Chapter 3 : Material Balances without Chemical Reaction¶
Example 3.1 Page 60¶
In [18]:
# 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,"."
Example 3.2 Page 61¶
In [2]:
# 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."
Example 3.3 Page 61¶
In [3]:
# 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,"."
Example 3.4 Page 62¶
In [4]:
# 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,"."
Example 3.5 Page 62¶
In [4]:
# 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[0][0]+X[1,0] # [kg] total mixture
Ms = M - m #[kg] mixed solvent
Mr = Ms/m
S1 = x[0]*.574+x[1]*.028 #[kg] water balance
S2 = x[0]*.316+x[1]*.096 #[kg] acetic acid balance
Qs = S1+S2
pS1 = (S1*100)/Qs
pS2 = 100-pS1
# Result
print "a) Upper layer = ",x[0]," kg \nand Lower layer = ",x[1],\
" \nb) mass ratio of the mixed solvent to the original mixture is ",Mr,\
" \nc) water mass percent = ",pS1,"\nand acetic acid mass percent = ",pS2,"."
Example 3.6 Page 63¶
In [6]:
# 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,"."
Example 3.7 Page 64¶
In [7]:
# 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."
Example 3.8 Page 64¶
In [3]:
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[0],"kg"
print " HNO3 = ",x[1]," kg"
print " H2SO4 = ",x[2]," kg."
Example 3.9 Page 65¶
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# 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
Example 3.11 Page 68¶
In [10]:
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[0]+s[1]+s[2]
B = s[3]+s[4]+s[5]
C = s[6]+s[7]+s[8]
# Result
print "Flowrates are : "
print " A = ",A," kg/h"
print " B = ",B," kg/h"
print " C = ",C," kg/h"
Example 3.12 Page 70¶
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%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,"."
Example 3.14 Page 73¶
In [13]:
# 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
Example 3.15 Page 76¶
In [14]:
# 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
Example 3.16 Page 78¶
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# 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]."
Example 3.17 Page 79¶
In [16]:
# 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
Example 3.18 Page 79¶
In [1]:
# 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
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