{
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"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 12 : Recycle Bypass Purge and the Industrial Application of Material Balance"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 12.1 Page No.349\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from numpy import matrix\n",
"\t\t\t\n",
"# Variables \n",
"F = 10000. ;\t\t\t#[lb/hr]\n",
"\t\t\t\n",
"NaOH_F = 40./100 ;\t \t\t#[wt. fraction]\n",
"NaOH_P1 = 95./100 ;\t \t\t#[wt. fraction of NaOH filter cake]\n",
"NaOH_P2 = (0.05 * 45)/100 ;\t\t\t#[wt. fraction of NaOH in NaOH soln.]\n",
"H2O_P2 = (0.05 * 55)/100 ;\t\t\t#[wt. fraction of H2O in NaOH soln.]\n",
"NaOH_R = 45./100;\t\t\t#[wt. fraction]\n",
"NaOH_G = 50./100;\t\t\t#[wt. fraction]\n",
"\t\t\t\n",
"P = (NaOH_F * F)/(NaOH_P1 + NaOH_P2) ;\t\t\t#[lb/hr]\n",
"\t\t\t\n",
"W = F-P ;\t\t\t# [lb/hr]\n",
"\n",
"# Calculations & Results\t\n",
"a = matrix([[NaOH_G,-NaOH_R],[1,-1]]) \t\t\t# matrix formed of coefficients of unknown\n",
"b = matrix([[F*NaOH_F],[P]]);\t\t\t# matrix formed by constant\n",
"a = a.I\n",
"x = a*b ;\t\t\t# matrix of solutions . x(1) = G, x(2) = R\n",
"G = x[0] ;\t\t\t# [lb/hr]\n",
"R = x[1] ;\t\t\t# [lb/hr]\n",
" \n",
"print '(a) Flow rate of water removed by evaporator is %.1f lb/hr'%W\n",
"print 'The recycle rate of the process is %.1f lb/hr'%R\n",
"\n",
"\t\t\t\n",
"NaOH_H = 45./100 ;\t\t\t#[wt fraction]\n",
"H2O_H = 55./100 ;\t\t\t#[wt fraction]\n",
"\n",
"a1 = matrix([[NaOH_G,-NaOH_H],[NaOH_G,-H2O_H]]) ;\t\t\t# matrix formed of coefficients of unknown\n",
"b1 = matrix([[((NaOH_P1+NaOH_P2)* P)],[(H2O_P2) * P]]);\t\t\t# matrix formed by constant\n",
"x1 = a1**-1\n",
"x1 = x1*b1 ;\t\t\t# matrix of solutions nw_G = x1(1);H = x1(2)\n",
"nw_G1 = x1[0] ;\t\t\t# [lb/hr]\n",
"H = x1[1];\t\t\t# [lb/hr]\n",
"\n",
"nw_F = (NaOH_H * H + (NaOH_P1 + NaOH_P2) * P)/NaOH_F ;\t\t\t#[lb/hr]\n",
"print ' (b) Total feed rate when filterate is not recycled is %.1f lb/hr'%nw_F\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) Flow rate of water removed by evaporator is 5886.9 lb/hr\n",
"The recycle rate of the process is 38868.9 lb/hr\n",
" (b) Total feed rate when filterate is not recycled is 53727.5 lb/hr\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 12.2 Page No.357\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"F_Bz = 100. ;\t\t\t# Fresh benzene feed / basis - [mol]\n",
"con_Bz = .95 ;\t\t\t# Fraction of conversion of benzene\n",
"sp_con = .20 ;\t\t\t# Fraction of single pass conversion\n",
"ex_H2 = .20 ;\t\t\t# Fraction of exces H2 used in fresh feed\n",
"R_Bz = 22.74 ;\t\t\t# Benzene in Recycle stream - [mol %]\n",
"R_H2 = 78.26 ;\t\t\t# H2 in Recycle stream - [mol %]\n",
"TLV_Bz = 0.5 ;\t\t\t# TLV value of benzene -[ppm]\n",
"TLV_C6H12 = 300. ;\t\t\t# TLV value of cyclohexane -[ppm]\n",
"TLV_H2 = 1000. ;\t\t\t# TLV value of H2 -[ppm]\n",
"\n",
"# Calculations\n",
"F_H2 = F_Bz*3*(1+ex_H2) ;\t\t\t# H2 in Feed - [mol]\n",
"F = F_Bz + F_H2 ;\t\t\t# Total feed - [mol] \n",
"\n",
"ex_r = con_Bz*F_Bz/(-(-1)) ;\t\t\t# Extent of reaction\n",
"\n",
"P_Bz = F_Bz -1*(ex_r) ;\t\t\t# Benzene in P ,by benzene balance - [mol]\n",
"P_H2 = F_H2 + -3*(ex_r) ;\t\t\t# H2 in P ,by H2 balance - [mol]\n",
"P_C6H12 = 0 + 1*(ex_r) ;\t\t\t# Cyclohexane in P ,by cyclohexane balance - [mol]\n",
"P = P_Bz + P_H2 + P_C6H12 ;\t\t\t# Total Product - [ mol]\n",
"\n",
"R = ((-(-ex_r))/(sp_con) - F_Bz)/(R_Bz/100.) ;\t\t\t# Recycle stream - [mol]\n",
"R_by_F = R/F ;\t\t\t# Ratio of R to F \n",
"\n",
"TLV = (P_Bz/P)*(1/TLV_Bz) + (P_H2/P)*(1./TLV_H2) + (P_C6H12/P)*(1/TLV_C6H12) ;\t\t\t# TLV (environmental index) \n",
"\n",
"# Results\n",
"print 'Ratio of R to F is %.2f .'%R_by_F\n",
"print ' TLV (environmental index) is %.3f .'%TLV\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ratio of R to F is 3.58 .\n",
" TLV (environmental index) is 0.059 .\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 12.3 Page No.359\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"RR = 8.33 ;\t\t\t# Recycle ratio\n",
"F = 100. ;\t\t\t# Overall feed/basis - [lb]\n",
"F_g = 0.40 ;\t\t\t# Fraction of glucose in overall feed \n",
"F_w = 0.60 ;\t\t\t# Fraction of water in overall feed \n",
"F_dash_f = 0.04 ; \t\t\t# Fraction of fructose in feed to reactor\n",
"P = F \t\t\t# By overall balance -[lb]\n",
"R = P/RR ;\t\t\t# Recycle stream - [lb]\n",
"P_w = (F_w * F)/ P ;\t\t\t# Fraction of water in product(P), by overall water balance\n",
"R_w = P_w ;\t\t\t#Fraction of water in recycle (R), since both R and P has same composition\n",
"\n",
"# Calculations\t\t\n",
"F_dash = F +R ;\t\t\t# Feed to reactor ,by total balance -[lb]\n",
"R_f = (F_dash*F_dash_f)/R ;\t\t\t# Fraction of fructose in recycle stream \n",
"R_g = 1 - (R_f + R_w) ;\t\t\t# Fraction of glucose in recycle stream\n",
"F_dash_g = (F*F_g + R*R_g)/F_dash ;\t\t\t# Fraction of glucose i feed to reactor\n",
"\n",
"f_con = ((F_dash*F_dash_g) - (R + P)*R_g)/(F_dash*F_dash_g) ;\t\t\t# Fraction of conversion of glucose in reactor\n",
"\n",
"# Results\n",
"print 'Fraction of conversion of glucose in reactor is %.2f .'%f_con\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Fraction of conversion of glucose in reactor is 0.93 .\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 12.4 Page No.362\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"F = 100. ;\t\t\t# Overall feed/basis - [kg]\n",
"F_com = 0.10 ;\t\t\t# Mass fraction of component in fresh feed \n",
"F_w = 0.90 ;\t\t\t# Mass fraction of water in fresh feed \n",
"P_w = 0.10 ;\t\t\t# Mass fraction of water in product\n",
"P_com = 0.90 ;\t\t\t#Mass fraction of component in product\n",
"F_dash_com = 0.03 ;\t\t\t#Mass fraction of component in feed to reactor\n",
"W_w = 1. ;\t\t\t# Mass fraction of water in W(waste)\n",
"C_con = .40 ;\t\t\t# Fraction of conversion of component in reactor\n",
"\n",
"# Calculations\n",
"P = F_com*F/P_com ;\t\t\t#By component balance- Product - [kg]\n",
"W = F - P ;\t\t\t# By overall balance - waste(W)- [kg]\n",
"\n",
"Rw = (F*F_com - F*F_com*C_con)/C_con ;\t\t\t# Mass of component in recycle(R) - [kg]\n",
"F_dash = ( F*F_com + Rw )/F_dash_com ;\t\t\t# By component balance - feed to reactor(F') -[kg]\n",
"R = F_dash - F ;\t\t\t # Recycle(R) - By total balance -[kg]\n",
"w = Rw/R ;\t\t\t # Mass fraction of component in recycle(R) \n",
"\n",
"# Results\n",
"print 'Recycle(R) stream- %.0f kg '%R\n",
"print ' Mass fraction of component in recycle(R)- %.4f .'%w\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Recycle(R) stream- 733 kg \n",
" Mass fraction of component in recycle(R)- 0.0205 .\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 12.5 Page No.367\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"F = 100. \t\t\t# Overall feed/basis - [kg]\n",
"F_n_C5H12 = 0.80 ;\t\t\t# Fraction of n_C5H12 in overall feed \n",
"F_i_C5H12 = 0.20 ;\t\t\t# Fraction of i_C5H12in overall feed \n",
"S_i_C5H12 = 1. ;\t\t\t# Fraction of i_C5H12 in isopentane stream\n",
"P_n_C5H12 = .90 ;\t\t\t# Fraction of n_C5H12 in overall product\n",
"P_i_C5H12 = .10 ;\t\t\t# Fraction of i_C5H12 in overall product\n",
"\n",
"# Calculations\n",
"P = (F*F_n_C5H12)/P_n_C5H12 ;\t\t\t#Product Material Balance of n_C5H12 -[kg]\n",
"S = F - P ;\t\t\t# Isopentane stream (S) from overall material balance - [kg]\n",
"\n",
"from numpy import matrix\n",
"a = matrix([[1,-1],[F_n_C5H12,-1]]) ;\t\t\t# Matrix of coefficients of unknown \n",
"b = matrix([[S],[0]]) ;\t\t\t# Matrix of constants\n",
"a = a.I\n",
"x = a*b \t\t\t# Matrix of solutions, x(1) = x , x(2) = y\n",
"\n",
"xf = x[0]/F ;\t \t\t# Fraction of butane-free gas going to isopentane tower \n",
"\n",
"# Results\n",
"print 'Fraction of butane-free gas going to isopentane tower is %.3f .'%xf\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Fraction of butane-free gas going to isopentane tower is 0.556 .\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
" Example 12.6 Page No.369\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"F = 100. ;\t\t\t# Overall feed/basis - [mole]\n",
"F_H2 = 0.673 ;\t\t\t# Mole fraction of H2 in overall feed \n",
"F_CO = 0.325 ;\t\t\t# Mole fraction of i_C5H12in overall feed \n",
"F_CH4 = .002 ;\t\t\t# Mole fraction of CH4 in overall feed \n",
"E_CH3OH = 1. ;\t\t\t# Mole fraction of CH3OH in Exit(E)\n",
"\t\n",
"z = .032 ;\n",
"CO_con = .18 ;\t\t\t# Fraction of conversion of CO in reactor\n",
"\n",
"# Calculations\n",
"#By using eqn.(c) and (d)\n",
"P = F_CH4*F/z ;\t\t\t# Purge stream - [mole]\n",
"\n",
"# Using eqn.(a) , (b) and (c)\n",
"x_plus_y = 1 - z ;\t\t\t# x + y \n",
"E = (F_H2*F + F_CO*F + 3*F_CH4*F - P*(x_plus_y + 3*z ))/3 ;\t\t\t# Exit stream - [mole]\n",
"\n",
"# By using eqn. (d)\n",
"y = ( F_CO*F - E )/P ;\t\t\t# Mole fraction of CO \n",
"\n",
"# By using eqn. (a)\n",
"x = 1 - z - y ;\t\t\t# Mole fraction of H2 \n",
"\n",
"# Lastly by using eqn.(e)\n",
"R = ( F_CO*F - P*y - F_CO*F*CO_con )/(y*CO_con) ;\t\t\t# Recycle steam - [mole]\n",
"\n",
"# Results\n",
"print 'Moles of recycle(R) per mole of feed(F) - %.4f '%(R/F)\n",
"print ' Moles of CH3OH(E) per mole of feed(F) - %.4f '%(E/F)\n",
"print ' Moles of Purge(P) per mole of feed(F) - %.4f '%(P/F);\n",
"print ' Composition of Purge '\n",
"print ' Component Mole fraction '\n",
"print ' H2 %.3f '%x\n",
"print ' CO %.3f '%y\n",
"print ' CH4 %.3f '%z\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Moles of recycle(R) per mole of feed(F) - 7.0556 \n",
" Moles of CH3OH(E) per mole of feed(F) - 0.3125 \n",
" Moles of Purge(P) per mole of feed(F) - 0.0625 \n",
" Composition of Purge \n",
" Component Mole fraction \n",
" H2 0.768 \n",
" CO 0.200 \n",
" CH4 0.032 \n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "code",
"collapsed": true,
"input": [],
"language": "python",
"metadata": {},
"outputs": []
}
],
"metadata": {}
}
]
}