# Chapter 6 : Stoichiometry and Unit Operations¶

### Example 6.1 Page 346¶

In [1]:
# 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 [2]:
# 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 [3]:
# 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 [4]:
# 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 [5]:
# 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 [6]:
# 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 [7]:
# 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 [8]:
# 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
Water waste              728.743068392
Stream                   1806.38358865


### Example 6.9 Page 367¶

In [1]:
# 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 [10]:
# 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 [11]:
# 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 [12]:
# 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 [13]:
# 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 [14]:
# 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 [15]:
# 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 [16]:
# 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 [17]:
# 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 [18]:
# 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 [19]:
# 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)

# 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 [20]:
# 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 [21]:
# 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 [22]:
# 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 [23]:
# 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 [24]:
# 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 [25]:
# 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