# Chapter 7 : Combustion¶

### Example 7.1 Page 434¶

In :
# solution

# Variables
# basis 100 kg as received coal
O2 = 18.04                      #kg
nH2 = 2.79-(O2/8)               #kg
print "a   Net H2 in coal = ",nH2," kg.    b   "
cbW = 1.128*18                  # kg
print "Combined water in the coal = ",cbW," kg.    c   "

# Calculation
# Dulong's formula
GCV1 = 33950*(50.22/100) + 144200*nH2/100 + 9400*.37/100            # kJ/kg

# Result
print "GCV by Dulongs formula = ",GCV1," kJ/kg.    d   "
tH2 = 1.395                     # kmol
wp = tH2*18 + 7
Hv = 2442.5*wp/100              # kJ/kg fuel
GCV2 = 23392*(1-.21-.07)        # as of received coal
NCV = GCV2-Hv
print "NCV of the coal = ",NCV," kJ/kg.    e   "
# Calderwood eq
# Total C = 5.88 + .00512(B-40.5S) +- .0053[80-100*(VM/FC)]**1.55
C = 5.88 + .00512*(7240.8-40.5*.37)+.0053*(80-56.52)**1.55
print "Total Carbon by Calderwood eq = ",C,"."

a   Net H2 in coal =  0.535  kg.    b
Combined water in the coal =  20.304  kg.    c
GCV by Dulongs formula =  17855.94  kJ/kg.    d
NCV of the coal =  16057.95325  kJ/kg.    e
Total Carbon by Calderwood eq =  43.5822593081 .


### Example 7.2 Page 436¶

In :
# solution

# Variables
# basis 1 kg crude oil
H2 = .125                   # kg   burnt

# Calculation
H2O = H2*18/2.
Lh = H2O*2442.5             #kJ
GCV = 45071
NCV = GCV-Lh                #kJ/kg oil

# Result
print "NCV = ",NCV," kJ/kg."

NCV =  42323.1875  kJ/kg.


### Example 7.3 Page 444¶

In :
# solution
# Variables

# basis 1 mol of gaseous propane
H2O = 4*18.0153             #g

# Calculation
NHV = 2219.17-(H2O*2442.5/1000.)

# Result
print "NHV = ",NHV," kJ/mol."

NHV =  2043.160519  kJ/mol.


### Example 7.4 Page 444¶

In :
# solution

# Variables
# basis 1 mol of natural gas
# using table 7.7
# Calculation
H2O = (2*.894+3*.05+.019+5*(.004+.006))*18              # g
Hv = H2O*2442.5/1000.
NCV1 = 945.16-Hv
GCV = 945.16*1000/18.132
NCV = NCV1*1000/18.132

# Result
print " GCV = ",GCV," kJ/kg.  NCV = ",NCV," kJ/kg."

 GCV =  52126.6269579  kJ/kg.  NCV =  47260.2164681  kJ/kg.


### Example 7.5 Page 451¶

In :
# solution

# Variables
# basis 100 kg fuel
O2req = 4.331*32                    # kg

# Calculation
rO2req = O2req/100
N2in = (79/21.)*4.331               # kmol
AIRreq = O2req+N2in*28              #kg
rAIRreq = AIRreq/100.
R = AIRreq/100.
AIRspld = R*2                       # kg/kg coal
O2spld = 4.331*2                    # kmol
N2spld = N2in*2
N2coal = 2.05/28                    # kmol
tN2 = N2spld+N2coal
moist = 1.395+(7/18.)               # kmol

# Result
print "a   Theoratical O2 requirement per unit mass of coal = ",rO2req,\
" kg.   b   Theoratical dry air requirement = ",rAIRreq," kg/kg coal."

a   Theoratical O2 requirement per unit mass of coal =  1.38592  kg.   b   Theoratical dry air requirement =  5.94790666667  kg/kg coal.


### Example 7.6 Page 452¶

In :
# solution

# Variables
# basis 100 kg of RFO
O2req = 9.786                   #kmol
N2req = (79/21.)*O2req          #kmol
AIRreq = O2req+N2req            #kmol
rAIRreq = AIRreq*29/100
AIRspld = AIRreq*1.25
rAIRspld = AIRspld/100

# Calculation
# using table 7.11 and 7.12
xSO2 = .07/(55.925+5.695)       # kmol SO2/kmol wet gas
vSO2 = xSO2*10**6               # ppm
mSO2 = 4.48*10**6/(1696.14+102.51)

# at 523.15 K and 100.7 kPa
V = ((55.925+5.695)*8.314*523.15)/100.7         # m**3
cSO2 = (4.48*10**6)/V                           # mg/m**3
#from fig 7.3
dp = 424.4                      #K

# Result
print "a   Theoretical air required = ",rAIRreq," kg/kg fuel.  "
print "b   Actual dry air supplied = ",rAIRspld," kg/kg fuel.  "
print "c   Concentration of SO2 = ",mSO2," mg/kg. "
print "d   Concentration of SO2 = ",vSO2," ppm vol/vol."
print "e   Concentration of SO2 if gases are discharged at 523.15K and100.7kPa = ",cSO2," mg/m**3. "
print "f   Dew Point of flue gas = ",dp," K."

a   Theoretical air required =  13.514  kg/kg fuel.
b   Actual dry air supplied =  0.5825  kg/kg fuel.
c   Concentration of SO2 =  2490.75695661  mg/kg.
d   Concentration of SO2 =  1135.99480688  ppm vol/vol.
e   Concentration of SO2 if gases are discharged at 523.15K and100.7kPa =  1683.25355648  mg/m**3.
f   Dew Point of flue gas =  424.4  K.


### Example 7.7 Page 454¶

In :
# solution

# Variables
# basis 100 kmol of dry flue gas
O2acntd = 11.4+4.2                  # kmol
O2avlbl = (21./79)*84.4              # kmol
O2excs = 4.2                        #kmol

# Calculation
O2unactd = O2avlbl-O2acntd
H2brnt = O2unactd*2
O2req = 11.4+O2unactd
pexcsAIR = O2excs*100/O2req
mH2brnt = H2brnt*2                  # kg
mCbrnt = 11.4*12
r = mCbrnt/mH2brnt

# Result
print "a   Percent excess air = ",pexcsAIR,".   b   In fuel C:H = ",r,"."

a   Percent excess air =  23.0320699708 .   b   In fuel C:H =  5.00333333333 .


### Example 7.8 Page 459¶

In :
# solution

# Variables
# basis 100 kg of bagasse fired in th boiler
#(a)
O2req = 2.02                    # kmol
N2in = (79/21)*O2req            # kmol
AIRreq = (O2req+N2in)*29        # kg
rAIR = AIRreq/100
print "a   Theoretical air required = ",rAIR," kg dry air/kg fuel.  \n  b   ",

# Calculation
# (b)
tflugas = 1.95/.1565            #/kmol
xcsO2N2 = tflugas - 1.95
x = (xcsO2N2-7.6)/4.76          # kmol
pxcsAIR = x*100/O2req

# Result
print "Percent excess air = ",pxcsAIR,". \n  c   ",
#(c)
pW = 100*.2677                  # kPa    partial p of water vap
# from fig 6.13
dp = 339.85                     #K
print "Dew Point of flue gas = ",dp,"K.  \n  d   ",
# (d)
# from appendix IV
hfw = 292.97                    #kJ/kg     enthalpy of feed water at 343.15 K
Hss = 3180.15                   # kJ/kg   enthalpy of super heated steam at 2.15 bar and 643.15K
Hgain = Hss - hfw
H6 = Hgain*2.6*100              # kJ    heat gained by water
H1 = 100*1030000.               # kJ
GCV = H6*100/H1
print "Thermal efficiency of the boiler = ",GCV,"."

a   Theoretical air required =  2.3432  kg dry air/kg fuel.
b    Percent excess air =  30.2652456295 .
c    Dew Point of flue gas =  339.85 K.
d    Thermal efficiency of the boiler =  0.728802718447 .


### Example 7.9 Page 465¶

In :
# solution

# Variables
# using mean heat capacity data Table 7.21
# basis 100 kmol of dry flue gas

# Calculation
H7 = 1.0875*100*30.31*(423.15-298.15)
H71 = 3633.654*(423.15-298.15)
fi7 = H71*3900*.7671/162.2              # kJ/h
fi1 = 3.9*1000*26170                    # kJ/h
# performing heat balance
Hsteamgen = 23546.07
eff = Hsteamgen*100/fi1                 # overall efficiency rate

# Result
print "Overall efficiency rate = ",eff," percent."

Overall efficiency rate =  0.0230701331531  percent.


### Example 7.10 Page 468¶

In :
# solution

# Variables
# basis 100 kg of fuel oil
O2req =  9.364              # kmol
N2in = (79/21.)*O2req
tN2 = N2in+.036
AIRreq = O2req*32 + tN2*28
rAIR = AIRreq/100.          # kg/kg
wp = 4.5                    # kmol
Hloss = 2442.8*wp*18/100    # kJ/kg fuel
NCV = 43540-Hloss
print "a   NCV = ",NCV," kJ/kg.  \n  b   Theoretical air required = ",rAIR," kg/kg fuel. \n   c   ",
H1 = 100*41561.33           # kJ

# Calculation
# from table 5.1
H71 = 1349.726*(1500-298.15)+252.924*10**-3 * ((1500**2-298.15**2)/2)+ \
257.436*10**-6*((1500**3-298.15**3)/3)-137.532*10**-9*((1500**4-298.15**4)/4)   # upto 1500 K
H711 = H1-H71               # above 1500K
# F(T) = {1500 to T} integr[1477.301+375.2710*10**-3T-91.2760*10**-6T**2+8.146*10**-9T**3]dT-2147118     (i)
# solving it for T = 2000
AFT = 2612.71               # K

# Result
print "When fluid is burnt with theoretical air AFT = ",AFT," K.    d   "
# with 30% excess air
O2spld = 9.364*1.3
xcsO2 = O2spld-O2req
N2in1 = (79/21.)*O2spld
tN21 = N2in1+.036
# now, using table 7.26, table 7.27 and eq(i)  we get
AFT1 = 2178.66              # K
# from fig 7.3
dp = 429.                   # K
# similarly for incomplete combustion we find
AFT2 = 2561.42              #K
print "When 30 percent excess air is supplied AFT = ",AFT1,\
" K.    d   Dew Point = ",dp," K.    e   For incomplete combustion AFT = ",AFT2," K."

a   NCV =  41561.332  kJ/kg.
b   Theoretical air required =  12.8699733333  kg/kg fuel.
c    When fluid is burnt with theoretical air AFT =  2612.71  K.    d
When 30 percent excess air is supplied AFT =  2178.66  K.    d   Dew Point =  429.0  K.    e   For incomplete combustion AFT =  2561.42  K.


### Example 7.11 Page 473¶

In :
# solution

# Variables
# basis 100 kg of fuel
# material balance of carbon
CO2 = 7.092+.047                #kmol   in flue gases
N2 = 11.94*7.139/7.01
O2 = 11.94*7.139/7.01
flue = CO2+N2+O2

# Calculation
# material balance of O2
O2air = 21*N2/79.
airin = N2+O2air
tO2in = O2air+.078              # O2 in burner
O2xcs = tO2in-9.864
# material balance of water vapour
moistfrmd = 5.45                # kmol    from combustion of H2
H = .0331                       # kmol/kmol of dry air        humidity at 100.7 kPa
moistair = H*104.482            #kmol
tmoist = moistfrmd+moistair
pxcsair = O2xcs*100/9.786
# now using table 7.32
H7 = 3391.203*(563.15-298.15)   #kJ
Ff = 400.                       # kg/h   fuel firing rate
tH = 2791.7-179.99              # kJ/kg    total heat supplied in boiler
fi5 = tH*4365                   # kJ/h
fi8 = 5.45*18*Ff*2403.5/100     # kJ/h
GCVf = 42260.                   #kJ/kg
fi1 = Ff*GCVf
Fdryair = 104.482*29*Ff/100
Cha = 1.006+1.84*.0205          # kJ/kg dry air K
fi3 = Fdryair*Cha*(308.15-298.15)
fi2 = Ff*1.758*(353.15-298.15)
BOILEReff1 = fi5*100/fi1
NCVf = GCVf-(18.0153/2.016)*.109*2442.8  # kJ/kg
BOILEReff2 = fi5*100/(Ff*NCVf)
r = 4365/Ff                     # steam:fuel
BOILERcapacity = fi5/2256.9

# Result
print " After performing material and thermal balance\
operations we get    Overall thermal efficiency of the boiler based on GCV\
of the fuel = ",BOILEReff1," percent.   Overall efficiency of the boiler based\
on NCV of the fuel = ",BOILEReff2," percent.   Steam to fuel ratio = ",r," \
at 16 bar.   Equivalent boiler capacity = ",BOILERcapacity," kg/h."

 After performing material and thermal balance operations we get    Overall thermal efficiency of the boiler based on GCV  of the fuel =  67.4403345362  percent.   Overall efficiency of the boiler based   on NCV of the fuel =  71.464013185  percent.   Steam to fuel ratio =  10.9125     at 16 bar.   Equivalent boiler capacity =  5051.22697062  kg/h.


### Example 7.12 Page 478¶

In :
# solution

# Variables
# basis 100 kmol of dry producer gas
C = 33*12.                  # kg
O2 = 18.5*32                #kg
H2 = 20*2.                  # kg
O2air = 21*51/79.           # kmol
COALgassified = 396/.672    # kg

# Calculation
O2coal = COALgassified*.061/32              # kmol
tO2 = O2coal + O2air
O2steam = 18.5-tO2          # kmol
H2steam = 2*O2steam         # kmol
H2fuel = 20-H2steam
dryproducergas = 100*22.41/COALgassified    # Nm**3/kg coal
Pw = 2.642                  # kPa
Ha = Pw/(100.7-Pw)          # kmol/kmol dry gas
water = Ha*100.
moistproducergas = (100+water)*22.41/COALgassified          # Nm**3/kg coal
dryair = (51*28+O2air*32)/COALgassified                     # kg/kg coal
tsteamsupplied = H2steam+water-(COALgassified*.026/18)      # kmol
steam = tsteamsupplied*18/COALgassified

# Result
print " a   Moistproducer gas obtained = ",moistproducergas," Nm**3/kg coal.  \n \
b   Air supplied = ",dryair," kg/kg coal gassified.  \n  c   Steam supplied = ",steam," kg/kg coal."

 a   Moistproducer gas obtained =  3.9053717744  Nm**3/kg coal.
b   Air supplied =  3.15945684695  kg/kg coal gassified.
c   Steam supplied =  0.289649027717  kg/kg coal.


### Example 7.13 Page 479¶

In :
# solution

# Variables
# solving by alternate method on page 483
# basis 100 kmol of dry producer gas
# using tables 7.38 and 7.39
fi7 = 6469.67*(833.15-298.15)*(27650/2672.)          # kJ/h

# Calculation
# heat output basis 1 kg of steam
# referring Appendix IV
H4 = 675.47-272.03                                  # kJ/kg
Ts = 463.                                           # K
h = 806.69                                          # kJ/kg
lambdav = 1977.4                                    # kJ/kg
Hss = 2784.1                                        # kJ/kg at Ts
i = 3045.6                                          # kJ/kg
H6 = i-Hss
fi4 = H4*7100                                       # kJ/h
fi5 = (Hss-675.47)*7100                             # kJ/h
fi6 = H6*7100                                       # kJ/h
recovery = fi4+fi5+fi6
BOILERcapacity = recovery*3600/2256.9               # kg/h
fi8 = 6125.47*(478.15-298.15)*(27650/2672.)         # kJ/h
hloss = fi7-fi4-fi5-fi6-fi8                         #/ kJ/h

# Result
print "Heat Balance of Waste Heat Boiler kJ/h  \nHeat Output  Steam rising  Economiser",fi4,\
"  \nSteam generator               ",fi5," \nSuper heater  ",fi6,\
"   \nHeat loss in flue gases   ",fi8,"  \nUnaccounted heat loss  ",hloss

Heat Balance of Waste Heat Boiler kJ/h
Heat Output  Steam rising  Economiser 2864424.0
Steam generator                14971273.0
Super heater   1856650.0
Heat loss in flue gases    11409604.8615
Unaccounted heat loss   4715492.33477