# Chapter 8 : Solving Material Balance Problems for Single Units without Reaction¶

### Example 8.1 Page no. 197¶

In [1]:
# Variables
# Basis : 1 min
d_w =  1.0 ;            # Density of aqueous solution-[g/cubic metre]
d_sol =  0.6 ;          # Density of organic solvent-[g/cubic metre]
n_un = 8 ;              # Number of unknowns in the given problem
n_ie  = 8 ;             # Number of independent equations

# Calculation and Results
d_o_f =  n_un-n_ie ;    # Number of degree of freedom
print 'Number of degree of freedom for the given system is  %i .'%d_o_f

# Material balance of Strep.
x =  (200*10+10*0-200*0.2)/10;        #[g]
print 'Strep per litre of solvent is  %.1f g .'%x

cnc = x/(1000*d_sol) ;                #[g Strep/g of S]
print 'Strep per gram of solvent is  %.4f g Strep/g of S .'%cnc

m_fr = cnc/(1+cnc) ;                  #Mass fraction
print 'Mass fraction of Strep is  %.3f g .'%m_fr

Number of degree of freedom for the given system is  0 .
Strep per litre of solvent is  196.0 g .
Strep per gram of solvent is  0.3267 g Strep/g of S .
Mass fraction of Strep is  0.246 g .


### Example 8.2 Page no. 199¶

In [2]:
# Variables
F_O2 = 0.21 ;         # fraction of O2 in feed(F)
F_N2 = 0.79 ;         # fraction of N2 in feed(F)
P_O2 = 0.25 ;         # fraction of O2 in product(P)
P_N2 = 0.75 ;         # fraction of N2 in  product(P)
F = 100 ;             # Feed - [g mol]
w = 0.80 ;            # Fraction of waste
W = w*F ;             # Waste -[g mol]

# Calculation
# By analysis for degree of freedom , DOF comes to be zero
P = F - W ;           # By overall balance - [g mol]
W_O2 = (F_O2*F - P*P_O2)/100       # Fraction of O2 in waste stream by O2 balance
W_N2 = (W - W_O2*100)/100 ;        #Fraction of N2 in waste stream

# Results
print 'Composition of Waste Stream'
print '  Component             Fraction in waste stream'
print '  O2                    %.2f'%W_O2
print '  N2                    %.2f'%W_N2

Composition of Waste Stream
Component             Fraction in waste stream
O2                    0.16
N2                    0.64


### Example 8.3 Page no. 202¶

In [3]:
# Variables
# Basis : 1 hr so F  = 1000 kg
F = 1000 ;        # feed rate-[kg/hr]
P = F/10.0 ;      # product mass flow rate -[kg/hr]

n_un = 9 ;        # Number of unknowns in the given problem
n_ie  = 9 ;       # Number of independent equations

# Calculation and Result
d_o_f =  n_un-n_ie ;        # Number of degree of freedom
print 'Number of degree of freedom for the given system is  %i .'%d_o_f

#  Overall mass balance: F  =  P+B
B =  F-P ;                  # bottom mass flow rate -[kg/hr]
print ' Bottom mass flow rate -              %.1f kg '%B

# Composition of bottoms by material balances
m_EtOH = 0.1*F-0.6*P ;      # By EtOH balance-[kg]
m_H2O = 0.9*F - 0.4*P ;     # By H2O balance-[kg]
total  = m_EtOH+m_H2O ;     #[kg]
f_EtOH = m_EtOH/total ;     # Mass fraction of EtOH
f_H2O = m_H2O/total ;       # Mass fraction of H2O

print ' Mass of EtOH in bottom -             %.1f kg '%m_EtOH
print ' Mass of H2O in bottom -              %.1f kg '%m_H2O
print ' Mass fraction of EtOH in bottom -    %.3f '%f_EtOH
print ' Mass fraction of H2O in bottom -     %.3f '%f_H2O

Number of degree of freedom for the given system is  0 .
Bottom mass flow rate -              900.0 kg
Mass of EtOH in bottom -             40.0 kg
Mass of H2O in bottom -              860.0 kg
Mass fraction of EtOH in bottom -    0.044
Mass fraction of H2O in bottom -     0.956


### Example 8.4 Page no. 205¶

In [4]:
from numpy import matrix

# Variables
# Given
A = 200 ;             # Mass of added solution [kg]
P_H2SO4 = .1863 ;     #Fraction of H2SO4 in P(Final solution)
P_H2O = .8137 ;       #Fraction of H2O in P(Final solution)
A_H2SO4 = .777 ;      #Fraction of H2SO4 in A(Added solution)
A_H2O = .223 ;        #Fraction of H2O in A(Added solution)
F_H2SO4 = .1243 ;     #Fraction of H2SO4 in F(Original solution)
F_H2O = .8757 ;       #Fraction of H2O in F(Original solution)

# Calculations
# P - F = A - By overall balance
a = matrix([[P_H2O,-F_H2O],[1,-1]]) ;      # Matrix of coefficient
b = matrix([[A*A_H2O],[A]]) ;              # Matrix of contants
a = a.I
x = a*b ;                                  # Matrix of solutions- P = x(1) and F = x(2)

#Results
print ' Original solution taken- %.0i kg'%x[1]
print '  Final solution or kilograms of battery acid formed- %.0i kg'%x[0]

 Original solution taken- 1905 kg
Final solution or kilograms of battery acid formed- 2105 kg


### Example 8.5 Page no. 207¶

In [6]:
from numpy import matrix

# Variables
# Given
W = 100.0 ;      # Water removed - [kg]
A_H2O = 0.80 ;   # Fraction of water in A(intial fish cake)
A_BDC = 0.20 ;   #  Fraction of BDC(bone dry cake) in B(final dry fish cake)
B_H2O = 0.40 ;   # Fraction of water in A(intial fish cake)
B_BDC = 0.60 ;   #  Fraction of BDC(bone dry cake) in B(final dry fish cake)

# Calculations
a = matrix([[A_H2O, -B_H2O],[1, -1]]) ;       # Matrix of coefficient
b = matrix([[W],[W]]) ;                       # Matrix of contants
a = a.I
x = a * b;                                    # Matrix of solutions- A = x(1) and B = x(2)

# Results
print 'Weight of the fish cake originally put into dryer -%.0i kg'%x[0]

Weight of the fish cake originally put into dryer -150 kg


### Example 8.6 Page no. 209¶

In [7]:
# Variables
# Composition of initial solution at 30 degree C
s_30 = 38.8 ;                      #  solublity of Na2CO3 at 30 degree C, by using the table for solublity of Na2CO3-[g Na2CO3/100 g H2O]
If_Na2CO3 =  s_30/(s_30+100) ;     # Initial mass fraction of Na2CO3
If_H2O = 1-If_Na2CO3 ;             # Initial mass fraction of H2O

# Composition of crystals
# Basis : 1g mol Na2CO3.10H2O
n_mol_Na2CO3 = 1 ;                 # Number of moles of Na2CO3
n_mol_H2O = 10. ;                  # Number of moles of H2O
mwt_Na2CO3 = 106. ;                # mol. wt of Na2CO3
mwt_H2O = 18. ;                    # mol. wt of H2O

# Calculation and Results
m_Na2CO3 = mwt_Na2CO3*n_mol_Na2CO3 ;        # Mass of Na2CO3
m_H2O = mwt_H2O*n_mol_H2O ;                 # Mass of H2O
Cf_Na2CO3 =  m_Na2CO3/(m_Na2CO3+m_H2O) ;    # mass fraction of Na2CO3
Cf_H2O = 1-Cf_Na2CO3 ;                      # mass fraction of H2O

n_un = 9. ;                     # Number of unknowns in the given problem
n_ie  = 9. ;                    # Number of independent equations
d_o_f =  n_un-n_ie ;            # Number of degree of freedom

print 'Number of degree of freedom for the given system is  %i .'%d_o_f

# Final composition of tank
#Basis :I = 10000 kg
# Material balance reduces to Accumulation  =  final -initial = in-out(but in = 0)
I = 10000. ;               #initial amount of saturated solution-[kg]
amt_C = 3000. ;            # Amount of crystals formed-[kg]
Fm_Na2CO3 = I*If_Na2CO3-amt_C*Cf_Na2CO3 ;             # Mass balance of Na2CO3
Fm_H2O = I*If_H2O-amt_C*Cf_H2O ;                      # Mass balance of H2O

#To find temperature,T
s_T =  (Fm_Na2CO3/Fm_H2O)*100 ;             # Solublity of Na2CO3 at temperature T
s_20 = 21.5 ;                               #Solublity of Na2CO3 at temperature 20 degree C ,from given table-[g Na2CO3/100 g H2O]

# Find T by interpolation
T =  30-((s_30-s_T)/(s_30-s_20))*(30-20) ;       # Temperature -[degree C]
print ' Temperature to which solution has to be cooled to get 3000 kg crystals is %.0f degree C .'%T

Number of degree of freedom for the given system is  0 .
Temperature to which solution has to be cooled to get 3000 kg crystals is 26 degree C .


### Example 8.7 Page no. 213¶

In [8]:
# Variables
# Write given data
B_in =  1.1 ;        # Flow rate in  of blood -[L/min]
B_out = 1.2;         # Flow rate out of blood -[L/min]
S_in = 1.7;          # Flow rate in  of solution -[L/min]

# Composition of input blood
B_in_CR = 2.72 ;     #[g/L]
B_in_UR = 1.16 ;     #[g/L]
B_in_U = 18 ;        #[g/L]
B_in_P = 0.77 ;      #[g/L]
B_in_K = 5.77 ;      #[g/L]
B_in_Na = 13.0 ;     #[g/L]
B_in_water = 1100 ;  #[mL/min]

# Composition of output blood
B_out_CR = 0.120 ;   #[g/L]
B_out_UR = 0.060;    #[g/L]
B_out_U = 1.51 ;     #[g/L]
B_out_P = 0.040 ;    #[g/L]
B_out_K = 0.120 ;    #[g/L]
B_out_Na = 3.21 ;    #[g/L]
B_out_water = 1200. ;     #[mL/min]

# Calculation and Result
n_un = 7. ;             # Number of unknowns in the given problem
n_ie  = 7. ;            # Number of independent equations
d_o_f =  n_un-n_ie ;    # Number of degree of freedom
print 'Number of degree of freedom for the given system is  %i .'%d_o_f

# Water balance in grams, assuming 1 ml is equivalent to 1 g
S_in_water = 1700. ;              #[ml/min]
S_out_water =  B_in_water+ S_in_water - B_out_water;
S_out = S_out_water/1000. ;       #[L/min]
print ' Flow rate of water in output solution is  %.2f L/min.'%S_out

# The component balance in grams for CR,UR,U,P,K and Na are
S_out_CR  =  (B_in*B_in_CR - B_out*B_out_CR)/S_out;
S_out_UR  =  (B_in*B_in_UR - B_out*B_out_UR)/S_out;
S_out_U  =  (B_in*B_in_U - B_out*B_out_U)/S_out;
S_out_P  =  (B_in*B_in_P - B_out*B_out_P)/S_out;
S_out_K  =  (B_in*B_in_K - B_out*B_out_K)/S_out;
S_out_Na  =  (B_in*B_in_Na - B_out*B_out_Na)/S_out;
print ' Component     Concentration(g/L) in output Dialysis solution   '
print ' UR            %.2f    '%S_out_UR
print ' CR            %.2f    '%S_out_CR
print ' U             %.2f    '%S_out_U
print ' P             %.2f    '%S_out_P
print ' K             %.2f    '%S_out_K
print ' Na            %.2f    '%S_out_Na

Number of degree of freedom for the given system is  0 .
Flow rate of water in output solution is  1.60 L/min.
Component     Concentration(g/L) in output Dialysis solution
UR            0.75
CR            1.78
U             11.24
P             0.50
K             3.88
Na            6.53

In [ ]: