# Chapter 12 : Recycle Bypass Purge and the Industrial Application of Material Balance¶

### Example 12.1 Page No.349¶

In [1]:
from numpy import matrix

# Variables
F = 10000. ;			#[lb/hr]

NaOH_F = 40./100 ;	    		#[wt. fraction]
NaOH_P1 = 95./100 ;	    		#[wt. fraction of NaOH filter cake]
NaOH_P2 = (0.05 *  45)/100 ;			#[wt. fraction of NaOH in NaOH soln.]
H2O_P2 = (0.05 *  55)/100 ;			#[wt. fraction of H2O in NaOH soln.]
NaOH_R = 45./100;			#[wt. fraction]
NaOH_G = 50./100;			#[wt. fraction]

P = (NaOH_F *  F)/(NaOH_P1 + NaOH_P2) ;			#[lb/hr]

W = F-P ;			# [lb/hr]

# Calculations & Results
a = matrix([[NaOH_G,-NaOH_R],[1,-1]]) 			# matrix formed of coefficients of unknown
b = matrix([[F*NaOH_F],[P]]);			# matrix formed by constant
a = a.I
x = a*b ;			# matrix of solutions . x(1) = G, x(2) = R
G = x[0] ;			# [lb/hr]
R = x[1] ;			# [lb/hr]

print '(a)  Flow rate of water removed by evaporator is  %.1f lb/hr'%W
print 'The recycle rate of the process is %.1f lb/hr'%R

NaOH_H = 45./100 ;			#[wt fraction]
H2O_H = 55./100 ;			#[wt fraction]

a1 = matrix([[NaOH_G,-NaOH_H],[NaOH_G,-H2O_H]]) ;			# matrix formed of coefficients of unknown
b1 = matrix([[((NaOH_P1+NaOH_P2)* P)],[(H2O_P2) *  P]]);			# matrix formed by constant
x1 = a1**-1
x1 = x1*b1 ;			# matrix of solutions nw_G = x1(1);H = x1(2)
nw_G1 = x1[0] ;			# [lb/hr]
H = x1[1];			# [lb/hr]

nw_F = (NaOH_H *  H + (NaOH_P1 + NaOH_P2) *  P)/NaOH_F ;			#[lb/hr]
print ' (b)  Total feed rate when filterate is not recycled is %.1f lb/hr'%nw_F

(a)  Flow rate of water removed by evaporator is  5886.9 lb/hr
The recycle rate of the process is 38868.9 lb/hr
(b)  Total feed rate when filterate is not recycled is 53727.5 lb/hr


### Example 12.2 Page No.357¶

In [2]:
# Variables
F_Bz = 100. ;			# Fresh benzene feed / basis - [mol]
con_Bz = .95 ;			# Fraction of conversion of benzene
sp_con = .20 ;			# Fraction of single pass conversion
ex_H2 = .20 ;			# Fraction of exces H2 used in fresh feed
R_Bz = 22.74 ;			# Benzene in Recycle stream - [mol %]
R_H2 = 78.26 ;			# H2 in Recycle stream - [mol %]
TLV_Bz = 0.5 ;			# TLV value of benzene -[ppm]
TLV_C6H12 = 300. ;			# TLV value of cyclohexane -[ppm]
TLV_H2 = 1000. ;			# TLV value of H2 -[ppm]

# Calculations
F_H2 = F_Bz*3*(1+ex_H2) ;			# H2 in Feed - [mol]
F = F_Bz + F_H2 ;			# Total feed - [mol]

ex_r = con_Bz*F_Bz/(-(-1)) ;			# Extent of reaction

P_Bz = F_Bz -1*(ex_r) ;			# Benzene in P ,by benzene balance - [mol]
P_H2 = F_H2 + -3*(ex_r) ;			# H2 in P ,by H2 balance - [mol]
P_C6H12 = 0 + 1*(ex_r) ;			# Cyclohexane in P ,by cyclohexane balance - [mol]
P = P_Bz + P_H2 + P_C6H12 ;			# Total Product - [ mol]

R = ((-(-ex_r))/(sp_con) - F_Bz)/(R_Bz/100.) ;			# Recycle stream - [mol]
R_by_F = R/F ;			# Ratio of R to F

TLV = (P_Bz/P)*(1/TLV_Bz) + (P_H2/P)*(1./TLV_H2) + (P_C6H12/P)*(1/TLV_C6H12) ;			# TLV (environmental index)

# Results
print 'Ratio of R to F is %.2f .'%R_by_F
print ' TLV (environmental index) is %.3f .'%TLV

Ratio of R to F is 3.58 .
TLV (environmental index) is 0.059 .


### Example 12.3 Page No.359¶

In [3]:
# Variables
RR = 8.33 ;			# Recycle ratio
F = 100. ;			# Overall feed/basis - [lb]
F_g = 0.40 ;			# Fraction of glucose in overall feed
F_w = 0.60 ;			# Fraction of water in overall feed
F_dash_f = 0.04 ; 			# Fraction of fructose in feed to reactor
P = F   			# By overall balance -[lb]
R = P/RR ;			# Recycle stream - [lb]
P_w = (F_w * F)/ P ;			# Fraction of water in product(P), by overall water balance
R_w = P_w ;			#Fraction of water in recycle (R), since both R and P has same composition

# Calculations
F_dash = F +R ;			# Feed to reactor ,by total balance -[lb]
R_f = (F_dash*F_dash_f)/R ;			# Fraction of fructose in recycle stream
R_g = 1 - (R_f + R_w) ;			# Fraction of glucose in recycle stream
F_dash_g = (F*F_g + R*R_g)/F_dash ;			# Fraction of glucose i feed to reactor

f_con = ((F_dash*F_dash_g) - (R + P)*R_g)/(F_dash*F_dash_g) ;			# Fraction of conversion of glucose in reactor

# Results
print 'Fraction of conversion of glucose in reactor is %.2f .'%f_con

Fraction of conversion of glucose in reactor is 0.93 .


### Example 12.4 Page No.362¶

In [4]:
# Variables
F = 100. ;			# Overall feed/basis - [kg]
F_com = 0.10 ;			# Mass fraction of component in fresh feed
F_w = 0.90 ;			# Mass fraction of water in fresh feed
P_w = 0.10 ;			# Mass fraction of water in product
P_com = 0.90 ;			#Mass fraction of component in product
F_dash_com = 0.03 ;			#Mass fraction of component in feed to reactor
W_w = 1. ;			# Mass fraction of water in W(waste)
C_con = .40 ;			# Fraction of conversion of component in reactor

# Calculations
P = F_com*F/P_com ;			#By component balance- Product - [kg]
W = F - P ;			# By overall balance - waste(W)- [kg]

Rw = (F*F_com - F*F_com*C_con)/C_con ;			# Mass of component in recycle(R) - [kg]
F_dash = ( F*F_com + Rw  )/F_dash_com ;			# By component balance - feed to reactor(F') -[kg]
R = F_dash - F ;			                    # Recycle(R) - By total balance -[kg]
w = Rw/R ;			                            # Mass fraction of component in recycle(R)

# Results
print 'Recycle(R) stream-  %.0f kg '%R
print ' Mass fraction of component in recycle(R)-   %.4f .'%w

Recycle(R) stream-  733 kg
Mass fraction of component in recycle(R)-   0.0205 .


### Example 12.5 Page No.367¶

In [5]:
# Variables
F = 100.        			# Overall feed/basis - [kg]
F_n_C5H12 = 0.80 ;			# Fraction of n_C5H12 in overall feed
F_i_C5H12 = 0.20 ;			# Fraction of i_C5H12in overall feed
S_i_C5H12 = 1. ;			# Fraction of i_C5H12 in isopentane stream
P_n_C5H12 = .90 ;			# Fraction of n_C5H12 in overall product
P_i_C5H12 = .10 ;			# Fraction of i_C5H12 in overall product

# Calculations
P = (F*F_n_C5H12)/P_n_C5H12 ;			#Product Material Balance of n_C5H12 -[kg]
S = F - P ;			# Isopentane stream (S) from overall material balance - [kg]

from numpy import matrix
a = matrix([[1,-1],[F_n_C5H12,-1]]) ;			# Matrix of coefficients of unknown
b = matrix([[S],[0]]) ;			# Matrix of constants
a = a.I
x = a*b         			# Matrix of solutions, x(1) = x , x(2) = y

xf = x[0]/F ;	    		# Fraction of butane-free gas going to isopentane tower

# Results
print 'Fraction of butane-free gas going to isopentane tower is %.3f .'%xf

Fraction of butane-free gas going to isopentane tower is 0.556 .


### Example 12.6 Page No.369¶

In [6]:
# Variables
F = 100. ;			# Overall feed/basis - [mole]
F_H2 = 0.673 ;			# Mole fraction of H2 in overall feed
F_CO = 0.325 ;			# Mole fraction of i_C5H12in overall feed
F_CH4 = .002 ;			# Mole fraction of CH4 in overall feed
E_CH3OH = 1. ;			#  Mole fraction of CH3OH in Exit(E)

z = .032 ;
CO_con = .18 ;			# Fraction of conversion of CO in reactor

# Calculations
#By using eqn.(c) and (d)
P = F_CH4*F/z ;			# Purge stream - [mole]

# Using eqn.(a) , (b) and (c)
x_plus_y = 1 - z ;			# x + y
E = (F_H2*F + F_CO*F + 3*F_CH4*F - P*(x_plus_y + 3*z ))/3 ;			# Exit stream - [mole]

# By using eqn. (d)
y = ( F_CO*F - E )/P ;			# Mole fraction of CO

# By using eqn. (a)
x = 1 - z - y ;			# Mole fraction of H2

# Lastly by using eqn.(e)
R = ( F_CO*F - P*y - F_CO*F*CO_con )/(y*CO_con) ;			# Recycle steam - [mole]

# Results
print 'Moles of recycle(R) per mole of feed(F) -                      %.4f '%(R/F)
print ' Moles of CH3OH(E) per mole of feed(F)   -                      %.4f '%(E/F)
print ' Moles of Purge(P) per mole of feed(F)   -                      %.4f '%(P/F);
print ' Composition of Purge '
print '  Component                 Mole fraction '
print '  H2                        %.3f '%x
print '  CO                        %.3f '%y
print '  CH4                       %.3f '%z

Moles of recycle(R) per mole of feed(F) -                      7.0556
Moles of CH3OH(E) per mole of feed(F)   -                      0.3125
Moles of Purge(P) per mole of feed(F)   -                      0.0625
Composition of Purge
Component                 Mole fraction
H2                        0.768
CO                        0.200
CH4                       0.032

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