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