# Chapter 10 : Material Balance with Chemical Reaction¶

## Example 10.1 pageno : 329¶

In :
# variables
mair = 500.                                 #kg
mCO2 = 55.                                  #kg
mCO = 15.                                   #kg

#C3H8 + 5O2 = 3CO2 + 4H20
MCO2 = 44
MCO = 28

# Calculation
NCO2 = mCO2 / MCO2
NCO = mCO / MCO
Mair = 29
Nair = mair / Mair
#carbon balance gives,
F = (NCO2 + NCO)/3
MC3H8 = 44.064
mC3H8 = MC3H8 * F

# Result
print "(a)mass of propane burnt = %.2f"%mC3H8,"kg"

#one mole of propane requres 5 moles of oxygen for combustion
NO2 = F * 5
Nairt = NO2 * 100 /21.                      #theoretical air required
Pexcess = (Nair - Nairt) * 100 / Nairt
print "(b)The percent excess air = %.2f"%Pexcess,"%"

#C3H8 + 7/2 * O2 = 3CO + 4H2O
NH2O = F * 4
#Taking oxygen balance, unburned oxygen is calculated,
#O2 supplied = O2 present in form of CO2, CO and H2O + unburned O2
Nunburnt = Nair * 21. / 100 - NCO2 - NCO/2 - NH2O/2.
NN2 = Nair * 79 / 100.
Ntotal = NCO2 + NCO + NH2O + NN2 + Nunburnt
PCO2 = NCO2 * 100 / Ntotal
PCO = NCO *100/ Ntotal
PH2O = NH2O *100/ Ntotal
PN2 = NN2 *100/ Ntotal
PO2 = Nunburnt *100 / Ntotal

print "(c)Percent composition of CO2 = %.2f"%PCO2,"%"
print "Percent composition of CO = %.2f"%PCO,"%"
print "Percent composition of H2O = %.2f"%PH2O,"%"
print "Percent composition of N2 = %.2f"%PN2,"%"
print "Percent composition of O2 = %.2f"%PO2,"%"

(a)mass of propane burnt = 26.23 kg
(b)The percent excess air = 21.66 %
(c)Percent composition of CO2 = 6.68 %
Percent composition of CO = 2.86 %
Percent composition of H2O = 12.73 %
Percent composition of N2 = 72.84 %
Percent composition of O2 = 4.88 %


## Example 10.2 pageno : 331¶

In :
# variables
Nflue = 100.                        #kmol
NCO2 = 14.84
NCO = 1.65
NO2 = 5.16
NN2 = 78.35
PCF = 85.                           #PERCENT CARBON IN FEED
PIF = 15.                           #PERCENT INERT IN FEED

# Calculation
#F - amount of coke charged, W - mass of coke left,W = 0.05F
NCflue = NCO2 + NCO
MC = 12.
mC = MC * NCflue

#carbon balance gives, F * PCF / 100 = W * PCF + mC
F = mC / ( PCF / 100 - 0.05*PCF / 100)
#let A kmol air supplied, taking N2 balance,
Nair = NN2 * 100/79.
NO2supplied = Nair - NN2
Ntheoretical = F * PCF / (100 * MC)
Pexcess = ( NO2supplied - Ntheoretical ) * 100 / ( Ntheoretical )

# Result
print "(a)Percentage excess air = %.2f"%Pexcess,"%"
mair = Nair * 29
m = mair / F                        #air supplied per kg of coke charged
print "(b)air supplied per kg of coke charged = %.2f"%m,"kg"
P = 100.                            #kPa
T = 500.                            #K
V = Nflue *22.4143*101.325 * T / (F * P * 273.15)
print "(c)volume of flue gas per kg of coke = %.2f"%V,"m**3"
W = 0.05*F
mCr = W * PCF/100.                  #carbon in refuse
mir = F * (1-PCF/100.)              #inert in refuse
mr = mCr + mir
C = mCr * 100 / mr
I = mir *100/ mr
print "(d)Carbon = %.2f"%C,"%"
print "Inert = %.2f"%I,"%"

(a)Percentage excess air = 19.99 %
(b)air supplied per kg of coke charged = 11.74 kg
(c)volume of flue gas per kg of coke = 16.96 m**3
(d)Carbon = 22.08 %
Inert = 77.92 %


## Example 10.3 pageno : 333¶

In :
# variables
Nflue = 100.                            #kmol
NCO2 = 9.
NCO = 2.
NO2 = 3.
NN2 = 86.
NCflue = NCO2 + NCO
MC = 12.
mC = MC * NCflue

# Calculation
#let A kmol air supplied, taking N2 balance,
Nair = NN2 * 100/79.
NO2supplied = Nair - NN2

# if CO in the flue gas was to be completely converted to CO2,
#then, the moles of oxygen present in the flue gas would be 3-1 =2kmol
Noexcess = NO2 - NCO/2.
Pexcess =  Noexcess * 100. / ( NO2supplied - Noexcess)

# Result
print "(a)Percentage excess air = %.2f"%Pexcess,"%"
NwaterO = NO2supplied - NCO2 - NCO/2 - NO2
NH2 = NwaterO*2
mH2 = NH2 * 2
xCF = 0.7
R = mC / mH2
print "(b)Ratio of carbon to hydrogen in the fuel = %.2f"%R
#let x be the amount of moisture in the feed, n it is
#given that 70% is carbon,therefore,
#0.7 = 3.32 / ( 1 + 3.32 + x )
x = R / xCF - 1 - R
mH = x * 2.016 / 18.016
mHtotal = mH + mH2
Rtotal = mC / mHtotal
print "(c)Ratio of carbon to total hydrogen in the fuel = %.2f"%Rtotal
ntotal = R + 1  +x
PH2 = 1*100./ntotal
PH2O = x * 100. / ntotal
print "(d)percentage of combustible hydrogen in the fuel = %.2f"%PH2,"%"
print "percentage of moisture in the fuel = %.2f"%PH2O,"%"
nH2Ototal = (PH2O + PH2 * 18.016 / 2.016)/100
print "(e)The mass of moisture in the flue gas per kg of fuel burned = %.3f"%nH2Ototal,"kg"

# note : answer may vary because of rounding error

(a)Percentage excess air = 9.59 %
(b)Ratio of carbon to hydrogen in the fuel = 3.35
(c)Ratio of carbon to total hydrogen in the fuel = 3.34
(d)percentage of combustible hydrogen in the fuel = 20.92 %
percentage of moisture in the fuel = 9.08 %
(e)The mass of moisture in the flue gas per kg of fuel burned = 1.960 kg


## Example 10.4 pageno : 335¶

In :
# variables
Nflue = 100.                            #kmoles
NCO2 = 9.05
NCO = 1.34
NO2 = 9.98
NN2 = 79.63
PCO2F = 9.2                             #% ( Feed )
PCOF = 21.3                             #%
PH2F = 18.                              #%
PCH4F = 2.5                             #%
PN2F = 49.                              #%

# Calculation
#Taking carbon balance,
F = (NCO2 + NCO )/ ( (PCO2F + PCOF + PCH4F)/100.)

#Nitrogen balance gives,
Nair = (NN2 - F*PN2F/(100) )* 100 / 79.
R = Nair/F

# Result
print "(a)molar Ratio of air to fuel = %.2f"%R
Oexcess = NO2 - NCO / 2
Pexcess = Oexcess *100/ (Nair*21./100 - Oexcess)
print "(b)Percent excess of air = %d"%Pexcess,"%"
NN2F = F * PN2F / 100
PN2F = NN2F *100/ NN2
print "(c)Percent of nitrogen in the flue gas that came from fuel = %.2f"%PN2F,"%"

(a)molar Ratio of air to fuel = 2.58
(b)Percent excess of air = 120 %
(c)Percent of nitrogen in the flue gas that came from fuel = 19.37 %


## Example 10.5 pageno : 337¶

In :
# variables
Nflue = 100.                        #kmole
NCO2 = 16.4
NCO = 0.4
NO2 = 2.3
NN2 = 80.9
PCF = 80.5                          #% ( Feed )
PO = 5.0                            #%
PHF = 4.6                           #%
PN = 1.1                            #%
Pash = 8.8                          #%

# Calculation
#Taking Carbon balance,
W = (NCO2 + NCO)*12. / (PCF / 100)
mCO2 = NCO2 * 44
mCO = NCO * 32
mO2 = NO2 * 28
mN2 = NN2 * 28.014
mtotal = mCO2 + mCO + mO2 + mN2
Mdryflue = mtotal * 100/ W

# Result
print "(a)The weight of dry gaseous products formed per 100 kg of coal fired = %.2f"%Mdryflue,"kg"
#taking nitrogen balance,
x = (mN2 - W*PN/100)/28.014
Noxygen = x * 21 / 79.
Nrequired = W * (PCF /12 + PHF/(2*2.016) - PO/32)/100
Pexcess = (Noxygen - Nrequired)*100/Nrequired
print "(b)Percent excess air supplied for combustion = %.2f"%Pexcess,"%"

# note: answer may vary because of rounding error.

(a)The weight of dry gaseous products formed per 100 kg of coal fired = 1223.92 kg
(b)Percent excess air supplied for combustion = 11.49 %


## Example 10.6 pageno : 338¶

In :
# variables
mcoal = 100.                    #kg
mC = 63.                        #kg
mH = 12.                        #kg
mO = 16.                        #kg
mash  = 9.                      #kg
mfixC = 39.                     #kg
mH2O = 10.                      #kg

# Calculation
mCvolatile = mC - mfixC
mHH2O = mH2O *2.016/18.016      #(mass of hydrogen in moisture)
mHvolatile = mH - mHH2O
mOH2O = mH2O - mHH2O
mOvolatile = mO - mOH2O
mtvolatile = mCvolatile + mHvolatile + mOvolatile
PC = mCvolatile * 100 / mtvolatile
PH = mHvolatile * 100 / mtvolatile
PO = mOvolatile * 100 / mtvolatile

# Result
print "(a)percent carbon in volatile matter = %.2f"%PC,"%"
print "   percent hydrogen in volatile matter = %.2f"%PH,"%"
print "   percent oxygen in volatile matter = %.2f"%PO,"%"

PCflue = 10.8                   #%
Pvflue = 9.0                    #%
Pashflue = 80.2                 #%

#taking ash balance, Wis the weight of the refuse,
W = mash *100. / Pashflue
mvflue = Pvflue * W /100.
mCflue = W * PCflue / 100
Ctflue = mCflue + mvflue * PC / 100 #total carbon in flue
Htflue = mvflue * PH / 100
Otflue = mvflue * PO / 100
PCflue = Ctflue *100/W
PHflue = Htflue *100/W
POflue = Otflue *100/W

print "(b)percent Carbon in refuse = %.2f"%PCflue,"%"
print "   percent Hydrogen in refuse = %.2f"%PHflue,"%"
print "   percent Oxygen in refuse = %.2f"%POflue,"%"
print "   percent Ash in refuse = %.2f"%Pashflue,"%"

Coalburnt = mcoal - W
NCburnt = (mC - Ctflue)/12
NHburnt = (mH - Htflue)/2.016
NOburnt = (mO - Otflue)/32
PCO2 = 80.                      #Percentage of carbon burnt
NCO2 = PCO2 * NCburnt / 100.
NCO = ( 1 - PCO2/100. )*NCburnt
Vair = 1000.                    #m**3
Nair = Vair / 22.4143
NN2 = Nair * 79 / 100.
NO2 = Nair * 21 / 100.
Ocompounds = NCO2 + NCO/2 + NHburnt/2               #Oxygen present in CO2,CO and H2O

#Oxygen balance gives free oxygen as,
Ofree = NO2 + mO/32 - Otflue/32 - Ocompounds
Ntotal = NN2 + Ofree + NCO2 + NCO                   #dry basis
PCO21 = NCO2 *100/Ntotal
PCO1 = NCO * 100/Ntotal
PO21 = Ofree * 100/Ntotal
PN21 = NN2 * 100/Ntotal
print "(c)percent CO2 in flue = %.2f"%PCO21,"%"
print "   percent CO in flue = %.2f"%PCO1,"%"
print "   percent O2 in flue = %.2f"%PO21,"%"
print "   percent N2 in flue = %.2f"%PN21,"%"
NOrequired = mC/12 + mH/(2.016*2) - mO/32
Oexcess = NO2 - NOrequired
Pexcess = Oexcess * 100 / NOrequired
print "(d)Percent excess air supplied = %.2f"%Pexcess,"%"
NH2Oflue = NHburnt
mH2O = NH2Oflue * 18.016
m = mH2O * 100/Ntotal
print "(e)mass of water vapour per 100 moles of dry flue gas = %.2f"%\
m,"g water vapour / 100kmol dry flue gas"

# note: answer may vary because of rounding error.

(a)percent carbon in volatile matter = 57.14 %
percent hydrogen in volatile matter = 25.91 %
percent oxygen in volatile matter = 16.95 %
(b)percent Carbon in refuse = 15.94 %
percent Hydrogen in refuse = 2.33 %
percent Oxygen in refuse = 1.53 %
percent Ash in refuse = 80.20 %
(c)percent CO2 in flue = 9.55 %
percent CO in flue = 2.39 %
percent O2 in flue = 5.53 %
percent N2 in flue = 82.53 %
(d)Percent excess air supplied = 21.26 %
(e)mass of water vapour per 100 moles of dry flue gas = 245.62 g water vapour / 100kmol dry flue gas


## Example 10.7 pageno : 343¶

In :
# variables
Pexcess = 20.                           #%
PSO3 = 5.                               #% ( Percent of sulphur burnt to SO3 )

#S + O2 = SO2
N = 1.                                  #kmol sulphur
Orequired = N                           #kmol

# Calculation
Osupplied = Orequired * ( 1 + Pexcess/100)
Nsupplied = Osupplied * 79/21.
NSO2 = (1-PSO3/100)*N
NSO3 = PSO3 * N /100.
Oconsumed = NSO2 + 3/2. * PSO3/100
Oremaining = Osupplied - Oconsumed
Ntotal = NSO2 + NSO3 + Oremaining + Nsupplied
PSO2 = NSO2 * 100 / Ntotal
PSO3 = NSO3 * 100 / Ntotal
PO2 = Oremaining * 100 / Ntotal
PN2 = Nsupplied * 100 / Ntotal

# Result
print "Percent SO2 in burner gas = %.2f"%PSO2,"%"
print "Percent SO3 in burner gas = %.2f"%PSO3,"%"
print "Percent O2 in burner gas = %.2f"%PO2,"%"
print "Percent N2 in burner gas = %.2f"%PN2,"%"

Percent SO2 in burner gas = 16.70 %
Percent SO3 in burner gas = 0.88 %
Percent O2 in burner gas = 3.08 %
Percent N2 in burner gas = 79.35 %


## Example 10.8 pageno : 343¶

In :
# variables
Nburner = 100.                      #kmol
NSO2b = 9.5                         #kmol
NO2b = 7.                           #kmol

# Calculation
NN2 = Nburner - NSO2b - NO2b
NOsupplied = NN2 * 21 / 79.         #Oxygen supplied

#4FeS2 + 11O2 = 2Fe2O3 + 8SO2
#4FeS2 + 15O2 = 2Fe2O3 + 8SO3
NOtotal = NO2b + NSO2b + NSO2b * 3 / 8.
NOunaccounted = NOsupplied - NOtotal
NSO31 = NOunaccounted * 8 /15
NStotal = NSO2b + NSO31
mS = NStotal * 32.064
Pburnt = 50.                        #% ( percentage of pyrites burnt )
mFeS2 = mS * 100/ Pburnt

# Result
print "(a)Total pyrites burnt = %.2f"%mFeS2,"kg"
NFeS2 = NStotal / 2
MFeS2 = 119.975
mFeS21 = MFeS2 * NFeS2
mgangue = mFeS2 - mFeS21
NFe2O3 = NFeS2 * Pburnt / 100
MFe2O3 = 159.694
mFe2O3 = MFe2O3 * NFe2O3
PSO3c = 2.5                         #% ( percentage sulphur as SO3 in cinder )
mc = 100.                           #kg ( basis )
NSO3 = PSO3c / 32.064
mSO3 = NSO3 * 80.064
mremaining = mc - mSO3              # ( Fe2O3 + gangue )

#x be the weight of the cinder
x = (mFe2O3 + mgangue)*100/mremaining
print "(b)weight of cinder produced = %.f"%x,"kg"
Slost = x * NSO3 / 100
PSlost = Slost *100/ NStotal
print "(c)Percent of total S lost in the cinder = %.2f"%PSlost,"%"
mSO3c = mSO3 * x / 100
NSO3b = NSO31 - Slost
P = NSO3b * 100 / NStotal
print "(d)Percentage of S charged that is present as SO3 in the burner gas = %.2f"%P,"%"

(a)Total pyrites burnt = 682.19 kg
(b)weight of cinder produced = 500 kg
(c)Percent of total S lost in the cinder = 3.66 %
(d)Percentage of S charged that is present as SO3 in the burner gas = 7.03 %


## Example 10.9 pageno : 345¶

In :
# variables
Ncgas = 100.                #kmol ( basis - SO3 free converter gas )
NSO2 = 4.5                  #kmol
NO2 = 7.5                   #kmol
NN2 = 88.0                  #kmol

# Calculation
NOsupplied = NN2 * 21./ 79
NOconverter = NO2 + NSO2
NOconsumed = NOsupplied - NOconverter           #(Oxygen consumed for SO3)
NSO3c = NOconsumed / 1.5
NStotal = NSO3c + NSO2
Nbgas = 100.                                    #kmol ( basis - SO3 free burner gas )
NSO21 = 15.                                     #%
NO21 = 5.                                       #%
NN21 = 80.                                      #%
NOburner = NO21 + NSO21
NOsupplied1 = NN21 * 21. / 79.
NOconsumed1 = NOsupplied1 - NOburner            #(Oxygen consumed for SO3)
NSO3b = NOconsumed1 / 1.5
NStotal1 = NSO3b + NSO21
mS = 100.                                       #kg ( basis - sulphur charged )
Pburned = 95.                                   #%
mburned = mS * Pburned / 100.
Nburned = mburned / 32.064

#let x be the SO3 free burner gas produced, then sulphur balance gives,
x = Nburned * Nbgas / NStotal1
NSO2b = NSO21 * x / 100
NO2b = NO21 * x / 100
NN2b = NN21 * x / 100
Ntotalb = NSO2b + NO2b + NN2b
NSO3b1 = NSO3b * x / 100

#let y be the no. of converter gas produced
y = Nburned * Ncgas / NStotal
NSO2c = NSO2 * y / 100
NO2c = NO2 * y / 100
NN2c = NN2 * y / 100
Ntotalc = NSO2c + NO2c + NN2c
NSO3c1 = NSO3c * y / 100
Nairsec = (NN2c - NN2b ) * 100 / 79.
P = 100.                                        #kPa
T = 300.                                        #K
V = Nairsec * 22.414 * 101.3 * T / (P * 273.15)

# Result
print "(a)The volume of secondary air at 100kPa and 300K = %.1f"%V,"m**3/h"
NSabsorbed = 95.                                #%
mSO3abs = NSabsorbed * NSO3c1 * 80.064 / 100

#let z be the amount of 98% H2SO4, therefore , 100% H2SO4 produced = z + mSO3abs
# taking SO3 balance
z = (mSO3abs - mSO3abs * 80.064 / 98.08) / ( 80.064 / 98.08 - 0.98 * 80.064/98.08)
print "(b)98%% H2SO4 required per hour = %.1f"%z,"kg"
w = z + mSO3abs
print "(c)100%% H2SO4 produced per hour = %.1f"%w,"kg"

# note: answer may vary because of rounding error

(a)The volume of secondary air at 100kPa and 300K = 208.2 m**3/h
(b)98% H2SO4 required per hour = 1592.1 kg
(c)100% H2SO4 produced per hour = 1733.6 kg


## Example 10.10 pageno : 349¶

In :
# variables
mlime = 5.                          #kg
mcoke = 1.                          #kg
PCaCO3l = 84.5                      #%
PMgCO3l = 11.5                      #%

# Calculation
NCaCO3l = PCaCO3l * mlime / (100.09*100)
NMgCO3l = PMgCO3l * mlime / (84.312*100)
mInertsl = mlime * ( 100 - PCaCO3l - PMgCO3l ) / 100
PCc = 76.                           #%
Pashc = 21.                         #%
Pwaterc = 3.                        #%
NCc = mcoke *  PCc /(100*12)
Nwaterc = mcoke * Pwaterc / ( 100 * 18.016 )
mash = Pashc * mcoke / 100

#CaCO3 + C + O2 = CaO + 2CO2
#MgCO3 + C + O2 = MgO + 2CO2
PCaCO3conv = 95.                    #(Percent calcination of CaCO3)
PMgCO3conv = 90.                    #(Percent calcination of MgCO3)
NCaO = PCaCO3conv * NCaCO3l / 100.
mCaO = NCaO * 56.08
NMgO = PMgCO3conv * NMgCO3l / 100.
mMgO = NMgO * 40.312
mCaCO3 = (NCaCO3l * (1-PCaCO3conv/100)*100.09)
mMgCO3 = (NMgCO3l * (1-PMgCO3conv/100)*84.312)
mtotal = mCaO + mMgO + mCaCO3 + mMgCO3 + mInertsl + mash
PCaO = mCaO * 100 / mtotal

# Result
print "The weight percent of CaO in the product leaving the kiln = %.2f"%PCaO,"%"

The weight percent of CaO in the product leaving the kiln = 70.83 %


## Example 10.11 pageno : 350¶

In :
# variables
R = 100.                        #kg ( basis - residue )
MCaSO4 = 136.144
MMgSO4 = 120.376
mCaSO4r = 9.                    #kg
mMgSO4r = 5.                    #kg
mH2SO4r = 1.2                   #kg
minertr = 0.5                   #kg
mCO2r = 0.2                     #kg
mH2O = 84.10                    #kg

# Calculation
NCaSO4 = mCaSO4r / MCaSO4
NMgSO4 = mMgSO4r / MMgSO4

#CaCO3 + H2SO4 = CaSO4 + H2O + CO2
#MgSO4 + H2SO4 = MgSO4 + H2O + CO2

mCaCO3 = NCaSO4 * 100.08
mMgCO3 = NMgSO4 * 84.312
mtotallime = minertr + mCaCO3 + mMgCO3
PCaCO3 = mCaCO3 * 100/ mtotallime
PMgCO3 = mMgCO3 *100/ mtotallime
Pinerts = minertr *100/ mtotallime

# Result
print "(a)Percentage of CaCO3 in limestone = %.2f"%PCaCO3,"%"
print "   Percentage of MgCO3 in limestone = %.2f"%PMgCO3,"%"
print "   Percentage of inerts in limestone = %.2f"%Pinerts,"%"

NH2SO4 = NCaSO4 + NMgSO4
mH2SO4 = NH2SO4 * 98.08
Pexcess = mH2SO4r * 100 / ( mH2SO4)
print "(b)The percentage excess of acid used = %.2f"%Pexcess,"%"

macidt = mH2SO4 + mH2SO4r
Pacidic = 12.                   #%
mwaterin = macidt * (100 - Pacidic)/ Pacidic
mwaterr = (NCaSO4 + NMgSO4)*18.016
mwatert = mwaterin + mwaterr
mvaporized = mwatert - mH2O
m = mvaporized * 100/mtotallime             #water vaporized per 100kg of limestone
print "(c)the mass of water vaporized per 100kg of limestone = %.2f"%m,"kg"
mCO2pr = (NCaSO4 + NMgSO4)*44
mCO2rel = mCO2pr - mCO2r
m1 = mCO2rel * 100 / mtotallime             #CO2 per 100kg of limestone
print "(d)the mass of CO2 per 100kg of limestone = %.2f"%m1,"kg"

# note : answer may vary because of rounding error

(a)Percentage of CaCO3 in limestone = 62.31 %
Percentage of MgCO3 in limestone = 32.98 %
Percentage of inerts in limestone = 4.71 %
(b)The percentage excess of acid used = 11.37 %
(c)the mass of water vaporized per 100kg of limestone = 38.25 kg
(d)the mass of CO2 per 100kg of limestone = 42.72 kg


## Example 10.12 pageno : 353¶

In :
# variables
macid = 1000.                   #kg ( basis - dilute phosphoric acid )
Mphacid = 97.998
P = 1.25                        #% ( dilute % )

# Calculation
mphacid = macid * P /100
Nphacid = mphacid / Mphacid

#1mole of phosphoric acid - 1mole of trisodium phosphate
NTSP = Nphacid
MTSP = 380.166
mTSP = NTSP * MTSP

# Result
print "(a)Maximum weight of TSP obtained = %.2f"%mTSP,"kg"
NCO2 = NTSP
Pwater = 6.27                   #kPa

#since gas is saturated with water vapour, vapour pressure = partial pressure
Nwater = NCO2 * Pwater / ( 100 - Pwater )
Ntotal = Nwater + NCO2
P = 100.                        #kPa
T = 310.                        #K
V = Ntotal * 101.3 * T *22.4143 / ( P * 273.15 )
print "(b)Volume of CO2 = %.2f"%V,"m**3"

(a)Maximum weight of TSP obtained = 48.49 kg
(b)Volume of CO2 = 3.51 m**3


## Example 10.13 pageno : 353¶

In :
# variables
mTSPd = 1000.                       #kg ( basis - 20% dilute TSP )
P = 20.                             #%

# Calculation
mTSP = mTSPd * P / 100
NTSP = mTSP / 163.974
msodaashd = NTSP * 106
mphacidd = NTSP * 97.998
mNaOHd = NTSP * 40.008
Pphacid = 85.                       #% (85% solution phosphoric acid)
PNaOH = 50.                         #% (50% solution NaOH)

#let x be the water in soda ash,
#taking water balance,
x = (mTSPd - mTSP) - mNaOHd * PNaOH /(100 - PNaOH) - mphacidd * (100 - Pphacid) / Pphacid
msodaash = msodaashd + x
C = msodaashd *100 / msodaash

# Result
print "(a)Concentration of soda ash solution = %.2f"%C,"%"
mphacid = mphacidd * 100 / Pphacid
R = msodaash / mphacid
print "(b)Weight ratio in which soda ash and commercial phosphoric acid are mixed = %.2f"%R

(a)Concentration of soda ash solution = 15.04 %
(b)Weight ratio in which soda ash and commercial phosphoric acid are mixed = 6.11


## Example 10.14 pageno : 355¶

In :
# variables
m = 1000.                       #kg ( basis - pig iron produced )

#let x be the iron ore charged and y be the amount of flux added and z be the weight of slag produced
PFepg = 95.                     #% ( Fe% in product )
PCpg = 4.                       #%
PSipg = 1.                      #%
PFech = 85.                     #% (Fe% in feed )
mcoke = 1000.                   #kg
PCcoke = 90.                    #%
PSicoke = 10.                   #%
PSislag = 60.                   #%
PSiflux = 5.                    #%
PCaCO3fx = 90.                  #%
PMgCO3fx = 5.                   #%
PCMslag = 40.                   #%

# Calculation
#iron balance gives,
x = PFepg * m *159.694 / ( PFech * 111.694)
#silicon balance gives,
#x*(100 - PFech)*28.086/(100*60.086)+mcoke*Psicoke*28.086/(100*60.086)+y*PSiflux*28.086/(100*60.086) = 10 + z*Psislag*28.086 / ( 100*60.086 )
#taking (CaO + MgO) balance
#y * ((PCaCO3fx)*56.88/(100*100.88)+(PMgCO3fx*40.312/(100*84.312))=z*PCMslag/100
#solving above 2 equations , we get
y = 403.31

# Result
print "the amount of flux required to produce 1000kg of pig iron = ",y,"kg"

the amount of flux required to produce 1000kg of pig iron =  403.31 kg


## Example 10.15 pageno : 357¶

In :
# variables
N = 100.                    #mol(basis - scrubber)
NNOs = 2.4                  #mol
NN2s = 92.                  #mol
NO2s = 5.6                  #mol
PNOs = 20.                  #% ( Percentage NO leaving scrubber)

# Calculation
NNOreac = NNOs * 100 / PNOs
#let x mol of nitroge be produced in the reaction, then the amount of N2 present in the air = NN2s - x mol - (1)
#4NH3 + 5O2 = 4NO + 6H2O
#4NH3 + 3O2 = 2N2 + 6H2O
#4moles of NO - 5 moles of O2, 2moles of N2 - 3 moles of O2
#Total oxygen used up, O  =  NNOreac * 5/4 + x*3/2
#total oxygen supplied, NOtotal= (O) + NO2s
#Nitrogen associated with O2 supplied NN2 = NOtotal*79/21 - (2)
#comparing 1 and 2,
x = 2.1835
#12moles NO requires 12moles ammonia, 1 mole N2 requires 2 mole ammonia
Nammonia = x*2 + NNOreac
Oreq = Nammonia * 5 / 4
Osupp = NNOreac * 5/4 + x*3/2 + NO2s
Pexcess = (Osupp - Oreq)*100/Oreq

# Result
print "(a)Percentage excess oxygen = %.1f"%Pexcess,"%"
fr = x * 2 / Nammonia
print "Fraction of ammonia taking part in side reaction = %.3f"%fr

(a)Percentage excess oxygen = 16.7 %
Fraction of ammonia taking part in side reaction = 0.267


## Example 10.16 pageno : 359¶

In :
# variables
m = 100.                        #kg (basis sodium nitrate reacted)
NNaNO3 = m/85

# Calculation
#2NaNO3 + H2SO4 = 2HNO3 + Na2SO4
mh2so4 = NNaNO3 * 98.08/2
mhno3 = NNaNO3*63.008
mna2so4 = NNaNO3 * 142.064 /2
Phno3 = 2.                      #%(percent nitric acid remaining in the cake)
mhno3cake = mhno3 * Phno3 / 100
Ph2so4 = 35.                    #%
Pwater = 1.5                    #%
mtotal = (mna2so4 + mhno3cake)*100/(100 - Ph2so4 - Pwater)
mwater = Pwater * mtotal / 100
mh2so4c = Ph2so4 * mtotal / 100
Pna2so4 = mna2so4 *100/mtotal
Phno3c = mhno3cake * 100 / mtotal

# Result
print "(a)Mass of Na2SO4 in the cake = %.3f"%mna2so4,"kg"
print "Mass of H2SO4 in the cake = %.3f"%mh2so4c,"kg"
print "Mass of HNO3 in the cake = %.4f"%mhno3,"kg"
print "Mass of water in the cake = %.3f"%mwater,"kg"
print "Percentage of Na2SO4 in the cake = %.3f"%Pna2so4,"%"
print "Percentage of H2SO4 in the cake = %.2f"%Ph2so4,"%"
print "Percentage of HNO3 in the cake = %.2f"%Phno3c,"%"
print "Percentage of water in the cake = %.2f"%Pwater,"%"

mh2so4req = mh2so4 + mh2so4c
P = 95.                         #% (95% dilute sulphuric acid)
w = mh2so4req * 100 / P
print "(b)Weight of 95%% sulphuric acid required = %.2f"%w,"kg"
mnitric = mhno3 - mhno3cake
print "(c)weight of nitric acid product obtained = %.2f"%mnitric,"kg"
mwaterd = w*(1-P/100)-mwater
print "(d)the water vapour tha tis distilled from the nitre cake = %.3f"%mwaterd,"kg"

# Note : The value of HNO3 is taken wrongly in the textbook while calculating mass of HNO3

(a)Mass of Na2SO4 in the cake = 83.567 kg
Mass of H2SO4 in the cake = 46.878 kg
Mass of HNO3 in the cake = 74.1271 kg
Mass of water in the cake = 2.009 kg
Percentage of Na2SO4 in the cake = 62.393 %
Percentage of H2SO4 in the cake = 35.00 %
Percentage of HNO3 in the cake = 1.11 %
Percentage of water in the cake = 1.50 %
(b)Weight of 95% sulphuric acid required = 110.08 kg
(c)weight of nitric acid product obtained = 72.64 kg
(d)the water vapour tha tis distilled from the nitre cake = 3.495 kg


## Example 10.17 pageno : 361¶

In :
# variables
m = 50.                         #kg ( basis - mass of brine charged )
#let x be the amount of NaCl in the brine
Pelect = 50.                    #% ( electrolyzed )

#2NaCl + 2H2O = 2NaOH + Cl2 + H2
#amount of NaCl reacted =x*Pelect/(100*58.45)kmol=x*Pelect/100kg   ( 1 )
#amount of water reacted = x * Pelect * 18.016 / ( 100 * 58.45 )kg ( 2 )
#Gases produced, Cl2 = x * Pelect / (100 * 58.45 * 2 )kmol = x * Pelect *71/ (100 * 58.45 * 2 )kg                                           ( 3 )
#H2 = x * Pelect / (100 * 58.45 * 2 )kmol = x * Pelect *2.016/ (100 * 58.45 * 2 )kg                                                         ( 4 )
Nwater = 0.03                   #mol water vapour/mol of gas

# Calculation
#water vapour present = Nwater * 2*(Cl2 + H2)kmol = Nwater * 2*(Cl2 + H2)*18.016 kg                                                         ( 5 )
#NaoH = x * Pelect * 40.008/ (100 * 58.45 )kg                      ( 6 )
#water = water in brine - water reacted - water present in gas     ( 7 )
#= (m - Pelect/100) - water reacted ( 2 ) - water present in the gas( 5 )
#Total weight of solution = NaCl ( 1 ) + NaOH ( 6 ) + Water ( 7 )
#since NaOH is 10 percent of the total weight, we have NaOH = 0.1 * total weight, from these we get,
x = 0.1 * 50 / (0.1* 0.3165 + 0.3422 )
NaOH = x * Pelect * 40.008/ (100 * 58.45 )
NaCl = x * Pelect / 100
water = 34.5032                 #kg
Pevap = 50.                     #NaOh percentage in solution leaving evaporator

#taking NaOH balance
mevap = NaOH * 100 / Pevap

# Result
print "(a)amount of 50%% NaOH solution produced = %.4f"%mevap,"kg"
Cl2 = x * Pelect *71/ (100 * 58.45 * 2 )                #kg
H2 = x * Pelect *2.016/ (100 * 58.45 * 2 )              #kg
print "(b)Chlorine produced = %.4f"%Cl2,"kg"
print "   Hydrogen produced = %.4f"%H2,"kg"
Pleav = 1.5                                             #% NaCl leaving the evaporator
NaClleav = mevap * Pleav / 100
mcrystal = NaCl - NaClleav
print "(c)Amount of NaCl crystallized = %.4f"%mcrystal,"kg/h"
mwaterleav = mevap - NaOH - NaClleav
Mwaterevap = water - mwaterleav
print "(d)Weight of water evaporated = %.4f"%Mwaterevap,"kg"

(a)amount of 50% NaOH solution produced = 9.1545 kg
(b)Chlorine produced = 4.0615 kg
Hydrogen produced = 0.1153 kg
(c)Amount of NaCl crystallized = 6.5499 kg/h
(d)Weight of water evaporated = 30.0633 kg


## Example 10.18 pageno : 364¶

In :
# variables
m = 100.                        #mol ( basis reactore exit gas )
#CH3OH + O2 = HCOOH + H2O
#CH3OH + O2 / 2 = HCHO + H2O
Nn2 = 64.49                     #mol
No2 = 13.88                     #mol
Nh2o = 5.31                     #mol
Nch3oh = 11.02                  #mol
Nhcho = 4.08                    #mol
Nhcooh = 1.22                   #mol

# Calculation
#x be the moles of methanol reacted, taking C balance, we get,
x = Nch3oh + Nhcho + Nhcooh
Pconv = Nhcho * 100 / x

# Result
print "(a)Percent conversion of formaldehyde = ",Pconv,"%"
Nair = Nn2 * 100 / 79.
R = Nair / x
print "(b)Ratio of air to methanol in the feed = %d"%R

(a)Percent conversion of formaldehyde =  25.0 %
(b)Ratio of air to methanol in the feed = 5


## Example 10.19 pageno : 367¶

In :
# variables
NA = 100.                       #mol ( basi - 100 mol A in the fresh feed )
Pconv = 95.                     #%

# Calculation
NApro = NA * (100 - Pconv)/100

#A = 2B + C
NB = NA * Pconv * 2 / 100
NC = NA * Pconv/100
PAent = 0.5                     #%
NAent = NApro * 100 / PAent
PBrec = 1.                      #%
NBent = NB * 100 / (100 - PBrec)
m = (NAent - NApro + NA)
conv = ((NAent - NApro + NA) - NAent)*100/(NAent - NApro + NA)

# Result
print "(a)single pass converion = %.2f"%conv,"%"
Nrecycled = (NAent - NApro) + (NBent - NB)
R = Nrecycled/NA
print "(b)recycle ratio = %.2f"%R

(a)single pass converion = 8.68 %
(b)recycle ratio = 9.97


## Example 10.20 pageno : 368¶

In :
# variables
m = 100.                #kg ( basis - sucrose solution as fresh feed )

#R - recycle reactor exit,
#let x be the weight fraction of sucrose and y be the weight fracton of inversion sugar in the recycle stream,
#for combined stream  fraction of Glucose + fructose = 0.04
#z be the weight fraction of sucrose in the combined stream entering the reactor
Psfeed = 25.            #% percent sucrose in fresh feed

#sucrose balance gives, 25 + R*x = (100+R)* z                        (A)
#Glucose + fructose balance, R * y = (100 + R )*0.04                 (B)
Sucrosecon = 71.7       #% sucrose consumed

# Calculation
#sucrose balance around the reactor,(100+R)z=0.717*(100+R)z+(100+R)x (C)
#From (C) , x = 0.283*z                                              (D)
#Amount converted to Glucose + fructose = 0.717 ( 100 + R )* z
# = 0.717 ( 100 + R )* z * 360.192 / 342.176 kg
#Glucose and fructose balance around the reactor,
#(100+R)*0.04 + 0.717(100+R)*z*360.192/342.176 = (100+R)*y           (E)
#Solving (E), y = 0.04 + 0.7548*z                                    (F)
#Solving, (A), (B), (C) and (F)
x = 0.06
y = 0.2
z = 0.212
R = 25

# Result
print "(a)Recycle flow = ",R,"kg"
print "(b)Combined concentration of Glucose and Fructose in the recycle stream = ",y*100,"%"

(a)Recycle flow =  25 kg
(b)Combined concentration of Glucose and Fructose in the recycle stream =  20.0 %


## Example 10.21 pageno : 370¶

In :
# variables
N = 1.                      #mol ( basis - combined feed )

#F - moles of fresh feed
Pinert = 0.5                #%
Pconv = 60.                  #%
P1inert = 2.                #%

# Calculation
NA1 = N * ( 1- P1inert/100. )
NA2 = NA1 * ( 1 - Pconv / 100. )
NB2 = NA1 - NA2
N1inert = N * P1inert / 100.
N2inert = N1inert

#Let R be the moles recycled and P be the moles purged
#W = R + P
W = NA2 + N2inert           #              (A)
PWinert = N2inert * 100/ ( NA2 + N2inert)

#component A balance, A fresh feed = A purge stream + A recycle stream
#F * 0.9 = P * 0.9515 + 0.588                      (B)
#inert balance at the point where fresh feed is mixed with the recycle,
#F*0.005 + R*0.0485 = 1* 0.02                      (C)
#Solving (A),(B) and (C)
F = 0.6552                  #mol
P = 0.0671                  #mol
R = 0.3448                  #mol

# Result
print "(a)moles of recycle stream = ",R,"mol"
print "(b)moles of purge stream = ",P,"mol"
NAconv = NA1 - NA2
NAf = F * (1 - Pinert / 100.)
Conv = NAconv *100./ NAf
print "(c)Overall conversion = %.1f"%Conv,"%"

(a)moles of recycle stream =  0.3448 mol
(b)moles of purge stream =  0.0671 mol
(c)Overall conversion = 90.2 %


## Example 10.22 pageno : 371¶

In :
# variables
N = 100.                            #moles ( Basis - Fresh feed )
Pconv = 20.                         #%
xco = 0.33
xh2 = 0.665
xch4 = 0.005

#R - recycle stream, P - purge stream
#x - mole fraction of CO in recycle stream ,
xch4r = 0.03

# Calculation
#CO = x, H2 = 1 - xch4r - CO = 0.97- x
#methane balance over the entire system,
P = xch4 * N / xch4r
#taking caron balance, 33.5 = M + P ( 0.03 + x )
#Hydrogen balance, 66.5 + 2*0.5 = 2M + P(2*0.03 + 0.97 - x)
#substituting P, M + 16.67x = 33.0 and 2M - 16.67x = 50.33
M = (33.0 + 50.33)/3
x = ((xco + xch4)*N - M ) / P - xch4r

#methanol balance,(xco*N+Rx) * Poncv/100 = M
R = (M*100 / Pconv - (xco*N))/x

# Result
print "(a)moles of recycle stream = %.1f"%R,"mol"
print "(b)moles of purge stream = %.2f"%P,"mol"
H2 = 1 - xch4r - x
print "(c)CH4 in purge stream = ",xch4r*100,"%"
print "CO in purge stream = ",x*100,"%"
print "hydrogen in purge stream = ",H2*100,"%"
print "(d)Methanol produced = %.2f"%M,"mol"

(a)moles of recycle stream = 337.9 mol
(b)moles of purge stream = 16.67 mol
(c)CH4 in purge stream =  3.0 %
CO in purge stream =  31.34 %
hydrogen in purge stream =  65.66 %
(d)Methanol produced = 27.78 mol