From 206d0358703aa05d5d7315900fe1d054c2817ddc Mon Sep 17 00:00:00 2001 From: Jovina Dsouza Date: Wed, 18 Jun 2014 12:43:07 +0530 Subject: adding book --- .../ch8.ipynb | 449 +++++++++++++++++++++ 1 file changed, 449 insertions(+) create mode 100644 Basic_Principles_And_Calculations_In_Chemical_Engineering/ch8.ipynb (limited to 'Basic_Principles_And_Calculations_In_Chemical_Engineering/ch8.ipynb') diff --git a/Basic_Principles_And_Calculations_In_Chemical_Engineering/ch8.ipynb b/Basic_Principles_And_Calculations_In_Chemical_Engineering/ch8.ipynb new file mode 100644 index 00000000..4d600c4f --- /dev/null +++ b/Basic_Principles_And_Calculations_In_Chemical_Engineering/ch8.ipynb @@ -0,0 +1,449 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 8 : Solving Material Balance Problems for Single Units without Reaction" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + " Example 8.1 Page no. 197\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Extraction of Streptomycin from a Fermentation Broth\n", + "# Solution\n", + "\n", + "# Variables\n", + "# Basis : 1 min\n", + "d_w = 1.0 ; # Density of aqueous solution-[g/cubic metre]\n", + "d_sol = 0.6 ; # Density of organic solvent-[g/cubic metre]\n", + "n_un = 8 ; # Number of unknowns in the given problem\n", + "n_ie = 8 ; # Number of independent equations\n", + "\n", + "# Calculation and Results\n", + "d_o_f = n_un-n_ie ; # Number of degree of freedom\n", + "print 'Number of degree of freedom for the given system is %i .'%d_o_f\n", + "\n", + "# Material balance of Strep.\n", + "x = (200*10+10*0-200*0.2)/10; #[g]\n", + "print 'Strep per litre of solvent is %.1f g .'%x\n", + "\n", + "cnc = x/(1000*d_sol) ; #[g Strep/g of S]\n", + "print 'Strep per gram of solvent is %.4f g Strep/g of S .'%cnc\n", + "\n", + "m_fr = cnc/(1+cnc) ; #Mass fraction\n", + "print 'Mass fraction of Strep is %.3f g .'%m_fr" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of degree of freedom for the given system is 0 .\n", + "Strep per litre of solvent is 196.0 g .\n", + "Strep per gram of solvent is 0.3267 g Strep/g of S .\n", + "Mass fraction of Strep is 0.246 g .\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + " Example 8.2 Page no. 199\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Separation of gases Using a Membrane\n", + "# Solution Fig. E8.2b\n", + "\n", + "# Variables\n", + "F_O2 = 0.21 ; # fraction of O2 in feed(F) \n", + "F_N2 = 0.79 ; # fraction of N2 in feed(F) \n", + "P_O2 = 0.25 ; # fraction of O2 in product(P)\n", + "P_N2 = 0.75 ; # fraction of N2 in product(P)\n", + "F = 100 ; # Feed - [g mol]\n", + "w = 0.80 ; # Fraction of waste\n", + "W = w*F ; # Waste -[g mol]\n", + "\n", + "# Calculation\n", + "# By analysis for degree of freedom , DOF comes to be zero \n", + "P = F - W ; # By overall balance - [g mol]\n", + "W_O2 = (F_O2*F - P*P_O2)/100 # Fraction of O2 in waste stream by O2 balance \n", + "W_N2 = (W - W_O2*100)/100 ; #Fraction of N2 in waste stream\n", + " \n", + "# Results \n", + "print 'Composition of Waste Stream' \n", + "print ' Component Fraction in waste stream' \n", + "print ' O2 %.2f'%W_O2 \n", + "print ' N2 %.2f'%W_N2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Composition of Waste Stream\n", + " Component Fraction in waste stream\n", + " O2 0.16\n", + " N2 0.64\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + " Example 8.3 Page no. 202\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Overall analysis for a continuous Distillation Column\n", + "# Solution\n", + "\n", + "# Variables\n", + "# Basis : 1 hr so F = 1000 kg\n", + "F = 1000 ; # feed rate-[kg/hr]\n", + "P = F/10.0 ; # product mass flow rate -[kg/hr]\n", + "\n", + "n_un = 9 ; # Number of unknowns in the given problem\n", + "n_ie = 9 ; # Number of independent equations\n", + "\n", + "# Calculation and Result\n", + "d_o_f = n_un-n_ie ; # Number of degree of freedom\n", + "print 'Number of degree of freedom for the given system is %i .'%d_o_f\n", + "\n", + "# Overall mass balance: F = P+B\n", + "B = F-P ; # bottom mass flow rate -[kg/hr]\n", + "print ' Bottom mass flow rate - %.1f kg '%B\n", + "\n", + "# Composition of bottoms by material balances\n", + "m_EtOH = 0.1*F-0.6*P ; # By EtOH balance-[kg]\n", + "m_H2O = 0.9*F - 0.4*P ; # By H2O balance-[kg]\n", + "total = m_EtOH+m_H2O ; #[kg]\n", + "f_EtOH = m_EtOH/total ; # Mass fraction of EtOH\n", + "f_H2O = m_H2O/total ; # Mass fraction of H2O\n", + "\n", + "print ' Mass of EtOH in bottom - %.1f kg '%m_EtOH\n", + "print ' Mass of H2O in bottom - %.1f kg '%m_H2O\n", + "print ' Mass fraction of EtOH in bottom - %.3f '%f_EtOH\n", + "print ' Mass fraction of H2O in bottom - %.3f '%f_H2O" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of degree of freedom for the given system is 0 .\n", + " Bottom mass flow rate - 900.0 kg \n", + " Mass of EtOH in bottom - 40.0 kg \n", + " Mass of H2O in bottom - 860.0 kg \n", + " Mass fraction of EtOH in bottom - 0.044 \n", + " Mass fraction of H2O in bottom - 0.956 \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + " Example 8.4 Page no. 205\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Mixing of Battery Acid\n", + "# Solution Fig E8.4\n", + "from numpy import matrix\n", + "\n", + "# Variables\n", + "# Given\n", + "A = 200 ; # Mass of added solution [kg] \n", + "P_H2SO4 = .1863 ; #Fraction of H2SO4 in P(Final solution)\n", + "P_H2O = .8137 ; #Fraction of H2O in P(Final solution)\n", + "A_H2SO4 = .777 ; #Fraction of H2SO4 in A(Added solution)\n", + "A_H2O = .223 ; #Fraction of H2O in A(Added solution)\n", + "F_H2SO4 = .1243 ; #Fraction of H2SO4 in F(Original solution)\n", + "F_H2O = .8757 ; #Fraction of H2O in F(Original solution)\n", + "\n", + "# Calculations\n", + "# P - F = A - By overall balance\n", + "a = matrix([[P_H2O,-F_H2O],[1,-1]]) ; # Matrix of coefficient\n", + "b = matrix([[A*A_H2O],[A]]) ; # Matrix of contants\n", + "a = a.I\n", + "x = a*b ; # Matrix of solutions- P = x(1) and F = x(2)\n", + "\n", + "#Results\n", + "print ' Original solution taken- %.0i kg'%x[1]\n", + "print ' Final solution or kilograms of battery acid formed- %.0i kg'%x[0]" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Original solution taken- 1905 kg\n", + " Final solution or kilograms of battery acid formed- 2105 kg\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + " Example 8.5 Page no. 207\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Drying\n", + "# Solution Fig E8.5\n", + "from numpy import matrix\n", + "\n", + "# Variables\n", + "# Given\n", + "W = 100.0 ; # Water removed - [kg]\n", + "A_H2O = 0.80 ; # Fraction of water in A(intial fish cake)\n", + "A_BDC = 0.20 ; # Fraction of BDC(bone dry cake) in B(final dry fish cake)\n", + "B_H2O = 0.40 ; # Fraction of water in A(intial fish cake)\n", + "B_BDC = 0.60 ; # Fraction of BDC(bone dry cake) in B(final dry fish cake)\n", + "\n", + "# Calculations\n", + "a = matrix([[A_H2O, -B_H2O],[1, -1]]) ; # Matrix of coefficient\n", + "b = matrix([[W],[W]]) ; # Matrix of contants\n", + "a = a.I\n", + "x = a * b; # Matrix of solutions- A = x(1) and B = x(2)\n", + "\n", + "# Results\n", + "print 'Weight of the fish cake originally put into dryer -%.0i kg'%x[0]" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Weight of the fish cake originally put into dryer -150 kg\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + " Example 8.6 Page no. 209\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Crystallizaton\n", + "# Solution\n", + "\n", + "# Variables\n", + "# Composition of initial solution at 30 degree C\n", + "s_30 = 38.8 ; # solublity of Na2CO3 at 30 degree C, by using the table for solublity of Na2CO3-[g Na2CO3/100 g H2O]\n", + "If_Na2CO3 = s_30/(s_30+100) ; # Initial mass fraction of Na2CO3\n", + "If_H2O = 1-If_Na2CO3 ; # Initial mass fraction of H2O\n", + "\n", + "# Composition of crystals\n", + "# Basis : 1g mol Na2CO3.10H2O\n", + "n_mol_Na2CO3 = 1 ; # Number of moles of Na2CO3\n", + "n_mol_H2O = 10. ; # Number of moles of H2O\n", + "mwt_Na2CO3 = 106. ; # mol. wt of Na2CO3\n", + "mwt_H2O = 18. ; # mol. wt of H2O\n", + "\n", + "# Calculation and Results\n", + "m_Na2CO3 = mwt_Na2CO3*n_mol_Na2CO3 ; # Mass of Na2CO3\n", + "m_H2O = mwt_H2O*n_mol_H2O ; # Mass of H2O\n", + "Cf_Na2CO3 = m_Na2CO3/(m_Na2CO3+m_H2O) ; # mass fraction of Na2CO3 \n", + "Cf_H2O = 1-Cf_Na2CO3 ; # mass fraction of H2O\n", + "\n", + "n_un = 9. ; # Number of unknowns in the given problem\n", + "n_ie = 9. ; # Number of independent equations\n", + "d_o_f = n_un-n_ie ; # Number of degree of freedom\n", + "\n", + "print 'Number of degree of freedom for the given system is %i .'%d_o_f\n", + "\n", + "# Final composition of tank\n", + "#Basis :I = 10000 kg\n", + "# Material balance reduces to Accumulation = final -initial = in-out(but in = 0)\n", + "I = 10000. ; #initial amount of saturated solution-[kg]\n", + "amt_C = 3000. ; # Amount of crystals formed-[kg]\n", + "Fm_Na2CO3 = I*If_Na2CO3-amt_C*Cf_Na2CO3 ; # Mass balance of Na2CO3\n", + "Fm_H2O = I*If_H2O-amt_C*Cf_H2O ; # Mass balance of H2O\n", + "\n", + "#To find temperature,T\n", + "s_T = (Fm_Na2CO3/Fm_H2O)*100 ; # Solublity of Na2CO3 at temperature T\n", + "s_20 = 21.5 ; #Solublity of Na2CO3 at temperature 20 degree C ,from given table-[g Na2CO3/100 g H2O]\n", + "\n", + "# Find T by interpolation\n", + "T = 30-((s_30-s_T)/(s_30-s_20))*(30-20) ; # Temperature -[degree C]\n", + "print ' Temperature to which solution has to be cooled to get 3000 kg crystals is %.0f degree C .'%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of degree of freedom for the given system is 0 .\n", + " Temperature to which solution has to be cooled to get 3000 kg crystals is 26 degree C .\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + " Example 8.7 Page no. 213\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Hemodialysis\n", + "# Solution\n", + "\n", + "# Variables\n", + "# Write given data\n", + "B_in = 1.1 ; # Flow rate in of blood -[L/min]\n", + "B_out = 1.2; # Flow rate out of blood -[L/min]\n", + "S_in = 1.7; # Flow rate in of solution -[L/min]\n", + "\n", + "# Composition of input blood\n", + "B_in_CR = 2.72 ; #[g/L]\n", + "B_in_UR = 1.16 ; #[g/L]\n", + "B_in_U = 18 ; #[g/L]\n", + "B_in_P = 0.77 ; #[g/L]\n", + "B_in_K = 5.77 ; #[g/L]\n", + "B_in_Na = 13.0 ; #[g/L]\n", + "B_in_water = 1100 ; #[mL/min]\n", + "\n", + "# Composition of output blood\n", + "B_out_CR = 0.120 ; #[g/L]\n", + "B_out_UR = 0.060; #[g/L]\n", + "B_out_U = 1.51 ; #[g/L]\n", + "B_out_P = 0.040 ; #[g/L]\n", + "B_out_K = 0.120 ; #[g/L]\n", + "B_out_Na = 3.21 ; #[g/L]\n", + "B_out_water = 1200. ; #[mL/min]\n", + "\n", + "# Calculation and Result\n", + "n_un = 7. ; # Number of unknowns in the given problem\n", + "n_ie = 7. ; # Number of independent equations\n", + "d_o_f = n_un-n_ie ; # Number of degree of freedom\n", + "print 'Number of degree of freedom for the given system is %i .'%d_o_f\n", + "\n", + "# Water balance in grams, assuming 1 ml is equivalent to 1 g\n", + "S_in_water = 1700. ; #[ml/min]\n", + "S_out_water = B_in_water+ S_in_water - B_out_water;\n", + "S_out = S_out_water/1000. ; #[L/min]\n", + "print ' Flow rate of water in output solution is %.2f L/min.'%S_out\n", + "\n", + "# The component balance in grams for CR,UR,U,P,K and Na are\n", + "S_out_CR = (B_in*B_in_CR - B_out*B_out_CR)/S_out;\n", + "S_out_UR = (B_in*B_in_UR - B_out*B_out_UR)/S_out;\n", + "S_out_U = (B_in*B_in_U - B_out*B_out_U)/S_out;\n", + "S_out_P = (B_in*B_in_P - B_out*B_out_P)/S_out;\n", + "S_out_K = (B_in*B_in_K - B_out*B_out_K)/S_out;\n", + "S_out_Na = (B_in*B_in_Na - B_out*B_out_Na)/S_out;\n", + "print ' Component Concentration(g/L) in output Dialysis solution '\n", + "print ' UR %.2f '%S_out_UR\n", + "print ' CR %.2f '%S_out_CR\n", + "print ' U %.2f '%S_out_U\n", + "print ' P %.2f '%S_out_P\n", + "print ' K %.2f '%S_out_K\n", + "print ' Na %.2f '%S_out_Na" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of degree of freedom for the given system is 0 .\n", + " Flow rate of water in output solution is 1.60 L/min.\n", + " Component Concentration(g/L) in output Dialysis solution \n", + " UR 0.75 \n", + " CR 1.78 \n", + " U 11.24 \n", + " P 0.50 \n", + " K 3.88 \n", + " Na 6.53 \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "code", + "collapsed": true, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit