.3f' %(\" \\n Mass of CCl4 adsorbed (g) = \",Mads)\n", + "Mc=Mads/Xstar\n", + "print '%s %.3f' %(\" \\n Mass of carbon Required (g) = \",Mc)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " No.of moles (mol) = 1.983\n", + " \n", + " Y0= (mol CCl4/mol) = 0.067\n", + " \n", + " Mass of CCl4 adsorbed to Carbon at equilibrium (g CCl4 ads/g C) = 0.054\n", + " \n", + " Mass of CCl4 adsorbed (g) = 20.071\n", + " \n", + " Mass of carbon Required (g) = 371.750\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 11, + "text": [ + "''" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Principles_of_Chemical_Processes/Chapter7.ipynb b/Elementary_Principles_of_Chemical_Processes/Chapter7.ipynb new file mode 100755 index 00000000..6fdddf6a --- /dev/null +++ b/Elementary_Principles_of_Chemical_Processes/Chapter7.ipynb @@ -0,0 +1,893 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:140795a2dc739119552fbce01ad304d2882d8bdd1339a30f3767fc2e9c80a840" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 7: Energy and Energy Balances" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2-1, page no. 318" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "ID=2.0 #cm\n", + "Vdot=2.0 #m^3/h\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "u=Vdot*100*100 /(math.pi*math.pow(ID/2,2) *3600)\n", + "mdot=Vdot*math.pow(10,3) /3600.\n", + "Ek=mdot*math.pow(u,2) /2\n", + "print '%s %.3f' %(\" \\n Ek (J/s) = \",Ek)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Ek (J/s) = 0.869\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 1, + "text": [ + "''" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2-2, page no. 318" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "g=9.81 #m/s^2\n", + "mdot=15.0 #Kg/s\n", + "z2=20.0 #m\n", + "z1=-220. #m\n", + "\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "Power=mdot*g*(z2-z1)\n", + "print '%s %d' %(\" \\n Power (J/s) = \",Power)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Power (J/s) = 35316\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 2, + "text": [ + "''" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3-1, page no. 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initilization of variables\n", + "Q1=2. #Kcal\n", + "Ek1=0#System is stationary\n", + "Ep1=0#No vertical displacement\n", + "W1=0#No moving boundaries\n", + "deltaU1=Q1-W1-Ek1-Ep1\n", + "deltaU1J=deltaU1*1000./0.23901 \n", + "print '%s %.2f' %(\"Energy gained by the gas in going from 25C to 100C is (J)\",deltaU1J)\n", + "W2=100. #J\n", + "Ek2=0 #The system is stationary at the initial and final states\n", + "Ep2=0#Assumed negligible by hypothesis\n", + "deltaU2=0#U depends only on T for an ideal gas and T doesn't change\n", + "Q2=deltaU2+W2+Ek2+Ep2\n", + "print '%s %.1f' %(\"Heat transferred to the gas while it expands and reequilibrates at 100C is (J)\",Q2)\n", + "raw_input(\"Press the Enter key to quit\")" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Energy gained by the gas in going from 25C to 100C is (J) 8367.85\n", + "Heat transferred to the gas while it expands and reequilibrates at 100C is (J) 100.0\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "Press the Enter key to quit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 3, + "text": [ + "''" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4-1, page no. 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "U=3800.0 #J/mol\n", + "P=1.0 #atm\n", + "Vcap=24.63 #L/mol\n", + "ndot=250.0 #Kmol/h\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "Hcap=U+P*Vcap*101.3\n", + "H=ndot*Hcap*math.pow(10,3)\n", + "print '%s %d' %(\" \\n Specific Enthalpy (J/mol) = \",Hcap)\n", + "print '%s %.3E' %(\"\\n Enthalpy of Helium (J/h) = \",H)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Specific Enthalpy (J/mol) = 6295\n", + "\n", + " Enthalpy of Helium (J/h) = 1.574E+09\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 4, + "text": [ + "''" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4-2, page no. 324" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "mdot=500.0/3600.#Kg/s\n", + "u1=60.0 #m/s\n", + "u2=360.0 #m/s\n", + "deltaZ=-5. #m\n", + "g=9.81 #m/s^2\n", + "Qdot= -10000.#Kcal/h\n", + "Ws=70.0 #KW\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "Ek=mdot*math.pow(10,-3)*(math.pow(u2,2)-math.pow(u1,2))/2\n", + "Ep=mdot*g*deltaZ/math.pow(10,3)\n", + "Qdot=Qdot/(0.239*3600.)\n", + "Hdot=Qdot-Ws-Ek-Ep\n", + "print '%s %f' %(\" \\n DeltaH (KW) = \",Hdot)\n", + "Hcap=Hdot/mdot\n", + "print '%s %f' %(\"\\n Specific Enthalpy (Kj/Kg) = \",Hcap)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " DeltaH (KW) = -90.365689\n", + "\n", + " Specific Enthalpy (Kj/Kg) = -650.632958\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 6, + "text": [ + "''" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5-1, page no. 326" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "H0=196.23 #Btu/lbm\n", + "H50=202.28 #Btu/lbm\n", + "Pfinal=51.99 #psia\n", + "Pinitial=18.90 #psia\n", + "Vfinal=1.920 #ft^3/lbm\n", + "Vinitial=4.969 #ft^3/lbm\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "deltaH=H0-H50\n", + "deltaU=deltaH+((Pfinal*Vfinal-Pinitial*Vinitial)*1.987/10.73)\n", + "print '%s %.3f' %(\" \\n change in Specific Enthalpy (Btu/lbm) = \",deltaH)\n", + "print '%s %.3f' %(\" \\n change in Specific Internal Energy (Btu/lbm) = \",deltaU)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " change in Specific Enthalpy (Btu/lbm) = -6.050\n", + " \n", + " change in Specific Internal Energy (Btu/lbm) = -4.956\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 8, + "text": [ + "''" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5-3, page no. 329" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "mdot=2000.0 #Kg/h\n", + "P=10.0 #bar\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"From Steam tables, \")\n", + "Hin=3201. #Kj/Kg\n", + "Hout=2675. #Kj/Kg\n", + "Ws= -mdot*(Hout-Hin)/3600.\n", + "print '%s %d' %(\"Work delivered by Turbine to surroundings (Kw) = \",Ws)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "From Steam tables, \n", + "Work delivered by Turbine to surroundings (Kw) = 292\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 9, + "text": [ + "''" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6-1, page no. 330" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "m1=120.0 #kg\n", + "m2=175.0 #kg\n", + "m3=295.0 #kg\n", + "ID=6.0 #cm\n", + "P=17.0 #bar\n", + "H1=125.7 #Kj/Kg\n", + "H2=271.9 #Kj/Kg\n", + "H3=2793.0 #Kj/kg\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "deltaH=m3*H3-m1*H1-m2*H2\n", + "print(\"From tables, Vdot=0.1166 m^3/kg\")\n", + "Vdot=0.1166 \n", + "A=math.pi*math.pow(ID/2,2) /math.pow(10,4)\n", + "u=m3*Vdot/(A*60)\n", + "Ek=m3*math.pow(u,2) /(2*math.pow(10,3))\n", + "Qdot=deltaH+Ek\n", + "print '%s %.3E' %(\"Heat required (Kj/min) = \",Qdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "From tables, Vdot=0.1166 m^3/kg\n", + "Heat required (Kj/min) = 7.673E+05\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 10, + "text": [ + "''" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6-2, page no. 331" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "basis=1.0 #Kg/s\n", + "x=0.6 #ethane\n", + "T1=150.0 #K\n", + "T2=250.0 #K\n", + "P=5.0 #bar\n", + "Hout1=434.5 #KJ/Kg\n", + "Hout2=130.2 #KJ/Kg\n", + "Hin1=314.3 #KJ/Kg\n", + "Hin2=30.0 #KJ/Kg\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "Qdot=basis*(x*Hout1+(1-x)*Hout2-x*Hin1-(1-x)*Hin2)\n", + "print '%s %.3f' %(\" \\n Heat required (KJ/Kg) = \",Qdot/basis)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Heat required (KJ/Kg) = 112.200\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 11, + "text": [ + "''" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6-3, page no. 332" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "m3=1150.0 #Kg/h\n", + "H3=2676.0 #KJ/Kg\n", + "H2=3074.0 #KJ/Kg\n", + "H1=3278.0 #KJ/Kg\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Mass balance on Water,\")\n", + "print(\"m3+m1=m2\")\n", + "print(\"Energy balance,\")\n", + "print(\"m3*H3+m1*H1=m2*H2\")\n", + "A=([[1,-1],[H2,-H1]])\n", + "b=([[m3],[m3*H3]])\n", + "C=numpy.dot(linalg.inv(A),b)\n", + "#here we solved two linear equations simultaneously.\n", + "m2=C[0,0]\n", + "m1=C[1,0]\n", + "print '%s %.3f' %(\" Input flowrate,m1 (Kg/h) = \",m1)\n", + "print '%s %.3f' %(\" \\n Output flowrate, m2 (Kg/h) = \",m2)\n", + "print(\"From tables,Vdot=3.11 m^3/Kg\")\n", + "Vdot=3.11\n", + "print '%s %.3f' %(\" Volumetric input flowrate (m^3/h) = \",m1*Vdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Mass balance on Water,\n", + "m3+m1=m2\n", + "Energy balance,\n", + "m3*H3+m1*H1=m2*H2\n", + " Input flowrate,m1 (Kg/h) = 2243.627\n", + " \n", + " Output flowrate, m2 (Kg/h) = 3393.627\n", + "From tables,Vdot=3.11 m^3/Kg\n", + " Volumetric input flowrate (m^3/h) = 6977.681\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 12, + "text": [ + "''" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7-1, page no. 334" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "Vdot=20.0 #L/min\n", + "P2=1.01325*100000. #atm\n", + "ID1=0.5 #cm\n", + "ID2=1.0 #cm\n", + "g=9.81 #m/s^2\n", + "deltaZ=50.0 #m\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "u1=Vdot*math.pow(10,4) /(math.pow(10,3) *60*math.pi*math.pow(ID1/(2),2))\n", + "u2=Vdot*math.pow(10,4) /(math.pow(10,3) *60*math.pi*math.pow(ID2/(2),2))\n", + "deltaP=-((u2*u2-u1*u1)/2 +g*deltaZ)*1000.\n", + "P1=P2-deltaP\n", + "print '%s %.3f' %(\" \\n P1 (Pa) = \",P1)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " P1 (Pa) = 456730.088\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 13, + "text": [ + "''" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7-2, page no. 335" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "deltaZ= -2.5 #ft\n", + "u1=0.0\n", + "D=50.0 #lbm/ft^3\n", + "F=0.80 #ft.lbf/lbm\n", + "V=5.0 #gal\n", + "g=32.174 #ft/s^2\n", + "ID=0.25 #in\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "u2=math.sqrt(2*32.174*(-F-g*deltaZ/32.174))\n", + "Vdot=u2*math.pi*math.pow(ID/(2),2) /144.\n", + "t=V*0.1337/(Vdot*60.)\n", + "print '%s %.3f' %(\"Total time taken (min) = \",t)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Total time taken (min) = 3.125\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 14, + "text": [ + "''" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7-3, page no. 336" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "Ws=1000000. #N.m/s\n", + "deltaP= -83.*1000. #N/m^2\n", + "g=9.81 #m/s^2\n", + "deltaZ= -103 #m\n", + "D=1000. #kg/m^3\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "mdot= -Ws/(deltaP/D + g*deltaZ)\n", + "print'%s %.3f' %(\" \\n Water flow rate (kg/s) = \",mdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Water flow rate (kg/s) = 914.553\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 15, + "text": [ + "''" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Principles_of_Chemical_Processes/Chapter8.ipynb b/Elementary_Principles_of_Chemical_Processes/Chapter8.ipynb new file mode 100755 index 00000000..34aa53f2 --- /dev/null +++ b/Elementary_Principles_of_Chemical_Processes/Chapter8.ipynb @@ -0,0 +1,1053 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:96d9d4935cd6f3c8e019e8d452a9980bb4f36e96fb606a03b9b38b11670ead8c" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8: Balances on Nonreactive Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1-1, page no. 362" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "basis=100. #mol/s\n", + "ninAcv=66.9 #mol/s\n", + "noutAcv=3.35 #mol/s\n", + "noutAcl=63.55 #mol/s\n", + "ninN2=33.1 #mol/s\n", + "noutN2=ninN2\n", + "HinAcv=35.7 #Kj/mol\n", + "HoutAcv=32.0 #Kj/mol\n", + "HinN2=1.16 #Kj/mol\n", + "HoutN2= -0.10 #Kj/mol\n", + "#Calculations and printing:\n", + "deltaH=noutAcv*HoutAcv+ noutN2*HoutN2-ninN2*HinN2-ninAcv*HinAcv\n", + "Q=deltaH\n", + "print '%s %.2f' %(\"Heat transferred from condenser to achieve required cooling and condensation at the rate of (kW) \",Q)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Heat transferred from condenser to achieve required cooling and condensation at the rate of (kW) -2322.84\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 2, + "text": [ + "''" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3-1, page no. 367" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "mass=200.0 #kg\n", + "Ti=20.0 #C\n", + "Tf=150.0 #C\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \\n \")\n", + "def Cv(T):\n", + " Cv=0.855+ T*9.42*math.pow(10,-4)\n", + " return Cv\n", + "\n", + "Ucap, err= scipy.integrate.quad(Cv,Ti,Tf) #integrate.quad is an inbult function used for definite integration\n", + "Q=mass*Ucap\n", + "print '%s %.3f' %(\"Heat Required (KJ) = \",Q)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n", + " \n", + "Heat Required (KJ) = 24311.820\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 3, + "text": [ + "''" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3-2, page no. 369" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "T1=20.0 #C\n", + "T2=100.0 #C\n", + "T3=90.0 #C\n", + "T4=30.0 #C\n", + "P=3.0 #bar\n", + "V=5.0 #L\n", + "R=0.08314 #L.bar/mol.K\n", + "T=363.0 #K\n", + "ndot=100.0 #mol/min\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \\n \")\n", + "print(\"part 1\")\n", + "def fun1(T):\n", + " fun1=0.02900+ T*0.2199*math.pow(10,-5) + math.pow(T,2) * 0.5723 *math.pow(10,-8) - math.pow(T,3) * 2.871 * math.pow(10,-12)\n", + " return fun1\n", + "\n", + "deltaH, err=scipy.integrate.quad(fun1,T1,T2) #scipy.integrate.quad is an inbult function used for definite integration\n", + "Qdot=ndot*deltaH\n", + "print '%s %.3f' %(\"Heat Transferred (KJ/min) = \",Qdot)\n", + "print(\"part2\")\n", + "def fun2(T):\n", + " fun2=0.02900+ T*0.2199*math.pow(10,-5) + math.pow(T,2) * 0.5723 *math.pow(10,-8) - math.pow(T,3) * 2.871 * math.pow(10,-12)-8.14* math.pow(10,-3)\n", + " return fun2\n", + "\n", + "deltaU, err2=scipy.integrate.quad(fun2,T3,T4) #scipy.integrate.quad is an inbult function used for definite integration\n", + "n=P*V/(R*T)\n", + "Q=n*deltaU\n", + "print '%s %.3f' %(\"Heat transferred (KJ) = \",Q)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n", + " \n", + "part 1\n", + "Heat Transferred (KJ/min) = 233.238\n", + "part2\n", + "Heat transferred (KJ) = -0.627\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 4, + "text": [ + "''" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3-3, page no. 371" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "T1=430.0 #C\n", + "T2=100.0 #C\n", + "ndot=15.0 #Kmol/min\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \\n \")\n", + "def fun(T):\n", + " fun=0.02894 + T* 0.4147 *math.pow(10,-5) + math.pow(T,2) * 0.3191 * math.pow(10,-8) - math.pow(T,3) * 1.965 * math.pow(10,-12)\n", + " return fun\n", + "\n", + "deltaH, err=scipy.integrate.quad(fun,T1,T2) #scipy.integrate.quad is an inbult function used for definite integration\n", + "Qdot=ndot*deltaH*math.pow(10,3) /60.\n", + "print '%s %.3f' %(\" \\n Rate of heat removal (KW) = \",Qdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n", + " \n", + " \n", + " Rate of heat removal (KW) = -2494.904\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 5, + "text": [ + "''" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3-4, page no. 373" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "x=0.6\n", + "T1=0.0\n", + "T2=400.0\n", + "ndot=150.0 #mol/h\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \\n \")\n", + "def fun1(T):\n", + " fun1=0.04937 + T*13.92*math.pow(10,-5) - math.pow(T,2) *5.816*math.pow(10,-8) + math.pow(T,3) *7.280 * math.pow(10,-12)\n", + " return fun1\n", + "\n", + "def fun2(T):\n", + " fun2=0.06803 +T*22.59*math.pow(10,-5) - math.pow(T,2) *13.11*math.pow(10,-8) + math.pow(T,3) *31.71 * math.pow(10,-12)\n", + " return fun2\n", + "\n", + "def fun(T):\n", + " fun=x*fun1(T)+ (1-x)*fun2(T)\n", + " return fun\n", + "\n", + "deltaH , err=scipy.integrate.quad(fun,T1,T2) #scipy.integrate.quad is an inbult function used for definite integration\n", + "print '%s %.3f' %(\" \\n Heat capacity of Mixture (KJ/mol) = \",deltaH)\n", + "Qdot=ndot*deltaH\n", + "print '%s %.3f' %(\" \\n Heat Required (KJ/h) = \",Qdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n", + " \n", + " \n", + " Heat capacity of Mixture (KJ/mol) = 34.890\n", + " \n", + " Heat Required (KJ/h) = 5233.495\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 6, + "text": [ + "''" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3-5, page no. 374" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "x=0.1 #CH4\n", + "T1=20.0\n", + "T2=300.0\n", + "Vdot=2000.0 #L/min\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \\n \")\n", + "ndot=Vdot/22.4\n", + "def fun(T):\n", + " fun=0.03431 + T*5.469*math.pow(10,-5) + math.pow(T,2) *0.3661*math.pow(10,-8) + math.pow(T,3) *11*math.pow(10,-12)\n", + " return fun\n", + "\n", + "H1 ,err=scipy.integrate.quad(fun,T1,T2) #scipy.integrate.quad is an inbult function used for definite integration\n", + "print '%s %.3f' %(\"H1 (Kj/mol) = \",H1)\n", + "print(\"From Tables H2= -0.15 Kj/mol , H3=8.17 Kj/mol\")\n", + "H2= -0.15\n", + "H3=8.17\n", + "Qdot=(ndot*x*H1+ ndot*(1-x)*(H3-H2))/60.\n", + "print '%s %.3f' %(\"Heat Input (KW) = \",Qdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n", + " \n", + "H1 (Kj/mol) = 12.112\n", + "From Tables H2= -0.15 Kj/mol , H3=8.17 Kj/mol\n", + "Heat Input (KW) = 12.945\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 7, + "text": [ + "''" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4-1, page no. 379" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "mdot=1500.0 #g/min\n", + "M=32.0 #g/mol\n", + "deltaHv=35.3 #Kj/mol\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \\n \")\n", + "Qdot=mdot*deltaHv/(M*60.)\n", + "print '%s %.3f' %(\"Rate of Heat transfer (KW) = \",Qdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n", + " \n", + "Rate of Heat transfer (KW) = 27.578\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 9, + "text": [ + "''" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4-2, page no. 379" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "deltaHv=28.85 #Kj/mol at 69 C\n", + "T1=25.0 #C\n", + "T2=69.0 #C\n", + "Cp=0.2163 #Kj/mol C\n", + "V=1.0 #L\n", + "P=7.0 #bar\n", + "D=0.659 #KG/L\n", + "M=86.17 #Kg\n", + "ndot=100.0 #mol/h\n", + "T3=300.0 #C\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"We have deltaHv at 69C hence We follow path ADG\")\n", + "deltaHA=Cp*(T2-T1) + V*(1.013-P)*M/(D*math.pow(10,4))\n", + "print '%s %.3f' %(\" \\n deltaHA (Kj/mol) =\",deltaHA)\n", + "deltaHD=deltaHv\n", + "print '%s %.3f' %(\" \\n deltaHD (Kj/Kg) =\",deltaHD)\n", + "def fun1(T):\n", + " fun1=0.13744 + T*40.85*math.pow(10,-5) - math.pow(T,2) *23.92*math.pow(10,-8) + math.pow(T,3) *57.66*math.pow(10,-12)\n", + " return fun1\n", + "\n", + "deltaHG, err=scipy.integrate.quad(fun1,T2,T3) #scipy.integrate.quad is an inbult function used for definite integration\n", + "print '%s %.3f' %(\" \\n deltaHG (KJ/mol) = \",deltaHG)\n", + "Qdot=ndot*(deltaHA+deltaHD+deltaHG)/3600\n", + "print '%s %.3f' %(\" \\n rate of Heat supply (KJ/mol) = \",Qdot)\n", + "print(\" \\n In this problem we neglected V*deltaP as it is negligible \")\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "We have deltaHv at 69C hence We follow path ADG\n", + " \n", + " deltaHA (Kj/mol) = 9.439\n", + " \n", + " deltaHD (Kj/Kg) = 28.850\n", + " \n", + " deltaHG (KJ/mol) = 47.149\n", + " \n", + " rate of Heat supply (KJ/mol) = 2.373\n", + " \n", + " In this problem we neglected V*deltaP as it is negligible \n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 10, + "text": [ + "''" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4-3, page no. 382" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "T1=337.9 #K\n", + "T2=473.0 #K\n", + "Tc=513.2 #K\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"using Trouton rule,\")\n", + "deltaHvT1=0.109*T1\n", + "print(\"In this case, trouton rule gives a better estimate\")\n", + "print(\"using Watson correction\")\n", + "deltaHvT2=36.8*math.pow(((Tc-T2)/(Tc-T1)),0.38)\n", + "print '%s %.3f' %(\"Estimated value using Trouton rule (Kj/mol) = \",deltaHvT1)\n", + "print '%s %.3f' %(\" \\n Estimated value using watson correction (Kj/mol) = \",deltaHvT2)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "using Trouton rule,\n", + "In this case, trouton rule gives a better estimate\n", + "using Watson correction\n", + "Estimated value using Trouton rule (Kj/mol) = 36.831\n", + " \n", + " Estimated value using watson correction (Kj/mol) = 21.029\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 11, + "text": [ + "''" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4-4, page no. 383" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "basis=1.0 #mol feed\n", + "x=0.684 #mole fraction Of B\n", + "y=0.4\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "A=([[1, 1],[x ,y]])\n", + "b=([[basis],[basis/2]])\n", + "C=numpy.dot(linalg.inv(A),b)\n", + "#Here We solved two linear equations simultaneously\n", + "nV=C[0,0]\n", + "nL=C[1,0]\n", + "H1=5.332\n", + "H2=6.340\n", + "H3=37.52\n", + "H4=42.93\n", + "Q=nV*x*H1 + nV*(1-x)*H2 + nL*y*H3 + nL*(1-y)*H4\n", + "print '%s %.3f' %(\" \\n Heat transferred (KJ) = \",Q)\n", + "print(\"The answer for this problem in Text is wrong\")\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Heat transferred (KJ) = 28.401\n", + "The answer for this problem in Text is wrong\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 1, + "text": [ + "''" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4-6, page no. 390" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "basis=1. #lbm dry air\n", + "Tin=80. #F Inlet temperature\n", + "Tout=51. #F Outlet temperature\n", + "\n", + "\n", + "\n", + "#Calculations and printing:\n", + "print(\"For 80F and 80 percent RH, ha=0.018 lbm H20/lbm DA H1=38.8 Btu/lbm DA\")\n", + "ha=0.018 #lbm H20/lbm DA\n", + "H1=38.8 #Btu/lbm DA\n", + "m1=basis*ha\n", + "print(\"\\n For 51F and saturated, ha=0.0079 lbm H20/lbm DA H2=20.9 Btu/lbm DA\")\n", + "ha2=0.0079 #lbm H20/lbm DA\n", + "H2=20.9 #Btu/lbm DA\n", + "m2=basis*ha2\n", + "print(\"\\n Balance on H2O\")\n", + "m3=m1-m2\n", + "Fraction=m3/m1\n", + "print '%s %.3f' %(\"\\n Fraction H2O condensed = \",Fraction)\n", + "H3=basis*(Tout-32.)\n", + "Q=basis*H2+m3*H3-basis*H1\n", + "print(\"\\n From psychrometric chart, Vh=13.0 ft^3/lbm DA\")\n", + "Vbasis=basis*13.0\n", + "Qdot=Q*1000./Vbasis\n", + "print '%s %.2f' %(\"Rate at which heat must be removed (Btu/min) = \",Qdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For 80F and 80 percent RH, ha=0.018 lbm H20/lbm DA H1=38.8 Btu/lbm DA\n", + "\n", + " For 51F and saturated, ha=0.0079 lbm H20/lbm DA H2=20.9 Btu/lbm DA\n", + "\n", + " Balance on H2O\n", + "\n", + " Fraction H2O condensed = 0.561\n", + "\n", + " From psychrometric chart, Vh=13.0 ft^3/lbm DA\n", + "Rate at which heat must be removed (Btu/min) = -1362.16\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 13, + "text": [ + "''" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.5-1, page no. 397" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "M1=35.6 #g/mol\n", + "M2=18.0 #g/mol\n", + "x=0.2\n", + "mdot=1000.0 #kg/h\n", + "T1=25.0 #C\n", + "T2=100.0 #C\n", + "T3=40.0 #C\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "nHCl=mdot*x*math.pow(10,3) /M1\n", + "nH2O=mdot*(1-x)*math.pow(10,3) /M2\n", + "def fun(T):\n", + " fun=29.13*math.pow(10,-3) - T*0.1341*math.pow(10,-5) +math.pow(T,2) *0.9715*math.pow(10,-8) - math.pow(T,3) *4.335*math.pow(10,-12)\n", + " return fun\n", + "\n", + "H1, err=scipy.integrate.quad(fun,T1,T2) #scipy.integrate.quad is an inbult function used for definite integration\n", + "r=nH2O/nHCl\n", + "print(\"From table B.11, deltaHa= -67.4 Kj/mol HCl\")\n", + "deltaHa= -67.4\n", + "y=nHCl/(nHCl+nH2O)\n", + "Cp=0.73*mdot*4.184/nHCl\n", + "deltaHb=Cp*(T3-T1)\n", + "H2=deltaHa+deltaHb\n", + "Qdot=nHCl*(H2-H1)\n", + "print '%s %.3E' %(\"Heat (kJ/h) = \",Qdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "From table B.11, deltaHa= -67.4 Kj/mol HCl\n", + "Heat (kJ/h) = -3.451E+05\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 14, + "text": [ + "''" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.5-2, page no. 400" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "x=0.05\n", + "y=0.4\n", + "mdot=1000.0 #lbm/h\n", + "Hf=10.0 #Btu/lbm\n", + "Hl= -17 #Btu/lbm\n", + "Hv=1138.0 #Btu/lbm\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"\\n sulphuric acid balance\")\n", + "m2=x*mdot/y\n", + "print(\"\\n Total Mass balance\")\n", + "m1=mdot-m2\n", + "Qdot=m1*Hv+m2*Hl-mdot*Hf\n", + "print '%s %.3f' %(\" Rate of Heat transfer (Btu/h) = \",Qdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "\n", + " sulphuric acid balance\n", + "\n", + " Total Mass balance\n", + " Rate of Heat transfer (Btu/h) = 983625.000\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 15, + "text": [ + "''" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.5-5, page no. 405" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "basis=100.0 #lbm/h\n", + "Hv=728.0 #Btu/lbm\n", + "Hl=45.0 #Btu/lbm\n", + "HF=100.0 #Btu/lbm\n", + "T=120.0 #F\n", + "xF=0.30\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"From figure 8.5-2, \")\n", + "xL=0.185\n", + "xV=0.89\n", + "mL=basis*((xV-xF)/(xV-xL))\n", + "mV=basis-mL\n", + "Qdot=mV*Hv + mL*Hl - basis*HF\n", + "print '%s %.3f' %(\"Rate of heat transfer (Btu/h) =\",Qdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "From figure 8.5-2, \n", + "Rate of heat transfer (Btu/h) = 5641.135\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 16, + "text": [ + "''" + ] + } + ], + "prompt_number": 16 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Principles_of_Chemical_Processes/Chapter9.ipynb b/Elementary_Principles_of_Chemical_Processes/Chapter9.ipynb new file mode 100755 index 00000000..230963f2 --- /dev/null +++ b/Elementary_Principles_of_Chemical_Processes/Chapter9.ipynb @@ -0,0 +1,822 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e681b00ee14cb4ba49bab5814ba58d8e293c97db9c72b39fdd9bbdd0d2223049" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 9: Balances on Reactive Processes" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.9-1, page no. 443" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "ndot=2400.0 #mol/s\n", + "Hr1= -2878 #Kj/mol\n", + "HvWater=44.0 #Kj/mol\n", + "HvButane=19.2 #Kj/mol\n", + "\n", + "#Calculations and print '%s %.3f' %ing :\n", + "\n", + "print (\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print (\"Part 1\")\n", + "E1= ndot/4\n", + "deltaH1=E1*Hr1\n", + "print '%s %.3E' %(\"enthalpy change (KJ/s) = \",deltaH1)\n", + "print (\"part2\")\n", + "Hr2=2*Hr1\n", + "E2=ndot/8\n", + "deltaH2=E2*Hr2\n", + "print '%s %.3E' %(\"Enthalpy change (kJ/s) = \",deltaH2)\n", + "print (\"part 3\")\n", + "Hr3=Hr1+5*HvWater+HvButane\n", + "deltaH3=E1*Hr3\n", + "print '%s %.3E' %(\"Enthalpy change (kJ/s) = \",deltaH3)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Part 1\n", + "enthalpy change (KJ/s) = -1.727E+06\n", + "part2\n", + "Enthalpy change (kJ/s) = -1.727E+06\n", + "part 3\n", + "Enthalpy change (kJ/s) = -1.583E+06\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 1, + "text": [ + "''" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1-2, page no. 445" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + " \n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "deltaHr= -420.8 #kj/mol\n", + "R=8.314\n", + "T=298.0 #K\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"From reaction, only gaseous are counted\")\n", + "left=1+2\n", + "right=1+1\n", + "deltaUr=deltaHr-R*T*(right-left)/math.pow(10,3)\n", + "print '%s %.3f' %(\"deltaUr (KJ/mol) = \",deltaUr)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "From reaction, only gaseous are counted\n", + "deltaUr (KJ/mol) = -418.322\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 2, + "text": [ + "''" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3-1, page no. 447" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "HCO2= -393.5 #KJ/mol\n", + "HH2O= -285.84 #KJ/mol\n", + "HC5H12= -173. #KJ/mol\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "Hr=5*HCO2+6*HH2O-HC5H12\n", + "print '%s %.3f' %(\" \\n Heat of the rxn (KJ/mol) = \",Hr)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Heat of the rxn (KJ/mol) = -3509.540\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 3, + "text": [ + "''" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4-1, page no. 449" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "Hethane= -1559.9 #Kj/mol\n", + "Hethene= -1411 #Kj/mol\n", + "Hhydrogen= -285.84 #Kj/mol\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "Hr=Hethane-Hethene-Hhydrogen\n", + "print '%s %.3f' %(\" \\n Heat of the rxn (KJ/mol) = \",Hr)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Heat of the rxn (KJ/mol) = 136.940\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 4, + "text": [ + "''" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5-1, page no. 453" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "nNH3=100.0 #mol/s\n", + "nO2in=200.0 #mol/s\n", + "H1=8.470 #Kj/mol\n", + "H3=9.570 #Kj/mol\n", + "T1=25.0\n", + "T2=300.0\n", + "Hr= -904.7 #Kj/mol\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "def fun(T):\n", + " fun=29.50*math.pow(10,-3)+ T*0.8188*math.pow(10,-5) - math.pow(T,2) * 0.2925 *math.pow(10,-8) + math.pow(T,3) * 0.3652 * math.pow(10,-12)\n", + " return fun\n", + "\n", + "H2, err=scipy.integrate.quad(fun,T1,T2) #scipy.integrate.quad is an inbuilt function which can calculate definite integrals\n", + "E=nNH3/4.\n", + "nO2out=nO2in-nNH3*5./4.\n", + "nNO=nNH3\n", + "nH2O=nNH3*6./4.\n", + "deltaH=E*Hr+(nO2out*H1+nNO*H2+nH2O*H3)\n", + "Qdot=deltaH\n", + "print '%s %.3f' %(\" \\n Heat Transferred (kW) = \",Qdot)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Heat Transferred (kW) = -19701.467\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 5, + "text": [ + "''" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5-2, page no. 454" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "import scipy\n", + "from scipy import integrate\n", + "\n", + "NinMethane=100.0 #mol\n", + "NinOxygen=100.0 #mol\n", + "NinNitrogen=376.0 #mol\n", + "NoutMethane=60.0 #mol\n", + "NoutOxygen=50.0 #mol\n", + "NoutNitrogen=376.0 #mol\n", + "NoutFormal=30.0 #mol\n", + "NoutCarbon=10.0 #mol\n", + "NoutWater=50.0 #mol\n", + "H1= -74.85 #Kj/mol\n", + "H2= 2.235 #Kj/mol\n", + "H3= 2.187 #Kj/mol\n", + "H5=3.758#Kj/mol\n", + "H6=3.655 #Kj/mol\n", + "H8= -393.5+4.75 #Kj/mol\n", + "H9= -241.83+4.27 #Kj/mol\n", + "T1=25.0 #C\n", + "T2=150.0 #C\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "def fun1(T):\n", + " fun1=34.31*math.pow(10,-3)+ T*5.469*math.pow(10,-5) + math.pow(T,2) * 0.3661 *math.pow(10,-8) - math.pow(T,3) * 11 * math.pow(10,-12)\n", + " return fun1\n", + "\n", + "HoutMethane, err=scipy.integrate.quad(fun1,T1,T2)\n", + "H4= -74.85 + HoutMethane\n", + "def fun2(T):\n", + " fun2=34.28*math.pow(10,-3)+ T*4.268*math.pow(10,-5) - math.pow(T,3) * 8.694 * math.pow(10,-12)\n", + " return fun2\n", + "\n", + "HoutFormal, err2=scipy.integrate.quad(fun2,T1,T2) #scipy.integrate.quadis an inbuilt function which can calculate definite integrals\n", + "H7= -115.90+ HoutFormal\n", + "deltaH=NoutWater*H9+NoutCarbon*H8+NoutFormal*H7+NoutNitrogen*H6+NoutOxygen*H5+NoutMethane*H4-NinNitrogen*H3-NinOxygen*H2-NinMethane*H1\n", + "Q=deltaH\n", + "print '%s %.3f' %(\" \\n Q (KJ) = \",Q)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " Q (KJ) = -15296.233\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 6, + "text": [ + "''" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5-4, page no. 458" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "basis=150.0 #mol/s\n", + "x=0.9\n", + "HinEthanol= -212.19\n", + "HinEthanone= -147.07\n", + "HoutEthanol= -216.81\n", + "HoutEthanone= -150.9\n", + "HoutHydrogen=6.595\n", + "NinEthanol=135.0\n", + "NinEthanone=15.0\n", + "Q=2440.0 #KW\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Carbon Balance\")\n", + "print(\"basis*x *2 + basis* (1-x) *2=2 n1+2 n2\")\n", + "print(\"Hydrogen Balance\")\n", + "print(\"basis * x *6 + basis * (1-x)*4 = 6 n1+4 n2+2 n3\")\n", + "print(\"Energy Balance\")\n", + "print(\"Q= HoutEthanol n1 HoutEthanone n2 + HoutHydrogen n3 -NinEthanol * HinEthanol -NinEthanone* HinEthanone\")\n", + "A=([[1, 1, 0],[3, 2, 1],[216.81, 150.9, -6.595]])\n", + "b=([[150],[435],[28412]])\n", + "C=numpy.dot(linalg.inv(A),b)\n", + "n1=C[0,0]\n", + "print '%s %.3f' %(\" \\n n1 (mol Ethanol/s) = \",n1)\n", + "n2=C[1,0]\n", + "print '%s %.3f' %(\" \\n n2 (mol Ethanone/s) = \",n2)\n", + "n3=C[2,0]\n", + "print '%s %.3f' %(\" \\n n3 (mol Hydrogen/s) = \",n3)\n", + "print(\"The solutions in the Text are Wrong\")\n", + "fraction=(NinEthanol-n1)/NinEthanol\n", + "print '%s %.3f' %(\"Fractional conversion of Ethanol = \",fraction)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Carbon Balance\n", + "basis*x *2 + basis* (1-x) *2=2 n1+2 n2\n", + "Hydrogen Balance\n", + "basis * x *6 + basis * (1-x)*4 = 6 n1+4 n2+2 n3\n", + "Energy Balance\n", + "Q= HoutEthanol n1 HoutEthanone n2 + HoutHydrogen n3 -NinEthanol * HinEthanol -NinEthanone* HinEthanone\n", + " \n", + " n1 (mol Ethanol/s) = 91.957\n", + " \n", + " n2 (mol Ethanone/s) = 58.043\n", + " \n", + " n3 (mol Hydrogen/s) = 43.043\n", + "The solutions in the Text are Wrong\n", + "Fractional conversion of Ethanol = 0.319\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 7, + "text": [ + "''" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5-5, page no. 460" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "print(\"from tables,\")\n", + "HfAcid= -1294. #Kj/mol\n", + "HfBase= -469.1 #Kj/mol\n", + "HfSalt= -1974. #Kj/mol \n", + "HfWater= -285.8 #Kj/mol\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"part 1\")\n", + "Hr=HfSalt+3*HfWater-HfAcid-3*HfBase\n", + "print '%s %.3f' %(\"Hr of the rxn (KJ/mol) = \",Hr)\n", + "print(\"part 2\")\n", + "deltaH=Hr*5./3.\n", + "print '%s %.3f' %(\"deltaH (KJ) = \",deltaH)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "from tables,\n", + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "part 1\n", + "Hr of the rxn (KJ/mol) = -130.100\n", + "part 2\n", + "deltaH (KJ) = -216.833\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 8, + "text": [ + "''" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5-6, page no. 461" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "x=0.1\n", + "y=0.2\n", + "MAcid=98.1\n", + "MS=32.0\n", + "MSalt=142.0\n", + "MBase=40.0\n", + "MWater=18.0\n", + "MNa=46.0\n", + "basis=1000.0 #g\n", + "T2=35.0\n", + "T1=25.0\n", + "T3=40.0\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "print(\"Using S balance, \")\n", + "m2=basis*x*MS*MSalt/(MAcid*MS)\n", + "print '%s %.3f' %(\" \\n m2 (g Na2SO4) = \",m2)\n", + "print(\"Using Na balance, \")\n", + "m1=2*MNa*m2*MBase/(y*MNa*MSalt)\n", + "print '%s %.3f' %(\" \\n m1 (g NaOH) = \",m1)\n", + "print(\"Total mass balance, \")\n", + "m3=basis+m1-m2\n", + "print '%s %.3f' %(\" \\n m3 (g H2O) = \",m3)\n", + "print '%s %.3f' %(\" \\n Mass of product solution = \",m2+m3)\n", + "m=m2+m3\n", + "Water=m2*2/MSalt\n", + "print '%s %.3f' %(\" \\n Water Formed in the reaction (mol H2O) = \",Water)\n", + "print(\"H2SO4(aq):\")\n", + "a1=basis*(1-x)/MWater\n", + "b1=basis*x/MAcid\n", + "rAcid=a1/b1\n", + "print '%s %.3f' %(\" \\n rAcid (mol Water/mol Acid) = \",rAcid)\n", + "print(\"NaOH(aq):\")\n", + "a2=m1*(1-y)/MWater\n", + "b2=m1*y/MBase\n", + "rBase=a2/b2\n", + "print '%s %.3f' %(\" \\n rBase (mol Water/mol Base) = \",rBase)\n", + "print(\"Na2SO4(aq):\")\n", + "a3=m3/MWater\n", + "b3=m2/MSalt\n", + "rSalt=a3/b3\n", + "print '%s %.3f' %(\" \\n rSalt (mol Water/mol Salt) =\",rSalt)\n", + "E=b1\n", + "print '%s %.3f' %(\" \\n Extent of reaction (mol) = \",E)\n", + "nHAcid=basis*3.85*(T3-T1)/1000\n", + "nHSalt=m*4.184*(T2-T1)/1000\n", + "nHBase=0\n", + "HfSalt= -1384\n", + "HfAcid= -884.6\n", + "HfBase= -468.1\n", + "HfWater= -285.84\n", + "deltaHr=HfSalt+ 2*HfWater - HfAcid - 2*HfBase\n", + "print '%s %.3f' %(\" \\n Entahlpy change in the rxn (KJ/mol) = \",deltaHr)\n", + "Q=E*deltaHr + (nHSalt-nHAcid-nHBase)\n", + "print '%s %.3f' %(\" \\n Q of the rxn (KJ) = \",Q)\n", + "print(\"The answer in the Text is wrong.\")\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + "Using S balance, \n", + " \n", + " m2 (g Na2SO4) = 144.750\n", + "Using Na balance, \n", + " \n", + " m1 (g NaOH) = 407.747\n", + "Total mass balance, \n", + " \n", + " m3 (g H2O) = 1262.997\n", + " \n", + " Mass of product solution = 1407.747\n", + " \n", + " Water Formed in the reaction (mol H2O) = 2.039\n", + "H2SO4(aq):\n", + " \n", + " rAcid (mol Water/mol Acid) = 49.050\n", + "NaOH(aq):\n", + " \n", + " rBase (mol Water/mol Base) = 8.889\n", + "Na2SO4(aq):\n", + " \n", + " rSalt (mol Water/mol Salt) = 68.833\n", + " \n", + " Extent of reaction (mol) = 1.019\n", + " \n", + " Entahlpy change in the rxn (KJ/mol) = -134.880\n", + " \n", + " Q of the rxn (KJ) = -136.342\n", + "The answer in the Text is wrong.\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 9, + "text": [ + "''" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.6-1, page no. 465" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "#Initialization of variables\n", + "import math\n", + "import numpy\n", + "from numpy import linalg\n", + "\n", + "x=0.85\n", + "H1= 802 #KJ/mol\n", + "H2= 1428 #kJ/mol\n", + "M1=16.0\n", + "M2=30.0\n", + "\n", + "#Calculations and printing :\n", + "\n", + "print(\" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\")\n", + "y1=x*16\n", + "y2=(1-x)*30\n", + "xCH4=y1/(y1+y2)\n", + "HHVMethane=(H1+ 2*44.013)/M1\n", + "HHVEthane=(H2+ 3*44.013)/M2\n", + "HHV=xCH4*HHVMethane + (1-xCH4)*HHVEthane\n", + "print '%s %.3f' %(\" \\n HHV of Fuel (KJ/g) = \",HHV)\n", + "raw_input('press enter key to exit')" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " All the values in the textbook are Approximated hence the values in this code differ from those of Textbook\n", + " \n", + " HHV of Fuel (KJ/g) = 54.725\n" + ] + }, + { + "name": "stdout", + "output_type": "stream", + "stream": "stdout", + "text": [ + "press enter key to exit\n" + ] + }, + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 10, + "text": [ + "''" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Elementary_Principles_of_Chemical_Processes/README.txt b/Elementary_Principles_of_Chemical_Processes/README.txt new file mode 100755 index 00000000..5cf37f06 --- /dev/null +++ b/Elementary_Principles_of_Chemical_Processes/README.txt @@ -0,0 +1,10 @@ +Contributed By: Chaitanya Potti +Course: btech +College/Institute/Organization: IITB +Department/Designation: Chemical engineering +Book Title: Elementary Principles of Chemical Processes +Author: Richard M. Felder & Ronald W. Rousseau +Publisher: John Wiley & Sons Inc +Year of publication: 2005 +Isbn: 047168757X +Edition: 3rd
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