# Chapter 3 : Material Balances without Chemical Reaction¶

### Example 3.1 Page 60¶

In :
# solution

# Variables
m = 1.               #[kg] feed water
m1 = 1200.           #[mg] dissolved solids in 1 kg feed water
m2 = 3500.           #[mg] max dissolved solid content

# Calculation
x = (m*m1)/m2        # [kg] blown down water

# Result
print "Percentage of feed water to be blown down is ",x,"."

Percentage of feed water to be blown down is  0.342857142857 .


### Example 3.2 Page 61¶

In :
# solution

# Variables
m = 100.                 #[kg] weak liquor (feed)
m1 = 4.                  #[kg] NaOH
p = .25

# Calculation
x = 4./p                 # water left
y = 100-16               # [kg] evaporated water

# Result
print "Amount of water that evaporated is ",y," kg."

Amount of water that evaporated is  84  kg.


### Example 3.3 Page 61¶

In :
# solution

# Variables
m = 100.                     #[kg] babul bark (basis)
m1 = 5.8                     #[kg] moisture
m2 = 12.6                    #[kg] Tannin
m3 = 8.3                     #[kg] soluble non tannin organic material
m4 = m-m1-m2-m3              # [kg] Lignin

# Calculation
# lignin content remains unaffected during leaching
m5 = 100-.92-.65             # [kg lignin/kg dry residue]
x = (m4*100)/m5              # [kg]
T1 = x*.0092                 #[kg] Tannin present in residue
T2 = m2 - T1                 # [kg] Tannin recovered
T = (T2/m2)*100

# Result
print "Percentage of Tannin recovered during leaching is ",T,"."

Percentage of Tannin recovered during leaching is  94.5625688387 .


### Example 3.4 Page 62¶

In :
# solution

# Variables
m = 1.                   #[kg] dry neem leaves (basis)
m1 = .01/100             #[kg] beta cartene content of leaves

# Calculation
Ex = (m1*100)/.41        #[kg] extract quantity
Tc1 = Ex*.155            #[kg] Alpha Tocopherol in the extract
Tc2 = .46/100            #[kg] Alpha Tocopherol in the neem leaves
R = (Tc1*100)/Tc2        # recovery of Alpha Tocopherol

# Result
print "a  mass of extract phase per kg of dry leaves is "\
,Ex," kg   \nb  percent recovery of Alpha Tocopherol is ",R,"."

a  mass of extract phase per kg of dry leaves is  0.0243902439024  kg
b  percent recovery of Alpha Tocopherol is  82.1845174973 .


### Example 3.5 Page 62¶

In :
# solution
from numpy import array, linalg

# Variables
m= 100.                  #[kg] original mixture (basis)
A = 27.8                 #[kg]
B = 72.2                 #[kg]

# let x and y be uper and lower layer amounts
# total mixture = (x+y) kg
# balancing A and B
X = array([[.075 ,.203],[.035, .673]])
d = array([27.8,72.2])

# Calculation
x = linalg.solve(X,d)
M = X+X[1,0]          # [kg] total mixture
Ms = M - m                  #[kg] mixed solvent
Mr = Ms/m
S1 = x*.574+x*.028            #[kg] water balance
S2 = x*.316+x*.096            #[kg] acetic acid balance
Qs = S1+S2
pS1 = (S1*100)/Qs
pS2 = 100-pS1

# Result
print "a)  Upper layer = ",x," kg \nand Lower layer = ",x,\
"  \nb)  mass ratio of the mixed solvent to the original mixture is ",Mr,\
"    \nc)  water mass percent = ",pS1,"\nand acetic acid mass percent = ",pS2,"."

a)  Upper layer =  93.4470832373  kg
and Lower layer =  102.421028361
b)  mass ratio of the mixed solvent to the original mixture is  -0.9989
c)  water mass percent =  58.9418250036
and acetic acid mass percent =  41.0581749964 .


### Example 3.6 Page 63¶

In :
# solution
# Variables
m = 170.                         #[Nm**3/h] air (basis)
m1 = 50*.99                     #[Nm**3/h] N2 content of the stream
m2 = 50*.01                      #[Nm**3/h]

# Calculation
N = m*.79-m1                     # [Nm**3/h] N2
O = m*.21-m2                     # [Nm**3/h] O2
V1 = N*100/(N+O)
V2 = O*100/(N+O)

# Result
print "Vol percent of N2 is ",V1," and Vol percent of O2 is ",V2,"."

Vol percent of N2 is  70.6666666667  and Vol percent of O2 is  29.3333333333 .


### Example 3.7 Page 64¶

In :
# solution
# Variables
m = 100.                     #[kg] SO3 free mixed acid (basis)
m1 = 55.                     #[kg] HNO3
m2 = 45.                     #[kg] H2SO4

# Calculation
# SO3 + H2O --> H2SO4
m3 = (80./18)*3              #[kg] SO3 equivalent to 3 kg of water
Q = m2+m3                    #[kg] oleum to be mixed
S = (m3/Q)*100               # strength of oleum
R = m1/Q

# Result
print "Strength of Oleum required is ",S," HNO3 and Oleum are required\
to be mixed in the proportion of ",R,":1."

Strength of Oleum required is  22.8571428571  HNO3 and Oleum are required to be mixed in the proportion of  0.942857142857 :1.


### Example 3.8 Page 64¶

In :
from numpy import array, linalg

# solution
# Variables
m = 1000.                #[kg] mixed acid (basis)
# doing overall mass balance, H2SO4 balance and HNO3 balance
A = array([[1, 1, 1],[.444, 0, .98],[.113, .9, 0]])
d = array([1000,600,320])

# Calculation
x = linalg.solve(A,d)

# Result
print "quantities of acids required are :"
print "    Spent  =  ",x,"kg"
print "    HNO3   =  ",x," kg"
print "    H2SO4  =  ",x," kg."

quantities of acids required are :
Spent  =   76.4139267072 kg
HNO3   =   345.961362536  kg
H2SO4  =   577.624710757  kg.


### Example 3.9 Page 65¶

In :
# solution
# Variables
l = 1.                      #[litre] water (basis)
Cl = 475.6                  #[mg]
m1 = (58.5/35.5)*Cl         #[mg] NaCl present in water
SO4 = 102.9                 #[mg] # SO4
m3 = (142./96)*SO4          #[mg] Na2SO4 present in water

# Calculation
# carbonates are present due to Na2CO3
# eq mass of CaCO3 = 50
# eq mass of Na2CO3 = 53
m4 = (53./50)*65.9          # [mg] Na2CO3 present in water

# NaHCO3 in water = bicarbonates - temporary hardness
m5 = 390.6-384              # [mg] NaHCO3 present as CaCO3
m6 = (84./50)*m5            # [mg] NaHCO3 present in water

# equivalent mass of Mg(HCO3)2 = 73.15
m7 = (m6/50.)*225
m8  = 384-225               #[mg] CaCO3 from Ca(HCO3)2
# equivalent mass of Ca(HCO3)2 is 81
m9 = (m8/50.)*159           #[mg] Ca(HCO3)2 present in water

# Result
print "Component analysis of raw water:"
print " Compound          mg/l "
print " Ca(HCO3)2       ",m9
print " Mg(HCO3)2         ",m7
print " NaHCO3          ",m6
print " Na2CO3          ",m4
print " NaCl            ",m1
print " Na2SO4          ",m3

Component analysis of raw water:
Compound          mg/l
Ca(HCO3)2        505.62
Mg(HCO3)2          49.896
NaHCO3           11.088
Na2CO3           69.854
NaCl             783.735211268
Na2SO4           152.20625


### Example 3.11 Page 68¶

In :
from numpy import array, linalg
# solution
# Variables
# basis : 1000 kg/h of feed
# balancing H2SO4, HNO3 and H2O in all the three product streams
M = array([[1, 0, 0, 1, 0, 0, 1, 0, 0],[0, 1, 0 ,0, 1, 0, 0, 1, 0],\
[0, 0, 1, 0, 0, 1, 0, 0, 1],[1, 0, 0, 0, 0, 0, 0, 0, 0],[0, 1, 0, 0, 0, 0, 0, 0, 0]\
,[0, 0, 1, 0, 0, 0, 0, 0, 0],[0 ,0, 0, 1, 0, 0, 0, 0, 0],[0, 0, 0, 0, 1, 0, 0, 0, 0],\
[0, 0, 0, 0, 0, 1, 0, 0, 0]])
v = array([400,100,500,4,94,60,16,6,400])

# Calculation
s = linalg.solve(M,v)
A = s+s+s
B = s+s+s
C = s+s+s

# Result
print "Flowrates are : "
print "   A = ",A," kg/h"
print "   B = ",B," kg/h"
print "   C = ",C," kg/h"

Flowrates are :
A =  158.0  kg/h
B =  422.0  kg/h
C =  420.0  kg/h


### Example 3.12 Page 70¶

In :
%matplotlib inline
from numpy import linspace
from matplotlib.pyplot import plot, show
# solution
# Variables
m = 100.                      # kg
x = linspace(70,110,5);
y = linspace(100,115,4);

# Calculation
y1 = 27.8/.203 - .075*x/.203
y2 = 72.2/.673 - .035*x/.673
x = linspace(70,110,5);
plot(x,y1)
plot(x,y2)
show()
x = 93.4;
y = 102.4;
M = x+y                      # [kg] total mixture
Ms = M - m                   #[kg] mixed solvent
Mr = Ms/m                    # mixed solvent/original mixture
S1 = x*.574+y*.028           #[kg] water balance
S2 = x*.316+y*.096           #[kg] acetic acid balance
Qs = S1+S2
pS1 = (S1*100)/Qs
pS2 = 100-pS1

# Result
print "a  Upper layer = ",x," kg and Lower layer = ",y
print "b  mass ratio of the mixed solvent to the original mixture is ",Mr
print "c  water mass percent = ",pS1," and acetic acid mass percent = ",pS2,"." a  Upper layer =  93.4  kg and Lower layer =  102.4
b  mass ratio of the mixed solvent to the original mixture is  0.958
c  water mass percent =  58.9403862931  and acetic acid mass percent =  41.0596137069 .


### Example 3.14 Page 73¶

In :
# solution
# Variables
#using table 2.7 on page no 75
Rg = 8124.*100/9448                 # recovery of glycerine
Lg = (16+83)*100./9448              # loss of glycerine in waste
Reg = 100-Rg-Lg                     # recycle of glycerine

# Calculation
m1 = 238/8124.                      # NaCl in product
m2 = Rg*12/100.                     # glycerine in product
m3 = m1+m2                          # total solute
n = m1*100/m3                       # NaCl percent in total solute

# Result
print "a  recovery percent of glycerine is ",Rg
print "   b  percent loss of glycerinr is ",Lg
print "   c  product contamination with respect to salt NaCl is ",n

a  recovery percent of glycerine is  85.9864521592
b  percent loss of glycerinr is  1.04784081287
c  product contamination with respect to salt NaCl is  0.283116033416


### Example 3.15 Page 76¶

In :
# solution

# Variables
f1 = 1.25                   #[m**3/s] fresh ambient air as feed (basis)
f2 = 5.806                  #[m**/s] air entering auditorium
v1 = 8.314*290/101.3        #[m**3/kmol] sp. vol. of moist air at 101.3 kPa and 290 K
na1 = f2*1000/v1            # [mol/s] molar flow rate of air entering auditorium
nw1 = 243.95*.0163/1.0163   # [mol/s]
na2 = 243.95 - nw1          #[mol/s] dry air flow
nw2 = 240.04*.0225          #[mol/s] moisture enterin air conditioning plant

# Calculation
# using table 3.8
m1 = (nw2-nw1)              #[kg/h] moisture removed in a c plant
m2 = na2-.0181              #[mol/s] moisture in air leaving auditorium
m3 = (m2-nw1)*18            # [kg/h] moisture added in auditorium
Vm2 = 8.314*308/101.3       # [m**3/kmol]
na3 = (f1/25.28)*1000       #[mol/s]
n4 = 5.40-1.925             #[mol/s] moisture in recycle stream
mr = 240.04-47.525          #[mol/s] molar flow rate of wet recycle stream
R = mr/na3

# Result
print "a  moisture removed in AC plant = ",m1
print "  b  moisture added in auditorium = ",m3
print "   c  recycle ratio of moles of air recycled per mole mole of fresh ambient air input = ",R

a  moisture removed in AC plant =  1.48829053429
b  moisture added in auditorium =  4249.92025923
c  recycle ratio of moles of air recycled per mole mole of fresh ambient air input =  3.89342336


### Example 3.16 Page 78¶

In :
# solution

# screen 1
# feed = N kg
# Oversize particle = NE1 kg
# Undersize particle = N-NE1

#screen 2
#feed = NE1+X kg
# Oversize particle = (NE1+X)*E2 kg
# Undersize particle = (NE1+X)(1-E2) kg

#screen 3
# feed = (NE1+X)*E2 kg
# Oversize particle = (NE1+X)*E2*E3 kg
# Undersize particle = (NE1+X)*E2*(1-E3) kg
print "Overall Efficiency = E1 E2 E3*100/[1-E11-E2+E2 E3]."

Overall Efficiency = E1 E2 E3*100/[1-E11-E2+E2 E3].


### Example 3.17 Page 79¶

In :
# solution
# Variables
print "a  "
F = 5000.                       #[kmol/h] feed (basis)

# Calculation
m1 = F*.47                      #[kmol/h] CO in F
m2 = F-m1                       #[kmol/h] H2 in F
m3 = m1*.932                    # CO in product stream
n2 = m3/.98                      #[kmol/h]

# Result
print "     Flow rate of product stream is ",n2," kmol/h.\nb  "
n2 = n2-m3                      #[kmol/h] H2 in CO stream
print "    Product H2 stream :   H2 = ",m2-n2," kmol/h   CO = ",m1-m3," kmol/h \nc  "
nH2 = 2697.39                   #[kmol/h]
nCO = 3000-nH2                  # [kmol/h]
n4 = m2+nH2
n5 = m1+nCO
n6 = n4+n5

print "    Composition of Mixed feed :   H2 = ",n4*100/n6,"   CO = ",n5*100/n6

a
Flow rate of product stream is  2234.89795918  kmol/h.
b
Product H2 stream :   H2 =  2605.30204082  kmol/h   CO =  159.8  kmol/h
c
Composition of Mixed feed :   H2 =  66.842375    CO =  33.157625


### Example 3.18 Page 79¶

In :
# solution

# Variables

# Overall balance
# F=R1+P2
# Balance across Module I
# F+R2 = R1+P1 ==> R1+P2+R2 = R1+P1
# balance across module II
# P1 = P2+R2
P2 = 5.                     #[m**3/h]
P1 = P2/.8                  #[m**3/h]
R2 = P1-P2                  #[m**3/h]
F = P1/.66 - R2             #[m**3/h]
R1 = F-P2                   #[m**3/h]

# Calculation
# Overall balance of DS in water
xR1 = (F*4200-P2*5.)/R1     #[mg/l]
xP1 = (P2*5)/(.015*P1)      # [mg/l]
xR2 = (P1*xP1-P2*5)/R2      #[mg/l]
m1 = F*4200+R2*xR2          #[g] DS mixeed in MF
C1 = m1/(F+R2)              # [mg/l]
m2 = R1*xR1                 #[g] DS in R1
r = m2*100/m1               # rejection in module in I
m3 = m1-m2                  #[g] DS in P1
C2 = m3/P1                  # [mg/l]
R = R2/F
R1 = P2*100/F

# Result
print "F = ",F
print " m**3/h R1 = ",R1
print " m**3/h P = ",P1+P2
print " m**3/h R2 = ",R2
print " m**3/h recycle ratio = ",R
print " rejection percentage of salt in module I = ",r

F =  8.2196969697
m**3/h R1 =  60.8294930876
m**3/h P =  11.25
m**3/h R2 =  1.25
m**3/h recycle ratio =  0.152073732719
rejection percentage of salt in module I =  95.3914154639

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