From 4a1f703f1c1808d390ebf80e80659fe161f69fab Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 28 Aug 2015 16:53:23 +0530 Subject: add books --- .../ch8.ipynb | 540 +++++++++++++++++++++ 1 file changed, 540 insertions(+) create mode 100755 Chemical_Engineering_Thermodynamics_by_T._E._Daubert/ch8.ipynb (limited to 'Chemical_Engineering_Thermodynamics_by_T._E._Daubert/ch8.ipynb') diff --git a/Chemical_Engineering_Thermodynamics_by_T._E._Daubert/ch8.ipynb b/Chemical_Engineering_Thermodynamics_by_T._E._Daubert/ch8.ipynb new file mode 100755 index 00000000..4c26dfce --- /dev/null +++ b/Chemical_Engineering_Thermodynamics_by_T._E._Daubert/ch8.ipynb @@ -0,0 +1,540 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:b2ee1c3d3e88f5241245b2aaba0096c6b15a6abe1de87d5ab25de8744d1af1f8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8 : Physical Equilibria Among Phases" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1 page : 293" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\t\t\n", + "# Variables\n", + "def func(C,phi):\n", + " return C+2-phi\n", + "\n", + "# Calculations and results\n", + "print (\"part a\")\n", + "print \"degrees of freedom = %d \"%(func(2,2))\n", + "print (\"part b\")\n", + "print \"degrees of freedom = %d \"%(func(3,2))\n", + "print (\"part c\")\n", + "print \"degrees of freedom = %d \"%(func(3,3))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "part a\n", + "degrees of freedom = 2 \n", + "part b\n", + "degrees of freedom = 3 \n", + "part c\n", + "degrees of freedom = 2 \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2 page : 297" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\t\t\n", + "# Variables\n", + "T = 95. \t\t\t#C\n", + "P = 1013. \t\t\t#kPa\n", + "Tc = 135. \t\t\t#C\n", + "Pc = 3648. \t\t\t#kPa\n", + "T0 = 273.1 \t\t\t#C\n", + "D = 0.3\n", + "P0 = 1800. \t\t\t#kPa\n", + "D2 = 0.42\n", + "\t\t\t\n", + "# Calculations\n", + "Zc = 0.283\n", + "Tr = (T+T0)/(Tc+T0)\n", + "Pr = P/Pc\n", + "phic = 0.88\n", + "phi2 = phic*10**(D*0.013)\n", + "Prd = P0/Pc\n", + "phi3 = 0.78\n", + "phi4 = phi3*10**(D2*0.013)\n", + "gl = phi2*P/(phi3*P0)\n", + "\t\t\t\n", + "# Results\n", + "print \"equation is gl = %.3f *y/x\"%(gl)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "equation is gl = 0.641 *y/x\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3 page :300" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\t\t\n", + "# Variables\n", + "ye = 0.434\n", + "Pt = 40.25 \t\t\t#kPa\n", + "xe = 0.616\n", + "Pe1 = 22.9 \t\t\t#kPa\n", + "Pe2 = 29.6 \t\t\t#kPa\n", + "\t\t\t\n", + "# Calculations\n", + "ge = ye*Pt/(xe*Pe1)\n", + "gb = (1-ye)*Pt/((1-xe)*Pe2)\n", + "E = math.log10(ge) *(1+ (1-xe)*math.log(gb) /(xe*math.log(ge)))**2\n", + "B = math.log10(gb) *(1+ xe/(1-xe) *math.log(ge) /math.log(gb))**2\n", + "xe2 = 0.4\n", + "xb2 = 0.6\n", + "lnge2 = E/(1+ E*xe2/(B*xb2))**2\n", + "lngb2 = B/(1+ B*xb2/(E*xe2))**2\n", + "ge2 = 10**(lnge2)\n", + "gb2 = 10**(lngb2)\n", + "Pt1 = ge2*Pe1\n", + "Pt2 = gb2*Pe2\n", + "\t\t\t\n", + "# Results\n", + "print \"Total pressure in case 1 = %.2f kPa and in case 2 = %.2f kPa\"%(Pt1, Pt2 )\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total pressure in case 1 = 40.78 kPa and in case 2 = 40.93 kPa\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4 page : 310" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy.linalg import solve\n", + "\t\t\t\n", + "# Variables\n", + "k4 = 1.8\n", + "k5 = 0.8\n", + "\t\t\t\n", + "# Calculations\n", + "A = [[k4, k5],[1, 1]]\n", + "b = [[1],[1]]\n", + "C = solve(A,b)\n", + "x4 = C[0]\n", + "x5 = C[1]\n", + "y4 = k4*x4\n", + "y5 = k5*x5\n", + "\t\t\t\n", + "# Results\n", + "print \"Vapor and liquid mole fractions of component 1 = %.2f and %.2f respectively\"%(y4,x4)\n", + "print \" Vapor and liquid mole fractions of component 2 = %.2f and %.2f respectively\"%(y5,x5)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Vapor and liquid mole fractions of component 1 = 0.36 and 0.20 respectively\n", + " Vapor and liquid mole fractions of component 2 = 0.64 and 0.80 respectively\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.5 page : 312" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\t\t\n", + "# Variables\n", + "v1 = 81. \t\t\t#cm**3/gmol\n", + "v2 = 97. \t\t\t#cm**3/gmol\n", + "d1 = 9.2 \t\t\t#(cal/cm**3)**0.5\n", + "d2 = 8.6 \t\t\t#(cal/cm**3)**0.5\n", + "R = 1.987\n", + "T = 373.1 \t\t\t#K\n", + "\t\t\t\n", + "# Calculations\n", + "d = 0.5*(d1+d2)\n", + "lng1 = v1*(d1-d)**2 /(R*T)\n", + "lng2 = v2*(d2-d)**2 /(R*T)\n", + "g1 = math.exp(lng1)\n", + "g2 = math.exp(lng2)\n", + "\t\t\t\n", + "# Results\n", + "print \"Activity coeffecients of components are %.3f and %.3f respectively\"%(g1,g2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Activity coeffecients of components are 1.010 and 1.012 respectively\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.6 page : 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\t\t\n", + "# Variables\n", + "xe = 0.3\n", + "xe2 = 0.9\n", + "Pe0 = 810.\n", + "Pa0 = 470.\n", + "ge = 1.85\n", + "ge2 = 1.05\n", + "ga = 1.15\n", + "ga2 = 3.\n", + "Pt = 820. \t\t\t#mm\n", + "Pt2 = 900. \t\t\t#mm\n", + "\t\t\t\n", + "# Calculations\n", + "ye = ge*xe*Pe0/Pt\n", + "ya = ga*(1-xe)*Pa0/Pt\n", + "yt = ye+ya\n", + "ye2 = ye/yt\n", + "ya2 = ya/yt\n", + "ye3 = ge2*xe2*Pe0/Pt2\n", + "ya3 = ga2*(1-xe2)*Pa0/Pt2\n", + "yt2 = ye+ya\n", + "ye4 = ye3/yt2\n", + "ya4 = ya3/yt2\n", + "\t\t\t\n", + "# Results\n", + "print \"In case 1, ye = %.3f and ya = %.3f\"%(ye2,ya2)\n", + "print \" In case 1, ye = %.3f and ya = %.3f\"%(ye4,ya4)\n", + "print ('The calculations of ya in case 1 in textbook is wrong. please use a calculator')\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "In case 1, ye = 0.543 and ya = 0.457\n", + " In case 1, ye = 0.842 and ya = 0.155\n", + "The calculations of ya in case 1 in textbook is wrong. please use a calculator\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.7 page : 326" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\t\t\n", + "# Variables\n", + "m1 = 121.\n", + "m2 = 18.\n", + "p1 = 0.0042\n", + "p2 = 0.0858\n", + "\t\t\t\n", + "# Calculations\n", + "massfrac = (p1*m1)/(p1*m1+p2*m2)\n", + "\t\t\t\n", + "# Results\n", + "print \"mass fractions of DMA and water are %.3f and %.3f respectively\"%(massfrac,1-massfrac)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "mass fractions of DMA and water are 0.248 and 0.752 respectively\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.9 page : 335" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import array\n", + "\t\t\t\n", + "# Variables\n", + "FR = 25.\n", + "FE = 19.\n", + "bf = 130. \t\t\t#kg\n", + "af = 85. \t\t\t#kg\n", + "\t\t\t\n", + "# Calculations\n", + "law = FR/FE\n", + "x1 = 45./150\n", + "x2 = 65./150\n", + "ER = 18.5/6\n", + "e = array([0.5, 0.1, 0.9])\n", + "r = array([0.28, 0.96, 0.04])\n", + "et = sum(e)\n", + "rt = sum(r)\n", + "ett = e/et\n", + "rtt = r/rt\n", + "\t\t\t\n", + "# Results\n", + "print \"the compositions of raffinate are \",\n", + "print (rtt)\n", + "print \"the compositions of extract are\",\n", + "print (ett)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the compositions of raffinate are [ 0.21875 0.75 0.03125]\n", + "the compositions of extract are [ 0.33333333 0.06666667 0.6 ]\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.10 page : 342" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\t\t\t\n", + "# Variables\n", + "v1 = 0.1316\n", + "v2 = 0.2941\n", + "x1 = 0.5\n", + "x2 = 0.2\n", + "x3 = 0.8 \n", + "d1 = 14.87\n", + "d2 = 16.34\n", + "\t\t\t\n", + "# Calculations and results\n", + "vm = x1*(v1+v2)\n", + "phi1 = x1*v1/vm\n", + "phi2 = (1-x1)*v2/vm\n", + "Hl1 = vm*phi1*phi2*(d1-d2)**2 *10**3\n", + "print (\"case 1\")\n", + "print \"enthalpy = %.1f kJ/mol\"%(Hl1)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "case 1\n", + "enthalpy = 98.2 kJ/mol\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.10b page : 343" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\t\t\n", + "# Variables\n", + "v1 = 0.1316\n", + "v2 = 0.2941\n", + "x1 = 0.5\n", + "x2 = 0.2\n", + "x3 = 0.8 \n", + "d1 = 14.87\n", + "d2 = 16.34\n", + "\t\t\t\n", + "# Calculations and results\n", + "vm = (1-x2)*v1+x2*v2\n", + "phi1 = (1-x2)*v1/vm\n", + "phi2 = (x2)*v2/vm\n", + "Hl2 = vm*phi1*phi2*(d1-d2)**2 *10**3\n", + "print (\"case 2\")\n", + "print \"enthalpy = %.1f kJ/mol\"%(Hl2)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "case 2\n", + "enthalpy = 81.5 kJ/mol\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.10c page : 343" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\t\t\t\n", + "# Variables\n", + "v1 = 0.1316\n", + "v2 = 0.2941\n", + "x1 = 0.5\n", + "x2 = 0.2\n", + "x3 = 0.8 \n", + "d1 = 14.87\n", + "d2 = 16.34\n", + "\t\t\t\n", + "# Calculations and results\n", + "vm = (1-x3)*v1+x3*v2\n", + "phi1 = (1-x3)*v1/vm\n", + "phi2 = (x3)*v2/vm\n", + "Hl3 = vm*phi1*phi2*(d1-d2)**2 *10**3\n", + "print (\"case 3\")\n", + "print \"enthalpy = %.1f kJ/mol\"%(Hl3)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "case 3\n", + "enthalpy = 51.2 kJ/mol\n" + ] + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit