In [1]:

```
# 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,"."
```

In [2]:

```
# 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."
```

In [3]:

```
# 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."
```

In [4]:

```
# 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."
```

In [5]:

```
# 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."
```

In [6]:

```
# 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."
```

In [7]:

```
# 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,"."
```

In [2]:

```
# 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,"."
```

In [9]:

```
# 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."
```

In [1]:

```
# 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."
```

In [1]:

```
# 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."
```

In [12]:

```
# 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."
```

In [15]:

```
# 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
```