{ "metadata": { "name": "", "signature": "sha256:db3a394cd915f3675da3a1ec91e93913eb152d495576bd0a47778fe0952f79fb" }, "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": [ "\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": [ "\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": [ "\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": [ "\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": [ "\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": [ "\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": [ "\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": {} } ] }