# Chapter 6 : Stoichiometry and Unit Operations¶

### Example 6.1 Page 346¶

In :
# solution

# Variables
# basis = 100kmol of feed
Benzene = 100*.72                               # kmol
Toulene = 100-Benzene                           #kmol
# use fig 6.1
# D = distillate, B = bottom
# F = B + D        (i)   overall material balance
xd = .995
xb = .03
xf = .72
# xd*D + xb*B = F*xf     (ii)    benzene balance
# solving (i) and (ii)
D = 71.5                                        #kmol
B = 28.5                                        #kmol
print "a   performing overall material balance for 100kmol of feed we get ",D,\
"kmol as distillate and ",B,"kml as bottom product.    b   "
# enthalpy balance
# use fig 6.2
R = 1.95

# Calculation
v = D*(1+R)                                     #kmol    total overhead vapours
To = 273.15                                     #K
# using fig 6.2
Ev = 42170.                                     #kJ/kmol    enthalpy of vapours overhead
El = 11370.                                     #kJ/kmol    enthalpy of liquid
E1 = Ev-El                                      # enthalpy removed in condenser
Hc = E1*v                                       # heat load of condenser
Hd = El*71.5
Hb = 18780*28.5
Hf = 44500*100
Hn = Hd+Hc+Hb-Hf                                # kJ    heat load of reboiler

# Result
print " performing overall enthalpy balance we get Heat load of condenser = ",Hc,\
"kJ/kmol and Heat load of reboiler = ",Hn,"kJ/kmol."

a   performing overall material balance for 100kmol of feed we get  71.5 kmol as distillate and  28.5 kml as bottom product.    b
performing overall enthalpy balance we get Heat load of condenser =  6496490.0 kJ/kmol and Heat load of reboiler =  3394675.0 kJ/kmol.


### Example 6.2 Page 349¶

In :
# solution

# Variables
# basis = 2000kg/h liquid feed rate
F = 2000/28.84                      #kmol/h

#D = distillate, W = residue flow rate
#N2 balance
# F*.79 = .999D + .422W    (i)
# 54.840 = D + .4224W      (ii)
# solving it
W = 25.118                          #kmol/h
D = 44.230                          #kmol/h
#using fig 6.4 and 6.5
# trial method is used for flash calculations
# Trial I

x = .75

# from fig 6.4
y = .8833

# from fig 6.5
Hl = 1083.65
Hv = 6071.7

# Calculation
Hf = .3*Hv+Hv*.7

# calculating we get Emix is not close to 2592.2kJ/kmol
#Trial II
x = .71
y = .859
Hl = 1085.6
Hv = 6118.6
Hf = .3*Hv+.7*Hl                     #kJ/kmol
# which is aproox equal to 2595.2kJ/kmol, so flashing will occur

# Result
print "composition of vapour liquid mix :  mol fraction N2 = ",x,\
" in liquid phase and ",y," in vapour phase."

composition of vapour liquid mix :  mol fraction N2 =  0.71  in liquid phase and  0.859  in vapour phase.


### Example 6.3 Page 353¶

In :
# solution

# Variables
# material balance
# V2 vapour mix is a ternary azeotrope in which all cyclohexane of D1 is recycled
# V2 stream
# Cyclohexane balance
# D1 = (.488/.024)*V2
# IPA in V2 = .206V2
# water in V2 = (1-.488-.206)*V2
# W2 stream
# IPA in W2 = (.23D1-.206V2)
# water in W2 = (1-.024-.23)*D1-.306V2
# W2 stream = 4.471V2 + 14.862V2
# D3 is an azeotrope containing 67.5 mol% IPA
# water in W3 stream = (1-.675)F
# basis = 100 kmol/h fresh feed
# W1+W3 = 100                (i)
# .998W1 + .001W3 = 67.5    (ii)
# solving it
W1 = 67.603                     #kmol/h
W3 = 32.397                     #kmol/h
IPA1 = W3*.001 # IPA in W3
#IPA2 = 4.471*V2 - .032   IPA in D3
#C-1 = F+D3 = F1
# water in D3 = 6.624V2 - .047-4.471V2+.032
# water in W3 = 14.862V2-2.153V2+.015
# solving them
V2 = 2.624                      #kmol/h

# Calculation
D3 = 2.153*V2-.015
D1 = 20.333*V2
F1 = 6.624*V2+99.953
R = 1.75*D1                     # = V1+V2-D1
V1 = 144.1
r = D3/100                      # recycle ratio

# Result
print "After performing overall material balance we get Reflux, R = ",R,\
"kmol/h and  recycle ratio = ",r," kmol/kmol fresh feed."

After performing overall material balance we get Reflux, R =  93.369136 kmol/h and  recycle ratio =  0.05634472  kmol/kmol fresh feed.


### Example 6.4 Page 355¶

In :
# solution

# variables
# basis 0.625 l/s of MEA solution
c = 3.2                         #M  conc of MEA
M = 61.                         # molar mass of MEA

# calculation
C = M*c                         #g/l  conc of MEA in sol
MEAin = c*.625*3600/1000.       # kmol/h
CO2diss = .166*7.2              #kmol/h    CO2 dissolved in lean MEA
v = 26.107                      #m**3  sp. vol of gas at 318K and 101.3kPa  (table 7.8)
qv = 1000/v                     #kmol/h
CO2in = qv*.104                 # moles of CO2 in inlet gas
CO2freegas = qv - CO2in

#outgoing has 4.5% CO2
GASout = CO2freegas/(1-.0455)   #kmol/h
CO2abs = qv-GASout
CO2 = CO2diss + CO2abs
CO2conc = CO2/MEAin             #kmol/kmol MEA

# Result
print "Concentration of dissolved CO2 in the solution leaving the tower = ",CO2conc,\
"kmol/kmol of MEA."

Concentration of dissolved CO2 in the solution leaving the tower =  0.492054702187 kmol/kmol of MEA.


### Example 6.5 Page 356¶

In :
# solution
# Variables
#(a)
print "a  ",
# basis 50000 m**3/h of gas mix at 295.5K 100kPa
v = 24.57                           #m**3/kmol sp vol of gas at 295.5K and 100kPa
n1 = 50000/v                        # kmol/h    flow of incoming gas

# Calculation
NO2in = n1*.0546
N2O4in = n1*.0214
N2in = n1-NO2in-N2O4in

#N2 is unaffected
n2 = 1880.34/.95                    #kmol/h   outgoing gas flow
# using tables 6.3 and 6.4 on page 357
NO2rem = NO2in - (n2*.0393)
N2O4rem = N2O4in - (n2*.0082)

# rxn (ii)
NaOHreac2 = 2*40*N2O4rem
NaNO2pro2 = 69*N2O4rem
NaNO3pro2 = 85*N2O4rem
H2Opro2 = 18*N2O4rem

# rxn (iii)
NO2reac3 = 3*n2*.0025
NaOHreac3 = 2*4.95*40
NaNO3pro3 = 2*4.95*85
H2Opro3 = 4.95*18
NO2abs2 = 33.33-NO2reac3
NaOHreac1 = 18.48*40
NaNO2pro1 = 69*NO2abs2/2
NaNO3pro1 = 85*NO2abs2/2
H2Opro1 = 18*NO2abs2/2
NaNO2t = NaNO2pro2 + NaNO2pro1
NaNO3t = NaNO3pro2+NaNO3pro3
H2Ot = H2Opro1+H2Opro2+H2Opro3
NaOHt = NaOHreac1+NaOHreac2+NaOHreac3
liq = 37500.                        #kg/h
NaOHin = liq*.236
NaOHout = NaOHin-NaOHt
moist = n2*.045*18
water = liq-NaOHin-H2Ot-moist       #kg/h

# Result
print "Composition of final liquor :"
print "Component          mi kg/h"
print " NaOH             ",NaOHout
print " NaNO2            ",NaNO2t
print " NaNO3            ",NaNO3t
print " H2O              ",water
print " b"

#(b)
#heat effect of scrubbing
#using tables 6.6 and 6.7
#fi1 = integ{59865.7+4545.8+10**-3 *T + 15266.3*10**-6*T**2-705.11*10**-9*T**3}
fi1 = -155941.3/3600                #kW
#similarly
fi2 = 75.778                        #kW
dH1 = (-346.303-450.1-285.83-(2*(-468.257)+2*33.18))/2          #kJ/mol NO2
dH2 = -346.303-450.1-285.83-(2*(-468.257)+9.16)                 #kJ/mol N2O4
dH3 = (2*(-450.1)-285.83+90.25-(2*(-468.257)+3*33.18))/3        # kJ/mol NO2
dHdil = -469.837-(-468.257)                                     #kJ/mol NaOH
fi3 = (dH1*1000*18.48+dH2*1000*27.32+dH3*1000*14.85+dHdil*1000*138.23)/3600         #kW
fi4 = -fi1+fi2+fi3
print "Heat efeet of scrubbing system = ",fi4," kW."

a   Composition of final liquor :
Component          mi kg/h
NaOH              5529.30076871
NaNO2             2522.73285015
NaNO3             3163.59293325
H2O               26299.5585151
b
Heat efeet of scrubbing system =  -2017.12645556  kW.


### Example 6.6 Page 361¶

In :
# solution

# Variables
#(a)
# basis 100 kg feed mix
# F = E +R = 100           (i)
xf = .475
xe = .82
xr = .14
#acetic acid balance
# xf*F = xe*E + xr*R      (ii)
#solving (i) & (ii)
E = 49.2                #kg
R = 50.8                #kg

# Calculation
a = R*xr                #kg  acetic acid leftover
b= (a/(xf*100))*100.

# Result
print " Acetic acid that remained unextracted = ",b," percent."

 Acetic acid that remained unextracted =  14.9726315789  percent.


### Example 6.7 Page 361¶

In :
# solution

# Variables
# referring to fig 6.9
#basis 1000kg/h halibut livers
F = 1000.               #kg/h
OILin = F*.257
Sin = F-OILin           # solid in the charge
U = .23*Sin

# Calculation
OILu = U*.128
Eu = U-OILu             # ether in underflow
R = OILin-OILu          #kg/h   recovery of oil
p = R*100/OILin
O = R/.7
Eo = O-R
Et = Eu+Eo

# Result
print " Flow rate of ether to the system = ",Et,\
" kg/h  and percentage of recovery oil = ",p,"."

 Flow rate of ether to the system =  249.7844  kg/h  and percentage of recovery oil =  91.4887470817 .


### Example 6.8 Page 362¶

In :
# solution

# Variables
F = 1000.                   #kg/h    Basis feed rate
# using fig 6.11
# W/A = 15.77/5.87
# A+F+W = 1000
# solving it
W = 15.77*F/21.64           #kg/h
A = F-W                     #kg/h
# material balance across C3
# R+R1 = D+W
# W/D = 19.31/1.81

# Calculation
# solving it
D = 1.81*W/19.31            #kg/h
M1 = D+W
# R1/R = 4.63/6.57
R1 = 4.63*793/11.2
R = M1-R1
# material balance across C2
m = .89                     # = E1/R1
# E = A+E1+R1 = A+M11
# M11/A = 15.6/3.97
M11 = 15.6*A/3.97
E = M11 + A
E1 = M11 - R1
# material balance across C1
# F+S = M = E+R
M = E+R
S = D+E1
AAloss = W*.4*100/(100*.3)
AArec = 100-AAloss

# Result
print " Summary :  "
print " Stream              Flow rate kg/h"
print " Feed                    ",F
print " Solvent                 ",S
print " Extract                 ",E
print " Raffinate               ",R
print " Acetic acid             ",A
print " Top layer from D1       ",E1
print " Bottom layer from D1    ",R1
print " Feed to C3              ",M1
print " Overhead from C3        ",D
print " Water waste             ",W
print " Stream                  ",M

 Summary :
Stream              Flow rate kg/h
Feed                     1000.0
Solvent                  806.383588647
Extract                  1337.15318679
Raffinate                469.230401854
Acetic acid              271.256931608
Top layer from D1        738.075719471
Bottom layer from D1     327.820535714
Feed to C3               797.050937568
Overhead from C3         68.307869176
Water waste              728.743068392
Stream                   1806.38358865


### Example 6.9 Page 367¶

In :
# solution

# Variables
# basis 100 kg free water
Na2SO4in = 32.              #kg
Win = 68.                   #kg

# Calculation
W1 = (180/142.)*32          #kg   water with Na2SO4
Wfree1 = Win-W1
GS1 = ((Na2SO4in+W1)*100)/Wfree1        #kg    glauber salt present in 100 kg free water
W2 = (180*19.4)/142.                    # water associated with Na2SO4 in final mother liquor
Wfree2 = 100-W2
GS2 = ((19.4+W2)/Wfree2)*100.
Y = GS1-GS2                 #kg
p = Y*100/GS1

# Result
print "Percent yield of glauber salt = ",p,"."

Percent yield of glauber salt =  77.9421927531 .


### Example 6.10 Page 368¶

In :
# solution

# Variables
# basis 100kg free water in original sol
# initial T = 353K
W1 = (126/120.3)*64.2                    #kg
Wfree1 = 100-W1
MS1 = ((64.20+W1)*100)/32.76             # MgSO4.7H2O in 100kg free water

# Calculation
# 4% of original sol evaporates
E = (MS1 + 100)*.04
Wfree2 = 100-E                           # free water in mother liquor
# at 303.15 K
W2 = (126/120.3)*40.8
Wfree3 = 100-W2
MS2 = (W2+40.80)*Wfree2/Wfree3           # crystals of MgSO4.7H2O
y = MS1-MS2                              #kg
q = 501.2*1000/284.6                     # quantity of original sol to be fed

# Result
print " Quantity if original solution to be fed to the crystallizer \
per 1000kg crystals of MgSO4.7H2O = ",q,"kg."

 Quantity if original solution to be fed to the crystallizer  per 1000kg crystals of MgSO4.7H2O =  1761.06816585 kg.


### Example 6.11 Page 370¶

In :
# solution

# Variables
# (a)
print "a   ",
# using fig 6.12
# peforming material balance at 290K
a1 = 5.76
b1 = 4.91

# Calculation and  Result
DCBs = b1*100/(a1+b1)                   # % of solid separated p-DCB
DCBr1 = DCBs*100/70.                    # recovery of p-DCB
print "Percentage recovery of p-DCB = ",DCBr1,".    b   "

#(b)
#at 255K
a2 = 5.76
b2 = 10.22
DCBs = b2*100/(a2+b2)
DCBr2 = (DCBs*100)/70
Ar = DCBr2-DCBr1
print "Additional recovery of p-DCB = ",Ar,"."

a    Percentage recovery of p-DCB =  65.738385326 .    b
Additional recovery of p-DCB =  25.6258199306 .


### Example 6.12 Page 371¶

In :
# solution

# Variables
F = 5000.                           #kg/h  solvent free mix fed to simple crystallization unit
B1 = 4000/157.5                     # kmol/h   p-NCB in feed
A1 = 1000/157.5                     # kmol/h   o-NCB in feed

# Calculation
# after crstallization mother liquor has 33.1 mol % B, A doesn't crstallizes
m = A1/(1-.331)                     # mother liquor entering extractive crytallization unit
B2 = m-A1

# optimizing solid flux
# dCt/dR = 1 - 2/R**3 = 0
R = 2**(1/3)
# referring fig 6.14
# overall material balance
# p-isomer (B)
# .98D + xT = 4000      (i)
# o-isomer (A)
# .02D + (1-.05-x)T = 1000      (ii)
# material balance around solvent recovery unit
# B
# 2.26Tx = .198G = xH         (iii)
# A
# 2.26T(.95-x) = .531G          (iv)
# solving above eq
T = 1337.6                          # kg/h
D = 3729                            # kg/h
G = 3939                            # kg/h
x = .258

# Result
#putting these values we get composition of various streams
print " Composition of various streams :  "
print " Component           T kg/h           D kg/h"
print "    A                 925.6            74.6 "
print "    B                 345.1            3654.9"
print "    C                  66.9             nil  "
print " Purity of top product = 69.2 percent A  Purity of bottom product \
= 98.0 percent  Make-up solvent = 66.9 kg/h."

 Composition of various streams :
Component           T kg/h           D kg/h
A                 925.6            74.6
B                 345.1            3654.9
C                  66.9             nil
Purity of top product = 69.2 percent A  Purity of bottom product = 98.0 percent  Make-up solvent = 66.9 kg/h.


### Example 6.13 Page 383¶

In :
# solution

# Variables
Pw1 = 12.84                 #Pa  v.p. of ice at 233.15K   (table 6.12)
P1 = 101325.                #Pa

# Calculation
Hm = (Pw1/(P1-Pw1))         # kmol/kmol dry air
P2 = 801325.                #Pa
Pw2 = P2*.0001267/(1+.0001267)
dp = -20.18 + 273.15        #K     from table 6.12

# Result
print "Dew Point = ",dp,"K."

Dew Point =  252.97 K.


### Example 6.14 Page 384¶

In :
# solution

# Variables
#Pa = v.p. at DP
Pw = 2.0624                 #kPa
P = 100.                    #kPa

# Calculation
Hm = Pw/(P-Pw)              # kmol water vapour / kmol dry air
H = .622*Hm                 # kg moisture/kg dry air

# at saturation, DB = WB = DP
Ps = 4.004                  #kPa
RH = Pw*100/Ps
Hs = (Ps/(P-Ps))*.622
s = H*100/Hs
Ch = 1.006+1.84*H           #kJ/kg dry air K
Vh = (.00073+.03448)*22.414*1.1062*1.0133           #m**3/kg dry air
# using fig 6.15
WB = 294.55                 #K
ias = 62.3                  # kJ/kg dry air
d = -.28                    # kJ/kg dry air
ia = ias + d

# Result
print "The absolute molar humidity = ",Hm,\
" kmol water vapour/kg dry air \nAbsolute humidity = ",H,\
" kg moisture/kg dry air percent RH = ",RH," percent saturation = ",s,\
" \nHumid heat = ",Ch," kJ/kg dry air K \nHumid volume = ",Vh," m**3/kg dry air."

The absolute molar humidity =  0.0210583065135  kmol water vapour/kg dry air
Absolute humidity =  0.0130982666514  kg moisture/kg dry air percent RH =  51.5084915085  percent saturation =  50.4873424594
Humid heat =  1.03010081064  kJ/kg dry air K
Humid volume =  0.88462068344  m**3/kg dry air.


### Example 6.15 Page 385¶

In :
# solution

# Variables
#basis 1kg of dry air entering the air washer
#from fig 6.15
H1 = 11.8                   #g/kg dry air
H2 = 17.76                  #g/kg dry air
H = H2-H1                   # moisture added during saturation
DB = 300.95                 #K
WB = 298.15                 #K
DP = 297.15                 #K

# Calculation
Ch = 1.006+1.84*.01776      #kJ/kg dry air K
dT = DB-DP
Hs = Ch*3.8
A = 25000.                  #m**3/h    actual air at 41 and 24 degree celcius
# again from fig 6.15
Vh = .9067                  #m**3/kg dry air
qm = A/Vh                   #kg dry air/h
fi = qm*Hs                  #kJ/h
P = 300.                    #kPa
lamda= 2163.2               #kJ/kg         by appendix IV.2
SC = fi/lamda               #kg/h     steam consumption at the heater

# Result
print " the moisture added to the air = ",H,\
" g/kg dry air  DB temp of final air = ",DB,\
"K  WB temp of final air = ",WB,\
"K  The heating load of the steam coil per kg dry air = ",fi,\
" kJ/h  Steam consumption = ",SC," kg/h."

 the moisture added to the air =  5.96  g/kg dry air  DB temp of final air =  300.95 K  WB temp of final air =  298.15 K  The heating load of the steam coil per kg dry air =  108828.110731  kJ/h  Steam consumption =  50.3088529638  kg/h.


### Example 6.16 Page 387¶

In :
# solution

# Variables
# M = E+B+W
Tav = (45+32.)/2 +273.15                #K   avg cooling water T
# using steam tables (Appendix A IV.1)
lamda = 2410.5                          #kJ/kg
E = 530/lamda                           #kg/s
Cl = 4.1868
Ti = 45+273.15                          #K
To = 32+273.15                          #K
fi = 530.                               # = mc*Cl*(Ti-To)
mc = 530/(Cl*(Ti-To))                   #kg/s
W = .3*mc/100                           #kg/s

# Calculation
# dissolved solid balance
# M*xm = (B+W)*xc
# 500*10**-6*M = (B+.0292)*2000*10**-6
# solving above eqs
B = .0441                               #kg/s
M = .2932                               #kg/s

#energy balance on cooling tower
# fi = ma*(i2-i1)
# i2-i1 = 11.042 kJ/kg dry air
# moisture balance
#E = ma(H2-H1)
H2 = .2199/48 + .0196
iws = 2546.2                            # Appendix IV
Ch1 = 1.006+1.84*.0196
i1 = 1.006*(297.45-273.15)+.0196*iws+1.042*(308.15-297.5)       # kJ/kg dry air
i2 = i1 + 11.04
Tdb = ((i2 - 1.006*(301.25-273.15)-iws*H2)/1.05)+301.25         # K

# Result
print "Air leaves th induced draft fan at ",Tdb," K."

Air leaves th induced draft fan at  307.583067857  K.


### Example 6.17 Page 389¶

In :
# solution

# Variables
# basis 1 kg dry air fed to tower
# from fig 6.16 we get
# at WB=330 K and DB=393 K
H1 = .0972                          # kg/kg dry air
DP = 325.15                         #K
# at 313 K
H2 = .0492                          # kg/kg dry air
H = H1-H2                           # moisture condensed in tower

# Calculation
Ch1 = 1.006 + 1.84*H1               # kJ/kg dry air
Ch2 = 1.006 + 1.84*H2
ia1 = 1.006*(325-273) + H1*2596 + 1.185*(393-325)       # enthalpy of entering air
ia2 = 1.006*(313-273) + H2*2574.4                       # enthalpy of outgoing air
i = ia1-ia2
qm = 2000/(1+H1)
fi1 = qm*i                          # heat loss rate
fi2 = 1.167*3600*4.1868*(323-305)   # heat gained by water
r = fi2*100/fi1

# Result
print "a   The heat loss rate rate from the hot air in the bed = ",fi1,\
" kW   b   The percentage heat recovery in hot water = ",r," percent."

a   The heat loss rate rate from the hot air in the bed =  397963.39774  kW   b   The percentage heat recovery in hot water =  79.5581997436  percent.


### Example 6.18 Page 390¶

In :
# solution

# Variables
# basis 800 kmol of inlet CS2-H2 mix
Pi = 106.7                      #kPa  Total Pressure
Pcs2i = 16.93                   # kPa
n = 800.                        # kmol

# Calculation
ncs2i = Pcs2i*n/Pi              # kmol
nh2i = n-ncs2i
Po = 101.325                    # kPa
Pcs2o = 6.19                    # kPa
nh2o = 673.1                    # kmol
ncs2o = Pcs2o*nh2o/(Po-Pcs2o)
ncs2a = ncs2i-ncs2o
mcs2a = ncs2a*76.1407           #kg
r = 600.                        # kg/h  design adsorption rate
Mi = n*r/mcs2a                  # kmol/h
Vi = Mi*22.843                  # m**3/h
mcs2ac = .32-.04                # kg  CS2 absorbed per kg BD activated carbon
qm = r*1.04/mcs2ac              # kg/h
C = ncs2o/nh2o                  # kmol CS2/kmol H2   = Pcs2/(P-Pcs2)
Pcs2 = 24.763                   # kPa
T = 281.5                       #K  by eq 5.24

# Result
print "a   Volumetric flowrate of entering mixture = ",Vi,\
" m**3/h   b   Mass flowrate of activated carbon = ",qm,\
" kg/h   c   Original mixture must be coole to ",T,\
" K at 405 kPa for achieving same concentration of the outlet mixture with adsorption."

a   Volumetric flowrate of entering mixture =  1732.08256916  m**3/h   b   Mass flowrate of activated carbon =  2228.57142857  kg/h   c   Original mixture must be coole to  281.5  K at 405 kPa for achieving same concentration of the outlet mixture with adsorption.


### Example 6.19 Page 391¶

In :
# solution

# Variables
# basis 4000 kg/h of NaOH produced
Cl2p = 35.5*2*4000/80               # kg/h
Mcl2 = Cl2p/71                      # kmol/h
P = 101.325                         # kPa
Pw = 2.0624                         # kPa

# Calculation
moist = (Pw/(P-Pw))*(18.0154/70.906)
Tmoist = Cl2p*moist                 # kg/h
# for 90% onc of acid
n = (10/18.0153)/(90/98.0776)       # kmol H2O/kmol acid
Q = 134477/(18.*(n+1.7983)**2)      #kJ/kg H2O   by eq (ii)
lambdav = 2459.                     # kJ/kg  (Appendix IV)
heatload = Q+lambdav
fi = heatload*18.74                 #kJ/h

# Result
print " The heat liberation rate in the tower = ",fi," kJ/h."

 The heat liberation rate in the tower =  70323.4193977  kJ/h.


### Example 6.20 Page 393¶

In :
# solution

# Variables
# basis 100 kmol of feed gas
# using table 5.1
Sniai = 20.6*29.5909+62*28.6105+4.1*20.7723+11.1*19.2494+2.2*25.6503
Snibi = (20.6*(-5.141)+62*1.0194+11.1*52.1135+2.2*33.4806)/1000
Snici = (20.6*13.1829+62*(-.1476)+11.1*11.973+2.2*.3518)/10**6
Snidi = (20.6*(-4.968)+62*.769+11.1*(-11.3173)+2.2*(-3.0832))/10**9
Hgas = Sniai*(283-263) + Snibi*(283**2-263**2)/2 + Snici*(283**3-263**3)/3 + Snidi*(283**4-263**4)/4  #kJ
Hnh3 = 1533.8                               #kJ

# Calculation
SniCmpi = (Hgas-Hnh3)/20                    # kJ/(K 97.8 kmol gas)  NH3 free gas
Go = 97.8/.99995                            #kmol
NH3a = (2.2-.005)*17                        # kg
F1 = NH3a/.04                               # flowrate of 4% NH3 solution
Water = F1-NH3a                             #kg
dT1 = Hgas/(Water*4.1868)                   # K
Twater = 307-dT1                            #K
Wvp = 2.116                                 #kPa
P = 5101.325                                #kPa
moist = Go*Wvp/(P-Wvp)                      # kg
W = Water + moist                           # total demineralised water
dTactual = Hgas/(W*4.1868)                  #K
# from table 5.59
dHf1 = -80.093                              #kJmol NH3  of 4% NH3 sol
dHf2 = -46.11                               #kJ/molNH3
H = dHf1-dHf2                               # heat of 4% NH3 sol
Hevl = -(H*NH3a*1000)/17.                   # total heat evolved

# in absorber gas is further heated from 283K to 291.4K
Hsol = Hevl-(2854.1*(291.4-283.15))         # kJ
# c 0f 4% NH3 sol = c of water = 4.1868 kJ/kg K
dT2 = Hsol/(F1*4.1868)
To = 291.4+dT2

# Result
print "a   Temp of feed water to absorber = ",Twater,\
"K.   b   Temp of aq NH3 sol leaving the absorber = ",To,"K."

a   Temp of feed water to absorber =  291.367208278 K.   b   Temp of aq NH3 sol leaving the absorber =  304.469504086 K.


### Example 6.21 Page 396¶

In :
# solution

# Variables
# basis : product rate of 100 kg/h
H1 = .036                   # kg moist/ kg dry solid
X1 = .25/.75                # kg /kg dry solid
X2 = .02/.98                # kg/kg dry solid

# Calculation
# moist balance
# ms*(X1-X2) = ma*(H2-H1)
To = 273.15                 #K
is1 = 1.43*(30-0)+X1*4.1868*30
is2 = 1.43*80+.0204*4.1868*80
Tdb = 393.15                #K
Tdp1 = 308.15               #K
iwb1 = 2565.4               #kJ/kg
Ch1 = 1.006+1.84*.036
ia1 = 1.006*(Tdp1-273.15)+H1*iwb1+Ch1*(Tdb-Tdp1)
H2 = .056
Tdp2 = 315.55
iwb2 = 2578.7
ia2 = 1.006*(Tdp2-273.15)+H2*iwb2+(1.006+1.84*H2)*(323.15-Tdp2)
ma = .085/(.056-.036)
iaa = 1.006*(Tdp1-273.15)+H1*iwb1
fi = 4.25*(218.68-iaa)      #kW
lambda_ = 2133.0
steam = fi/lambda_          # kg/h

# Result
print "a   Flowrate of incoming air on dry basis = ",ma,\
" kg/s   b   Humidity of air leaving the drier = ",H2,\
" kg/kg dry air.   c   Steam consumption in the heater = ",steam," kg/h."

a   Flowrate of incoming air on dry basis =  4.25  kg/s   b   Humidity of air leaving the drier =  0.056  kg/kg dry air.   c   Steam consumption in the heater =  0.181547726207  kg/h.


### Example 6.22 Page 398¶

In :
# solution

# Variables
# basis cloth speed = 1.15 m/s
prod = 1.15*1.2*3600*.095
moisti = .90                    # kg/kg bone dry cloth
moisto = .06
evp = 471.96*(moisti-moisto)

# using fig 6.15 and 6.16
H1 = .01805
H2 = .0832
dH = H2-H1
qm1 = evp/dH                    # kg dry air/h
Vh = .8837                      #m**3/kg dry air
qv = qm1*Vh
DP1 = 296.5                     #K
DP2 = 322.5                     #K
lambdaV2 = 2384.1               #kJ/kg
To = 273.15                     #K

# Calculation
fi1 = prod*1.256*(368-303)+prod*.06*(368-303)*4.1868                # kJ/h
fi2 = evp*(322.5-303.15)+evp*lambdaV2                               #kJ/h
ia1 = 1.006*(303.15-273.15)+2556.4*.01805                           #kJ/kg dry air
ia2 = 1.006*(322.8-273.15)+2591.5*.0832+(1.006+1.84*.0832)*(393-328.8)
fi2 = ia2-ia1
hlost = fi2-fi1                 # kJ/h
# using Appendix IV
h = 720.94                      #kJ/kg
lambdav = 2046.5                # kJ/kg
steami = (h+lambdav)*885        # kJ/h
fi4 = h*885                     #kJ/h
qm2 = 885/evp

# Result
print "a   Bone dry production of the dryer = ",prod,\
" kg/h.   b   The evaporation taking place in the dryer = ",evp,\
" kg/h.   c   The air circulation rate = ",qv," m**3/h."

a   Bone dry production of the dryer =  471.96  kg/h.   b   The evaporation taking place in the dryer =  396.4464  kg/h.   c   The air circulation rate =  5377.43182932  m**3/h.


### Example 6.23 Page 401¶

In :
# solution

# Variables
# basis : weak liquor flowrate = 1060 kg/h
s1 = 1060*.04           #kg/h   solids in weak liquor
liqr = s1/.25           #  kg/h  conc liquor leaving 4th effect
evp = 1060-liqr         # kg/h
lambdas = 2046.3        # kJ/kg
Wf = 1060.              # kg/h
C1f = 4.04
T1 = 422.6
Tf = 303
lambdav1 = 2114.4

# Calculation
# enthalpy balance of 1st effect
# Ws*lambdas = Wf*C1f*(T1-Tf) + (Wf-W1)*2114.4
#putting values we get
# Ws = 1345.57 - 1.033*W1
# 2nd effect
# W1 = 531.38+.510*W2
# 3rd effect
# W1 - 1.990*W2 = -1.027*W3
# 4th effect
# W2 - 1.983*W3 = -176.84
#solving above eqs
W1 = 862.               # kg/h
W2 = 648.2              # kg/h
W3 = 416.7              # kg/h
Ws = 455.2              # kg/h
eco = evp/Ws            # kg evaporation/kg steam
spcon = 1/eco           # kg steam/kg evaporation

# Result
print "Specific heat consumption of the system is ",spcon," kg steam/kg evaporation."

Specific heat consumption of the system is  0.511230907457  kg steam/kg evaporation.


### Example 6.24 Page 403¶

In :
# solution

# Variables
Fspd1 = 4300.                   # kg/h
Bcrtn = Fspd1*600*10**-6        # kg/h
Fspd2 = Bcrtn/.00645            # kg/h
evp1 = Fspd1-Fspd2
Fspd3 = Bcrtn/.057
evp2 = Fspd2-Fspd3
C3 = Bcrtn/.4
evp3 = Fspd3-C3

# Calculation
fi1 = Fspd1*2.56*(468.15-373.15)+3900*450               # kJ/h
fi2 = Fspd2*2.56*(463.15-468.15)+354.737*450            # kJ/h
fi3 = Fspd3*2.56*(453.15-463.15)+38.813*450             # kJ/h
fi = fi1+fi2+fi3
mt = fi/(2.95*(503.15-478.15))                          # kg/h
qt = mt/.71                                             # l/h
mccw1 = 1755000/(8*4.1868)                              # kg/h
mccw2 = mccw1*.9
dT2 = 159632/(mccw2*4.1868)
mccw3 = mccw1-mccw2
dT3 = 17466/(mccw3*4.1868)
dT = (1755000+159632+17466)/(mccw1*4.1868)
Fw = 1932098/(8*4.1868)                                 # kg/h

# Result
print "By mass balance, required cooling water flow in external cooler = ",Fw,\
" kg/h.By enthalpy balance, overall rise in CCW temperature = ",dT," K."

By mass balance, required cooling water flow in external cooler =  57684.2098978  kg/h.By enthalpy balance, overall rise in CCW temperature =  8.80728433048  K.


### Example 6.25 Page 405¶

In :
# solution

# Variables
#stream M2
Vcaco3M2 = .349/2.711
VliqrM2 = .651/1.167
VslryM2 = Vcaco3M2+VliqrM2
spgM2 = 1/VslryM2
FsM2 = 2.845*3600*spgM2
sM2 = FsM2*.349                             # kg/h
liqrM2 = FsM2*.651
Na2OM2 = liqrM2*.1342/1.167

# Calculation
#stream O2
FsO2 = 14.193*3600*1.037                    # kg/h
sO2 = FsO2*.0003
liqrO2 = FsO2-sO2
Na2OO2 = liqrO2*.0272/1.037
#stream M1
VM1 = .194/2.711 + .806/1.037               # l
spgM1 = 1/VM1
FsM1 = 5206.9/.194
liqrM1 = FsM1 - 5206.9
Na2OM1 = liqrM1*.0252/1.034
# stream O1
FsO1 = FsO2+FsM1-FsM2
sO1 = FsO1*.0002
liqrO1 = FsO1 - sO1
Na2OO1 = liqrO1*.0096/1.014
# stream W
VW = .037/2.711 + .963
spgW = 1/VW
FsW = 14.977*3600*spgW
sW = FsW*.037
liqrW = FsW-sW
Na2OW = liqrW*.0024
# stream Mo
VMo = .402/2.711 + .598/1.022
spgMo = 1/VMo
FsMo = 3.627*3600*spgMo
sMo = FsMo*.402
liqrMo = FsMo - sMo
Na2OMo = liqrMo*.0162/1.022

# Result
print " Material balance thickener   ITEM    STREAM, kg/h   M2   O2   M1\
O1    W   Mo Slurry    ",FsM2,"      ",FsO2,"      ",FsM1,"      ",FsO1,\
"      ",FsW,"      ",FsMo," Suspended solids    ",sM2,"      ",sO2,\
"      ",sM2,"      ",sO1,"      ",sW,"      ",sMo," Liquor   ",liqrM2,\
"      ",liqrO2,"      ",liqrM1,"      ",liqrO1,"        ",liqrW,"      ",liqrMo,\
"  Na2O                ",Na2OM2,"      ",Na2OO2,"      ",Na2OM1,"      ",Na2OO1,"      ",Na2OW,\
"       ",Na2OMo

 Material balance thickener   ITEM    STREAM, kg/h   M2   O2   M1     O1    W   Mo Slurry     14917.5166812        52985.3076        26839.6907216        64907.4816404        55206.3736997        17803.3637049  Suspended solids     5206.21332174        15.89559228        5206.21332174        12.9814963281        2042.63582689        7156.95220938  Liquor    9711.30335947        52969.4120077        21632.7907216        64894.5001441          53163.7378728        10646.4114955   Na2O                 1116.75827836        1389.36162643        527.220818361        614.385800181        127.592970895         168.759164606